Question

Calculate $O{H}^{-}$ concentration at the equivalent point when a solution of $0.1M$ acetic acid is titrated with a solution of $0.1M$ NaOH $K_2$ for the acid $=1.9\times {10}^{-3}$

Answer 1

${CH}_{3}COOH+NaOH\to H_{2}O+{CH}_{3}COONa$

${CH}_{3}COONa+H_{2}O\to {CH}_{3}COOH+{OH}^{(-)}$

$\left[H^{+}\right]=\sqrt{\frac{K_w{K}_a}{C}}C=\frac{0.1\times V}{2V}=0.05$

$\Rightarrow \sqrt{\frac{{10}^{-14}\times 1.9\times {10}^{-3}}{0.05}}=\sqrt{{10}^{-15}\times \frac{1.9}{5}}=\sqrt{\frac{19}{5}\times {10}^{-16}}={10}^{-8}\times 1.94$

$pH=7.71$

$pOH=14-pH=6.289$

$\left[{OH}^{-}\right]=0.51\times {10}^{-6}$