Question Encodings

Here are some notes on the encodings:

deviations from the English are marked ;; !*! in the .km files
Our EnglishName-to-ChemicalFormula code can't do "oxygen"->O2 or "nitrogen gas"->N2, so we've done that translation manually throughout.
(should lose fidelity points)
The EnglishName-to-ChemicalFormula stuff requires that all questions (QF*.km) be loaded into KM after everything else.
(should not lose fidelity points)
There are several methods in the EnglishName-to-ChemicalFormula code: Compute-Compound-from-Chemical-Name, Compute-Compound-from-Common-Name, Compute-Element-from-Name, etc. In encoding the questions we've chosen the appropriate method in each instance. It would be possible to make that choice automatically, since in general only one of the available methods will return a result for a given English name (for example, Compute-Compound-from-Chemical-Name won't return a result for "sodium", but Compute-Element-from-Name will).
(should not lose fidelity points)
In many questions we call the EnglishName-to-ChemicalFormula stuff in the global environment (outside the question definition) for the sake of efficiency. In order to reference the Chemical-Entity returned, we need to unify it with a Chemical-Entity that has a name. In all cases, it works to unify with the most general: Chemical-Entity. But for efficiency we often unify with an instance of something more specific (like Ionic-Compound or Acid-Compound). That more specific class is what's returned by the Compute- method, so it's not Chemistry knowledge that allows us to state the more specific class. This more specific class could be determined automatically (an instance of (the instance-of of (the output of (a Compute-...)))).
(should not lose fidelity points)

Look at the question encodings here)

Question Answers

QF1

Which of the given compounds will produce a gas when HCl is added to the solid compound? HCl and CaCO3 react to form a gas :.
- By definition, a reaction between an acid and a carbonate is an acid-base exchange reaction (with gas).
- Therefore, this is an acid-base exchange reaction (with gas).
- When an acid reacts with a carbonate, the products are water, CO2(g), and a salt formed by the acid's anion and the bicarbonate metal.
- The products of the reaction of HCl and CaCO3 are thus Ca(Cl)2, H2O, and CO2.
The following compound(s) will produce a gas when HCl is added to the solid compound: (b) CaCO3(s).

QF2

When lithium metal is reacted with nitrogen gas, under proper conditions, what is the product?.
- Metals and nonmetals react to form salts in which the cation is formed from the metal and the anion is formed from the nonmetal.
- The product of the reaction of Li and N2 is thus (Li)3N (an ionic compound consisting of Li+ and N 3-).
When lithium metal is reacted with nitrogen gas, under proper conditions, the product is: (d) Li3N.

QF3

What are the products of the given decomposition reaction?.
- By definition, oxidation-reduction reactions occur when electrons are transferred from the atom that is oxidized to the atom that is reduced. We need to look for changes in the oxidation states of the elements in the reaction.
  - In the reactants, the oxidation state(s) of the element Na is/are (1). In the product, the oxidation state(s) is/are (0).
  - Therefore, the reaction causes a change in oxidation state.
- Therefore, this is an oxidation reduction reaction.
- By definition, a Binary Ionic-Compound Decomposition Reaction occurs when a binary ionic compound is heated.
- Therefore, this reaction is a Binary-Ionic Compound Decomposition reaction.
- In general, a Binary Ionic-Compound Decomposition Reaction converts a binary ionic-compound into basic elements.
- In this reaction, NaN3 reacts to produce Na and N2.
The products of the decomposition reaction are: (d) Sodium and nitrogen gas.

QF4

When calcium carbonate is heated it decomposes forming what?.
- By definition, a reaction with exactly one input and multiple products is a decomposition reaction.
- Therefore, this is a decomposition reaction.
- By definition, a Metal Carbonate Decomposition Reaction occurs when a metal carbonate is heated.
- Therefore, this reaction is a Metal-Carbonate decomposition reaction.
- In general, the reaction produces a metal oxide (an ionic compound) along with carbon dioxide.
- In this reaction, CaCO3 reacts to produce CaO.
When calcium carbonate is heated it decomposes forming: (c) Solid CaO and CO2 gas.

QF6

Which solution has the highest conductivity?.
- Finding the solution with the highest conductivity.
  - Conductivity depends on the electrolyte status of a chemical.
    - Weak electrolytes have low conductivity.
      - Checking the electrolyte status of NH3.
        
        Weak acids and bases are weak electrolytes.
        NH3 is a weak base and is therefore a weak electrolyte.
      - NH3 is thus a weak electrolyte.
    - NH3 is a weak electrolyte and thus has low conductivity.
  - Therefore, an aqueous solution of NH3 has low-conductivity.
  - Conductivity depends on the electrolyte status of a chemical.
    - Strong electrolytes have high conductivity.
      - Checking the electrolyte status of NaOH.
        
        Strong acids and bases are strong electrolytes.
        NaOH is a strong base and is therefore a strong electrolyte.
      - NaOH is thus a strong electrolyte.
    - NaOH is a strong electrolyte and thus has high conductivity.
    - The conductivity of a solution is proportional to the sum of the concentrations of all ions in the solution.
      - If a solute is a strong electrolyte, the concentration of ions is maximal.
        
        Checking the electrolyte status of NaOH.
        
        Strong acids and bases are strong electrolytes.
        NaOH is a strong base and is therefore a strong electrolyte.
        
        NaOH is thus a strong electrolyte.
      - The concentration of ions in NaOH is 1.00 molar.
    - The conductivity of NaOH is thus proportional to 1.00.
  - Therefore, an aqueous solution of NaOH has high-conductivity.
  - Conductivity depends on the electrolyte status of a chemical.
    - Strong electrolytes have high conductivity.
      - Checking the electrolyte status of (Na)3PO4.
        
        Soluble salts are strong electrolytes.
        (Na)3PO4 is a soluble salt and is therefore a strong electrolyte.
      - (Na)3PO4 is thus a strong electrolyte.
    - (Na)3PO4 is a strong electrolyte and thus has high conductivity.
    - The conductivity of a solution is proportional to the sum of the concentrations of all ions in the solution.
      - If a solute is a strong electrolyte, the concentration of ions is maximal.
        
        Checking the electrolyte status of (Na)3PO4.
        
        Soluble salts are strong electrolytes.
        (Na)3PO4 is a soluble salt and is therefore a strong electrolyte.
        
        (Na)3PO4 is thus a strong electrolyte.
      - The concentration of ions in (Na)3PO4 is 2.00 molar.
    - The conductivity of (Na)3PO4 is thus proportional to 2.00.
  - Therefore, an aqueous solution of (Na)3PO4 has high-conductivity.
  - Conductivity depends on the electrolyte status of a chemical.
    - Strong electrolytes have high conductivity.
      - Checking the electrolyte status of HCl.
        
        Strong acids and bases are strong electrolytes.
        HCl is a strong acid and is therefore a strong electrolyte.
      - HCl is thus a strong electrolyte.
    - HCl is a strong electrolyte and thus has high conductivity.
    - The conductivity of a solution is proportional to the sum of the concentrations of all ions in the solution.
      - If a solute is a strong electrolyte, the concentration of ions is maximal.
        
        Checking the electrolyte status of HCl.
        
        Strong acids and bases are strong electrolytes.
        HCl is a strong acid and is therefore a strong electrolyte.
        
        HCl is thus a strong electrolyte.
      - The concentration of ions in HCl is 1.00 molar.
    - The conductivity of HCl is thus proportional to 1.00.
  - Therefore, an aqueous solution of HCl has high-conductivity.
  - Conductivity depends on the electrolyte status of a chemical.
    - Weak electrolytes have low conductivity.
      - Checking the electrolyte status of HCN.
        
        Weak acids and bases are weak electrolytes.
        HCN is a weak acid and is therefore a weak electrolyte.
      - HCN is thus a weak electrolyte.
    - HCN is a weak electrolyte and thus has low conductivity.
  - Therefore, an aqueous solution of HCN has low-conductivity.
- The solution with the highest conductivity is: (Na)3PO4.
The answer is: (c) 0.5M Na3PO4.

QF7

Which of the given chemicals is a non-electrolyte?.
- Checking the electrolyte status of NaCl.
  - Soluble salts are strong electrolytes.
  - NaCl is a soluble salt and is therefore a strong electrolyte.
- NaCl is thus a strong electrolyte.
- Checking the electrolyte status of HC2H3O2.
  - Weak acids and bases are weak electrolytes.
  - HC2H3O2 is a weak acid and is therefore a weak electrolyte.
- HC2H3O2 is thus a weak electrolyte.
- Checking the electrolyte status of NH3.
  - Weak acids and bases are weak electrolytes.
  - NH3 is a weak base and is therefore a weak electrolyte.
- NH3 is thus a weak electrolyte.
- Checking the electrolyte status of Ba(OH)2.
  - Strong acids and bases are strong electrolytes.
  - Ba(OH)2 is a strong base and is therefore a strong electrolyte.
- Ba(OH)2 is thus a strong electrolyte.
The following are non-electrolytes: (e) CH3CH2OH.

QF8

Which of the given combinations would produce a precipitate?.
- By definition, oxidation-reduction reactions occur when electrons are transferred from the atom that is oxidized to the atom that is reduced. We need to look for changes in the oxidation states of the elements in the reaction.
  - In the reactants, the oxidation state(s) of the element Ag is/are (2 1). In the product, the oxidation state(s) is/are (2 1).
  - Therefore, the reaction causes a change in oxidation state.
- Therefore, this is an oxidation reduction reaction.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - AgCl contains Ag+ and Cl-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ag+ and Cl- is insoluble in water.
  - Therefore, AgCl is insoluble in water.
- Therefore, AgCl is a precipitate.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - KNO3 contains K+ and NO3-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and NO3- is soluble in water.
  - Therefore, KNO3 is soluble in water.
- Therefore, KNO3 is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- KNO3 is made up of NO3- and K+, and is therefore a salt.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - Ca(Cl)2 contains Ca 2+ and Cl-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ca 2+ and Cl- is soluble in water.
  - Therefore, Ca(Cl)2 is soluble in water.
- Therefore, Ca(Cl)2 is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- Ca(Cl)2 is made up of Cl- and Ca 2+, and is therefore a salt.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - Al(NO3)3 contains Al 3+ and NO3-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Al 3+ and NO3- is soluble in water.
  - Therefore, Al(NO3)3 is soluble in water.
- Therefore, Al(NO3)3 is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- Al(NO3)3 is made up of NO3- and Al 3+, and is therefore a salt.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- KC2H3O2 is made up of C2H3O2- and K+, and is therefore a salt.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - KCl contains K+ and Cl-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and Cl- is soluble in water.
  - Therefore, KCl is soluble in water.
- Therefore, KCl is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- KCl is made up of Cl- and K+, and is therefore a salt.
The following combinations would produce a precipitate: (b) 1M AgNO3 solution and 1M sodium chloride.

QF9

A solution of nickel nitrate and sodium hydroxide are mixed together. Which of the given statements is true?.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Ni(NO3)2 and NaOH are thus Ni(OH)2 and NaNO3.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - Ni(OH)2 contains Ni 2+ and OH-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ni 2+ and OH- is insoluble in water.
  - Therefore, Ni(OH)2 is insoluble in water.
- Therefore, Ni(OH)2 is a precipitate.
The following statement is true: (d) Nickel hydroxide will precipitate.

QF10

The correct net ionic equation for the given reaction would include which of the following species?.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Pb(NO3)2 contains Pb 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Pb 2+ and NO3- is soluble in water.
    - Therefore, Pb(NO3)2 is soluble in water.
    - Checking the electrolyte status of Pb(NO3)2.
      - Soluble salts are strong electrolytes.
      - Pb(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Pb(NO3)2 is thus a strong electrolyte.
    - NaCl contains Na+ and Cl-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and Cl- is soluble in water.
    - Therefore, NaCl is soluble in water.
    - Checking the electrolyte status of NaCl.
      - Soluble salts are strong electrolytes.
      - NaCl is a soluble salt and is therefore a strong electrolyte.
    - NaCl is thus a strong electrolyte.
    - Checking the electrolyte status of Pb(Cl)2.
      - .
      - Pb(Cl)2 is not an electrolyte.
    - Pb(Cl)2 is thus a non electrolyte.
    - NaNO3 contains Na+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and NO3- is soluble in water.
    - Therefore, NaNO3 is soluble in water.
    - Checking the electrolyte status of NaNO3.
      - Soluble salts are strong electrolytes.
      - NaNO3 is a soluble salt and is therefore a strong electrolyte.
    - NaNO3 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore Pb 2+ + 2NO3- + 2Na+ + 2Cl- --> Pb(Cl)2 + 2Na+ + 2NO3-.
- The net ionic equation for this reaction is therefore Pb 2+ + 2Cl- --> Pb(Cl)2.
The correct net ionic equation for the given reaction would include the following species: (c) Pb2+, Cl-, and PbCl2.

QF11

Which of the given compounds is insoluble in water?.
- Pb(NO3)2 contains Pb 2+ and NO3-.
  - .
  - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Pb 2+ and NO3- is soluble in water.
- Therefore, Pb(NO3)2 is soluble in water.
- (Li)2CO3 contains Li+ and CO3 2-.
  - .
  - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Li+ and CO3 2- is soluble in water.
- Therefore, (Li)2CO3 is soluble in water.
- Ba(OH)2 contains Ba 2+ and OH-.
  - .
  - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ba 2+ and OH- is soluble in water.
- Therefore, Ba(OH)2 is soluble in water.
- BaSO4 contains Ba 2+ and SO4 2-.
  - .
  - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ba 2+ and SO4 2- is insoluble in water.
- Therefore, BaSO4 is insoluble in water.
The following compounds are insoluble in water: (e) BaSO4.

