Challenge Exam

Project Halo

Multiple Choice

Answer all questions. Also, provide a short justification statement for your choice. Points will be given for the correct answer and for the justification.

  1. Which of the following compounds will produce a gas when HCl is added to the solid compound? HCl is a strong acid producing a yellow-green colored gas above the acid solution.

      1. Ba(OH)2 (s)

      2. CaCO3 (s)

      3. CuSO4 (s)

      4. Na3PO4(s)

      5. NaCl(s)

    1. When lithium metal is reacted with nitrogen gas, under proper conditions, the product is:

    1. no reaction occurs

    2. LiN

    3. Li2N

    4. Li3N

    5. LiN3

    1. Sodium azide is used in air bags to rapidly produce gas to inflate the bag. The products of the decomposition reaction are:

(a) Na and water.

    1. Ammonia and sodium metal.

    2. N2 and O2

    3. Sodium and nitrogen gas.

    4. Sodium oxide and nitrogen gas.

    1. When calcium carbonate is heated it decomposes forming:

Calcium carbonate reacts with acids to produce gas

      1. Solid Ca and CO2 gas

      2. Gaseous CaCO and CO2 gas

      3. Solid CaO and CO2 gas

      4. Gaseous Ca and CO2

      5. Solid CaO and liquid CO2

    1. The most likely products for the reaction of NH3 with oxygen are:

Oxygen is reactive with many chemical compounds while nitrogen gas is very unreactive.

      1. NO and water

(b) N2, H2, and H2O

(c) N2 and water

(d) N2O5 and water

(e) H2 and NO

6. Which solution has the highest conductivity?

  1. 0.5M NH3

  2. 0.5M NaOH

  3. 0.5M Na3PO4

  4. 0.5M HCl

  5. 0.5M HCN

7. Which of the following is a non-electrolyte?

(a) NaCl


(c) NH3

(d) Ba(OH)2

(e) CH3CH2OH

8. Which of the following combinations would produce a precipitate?

    1. Na metal and water.

    2. 1M AgNO3 solution and 1M sodium chloride.

    3. 1M KCl and 1M Ca(NO3)2.

    4. 0.5M Al(CH3COO)3 and 1M KNO3.

    5. 0.5M Hydrochloric acid and 0.5M KOH.

9. A solution of nickel nitrate and sodium hydroxide are mixed together. Which of the following statements is true?

(a) A precipitate will not form

(b) A precipitate of sodium nitrate will be produced.

(c) Nickel hydroxide and sodium nitrate will precipitate.

(d) Nickel hydroxide will precipitate.

(e) Hydrogen gas is produced from the sodium hydroxide.

10. Pb(NO3)2 + 2 NaCl " PbCl2 + 2 NaNO3

The correct net ionic equation for this reaction would include which of the following species?

  1. all of them

  2. Only Pb(NO3)2 and PbCl2

  3. Pb2+ , Cl-, and PbCl2

  4. Na+, NO3-, and NaNO3

(e) NaCl, Pb(NO3)2, and PbCl2

11. Which of the following compounds is insoluble in water?

(a) Pb(NO3)2

      1. Li2CO3

      2. (NH4)3PO4

      3. Ba(OH)2

      4. BaSO4

12. The spectator ions in the reaction of barium nitrate with sodium sulfate are:

(a) Sodium ions and barium ions.

(b) Sodium ions and sulfate ions.

(c) Nitrate ions and sulfate ions.

(d) Sodium ions and nitrate ions.

(e) These are not ionic compounds so there are no spectator ions.

13. H2Te(g) + 4 O2F2(g) " TeF6(g) + 2 HF(g) + 4 O2(g)

Which of the following is true regarding the reaction represented above?

  1. The oxidation number of O change from 2 to 0.

  2. The oxidation number of H changes.

  3. The oxidation number of F changes from +1 to -1.

  4. The oxidation number of Te changes from +6 to -2.

  5. There are no changes in oxidation states or the above answers are not correct.

14. When methane, CH4, gas reacts with oxygen, the following changes occur

  1. Carbon dioxide is formed and the oxidation number of oxygen remain unchanged.

  2. Carbon dioxide and water are formed and the oxidation number of oxygen remains unchanged.

  3. Carbon dioxide and water are formed and the oxidation number of oxygen changes from -2 to zero.

  4. Carbon monoxide and water are formed and the oxidation number of carbon changes from -4 to +4.

  5. Carbon dioxide and water are formed and the oxidation number of oxygen changes from zero to -2.

15. Lithium, a very reactive metal with water, is above zinc, a metal used in galvanizing, in the activity series. This means that:

  1. Zinc will react with lithium ions to produce lithium metal.

  2. Lithium metal will react with zinc ions to produce zinc metal.

  3. Zinc is oxidized.

  4. One mole of lithium reacts with one mole of zinc ions.

  5. No reaction will take place between lithium metal and zinc ions.

16. PbS (s) + 4 H2O2 (aq) " PbSO4 (s) + 4 H2O (l)

In the reaction above, which species have a change in oxidation number?

  1. Pb and S

  2. Pb and O

  3. S and O

  4. S and H

  5. O and H

17. V2O5 + Cl2 " VOCl and O2

The absolute change in oxidation number of vanadium in the reaction above is:

(a) +1

(b) +3

(c) -2

(d) -3

(e) +4

18. HCO3- + H2O D H2CO3 + OH-

In the equilibrium represented above, the species that act as acids include which of the following?

