Acid-Base Equilibrium Calculations Practice Questions

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.

Solution
$Mg(OH)_2\;(s)+$ $2HCl\;(aq)⟶$ $Mg^{2+}\;(aq)+2Cl^-$ $(aq)+H_2O\;(l)$
BB BA CB CA

Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.

What is the ionization constant at 25 °C for the weak acid $CH_3NH_3^+$, the conjugate acid of the weak base $CH_3NH_2,\;K_w=4.4×10^{-4}$.

Solution$$K_a=2.3×10^{-11}$$

What is the ionization constant at 25 °C for the weak acid $(CH_3)_2NH_2^+$,the conjugate acid of the weak base $(CH_3)_2NH,\quad K_b=5.9×10^{-4}$?

Which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, $(CH_3)_2NH_2^+$ or $CH_3NH_3^+$, is the stronger acid?

Solution

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 $CH_3NH_3^+$

Which is the stronger acid, $NH_4^+$ or HBrO?

Which is the stronger base, (CH3)3N or $H_2BO_3^-$?

Solution

triethylamine

Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN, is the stronger base?

Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.

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

Solution

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+.

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

Solution

(b) The addition of HCl

Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?

(a) addition of NaOH

(b) addition of HCl

(c) addition of NaF

What is the effect on the concentrations of $NO_2^-$, $HNO_2$ and $OH^-$ when the following are added to a solution of $KNO_2$ in water:

(a) HCl

(b) HNO2

(c) NaOH

(d) NaCl

(e) KNO

Solution

(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 $NO_2^-$ ions. (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of $NO_2^-$ 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 $NO_2^-$ ions and decreasing the concentrations of HNO2. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO2 adds $NO_2^-$ ions and shifts the equilibrium to the right, increasing the HNO2 and OH ion concentrations.

What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?

(a) HCl

(b) KF

(c) NaCl

(d) KOH

(e) HF

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?

Solution

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.

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

(a) $CH_3CO_2H$: $[H_3O^+]=1.34×10^{-3}\;M$;

$[CH_3CO_2^-]=1.34×10^{-3}\;M$;

$[CH_3CO_2H] = 9.866×10^{-2}\;M$;

(b) $ClO^-$: $[OH^−] = 4.0×10^-4\;M$;

[HClO] = $2.38×10^-4\;M$

$[ClO^-] = 0.273\;M$;

(c) $HCO_2H$: $[HCO_2H] = 0.524\;M$;

$[H_3O^+]= 9.8×10^{-3}\;M$ $[HCO_2^-]= 9.8×10^{-3}\;M$

(d) $C_6H_5NH_3^+\;:\;[C_6H_5NH_3^+]=0.233\;M$;

$[C_6H_5NH_2] = 2.3×10^{-3}\;M$;

$[H_3O^+]= 2.3×10^{-3}\;M$

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

(a) $NH_3:\;[OH^−] = 3.1×10^{-3}\;M$;

$[NH_4^+]= 3.1×10^{-3}\;M$;

$[NH_3] = 0.533\;M$;

(b) $HNO_2\;:\;[H_3O^+]=0.011\;M$;

$[NO_2^-]= 0.0438\;M$;

$[HNO_2] = 1.07\;M$;

(c) $(CH_3)_3N:\;[(CH_3)_3N] = 0.25\;M$; $[(CH_3)_3NH^+] = 4.3×10^{-3}\;M$;

$[OH^−] = 3.7×10^{-3}\;M$;

(d) $NH_4^+\;:\;[NH_4^+]=0.100\;M$;

$[NH_3] = 7.5×10^{-6}\;M$;

$[H_3O^+] = 7.5×10^{-6}\;M$;
Solution

(a) $K_b=1.8×10^{-5}$;

(b) $K_a=4.5×10^{-4}$; (c) $K_b=6.4×10^{-5}$; (d) $K_a=5.6×10^{-10}$;

Determine $K_b$ for the nitrite ion, $NO_2^-$. In a 0.10-M solution this base is 0.0015% ionized.

Determine $K_a$ for hydrogen sulfate ion, $HSO_4^-$. In a 0.10-M solution the acid is 29% ionized.