QF12

What are the spectator ions in the given reaction?.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of (Na)2SO4 and Ba(NO3)2 are thus NaNO3 and BaSO4.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - (Na)2SO4 contains Na+ and SO4 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and SO4 2- is soluble in water.
    - Therefore, (Na)2SO4 is soluble in water.
    - Checking the electrolyte status of (Na)2SO4.
      - Soluble salts are strong electrolytes.
      - (Na)2SO4 is a soluble salt and is therefore a strong electrolyte.
    - (Na)2SO4 is thus a strong electrolyte.
    - Ba(NO3)2 contains Ba 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ba 2+ and NO3- is soluble in water.
    - Therefore, Ba(NO3)2 is soluble in water.
    - Checking the electrolyte status of Ba(NO3)2.
      - Soluble salts are strong electrolytes.
      - Ba(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Ba(NO3)2 is thus a strong electrolyte.
    - NaNO3 contains Na+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and NO3- is soluble in water.
    - Therefore, NaNO3 is soluble in water.
    - Checking the electrolyte status of NaNO3.
      - Soluble salts are strong electrolytes.
      - NaNO3 is a soluble salt and is therefore a strong electrolyte.
    - NaNO3 is thus a strong electrolyte.
    - Checking the electrolyte status of BaSO4.
      - .
      - BaSO4 is not an electrolyte.
    - BaSO4 is thus a non electrolyte.
  - The complete ionic equation for this reaction is therefore 2Na+ + SO4 2- + Ba 2+ + 2NO3- --> 2Na+ + 2NO3- + BaSO4.
- The net ionic equation for this reaction is therefore SO4 2- + Ba 2+ --> BaSO4.
The spectator ions in the given reaction are: (d) Sodium ions and nitrate ions.

QF13

Which of the given statements is true regarding the reaction given above?.
- Computing the oxidation states for H2Te.
  - Finding rules for computing the oxidation numbers for the atoms in H2Te.
    - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
    - H2Te is not a metal hydride compound, so the oxidation number of H is +1.
  - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of Te in H2Te can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of Te in H2Te is thus 0 - (2 / 1) = -2.
- Therefore, the oxidation states for H2Te are H:1 and Te:-2.
- Computing the oxidation states for O2F2.
  - Finding rules for computing the oxidation numbers for the atoms in O2F2.
    - The oxidation number of Fluorine is always -1.
    - Therefore, the oxidation number of F in O2F2 is -1.
  - Found rules to determine that: the oxidation number of F is -1. The oxidation state O(x) of O in O2F2 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of O in O2F2 is thus 0 - (-2 / 2) = 1.
- Therefore, the oxidation states for O2F2 are F:-1 and O:1.
- Computing the oxidation states for TeF6.
  - Finding rules for computing the oxidation numbers for the atoms in TeF6.
    - The oxidation number of Fluorine is always -1.
    - Therefore, the oxidation number of F in TeF6 is -1.
  - Found rules to determine that: the oxidation number of F is -1. The oxidation state O(x) of Te in TeF6 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of Te in TeF6 is thus 0 - (-6 / 1) = 6.
- Therefore, the oxidation states for TeF6 are F:-1 and Te:6.
- Computing the oxidation states for HF.
  - Finding rules for computing the oxidation numbers for the atoms in HF.
    - The oxidation number of Fluorine is always -1.
    - Therefore, the oxidation number of F in HF is -1.
    - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
    - HF is not a metal hydride compound, so the oxidation number of H is +1.
  - Found rules to determine that: the oxidation number of F is -1 and the oxidation number of H is 1.
- Therefore, the oxidation states for HF are F:-1 and H:1.
- Computing the oxidation states for O2.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of O2 is 0.
- Therefore, the oxidation states for O2 are O:0.
The following statements are true: (e) There are no changes in oxidation states or the above answers are not correct.

QF14

When methane, CH4, gas reacts with oxygen, what changes occur?.
- By definition, oxidation-reduction reactions occur when electrons are transferred from the atom that is oxidized to the atom that is reduced. We need to look for changes in the oxidation states of the elements in the reaction.
  - In the reactants, the oxidation state(s) of the element O is/are (0). In the product, the oxidation state(s) is/are (-2).
  - Therefore, the reaction causes a change in oxidation state.
- Therefore, this is an oxidation reduction reaction.
- By definition, a Hydrocarbon Combustion Reaction occurs when a hydrocarbon reacts with oxygen in air.
- Therefore, this reaction is a hydrocarbon combustion reaction.
- A hydrocarbon combustion reaction produces water and carbon dioxide.
- In this reaction CH4 and O2 react to produce H2O and CO2.
- Computing the oxidation states for CH4.
  - Finding rules for computing the oxidation numbers for the atoms in CH4.
    - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
    - CH4 is not a metal hydride compound, so the oxidation number of H is +1.
  - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of C in CH4 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of C in CH4 is thus 0 - (4 / 1) = -4.
- Therefore, the oxidation states for CH4 are H:1 and C:-4.
- Computing the oxidation states for O2.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of O2 is 0.
- Therefore, the oxidation states for O2 are O:0.
- Computing the oxidation states for the ionic parts of H2O.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of H+ is 1.
  - Finding rules for computing the oxidation numbers for the atoms in OH-.
    - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
    - OH- is not a metal hydride compound, so the oxidation number of H is +1.
  - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of O in OH- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of O in OH- is thus -1 - (1 / 1) = -2.
- Therefore, the oxidation states for H2O are H:1 and O:-2.
- Computing the oxidation states for CO2.
  - Finding rules for computing the oxidation numbers for the atoms in CO2.
  - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of C in CO2 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of C in CO2 is thus 0 - (-4 / 1) = 4.
- Therefore, the oxidation states for CO2 are O:-2 and C:4.
When methane, CH4, gas reacts with oxygen, the following changes occur: (e) Carbon dioxide and water are formed and the oxidation number of oxygen changes from zero to -2.

QF15

Given lithium and zinc:.
- By definition, a reaction between a metal and a salt is a Metal-Salt displacement reaction.
- Therefore, this reaction is a Metal-Salt displacement reaction.
- By definition, oxidation-reduction reactions occur when electrons are transferred from the atom that is oxidized to the atom that is reduced. We need to look for changes in the oxidation states of the elements in the reaction.
  - In the reactants, the oxidation state(s) of the element Zn is/are (2). In the product, the oxidation state(s) is/are (0).
  - Therefore, the reaction causes a change in oxidation state.
- Therefore, this is an oxidation reduction reaction.
- By definition, a reaction between a metal and a salt is a Metal-Salt displacement reaction.
- Therefore, this reaction is a Metal-Salt displacement reaction.
- Whether a reaction occurs depends on the activity series of the reactants.
  - The activity series can be used to predict the outcome of reactions between metals and metal salts or acids.
  - Based on the activity series, Li is more active than Zn 2+.
- Therefore, the reaction occurs, and it produces Zn and LiNO3.
(b) Lithium metal will react with zinc ions to produce zinc metal.

QF16

Which of the given species has a change in oxidation state?
- Finding changes in oxidation states.
  - Computing the oxidation states for the ionic parts of PbS.
    - For monatomic ions the oxidation number is the charge on the ion.
    - Therefore, the oxidation number of S 2- is -2.
    - For monatomic ions the oxidation number is the charge on the ion.
    - Therefore, the oxidation number of Pb 2+ is 2.
  - Therefore, the oxidation states for PbS are S:-2 and Pb:2.
  - Computing the oxidation states for H2O2.
    - Finding rules for computing the oxidation numbers for the atoms in H2O2.
      - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
      - H2O2 is not a metal hydride compound, so the oxidation number of H is +1.
    - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of O in H2O2 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of O in H2O2 is thus 0 - (2 / 2) = -1.
  - Therefore, the oxidation states for H2O2 are H:1 and O:-1.
  - Computing the oxidation states for the ionic parts of PbSO4.
    - Finding rules for computing the oxidation numbers for the atoms in SO4 2-.
    - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of S in SO4 2- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of S in SO4 2- is thus -2 - (-8 / 1) = 6.
    - For monatomic ions the oxidation number is the charge on the ion.
    - Therefore, the oxidation number of Pb 2+ is 2.
  - Therefore, the oxidation states for PbSO4 are O:-2, S:6, and Pb:2.
  - Computing the oxidation states for the ionic parts of H2O.
    - For monatomic ions the oxidation number is the charge on the ion.
    - Therefore, the oxidation number of H+ is 1.
    - Finding rules for computing the oxidation numbers for the atoms in OH-.
      - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
      - OH- is not a metal hydride compound, so the oxidation number of H is +1.
    - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of O in OH- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of O in OH- is thus -1 - (1 / 1) = -2.
  - Therefore, the oxidation states for H2O are H:1 and O:-2.
- The following species have a change in oxidation state: S and O.
The answer is: (c) S and O.

QF17

What is the absolute change in oxidation number of vanadium in the given reaction?.
- Finding the absolute change in oxidation number of vanadium in the given reaction.
  - Computing the oxidation states for V2O5.
    - Finding rules for computing the oxidation numbers for the atoms in V2O5.
    - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of V in V2O5 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of V in V2O5 is thus 0 - (-10 / 2) = 5.
  - Therefore, the oxidation states for V2O5 are O:-2 and V:5.
  - Computing the oxidation states for the ionic parts of V2O5.
  - Therefore, the oxidation states for V2O5 are O:-2 and V:5.
  - Computing the oxidation states for Cl2.
    - The oxidation number of an element is 0.
    - Therefore, the oxidation number of Cl2 is 0.
  - Therefore, the oxidation states for Cl2 are Cl:0.
  - Computing the oxidation states for the ionic parts of VOCl.
    - For monatomic ions the oxidation number is the charge on the ion.
    - Therefore, the oxidation number of V+ is 1.
    - Finding rules for computing the oxidation numbers for the atoms in ClO-.
    - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of Cl in ClO- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of Cl in ClO- is thus -1 - (-2 / 1) = 1.
  - Therefore, the oxidation states for VOCl are V:1, O:-2, and Cl:1.
  - Computing the oxidation states for O2.
    - The oxidation number of an element is 0.
    - Therefore, the oxidation number of O2 is 0.
  - Therefore, the oxidation states for O2 are O:0.
- The absolute change in oxidation number of vanadium in the given reaction is: 5 - 1 = 4.
The absolute change in oxidation number of vanadium in the given reaction is: (e) +4.

QF18

In the given equilibrium, the species that act as acids include which of the following? I. HCO3- II. H2O III. H2CO3.
- In an acid-base reaction, a chemical that gains a proton acts as a base.
- HCO3- gains a proton and is thus acting as a base.
- In an acid-base reaction, a chemical that loses a proton acts as an acid.
- H2O loses a proton and is thus acting as an acid.
- In an acid-base reaction, a chemical that loses a proton acts as an acid.
- H2CO3 loses a proton and is thus acting as an acid.
In the given equilibrium, the species that act as acids include: (e) II and III.

QF19

What is the correct acid/conjugate base pair for the given reaction?.
- Is the conjugate base of CHO2- H2O?.
  - In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
  - Therefore, the conjugate base of CHO2- is CO2 2-.
- No, the conjugate base of CHO2- is not H2O.
- Is the conjugate base of H3O+ H2O?.
  - In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
  - Therefore, the conjugate base of H3O+ is H2O.
- Yes, the conjugate base of H3O+ is H2O.
- Is the conjugate base of H3O+ HCHO2?.
  - In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
  - Therefore, the conjugate base of H3O+ is H2O.
- No, the conjugate base of H3O+ is not HCHO2.
- Is the conjugate base of HCHO2 H2O?.
  - In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
  - Therefore, the conjugate base of HCHO2 is CHO2-.
- No, the conjugate base of HCHO2 is not H2O.
- Is the conjugate base of CHO2- HCHO2?.
  - In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
  - Therefore, the conjugate base of CHO2- is CO2 2-.
- No, the conjugate base of CHO2- is not HCHO2.
The correct acid/conjugate base pair is: (b) H3O+/H2O.

QF20

What is the pH of a 1.0M solution of HCl?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - For strong acids, the concentration of H+ is equal to the concentration of acid.
      - The equilibrium concentration of strong acid reactants in equilibrium reactions is zero.
      - The equilibrium concentration of HCl is thus 0 molar.
    - HCl is a strong acid, so [H+] = 1.00 molar.
  - The pH of a solution of HCl is therefore -log10(1.00) = 0.00e+0.
- The pH of HCl solution is therefore 0.00e+0.
The pH of a 1.0M solution of HCl is: (c) 0.0.

QF22

Identify the species that is the strongest acid in the given reaction.
- In an acid-base reaction, the weakest acid and base are those on the side of the reaction where equilibrium lies, and thus the strongest acid and base lie on the opposite side of the equilibrium.
  - In a chemical reaction, the side on which equilibrium lies is related to the equilibrium constant as follows: if Keq > 1, the equilibrium lies to the right; whereas if Keq < 1, the equilibrium lies to the left.
  - The equilibrium constant, Keq, is < 1, therefore, the equilibrium lies to the left.
- Therefore, the strongest acid and base are HF and C7H5O2-.
The species that is the strongest acid is: (a) HF.

QF23

For the given reaction, what is the equilibrium constant?.
- The overall equilibrium constant Keq of a combined reaction is equal to the product of the equilibrium constants for each individual reaction.
- Keq = 4.30e-7 * 5.60e-11 = 2.41e-17.
For the given reaction, the equilibrium constant is (c) 2.4 x 10^-17.

QF24

For the given acid, what is the ionization constant?.
- In a hydrolysis reaction the acid dissociation constant Ka is equal to the equilibrium constant Keq.
  - The equilibrium constant Keq of a reaction must be computed from the Keq expression.
    - The equilibrium constant expression for an equilibrium reaction is a ratio of product concentrations and reactant concentrations.
      - For the hydrolysis of an acid (aHX + H2O <--> cX + dH3O+) the equilibrium expression is {[X]^c [H3O+]^d}/{[HX]^a}.
      - The equilibrium expression for the hydrolysis is thus {[H3O+][C2H3O2-]}/{[HC2H3O2]}.
    - .
    - In the hydrolysis of an acid, Keq = {[H+]^2}/{[HC2H3O2]initial - [H+]}.
      - In the hydrolysis of HC2H3O2, [H+] can be computed from pH.
        