  1. HCO3-

  2. H2O

  3. H2CO3

  1. II only

  2. III only

  3. I and II

  4. I and III

(e) II and III

19. CHO2- + H3O+ D HCHO2 + H2O

The correct acid/conjugate base pair is

  1. CHO2-/H2O

  2. H3O+/H2O

  3. H3O+/HCHO2

  4. HCHO2/H2O

  5. CHO2-/HCHO2

20. The pH of a 1.0M solution of HCl is:

  1. 1.0

  2. 0.1

  3. 0.0

  4. less than zero

  5. between 0 and 1

21. The pOH of a solution containing 2.250 g of LiOH, a compound used in space craft and submarines to remove excess carbon dioxide from the atmosphere, in 250.0 mL of solution is:

(a) 0.425

(b) 13.58

(c) 0.376

(d) 13.62

(e) 0.954

22. C6H5COOH (aq) + F- (aq) D HF (aq) + C6H5COO- (aq)

The equilibrium constant is less than 0.10. HF and benzoic acid are weak acids. Benzoic acid is a colorless solid with a melting point of 1220C. Identify the species that is the strongest acid.

(a) HF

(b) C6H5COOH

(c) F-

(c) C6H5COO-

(d) H2O

23. H2CO3 + 2 H2O D 2 H3O+ + CO32-

Carbonic acid is a diprotic acid with K1 = 4.3 x 10-7 and

K2 = 5.6 x 10-11. For the reaction above, what is the equilibrium constant?

Carbon dioxide reacts with water to produce an acid.

  1. 4.3 x 10-7

  2. 5.6 x 10-11

  3. 2.4 x 10-17

  4. 2.3 x 10-8

  5. 1.8 x 10-4

24. A 0.3M solution of acetic acid has a pH of 2.63. The ionization constant of this acid is

  1. 1.8 x 10-5

  2. 7.0 x 10-4

  3. 1.1 x 10-6

  4. 7.8 x 10-3

  5. 1.9 x 10-6

25. What is the pH of a 0.05M solution of hypochlorous acid?

(For HOCl, Ka = 3.0 x 10-8). Hypochlorous acid is colorless.

(a) 8

(b) 10

(c) Between 7 and 10

(d) Between 4 and 7

(e) 3

26. Acetic acid, CH3COOH, has an equilibrium constant of 1.8 x 10-5. A 0.125M acetic acid would have what percent of the acid dissociated at 250C?

(a) 1.0%

(b) 0.2%

(c) 1.2%

(d) 9.6%

(e) 4%

____________checked after here?

27. The concentration of hydronium ions in a 0.075 solution of acetic acid is?

(For acetic acid, Ka = 1.8 x 10-5)

  1. 1.16 x 10-3M

  2. 1.35 x 10-6M

  3. 2.4 x 10-4M

  4. 0.25M

  5. 8.5 x 10-3M

28. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.500L to form a solution with a pH of 1.85?

(a) 0.147 moles

(b) 0.294 moles

(c) 0.048 moles

(d) 0.024 moles

(e) 3.41 moles

29. A particular sample of vinegar has a pH of 2.90. Assuming that acetic acid is the only acid that vinegar contains (Ka = 1.8 x 10-5), calculate the molar concentration of acetic acid in the sample?

(a) 0.088 M

(b) 0.126 M

(c) 1.26 x 10-3 M

(d) 0.890 M

(e) 0.014 M

30. A solution of acetic acid is 1.34% ionized. Ka = 1.8 x 10-5 for acetic acid. The molar concentration of the solution is:

(a) 1.0 M

(b) 0.134 M

(c) 0.0134 M

(d) 1.34 x 10-3 M

(e) lack enough information or no correct answer available.

31. Hydrofluoric acid is a weak acid, Ka = 6.8 x 10-4, and yet it is considered to be a very reactive compound. For example, HF dissolves glass. The major reason that it is considered highly reactive is:

(a) It is an acid.

(b) It forms H3O+.

(c) It dissociates.

(d) It readily forms very stable fluoride compounds.

(e) It is a weak electrolyte.

32. The [S2-] in a 0.10 M solution of H2S is: (For H2S Ka1 = 9.1 x 10-8, Ka2 = 1.2 x 10-15)

(a) 9.1 x 10-8 M

(b) 1.2 x 10-15 M

(c) 9.5 x 10-5 M

(d) 3.5 x 10-7 M

(e) 9.1 x 10-9 M

33. A 0.5M solution of a weak base B has a pH of 11.25. The pKb value for this base is

(a) 1.2 x 10-11

(b) 6.3 x 10-23

(c) 3.6 x 10-3

(d) 6.3 x 10-6

(e) Not enough information to answer the question.

This seems odd, I got that the Kb=6.3x10-6, so of course my pKb is very different????

34. The acid ionization constant for benzoic acid, C6H5COOH, is 6.3 x 10-5. The conjugate base of this acid has an ionization constant of

(a) 6.3 x 10-5

(b) 1 x 10-14

(c) 1.59 x 10-10

(d) 6.3 x 10-19

(e) 6.3 x 109

35. Which of the following is the correct expression for the hydrolysis of sodium formate, NaCOOH? Formic acid reacts with sodium hydroxide to form sodium formate.

(a) K = [HCOO-]/{[HCOOH][OH-]}

(b) K = [HCOO-]/{[HCOOH][H3O+]}

(c) K = [HCOOH]/{[HCOO-][OH-]}

(d) K = [HCOOH][OH-]/[HCOO-]

(e) K = [HCOOH][OH-]/{[HCOO-][H2O]}

36. Which of the following salts, when dissolved in water to produce a 1.0 molar solution, will produce a solution with a pH greater than 7?

(a) Sodium chloride

(b) Ammonium nitrate

(c) Potassium acetate

(d) Calcium sulfate

(e) Sodium nitrate

37. Which of the following species forms an acid when added to water?

(a) NH4NO3

(b) NH3

(c) KCl

(d) NaNO3

(e) Ca(CH3COO)2

check this??

