Acid-Base Calculations 1

Determination of K_a from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar that provides its sour taste. At equilibrium, a solution contains [CH₃CO₂H] = 0.0787 M and $[H_3O^+]=[CH_3CO_2^-]=0.00118\;M$

What is the value of K_a for acetic acid?

Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the K_a for acetic acid.

$$CH_3CO_2H(aq)+H_2O(l)⇌H_3O^+(aq)+CH_3CO_2^-(aq)$$ $$K_a=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}=\frac{(0.00118)(0.00118)}{0.0787}=1.77×10^{-5}$$

Check Your Learning

The $HSO_4^-$ ion, weak acid used in some household cleansers:

$$HSO_4^-(aq)+H_2O(l)⇌H_3O^+(aq)+SO_4^{2-}(aq)$$

What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: $[H_3O^+]=0.027\;M$; $[HSO_4^-]=0.29\;M$; and $[SO_4^{2-}]=0.13\;M$?

 

Answer:

K_a for $HSO_4^-=1.2×10^{-2}$

Determination of K_b from Equilibrium Concentrations
Caffeine, C₈H₁₀N₄O₂ is a weak base. What is the value of K_b for caffeine if a solution at equilibrium has [C₈H₁₀N₄O₂] = 0.050 M, $[C_8H_{10}N_4O_2H^+]=5.0×10^{-3}\;M$, and $[OH^-]=2.5×10^{-3}\;M$?

Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the K_b for caffeine.

$$C_8H_{10}N_4O_2(aq)+H_2O(l)⇌C_8H_{10}N_4O_2H^+(aq)+OH^-(aq)$$ $$K_b=\frac{[C_8H_{10}N_4O_2H^+][OH^-]}{[C_8H_{10}N_4O_2]}=\frac{(5.0×10^{-3})(2.5×10^{-3})}{0.050}=2.5×10^{-4}$$

Check Your Learning

What is the equilibrium constant for the ionization of the $HPO_4^{2-}$ ion, a weak base

$$HPO_4^{2-}(aq)+H_2O(l)⇌H_2PO_4^-(aq)+OH^-(aq)$$

if the composition of an equilibrium mixture is as follows: $[OH^-] = 1.3×10^{-6}\;M$; $[H_2PO_4^-]=0.042\;M$; and $[HPO_4^{2-}]=0.341\;M$?

 

Answer:

$K_b$ for $HPO_4^{2-}=1.6×10^{-7}$