CHAPTER 15
Acids and Bases
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
15.1. See labels below reaction:
H2CO3 is the proton donor (Brønsted-Lowry acid) on the left, and HCN is the proton donor
(Brønsted-Lowry acid) on the right. The CN− and HCO3− ions are proton acceptors (Brønsted-
Lowry bases). HCN is the conjugate acid of CN−.
15.2. Part a involves molecules with all single bonds; part b does not, so bonds are drawn in.
a.
b.
15.3. The HC2H3O2 is a stronger acid than H2S, and HS− is a stronger base than the C2H3O2− ion. The
equilibrium favors the weaker acid and weaker base; therefore, the reactants are favored.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 15: Acids and Bases 545
15.4. a. PH3
b. HI
c. H2SO3
d. H3AsO4
e. HSO4−
15.5. A 0.125 M solution of Ba(OH)2, a strong base, ionizes completely to yield 0.125 M Ba2+ ion and 2
0.125 M, or 0.250 M, OH− ion. Use the Kw equation to calculate the [H3O+].
Kw 1.0 1014
[H3O+] = = = 4.00 10−14 = 4.0 10−14 M
[OH - ] (0.250)
15.6. Use the Kw equation to calculate the [H3O+].
Kw 1.0 1014
[H3O+] = = = 1.00 10−9 = 1.0 10−9 M
[OH - ] 1.0 10 5
Since the [H3O+] concentration is less than 1.0 10−7, the solution is basic.
15.7. Calculate the negative log of the [H3O+]:
pH = −log [H3O+] = −log (0.045) = 1.346 = 1.35
15.8. Calculate the pOH of 0.025 M OH−, and then subtract from 14.00 to find pH:
pOH = −log [OH−] = −log (0.025) = 1.602
pH = 14.00 − 1.602 = 12.397 = 12.40
15.9. Because pH = 3.16, by definition log [H3O+] = −3.16. Enter this on the calculator and convert to
the antilog (number) of −3.16.
[H3O+] = antilog (−3.16) = 10−3.16 = 6.91 10−4 = 6.9 10−4 M
15.10. Find the pOH by subtracting the pOH from 14.00. Then enter −3.40 on the calculator to convert
to the antilog (number) corresponding to −3.40.
pOH = 14.00 − 10.6 = 3.40
[H3O+] = antilog (−3.40) = 10−3.40 = 3.98 10−4 = 4 10−4 M
■ ANSWERS TO CONCEPT CHECKS
15.1. In any aqueous solution, you should consider the autoionization of water. And because we have a
solution of a weak acid in water, you should also consider the equilibrium between this acid and
water. Here are the two equilibria:
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
HCHO2(aq) + H2O(l) CHO2−(aq) + H3O+(aq)
The species present in these equilibria are H2O(l), H3O+(aq), OH−(aq), HCHO2(aq), and
CHO2−(aq).
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 546 Chapter 15: Acids and Bases
15.2. The stronger acid gives up its proton more readily, and therefore, its conjugate base ion holds on
to a proton less strongly. In other words, the stronger acid has the weaker conjugate base.
Because formic acid is the stronger acid, the formate ion is the weaker base. Acetate ion is the
stronger base.
15.3. Look at each solution, and determine whether it is acidic, basic, or neutral. In solution A, the
numbers of H3O+ and OH− ions are equal, so the solution is neutral. For solution B, the number of
H3O+ ions is greater than the number of OH− ions, so the solution is acidic. In solution C, the
number of H3O+ ions is less than the number of OH− ions, so the solution is basic. Therefore, the
ranking from most acidic to least acidic (most basic) is B > A > C.
15.4. In order to answer this problem qualitatively, it is essential that all the solutions have the same
solute concentrations. Bases produce solutions of pH greater than 7, whereas acids produce
solutions of pH less than 7. NH3 and NaOH are bases, and HCl and HC2H3O2 are acids. NaOH is
a stronger base than NH3, so the NaOH solution would have the highest pH, followed by the NH3
solution. HC2H3O2 is a much weaker acid than HCl, so the HC2H3O2 solution would have a higher
pH than the HCl solution. Therefore, the ranking from highest to lowest pH for solutions with the
same solute concentrations is NaOH > NH3 > HC2H3O2 > HCl.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
15.1. You can classify these acids using the information in Section 15.1. Also recall that all diatomic
acids of Group 7A halides are strong except for HF.
a. Weak
b. Weak
c. Strong
d. Strong
e. Weak
f. Weak
15.2. In Section 15.1, we are told that all neutralizations involving strong acids and bases evolve
55.90 kJ of heat per mole of H3O+. Thus, the thermochemical evidence for the Arrhenius concept
is based on the fact that when 1 mol of any strong acid (1 mol H3O+) is neutralized by 1 mole of
any strong base (1 mol OH−), the heat of neutralization is always the same (H = −55.90
kJ/mol).
15.3. A Brønsted-Lowry acid is a molecule or ion that donates an H+ ion (proton donor) to a base in a
proton-transfer reaction. A Brønsted-Lowry base is a molecule or ion that accepts an H+ ion
(proton acceptor) from an acid in a proton-transfer reaction. An example of an acid-base
equation:
HF(aq) + NH3(aq) NH4+(aq) + F -(aq)
acid base acid base
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Acids and Bases
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
15.1. See labels below reaction:
H2CO3 is the proton donor (Brønsted-Lowry acid) on the left, and HCN is the proton donor
(Brønsted-Lowry acid) on the right. The CN− and HCO3− ions are proton acceptors (Brønsted-
Lowry bases). HCN is the conjugate acid of CN−.
