A volume of of a solution which is in the acid HA () and in HB () is titrated with solution. The pH — Ionic Equilibrium Chemistry Question
Question
A volume of $50\text{ ml}$ of a solution which is $0.05\text{ M}$ in the acid HA ($pK_a = 3.80$) and $0.08\text{ M}$ in HB ($pK_a = 8.20$) is titrated with $0.2\text{ M} – \text{NaOH}$ solution. The pH of solution at the first equivalent point is ($\log 2 = 0.3, \log 1.6 = 0.2$)
💡 Solution & Explanation
The first equivalence point corresponds to the complete neutralization of the stronger weak acid, HA. Moles of HA = $50 \times 0.05 = 2.5\text{ mmol}$, requiring $2.5 / 0.2 = 12.5\text{ ml}$ of NaOH. Total volume = $62.5\text{ ml}$. At this point, the solution contains weak base $\text{A}^-$ and weak acid HB. $[\text{A}^-] = 2.5 / 62.5 = 0.04\text{ M}$ and $[\text{HB}] = (50 \times 0.08) / 62.5 = 4 / 62.5 = 0.064\text{ M}$. For this amphiprotic-like mixture, $[\text{H}^+] = \sqrt{K_a(\text{HA}) K_a(\text{HB}) \frac{[\text{HB}]}{[\text{A}^-]}}$. $\text{pH} = \frac{1}{2}[pK_a(\text{HA}) + pK_a(\text{HB}) - \log(\frac{[\text{HB}]}{[\text{A}^-]})] = \frac{1}{2}[3.80 + 8.20 - \log(1.6)] = \frac{1}{2}[12.0 - 0.2] = \frac{11.8}{2} = 5.9$. Therefore, correct answer is D.