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Chapter 14: Acid-Base Equilibria

69 problems · 1 figure

Example 14.1 §14.1 source ↗

The ion product of water at 80 °C is 2.4 × 10−13. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?

Answer

[H3O+] = [OH−] = 4.9 × 10−7 M

Example 14.2 §14.1 source ↗

What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?

Answer

[H3O+] = 1 × 10−11 M

Example 14.3 §14.1 source ↗

Write separate equations representing the reaction of H2PO4−

(a) as a base with HBr

(b) as an acid with OH−

Answer

(a) H2PO4−(aq)+ HBr(aq)⇌ H3PO4(aq)+ Br−(aq); (b) H2PO4−(aq)+ OH−(aq)⇌ HPO42−(aq)+ H2O(l)

Example 14.4 §14.2 source ↗

Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water: CO2(aq)+H2O(l)⇌H2CO3(aq)

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

5.70

Example 14.5 §14.2 source ↗

Calculate the hydronium ion concentration of a solution with a pH of −1.07.

Answer

12 M

Example 14.6 §14.2 source ↗

The hydronium ion concentration of vinegar is approximately 4 × 10−3 M. What are the corresponding values of pOH and pH?

Answer

pOH = 11.6, pH = 2.4

Example 14.7 §14.3 source ↗

Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.

Answer

1.3% ionized

Example 14.8 §14.3 source ↗

Determine the relative acid strengths of NH4+ and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as 4.9 × 10−10. The ionization constant of NH4+ is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 × 10−5.

Answer

NH4+ is the slightly stronger acid (Ka for NH4+ = 5.6 × 10−10).

Example 14.9 §14.3 source ↗

The HSO4− ion, weak acid used in some household cleansers: HSO4−(aq)+H2O(l)⇌H3O+(aq)+SO42−(aq)

What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: [H3O+] = 0.027 M; [HSO4−]= 0.29M; and [SO42−]= 0.13M?

Answer

Ka for HSO4− = 1.2 × 10−2

Example 14.10 §14.3 source ↗

What is the equilibrium constant for the ionization of the HPO42− ion, a weak base HPO42−(aq)+H2O(l)⇌H2PO4−(aq)+OH−(aq)

if the composition of an equilibrium mixture is as follows: [OH−] = 1.3 × 10−6 M; [H2PO4−]= 0.042M; and [HPO42−]= 0.341M?

Answer

Kb for HPO42−= 1.6×10−7

Example 14.12 §14.3 source ↗

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic acid, CH3CO2H? CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)Ka=1.8×10−5

Answer

percent ionization = 1.3%

Example 14.13 §14.3 source ↗

Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5.

Answer

7.56 × 10−4 M, 2.33%

Example 14.14 §14.3 source ↗

Calculate the pH in a 0.010-M solution of caffeine, a weak base: C8H10N4O2(aq)+H2O(l)⇌C8H10N4O2H+(aq)+OH−(aq)Kb=2.5×10−4

Answer

pH 11.16

Example 14.15 §14.4 source ↗

What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH4NO3, a salt composed of the ions NH4+ and NO3−. Which is the stronger acid C6H5NH3+ or NH4+?

Answer

[H3O+] = 7.5 × 10−6 M; C6H5NH3+ is the stronger acid.

Example 14.16 §14.4 source ↗

What is the pH of a 0.083-M solution of NaCN?

Answer

11.11

Example 14.17 §14.4 source ↗

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) K2CO3

(b) CaCl2

(c) KH2PO4

(d) (NH4)2CO3

Answer

(a) basic; (b) neutral; (c) acidic; (d) basic

Example 14.18 §14.4 source ↗

What is [Al(H2O)5(OH)2+] in a 0.15-M solution of Al(NO3)3 that contains enough of the strong acid HNO3 to bring [H3O+] to 0.10 M?

