Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.
Chapter 14: Acid-Base Equilibria
69 problems · 1 figureExample 14.1 §14.1 source ↗
Answer
Example 14.2 §14.1 source ↗
Answer
Example 14.3 §14.1 source ↗
(a) as a base with HBr
(b) as an acid with OH−
Answer
Example 14.4 §14.2 source ↗
Air-saturated water has a hydronium ion concentration caused by the dissolved CO2 of 2.0 10−6 M, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.
Answer
Example 14.5 §14.2 source ↗
Answer
Example 14.6 §14.2 source ↗
Answer
Example 14.7 §14.3 source ↗
Answer
Example 14.8 §14.3 source ↗
Answer
Example 14.9 §14.3 source ↗
What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: = 0.027 M; and
Answer
Example 14.10 §14.3 source ↗
if the composition of an equilibrium mixture is as follows: [OH−] = 1.3 10−6 M; and
Answer
Example 14.12 §14.3 source ↗
Answer
Example 14.13 §14.3 source ↗
Answer
Example 14.14 §14.3 source ↗
Answer
Example 14.15 §14.4 source ↗
Answer
Example 14.16 §14.4 source ↗
Answer
Example 14.17 §14.4 source ↗
(a) K2CO3
(b) CaCl2
(c) KH2PO4
(d) (NH4)2CO3
Answer
Example 14.18 §14.4 source ↗
Answer
Example 14.19 §14.5 source ↗
Answer
Example 14.20 §14.6 source ↗
Answer
Initial pH of 1.8 10−5 M HCl; pH = −log[H3O+] = −log[1.8 10−5] = 4.74
Moles of H3O+ in 100 mL 1.8 10−5 M HCl; 1.8 10−5 moles/L 0.100 L = 1.8 10−6
Moles of H3O+ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L 0.0010 L = 1.0 10−4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:
Example 14.21 §14.7 source ↗
Answer
Example 14.22 §14.7 source ↗
Answer
14-1 §14.1 source ↗
Answer
14-3 §14.1 source ↗
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
(a)
(b) HCl
(c) NH3
(d) CH3CO2H
(e)
(f)
Answer
14-5 §14.1 source ↗
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
(a) H2O
(b) OH−
(c) NH3
(d) CN−
(e) S2−
(f)
Answer
14-7 §14.1 source ↗
What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) OH−
(b) H2O
(c)
(d) NH3
(e)
(f) H2O2
(g) HS−
(h)
Answer
14-9 §14.1 source ↗
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Answer
14-11 §14.1 source ↗
Answer
14-13 §14.1 source ↗
State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
(a)
(b)
(c) Br−
(d)
(e)
Answer
14-15 §14.2 source ↗
Answer
14-17 §14.2 source ↗
Answer
14-19 §14.2 source ↗
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
(a) 0.000259 M HClO4
(b) 0.21 M NaOH
(c) 0.000071 M Ba(OH)2
(d) 2.5 M KOH
Answer
14-21 §14.2 source ↗
Answer
14-23 §14.2 source ↗
Answer
14-25 §14.2 source ↗
Answer
14-27 §14.3 source ↗
Answer
14-29 §14.3 source ↗
Answer
14-31 §14.3 source ↗
Answer
14-33 §14.3 source ↗
Answer
14-35 §14.3 source ↗
Answer
14-37 §14.3 source ↗
Answer
14-39 §14.3 source ↗
Answer
14-41 §14.3 source ↗
Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
(a) or
(b) NH3 or H2O
(c) PH3 or HI
(d) NH3 or PH3
(e) H2S or HBr
Answer
14-43 §14.3 source ↗
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
(a) acidity: NaHSO3, NaHSeO3, NaHSO4
(b) basicity:
(c) acidity: HOCl, HOBr, HOI
(d) acidity: HOCl, HOClO, HOClO2, HOClO3
(e) basicity: HS−, HTe−,
(f) basicity: BrO−,
Answer
14-45 §14.3 source ↗
Answer
14-47 §14.3 source ↗
Answer
14-48 §14.3 source ↗
Which of the following will increase the percent of NH3 that is converted to the ammonium ion in water?
