Question

Calculate ${OH}^{-}$ concentration at the equivalent point when a solution of $0.1M$ acetic acid is titrated with a solution of $0.1M$ $NaOH$. $K_a$ for the acid $=1.9\times {10}^{-5}$

• A. $5.12\times {10}^{-6}M$
• B. $5.12\times {10}^{-4}M$
• C. $5.12\times {10}^{-3}M$
• D. $5.12\times {10}^{-2}M$

Answer 1

Answer: (A)

$5.12\times {10}^{-6}M$

Assume 1 L of 0.1 M acetic acid is neutralized with 0.1 M NaOH solution.

Volume of NaOH solution required will be 1 L.

The concentration of sodium acetate will be $\frac{0.1}{2}=0.05$ M

The expression for the hydrolysis of salt is as shown.

${CH}_{3}COOH+{OH}^{\ominus }⟶{CH}_{3}CO{O}^{-}+H_{2}O$ (WASB)

$pH=\frac{1}{2}[p{K}_w+p{K}_a+\log C]$

$pH=0.5[14-log1.9\times {10}^{-5}+log0.05]=8.71$

$pOH=14-pH=14-8.71=5.29$

$\left[O{H}^{-}\right]={10}^{-pOH}={10}^{-5.29}=5.12\times {10}^{-6}M$