6.E: Acid-Base Equilibrium (Exercises)

6.1: Brønsted-Lowry Acids and Bases

Exercise \(\PageIndex{1}\)

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

Answer

One example for NH₃ as a conjugate acid: \(\ce{NH2-}(aq) + \ce{H3O+}(aq) ⟶ \ce{NH3}(aq) + \ce{H2O}(l)\); as a conjugate base: \(\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l)\)

Exercise \(\PageIndex{2}\)

Show by suitable reaction equations that each of the following species can act as a Brønsted-Lowry acid:

1. HCl (aq)
2. CH₃CO₂H (aq)
3. \(\ce{NH4+}\) (aq)
4. \(\ce{HSO4-}\) (aq)

Answer

a.\(\ce{HCl}(aq) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{Cl-}(aq)\)

b. \(\ce{CH3CO2H}(aq) + \ce{H2O}(l) ⟶ \ce{CH3CO2-}(aq) + \ce{H3O+}(aq)\)

c. \(\ce{NH4+}(aq) + \ce{H2O}(l) ⟶ \ce{NH3}(aq) + \ce{H3O+}(aq)\)

d. \(\ce{HSO4-}(aq) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{SO42-}(aq)\)

Exercise \(\PageIndex{3}\)

Show by suitable reaction equations that each of the following species can act as a Brønsted-Lowry base:

1. H₂O (l)
2. NH₃ (aq)
3. S²⁻ (aq)
4. \(\ce{H2PO4-}\) (aq)

Answer

a. \(\ce{H2O}(l) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{OH-}(aq)\)

b.\(\ce{NH3}(aq) + \ce{H2O}(l) ⟶ \ce{NH4+}(aq) + \ce{OH-}(aq)\)

c. \(\ce{S2-}(aq) + \ce{H2O}(l) ⟶ \ce{HS-}(aq) + \ce{OH-}(aq)\)

d. \(\ce{H2PO4-}(aq) + \ce{H2O}(l) ⟶ \ce{H3PO4}(aq) + \ce{OH-}(aq)\)

Exercise \(\PageIndex{4}\)

Decide if each of the following are a conjugate acid-base pair:

a. H₂O (l), H₃O⁺(aq)

b. OH^- (aq), H₃O⁺ (aq)

c. NH₃ (aq), OH^-(aq)

d. H₂CO₃ (aq), HCO₃^- (aq)

Answer

a. Yes, H₂O (l) is the base and H₃O⁺ is the conjugate acid.

b. No.

c. No.

d. Yes, H₂CO₃ s the acid and HCO₃^- is the conjugate base.

Exercise \(\PageIndex{5}\)

What is the conjugate acid of each of the following? What is the conjugate base of each?

1. \(\ce{OH-}\)
2. H₂O
3. \(\ce{HCO3-}\)
4. NH₃
5. \(\ce{HSO4-}\)
6. H₂O₂
7. HS⁻
8. \(\ce{H5N2+}\)

Answer

H₂O, O²⁻; H₃O⁺, \(\ce{OH^-}\) ; H₂CO₃, \(\ce{CO3^2-}\); \(\ce{NH4+}\), \(\ce{NH2-}\); H₂SO₄, \(\ce{SO4^2-}\); \(\ce{H3O2+}\), \(\ce{HO2-}\); H₂S; S²⁻; \(\ce{H6N2^2+}\), H₄N₂

Exercise \(\PageIndex{6}\)

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:

1. \(\ce{HNO3}(aq) + \ce{H2O}(l) ⟶ \ce{H3O+}(aq) + \ce{NO3-}(aq)\)
2. \(\ce{CN-}(aq) + \ce{H2O}(l) ⟶ \ce{HCN}(aq) + \ce{OH-}(aq)\)
3. \(\ce{H2SO4}(aq) + \ce{Cl-}(aq) ⟶ \ce{HCl}(aq) + \ce{HSO4-}(aq)\)
4. \(\ce{HSO4-}(aq) + \ce{OH-}(aq) ⟶ \ce{SO42-}(aq) + \ce{H2O}(l)\)
5. \(\ce{O2-}(aq) + \ce{H2O}(l) ⟶ \ce{2OH-}(aq)\)
6. \(\ce{H2S}(aq) + \ce{NH2-}(aq) ⟶ \ce{HS-}(aq) + \ce{NH3}(aq)\)