Answer

2.1 × 10−5 M

Example 14.19 §14.5 source ↗

The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate [H3O+], [HS−], and [S2−] in the solution: H2S(aq)+H2O(l)⇌H3O+(aq)+HS−(aq)Ka1=8.9×10−8HS−(aq)+H2O(l)⇌H3O+(aq)+S2−(aq)Ka2=1.0×10−19

Answer

[H2S] = 0.1 M; [H3O+] = [HS−] = 0.000094 M; [S2−] = 1 × 10−19 M

Example 14.20 §14.6 source ↗

Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 × 10−5 M HCl solution from 4.74 to 3.00.

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:

pH=−log[H3O+]=−log(total molesH3O+total volume)=−log(1.0×10−4mol+1.8×10−6mol101mL(1L1000mL))=3.00

Example 14.21 §14.7 source ↗

Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO3(aq) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.

Answer

0.00: 1.000; 15.0: 1.5111; 25.0: 7; 40.0: 12.523

Example 14.22 §14.7 source ↗

Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH(aq) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL.

Answer

0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.097

14-1 §14.1 source ↗

Write equations that show NH3 as both a conjugate acid and a conjugate base.

Answer

One example for NH3 as a conjugate acid: NH2−+ H+⟶ NH3; as a conjugate base: NH4+(aq)+ OH−(aq)⟶ NH3(aq)+ H2O(l)

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) H3O+

(b) HCl

(c) NH3

(d) CH3CO2H

(e) NH4+

(f) HSO4−

Answer

(a) H3O+(aq)⟶ H+(aq)+ H2O(l); (b) HCl(aq)⟶ H+(aq)+ Cl−(aq); (c) NH3(aq)⟶ H+(aq)+ NH2−(aq); (d) CH3CO2H(aq)⟶ H+(aq)+ CH3CO2−(aq); (e) NH4+(aq)⟶ H+(aq)+ NH3(aq); (f) HSO4−(aq)⟶ H+(aq)+ SO42−(aq)

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) H2PO4−

Answer

(a) H2O(l)+ H+(aq)⟶ H3O+(aq); (b) OH−(aq)+ H+(aq)⟶ H2O(l); (c) NH3(aq)+ H+(aq)⟶ NH4+(aq); (d) CN−(aq)+ H+(aq)⟶ HCN(aq); (e) S2−(aq)+ H+(aq)⟶ HS−(aq); (f) H2PO4−(aq)+ H+(aq)⟶ H3PO4(aq)

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) HCO3−

(d) NH3

(e) HSO4−

(f) H2O2

(g) HS−

(h) H5N2+

Answer

(a) H2O, O2−; (b) H3O+, OH−; (c) H2CO3, CO32−; (d) NH4+, NH2−; (e) H2SO4, SO42−; (f) H3O2+, HO2−; (g) H2S; S2−; (h) H6N22+, H4N2

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) HNO3+ H2O⟶ H3O++ NO3−

(b) CN−+ H2O⟶ HCN+ OH−

(c) H2SO4+ Cl−⟶ HCl+ HSO4−

(d) HSO4−+ OH−⟶ SO42−+ H2O

(e) O2−+ H2O⟶ 2OH−

(f) [Cu(H2O)3(OH)]++ [Al(H2O)6]3+⟶ [Cu(H2O)4]2++ [Al(H2O)5(OH)]2+

(g) H2S+ NH2−⟶ HS−+ NH3

Answer

The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) HNO3(BA), H2O(BB), H3O+(CA), NO3−(CB); (b) CN−(BB), H2O(BA), HCN(CA), OH−(CB); (c) H2SO4(BA), Cl−(BB), HCl(CA), HSO4−(CB); (d) HSO4−(BA), OH−(BB), SO42−(CB), H2O(CA); (e) O2−(BB), H2O(BA) OH−(CB and CA); (f) [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB); (g) H2S(BA), NH2−(BB), HS−(CB), NH3(CA)

14-11 §14.1 source ↗

What are amphiprotic species? Illustrate with suitable equations.