38. Which of the following is in correct order of increasing acidity?

(a) HClO, HClO4, HClO2, HClO3

(b) HClO4, HClO3, HClO2, HClO

(c) HClO4, HClO, HClO2, HClO3

(d) HClO, HClO2, HClO3, HClO4

(e) HCl, HClO, HClO2, HClO4

39. Lewis acids are sometimes used in chemical reactions to increase the speed of the reaction. Identify the Lewis acid in the following list.

(a) KNO3

(b) Ba(OH)2

(c) NH3

(d) Cu2+

(e) CO2

40. FeBr3 (s) + Br- (aq) D FeBr4- (aq)

For the reaction above, the Lewis acid is

  1. This reaction does not contain a Lewis acid

  2. FeBr3

  3. Br-

  4. FeBr4-

(e) Both FeBr3 and FeBr4-

41. BH3 is an electron deficient compound and thus considered to be a Lewis acid. The reaction between BH3 and ammonia gas would produce

(a) B, H2, and N2.

(b) BH3NH3

(c) BHNH3 and H2.

(d) NH4+ and BH2-

(e) B2H6 and N2H2

Questions 42-45 refer to aqueous solutions containing equimolar ratios of the following pairs of substances.

  1. CH3COOH and NaCH3COO

  2. HClO4 and NaCl

  3. KOH and HCl

  4. NH3 and KOH

  5. CH3NH2 and CH3NH3Cl

42. The solution with the lowest pH.


43. The most nearly neutral solution.


44. A buffer with a pH > 7


  1. A buffer with a pH < 7


46. A solution prepared to be initially 0.5M in CH3COOH and 1M in CH3COONa is:

  1. A solution with a pH less than 7 that is not a buffer solution.

  2. A buffer solution with a pH between 4 and 7.

  3. A buffer solution with a pH between 7 and 10.

  4. A solution with a pH greater than 7 that is not a buffer solution.

  5. A solution with a pH of 7.

47. One measure of the buffer capacity of an acid buffer solution is:

(a) The ratio of acid to its conjugate base.

(b) The value of pKa.

(c) Equal to its equilibrium constant.

(d) The pH of the solution.

(e) The ratio of pH to pKa.

48. 1.0 L of a buffer formed by mixing 0.25 moles of ammonia solution with 0.25 moles of ammonium nitrate has a pH of (For ammonia, Kb = 1.8 x 10-5)

(a) 4.24

(b) 4.74

(c) 5.35

(d) 8.65

(e) 9.26

49. How many moles of sodium cyanide must be added to 200.0 mL of a 0.100 M solution of HCN to give a solution whose pH is 9.22? Sodium cyanide, when dissolved in acid produces hydrogen cyanide. Ka = 4.9 x 10-10 for HCN

(a) 0.016 moles

(b) 0.040 moles

(c) 0.010 moles

(d) 0.0002 moles

(e) None of the above

50. A students is asked to make a buffer solution with a pH between 6 and 7, using equimolar quantities of chemicals. The proper choice of compounds is:

HF Ka = 6.8 x 10-4, HCN Ka = 4.9 x 10-10, CH3COOH Ka = 1.8 x 10-5

H2CO3 Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11, Kb (NH3) = 1.8 x 10-5

    1. Acetic acid and sodium acetate

    2. HF and NaF

    3. HCN and KCN

    4. H2CO3 and NaHCO3

    5. NH3 and NH4Cl

Detailed Answers

1. Balance the following reactions, and indicate whether they are examples of combustion, decomposition, or combination

(a) C4H10 + O2 " CO2 + H2O

2C4H10 + 13O2 ® 8CO2 + 10H2O


(b) KClO3 " KCl + O2

2KClO3 ® 2KCl + 3O2


(c) CH3CH2OH + O2 " CO2 + H2O

2CH3CH2OH + 7O2 ® 4CO2 + 6H2O


(d) P4 + O2 " P2O5

P4 + 5O2 ® 2P2O5


(e) N2O5 + H2O " HNO3

N2O5 + H2O ® 2HNO3


  1. Give an explanation and representative example for each of the following reaction types.

  1. Precipitation

Reaction occurs when two aqueous solutions of ionic compounds are mixed and produce an insoluble product, such as:

Mg(NO3)2 (aq) + 2NaOH (aq) ® Mg(OH)2 (s) + 2NaNO3 (aq)

Or : Mg2+ (aq) + OH- (aq) ® Mg(OH)2 (s) net ionic equation

  1. Oxidation

  1. combustion

  2. metathesis

  3. complete net ionic equation

  1. Complete and balance the following chemical reactions:

    1. Lithium metal is heated in the presence of nitrogen gas.

    2. Sodium iodide solution is added to silver nitrate solution.

    3. Potassium chlorate is heated.

    4. C2H4 reacts with oxygen gas.

(e) Solid aluminum hydroxide is added to a solution of perchloric acid

4. Answer five of the eight options in this part. Give the formulas to show the reactants and the products for FIVE of the following chemical reactions. Each of the reactions occurs in aqueous solution unless otherwise indicated. Write the reactions as correct net ionic equations. Balance the equations.

(a) Silver nitrate reacts with calcium bromide.

(b) Solid calcium carbonate is heated to temperatures above 300oC.

(c) Copper hydroxide reacts with perchloric acid.

(d) Lead nitrate reacts with potassium hydroxide.

(e) Sodium oxide is added to water.

(f) Methylamine gas is bubbled into distilled water.

(g) Copper nitrate reacts with hydrogen sulfide gas.

(h) Lithium metal is added to a 0.5 M solution of zinc nitrate.

5. Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced net ionic equation for each reaction.