(a) addition of NaOH
(b) addition of HCl
(c) addition of NH4Cl
Answer
14-50 §14.3 source ↗
What is the effect on the concentrations of HNO2, and OH− when the following are added to a solution of KNO2 in water:
(a) HCl
(b) HNO2
(c) NaOH
(d) NaCl
(e) KNO2
Answer
14-52 §14.3 source ↗
Answer
14-54 §14.3 source ↗
From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.
(a) NH3: [OH−] = 3.1 10−3 M;
= 3.1 10−3 M;
[NH3] = 0.533 M;
(b) HNO2: = 0.011 M;
= 0.0438 M;
[HNO2] = 1.07 M;
(c) (CH3)3N: [(CH3)3N] = 0.25 M;
[(CH3)3NH+] = 4.3 10−3 M;
[OH−] = 3.7 10−3 M;
(d) = 0.100 M;
[NH3] = 7.5 10−6 M;
[H3O+] = 7.5 10−6 M
Answer
14-56 §14.3 source ↗
Answer
14-58 §14.3 source ↗
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
(a) HTe− (as a base)
(b)
(c) (as a base)
(d) (as a base)
(e)
(f) (as a base)
Answer
14-60 §14.3 source ↗
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.
(a) 0.0092 M HClO, a weak acid
(b) 0.0784 M C6H5NH2, a weak base
(c) 0.0810 M HCN, a weak acid
(d) 0.11 M (CH3)3N, a weak base
(e) 0.120 M a weak acid, Ka = 1.6 10−7
Answer
Solving for x gives 1.63 10−5 M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [ClO–] = 1.6 10−5 M
[HClO–] = 0.0092 M
[OH−] = 6.1 10−10 M;
(b)
Solving for x gives 5.81 10−6 M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
= [OH−] = 5.8 10−6 M
[C6H5NH2] = 0.0784 M
[H3O+] = 1.7 10−9 M;
(c)
Solving for x gives 6.30 10−6 M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [CN−] = 6.3 10−6 M
[HCN] = 0.0810 M
[OH−] = 1.6 10−9 M;
(d)
Solving for x gives 2.63 10−3 M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH3)3NH+] = [OH−] = 2.6 10−3 M
[(CH3)3N] = 0.11 M
[H3O+] = 3.8 10−12 M;
(e)
Solving for x gives 1.39 10−4 M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H2O)5(OH)+] = [H3O+] = 1.4 10−4 M
= 0.120 M
[OH−] = 7.2 10−11 M
14-62 §14.3 source ↗
Answer
14-64 §14.3 source ↗
Answer
14-66 §14.3 source ↗
Answer
14-68 §14.3 source ↗
Answer
14-70 §14.4 source ↗
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) FeCl3
(b) K2CO3
(c) NH4Br
(d) KClO4
Answer
14-72 §14.5 source ↗
Answer
14-74 §14.5 source ↗
Answer
14-76 §14.5 source ↗
The ion HTe− is an amphiprotic species; it can act as either an acid or a base.
(a) What is Ka for the acid reaction of HTe− with H2O?
(b) What is Kb for the reaction in which HTe− functions as a base in water?
(c) Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe−] in a 0.10 M solution of H2Te.
Answer
(b)
(c)
Solving for x gives 1.5 10−11 M. Therefore, compared with 0.014 M, this value is negligible (1.1 10−7%).
14-78 §14.6 source ↗
Answer
Excess base is removed by the reaction:
14-80 §14.6 source ↗
Answer
14-82 §14.6 source ↗
Answer
14-84 §14.6 source ↗
What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
(a) HCl
(b) KCH3CO2
(c) NaCl
(d) KOH
(e) CH3CO2H
Answer
14-86 §14.6 source ↗
Answer
14-88 §14.6 source ↗
Answer
14-90 §14.6 source ↗
A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 10−5 as Ka for acetic acid.
(a) What is the pH of the solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer?
Answer
14-92 §14.7 source ↗
Answer
14-94 §14.7 source ↗
Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 10−5) with 0.100 M KOH.
(a) no KOH added
(b) 20 mL of KOH solution added
(c) 39 mL of KOH solution added
(d) 40 mL of KOH solution added
(e) 41 mL of KOH solution added
Answer
Loading chapter 15…
Problems and answers are adapted from Chemistry: Atoms First 2e by Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold and Richard Langley, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.