Chemistry 106, Chapter 13 Exercises
1. Give the formula of the conjugate acid of the following.
a. OH- b. HPO42- c. NH3 d. F- e. Zn(H2O)2(OH)2
2. Give the formula of the conjugate base of the following
a. HCO3- b. H2SO3 c. HNO2 d. CH3COOH e. Cu(H2O)(OH)3Answers for 1 and 2:
1a. H2O b. H2PO42a. CO32- b. HSO3-
c. NH4+
c. NO2-
d. HF e. Zn(H2O)3(OH)+
d. CH3COO- e. Cu(OH)42-
3. Complete the following acid-base reactions. Label the acids and bases. What are the conjugate
acid/base pairs?
H3O+ + CN- 
HNO2 + OH- 
CN- + H2O 
HCO3- + H3O+ 
HC2H3O2 + HS- 
NH3 + H3O+ 
NH3 + H2O 
CH3COOH + H2O
HCHO2 + H2O 
Answers:
H3O+ + CN-  H2O + HCN
acid
base base acid
H3O+/ H2O, CN-/HCN
HNO2 + OH-  NO2- + H2O
acid
base base acid
HNO2/NO2-, OH-/H2O
CN- + H2O  HCN + OHbase acid
acid
base
CN-/HCN, H2O/OH-
HCO3- + H3O+  H2CO3 + H2O
base
acid
acid
base
HCO3-/H2CO3. H3O+/H2O
HC2H3O2 + HS-  C2H3O2- + H2S
acid
base base
acid
HC2H3O2/C2H3O2-, HS-/H2S
NH3 + H3O+  NH4+ + H2O
base acid
acid
base
NH3/NH4+, H3O+/H2O
NH3 + H2O  NH4+ + OHbase acid
acid
base
NH3/NH4+, H2O/OH-
CH3COOH + H2O CH3COO- + H3O+
acid
base base
acid CH3COOH/CH3COO-, H2O/H3O+
HCHO2 + H2O  CHO2- + H3O+
acid
base base
acid
HCHO2/CHO2-, H2O/H3O+
4. Find the pH and pOH of solutions with the following [H+]. Classify each as acidic or basic.
a. 6.0 M b. 0.33 M c. 4.6 x 10-8 M d. 7.2 x 10-14 M
Ans. a. pH = -0.78, pOH = 14.78 b. pH = 0.48, pOH = 13.52 c. pH = 7.34, pOH = 6.66 d. pH
= 13.14, pOH = 0.86
5. Calculate [H+] and [OH-] in solutions with the following pH
a. 4.0 b. 8.52 c. 0.00 d. 12.60
Ans. a. 1 x 10-4, 1 x 10-10 b. 3 x 10-9, 3.3 x 10-6 c. 1, 1 x 10-14
d. 2.5 x 10-13, 4.0 x 10-2
6. Solution 1 has [H+] = 1.7 x 10-2. Solution 2 has [H+] = 4.3 x 10-4. Which solution is more acidic?
Which has the higher pH? Ans. Solution 1 is more acidic and solution 2 would have the higher pH
7. Find [H+] and pH for a solution containing 22 g of HBr in 479 mL of solution. The molar mass of
HBr is 80.9 g. Ans. [H+] = 0.57 M, pH = 0.24
8. Find [OH-], [H+], pH, and pOH for the following solutions:
a. 0.27 M Sr(OH)2. Ans. [OH-] = 0.54 M, [H+] = 1.9 x 10-14, pH = 13.73, pOH = 0.27
b. A solution containing 13.6 g KOH in 2.50 L of solution. The molar mass of KOH is 56.1 g.
Ans [OH-] = 0.0970 M, [H+] = 1.0 x 10-13 M, pH = 13.0, pOH = 1
c. A solution made by diluting 45.0 mL of 0.0921 M Ba(OH)2 to a volume of 350.0 mL.
