Question

When $0.2M$ acetic acid is neutralised with $0.1M$ $NaOH$ in $0.5$ litre of water the resulting solution is slightly alkaline. Calculate the pH of the resulting solution. $K_a$ for ${CH}_{3}COOH=1.8\times {10}^{-5}.$

Answer 1

$0.2M$ acetic acid will form $0.2M$ ${CH}_{3}COONa$ and $0.5$ litre of water. Hence, concentration of sodium acetate $\left[{CH}_{3}COONa\right]=0.1mol{L}^{-1}$

${CH}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons {CH}_{3}COOH+{OH}^{-}$

$C(1-x)$ $CxCx$

$K_h=\frac{C{x}^2}{(1-x)}=C{x}^2$

$(1-x)\to 1$

$K_h=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}$

So, $K_h=C{x}^2=5.5\times {10}^{-10}$

or $x^2=55\times {10}^{-10}$

or $x=7.42\times {10}^{-5}$

$\left[{OH}^{-}\right]=Cx=7.42\times {10}^{-5}\times 0.1=7.42\times {10}^{-6}M$

$\left[H^{+}\right]=\frac{K_w}{\left[{OH}^{-}\right]}=1.3477\times {10}^{-9}M$

$pH=-\log \left[H^{+}\right]=8.87$