A solution of weak acid was titrated with base NaOH. The equivalence point was reached when of have — Ionic Equilibrium Chemistry Question
Question
A solution of weak acid was titrated with base NaOH. The equivalence point was reached when $36.12\text{ ml}$ of $0.1\text{ M} – \text{NaOH}$ have been added. Now, $18.06\text{ ml}$ $0.1\text{ M} – \text{HCl}$ was added to the titrated solution, the pH was found to be $5.0$. The $pK_a$ of acid is
💡 Solution & Explanation
At the equivalence point, the solution contains purely the salt of the weak acid. Adding $18.06\text{ ml}$ of $0.1\text{ M HCl}$ (exactly half the volume and moles of the original $36.12\text{ ml}$ of NaOH base used) neutralizes half of the salt back into the weak acid. This forms a perfectly equimolar buffer solution where $[\text{Salt}] = [\text{Acid}]$. According to the Henderson-Hasselbalch equation, $\text{pH} = pK_a$ when $[\text{Salt}] = [\text{Acid}]$. Since $\text{pH} = 5.0$, $pK_a = 5.0$. Therefore, correct answer is 5.