        Since the pH is known, [H+] = 10^(-pH).
        Therefore, [H+] = 10^(-2.63e+0) = 2.34e-3 molar.
      - In the hydrolysis of HC2H3O2, [H+] = 2.34e-3 molar.
    - The equilibrium constant of the hydrolysis is thus Keq = {(2.34e-3)^2}/{0.300 - 2.34e-3} = 1.85e-5.
  - Keq = 1.85e-5.
- Therefore, Ka = 1.85e-5.
For the given acid, the ionization constant is (a) 1.8 x 10^-5.

QF25

What is the pH of a 0.05M solution of hypochlorous acid?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HClO, [H+] can be computed from the formula: Ka = [H+]^2 / ([HClO]initial - [H+]).
      - Given that Ka = 3.00e-8 and [HClO] = 0.0500, solving for [H+] gives 3.87e-5.
    - Therefore, [H+] = 3.87e-5 molar.
  - The pH of a solution of HClO is therefore -log10(3.87e-5) = 4.41.
- The pH of HClO solution is therefore 4.41.
The pH of a 0.05M solution of hypochlorous acid is: (d) Between 4 and 7.

QF26

A 0.125M acetic acid would have what percent of the acid dissociated at 25oC?.
- For a weak acid whose concentration and dissociation constant Ka are known, the percent dissociation is ([H+]/[HC2H3O2]) x 100.
  - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
  - Given that Ka = 1.80e-5 and [HC2H3O2] = 0.125, solving for [H+] gives 1.50e-3.
- The percent dissociation of the weak acid HC2H3O2 is thus (1.50e-3/0.125) x 100 = 1.20%.
At 25oC, a 0.125M acetic acid would have the following percentage of acid dissociated: (c) 1.2%.

QF27

What is the concentration of hydronium ions in a 0.075 solution of acetic acid?.
- [H+] must be computed.
  - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
  - Given that Ka = 1.80e-5 and [HC2H3O2] = 0.0750, solving for [H+] gives 1.16e-3.
- Therefore, [H+] = 1.16e-3 molar.
The concentration of hydronium ions in a 0.075 solution of acetic acid is: (a) 1.16 x 10^-3 M.

QF28

How many moles of HF must be present in 0.500L to form a solution with a pH of 1.85?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HF, [H+] can be computed from the formula: Ka = [H+]^2 / ([HF]initial - [H+]).
      - Given that Ka = 6.80e-4 and [HF] = 0.294, solving for [H+] gives 0.0141.
    - Therefore, [H+] = 0.0141 molar.
  - The pH of a solution of HF is therefore -log10(0.0141) = 1.85.
- The pH of HF solution is therefore 1.85.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HF, [H+] can be computed from the formula: Ka = [H+]^2 / ([HF]initial - [H+]).
      - Given that Ka = 6.80e-4 and [HF] = 0.588, solving for [H+] gives 0.0200.
    - Therefore, [H+] = 0.0200 molar.
  - The pH of a solution of HF is therefore -log10(0.0200) = 1.70.
- The pH of HF solution is therefore 1.70.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HF, [H+] can be computed from the formula: Ka = [H+]^2 / ([HF]initial - [H+]).
      - Given that Ka = 6.80e-4 and [HF] = 0.0960, solving for [H+] gives 8.08e-3.
    - Therefore, [H+] = 8.08e-3 molar.
  - The pH of a solution of HF is therefore -log10(8.08e-3) = 2.09.
- The pH of HF solution is therefore 2.09.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HF, [H+] can be computed from the formula: Ka = [H+]^2 / ([HF]initial - [H+]).
      - Given that Ka = 6.80e-4 and [HF] = 0.0480, solving for [H+] gives 5.71e-3.
    - Therefore, [H+] = 5.71e-3 molar.
  - The pH of a solution of HF is therefore -log10(5.71e-3) = 2.24.
- The pH of HF solution is therefore 2.24.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HF, [H+] can be computed from the formula: Ka = [H+]^2 / ([HF]initial - [H+]).
      - Given that Ka = 6.80e-4 and [HF] = 6.82, solving for [H+] gives 0.0681.
    - Therefore, [H+] = 0.0681 molar.
  - The pH of a solution of HF is therefore -log10(0.0681) = 1.17.
- The pH of HF solution is therefore 1.17.
To form a solution with a pH of 1.85 the following number of moles of HF must be present in 0.500L: (a) 0.147 moles.

QF29

What is the molar concentration of acetic acid in the given sample of vinegar?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
      - Given that Ka = 1.80e-5 and [HC2H3O2] = 0.0880, solving for [H+] gives 1.26e-3.
    - Therefore, [H+] = 1.26e-3 molar.
  - The pH of a solution of HC2H3O2 is therefore -log10(1.26e-3) = 2.90.
- The pH of HC2H3O2 solution is therefore 2.90.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
      - Given that Ka = 1.80e-5 and [HC2H3O2] = 0.126, solving for [H+] gives 1.51e-3.
    - Therefore, [H+] = 1.51e-3 molar.
  - The pH of a solution of HC2H3O2 is therefore -log10(1.51e-3) = 2.82.
- The pH of HC2H3O2 solution is therefore 2.82.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
      - Given that Ka = 1.80e-5 and [HC2H3O2] = 1.26e-3, solving for [H+] gives 1.51e-4.
    - Therefore, [H+] = 1.51e-4 molar.
  - The pH of a solution of HC2H3O2 is therefore -log10(1.51e-4) = 3.82.
- The pH of HC2H3O2 solution is therefore 3.82.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
      - Given that Ka = 1.80e-5 and [HC2H3O2] = 0.890, solving for [H+] gives 4.00e-3.
    - Therefore, [H+] = 4.00e-3 molar.
  - The pH of a solution of HC2H3O2 is therefore -log10(4.00e-3) = 2.40.
- The pH of HC2H3O2 solution is therefore 2.40.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
      - Given that Ka = 1.80e-5 and [HC2H3O2] = 0.0140, solving for [H+] gives 5.02e-4.
    - Therefore, [H+] = 5.02e-4 molar.
  - The pH of a solution of HC2H3O2 is therefore -log10(5.02e-4) = 3.30.
- The pH of HC2H3O2 solution is therefore 3.30.
The molar concentration of acetic acid in the given sample of vinegar is: (a) 0.088 M.

QF30

What is the molar concentration of the given solution?.
The answer is: (e) lack enough information or no correct answer available.

QF32

What is the [S 2-] in a 0.10 M solution of H2S?.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of (H)2S, the concentration change constant is 2.22e-8.
    - The equilibrium concentration of H3O+ is thus 2.22e-8 molar.
  - In the hydrolysis of (H)2S, the concentration change constant is 2.22e-8.
- The equilibrium concentration of S 2- is thus 2.22e-8 molar.
The [S 2-] in a 0.10 M solution of H2S is: (e) 9.1 x 10^-9 M.

QF33

What is the Kb value for the given base?.
- The equilibrium constant Keq of a reaction must be computed from the Keq expression.
  - The equilibrium constant expression for an equilibrium reaction is a ratio of product concentrations and reactant concentrations.
    - For the hydrolysis of a base (aX + H2O <--> cHX + dOH-) the equilibrium expression is {[HX]^c [OH-]^d}/{[X]^a}.
    - The equilibrium expression for the hydrolysis is thus {[OH-][HX]}/{[X]}.
  - .
  - In the hydrolysis of an acid, Keq = {[OH-]^2}/{[X] - [OH-]}.
    - In the hydrolysis of X, [OH-] can be computed from pH and [H+].
      - Since the pH is known, [OH-] = [H+]/(10^-14).
        
        Since the pH is known, [H+] = 10^(-pH).
        Therefore, [H+] = 10^(-1.12e+1) = 5.62e-12 molar.
      - In this equilibrium reaction, [OH-] = 1.78e-3 molar.
    - In the hydrolysis of X, [OH-] = 1.78e-3 molar.
  - The equilibrium constant of the hydrolysis is thus Keq = {(1.78e-3)^2}/{0.500 - 1.78e-3} = 6.35e-6.
- Keq = 6.35e-6.
The Kb value for the given base is: (d) 6.3 x 10^-6.

QF34

What is the ionization constant of the conjugate base of benzoic acid?.
- In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
- Therefore, the conjugate base of HC7H5O2 is C7H5O2-.
The conjugate base of benzoic acid has an ionization constant of: (c) 1.59 x 10e-10.

QF35

What is the correct expression for the hydrolysis of sodium formate?.
- The equilibrium constant expression for an equilibrium reaction is a ratio of product concentrations and reactant concentrations.
  - For the hydrolysis of an acid (aHX + H2O <--> cX + dH3O+) the equilibrium expression is {[X]^c [H3O+]^d}/{[HX]^a}.
  - The equilibrium expression for the hydrolysis is thus {[OH-][HCHO2]}/{[CHO2-]}.
- .
The answer is: (d) K = [HCOOH][OH-]/[HCOO-].

QF37

Which of the given species forms an acid when added to water?.
The answer is: (e) Ca(CH3COO)2.

QF38

Which of the given sequences is in correct order of increasing acidity?.
- For oxyacids (acids of the form HxYOz) with the same central (non-oxygen) atom Y, acid strength increases with the number of oxygen atoms attached to Y.
- HClO2 is an oxyacid with the same central atom as the oxyacid HClO but with a greater number of oxygen atoms; HClO2 is therefore the stronger oxyacid.
- For oxyacids (acids of the form HxYOz) with the same central (non-oxygen) atom Y, acid strength increases with the number of oxygen atoms attached to Y.
- HClO3 is an oxyacid with the same central atom as the oxyacid HClO2 but with a greater number of oxygen atoms; HClO3 is therefore the stronger oxyacid.
- For oxyacids (acids of the form HxYOz) with the same central (non-oxygen) atom Y, acid strength increases with the number of oxygen atoms attached to Y.
- HClO4 is an oxyacid with the same central atom as the oxyacid HClO3 but with a greater number of oxygen atoms; HClO4 is therefore the stronger oxyacid.
The following sequences are in correct order of increasing acidity: (d) HClO, HClO2, HClO3, HClO4.

QF39

Identify the Lewis acid in the given list.
- Only cations can be Lewis acids; if an cation is a metal, its strength as a Lewis acid increases with the cation charge.
- Cu 2+ is a metal cation with charge 2 and will be a stronger Lewis acid than a metal cation with charge < 2.
The Lewis acid in the given list is: (d) Cu 2+.

QF40

What is the Lewis acid in the given reaction?.
The answer is: (a) This reaction does not contain a Lewis acid.

QF41

What is the product of the reaction between BH3 and ammonia gas?.
The answer is: I don't know.

QF43

Because it results in a complete acid-base neutralization, the most nearly neutral solution is: (c) KOH and HCl.
- In a complete acid-base neutralization, the result is water and a salt consisting of the cation of the base reactant and the anion of the acid reactant.
- The result of the reaction of HCl and KOH is thus H2O and KCl.
.

QF45

Because it is an acidic buffer, the buffer with a pH < 7 is: (a) CH3COOH and NaCH3COO.
.

QF46

Which statements are true of the given solution?.
The given solution is: (b) A buffer solution with a pH between 4 and 7 and (e) A solution with a pH of 7.

QF47

One measure of the buffer capacity of an acid buffer solution is:.
No idea.

QF49

How many moles of sodium cyanide must be added to the given solution to produce a solution with pH = 9.22?.
The number of moles of sodium cyanide that must be added to the given solution is: (e) None of the above.

QF50

What compounds make a buffer solution with a pH between 6 and 7?.
The proper choice of compounds is: (b) HF and NaF and (d) H2CO3 and NaHCO3.

QFD1a

Balance the given reaction and indicate what kind of reaction it is.
- By definition, a Hydrocarbon Combustion Reaction occurs when a hydrocarbon reacts with oxygen in air.
- Therefore, this reaction is a hydrocarbon combustion reaction.
- Given the unbalanced equation: C4H10 + O2 -> CO2 + H2O.
- The balanced equation is: 2C4H10 + 13O2 -> 8CO2 + 10H2O.
The balanced equation is: 2C4H10 + 13O2 --> 8CO2 + 10H2O.

QFD1b

Balance the given reaction and indicate what kind of reaction it is.
- By definition, a reaction with exactly one input and multiple products is a decomposition reaction.
- Therefore, this is a decomposition reaction.
- Given the unbalanced equation: KClO3 -> KCl + O2.
- The balanced equation is: 2KClO3 -> 2KCl + 3O2.
The balanced equation is: 2KClO3 --> 2KCl + 3O2.

QFD1c

Balance the given reaction and indicate what kind of reaction it is.
- By definition, a Hydrocarbon Combustion Reaction occurs when a hydrocarbon reacts with oxygen in air.
- Therefore, this reaction is a hydrocarbon combustion reaction.
- Given the unbalanced equation: C2H6O + O2 -> H2O + CO2.
- The balanced equation is: C2H6O + 3O2 -> 3H2O + 2CO2.
The balanced equation is: C2H6O + 3O2 --> 3H2O + 2CO2.

QFD1d

Balance the given reaction and indicate what kind of reaction it is.
- By definition, a reaction with multiple reactants and exactly one result is a combination reaction.
- Therefore, this is a combination reaction.
- Given the unbalanced equation: P4 + O2 -> P2O5.
- The balanced equation is: P4 + 5O2 -> 2P2O5.
The balanced equation is: P4 + 5O2 --> 2P2O5.