Ans. [OH-] = 0.0237 M, [H+] = 4.2 x 10-13 M, pH = 12.37, pOH = 1.63
d. A solution made by dissolving 4.68 g of NaOH in 635 mL of solution. The molar mass of NaOH
is 40.0 g. Ans. [OH-] = 0.184 M, [H+] = 5.4 x 10-14 M ,pH = 13.26, pOH = 0.74
Weak Acids, Ionization Equations, Ka and pKa
9. Using the Bronsted-Lowry model, write equations to show how the following behave as weak acids in
water.
a. H2S b. HPO42- c. HClO2 d. CH3COOH e. NH4+ f. HSO4Answers:
a. H2S  H+ + HS- or H2S + H2O  H3O+ + HSb. HPO42-  H+ + PO43- or HPO42- + H2O  H3O+ + PO43c. HClO2  H+ + ClO2etc.
d. CH3COOH  CH3COO- + H+
e. NH4+  H+ + NH3
f. HSO4-  H+ + SO4210. Calculate Ka for the weak acids that have the following pKa values
a. 3.9 b. 10.12 c. 13.07
Ans. a. 1.3 x 10-4 b. 7.6 x 10-11 c. 8.5 x 10-14
11. Calculate pKa for the weak acids that have the following Ka values.
a. 1.8 x 10-4 b. 6.8 x 10-8 c. 4.9 x 10-11
Ans. a. 3.74 b. 7.17
c. 10.31
Weak Acids, Calculating pH
12. Butyric acid, HC4H7O2, is responsible for the odor of rancid butter and cheese. Its Ka is 1.51 x 10-5.
Calculate [H+] and pH for the following solutions of butyric acid.
a. 0.279 mol in 1.30 L of solution. Ans. 1.80 x 10-3, 2.74
b. 13.5 g in 1.30 L of solution (molar mass = 88 g). Ans. 1.33 x 10-3, 2.88
13. Penicillin is a weak acid. Its Ka is 1.7 x 10-3. Calculate [H+] and pH for the following solutions of
penicillin.
a. 0.187 mol in 725 mL of solution. Ans 2.1 x 10-2, 1.68
b. 127 g in 725 mL of solution (molar mass = 356 g). Ans 2.9 x 10-2, 1.54
14. Uric acid, HC5H3O3N4, can accumulate in the joints causing severe pain and the condition called
gout. Ka for uric acid is 5.1 x 10-6. Calculate the following for a 0.894 M solution of uric acid.
a.[H+] b. [OH-] c. pH d. % ionization
Ans. a. 2.1 x 10-3 b. 4.8 x 10-12 c. 2.68 d. 0.23%
15. Barbituric acid, HC4H3N2O3, is used in the manufacture of some sedatives. Its Ka is 1.1 x 10-4.
Calculate the following for a 0.673 M solution of barbituric acid.
a.[H+] b. [OH-] c. pH d. % ionization
Ans. a. 0.0086 b. 1.2 x 10-12 c. 2.07 d. 1.3%
Weak Acids, Calculating Ka
16. Caproic acid HC6H11O2, is found in coconut oil and is used in making artificial flavors. A solution
is made by dissolving 0.450 mol of caproic acid in 2.0 L of solution. The solution has [H+] = 1.7 x
10-3 M. What is Ka for caproic acid? Ans. 1.3 x 10-5
17. Ascorbic acid, HC6H7O6, also known as vitamin C, is a weak acid. A 100.0 mL solution contains
2.00 g of ascorbic acid. The solution has a pH = 2.54. The molar mass of ascorbic acid is 176 g.
