of aqueous solution of acetic acid is mixed with of at . If of is added to the above solution, deter — Ionic Equilibrium Chemistry Question
Question
$500 \text{ mL}$ of $0.2 \text{ M }$ aqueous solution of acetic acid is mixed with $500 \text{ mL}$ of $0.2 \text{ M } HCl$ at $25^\circ C$. If $6 \text{ g}$ of $NaOH$ is added to the above solution, determine final pH. Assume there is no change in volume on mixing. $K_a$ of acetic acid is $1.75 \times 10^{-5} \text{ M}$. (Given $\log 1.75 = 0.24$)
💡 Solution & Explanation
Total volume is $1 \text{ L}$. Moles of $CH_3COOH = 0.5 \times 0.2 = 0.1 \text{ mol}$. Moles of $HCl = 0.5 \times 0.2 = 0.1 \text{ mol}$. Moles of $NaOH$ added $= 6 / 40 = 0.15 \text{ mol}$. The strong base $NaOH$ first neutralizes the strong acid $HCl$: $0.1 \text{ mol}$ of $NaOH$ consumes $0.1 \text{ mol}$ $HCl$. Remaining $NaOH = 0.15 - 0.1 = 0.05 \text{ mol}$. This then neutralizes $CH_3COOH$: $0.05 \text{ mol } NaOH$ reacts with $0.05 \text{ mol } CH_3COOH$ to form $0.05 \text{ mol } CH_3COONa$. Remaining unreacted $CH_3COOH = 0.1 - 0.05 = 0.05 \text{ mol}$. The solution is a buffer with $[Salt] = [Acid]$. $pH = pK_a = -\log(1.75 \times 10^{-5}) = 5 - \log 1.75 = 5 - 0.24 = 4.76$.