QFD1e

Balance the given reaction and indicate what kind of reaction it is.
- By definition, a reaction with multiple reactants and exactly one result is a combination reaction.
- Therefore, this is a combination reaction.
- Given the unbalanced equation: N2O5 + H2O -> HNO3.
- The balanced equation is: N2O5 + H2O -> 2HNO3.
The balanced equation is: N2O5 + H2O --> 2HNO3.

QFD2a

The following is a representative example of a precipitation reaction: Ca(OH)2 + CO2 --> CaCO3 + H2O.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - CaCO3 contains Ca 2+ and CO3 2-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ca 2+ and CO3 2- is insoluble in water.
  - Therefore, CaCO3 is insoluble in water.
- Therefore, CaCO3 is a precipitate.
.

QFD2b

The following is a representative example of an oxidation reaction: 2Fe 3+ + 2I- --> 2Fe 2+ + I.
- By definition, oxidation-reduction reactions occur when electrons are transferred from the atom that is oxidized to the atom that is reduced. We need to look for changes in the oxidation states of the elements in the reaction.
  - In the reactants, the oxidation state(s) of the element Fe is/are (3). In the product, the oxidation state(s) is/are (2).
  - Therefore, the reaction causes a change in oxidation state.
- Therefore, this is an oxidation reduction reaction.
.

QFD2c

The following is a representative example of a combustion reaction: 3O2 + 2CH4O --> 4H2O + 2CO2.
- By definition, a Hydrocarbon Combustion Reaction occurs when a hydrocarbon reacts with oxygen in air.
- Therefore, this reaction is a hydrocarbon combustion reaction.
- A hydrocarbon combustion reaction produces water and carbon dioxide.
- In this reaction O2 and CH4O react to produce H2O and CO2.
.

QFD2d

The following is a representative example of a metathesis reaction: Ni(Cl)2 + (Na)2S --> NiS + 2NaCl.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Ni(Cl)2 and (Na)2S are thus NiS and NaCl.
.

QFD2e

The following is a representative example of a reaction, its net ionic equation, and its complete ionic equation: HSO4- + H2O --> H3O+ + SO4 2-.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Checking the electrolyte status of H2O.
    - H2O is thus a non electrolyte.
  - The complete ionic equation for this reaction is therefore HSO4- + H2O --> H3O+ + SO4 2-.
- The net ionic equation for this reaction is therefore HSO4- + H2O --> H3O+ + SO4 2-.
.

QFD3a

Complete and balance the given chemical reaction.
- Metals and nonmetals react to form salts in which the cation is formed from the metal and the anion is formed from the nonmetal.
- The product of the reaction of Li and N2 is thus (Li)3N (an ionic compound consisting of Li+ and N 3-).
- Given the unbalanced equation: Li + N2 -> (Li)3N.
- The balanced equation is: 6Li + N2 -> 2(Li)3N.
The final answer is: 6Li + N2 --> 2(Li)3N.

QFD3b

Complete and balance the given chemical reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of NaI and AgNO3 are thus NaNO3 and AgI.
- Given the unbalanced equation: NaI + AgNO3 -> NaNO3 + AgI.
- The balanced equation is: NaI + AgNO3 -> NaNO3 + AgI.
The final answer is: NaI + AgNO3 --> NaNO3 + AgI.

QFD3c

Complete and balance the given chemical reaction.
The final answer is: -->.

QFD3d

Complete and balance the given chemical reaction.
- A hydrocarbon combustion reaction produces water and carbon dioxide.
- In this reaction C2H4 and O2 react to produce H2O and CO2.
- Given the unbalanced equation: C2H4 + O2 -> H2O + CO2.
- The balanced equation is: C2H4 + 3O2 -> 2H2O + 2CO2.
The final answer is: C2H4 + 3O2 --> 2H2O + 2CO2.

QFD3e

Complete and balance the given chemical reaction.
- In a complete acid-base neutralization, the result is water and a salt consisting of the cation of the base reactant and the anion of the acid reactant.
- The result of the reaction of HClO4 and Al(OH)3 is thus H2O and Al(ClO4)3.
- Given the unbalanced equation: Al(OH)3 + HClO4 -> H2O + Al(ClO4)3.
- The balanced equation is: Al(OH)3 + 3HClO4 -> 3H2O + Al(ClO4)3.
The final answer is: Al(OH)3 + 3HClO4 --> 3H2O + Al(ClO4)3.

QFD4a

Show the net ionic equation and the balanced equation for the given chemical reation.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Ca(Br)2 and AgNO3 are thus Ca(NO3)2 and AgBr.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Ca(Br)2 contains Ca 2+ and Br-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ca 2+ and Br- is soluble in water.
    - Therefore, Ca(Br)2 is soluble in water.
    - Checking the electrolyte status of Ca(Br)2.
      - Soluble salts are strong electrolytes.
      - Ca(Br)2 is a soluble salt and is therefore a strong electrolyte.
    - Ca(Br)2 is thus a strong electrolyte.
    - AgNO3 contains Ag+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ag+ and NO3- is soluble in water.
    - Therefore, AgNO3 is soluble in water.
    - Checking the electrolyte status of AgNO3.
      - Soluble salts are strong electrolytes.
      - AgNO3 is a soluble salt and is therefore a strong electrolyte.
    - AgNO3 is thus a strong electrolyte.
    - Ca(NO3)2 contains Ca 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ca 2+ and NO3- is soluble in water.
    - Therefore, Ca(NO3)2 is soluble in water.
    - Checking the electrolyte status of Ca(NO3)2.
      - Soluble salts are strong electrolytes.
      - Ca(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Ca(NO3)2 is thus a strong electrolyte.
    - Checking the electrolyte status of AgBr.
  - .
  - AgBr is not an electrolyte.
- AgBr is thus a non electrolyte.
The complete ionic equation for this reaction is therefore Ca 2+ + 2Br- + 2Ag+ + 2NO3- --> Ca 2+ + 2NO3- + 2AgBr.

The net ionic equation for this reaction is therefore 2Br- + 2Ag+ --> 2AgBr.

Given the unbalanced equation: Ca(Br)2 + AgNO3 -> Ca(NO3)2 + AgBr.

The balanced equation is: Ca(Br)2 + 2AgNO3 -> Ca(NO3)2 + 2AgBr.

The net ionic equation is 2Br- + 2Ag+ --> 2AgBr and the balanced equation is Ca(Br)2 + 2AgNO3 --> Ca(NO3)2 + 2AgBr.

QFD4b

Show the net ionic equation and the balanced equation for the given chemical reation.
- In general, the reaction produces a metal oxide (an ionic compound) along with carbon dioxide.
- In this reaction, CaCO3 reacts to produce CaO.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Checking the electrolyte status of CaCO3.
      - .
      - CaCO3 is not an electrolyte.
    - CaCO3 is thus a non electrolyte.
    - CaO contains Ca 2+ and O 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ca 2+ and O 2- is not-known-to-be-soluble in water.
    - Therefore, CaO is not-known-to-be-soluble in water.
    - Checking the electrolyte status of CaO.
      - Soluble salts are strong electrolytes.
      - CaO is a soluble salt and is therefore a strong electrolyte.
    - CaO is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore CaCO3 --> CO2 + Ca 2+ + O 2-.
- The net ionic equation for this reaction is therefore CaCO3 --> CO2 + Ca 2+ + O 2-.
- Given the unbalanced equation: CaCO3 -> CO2 + CaO.
- The balanced equation is: CaCO3 -> CO2 + CaO.
The net ionic equation is CaCO3 --> CO2 + Ca 2+ + O 2- and the balanced equation is CaCO3 --> CO2 + CaO.

QFD4d

Show the net ionic equation and the balanced equation for the given chemical reation.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Pb(NO3)2 and KOH are thus Pb(OH)2 and KNO3.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Pb(NO3)2 contains Pb 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Pb 2+ and NO3- is soluble in water.
    - Therefore, Pb(NO3)2 is soluble in water.
    - Checking the electrolyte status of Pb(NO3)2.
      - Soluble salts are strong electrolytes.
      - Pb(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Pb(NO3)2 is thus a strong electrolyte.
    - KOH contains K+ and OH-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and OH- is soluble in water.
    - Therefore, KOH is soluble in water.
    - Checking the electrolyte status of KOH.
      - Soluble salts are strong electrolytes.
      - KOH is a soluble salt and is therefore a strong electrolyte.
    - KOH is thus a strong electrolyte.
    - Checking the electrolyte status of Pb(OH)2.
      - .
      - Pb(OH)2 is not an electrolyte.
    - Pb(OH)2 is thus a non electrolyte.
    - KNO3 contains K+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and NO3- is soluble in water.
    - Therefore, KNO3 is soluble in water.
    - Checking the electrolyte status of KNO3.
      - Soluble salts are strong electrolytes.
      - KNO3 is a soluble salt and is therefore a strong electrolyte.
    - KNO3 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore Pb 2+ + 2NO3- + 2K+ + 2OH- --> Pb(OH)2 + 2K+ + 2NO3-.
- The net ionic equation for this reaction is therefore Pb 2+ + 2OH- --> Pb(OH)2.
- Given the unbalanced equation: Pb(NO3)2 + KOH -> Pb(OH)2 + KNO3.
- The balanced equation is: Pb(NO3)2 + 2KOH -> Pb(OH)2 + 2KNO3.
The net ionic equation is Pb 2+ + 2OH- --> Pb(OH)2 and the balanced equation is Pb(NO3)2 + 2KOH --> Pb(OH)2 + 2KNO3.

QFD4g

Show the net ionic equation and the balanced equation for the given chemical reation.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Cu(NO3)2 and (H)2S are thus CuS and HNO3.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Cu(NO3)2 contains Cu 2+ and NO3-. .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Cu 2+ and NO3- is soluble in water.
  - Therefore, Cu(NO3)2 is soluble in water.
  - Checking the electrolyte status of Cu(NO3)2.
    - Soluble salts are strong electrolytes.
    - Cu(NO3)2 is a soluble salt and is therefore a strong electrolyte.
  - Cu(NO3)2 is thus a strong electrolyte.
  - Checking the electrolyte status of CuS.
    - .
    - CuS is not an electrolyte.
  - CuS is thus a non electrolyte.
- The complete ionic equation for this reaction is therefore Cu 2+ + 2NO3- + (H)2S --> CuS + 2HNO3.
The net ionic equation for this reaction is therefore Cu 2+ + 2NO3- + (H)2S --> CuS + 2HNO3.
Given the unbalanced equation: Cu(NO3)2 + (H)2S -> CuS + HNO3.
The balanced equation is: Cu(NO3)2 + (H)2S -> CuS + 2HNO3.

The net ionic equation is Cu 2+ + 2NO3- + (H)2S --> CuS + 2HNO3 and the balanced equation is Cu(NO3)2 + (H)2S --> CuS + 2HNO3.

QFD4h

Show the net ionic equation and the balanced equation for the given chemical reation.
- Whether a reaction occurs depends on the activity series of the reactants.
  - The activity series can be used to predict the outcome of reactions between metals and metal salts or acids.
  - Based on the activity series, Li is more active than Zn 2+.
- Therefore, the reaction occurs, and it produces Zn and LiNO3.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Zn(NO3)2 contains Zn 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Zn 2+ and NO3- is soluble in water.
    - Therefore, Zn(NO3)2 is soluble in water.
    - Checking the electrolyte status of Zn(NO3)2.
      - Soluble salts are strong electrolytes.
      - Zn(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Zn(NO3)2 is thus a strong electrolyte.
    - LiNO3 contains Li+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Li+ and NO3- is soluble in water.
    - Therefore, LiNO3 is soluble in water.
    - Checking the electrolyte status of LiNO3.
      - Soluble salts are strong electrolytes.
      - LiNO3 is a soluble salt and is therefore a strong electrolyte.
    - LiNO3 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore 2Li + Zn 2+ + 2NO3- --> Zn + 2Li+ + 2NO3-.
- The net ionic equation for this reaction is therefore 2Li + Zn 2+ --> Zn + 2Li+.
- Given the unbalanced equation: Li + Zn(NO3)2 -> Zn + LiNO3.
- The balanced equation is: 2Li + Zn(NO3)2 -> Zn + 2LiNO3.
The net ionic equation is 2Li + Zn 2+ --> Zn + 2Li+ and the balanced equation is 2Li + Zn(NO3)2 --> Zn + 2LiNO3.

QFD5a

Identify precipitates and give the net ionic equation for the given reaction.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Sn(NO3)2 and NaOH are thus Sn(OH)2 and NaNO3.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - Sn(OH)2 contains Sn 2+ and OH-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Sn 2+ and OH- is insoluble in water.
  - Therefore, Sn(OH)2 is insoluble in water.
- Therefore, Sn(OH)2 is a precipitate.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - NaNO3 contains Na+ and NO3-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and NO3- is soluble in water.
  - Therefore, NaNO3 is soluble in water.
- Therefore, NaNO3 is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- NaNO3 is made up of NO3- and Na+, and is therefore a salt.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Sn(NO3)2 contains Sn 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Sn 2+ and NO3- is soluble in water.
    - Therefore, Sn(NO3)2 is soluble in water.
    - Checking the electrolyte status of Sn(NO3)2.
      - Soluble salts are strong electrolytes.
      - Sn(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Sn(NO3)2 is thus a strong electrolyte.
    - NaOH contains Na+ and OH-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and OH- is soluble in water.
    - Therefore, NaOH is soluble in water.
    - Checking the electrolyte status of NaOH.
      - Strong acids and bases are strong electrolytes.
      - NaOH is a strong base and is therefore a strong electrolyte.
    - NaOH is thus a strong electrolyte.
    - Checking the electrolyte status of Sn(OH)2.
      - .
      - Sn(OH)2 is not an electrolyte.
    - Sn(OH)2 is thus a non electrolyte.
    - NaNO3 contains Na+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and NO3- is soluble in water.
    - Therefore, NaNO3 is soluble in water.
    - Checking the electrolyte status of NaNO3.
      - Soluble salts are strong electrolytes.
      - NaNO3 is a soluble salt and is therefore a strong electrolyte.
    - NaNO3 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore Sn 2+ + 2NO3- + 2Na+ + 2OH- --> Sn(OH)2 + 2Na+ + 2NO3-.
- The net ionic equation for this reaction is therefore Sn 2+ + 2OH- --> Sn(OH)2.
For the given reaction, the resulting Sn(OH)2 is a precipitate; NaNO3 is not a precipitate; the net ionic equation is: Sn 2+ + 2OH- --> Sn(OH)2.