  1. Sn(NO3)2 and NaOH

  2. NaOH and K2SO4

  3. Na2S and Cu(CH3COO)2

    1. For the following, indicate the oxidation number of each element, which species is reduced, and which is oxidized.

(a) Ni + Cl2 " NiCl2

(b) 3Fe(NO3)2 + 2Al " 3Fe + 2Al(NO3)3

(c) Cl2 + 2NaI " I2 + 2NaCl

(d) PbS + 4H2O2 " PbSO4 + 4H2O

    1. Use the following reactions to prepare an activity series for the unknown species X1, X2, and X3.

X1(s) + X2+(aq) " X1+(aq) + X2(s)

X1(s) + X3+(aq) " X1+(aq) + X3(s)

X3(s) + X2+(aq) " X3+(aq) + X2(s)

(b) Which species is the most reactive in relation to the other X species?

  1. Which species most readily undergoes oxidation?

    1. Acetone, CH3COCH3, is a nonelectrolyte; hypochlorous acid is a weak electrolyte; and ammonium chloride is a strong electrolyte. (Ka for hypochlorous acid is 3.0 x 10-8)

      1. What are the solute species present in a solution of each of the compounds?

CH3COCH3, as a nonelectrolyte, is the species present.

Hypochlorous acid, HClO, as a weak electrolyte, exists in equilibrium with ClO- and H+

Ammonium chloride, NH4Cl, as a strong electrolyte, dissociates to NH4+, Cl- and then NH4+ exists in aqueous media in equilibrium with NH4OH and H+.

(b) If 0.2 moles of each compound are dissolved in solution (compounds are not mixed together in this question), how many moles of each solute species are there in each solution? Be sure to identify the species and the amount.

Assume 1 L of solution:

CH3COCH3 should have 0.2 moles of itself in solution.

HClO « ClO- + H+

Ka = x2/0.2 M = 3.0 x 10-8

x = [H+] = [ClO-] = 7.74 x 10-5 M

so that should be 7.74 x 10-5 moles of H+ and ClO- and

0.19992 moles of HClO.

NH4Cl « NH4+ + Cl-

Complete dissociation should mean there are 0.2 moles of each NH4+ and Cl-. However, if NH4+ equilibrates futher&.

NH4+ « NH4OH + H+ but I think this is beyond the problem since they didnt give us the Kb for ammonium.

9. Write balanced molecular and net ionic equations for the reactions of (assume all are aqueous solutions)

(a) Hydrochloric acid with sodium oxide to produce water and sodium chloride.

Balanced molecular : 2HCl + Na2O ® H2O + 2NaCl

Balanced net ionic : 2H+ (aq) + Na2O (s) ® H2O (l) + 2Na+ (aq)

(b) Calcium chloride with sodium carbonate to produce calcium carbonate and

sodium chloride.

Molecular: CaCl2 + Na2CO3 ® CaCO3 + 2NaCl

Net ionic: Ca2+ (aq) + CO32- (aq) ® CaCO3 (S)

(c) Hydrobromic acid with magnesium to produce magnesium bromide and


Molecular: 2HBr (aq) + Mg (metal) ® MgBr2 (aq) + H2 (g)

Net ionic: 2H+ (aq) + Mg (metal) ® Mg2+ (aq) + H2 (g)

(d) Copper nitrate with sodium sulfide.

Molecular: Cu(NO3)2 (aq) + Na2S (aq) ® CuS (s) +2NO3- (aq) + 2Na+ (aq)

Ionic: Cu2+ (aq) + S2- (aq) ® CuS (s)

10. HCl, H2SO4, HClO4, and HNO3 are all examples of strong acids and are 100% ionized in water. This is known as the “leveling effect” of the solvent. Explain how you would establish the relative strengths of these acids. That is, how would you answer a question such as “which of these acids is the strongest?”

see response from Vulcan chemist

11. The acid ionization constant, Ka for propanoic acid, C2H5COOH, is 1.3 x 10-5.

(a) Calculate the hydrogen ion concentration, [H+], in a 0.35-molar solution of propanoic acid.

C2H5COOH ® C2H5COO- + H+

Ka = x2/0.35 M = 1.3 x 10-5

x = [H+] = [C2H5COO-] = 2.13 x 10-3 M

(b) Calculate the percentage of propanoic acid molecules that are ionized in the solution in part (a).

%ionized = [C2H5COO-] / [C2H5COOH] 2.13 x 10-3/0.35 = 6.09x10-3 or 0.61%

(c) What is the ratio of the concentration of propanoate ion, C2H5COO-, to that of propanoic acid in a buffer solution with a pH of 4.90?

C2H5COOH « C2H5COO- + H+

pH = -log[H+] so [H+] = 1.26 x 10-5 M

ratio: 1.26 x 10-5/0.35 = 3.6x10-5 /////?????

(d) In a 100.0 milliliter sample of a different buffer solution, the propanoic acid concentration is 0.35-molar and the sodium propanoate concentration is 0.50-molar. To this buffer solution, solid NaOH is added. Would the pH increase or decrease?

The pH should increase because a strong base (high pH) is added to bring the pH up. The NaOH should neutralized some of the acid thereby raising the pH.

12. The ionization constant for dimethylamine, (CH3)2NH, is 5.4 x 10-4. Dimethyl amine is a substituted ammonia compound.

(a) Calculate the pH and pOH of a 0.125 M solution of dimethylamine.

(CH3)2NH « (CH3)2NH2+ + OH-

Kb = x2/0.125 M = 5.4 x 10-4

x = [OH-] = [(CH3)2NH2+] = 8.22 x 10-3 M

so the pOH = -log[OH-] = 2.08

so the pH = 14 – pOH = 11.91

(b) 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.