15.2. Part a involves molecules with all single bonds; part b does not, so bonds are drawn in.
a.
b.
15.3. The HC2H3O2 is a stronger acid than H2S, and HS− is a stronger base than the C2H3O2− ion. The
equilibrium favors the weaker acid and weaker base; therefore, the reactants are favored.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 15: Acids and Bases 545
15.4. a. PH3
b. HI
c. H2SO3
d. H3AsO4
e. HSO4−
15.5. A 0.125 M solution of Ba(OH)2, a strong base, ionizes completely to yield 0.125 M Ba2+ ion and 2
0.125 M, or 0.250 M, OH− ion. Use the Kw equation to calculate the [H3O+].
Kw 1.0 1014
[H3O+] = = = 4.00 10−14 = 4.0 10−14 M
[OH - ] (0.250)
15.6. Use the Kw equation to calculate the [H3O+].
Kw 1.0 1014
[H3O+] = = = 1.00 10−9 = 1.0 10−9 M
[OH - ] 1.0 10 5
Since the [H3O+] concentration is less than 1.0 10−7, the solution is basic.
15.7. Calculate the negative log of the [H3O+]:
pH = −log [H3O+] = −log (0.045) = 1.346 = 1.35
15.8. Calculate the pOH of 0.025 M OH−, and then subtract from 14.00 to find pH:
pOH = −log [OH−] = −log (0.025) = 1.602
pH = 14.00 − 1.602 = 12.397 = 12.40
15.9. Because pH = 3.16, by definition log [H3O+] = −3.16. Enter this on the calculator and convert to
the antilog (number) of −3.16.
[H3O+] = antilog (−3.16) = 10−3.16 = 6.91 10−4 = 6.9 10−4 M
15.10. Find the pOH by subtracting the pOH from 14.00. Then enter −3.40 on the calculator to convert
to the antilog (number) corresponding to −3.40.
pOH = 14.00 − 10.6 = 3.40
[H3O+] = antilog (−3.40) = 10−3.40 = 3.98 10−4 = 4 10−4 M
■ ANSWERS TO CONCEPT CHECKS
15.1. In any aqueous solution, you should consider the autoionization of water. And because we have a
solution of a weak acid in water, you should also consider the equilibrium between this acid and
water. Here are the two equilibria:
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
HCHO2(aq) + H2O(l) CHO2−(aq) + H3O+(aq)
The species present in these equilibria are H2O(l), H3O+(aq), OH−(aq), HCHO2(aq), and
CHO2−(aq).
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 546 Chapter 15: Acids and Bases
15.2. The stronger acid gives up its proton more readily, and therefore, its conjugate base ion holds on
to a proton less strongly. In other words, the stronger acid has the weaker conjugate base.
Because formic acid is the stronger acid, the formate ion is the weaker base. Acetate ion is the
stronger base.
15.3. Look at each solution, and determine whether it is acidic, basic, or neutral. In solution A, the
numbers of H3O+ and OH− ions are equal, so the solution is neutral. For solution B, the number of
H3O+ ions is greater than the number of OH− ions, so the solution is acidic. In solution C, the
number of H3O+ ions is less than the number of OH− ions, so the solution is basic. Therefore, the
ranking from most acidic to least acidic (most basic) is B > A > C.
15.4. In order to answer this problem qualitatively, it is essential that all the solutions have the same
solute concentrations. Bases produce solutions of pH greater than 7, whereas acids produce
solutions of pH less than 7. NH3 and NaOH are bases, and HCl and HC2H3O2 are acids. NaOH is
a stronger base than NH3, so the NaOH solution would have the highest pH, followed by the NH3
solution. HC2H3O2 is a much weaker acid than HCl, so the HC2H3O2 solution would have a higher
pH than the HCl solution. Therefore, the ranking from highest to lowest pH for solutions with the
same solute concentrations is NaOH > NH3 > HC2H3O2 > HCl.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
15.1. You can classify these acids using the information in Section 15.1. Also recall that all diatomic
acids of Group 7A halides are strong except for HF.
a. Weak
b. Weak
c. Strong
d. Strong
e. Weak
f. Weak
15.2. In Section 15.1, we are told that all neutralizations involving strong acids and bases evolve
55.90 kJ of heat per mole of H3O+. Thus, the thermochemical evidence for the Arrhenius concept
is based on the fact that when 1 mol of any strong acid (1 mol H3O+) is neutralized by 1 mole of
any strong base (1 mol OH−), the heat of neutralization is always the same (H = −55.90
kJ/mol).
15.3. A Brønsted-Lowry acid is a molecule or ion that donates an H+ ion (proton donor) to a base in a
proton-transfer reaction. A Brønsted-Lowry base is a molecule or ion that accepts an H+ ion
(proton acceptor) from an acid in a proton-transfer reaction. An example of an acid-base
equation:
HF(aq) + NH3(aq) NH4+(aq) + F -(aq)
acid base acid base
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