Answer

Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2O. As an acid: H2O(aq)+ NH3(aq)⇌ NH4+(aq)+ OH−(aq). As a base: H2O(aq)+ HCl(aq)⇌ H3O+(aq)+ Cl−(aq)

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) NH3

(b) HPO42−

(c) Br−

(d) NH4+

(e) AsO43−

Answer

amphiprotic: (a) NH3+ H3O+⟶ NH4OH+ H2O, NH3+ OCH3−⟶ NH2−+ CH3OH; (b) HPO42−+ OH−⟶ PO43−+ H2O, HPO42−+ HClO4⟶ H2PO4−+ ClO4−; not amphiprotic: (c) Br−; (d) NH4+; (e) AsO43−

14-15 §14.2 source ↗

Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 × 10−7 M. Kw is 2.9 × 10−14 at 40 °C.

Answer

In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O+] = [OH−] = (2.910 × 10−14)1/2 = 1.7 × 10−7.

14-17 §14.2 source ↗

The ionization constant for water (Kw) is 9.311 × 10−14 at 60 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 60 °C.

Answer

x = 3.051 × 10−7 M = [H3O+] = [OH−]; pH = −log3.051 × 10−7 = −(−6.5156) = 6.5156; pOH = pH = 6.5156

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

(a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pOH = −0.40; pH = 14.4

14-21 §14.2 source ↗

What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?

Answer

[H3O+] = 3.0 × 10−7 M; [OH−] = 3.3 × 10−8 M

14-23 §14.2 source ↗

Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 14.2 for useful information.
Figure 14.2 in the book ↗
The pH and pOH scales represent concentrations of H3O+ and OH−, respectively. The pH and pOH values of some common substances at 25 °C are shown in this chart.

Answer

[H3O+] = 1 × 10−2 M; [OH−] = 1 × 10−12 M

14-25 §14.2 source ↗

The hydroxide ion concentration in household ammonia is 3.2 × 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?

Answer

[OH−] = 3.1 × 10−12 M

14-27 §14.3 source ↗

Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.

Answer

The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH−, which causes the solution to be basic.

14-29 §14.3 source ↗

The odor of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.

Answer

[H2O] > [CH3CO2H] > [H3O+] ≈ [CH3CO2−] > [OH−]

14-31 §14.3 source ↗

Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.

Answer

The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.

14-33 §14.3 source ↗

Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.

Answer

Mg ( OH ) 2 ( s ) + 2HCl ( a q ) ⟶ Mg 2+ ( a q ) + 2 Cl − ( a q ) + 2H 2 O ( l ) BB BA CB CA

14-35 §14.3 source ↗

What is the ionization constant at 25 °C for the weak acid CH3NH3+, the conjugate acid of the weak base CH3NH2, Kb = 4.4 × 10−4.

Answer

Ka= 2.3×10−11

14-37 §14.3 source ↗

Which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, (CH3)2NH2+ or CH3NH3+, is the stronger acid?

Answer

The stronger base or stronger acid is the one with the larger Kb or Ka, respectively. In these two examples, they are (CH3)2NH and CH3NH3+.

14-39 §14.3 source ↗

Which is the stronger base, (CH3)3N or H2BO3−?

Answer

triethylamine

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) HSO4− or HSeO4−

(b) NH3 or H2O

(c) PH3 or HI

(d) NH3 or PH3

(e) H2S or HBr

Answer

(a) HSO4−; higher electronegativity of the central ion. (b) H2O; NH3 is a base and water is neutral, or decide on the basis of Ka values. (c) HI; PH3 is weaker than HCl; HCl is weaker than HI. Thus, PH3 is weaker than HI. (d) PH3; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.

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: BrO2−, ClO2−, IO2−

(c) acidity: HOCl, HOBr, HOI

(d) acidity: HOCl, HOClO, HOClO2, HOClO3

(e) basicity: NH2−, HS−, HTe−, PH2−

(f) basicity: BrO−, BrO2−, BrO3−, BrO4−

Answer

(a) NaHSeO3 < NaHSO3 < NaHSO4; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) ClO2−< BrO2−< IO2−; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (d) HOCl < HOClO < HOClO2 < HOClO3; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) HTe−< HS−<<PH2−< NH2−; PH2− and NH2− are anions of weak bases, so they act as strong bases toward H+. HTe− and HS− are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) BrO4−< BrO3−< BrO2−< BrO−; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.