Is this really the correct compound? Don’t they mean (CH3)2NH2NO3 ? Let’s assume that is what is ment:

Net ionic eq: (CH3)2NH2+ « (CH3)2NH + H+

Kw = KaKb = 1x10-14, so Ka = Kw/Kb = 1x10-14/5.4x10-4 = 1.85 x 10-11

(c) Determine the pH of a solution made by adding 0.0250 mole of solid dimethylammonium nitrate to 250.0 mL of a 0.350 M solution of dimethylamine. Assume no volume change occurred.

(CH3)2NH « (CH3)2NH2+ + OH-

calc [(CH3)2NH2+]initial = 0.025mol/0.25L = 0.1 M

pH = pKa +log[base]/[acid] = -log(1.85x10-11) + log(0.35M/0.1M) = 11.28

(d) Would you need to add LiOH or HClO4 (state clearly which you choose) to (c) to produce a solution that has a pH of 11.00? Assume that no volume change occurred.

Need to add HClO4 to bring the pH down from 11.28 to 11.00.

      1. Boric acid, H3BO3, is a very weak acid with a Ka = 5.8 x 10-10. Boric acid is a colorless compound.

    1. Calculate the pH of a 0.0010 M solution of boric acid.

Ka = x2/0.001M = 5.8x10-10 so x = [H+] = 7.6x10-7 M

So pH = -log[H+] = 6.11

This has a low enough concentration of compound that it may require the quadratic equation????

    1. What is the pH of a buffer solution produced by mixing 0.010 moles of sodium borate with 100.0 mL of 0.0150 moles of boric acid. Assume that no volume change occurred.

Ka = (0.1+x)x/0.15 M = 5.8x10-10 so x = [H+] = 8.7x10-10 M

So pH = -log[H+] = 9.06

    1. Would the pH increase or decrease if 20.0 of 0.05 M NaOH was added to the solution in part (b)? Explain your answer.

The pH should increase. All of the NaOH should be consumed in the reaction, however, this will decrease the amount of H+ present thereby increasing the pH. ???????

    1. Explain whether the pH of solution (b) would increase or decrease if a small volume of 0.50 M HCl was added to the buffer solution.

The pH should not change, that is the purpose of a buffer. ??????

    1. The acid dissociation constant for benzoic acid (C6H5COOH) is 6.3 x 10-5.

(a) Calculate the equilibrium concentrations of H3O+, C6H5COO-, and C6H5COOH in a 0.05 M solution of the acid.

Ka = x2/0.05 M = 6.3x10-5, so x = [H3O+] = [C6H5COO-] = 1.77x10-3 M

[C6H5COOH] = 0.05 M 1.77x10-3 = 0.048 M

this problem appears to require the quadratic equation. recalculate!

(b) Write the formula for the conjugate base of benzoic acid and determine its equilibrium constant.

C6H5COO-, Kw = KaKb = 1x10-14, so Kb = Kw/Ka = 1x10-14/6.3x10-5 = 1.58 x 10-10

(c) What is the pH of a 1.0 M solution of benzoic acid?

Ka = x2/1.0 M = 6.3x10-5, so x = [H+] = 7.94x10-3 M

So pH = -log[H+] = 1.02 this seems too low. We probably need the quadratic eq.

(d) What would be the percentage of ionization of benzoic acid in a solution containing 0.120 moles of sodium benzoate in 250.0 mL of 0.375 M benzoic acid. Assume that no volume change occurred.

C6H5COOH « C6H5COO- + H+

0.375M 0.12mol/0.25L = 0.48M

Ka = (0.48 +x)x/(0.375-x) = 6.3x10-5,

Assuming x is much less than 0.375 do the calc leaving it out of the denominator. So, x = [H+] = [C6H5COO-] = 4.92x10-5 M

So, %ionization = [C6H5COO-]/[C6H5COOH] = 4.92x10-5/0.375 = 0.013%

    1. The Kw value for water is dependent on temperature. At 00C the value is

1.2 x 10-15. Nitrous acid Ka = 4.5 x 10-4.

(a) Calculate the pH and pOH of a 0.050 M solution of nitrous acid

(Ka = 4.5 x 10-4) at 00C. Assume that the acid ionization constant does not change with temperature.

HNO2 « NO2- + H+

Ka = x2/0.050 M = 4.5 x 10-4

x = [H+] = [NO2-] = 4.74 x 10-3 M

so the pH = -log[H+] = 2.23

Kw = KaKb = 1.2x10-15

-log[H+] + (-log[OH-]) = -logKw

so pOH = -log[OH-] = -logKw (-log[H+]) = 12.59

(b) What would the pH value be for a buffer solution made by adding 0.125 moles of calcium nitrite to 500.0 mL of a 0.500 M solution of nitrous acid? Assume no volume change occurred.

pH = pKa + log[base]/[acid]

and Nitrous acid Ka = 4.5 x 10-4

so, pH = -log(4.5x10-4) + log[0.125/0.5]/[0.5] = 3.04

(c) Would this solution have the maximum buffer capacity possible? If not, how many moles of calcium nitrate should be added to the 500.0 mL of 0.500 M acid to produce a buffer with maximum buffer capacity?

No, we need equimolar amount of the acid and conjugate base.

So, 0.25 total moles of calcium nitrate are needed for max buffer capacity.

    1. Carbonic acid is a diprotic acid with Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11.

    1. What is the ratio of HCO3- to H2CO3 in blood of pH 7.4?

pH = pKa + log[base]/[acid]

so, 7.4 - (-log4.3x10-7) = log [base]/[acid]

    1. so, [base]/[acid] = 1.08

    2. What is the ratio of HCO3- to H2CO3 in a patient with acidocis and a pH of 7.1?

so, 7.1 - (-log4.3x10-7) = log [base]/[acid]

    1. so, [base]/[acid] = 5.41 this answer doesnt make sense, it should have a lower base/acid ratio if the pH is lower, right?

