Exactly of is carefully titrated against a solution. What will be the exact pH of the resulting mixt — Ionic Equilibrium Chemistry Question
Question
Exactly $50 \text{ mL}$ of $0.2 \text{ M } CH_3COOH$ is carefully titrated against a $0.2 \text{ M } NaOH$ solution. What will be the exact pH of the resulting mixture after precisely $25 \text{ mL}$ of the $NaOH$ solution has been added? (Given the precise dissociation constant $pK_a$ for $CH_3COOH = 4.74$).
💡 Solution & Explanation
The initial millimoles of $CH_3COOH = 50 \text{ mL} \times 0.2 \text{ M} = 10 \text{ mmol}$. The millimoles of $NaOH$ added = $25 \text{ mL} \times 0.2 \text{ M} = 5 \text{ mmol}$. The $NaOH$ completely neutralizes $5 \text{ mmol}$ of $CH_3COOH$ to produce $5 \text{ mmol}$ of $CH_3COONa$. The remaining unreacted $CH_3COOH$ is $10 - 5 = 5 \text{ mmol}$. Because the solution now contains a weak acid and its conjugate base in strictly equal molar amounts ($[Salt] = [Acid]$), it functions as an ideal buffer at its maximum capacity. By the Henderson-Hasselbalch equation, $pH = pK_a + \log\frac{5}{5} = pK_a$. Therefore, the $pH = 4.74$.