$K_a$for $\mathrm{HCN}$is $5\times {10}^{-10}$at ${25}^{\circ }\mathrm{C}$. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is (log 2 = 0.3) :

$\begin{array}{l}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\text{ Salt }\right]}{\left[\text{ Acid }\right]}\end{array}$

$={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{KCN}\right]}{\left[\mathrm{HCN}\right]}$                     ...(i)

The needed amount of KCN solution should be VmL.

$\therefore \left[\mathrm{KCN}\right]=\frac{5\times V}{V+10}$and $\left[\mathrm{HCN}\right]=\frac{10\times 2}{V+10}$

now, from equivalence

$\begin{array}{l}\mathrm{pH}=-\log \left(5\times {10}^{-10}\right)+\log \left[\frac{5\times \mathrm{V}}{\mathrm{V}+10}/\frac{10\times 2}{\mathrm{V}+10}\right]\\ 9=-\log \left(5\times {10}^{-10}\right)+\log \frac{\mathrm{V}}{4}\end{array}$

On solving, $V=1.99\approx 2\mathrm{mL}$.