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), \(\ce{NO3- (CB)}\)

b. CN⁻(BB), H₂O(BA), HCN(CA), \(\ce{OH^-}\) (CB)

c. H₂SO₄(BA), Cl⁻(BB), HCl(CA), \(\ce{HSO4- (CB)}\)

d. \(\ce{HSO4- (BA)}\), OH^-(BB), \(\ce{SO4^2- (CB)}\), H₂O(CA)

e. O²⁻(BB), H₂O(BA) \(\ce{OH^-}\) (CB and CA)

f. H₂S(BA), \(\ce{NH2- (BB)}\), HS⁻(CB), NH₃(CA)

Exercise \(\PageIndex{7}\)

What is the [H₃O⁺] if an aqueous solution has an [OH^-] concentration of 1.65 x 10^-5 at 100^oC? K_w at 100^oC is 5.6 × 10⁻¹³

Answer

3.39 x 10^-8 M

Exercise \(\PageIndex{8}\)

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: \(\ce{H2O}(l) + \ce{NH3}(aq) \rightleftharpoons \ce{NH4+}(aq) + \ce{OH-}(aq)\). As a base: \(\ce{H2O}(l) + \ce{HCl}(aq) \rightleftharpoons \ce{H3O+}(aq) + \ce{Cl-}(aq)\)

Exercise \(\PageIndex{9}\)

State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.

1. NH₃
2. \(\ce{HPO4-}\)
3. Br⁻
4. \(\ce{NH4+}\)
5. \(\ce{ASO4^3-}\)

Answer

amphiprotic: \(\ce{NH3 + H3O+ ⟶ NH4OH + H2O}\), \(\ce{NH3 + OCH3- ⟶ NH2- + CH3OH}\); \(\ce{HPO4^2- + OH- ⟶ PO4^3- + H2O}\), \(\ce{HPO4^2- + HClO4 ⟶ H2PO4- + ClO4-}\); not amphiprotic: Br⁻; \(\ce{NH4+}\); \(\ce{AsO4^3-}\)

6.2: pH and pOH

Exercise \(\PageIndex{10}\)

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

Answer

In a neutral solution [H₃O⁺] = [OH⁻]. At 40 °C,

[H₃O⁺] = [OH⁻] = (2.910⁻¹⁴)^1/2 = 1.7 × 10⁻⁷.

Exercise \(\PageIndex{11}\)

The ionization constant for water (K_w) is 2.90 × 10⁻¹⁴ at 40 °C. Calculate [H₃O⁺], [OH⁻], pH, and pOH for pure water at 40 °C.

Answer

x = 1.70 × 10⁻⁷ M = [H₃O⁺] = [OH⁻]

pH = -log (1.70 × 10⁻⁷ ) = −(−6.769) = 6.769

pOH = pH = 6.769

Exercise \(\PageIndex{12}\)

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

1. 0.000259 M HClO₄
2. 0.21 M NaOH
3. 0.000071 M Ba(OH)₂
4. 2.5 M KOH

Answer

pH = 3.587; pOH = 10.413; pH = 0.68; pOH = 13.32; pOH = 3.85; pH = 10.15; pH = −0.40; pOH = 14.4

Exercise \(\PageIndex{13}\)

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

Exercise \(\PageIndex{14}\)

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

6.3: Relative Strengths of Acids and Bases

Exercise \(\PageIndex{15}\)

1. Arrange these acids in order of increasing strength.
   • acid A: pKa=1.52
   • acid B: pKa=6.93
   • acid C: pKa=3.86

Given solutions with the same initial concentration of each acid, which would have the highest percent ionization?

Answer

Acids in order of increasing strength: acidB<acidC<acidA.

Given the same initial concentration of each acid, the highest percent of ionization is acid A because it is the strongest acid.

Exercise \(\PageIndex{16}\)

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] > \(\ce{[H3O+]}\) ≈ \(\ce{[CH3CO2- ]}\) > [OH⁻]

Exercise \(\PageIndex{17}\)

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

\(\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}\)

Exercise \(\PageIndex{18}\)

What is the ionization constant at 25 °C for the weak acid \(\ce{CH3NH3+}\), the conjugate acid of the weak base CH₃NH₂, K_b = 4.4 × 10⁻⁴.

Answer

\(K_\ce{a}=2.3×10^{−11}\)

Exercise \(\PageIndex{19}\)

Which base, CH₃NH₂ or (CH₃)₂NH, is the strongest base? Which conjugate acid, \(\ce{(CH3)2NH2+}\) or (CH₃)₂NH, is the strongest acid?

Answer

The strongest base or strongest acid is the one with the larger K_b or K_a, respectively. In these two examples, they are (CH₃)₂NH and \(\ce{CH3NH3+}\).

Exercise \(\PageIndex{20}\)

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

Answer

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.