What is Ka for ascorbic acid? Ans. 7.6 x 10-5
18. Barbituric acid, HC4H3N2O3, is used to prepare barbiturates, a class of drugs used as sedatives. A
325-mL solution contains 9.00 g of barbituric acid. The solution has a pH = 2.34. The molar mass
of barbituric acid is 128 g. Find Ka. Ans. 1 x 10-4
Polyprotic Acids
19. Phthalic acid, H2C8H4O4, is a diprotic acid. It is used to make phenolphthalein indicator. Ka1 =
0.0012, Ka2 = 3.9 x 10-6. For a 2.9 M solution of phthalic acid calculate the pH, [HC8H4O4-], and
[C8H4O42-]. Ans. pH = 1.23, [HC8H4O4-] = 5.9 x 10-2, [C8H4O42-] = 3.9 x 10-6
20. Ascorbic acid, H2C6H6O6 is a diprotic acid with Ka1= 7.9 x 10-5 and Ka2 = 1.6 x 10-12. For a 0.63 M
solution calculate pH, [HC6H6O6-], and [C6H6O62-]. Ans. pH = 2.15, [HC6H6O6-] = 7.1 x 10-3M ,
[C6H6O62-] = 1.6 x 10-12M
Weak Bases, Ionization Equations
21. Using the Bronsted-Lowry model, write equations to show how the following behave as weak bases
in water.
a. NH3 b. HSO4- c. NO2- d. F- e. C6H5NH2 f. HPO42-
Answers
a. NH3 + H2O  NH4+ + OHb. HSO4- + H2O  H2SO4 + OHc. NO2- + H2O  HNO2 + OHd. F- + H2O  HF + OHe. C6H5NH2 + H2O  C6H5NH3+ + OHf. HPO42- + H2O  H2PO4- + OHWeak Bases, Calculations
22. Codeine (Cod), a powerful and addictive painkiller, is a weak base. Its conjugate acid has a Ka of
1.2 x 10-8. What is Kb for codeine? What is the pH of a 0.0020 M solution of codeine?
Ans. Kb = 8.3 x 10-7, pH = 9.62
23. Consider the weak base sodium acrylate, NaC3H3O2, which has a molar mass of 94 g. Its conjugate
acid has a Ka of 5.5 x 10-5. What is Kb for sodium acrylate? What is the pH of an 835 mL solution
containing 1.61 g of NaC3H3O2?
Ans. 1.8 x 10-10, pH = 8.24
24. The pH of a household ammonia cleaning solution is 11.68. How many grams of ammonia are
needed in a 1.25 L solution to give the same pH? Kb = 1.8 x10-5 and the molar mass is 17 g.
Ans. 27 g
25. A solution of sodium cyanide, NaCN, has a pH of 12.10. How many grams of sodium cyanide are
needed in 425 mL of solution to give the same pH? Kb = 1.7 x 10-5 and the molar mass is 49 g.
Ans. 2.0 x 102 g
Salt Solutions
26. For the following salts, state whether 1 M solutions would be acidic, basic, or neutral. For those that
are acidic or basic, write equations showing the reaction with water.
a. K2CO3 b. NH4Br c. NaNO2 d. FeCl3 e. BaI2 f. Cu(NO3)2 g. K3PO4 h. Sr(ClO4)2
i. NH4F (Ka for NH4+ = 5.6 x 10-10, Kb for F- = 1.4 x 10-11)
j. Na2HPO4 (Ka for HPO42- = 4.5 x 10-13, Kb for HPO42- = 1.6 x 10-7)
Answers:
a. basic; CO32- + H2O  HCO3- + OHb. acidic; NH4+ + H2O  NH3 + H3O+
c. basic; NO2- + H2O  HNO2 + OHd. acidic; Fe3+ 6H2O  Fe(H2O)63+  Fe(H2O)5OH2+ + H+
e. neutral
f. acidic; Cu2+ + 4H2O  Cu(H2O)42+  Cu(H2O)3(OH)+ + H+
g. basic; PO43- + H2O  HPO42- + OHh. neutral
i. acidic; NH4+ + H2O  NH3 + H3O+
j. basic; HPO42- + H2O  H2PO4- + OH-