Equilibria: Question 4

Worked solution

Part (a): pH of the weak acid

Lactic acid dissociates partially: $\text{CH}_3\text{CH(OH)COOH} \rightleftharpoons \text{CH}_3\text{CH(OH)COO}^- + \text{H}^+$ $K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH(OH)COO}^-]}{[\text{CH}_3\text{CH(OH)COOH}]}$

Assumption: since $K_a$ is small, dissociation is small, so the equilibrium concentration of undissociated acid is approximately equal to the initial concentration, $[\text{CH}_3\text{CH(OH)COOH}]_{eq}\approx0.200\ \text{mol dm}^{-3}$; also $[\text{H}^+]\approx[\text{CH}_3\text{CH(OH)COO}^-]$, since they are produced in a 1:1 ratio and any $\text{H}^+$ from water is negligible.

$K_a \approx \frac{[\text{H}^+]^2}{0.200}$ $[\text{H}^+]^2 = K_a\times0.200 = 1.4\times10^{-4}\times0.200 = 2.8\times10^{-5}$ $[\text{H}^+] = \sqrt{2.8\times10^{-5}} = 5.29\times10^{-3}\ \text{mol dm}^{-3}\ \text{(3 s.f.)}$ $\text{pH} = -\log(5.29\times10^{-3}) = 2.28\ \text{(2 d.p.)}$

(Check: recomputing independently, $1.4\times0.200=0.28$, so $K_a\times C=2.8\times10^{-5}=28\times10^{-6}$; $\sqrt{28\times10^{-6}}=\sqrt{28}\times10^{-3}\approx5.29\times10^{-3}\ \text{mol dm}^{-3}$; $-\log(5.29\times10^{-3})=3-\log(5.29)=3-0.7236=2.276\approx2.28$, consistent. The dissociation fraction is only $[\text{H}^+]/C=5.29\times10^{-3}/0.200\approx2.6\%$, small enough that the approximation $[\text{CH}_3\text{CH(OH)COOH}]_{eq}\approx0.200$ is justified.)

Part (b): pH of the buffer solution

For a buffer of a weak acid and its conjugate base, the Henderson-Hasselbalch equation applies: $\text{pH}=pK_a+\log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$

First find $pK_a$: $pK_a=-\log(1.4\times10^{-4})=3.854\ \text{(3 d.p.)}$

Then: $\text{pH}=3.854+\log\left(\frac{0.150}{0.100}\right)=3.854+\log(1.5)=3.854+0.176=4.03$

(Check by the direct equilibrium route: the salt supplies $[\text{CH}_3\text{CH(OH)COO}^-]\approx0.150\ \text{mol dm}^{-3}$, and the acid is only slightly dissociated so $[\text{CH}_3\text{CH(OH)COOH}]\approx0.100\ \text{mol dm}^{-3}$. Rearranging $K_a=\dfrac{[\text{H}^+][\text{salt}]}{[\text{acid}]}$ for $[\text{H}^+]$: $[\text{H}^+]=K_a\times\dfrac{[\text{acid}]}{[\text{salt}]}=1.4\times10^{-4}\times\dfrac{0.100}{0.150}=9.33\times10^{-5}\ \text{mol dm}^{-3}$, so $\text{pH}=-\log(9.33\times10^{-5})=4.03$. The two methods agree.)

Part (c): Why the buffer resists the pH change

The buffer contains a substantial reservoir of both the weak acid ($\text{CH}_3\text{CH(OH)COOH}$) and its conjugate base ($\text{CH}_3\text{CH(OH)COO}^-$, supplied by the sodium lactate). The $\text{H}^+$ ions added from the hydrochloric acid are mostly removed by reacting with the conjugate base: $\text{CH}_3\text{CH(OH)COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH(OH)COOH}$

Because this converts only a small fraction of the comparatively large amount of lactate ions present into lactic acid molecules, the ratio $[\text{salt}]/[\text{acid}]$ changes only slightly, and since pH depends on the logarithm of this ratio, the pH itself changes only very slightly.

Final answers

• (a) $\text{pH}=2.28$
• (b) $\text{pH}=4.03$
• (c) Added $\text{H}^+$ is largely consumed by the conjugate base $\text{CH}_3\text{CH(OH)COO}^-$, leaving the salt:acid ratio, and so the pH, almost unchanged