Assume we can neglect the contribution of water to the equilibrium concentration of H₃O⁺.

Exercise \(\PageIndex{21}\)

What is the effect on the concentrations of \(\ce{NO2-}\), HNO₂, and OH⁻ when the following are added to a solution of KNO₂ in water:

1. HCl
2. HNO₂
3. NaOH
4. NaCl
5. KNO

The equation for the equilibrium is:

\[\ce{NO2-}(aq)+\ce{H2O}(l)⇌\ce{HNO2}(aq)+\ce{OH-}(aq)\]

Answer

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 \(\ce{NO2-}\) ions. Adding HNO₂ increases the concentration of HNO₂ and shifts the equilibrium to the left, increasing the concentration of \(\ce{NO2-}\) ions and decreasing the concentration of OH⁻ ions. Adding NaOH adds OH⁻ ions, which shifts the equilibrium to the left, increasing the concentration of \(\ce{NO2-}\) ions and decreasing the concentrations of HNO₂. Adding NaCl has no effect on the concentrations of the ions. Adding KNO₂ adds \(\ce{NO2-}\) ions and shifts the equilibrium to the right, increasing the HNO₂ and OH⁻ ion concentrations.

Exercise \(\PageIndex{22}\)

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.

Exercise \(\PageIndex{23}\)

From the equilibrium concentrations given, calculate K_a for each of the weak acids and K_b for each of the weak bases.

NH₃: [OH⁻] = 3.1 × 10⁻³ M;

\(\ce{[NH4+]}\) = 3.1 × 10⁻³ M;

[NH₃] = 0.533 M;

HNO₂: \(\ce{[H3O+]}\) = 0.011 M;

\(\ce{[NO2- ]}\) = 0.0438 M;

[HNO₂] = 1.07 M;

(CH₃)₃N: [(CH₃)₃N] = 0.25 M;

[(CH₃)₃NH⁺] = 4.3 × 10⁻³ M;

[OH⁻] = 4.3 × 10⁻³ M;

\(\ce{NH4+ : [NH4+]}\) = 0.100 M;

[NH₃] = 7.5 × 10⁻⁶ M;

[H₃O⁺] = 7.5 × 10⁻⁶ M

Answer

\(K_\ce{b}=1.8×10^{−5};\) \(K_\ce{a}=4.5×10^{−4};\) \(K_\ce{b}=7.4×10^{−5};\) \(K_\ce{a}=5.6×10^{−10}\)

Exercise \(\PageIndex{24}\)

Determine K_a for hydrogen sulfate ion, \(\ce{HSO4-}\). In a 0.10-M solution the acid is 29% ionized.

Answer

\(K_\ce{a}=1.2×10^{−2}\)

Exercise \(\PageIndex{25}\)

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. Ionization constants can be found in the text or Resource Tables.

1. 0.0092 M HClO, a weak acid
2. 0.0784 M C₆H₅NH₂, a weak base
3. 0.0810 M HCN, a weak acid
4. 0.11 M (CH₃)₃N, a weak base
5. 0.120 M \(\ce{Fe(H2O)6^2+}\) a weak acid, K_a = 1.6 × 10⁻⁷

Answer

a. \(\ce{\dfrac{[H3O+][ClO- ]}{[HClO]}}=\dfrac{(x)(x)}{(0.0092−x)}≈\dfrac{(x)(x)}{0.0092}=3.5×10^{−8}\)

Solving for x gives 1.79 × 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] = 1.8 × 10⁻⁵ M[HClO] = 0.00092 M [OH⁻] = 5.6 × 10⁻¹⁰ M;

b. \(\ce{\dfrac{[C6H5NH3+][OH- ]}{[C6H5NH2]}}=\dfrac{(x)(x)}{(0.0784−x)}≈\dfrac{(x)(x)}{0.0784}=4.6×10^{−10}\)

Solving for x gives 6.01 × 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: \(\ce{[CH3CO2- ]}\) = [OH⁻] = 6.0 × 10⁻⁶ M [C₆H₅NH₂] = 0.00784 [H₃O⁺] = 1.7× 10⁻⁹ M

c. \(\ce{\dfrac{[H3O+][CN- ]}{[HCN]}}=\dfrac{(x)(x)}{(0.0810−x)}≈\dfrac{(x)(x)}{0.0810}=4×10^{−10}\) Solving for x gives 5.69 × 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⁻] = 5.7 × 10⁻⁶ M [HCN] = 0.0810 M [OH⁻] = 1.8 × 10⁻⁹ M

d. \(\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{(0.11−x)}≈\dfrac{(x)(x)}{0.11}=7.4×10^{−5}\) Solving for x gives 2.85 × 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⁻] = 2.9 × 10⁻³ M [(CH₃)₃N] = 0.11 M [H₃O⁺] = 3.5 × 10⁻¹² M

e. \(\ce{\dfrac{[Fe(H2O)5(OH)+][H3O+]}{[Fe(H2O)6^2+]}}=\dfrac{(x)(x)}{(0.120−x)}≈\dfrac{(x)(x)}{0.120}=1.6×10^{−7}\) 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 \(\ce{[Fe(H2O)6^2+]}\) = 0.120 M [OH⁻] = 7.2 × 10⁻¹¹ M