QFD5b

Identify precipitates and give the net ionic equation for the given reaction.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of (K)2SO4 and NaOH are thus KOH and (Na)2SO4.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - KOH contains K+ and OH-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and OH- is soluble in water.
  - Therefore, KOH is soluble in water.
- Therefore, KOH is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- KOH is made up of OH- and K+, and is therefore a salt.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - (Na)2SO4 contains Na+ and SO4 2-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and SO4 2- is soluble in water.
  - Therefore, (Na)2SO4 is soluble in water.
- Therefore, (Na)2SO4 is not a precipitate.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- (Na)2SO4 is made up of SO4 2- and Na+, and is therefore a salt.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - (K)2SO4 contains K+ and SO4 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and SO4 2- is soluble in water.
    - Therefore, (K)2SO4 is soluble in water.
    - Checking the electrolyte status of (K)2SO4.
      - Soluble salts are strong electrolytes.
      - (K)2SO4 is a soluble salt and is therefore a strong electrolyte.
    - (K)2SO4 is thus a strong electrolyte.
    - NaOH contains Na+ and OH-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and OH- is soluble in water.
    - Therefore, NaOH is soluble in water.
    - Checking the electrolyte status of NaOH.
      - Strong acids and bases are strong electrolytes.
      - NaOH is a strong base and is therefore a strong electrolyte.
    - NaOH is thus a strong electrolyte.
    - KOH contains K+ and OH-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing K+ and OH- is soluble in water.
    - Therefore, KOH is soluble in water.
    - Checking the electrolyte status of KOH.
      - Soluble salts are strong electrolytes.
      - KOH is a soluble salt and is therefore a strong electrolyte.
    - KOH is thus a strong electrolyte.
    - (Na)2SO4 contains Na+ and SO4 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and SO4 2- is soluble in water.
    - Therefore, (Na)2SO4 is soluble in water.
    - Checking the electrolyte status of (Na)2SO4.
      - Soluble salts are strong electrolytes.
      - (Na)2SO4 is a soluble salt and is therefore a strong electrolyte.
    - (Na)2SO4 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore 2K+ + SO4 2- + 2Na+ + 2OH- --> 2K+ + 2OH- + 2Na+ + SO4 2-.
- The net ionic equation for this reaction is therefore-->.
For the given reaction, KOH is not a precipitate; (Na)2SO4 is not a precipitate; the net ionic equation is:-->. For the given reaction, KOH is not a precipitate; (Na)2SO4 is not a precipitate; the net ionic equation is degenerate.

QFD5c

Identify precipitates and give the net ionic equation for the given reaction.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of (Na)2S and Cu(C2H3O2)2 are thus NaC2H3O2 and CuS.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- NaC2H3O2 is made up of C2H3O2- and Na+, and is therefore a salt.
- By definition, the result of a reaction is a precipitate if it is insoluble in water.
  - CuS contains Cu 2+ and S 2-.
    - .
    - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Cu 2+ and S 2- is insoluble in water.
  - Therefore, CuS is insoluble in water.
- Therefore, CuS is a precipitate.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - (Na)2S contains Na+ and S 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and S 2- is soluble in water.
    - Therefore, (Na)2S is soluble in water.
    - Checking the electrolyte status of (Na)2S.
      - Soluble salts are strong electrolytes.
      - (Na)2S is a soluble salt and is therefore a strong electrolyte.
    - (Na)2S is thus a strong electrolyte.
    - Cu(C2H3O2)2 contains Cu 2+ and C2H3O2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Cu 2+ and C2H3O2- is not-known-to-be-soluble in water.
    - Therefore, Cu(C2H3O2)2 is not-known-to-be-soluble in water.
    - Checking the electrolyte status of Cu(C2H3O2)2.
      - Soluble salts are strong electrolytes.
      - Cu(C2H3O2)2 is a soluble salt and is therefore a strong electrolyte.
    - Cu(C2H3O2)2 is thus a strong electrolyte.
    - NaC2H3O2 contains Na+ and C2H3O2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and C2H3O2- is not-known-to-be-soluble in water.
    - Therefore, NaC2H3O2 is not-known-to-be-soluble in water.
    - Checking the electrolyte status of NaC2H3O2.
      - Soluble salts are strong electrolytes.
      - NaC2H3O2 is a soluble salt and is therefore a strong electrolyte.
    - NaC2H3O2 is thus a strong electrolyte.
    - Checking the electrolyte status of CuS.
      - .
      - CuS is not an electrolyte.
    - CuS is thus a non electrolyte.
  - The complete ionic equation for this reaction is therefore 2Na+ + S 2- + Cu 2+ + 2C2H3O2- --> 2Na+ + 2C2H3O2- + CuS.
- The net ionic equation for this reaction is therefore S 2- + Cu 2+ --> CuS.
For the given reaction, NaC2H3O2 is not a precipitate; the resulting CuS is a precipitate; the net ionic equation is S 2- + Cu 2+ --> CuS.

QFD6a

Indicate the oxidation number of each element in the given reaction.
- Computing the oxidation states for Ni.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of Ni is 0.
- Therefore, the oxidation states for Ni are Ni:0.
- Computing the oxidation states for the ionic parts of Ni.
- Therefore, the oxidation states for Ni are Ni:0.
- Computing the oxidation states for Cl2.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of Cl2 is 0.
- Therefore, the oxidation states for Cl2 are Cl:0.
- Computing the oxidation states for NiCl2.
  - Finding rules for computing the oxidation numbers for the atoms in NiCl2.
    - A halogen (other than Fluorine) within a compound that does not contain oxygen or another halogen has an oxidation number of -1.
    - NiCl2 does not contain oxygen or another halogen, so the oxidation number of the halogen Cl is -1.
  - Found rules to determine that: the oxidation number of Cl is -1. The oxidation state O(x) of Ni in NiCl2 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of Ni in NiCl2 is thus 0 - (-2 / 1) = 2.
- Therefore, the oxidation states for NiCl2 are Cl:-1 and Ni:2.
- Computing the oxidation states for the ionic parts of NiCl2.
- Therefore, the oxidation states for NiCl2 are Cl:-1 and Ni:2.
The oxidation numbers are Ni:0 Cl:0 (reactants); Cl:-1 Ni:2 (products).

QFD6b

Indicate the oxidation number of each element in the given reaction.
- Computing the oxidation states for the ionic parts of Fe(NO3)2.
  - Finding rules for computing the oxidation numbers for the atoms in NO3-.
  - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of N in NO3- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of N in NO3- is thus -1 - (-6 / 1) = 5.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Fe 2+ is 2.
- Therefore, the oxidation states for Fe(NO3)2 are O:-2, N:5, and Fe:2.
- Computing the oxidation states for Al.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of Al is 0.
- Therefore, the oxidation states for Al are Al:0.
- Computing the oxidation states for the ionic parts of Al.
- Therefore, the oxidation states for Al are Al:0.
- Computing the oxidation states for Fe.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of Fe is 0.
- Therefore, the oxidation states for Fe are Fe:0.
- Computing the oxidation states for the ionic parts of Fe.
- Therefore, the oxidation states for Fe are Fe:0.
- Computing the oxidation states for the ionic parts of Al(NO3)3.
  - Finding rules for computing the oxidation numbers for the atoms in NO3-.
  - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of N in NO3- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of N in NO3- is thus -1 - (-6 / 1) = 5.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Al 3+ is 3.
- Therefore, the oxidation states for Al(NO3)3 are O:-2, N:5, and Al:3.
The oxidation numbers are O:-2 N:5 Fe:2 Al:0 (reactants); Fe:0 O:-2 N:5 Al:3 (products).

QFD6c

Indicate the oxidation number of each element in the given reaction.
- Computing the oxidation states for Cl2.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of Cl2 is 0.
- Therefore, the oxidation states for Cl2 are Cl:0.
- Computing the oxidation states for NaI.
  - Finding rules for computing the oxidation numbers for the atoms in NaI.
    - Alkali metal atoms within compounds have an oxidation number of +1.
    - Therefore, the alkali metal atoms (Na) in NaI have an oxidation number of +1.
    - A halogen (other than Fluorine) within a compound that does not contain oxygen or another halogen has an oxidation number of -1.
    - NaI does not contain oxygen or another halogen, so the oxidation number of the halogen I is -1.
  - Found rules to determine that: the oxidation number of Na is 1 and the oxidation number of I is -1.
- Therefore, the oxidation states for NaI are Na:1 and I:-1.
- Computing the oxidation states for the ionic parts of NaI.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Na+ is 1.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of I- is -1.
- Therefore, the oxidation states for NaI are Na:1 and I:-1.
- Computing the oxidation states for I2.
  - The oxidation number of an element is 0.
  - Therefore, the oxidation number of I2 is 0.
- Therefore, the oxidation states for I2 are I:0.
- Computing the oxidation states for NaCl.
  - Finding rules for computing the oxidation numbers for the atoms in NaCl.
    - Alkali metal atoms within compounds have an oxidation number of +1.
    - Therefore, the alkali metal atoms (Na) in NaCl have an oxidation number of +1.
    - A halogen (other than Fluorine) within a compound that does not contain oxygen or another halogen has an oxidation number of -1.
    - NaCl does not contain oxygen or another halogen, so the oxidation number of the halogen Cl is -1.
  - Found rules to determine that: the oxidation number of Na is 1 and the oxidation number of Cl is -1.
- Therefore, the oxidation states for NaCl are Na:1 and Cl:-1.
- Computing the oxidation states for the ionic parts of NaCl.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Na+ is 1.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Cl- is -1.
- Therefore, the oxidation states for NaCl are Na:1 and Cl:-1.
The oxidation numbers are Cl:0 Na:1 I:-1 (reactants); I:0 Na:1 Cl:-1 (products).

QFD6d

Indicate the oxidation number of each element in the given reaction.
- Computing the oxidation states for the ionic parts of PbS.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of S 2- is -2.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Pb 2+ is 2.
- Therefore, the oxidation states for PbS are S:-2 and Pb:2.
- Computing the oxidation states for H2O2.
  - Finding rules for computing the oxidation numbers for the atoms in H2O2.
    - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
    - H2O2 is not a metal hydride compound, so the oxidation number of H is +1.
  - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of O in H2O2 can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of O in H2O2 is thus 0 - (2 / 2) = -1.
- Therefore, the oxidation states for H2O2 are H:1 and O:-1.
- Computing the oxidation states for the ionic parts of PbSO4.
  - Finding rules for computing the oxidation numbers for the atoms in SO4 2-.
  - No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of S in SO4 2- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of S in SO4 2- is thus -2 - (-8 / 1) = 6.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of Pb 2+ is 2.
- Therefore, the oxidation states for PbSO4 are O:-2, S:6, and Pb:2.
- Computing the oxidation states for the ionic parts of H2O.
  - For monatomic ions the oxidation number is the charge on the ion.
  - Therefore, the oxidation number of H+ is 1.
  - Finding rules for computing the oxidation numbers for the atoms in OH-.
    - Hydrogen has an oxidation number of -1 in a metal hydride compound, otherwise it is +1.
    - OH- is not a metal hydride compound, so the oxidation number of H is +1.
  - Found rules to determine that: the oxidation number of H is 1. The oxidation state O(x) of O in OH- can be computed from the formula O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of O in OH- is thus -1 - (1 / 1) = -2.
- Therefore, the oxidation states for H2O are H:1 and O:-2.
The oxidation numbers are S:-2 Pb:2 H:1 O:-1 (reactants); O:-2 S:6 Pb:2 H:1 (products).

QFD8a

What are the solute species present in a solution of each of the given compounds?.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of an acid, the acid loses an H+.
  - The product of hydrolysis of HClO is therefore H3O+ and ClO-.
- The result of this hydrolysis is therefore H3O+ and ClO-.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of a base, the base gains an H+.
  - The product of hydrolysis of NH4Cl is therefore OH- and HNH4Cl+.
- The result of this hydrolysis is therefore OH- and HNH4Cl+.
The solute species are C3H6O, H3O+, ClO-, OH-, and HNH4Cl+.

QFD8b

What are the quantities of the solute species?.
- To find the quantity of multiply its concentration and volume:.
- 0.200 molar * 1.00 liter = 0.200 mole.
- The concentration of a chemical can be computed from its quantity and volume.
  - To find the concentration of a Chemical in a mixture, divide the quantity of the chemical by the volume of the mixture:.
  - / =.
- The concentration of NH4Cl is thus 0.200 molar.
- To find the quantity of multiply its concentration and volume:.
- 0.200 molar * 1.00 liter = 0.200 mole.
- The equilibrium concentration of a reactant in an equilibrium reaction is equal to its initial concentration minus the product of multiplying its coefficient by the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of HClO, the concentration change constant is 7.75e-5.
    - The equilibrium concentration of H3O+ is thus 7.75e-5 molar.
  - In the hydrolysis of HClO, the concentration change constant is 7.75e-5.
- The equilibrium concentration of HClO is thus 0.200 molar.
- To find the quantity of multiply its concentration and volume:.
- 0.200 molar * 1.00 liter = 0.200 mole.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
  - In the hydrolysis of HClO, the concentration change constant is 7.75e-5.
- The equilibrium concentration of H3O+ is thus 7.75e-5 molar.
- To find the quantity of multiply its concentration and volume:.
- 7.75e-5 molar * 1.00 liter = 7.75e-5 mole.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of HClO, the concentration change constant is 7.75e-5.
    - The equilibrium concentration of H3O+ is thus 7.75e-5 molar.
  - In the hydrolysis of HClO, the concentration change constant is 7.75e-5.
- The equilibrium concentration of ClO- is thus 7.75e-5 molar.
- To find the quantity of multiply its concentration and volume:.
- 7.75e-5 molar * 1.00 liter = 7.75e-5 mole.
The quantities of the solute species are: : 0.200; HClO: 0.200; H3O+: 7.75e-5; ClO-: 7.75e-5; NH4Cl: 0.200;.

