## Q. 18.7

Finding $K_a$ of a Weak Acid from the Solution pH

Problem Phenylacetic acid ($C_6H_5CH_2COOH$, simplified here to HPAc; see model) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.12 M HPAc is 2.62. What is the $K_a$ of phenylacetic acid?

## Verified Solution

Plan We are given $[HPAc]_{\text{init}}$ (0.12 M) and the pH (2.62) and must find $K_a$. As always, we first write the equation for HPAc dissociation and the expression for $K_a$ to see which values we need. We set up a reaction table (Table 1) and use the given pH to find $[H_3O^+]$, which equals $[PAc^−]$ and $[HPAc]_{\text{dissoc}}$ (we assume that $[H_3O^+]_{\text{from }H_2O}$ is negligible). To find [HPAc], we assume that, because it is a weak acid, very little dissociates, so $[HPAc]_{\text{init}} − [HPAc]_{\text{dissoc}} = [HPAc] ≈ [HPAc]_{\text{init}}$. We make these assumptions, substitute the equilibrium values, solve for $K_a$, and then check the assumptions using the 5% rule (Sample Problem 17.9).

Solution Writing the dissociation equation and $K_a$ expression:

$HPAc(aq) + H_2O(l) \xrightleftharpoons[]{} H_3O^+(aq) + PAc^−(aq) K_a = \frac{[H_3O^+][PAc^−]}{[HPAc]}$

Setting up a reaction table (Table 1), with $x = [HPAc]_{\text{dissoc}} = [H_3O^+]_{\text{from HPAc}} = [PAc^−]$:

Calculating $[H_3O^+]$:
$[H_3O^+] = 10^{−pH} = 10^{−2.62} = 2.4×10^{−3} M$
Making the assumptions:

1. The calculated $[H_3O^+] (2.4×10^{−3} M) >> [H_3O^+]_{\text{from }H_2O} (1.0×10^{−7} M)$, so we assume that $[H_3O^+] ≈ [H_3O^+]_{\text{from HPAc}} = [PAc^−] = x$ (the change in [HPAc], or $[HPAc]_{\text{dissoc}}$).
2. HPAc is a weak acid, so we assume that [HPAc] = 0.12 M − x ≈ 0.12 M.
Solving for the equilibrium concentrations:

$x ≈ [H_3O^+] = [PAc^−] = 2.4×10^{−3} M$
$[HPAc] = 0.12 M − x = 0.12 M − (2.4×10^{−3} M) ≈ 0.12 M \text{(to 2 sf)}$

Substituting these values into $K_a$:
$K_a = \frac{[H_3O^+][PAc^−]}{[HPAc]} ≈ \frac{(2.4×10^{−3})(2.4×10^{−3})}{0.12} = 4.8×10^{−5}$
Checking the assumptions by finding the percent error in concentration:

1. For $[H_3O^+]_{\text{from }H_2O}: \frac{1×10^{−7} M }{ 2.4×10^{−3} M} × 100 = 4×10^{−3}$% (<5%; assumption is justified.)

2. For $[HPAc]_{\text{dissoc}}: \frac{2.4×10^{−3} M}{0.12 M} × 100 = 2.0$% (<5%; assumption is justified.)

Check The $[H_3O^+]$ makes sense: pH 2.62 should give $[H_3O^+]$ between $10^{−2}$ and $10^{−3}$ M. The $K_a$ calculation also seems in the correct range: $(10^{−3})^2/10^{−1} = 10^{−5}$, and this value seems reasonable for a weak acid.

Table 1

 Concentration (M) $\mathbf{HPAc(aq) + H_2O(l) \xrightleftharpoons[]{} H_3O^+(aq) + PAc^−(aq)}$ Initial 0.12                               —                                   0                                      0 Change −x                                  —                                  +x                                   +x Equilibrium 0.12 − x                        —                                   x                                       x