Exercise \(\PageIndex{26}\)

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

Exercise \(\PageIndex{27}\)

The pH of a 0.15-M solution of \(\ce{HSO4-}\) is 1.43. Determine K_a for \(\ce{HSO4-}\) from these data.

Answer

\(K_\ce{a}=1.2×10^{−2}\)

Exercise \(\PageIndex{28}\)

The pH of a solution of household ammonia, a 0.950 M solution of NH_3, is 11.612. Determine K_b for NH₃ from these data.

Answer

\(K_\ce{b}=1.77×10^{−5}\)

6.4: Diprotic & Polyprotic Acids

Exercise \(\PageIndex{29}\)

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⁺],[H₃O⁺], [OH⁻], [H₂CO₃], [HCO₃⁻],[HCO₃⁻], [CO₃²⁻]?[CO₃²⁻]? No calculations are needed to answer this question.

Answer

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

Exercise \(\PageIndex{30}\)

Nicotine, C₁₀H₁₄N₂, is a base that will accept two protons (K₁ = 7 × 10⁻⁷, K₂ = 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 \(\ce{[C10H14N2H2^2+]}\) = 1.4 × 10⁻¹¹ M [OH⁻] = 1.9 × 10⁻⁴ M [H₃O⁺] = 5.3 × 10⁻¹¹ M

Exercise \(\PageIndex{31}\)

The ion HTe⁻ is an amphiprotic species; it can act as either an acid or a base.

K_a2 for the reation of HTe⁻ with H₂O is 1.5 x 10^-11

What is K_b for the reaction in which HTe⁻ functions as a base in water is 4.3 x 10^-12

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

\[\frac{\left[\mathrm{Te}^{2-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HTe}^{-}\right]}=\frac{(x)(0.0141+x)}{(0.0141-x)} \approx \frac{(x)(0.0141)}{0.0141}=1.5 \times 10^{-11} \notag\]

Solving for x gives 1.5 × 10⁻¹¹ M. Therefore, compared with 0.014 M, this value is negligible (1.1 ×× 10⁻⁷%).

6.5 Hydrolysis of Salt Solutions

Exercise \(\PageIndex{32}\)

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

1. FeCl₃
2. K₂CO₃
3. NH₄Br
4. KClO₄

Answer

a. acidic

b. basic

c. acidic

d. neutral

Exercise \(\PageIndex{33}\)

1. Why are aqueous solutions of salts such as CaCl₂ neutral? Why is an aqueous solution of NaNH₂ basic?

Answer

Aqueous solutions of salts such as CaCl₂ are neutral because it is created from hydrochloric acid (a strong acid) and calcium hydroxide (a strong base). An aqueous solution of NaNH₂ is basic because it can deprotonate alkynes, alcohols, and a host of other functional groups with acidic protons such as esters and ketones.

Exercise \(\PageIndex{34}\)

Predict whether the aqueous solutions of the following are acidic, basic, or neutral.

1. Li₃N
2. NaH
3. KBr
4. C₂H₅NH₃Cl

Answer

a. Li₃N is a base because the lone pair on the nitrogen can accept a proton.

b. NaH is a base because the hydrogen has a negative charge with a lone pai that can accept a proton.

c. KBr is neutral because it is formed from HBr (a strong acid) and KOH (a strong base) so no reaction.

d. C₂H₅NH₃Cl is acidic because it can donate a proton.

(make sure you can write out reactions to explain each answer!)

6.5 Lewis Acids & Bases

Exercise \(\PageIndex{35}\)

Identify the nature of each of the following as either a Lewis Acid or a Lewis Base:

a. NH₃

b. H₂O

c. SO₂

Answer

a. NH₃ is a lewis base because nitrogen has a lone pair of electrons to "donate."

b. H₂O is a lewis base because oxygen has two lone pairs of electrons to "donate."

c. SO₂ is a lewis acid because sulfur has an unfilled octet and thus is able to accept a pair of electrons.