QFD9a

Show the balanced molecular and net ionic equations for the given reaction.
- Given the unbalanced equation: HCl + (Na)2O -> (H)2O + NaCl.
- The balanced equation is: 2HCl + (Na)2O -> (H)2O + 2NaCl.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - (Na)2O contains Na+ and O 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and O 2- is not-known-to-be-soluble in water.
    - Therefore, (Na)2O is not-known-to-be-soluble in water.
    - Checking the electrolyte status of (Na)2O.
      - Soluble salts are strong electrolytes.
      - (Na)2O is a soluble salt and is therefore a strong electrolyte.
    - (Na)2O is thus a strong electrolyte.
    - NaCl contains Na+ and Cl-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and Cl- is soluble in water.
    - Therefore, NaCl is soluble in water.
    - Checking the electrolyte status of NaCl.
      - Soluble salts are strong electrolytes.
      - NaCl is a soluble salt and is therefore a strong electrolyte.
    - NaCl is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore 2HCl + 2Na+ + O 2- --> (H)2O + 2Na+ + 2Cl-.
- The net ionic equation for this reaction is therefore 2HCl + O 2- --> (H)2O + 2Cl-.
The final answer is: the balanced equation is 2HCl + (Na)2O --> (H)2O + 2NaCl and the net ionic equation is 2HCl + O 2- --> (H)2O + 2Cl-.

QFD9b

Show the balanced molecular and net ionic equations for the given reaction.
- Given the unbalanced equation: Ca(Cl)2 + (Na)2CO3 -> CaCO3 + NaCl.
- The balanced equation is: Ca(Cl)2 + (Na)2CO3 -> CaCO3 + 2NaCl.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Ca(Cl)2 contains Ca 2+ and Cl-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Ca 2+ and Cl- is soluble in water.
    - Therefore, Ca(Cl)2 is soluble in water.
    - Checking the electrolyte status of Ca(Cl)2.
      - Soluble salts are strong electrolytes.
      - Ca(Cl)2 is a soluble salt and is therefore a strong electrolyte.
    - Ca(Cl)2 is thus a strong electrolyte.
    - (Na)2CO3 contains Na+ and CO3 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and CO3 2- is soluble in water.
    - Therefore, (Na)2CO3 is soluble in water.
    - Checking the electrolyte status of (Na)2CO3.
      - Soluble salts are strong electrolytes.
      - (Na)2CO3 is a soluble salt and is therefore a strong electrolyte.
    - (Na)2CO3 is thus a strong electrolyte.
    - Checking the electrolyte status of CaCO3.
      - .
      - CaCO3 is not an electrolyte.
    - CaCO3 is thus a non electrolyte.
    - NaCl contains Na+ and Cl-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and Cl- is soluble in water.
    - Therefore, NaCl is soluble in water.
    - Checking the electrolyte status of NaCl.
      - Soluble salts are strong electrolytes.
      - NaCl is a soluble salt and is therefore a strong electrolyte.
    - NaCl is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore Ca 2+ + 2Cl- + 2Na+ + CO3 2- --> CaCO3 + 2Na+ + 2Cl-.
- The net ionic equation for this reaction is therefore Ca 2+ + CO3 2- --> CaCO3.
The final answer is: the balanced equation is Ca(Cl)2 + (Na)2CO3 --> CaCO3 + 2NaCl and the net ionic equation is Ca 2+ + CO3 2- --> CaCO3.

QFD9c

Show the balanced molecular and net ionic equations for the given reaction.
- Given the unbalanced equation: Mg + HBr -> H2 + Mg(Br)2.
- The balanced equation is: Mg + 2HBr -> H2 + Mg(Br)2.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Mg(Br)2 contains Mg 2+ and Br-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Mg 2+ and Br- is soluble in water.
    - Therefore, Mg(Br)2 is soluble in water.
    - Checking the electrolyte status of Mg(Br)2.
      - Soluble salts are strong electrolytes.
      - Mg(Br)2 is a soluble salt and is therefore a strong electrolyte.
    - Mg(Br)2 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore Mg + 2HBr --> H2 + Mg 2+ + 2Br-.
- The net ionic equation for this reaction is therefore Mg + 2HBr --> H2 + Mg 2+ + 2Br-.
The final answer is: the balanced equation is Mg + 2HBr --> H2 + Mg(Br)2 and the net ionic equation is Mg + 2HBr --> H2 + Mg 2+ + 2Br-.

QFD9d

Show the balanced molecular and net ionic equations for the given reaction.
- Given the unbalanced equation: Cu(NO3)2 + (Na)2S -> CuS + NaNO3.
- The balanced equation is: Cu(NO3)2 + (Na)2S -> CuS + 2NaNO3.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Cu(NO3)2 contains Cu 2+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Cu 2+ and NO3- is soluble in water.
    - Therefore, Cu(NO3)2 is soluble in water.
    - Checking the electrolyte status of Cu(NO3)2.
      - Soluble salts are strong electrolytes.
      - Cu(NO3)2 is a soluble salt and is therefore a strong electrolyte.
    - Cu(NO3)2 is thus a strong electrolyte.
    - (Na)2S contains Na+ and S 2-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and S 2- is soluble in water.
    - Therefore, (Na)2S is soluble in water.
    - Checking the electrolyte status of (Na)2S.
      - Soluble salts are strong electrolytes.
      - (Na)2S is a soluble salt and is therefore a strong electrolyte.
    - (Na)2S is thus a strong electrolyte.
    - Checking the electrolyte status of CuS.
      - .
      - CuS is not an electrolyte.
    - CuS is thus a non electrolyte.
    - NaNO3 contains Na+ and NO3-.
      - .
      - According to Table 4.1 of Brown, LeMay and Bursten (2003), an Ionic Compound containing Na+ and NO3- is soluble in water.
    - Therefore, NaNO3 is soluble in water.
    - Checking the electrolyte status of NaNO3.
      - Soluble salts are strong electrolytes.
      - NaNO3 is a soluble salt and is therefore a strong electrolyte.
    - NaNO3 is thus a strong electrolyte.
  - The complete ionic equation for this reaction is therefore Cu 2+ + 2NO3- + 2Na+ + S 2- --> CuS + 2Na+ + 2NO3-.
- The net ionic equation for this reaction is therefore Cu 2+ + S 2- --> CuS.
The final answer is: the balanced equation is Cu(NO3)2 + (Na)2S --> CuS + 2NaNO3 and the net ionic equation is Cu 2+ + S 2- --> CuS.

QFD11a

Calculate the hydrogen ion concentration in a 0.35-molar solution of propanoic acid.
- [H+] must be computed.
  - In the hydrolysis of HC3H5O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC3H5O2]initial - [H+]).
  - Given that Ka = 1.30e-5 and [HC3H5O2] = 0.350, solving for [H+] gives 2.13e-3.
- Therefore, [H+] = 2.13e-3 molar.
The hydrogen ion concentration in a 0.35-molar solution of propanoic acid is 2.13e-3.

QFD11b

Calculate the percentage of propanoic acid molecules that are ionized in the given solution.
- For a weak acid whose concentration and dissociation constant Ka are known, the percent dissociation is ([H+]/[HC3H5O2]) x 100.
  - In the hydrolysis of HC3H5O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC3H5O2]initial - [H+]).
  - Given that Ka = 1.30e-5 and [HC3H5O2] = 0.350, solving for [H+] gives 2.13e-3.
- The percent dissociation of the weak acid HC3H5O2 is thus (2.13e-3/0.350) x 100 = 0.609%.
The percentage of propanoic acid molecules that are ionized in the given solution is: 0.609%.

QFD12a

Calculate the pH and pOH of a 0.125M solution of dimethylamine.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+], where [H+] = [OH-]/10^-14.
    [OH-] must be computed.
    - In the hydrolysis of NH4(CH3)2, [H+] can be computed from the formula: Ka = [H+]^2 / ([NH4(CH3)2]initial - [H+]).
    - Given that Ka = 5.40e-4 and [NH4(CH3)2] = 0.125, solving for [H+] gives 8.22e-3.
  - In this equilibrium reaction, [OH-] = 8.22e-3 molar.
- The pH of NH4(CH3)2 solution is therefore -log10(8.22e-3/10^-14) = 11.91.
The pH of NH4(CH3)2 solution is therefore 11.91.

The pH is 11.91 and the pOH is 2.09.

QFD12b

Write the correctly balanced net ionic equation for the reaction that occurs when (CH3)2NHNO3 is dissolved in water and calculate the numerical value of the equilibrium constant for the reaction.
- The net ionic equation of a reaction is the complete ionic equation with ions that appear in both reactants and products omitted.
  - The complete ionic equation of a reaction is the chemical equation with all soluble strong electrolytes shown as ions.
    - Checking the electrolyte status of H2O.
    - H2O is thus a non electrolyte.
  - The complete ionic equation for this reaction is therefore (CH3)2NH2NO3 2+ + 0H2O --> 0H3O+ + (CH3)2NHNO3+.
- The net ionic equation for this reaction is therefore (CH3)2NH2NO3 2+ + 0H2O --> 0H3O+ + (CH3)2NHNO3+.
The correctly balanced net ionic equation for the reaction that occurs when (CH3)2NHNO3 is dissolved in water is (CH3)2NH2NO3 2+ + 0H2O --> 0H3O+ + (CH3)2NHNO3+. The numerical value of the equilibrium constant for the reaction is .

QFD13a

Calculate the pH of a 0.0010 M solution of boric acid.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of (H)3BO3, [H+] can be computed from the formula: Ka = [H+]^2 / ([(H)3BO3]initial - [H+]).
      - Given that Ka = 5.80e-10 and [(H)3BO3] = 1.00e-3, solving for [H+] gives 8.73e-4.
    - Therefore, [H+] = 8.73e-4 molar.
  - The pH of a solution of (H)3BO3 is therefore -log10(8.73e-4) = 3.06.
- The pH of (H)3BO3 solution is therefore 3.06.
The pH of a 0.0010 M solution of boric acid is 3.06.

QFD14a

Calculate the equilibrium concentrations of H3O+, C6H5COO- and C6H5COOH in the given solution.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
  - In the hydrolysis of HC7H5O2, the concentration change constant is 1.77e-3.
- The equilibrium concentration of H3O+ is thus 1.77e-3 molar.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of HC7H5O2, the concentration change constant is 1.77e-3.
    - The equilibrium concentration of H3O+ is thus 1.77e-3 molar.
  - In the hydrolysis of HC7H5O2, the concentration change constant is 1.77e-3.
- The equilibrium concentration of C7H5O2- is thus 1.77e-3 molar.
- The equilibrium concentration of a reactant in an equilibrium reaction is equal to its initial concentration minus the product of multiplying its coefficient by the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of HC7H5O2, the concentration change constant is 1.77e-3.
    - The equilibrium concentration of H3O+ is thus 1.77e-3 molar.
  - In the hydrolysis of HC7H5O2, the concentration change constant is 1.77e-3.
- The equilibrium concentration of HC7H5O2 is thus 0.0482 molar.
The equilibrium concentrations of H3O+, C6H5COO- and C6H5COOH are: 1.77e-3, 1.77e-3 and 0.0482.

QFD14b

Write the formula for the conjugate base of benzoic acid and determine its equilibrium constant.
- In acid-base equilibrium reactions, both the forward and the backward reaction involve proton transfers, so the reactant base is the conjugate base of the product acid and the product base is the conjugate base of the reactant acid.
- Therefore, the conjugate base of HC7H5O2 is C7H5O2-.
The kb of C7H5O2- = 1.59e-10.

QFD14c

What is the pH of a 1.0 M solution of benzoic acid?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HC7H5O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC7H5O2]initial - [H+]).
      - Given that Ka = 6.30e-5 and [HC7H5O2] = 1.00, solving for [H+] gives 7.94e-3.
    - Therefore, [H+] = 7.94e-3 molar.
  - The pH of a solution of HC7H5O2 is therefore -log10(7.94e-3) = 2.10.
- The pH of HC7H5O2 solution is therefore 2.10.
The pH of a 1.0 M solution of benzoic acid is: 2.10.

QFD14d

What would be the percentage of ionization of benzoic acid in a solution of containing 0.120 moles of sodium benzoate in 250.o mL of 0.375 M benzoic acid?.
- For a weak acid whose concentration and dissociation constant Ka are known, the percent dissociation is ([H+]/[HC7H5O2]) x 100.
  - In the hydrolysis of HC7H5O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC7H5O2]initial - [H+]).
  - Given that Ka = 6.30e-5 and [HC7H5O2] = 0.375, solving for [H+] gives 4.86e-3.
- The percent dissociation of the weak acid HC7H5O2 is thus (4.86e-3/0.375) x 100 = 1.30%.
The percentage of ionization is: 1.30%.

QFD15a

Calculate the pH and pOH of a 0.050 M solution of nitrous acid.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HNO2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HNO2]initial - [H+]).
      - Given that Ka = 4.50e-4 and [HNO2] = 0.0500, solving for [H+] gives 4.74e-3.
    - Therefore, [H+] = 4.74e-3 molar.
  - The pH of a solution of HNO2 is therefore -log10(4.74e-3) = 2.32.
- The pH of HNO2 solution is therefore 2.32.
The pH is 2.32 and the pOH is 11.68.