17. The overall dissociation of carbonic acid, H2CO3, is represented below. The overall dissociation constant is also indicated.

H2CO3 D 2 H+ + CO32- K = 2.41 x 10-17

  1. Write the equation for the complete neutralization of carbonic acid using sodium hydroxide.

H2CO3 + 2OH- « CO32- + 2H2O

  1. Give the equations representing the first and second dissociations of carbonic acid. Calculate the value of the first dissociation constant, K1, for oxalic acid if the value of the second dissociation constant, K2, is 5.6 x 10-11.

H2CO3 « H+ + HCO3- Ka1 = ?

HCO3- « H+ + CO32- Ka2 = 5.6 x 10-11

Ktotal = Ka1Ka2 , so Ka1 = Ktotal/Ka2 = 4.3 x 10-7

  1. Determine the pH and concentration of bicarbonate ion present in a 0.250 M solution of carbonic acid. What is the concentration of carbonate ion in this solution and what approximations were made in this calculation?

H2CO3 « H+ + HCO3- Ka1 = 4.3 x 10-7

Ka = x2/0.25 M = 4.3 x 10-7

x = [H+] = [HCO3-] = 4.28 x 10-4 M

so the pH = -log[H+] = 3.48

HCO3- « H+ + CO32- Ka2 = 5.6 x 10-11

Ka = (4.28 x 10-4 +y)y/4.28 x 10-4 = 5.6 x 10-11

y = [CO32-] = 5.6 x 10-11 M

we approximated that y was very small relative to the starting 4.28x10-4 M concentration of bicarbonate.

(d) Calculate the value of the equilibrium constant, Kb, for the reaction that occurs when solid Na2CO3 is dissolved in water.

Na2CO3 + H2O « HCO3- + OH- Kb = ?

Kb = [HCO3-] [OH-] / [CO32-]

Kw = KaKb = 1x10-14, so Kb = Kw/Ka1 = 1x10-14/4.3 x 10-7= 2.32 x 10-8

18. Ascorbic acid, H2C6H6O6, is a diprotic acid with a Ka1 value of 8.9 x 10-5. The pH of a 0.125 M solution of ascorbic acid is 2.48 and the concentration of C6H6O62- is 1.6 x 10-12 M. Determine the value of Ka2.

H2C6H6O6 « HC6H6O6- + H+ Ka1 = 8.9 x 10-5

HC6H6O6- « C6H6O62- + H+ Ka2 = ?

Total: H2C6H6O6 « C6H6O62- + 2H+ Kall =

[0.125] [1.6 x 10-12] pH=2.48

so [H+] = 3.311 x 10-3

so Kall = [C6H6O62-][ H+] / [H2C6H6O6]

so Kall = [1.6 x 10-12][3.311 x 10-3] / [0.125] = 4.238 x 10-14

and Kall = Ka1 x Ka2, therefore Ka2 = 4.76 x 10-10

(b) What is the pH of a 0.0010 M solution of ascorbic acid?

H2C6H6O6 « HC6H6O6- + H+

Ka1 = [HC6H6O6-][ H+] / [H2C6H6O6] = 8.9 x 10-5

Ka1 = x2 / [0.001M] = 8.9 x 10-5

[ H+] = x = 2.98 x 10-4

pH = -log[H+] = 3.52

    1. Citric acid, H3C6H5O7, is a triprotic acid that reacts with iron. The acid equilibrium constants are, Ka1 = 7.4 x 10-4, Ka2 = 1.7 x 10-5, and Ka3 = 4.0 x 10-7.

  1. Determine the equilibrium constant for the reaction

H3C6H5O7 + 3 H2O « 3 H3O+ + C6H5O73-

Kall = Ka1 x Ka2 x Ka3 = 5.032 x 10-15

  1. What is the molar concentration of each of the following species in a 0.275 M solution (assume the pH is determined by the first equilibrium).

    1. H2C6H5O7-

H3C6H5O7 « H+ + H2C6H5O7-

Ka1 = x2/0.275 M = 7.4 x 10-4

[H+] = [H2C6H5O7-] = x = 1.43 x 10-2 M

    1. HC6H5O72-

H2C6H5O7- « H+ + HC6H5O72-

Ka2 = (1.43 x 10-2)y/(1.43 x 10-2 M) = 1.7 x 10-5

[HC6H5O72-] = y = 1.7 x 10-5 M

    1. C6H5O73-

HC6H5O72- « H+ + C6H5O73-

Ka3 = (1.43 x 10-2)z/(1.7 x 10-5 M) = 4.0 x 10-7

[C6H5O73-] = z = 4.75 x 10-10 M

    1. Order the following species from the strongest acid to the weakest and support your answer with a rationale based on acid-base behavior, structure, and chemical properties.

(a) HOF

(b) HOCl

(c) HOBr

  1. HOI

    1. Metal ions in solution do not exist as free ions. Rather, they are bonded to water molecules to form a hydrated ion. For example, Fe(H2O)63+. These hydrated ions act as acids according to the equilibrium:

Fe(H2O)63+ + H2O D Fe(H2O)5OH+ + H3O+ Ka = 2 x 10-3

  1. What is the pH of a 0.0125 M solution of Fe(NO3)3?

Assume that there should be the same amount of Fe(H2O)63+ as Fe(NO3)3 so:

Ka = x2/0.0125 = 2 x 10-3

X = [H+] = 5 x 10-3

  1. Cu2+ ions in water have a Ka of 1 x 10-8 and are thus less acidic than Fe3+ solutions. What accounts for this difference?

