A volume of 18 ml of mixture of acetic acid and sodium acetate required 6 ml of for neutralization o — Ionic Equilibrium Chemistry Question
Question
A volume of 18 ml of mixture of acetic acid and sodium acetate required 6 ml of $0.1 \text{ M} – \text{NaOH}$ for neutralization of the acid and 12 ml of $0.1 \text{ M} – \text{HCl}$ reaction with salt separately. If $pK_a$ of acetic acid is 4.75, what is the pH of the mixture? ($\log 2 = 0.3$)
💡 Solution & Explanation
The moles of acid present in the mixture can be found from the NaOH titration: $n_{\text{acid}} = 6 \text{ ml} \times 0.1 \text{ M} = 0.6 \text{ mmol}$. The moles of salt present can be found from the HCl titration: $n_{\text{salt}} = 12 \text{ ml} \times 0.1 \text{ M} = 1.2 \text{ mmol}$. Using the Henderson-Hasselbalch equation: $\text{pH} = pK_a + \log(\text{Salt} / \text{Acid}) = 4.75 + \log(1.2 / 0.6) = 4.75 + \log 2 = 4.75 + 0.3 = 5.05$. Therefore, correct answer is A.