QFD16a

What is the ratio of HCO3- to H2CO3 in blood of pH 7.4?.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of (H)2CO3, the concentration change constant is 3.98e-8.
    - The equilibrium concentration of H3O+ is thus 3.98e-8 molar.
  - In the hydrolysis of (H)2CO3, the concentration change constant is 3.98e-8.
- The equilibrium concentration of HCO3- is thus 3.98e-8 molar.
The ratio of HCO3- to H2CO3 in blood of pH 7.4 is 3.98e-8 :.

QFD16b

What is the ratio of HCO3- to H2CO3 in blood of pH 7.1?.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of (H)2CO3, the concentration change constant is 7.94e-8.
    - The equilibrium concentration of H3O+ is thus 7.94e-8 molar.
  - In the hydrolysis of (H)2CO3, the concentration change constant is 7.94e-8.
- The equilibrium concentration of HCO3- is thus 7.94e-8 molar.
The ratio of HCO3- to H2CO3 in blood of pH 7.1 is 7.94e-8 :.

QFD17a

Show the equation for the complete neutralization of carbonic acid using sodium hydroxide.
- Given the unbalanced equation: (H)2CO3 + NaOH -> H2O + (Na)2CO3.
- The balanced equation is: (H)2CO3 + 2NaOH -> 2H2O + (Na)2CO3.
The final answer is: (H)2CO3 + 2NaOH --> 2H2O + (Na)2CO3.

QFD17b

Give the equations representing the first and second dissociations of carbonic acid; give the value of the first dissociation constant, K1.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of an acid, the acid loses an H+.
  - The product of hydrolysis of (H)2CO3 is therefore H3O+ and HCO3-.
- The result of this hydrolysis is therefore H3O+ and HCO3-.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of an acid, the acid loses an H+.
  - The product of hydrolysis of HCO3- is therefore H3O+ and CO3 2-.
- The result of this hydrolysis is therefore H3O+ and CO3 2-.
The equations are: (H)2CO3 + H2O --> H3O+ + HCO3- and HCO3- + H2O --> H3O+ + CO3 2- ; the first dissociation constant, K1 = 4.30e-7.

QFD17c

Determine the pH, the concentration of bicarbonate ion and the concentration of carbonate ion in the given solution.
- Determine the pH and concentration of bicarbonate ion in the given solution.
  - The pH of a chemical depends on the concentration of H+.
    - The pH can be computed as -log10[H+].
      - [H+] must be computed.
        
        In the hydrolysis of HHCO3, [H+] can be computed from the formula: Ka = [H+]^2 / ([HHCO3]initial - [H+]).
        Given that Ka = 1 and [HHCO3] = 0.250, solving for [H+] gives 0.500.
      - Therefore, [H+] = 0.500 molar.
    - The pH of a solution of HHCO3 is therefore -log10(0.500) = 0.301.
  - The pH of HHCO3 solution is therefore 0.301.
  - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
    - In the hydrolysis of HHCO3, [H+] can be computed from the formula: Ka = [H+]^2 / ([HHCO3]initial - [H+]).
    - Given that Ka = 1 and [HHCO3] = 0.250, solving for [H+] gives 0.500.
  - The equilibrium concentration of HCO3- is thus 0.500 molar.
- HHCO3 solution: pH = 0.301; [HCO3-] = 0.500.
- What is the concentration of carbonate ion in the given solution?.
  - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
    - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
        
        In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
        In the hydrolysis of (H)2CO3, the concentration change constant is 1.82e-6.
      - The equilibrium concentration of H3O+ is thus 1.82e-6 molar.
    - In the hydrolysis of (H)2CO3, the concentration change constant is 1.82e-6.
  - The equilibrium concentration of CO3 2- is thus 1.82e-6 molar.
- [CO3 2-] = 1.82e-6.
HHCO3 solution: pH = 0.301; [HCO3-] = 0.500; [CO3 2-] = 1.82e-6.

QFD17d

Calculate the value of the equilibrium constant, Kb, for the given reaction.
- The equilibrium constant expression for an equilibrium reaction is a ratio of product concentrations and reactant concentrations.
  - For the hydrolysis of a base (aX + H2O <--> cHX + dOH-) the equilibrium expression is {[HX]^c [OH-]^d}/{[X]^a}.
  - The equilibrium expression for the hydrolysis is thus {[OH-][H(Na)2CO3+]}/{[(Na)2CO3]}.
- .
Kb =.

QFD18a

Determine the value of Ka2 for the given acid.
The value of Ka2 for the given acid is: 1.60e-12.

QFD18b

What is the pH of a 0.0010 M solution of ascorbic acid?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HHC6H6O6, [H+] can be computed from the formula: Ka = [H+]^2 / ([HHC6H6O6]initial - [H+]).
      - Given that Ka = 8.90e-5 and [HHC6H6O6] = 1.00e-3, solving for [H+] gives 2.98e-4.
    - Therefore, [H+] = 2.98e-4 molar.
  - The pH of a solution of HHC6H6O6 is therefore -log10(2.98e-4) = 3.53.
- The pH of HHC6H6O6 solution is therefore 3.53.
The pH of a 0.0010 M solution of ascorbic acid is: pH = 3.53.

QFD19a

What is the equilibrium constant for the given reaction?.
- The overall equilibrium constant Keq of a combined reaction is equal to the product of the equilibrium constants for each individual reaction.
- Keq = 7.40e-4 * 1.70e-5 * 4.00e-7 = 5.03e-15.
Therefore, the equilibrium constant for the reaction is: 5.03e-15.

QFD19b

What is the molar concentration of each of the given species?.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
      - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
      - In the hydrolysis of (H)3C6H5O7, the concentration change constant is 0.0143.
    - The equilibrium concentration of H3O+ is thus 0.0143 molar.
  - In the hydrolysis of (H)3C6H5O7, the concentration change constant is 0.0143.
- The equilibrium concentration of (H)2C6H5O7- is thus 0.0143 molar.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction. In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In the hydrolysis of (H)2C6H5O7-, the concentration change constant is 4.92e-4.
  - The equilibrium concentration of H3O+ is thus 4.92e-4 molar.
- In the hydrolysis of (H)2C6H5O7-, the concentration change constant is 4.92e-4.
The equilibrium concentration of HC6H5O7 2- is thus 4.92e-4 molar.
In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
- In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
  - In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
    - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
    - In the hydrolysis of HC6H5O7 2-, the concentration change constant is 1.40e-5.
  - The equilibrium concentration of H3O+ is thus 1.40e-5 molar.
- In the hydrolysis of HC6H5O7 2-, the concentration change constant is 1.40e-5.
The equilibrium concentration of C6H5O7 3- is thus 1.40e-5 molar.

Therefore, the molar concentrations for items i-iii are the following: 0.0143 4.92e-4 1.40e-5.

QFD20

What are the pH values for the given solutions?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HClO, [H+] can be computed from the formula: Ka = [H+]^2 / ([HClO]initial - [H+]).
      - Given that Ka = 3.00e-8 and [HClO] = 1.00, solving for [H+] gives 1.73e-4.
    - Therefore, [H+] = 1.73e-4 molar.
  - The pH of a solution of HClO is therefore -log10(1.73e-4) = 3.76.
- The pH of HClO aqueous-solution is therefore 3.76.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HBrO, [H+] can be computed from the formula: Ka = [H+]^2 / ([HBrO]initial - [H+]).
      - Given that Ka = 2.00e-9 and [HBrO] = 1.00, solving for [H+] gives 4.47e-5.
    - Therefore, [H+] = 4.47e-5 molar.
  - The pH of a solution of HBrO is therefore -log10(4.47e-5) = 4.35.
- The pH of HBrO aqueous-solution is therefore 4.35.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - [H+] must be computed.
      - In the hydrolysis of HIO, [H+] can be computed from the formula: Ka = [H+]^2 / ([HIO]initial - [H+]).
      - Given that Ka = 2.00e-11 and [HIO] = 1.00, solving for [H+] gives 4.47e-6.
    - Therefore, [H+] = 4.47e-6 molar.
  - The pH of a solution of HIO is therefore -log10(4.47e-6) = 5.35.
- The pH of HIO aqueous-solution is therefore 5.35.
The pH values for the given solutions are: HClO aqueous-solution : pH = 3.76 ; HBrO aqueous-solution : pH = 4.35 ; HIO aqueous-solution : pH = 5.35 ;.

QFD21a

How does the pH of the given solution change?.
The pH decreases.

QFD21b

Why is Fe 3+ more acidic than Cu 2+ ions in water?.
- Only cations can be Lewis acids; if an cation is a metal, its strength as a Lewis acid increases with the cation charge.
- Cu 2+ is a metal cation with charge 2 and will be a stronger Lewis acid than a metal cation with charge < 2.
- Only cations can be Lewis acids; if an cation is a metal, its strength as a Lewis acid increases with the cation charge.
- Fe 3+ is a metal cation with charge 3 and will be a stronger Lewis acid than a metal cation with charge < 3.
Fe 3+ is more acidic.

QFD22

Determine the pH of the given solution and explain whether it is reasonable.
- A solution with pH > 7 is basic.
  - The pH of a chemical depends on the concentration of H+.
    - The pH can be computed as -log10[H+].
      - [H+] must be computed.
        
        In the hydrolysis of HC6H5O, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC6H5O]initial - [H+]).
        Given that Ka = 1.30e-10 and [HC6H5O] = 1.00e-5, solving for [H+] gives 3.61e-8.
      - Therefore, [H+] = 3.61e-8 molar.
    - The pH of a solution of HC6H5O is therefore -log10(3.61e-8) = 7.44.
  - The pH of HC6H5O aqueous-solution is therefore 7.44.
- The pH of HC6H5O aqueous-solution = 7.44; HC6H5O aqueous-solution is thus basic.
The pH of the given solution is: pH = 7.44; the weak acid solution is thus basic.

QFD24

Determine the pH of each of the given salts.
The pH values for the given salts are: NaCl salt-solution: 7; NaClO salt-solution: 7; CH3NH3Cl salt-solution: 7; KIO3 salt-solution: 7; C2H8N2O3 aqueous-solution:;.

QFD25

Explain why a solution of HClO4 and NaClO4 cannot act as a buffer solution while a solution of H2CO3 and NaHCO3 can.
- By definition, an ionic compound is a monoprotic acid if it yields one H+ per molecule of acid.
- Because each molecule of HClO4 contains one H+ ion, it is a monoprotic acid.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- NaClO4 is made up of ClO4- and Na+, and is therefore a salt.
- By definition, an ionic compound is a polyprotic acid if it yields more than one H+ per molecule of acid.
- Because each molecule of (H)2CO3 contains multiple H+ ion, it is a polyprotic acid.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- NaHCO3 is made up of HCO3- and Na+, and is therefore a salt.
(H)2CO3 and NaHCO3 salt-solution is a buffer.

QFFF2

What's the conductivity for tap water and pure water?.
- Non-electrolytes have no conductivity.
  - Checking the electrolyte status of H2O.
    - .
    - H2O is not an electrolyte.
  - H2O is thus a non electrolyte.
- H2O is a non-electrolyte and thus has no conductivity.
- Strong electrolytes have high conductivity.
  - Checking the electrolyte status of (H)2SiF6.
    - Strong acids and bases are strong electrolytes.
    - (H)2SiF6 is a strong acid and is therefore a strong electrolyte.
  - (H)2SiF6 is thus a strong electrolyte.
- (H)2SiF6 is a strong electrolyte and thus has high conductivity.
Pure water has no-conductivity ; tap water has high-conductivity ; .

QFFF3

Where do the most easily oxidized metals and least easily oxidized metals occur in the periodic table?.
Based on the activity series and comparing metals between groups (i.e. IIB, IVA, etc..), we can conclude the following: The most easily oxidized metals are the alkali-metal-element in the periodic table. The least easily oxidized metals are the alkaline-earth-metal-element in the periodic table.

QFFF6

Why is the result of mixing solutions of barium nitrate and Ag2SO4 a non-conductor of electricity?.
- By definition, oxidation-reduction reactions occur when electrons are transferred from the atom that is oxidized to the atom that is reduced. We need to look for changes in the oxidation states of the elements in the reaction.
  - In the reactants, the oxidation state(s) of the element S is/are (4). In the product, the oxidation state(s) is/are (6).
  - Therefore, the reaction causes a change in oxidation state.
- Therefore, this is an oxidation reduction reaction.
- By definition, a reaction involving ionic reactants is a metathesis reaction.
- Therefore, this reaction is a metathesis reaction.
- In a metathesis reaction, the cation of each reactant is combined with the anion of the other reactant.
- The products of a metathesis reaction of Ba(NO3)2 and (Ag)2SO4 are thus BaSO4 and AgNO3.
- Conductivity depends on the electrolyte status of a chemical.
  - Non-electrolytes have no conductivity.
    - Checking the electrolyte status of BaSO4.
      - .
      - BaSO4 is not an electrolyte.
    - BaSO4 is thus a non electrolyte.
  - BaSO4 is a non-electrolyte and thus has no conductivity.
- Therefore, an aqueous solution of BaSO4 has no-conductivity.
- Conductivity depends on the electrolyte status of a chemical.
  - Strong electrolytes have high conductivity.
    - Checking the electrolyte status of AgNO3.
      - Soluble salts are strong electrolytes.
      - AgNO3 is a soluble salt and is therefore a strong electrolyte.
    - AgNO3 is thus a strong electrolyte.
  - AgNO3 is a strong electrolyte and thus has high conductivity.
- Therefore, an aqueous solution of AgNO3 has high-conductivity.
The conductivity of the ionic compounds in the resulting solution are as follows, BaSO4 : no-conductivity ; AgNO3 : high-conductivity ;.