Cu2+ is below Fe2+ and therefore Fe3+ in the activity table. This would suggest that Cu2+ will accept electrons more readily than Fe3+ and therefore be less acidic. Cu2+ would be slower to accept protons accordingly.

    1. Phenol, C6H5OH, is a very weak acid with an acid equilibrium constant of Ka = 1.3 x 10-10. Determine the pH of a very dilute, 1 x 10-5 M, solution of phenol. Is the value acceptable? If not, give a possible explanation for the unreasonable pH value.

Ka = x2/1 x 10-5 = 1.3 x 10-10

X = [H+] = 3.6 x 10-8

So pH = -log[H+] = 7.44

This value seems unreasonable since the pH is above 7 and we know even weak acids should give a pH value below 7.

    1. Determine the pH of a buffer that is 0.240 M sodium bicarbonate and 0.375 M

sodium carbonate. Ka1= 4.3 x 10-7, Ka2 = 5.6 x 10-11

Ka2 = (0.375 - x)(x)/(0.24 + x) = 5.6 x 10-11 (assume x is so small it’s negligible)

X = [H+] = 3.58 x 10-11

So pH = -log[H+] = 10.4

  1. 0.075 moles of carbonic acid is added to 1.0 L of this buffer solution. Write the reaction that occurs. Without using a calculator, indicate whether the pH of the solution increases, decreases or stays the same, Explain your answer.

H2CO3 « HCO3- + H+

HCO3- « CO32- + H+ our buffer

H2CO3 « CO32- + 2H+ overall.

The pH should decrease because as we added carbonic acid to the left of the equilibrium, the equilibrium should shift to the right producing more H+.

    1. Determine the pH of each of the following salts when added to water to produce a 0.1000 M solution: (Ka = 1.1 x 10-2 for HOCl, Ka = 1.7 x 10-1 for HIO3, Kb = 4.4 x 10-4 for CH3NH2, Kb = 6.4 x 10-4 for CH3CH2NH2)

  1. NaCl

pH = 7. No OH- or H+ is produced.

  1. NaClO2


  1. CH3NH3Cl

CH3NH3+ + H2O « CH3NH2 + H3O+

Kw = KaKb = 1 x 10-14 , so Ka = Kw/Kb = 2.27 x 10-11

Ka = x2/0.1 = 2.27 x 10-11

x = [H+] = 1.5 x 10-6

So pH = -log[H+] = 5.82

  1. KIO3

HIO3 « IO3- + H+ Ka = 1.7 x 10-1

IO3- + H2O « HIO3 + OH-

Kw = KaKb = 1 x 10-14 , so Kb = Kw/Ka = 5.88 x 10-14

Kb = x2/0.1 = 5.88 x 10-14

x = [OH-] = 7.66 x 10-8

So pOH = -log[OH-] = 7.11

So pH = 14 pOH = 6.88

Humm, this doesnt make sense, I think it should be a pH above 7.

  1. CH3CH2NH3NO3

CH3CH2NH3+ « CH3CH2NH2 + H+

Kw = KaKb = 1 x 10-14 , so Ka = Kw/Kb = 1.56 x 10-11

Ka = x2/0.1 = 1.56 x 10-11

x = [H+] = 1.25 x 10-6

So pH = -log[H+] = 5.90

    1. Explain why a solution of HClO4 and NaClO4 cannot act as a buffer solution while a solution of H2CO3 and NaHCO3 can.

Strong acids cannot be part of a buffer, HClO4 is a strong acid.

However, H2CO3 is a weak acid and its conjugate pair, this would would well as a buffer.

Free Form

1. What is the difference between the subscript 3 in HNO3 and a coefficient 3 in front of HNO3?

The subscript 3 indicates the number of atoms of the corresponding element that are part of the molecule (in this case, 3 oxygen atoms are bound in one HNO3 molecule). The coefficient 3 indicates the number of molecules of HNO3 that are present or participating in a reaction.

2. Pure water is a poor conductor of electricity, yet ordinary tap water is a good conductor. Account for this difference.

Pure water is a molecular compound with no ions (therefore no electrolytes) of any sort present (no conductors present). However, ordinary tap water has many dissolved ions present, these electrolytes allow for conductivity.

3. Where, in general, do the most easily oxidized metals occur in the periodic table? Where do the least easily oxidized metals occur in the periodic table?

The left side of the periodic table has the most easily oxidized materials. These elements readily give up electrons to go to the +1 oxidation state.

The least easily oxidized metals are on the right side.

4. The metal cadmium tends to form Cd2+ ions. The following observations are made: (i) When a strip of zinc metal is placed in CdCl2 (aq), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in Ni(NO3)2 (aq), nickel metal is deposited.

    1. Write net ionic equations to explain each of the observations made above.

    2. What can you conclude about the position of cadmium in the activity series?

    3. When calcium metal is place in CdCl2(aq), cadmium metal is deposited on the strip. Is calcium above or below cadmium in the activity series?

5. How would you explain the following observations: copper metal reacts vigorously with 6 M nitric acid, yet reacts slowly with HCl.

  1. A solution of barium nitrate is a strong electrolyte and therefore conducts electricity. When a solution of Na2SO4 is added to the barium nitrate solution the conduction of electricity decreases until it no longer conducts. Explain this event.

BaNO3 is soluble and ionizes to conduct electricity.

When Na2SO4 is added, a reaction occurs and BaSO4 forms, this precipitates thereby removing ions and thereby reducing conductivity.

Why would it no longer conduct if there are still the spectator ions present?