Solution

$K_a=1.2×10^{-2}$

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) F

(b) $NH_4^+$

(c) $AsO_4^{3-}$

(d) $(CH_3)_2NH_2^+$

(e) $NO_2^-$

(f) $HC_2O_4^-$ (as a base)

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) $(CH_3)_3NH^+$

(c) $HAsO_4^{2-}$ (as a base)

(d) $HO_2^-$ (as a base)

(e) $C_6H_5NH_3^+$

(f) $HSO_3^-$ (as a base)

Solution

(a) $K_b=4.3×10^{-12}$

(b) $K_a=1.6×10^{10}$ (c) $K_b=5.9×10^{8}$ (d) $K_b=4.2×10^{-3}$ (e) $K_b=2.3×10^{5}$ (f) $K_b=6.3×10^{-13}$

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 $Fe(H_2O)_6^{2+}$ a weak acid, $K_a = 1.6×10^{-7}$

Solution

(a) $\frac{[H_3O^+][ClO^-]}{[HClO]}=\frac{(x)(x)}{(0.0092-x)}≈\frac{(x)(x)}{0.0092}=2.9×10^{-8}$

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) $\frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]}=\frac{(x)(x)}{(0.0784-x)}≈\frac{(x)(x)}{0.0784}=4.3×10^{-10}$ 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: $[CH_3CO_2^-]=[OH^-]=5.8×10^{-6}\;M$; $[C_6H_5NH_2] = 0.00784$ $[H_3O^+] = 1.7×10^{-9}\;M$; (c) $\frac{[H_3O^+][CN^-]}{[HCN]}=\frac{(x)(x)}{(0.0810-x)}≈\frac{(x)(x)}{0.0810}=4.9×10^{-10}$ 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: $[H_3O^+] = [CN^-] = 6.3×10^{-6}\;M$; [HCN] = 0.0810 M $[OH^-] = 1.6×10^{-9}\;M$; (d) $\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}=\frac{(x)(x)}{(0.11-x)}≈\frac{(x)(x)}{0.11}=6.3×10^{-5}$ 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: $[(CH_3)_3NH^+]=[OH^-]=2.63×10^-3\;M$ $[(CH_3)_3N]=0.11\;M$ $[H_3O^+] = 3.8×10^{-12}\;M$ (e) $\frac{[Fe(H_2O)_5(OH)^+][H_3O^+]}{[Fe(H_2O)_6^{2+}]}=\frac{(x)(x)}{(0.120-x)}≈\frac{(x)(x)}{0.120}=1.6×10^{-7}$ 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: $[Fe(H_2O)_5(OH)^+]=[H_3O^+]=1.4×10^{-4}\;M$ $[Fe(H_2O)_6^{2+}]=0.120\;M$ $[OH^-]= 7.2×10^{-11}\;M$

Propionic acid, $C_2H_5CO_2H$ ($K_a=1.34×10^{-5}$), is used in the manufacture of calcium propionate, a food preservative. What is the pH of a 0.698-M solution of $C_2H_5CO_2H$?

The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after strenuous exercise, is $1.36×10^{-4}$. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?

Nicotine, $C_{10}H_{14}N_2$, is a base that will accept two protons ($K_{b1}=7×10^{-7}$, $K_{b2}=1.4×10^{-11}$). What is the concentration of each species present in a 0.050-M solution of nicotine?

Solution

$[C_{10}H_{14}N_2]=0.049\;M$

$[C_{10}H_{14}N_2H^+]=1.9×10^{-4}\;M$; $[C_{10}H_{14}N_2H_2^{2+}]=1.4×10^{-11}\;M$; $[OH^-]=1.9×10^{-4}\;M$; $[H_3O^+]=5.3×10^{-11}\;M$

The pH of a 0.23-M solution of HF is 1.92. Determine Ka for HF from these data.

The pH of a 0.15-M solution of $HSO_4^-$ is 1.43. Determine Ka for$HSO_4^-$ from these data.

Solution

$K_a=1.2×10^{-2}$

The pH of a 0.10-M solution of caffeine is 11.70. Determine Kb for caffeine from these data: $$C_8H_{10}N_4O_2\;(aq)+H_2O\;(l)⇌C_8H_{10}N_4O_2H^+\;(aq)+OH^-\;(aq)$$

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.

Solution

$K_b=1.77×10^{-5}$