QFFF9

The process of the autoionization of water can be described as follows:.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of an acid, the acid loses an H+.
  - The product of hydrolysis of H2O is therefore H3O+ and OH-.
- The result of this hydrolysis is therefore H3O+ and OH-.
Therefore, the following chemical equation illustrates the autoionization of water: 2H2O --> H3O+ + OH-.

QFFF9a

What is the chemical equation for the autoionization of liquid ammonia?.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - The hydrolysis of NH3 is exceptional: although it contains H, it gains another H in hydrolysis.
  - The product of hydrolysis of NH3 is therefore OH- and NH4+.
- The result of this hydrolysis is therefore OH- and NH4+.
The following chemical equation illustrates the autoionization of liquid ammonia: NH3 --> OH- + NH4+.

QFFF9c

What is the chemical equation for the autoionization of liquid HCN?.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of an acid, the acid loses an H+.
  - The product of hydrolysis of HCN is therefore H3O+ and CN-.
- The result of this hydrolysis is therefore H3O+ and CN-.
The following chemical equation illustrates the autoionization of liquid HCN: HCN --> H3O+ + CN-.

QFFF11

What is the difference in the pH values of the given solutions?.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - For strong acids, the concentration of H+ is equal to the concentration of acid.
      - The equilibrium concentration of strong acid reactants in equilibrium reactions is zero.
      - The equilibrium concentration of H3O+ is thus 0 molar.
    - H3O+ is a strong acid, so [H+] = 0.100 molar.
  - The pH of a solution of H3O+ is therefore -log10(0.100) = 1.00.
- The pH of H3O+ solution is therefore 1.00.
- The pH of a chemical depends on the concentration of H+.
  - The pH can be computed as -log10[H+].
    - For strong acids, the concentration of H+ is equal to the concentration of acid.
      - The equilibrium concentration of strong acid reactants in equilibrium reactions is zero.
      - The equilibrium concentration of H3O+ is thus 0 molar.
    - H3O+ is a strong acid, so [H+] = 1.00e-4 molar.
  - The pH of a solution of H3O+ is therefore -log10(1.00e-4) = 4.00.
- The pH of H3O+ solution is therefore 4.00.
The pH of B is 4.00 times the pH of A.

QFFF12

What's the concentration of H3O-Plus for concentrated acid and diluted acid? In what conditions would we include its contribution? Using HC7H5O2 Acid as an example.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
  - In the hydrolysis of HC7H5O2, the concentration change constant is 8.06e-3.
- The equilibrium concentration of H3O+ is thus 8.06e-3 molar.
- In hydrolysis and complete dissociation reactions that do not involve strong acid reactants, the equilibrium concentrations of the products are equal to the concentration change constant for the reaction.
  - In hydrolysis and complete dissociation reactions, the concentration of H3O+ is equal to the concentration change constant of the reaction.
  - In the hydrolysis of HC7H5O2, the concentration change constant is 2.55e-4.
- The equilibrium concentration of H3O+ is thus 2.55e-4 molar.
The percentage of H3O-Plus relative to original acid concentrations for the acids are Concentrated HC7H5O2 acid= 8.06e-3 ; Diluted HC7H5O2 acid= 0.255 ; Since the percentage of H3O-Plus is larger for diluted acids than concentrated acids, thus it is necessary to include its contribution when using diluted acids.

QFFF13

What are the percentage ionization for the concentrated and diluted solutions of HBrO acid?.
- For a weak acid whose concentration and dissociation constant Ka are known, the percent dissociation is ([H+]/[HBrO]) x 100.
  - In the hydrolysis of HBrO, [H+] can be computed from the formula: Ka = [H+]^2 / ([HBrO]initial - [H+]).
  - Given that Ka = 2.00e-9 and [HBrO] = 1.00e-3, solving for [H+] gives 1.41e-6.
- The percent dissociation of the weak acid HBrO is thus (1.41e-6/1.00e-3) x 100 = 0.141%.
- For a weak acid whose concentration and dissociation constant Ka are known, the percent dissociation is ([H+]/[HBrO]) x 100.
  - In the hydrolysis of HBrO, [H+] can be computed from the formula: Ka = [H+]^2 / ([HBrO]initial - [H+]).
  - Given that Ka = 2.00e-9 and [HBrO] = 0.100, solving for [H+] gives 1.41e-5.
- The percent dissociation of the weak acid HBrO is thus (1.41e-5/0.100) x 100 = 0.0141%.
The percent ionization of dilute solution is 0.141 ; the percent ionization of concentrated solution is 0.0141 ;.

QFFF14

Explain the acid strength trends for the following, HCl > HF, H2S > H2O, CH4 < NH3, NH3 << H2O, H2O < HF.
- Checking the strength of acids...
- HCl is known to be a stronger acid than HF.
- Checking the strength of acids...
- (H)2S is known to be a weak acid, but the strength of H2O is negligible; (H)2S is therefore the stronger acid.
- Checking the strength of bases...
- NH3 is known to be a weak base, but the strength of H2O is negligible; NH3 is therefore the stronger base.
- Checking the strength of acids...
- HF is known to be a weak acid, but the strength of H2O is negligible; HF is therefore the stronger acid.
The explanations for the acid strength trends are as follows. HCl > HF is true ; H2S > H2O is true ; NH3 > CH4 is false ; H2O >> NH3 is false ; HF > H2O is true ; .

QFFF15

Does the pH increase, decrease, or remain the same when each of the following is added:.
(A) What is the effect on pH, when Ca(C7H5O2)2 is added to a solution of HC7H5O2.
- By definition, an ionic compound is a monoprotic acid if it yields one H+ per molecule of acid.
- Because each molecule of HC7H5O2 contains one H+ ion, it is a monoprotic acid.
- A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
- Ca(C7H5O2)2 is made up of C7H5O2- and Ca 2+, and is therefore a salt. The pH decreases when these two are combined.
- (B) What is the effect on pH, when (C5H5NH)(NO3) is added to a solution of C5H5N.
- The pH decreases when these two are combined.
- (C) What is the effect on pH, when Ammonia is added to a solution of Hydrochloric Acid.
  - By definition, an ionic compound is a monoprotic acid if it yields one H+ per molecule of acid.
  - Because each molecule of HCl contains one H+ ion, it is a monoprotic acid.
- The pH remains the same when these two are combined.
- (D) What is the effect on pH, when Sodium Hydrogen Carbonate is added to a solution of Carbonic Acid.
  - By definition, an ionic compound is a polyprotic acid if it yields more than one H+ per molecule of acid.
  - Because each molecule of (H)2CO3 contains multiple H+ ion, it is a polyprotic acid.
  - A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
  - NaHCO3 is made up of HCO3- and Na+, and is therefore a salt.
- The pH increases when these two are combined.
- (E) What is the effect on pH, when Sodium Perchlorate is added to a solution of Sodium Hydroxide.
  - A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
  - NaOH is made up of OH- and Na+, and is therefore a salt.
  - A chemical is a salt if it is made up of a cation that comes from a base and an anion that comes from an acid.
  - NaClO4 is made up of ClO4- and Na+, and is therefore a salt.
- The pH remains the same when these two are combined.
.

QFFF16

Explain why solutions of CH3COOH are acidic while solutions of NaCH3COO are basic.
- In an acid-base reaction, a chemical that loses a proton acts as an acid.
- HC2H3O2 loses a proton and is thus acting as an acid.
- In an acid-base reaction, a chemical that gains a proton acts as a base.
- NaC2H3O2 gains a proton and is thus acting as a base.
HC2H3O2 is acidic; NaC2H3O2 is basic.

QFFF17a

An example of a strong electrolyte that is neither an acid nor a base is: Ba(Cl)2.
- Checking the electrolyte status of Ba(Cl)2.
  - Soluble salts are strong electrolytes.
  - Ba(Cl)2 is a soluble salt and is therefore a strong electrolyte.
- Ba(Cl)2 is thus a strong electrolyte.
.

QFFF17b

An example of a Lewis acid is: Ag+.
- Only cations can be Lewis acids; if an cation is a metal, its strength as a Lewis acid increases with the cation charge.
- Ag+ is a metal cation with charge 1 and will be a stronger Lewis acid than a metal cation with charge < 1.
.

QFFF17c

An example of a weak acid is: (H)2C2O4.
This acid is known to have a weak intensity.

QFFF17d

A buffer solution with a pH known to be close to 7.0 is: blood.
.

QFFF18

Explain why the acid strength of oxyacids increases with increasing number of oxygen atoms.
- For oxyacids (acids of the form HxYOz) with the same central (non-oxygen) atom Y, acid strength increases with the number of oxygen atoms attached to Y.
- HClO4 is an oxyacid with the same central atom as the oxyacid HClO but with a greater number of oxygen atoms; HClO4 is therefore the stronger oxyacid.
Compared HClO and HClO4.

QFFF19

Is there a Lewis acid that is not a Bronsted-Lowry acid?.
Ag+ is a Lewis acid but not a Bronsted-Lowry acid.

QFFF20

For the given reaction does the equilibrium lie to the left or to the right?.
- In acid/base equilibrium reactions, the reaction proceeds in the direction of the side where equilibrium lies.
  - In a chemical reaction, equilibrium lies on the side opposite the strong acid and base.
    - Checking the strength of acids...
    - HF is known to be a stronger acid than NH4+.
    - Checking the strength of bases...
    - NH3 is known to be a stronger base than F-.
  - The strong acid and base are reactants in the reaction, so equilibrium lies to the right.
- Therefore, this reaction proceeds to the right.
For the given reaction, the equilibrium lies to the right.

QFFF21

Using properly balanced equilibrium reactions, show that HCO3- is amphoteric.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of an acid, the acid loses an H.
  - The product of hydrolysis of HCO3- is therefore H3O+ and CO3 2-.
- The result of this hydrolysis is therefore H3O+ and CO3 2-.
- In the hydrolysis of a substance the substance either gains a proton from water or loses one to water.
  - In the hydrolysis of a base, the base gains an H+.
  - The product of hydrolysis of HCO3 is therefore OH- and HHCO3.
- The result of this hydrolysis is therefore OH- and HHCO3.
HCO3- + H2O --> H3O+ + CO3 2-; HCO3- + H2O --> H3O+ + CO3 2-.

QFFF23

Explain why the solution resulting from a strong acid and a weak base is acidic even though all of the base and acid have reacted.
The pH of a NH4Cl solution resulting from HCl(Strong Acid) and NH3(weak base) reaction is 7.

QFFF24

Benzoic acid, C6H5COOH, contains six hydrogen atoms, yet it is a monoprotic acid. This is because:.
- By definition, an ionic compound is a monoprotic acid if it yields one H+ per molecule of acid.
- Because each molecule of HC7H5O2 contains one H+ ion, it is a monoprotic acid.
Therefore, C6H5COOH is a monoprotic acid.

QFFF25

How does the reaction shift when 0.200 moles of solud sodium acetate are added to a 0.700M solution of acetic acid?.
- In acid/base equilibrium reactions, the reaction proceeds in the direction of the side where equilibrium lies.
  - In a chemical reaction, the side on which equilibrium lies is related to the equilibrium constant as follows: if Keq > 1, the equilibrium lies to the right; whereas if Keq < 1, the equilibrium lies to the left.
    - The equilibrium constant Keq of a reaction must be computed from the Keq expression.
      - The equilibrium constant expression for an equilibrium reaction is a ratio of product concentrations and reactant concentrations.
        
        For the hydrolysis of an acid (aHX + H2O <--> cX + dH3O+) the equilibrium expression is {[X]^c [H3O+]^d}/{[HX]^a}.
        The equilibrium expression for the hydrolysis is thus {[H3O+][C2H3O2-]}/{[HC2H3O2]}.
      - .
      - In the hydrolysis of an acid, Keq = {[H+]^2}/{[HC2H3O2]initial - [H+]}.
        
        In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
        Given that Ka = 1.80e-5 and [HC2H3O2] = 0.700, solving for [H+] gives 3.55e-3.
      - The equilibrium constant of the hydrolysis is thus Keq = {(3.55e-3)^2}/{0.700 - 3.55e-3} = 1.81e-5.
    - Keq = 1.80e-5.
  - The equilibrium constant, Keq, is < 1, therefore, the equilibrium lies to the left.
- Therefore, this reaction proceeds to the left.
- In acid/base equilibrium reactions, the reaction proceeds in the direction of the side where equilibrium lies.
  - In a chemical reaction, the side on which equilibrium lies is related to the equilibrium constant as follows: if Keq > 1, the equilibrium lies to the right; whereas if Keq < 1, the equilibrium lies to the left.
    - The equilibrium constant Keq of a reaction must be computed from the Keq expression.
      - The equilibrium constant expression for an equilibrium reaction is a ratio of product concentrations and reactant concentrations.
        
        For the hydrolysis of an acid (aHX + H2O <--> cX + dH3O+) the equilibrium expression is {[X]^c [H3O+]^d}/{[HX]^a}.
        The equilibrium expression for the hydrolysis is thus {[H3O+][C2H3O2-]}/{[HC2H3O2]}.
      - .
      - In the hydrolysis of an acid, Keq = {[H+]^2}/{[HC2H3O2]initial - [H+]}.
        
        In the hydrolysis of HC2H3O2, [H+] can be computed from the formula: Ka = [H+]^2 / ([HC2H3O2]initial - [H+]).
        Given that Ka = 1.80e-5 and [HC2H3O2] = 0.700, solving for [H+] gives 3.55e-3.
      - The equilibrium constant of the hydrolysis is thus Keq = {(3.55e-3)^2}/{0.700 - 3.55e-3} = 1.81e-5.
    - Keq = 1.80e-5.
  - The equilibrium constant, Keq, is < 1, therefore, the equilibrium lies to the left.
- Therefore, this reaction proceeds to the left.
For the initial solution of 0.700M of acetic acid, the direction of the equilibrium reaction lies to the left. For the resulting solution after adding 0.200 moles of solid sodium acetate to the initial solution, the direction of the equilibrium reaction lies to the left.