  1. An instructor handed you a bottle and indicated that the contents were one of the following: silver acetate, calcium acetate, and lead nitrate. The instructor asked you to identify the contents of the bottle. Using only solubility as a guide, indicate what chemicals you would use to test the unknown, and what the expected results would be for each of the possibilities.

React each solution with KCl:

Silver acetate would react with Cl- compound to produce a AgCl precipitate.

Calcium chloride would not precipiate.

PbCl2 would also precipiate.

React with Na2SO4 :

Silver sulfate should precipitate.

Calcium sulfate should not precipitate.

PbSO4 should precipitate.

React with




8. Although nitric acid and phosphoric acid have very different properties as pure substances, their aqueous solutions possess many common properties. List some general properties of these solutions and explain their common behavior in terms of the species present.

9. Write a chemical equation that illustrates the autoionization of water.

Occasionally, chemists use solvents other than water or organic solvents. Like water, there solvents may undergo autoionization. Write chemical equations illustrating the autoionization of the following solvents:

  1. liquid ammonia

  2. liquid sulfur dioxide

  3. liquid HCN

10. Explain why the concentration of water, [H2O], is not included in the equilibrium


11. Consider two solutions, acid solution A and solution B. The [H3O+] in solution A is

1000 times greater than that in solution B. What is the difference in the pH values of

the two solutions?

12. Explain why we normally ignore the [H3O+] contributed by water to an acidic

solution. Under what conditions would we need to include its contribution?

The contribution from water is so small that it’s usually negligible relative to the concentration of the acid/base in question. However, under very dilute conditions, the contribution from water can become significant.

13. Explain why the percent of ionization is larger for very dilute solutions than it is for

concentrated solutions.

The percent ionization is larger for dilute sol

The absolute amount ionized is similar, but if the solution is dilute, this becomes a greater proportion of the total.

The concentration of the ionized material is small, however, in a dilute solution it becomes a greater percentage of the

This answer needs work.

14. Explain the acid strength trend observed for the following:

    1. HCl > HF and H2S > H2O

    2. CH4 < NH3 << H2O < HF

15. Indicate whether the pH increase, decreases, or remains the same when each of the

following is added:

  1. Ca(C7H5O2)2 to a solution of HC7H5O2.

  2. pH should decrease.

  3. Pyridinium nitrate (C5H5NH)(NO3), to a solution of pyridine, C5H5N.

  4. pH decreases.

  5. Ammonia to a solution of hydrochloric acid.

  6. pH remains the same until HCl is consumed whereby pH will increase.

  7. Sodium hydrogen carbonate to a solution of carbonic acid.

  8. pH should increase.

  9. Sodium perchlorate to a solution of sodium hydroxide.

  10. pH remains the same

16. How do you explain the fact that solutions of CH3COOH are acidic while solutions of

NaCH3COO are basic?

17. Give an example for each of the following and a rationale for your choice.

  1. A strong electrolyte that is not an acid or base.

NaCl is a strong electrolyte; it ionizes completely. But the ions are the product of a strong acid and a strong base and therefore it is completely neutral itself.

  1. A Lewis acid.

(c) A weak organic acid.

(d) A buffer solution with a pH close to 7.0

18. Explain why the acid strength of oxyacids increases with increasing number of

oxygen atoms.

19. Is it correct to say that all Lewis acids are Bronsted-Lowry acids? Explain

20. For the reaction HF + NH3 D F- + NH4 does the equilibrium lie

predominantly to the left or to the right. Explain your answer. Base your answer on quantitative acid equilibrium expressions and not qualitative Le Chatelier arguments.

Ka for HF is 6.8 x 10-4 and Kb for NH3 is 1.8 x 10-5.

To the right. HF ® H+ + F- equilibrium would lie to the right to give up a proton based on the Ka being less than 1.

NH3 ® NH4+ + OH- also lies to the right since the Kb is less than 1. Therefore the reaction equilibrium should also lie to the right.

21. HCO3- is amphoteric. Illustrate this fact using properly balanced equilibrium


HCO3- + H2O « H2CO3 + OH-

and HCO3- + H2O « CO32- + H3O+

22. When we solve equilibrium expressions for the [H3O+] approximations are often made to reduce the complexity of the equation thus making it easier to solve. Why can we make these approximations? Would these approximations ever lead to significant errors in the answer? If so give an example of an equilibrium problem that would require use of the quadratic equation.

We can make the approximations most of the time since the amount of acid that dissociates is usually very small compared to the initial concentration.

Yes, it could lead to errors if the two concentrations are very similar or even greater than 5% of the initial concentration. More dilute solutions are more likely to cause difficulties. For example, HF ionization is found to be 7.9% for a 0.1 M solution and 23% for a 0.01 M solution.

23. 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 complete reaction of a equimolar amounts of a strong acid with a weak base should produce a weak acid. This weak acid will then establish an equilibrium with its conjugate base and H+. Thus, the solution is acidic.

24. Benzoic acid, C6H5COOH, contains six hydrogen atoms, yet it is a monoprotic acid.


There is only one ionizable proton. The other five hydrogen atoms are strongly covalently linked to carbon atoms and are very difficult to remove. These five hydrogen atoms do not ionize at all in aqueous media.

25. When 0.200 moles of solid sodium acetate are added to a 0.700 M solution of acetic acid the equilibrium is disturbed. Explain how the reaction shifts, that is to the right or to the left, to re-establish equilibrium. Base your answer on equilibrium expressions before and after addition.


Before the addition of sodium acetate, there is an equilibrium that is represented above. It lies to the right (Ka = 1.8 x 10-5) . However, if acetate is added, it will shift the equilibrium to the left, removing some H+ in the process.

This answer may not be complete base on the yellow highlighted qualifier above.