Question

$0.1M$ $NaOH$ is titrated with $0.1M$ $HA$ till the end point; $K_a$ for $HA$ is $5.6\times {10}^{-6}$ and degree of hydrolysis is less compared to $1$. Calculate pH of the resulting solution at the end point.

Answer 1

Neutralisation of $HA$ with $NaOH$ may be given as
$HA+NaOH⟶NaA+H_{2}O$

Concentration of salt will be $\frac{0.1}{2}M$, i.e., $0.05M$, since volume will be double.

pH of the salt after hydrolysis may be calculated as,

$pH=\frac{1}{2}[p{K}_w+p{K}_a+\log C]........\left(i\right)$
$p{K}_w=14$

$p{K}_a=-\log K_a=-\log 5.6\times {10}^{-6}=5.1258$

$\log C=\log 0.05=-1.3010$

Substituting the values of $p{K}_w,p{K}_a$ and $\log C$ in eq. (i) we get,

$pH=\frac{1}{2}[14+5.2518-1.3010]=8.9754$