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Q.
$K_a$ for $HCN$ is $5 \times 10^{-10}$ at $25^{\circ}C$. For maintaining a constant $pH\, = \,9$, the volume of $5\, M \,KCN$ solution required to be added to $10 \,mL$ of $2\, M HCN$ solution is
$pH = pK_{a}+ log \frac{\left[Salt\right]}{\left[Acid\right]}$
$\quad\quad=pK_{a}+ log \frac{\left[KCN\right]}{\left[HCN\right]}\quad\quad\quad\quad\quad...\left(i\right)$
Let the volume of KCN solution required be V mL
$\therefore \quad\left[KCN\right]=\frac{5\times V}{V+10} and \left[HCN\right] = \frac{10\times2}{V +10}$
Now from eqn. $\left(i\right),$
$pH=- log \left(5\times10^{-10}\right)+log \left[\frac{5\times V}{V+10} /\frac{10\times2}{V+10}\right]$
$9 = -log\left(5\times10^{-10}\right)+log \frac{V}{4}$
On solving, $V = 1.99 \approx 2 mL$