14-45 §14.3 source ↗

The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxyl group (−CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).

Answer

[ H 2 O]> [ C 6 H 4 OH( CO 2 H)]> [H + ]0> [C 6 H 4 OH ( CO 2 ) − ]≫[ C 6 H 4 O ( CO 2 H ) − ]> [ OH − ]

14-47 §14.3 source ↗

What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?

Answer

1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H3O+.

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

(b) The addition of HCl

14-50 §14.3 source ↗

What is the effect on the concentrations of NO2−, 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

(a) Adding HCl will add H3O+ ions, which will then react with the OH− ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentration of NO2− ions. (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of NO2− ions and decreasing the concentration of OH− ions. (c) Adding NaOH adds OH− ions, which shifts the equilibrium to the left, increasing the concentration of NO2− ions and decreasing the concentrations of HNO2. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO2 adds NO2− ions and shifts the equilibrium to the right, increasing the HNO2 and OH− ion concentrations.

14-52 §14.3 source ↗

Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?

Answer

This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H3O+] produced by the stronger acid.

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;
[NH4+] = 3.1 × 10−3 M;

[NH3] = 0.533 M;

(b) HNO2: [H3O+] = 0.011 M;
[NO2−] = 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) NH4+: [NH4+] = 0.100 M;

[NH3] = 7.5 × 10−6 M;
[H3O+] = 7.5 × 10−6 M

Answer

(a) Kb= 1.8×10−5; (b) Ka= 4.5×10−4; (c) Kb= 6.4×10−5; (d) Ka= 5.6×10−10

14-56 §14.3 source ↗

Determine Ka for hydrogen sulfate ion, HSO4−. In a 0.10-M solution the acid is 29% ionized.

Answer

K a = 1.2 × 10 −2

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) (CH3)3NH+

(c) HAsO42– (as a base)

(d) HO2− (as a base)

(e) C6H5NH3+

(f) HSO3− (as a base)

Answer

(a) Kb= 4.3×10−12 (b) Ka= 1.6×10−10 (c) Kb=5.9×10−8 (d) Kb=4.2×10−3 (e) Kb=2.3×10−5 (f) Kb= 6.3×10−13

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 Fe (H2O)6 2+ a weak acid, Ka = 1.6 × 10−7

Answer

(a) [H3O+] [ClO−][HClO]= (x)(x)(0.0092−x)≈ (x)(x)0.0092= 2.9×10−8
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) [C6 H5NH3 +][OH−][C6H5NH2]= (x)(x)(0.0784−x)≈ (x)(x)0.0784= 4.3×10−10
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:
[C6H5 NH3+] = [OH−] = 5.8 × 10−6 M
[C6H5NH2] = 0.0784 M
[H3O+] = 1.7× 10−9 M;
(c) [H3O+][CN−][HCN]= (x)(x)(0.0810−x)≈ (x)(x)0.0810= 4.9×10−10
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) [(CH3)3NH+][OH−][(CH3)3N]= (x)(x)(0.11−x)≈ (x)(x)0.11= 6.3×10−5
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) [Fe(H2O)5(OH)+][H3O+][Fe(H2O)62+]= (x)(x)(0.120−x)≈ (x)(x)0.120= 1.6×10−7
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
[Fe(H2O)62+] = 0.120 M
[OH−] = 7.2 × 10−11 M

14-62 §14.3 source ↗

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm3, what is the pH?

Answer

pH = 2.41

14-64 §14.3 source ↗

Nicotine, C10H14N2, is a base that will accept two protons (Kb1 = 7 × 10−7, Kb2 = 1.4 × 10−11). What is the concentration of each species present in a 0.050-M solution of nicotine?

Answer

[C10H14N2] = 0.049 M; [C10H14N2H+] = 1.9 × 10−4 M; [C10H14N2H22+] = 1.4 × 10−11 M; [OH−] = 1.9 × 10−4 M; [H3O+] = 5.3 × 10−11 M

14-66 §14.3 source ↗

The pH of a 0.15-M solution of HSO4− is 1.43. Determine Ka for HSO4− from these data.

