Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{pH}=\mathrm{p}{K}_a+\log \frac{[\text{ Salt }]}{[\text{ Acid }]}$
Let the volume of $\mathrm{KCN}$ solution required be $V\mathrm{mL}$ $\therefore [\mathrm{KCN}]=\frac{5\times V}{V+10}$ and $[\mathrm{HCN}]=\frac{10\times 2}{V+10}$ Now from eqn. (i), $\begin{array}{rl}& \mathrm{pH}=-\log \left(5\times 10^{-10}\right)+\log \left[\frac{5\times V}{V+10}/\frac{10\times 2}{V+10}\right]\\ & 9=-\log \left(5\times 10^{-10}\right)+\log \frac{V}{4}\end{array}$ On solving, $V=1.99\approx 2\mathrm{mL}$