To 20 ml of solution, 3 ml of 1 M acetic acid solution is added. Is the solution now neutral, acidic — Ionic Equilibrium Chemistry Question
Question
To 20 ml of $0.1 \text{ M} – \text{NaOH}$ solution, 3 ml of 1 M acetic acid solution is added. Is the solution now neutral, acidic or alkaline? How much more of the acetic acid solution we add to produce a change of $\text{pH} = 0.3 \text{ unit}$? ($pK_a$ for $\text{CH}_3\text{COOH} = 4.74$, $\log 2 = 0.3$)
💡 Solution & Explanation
Initial moles of NaOH = $20 \times 0.1 = 2 \text{ mmol}$. Initial moles of $\text{CH}_3\text{COOH}$ = $3 \times 1 = 3 \text{ mmol}$. Acid is in excess ($1 \text{ mmol}$ remains) and $2 \text{ mmol}$ of salt is formed, creating an acidic buffer. $\text{pH}_1 = pK_a + \log(\text{Salt}/\text{Acid}) = 4.74 + \log(2/1) = 4.74 + 0.3 = 5.04$ (Acidic). To change the pH by 0.3 units using the acid, the pH must drop to $4.74$. $\text{pH}_2 = 4.74 = pK_a + \log(2/n_{\text{acid}}) \implies 4.74 = 4.74 + \log(2/n_{\text{acid}}) \implies \log(2/n_{\text{acid}}) = 0 \implies n_{\text{acid}} = 2 \text{ mmol}$. Since 1 mmol is currently present, 1 more mmol is needed. From a 1 M solution, this equals 1 ml. Therefore, correct answer is C.