Answer

K a = 1.2 × 10 −2

14-68 §14.3 source ↗

The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.

Answer

K b = 1.77× 10 −5

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

(a) acidic; (b) basic; (c) acidic; (d) neutral

14-72 §14.5 source ↗

Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid: [H3O+], [OH−], [H2CO3], [HCO3−], [CO32−]? No calculations are needed to answer this question.

Answer

[H3O+] and [HCO3−] are practically equal

14-74 §14.5 source ↗

Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.
C6H4(CO2H)2(aq)+H2O(l)⇌H3O+(aq)+C6H4(CO2H)(CO2)−(aq)Ka =1.1×10−3 C6H4(CO2H)(CO2)–(aq)+H2O(l)⇌H3O+(aq)+C6H4(CO2)22−(aq)Ka=3.9×10−6

Answer

[C6H4(CO2H)2] 7.2 × 10−3 M, [C6H4(CO2H)(CO2)−] = [H3O+] 2.8 × 10−3 M, [C6H4(CO2)22−]3.9 × 10−6 M, [OH−] 3.6 × 10−12 M

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

(a) Ka2= 1.5×10−11;
(b) Kb= 4.3×10−12;
(c) [Te2−][H3O+][HTe−]= (x)(0.0141+x)(0.0141−x)≈ (x)(0.0141)0.0141= 1.5×10−11
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 ↗

Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H3PO4 and a salt of its conjugate base NaH2PO4.

Answer

Excess H3O+ is removed primarily by the reaction: H3O+(aq)+ H2PO4−(aq)⟶ H3PO4(aq)+ H2O(l)
Excess base is removed by the reaction: OH−(aq)+ H3PO4(aq)⟶ H2PO4−(aq)+ H2O(l)

14-80 §14.6 source ↗

What is [H3O+] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2?
CH3CO2H(aq)+ H2O(l)⇌ H3O+(aq)+ CH3CO2−(aq) Ka= 1.8×10−5

Answer

[H3O+] = 1.5 × 10−4 M

14-82 §14.6 source ↗

What is [OH−] in a solution of 0.125 M CH3NH2 and 0.130 M CH3NH3Cl?
CH3NH2(aq)+ H2O(l)⇌ CH3NH3+(aq)+ OH−(aq) Kb= 4.4×10−4

Answer

[OH−] = 4.2 × 10−4 M

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

(a) The added HCl will increase the concentration of H3O+ slightly, which will react with CH3CO2− and produce CH3CO2H in the process. Thus, [CH3CO2−] decreases and [CH3CO2H] increases. (b) The added KCH3CO2 will increase the concentration of [CH3CO2−] which will react with H3O+ and produce CH3CO2 H in the process. Thus, [H3O+] decreases slightly and [CH3CO2H] increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce OH− ions, which will react with the H3O+, thus reducing [H3O+]. Some additional CH3CO2H will dissociate, producing [CH3CO2−] ions in the process. Thus, [CH3CO2H] decreases slightly and [CH3CO2−] increases. (e) The added CH3CO2H will increase its concentration, causing more of it to dissociate and producing more [CH3CO2−] and H3O+ in the process. Thus, [H3O+] increases slightly and [CH3CO2−] increases.

14-86 §14.6 source ↗

What will be the pH of a buffer solution prepared from 0.20 mol NH3, 0.40 mol NH4NO3, and just enough water to give 1.00 L of solution?

Answer

pH = 8.95

14-88 §14.6 source ↗

How much solid NaCH3CO2•3H2O must be added to 0.300 L of a 0.50-M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)

Answer

37 g (0.27 mol)

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

(a) pH = 5.222; (b) The solution is acidic. (c) pH = 5.220

14-92 §14.7 source ↗

Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.

Answer

At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.

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

(a) pH = 2.50; (b) pH = 4.01; (c) pH = 5.60; (d) pH = 8.35; (e) pH = 11.08

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Problems and answers are adapted from Chemistry 2e by Paul Flowers, Klaus Theopold, Richard Langley and William R. Robinson, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/chemistry-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.