Question

Calculate the pH at the equivalence point when a solution of $0.1M$ acetic acid is titrated with a solution of $0.1M$ sodium hydroxide.
$K_a$ for acetic acid $=1.9\times {10}^{-5}$

Answer 1

Concentration of sodium acetate $=\frac{0.1}{2}=0.05M$ at equal volumes of acid and base will be used.
The equilibrium is,
${CH}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons {CH}_{3}COOH+{OH}^{-}$
$C(1-x)$ $CxCx$
where $x$, is the degree of hydrolysis and
$K_h=\frac{C{x}^2}{(1-x)}$
we know that
$K_h=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{1.9\times {10}^{-5}}=5.26\times {10}^{-10}$
$K_h=C{x}^2$
$5.26\times {10}^{-10}=0.05\times x^2$
or $x=1.025\times {10}^{-4}$
$\left[{OH}^{-}\right]=Cx=1.025\times {10}^{-4}\times 0.05=5.125\times {10}^{-6}M$
$\left[H^{+}\right]=\frac{1\times {10}^{-14}}{5.125\times {10}^{-6}}=1.95\times {10}^{-9}M$
$pH=-\log \left[H^{+}\right]=-\log (1.95\times {10}^{-9})=8.71$