USNCO question - Finding pH of salt solution from pKa of its parent acid The question from 2017 USNCO is as below:

What is the $\mathrm{pH}$ of a $\pu{0.25 M}$ solution of $\ce{NaCN}$? (The $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is $9.21$.)
(A) $4.91$
(B) $8.61$
(C) $11.30$
(D) $13.40$

The answer sheet states (C).
My understandings/reasoning:

salt hydrolysis: solution of salt derived from strong base and weak acid is basic;
$\ce{NaCN}$ is derived from the base $\ce{NaOH}$ (strong) and acid $\ce{HCN}$ (weak according to given $\mathrm{p}K_\mathrm{a}$);
Solution is basic, so (A) is incorrect.

However, I am unable to proceed from here.
Sodium cyanide is a salt formed by a strong base and a weak acid, so the reduced ionic equation of hydrolysis is

$$\ce{CN- + H2O &lt;=&gt;[$K_\mathrm{h}$] HCN + OH-}$$

$$
\begin{align}
K_\mathrm{h} &amp;= \left.\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]} \qquad \right|\cdot \frac{[\ce{H+}]}{[\ce{H+}]}\\
&amp;= \frac{[\ce{HCN}][\ce{OH-}][\ce{H+}]}{[\ce{CN-}][\ce{H+}]}\\
&amp;=\frac{K_\mathrm{w}}{K_\mathrm{a}} \label{eq:1}\tag{1}
\end{align}
$$

Sodium cyanide is highly soluble salt, so one can neglect auto-protolysis of water, and also express hydrolysis constant (assuming $[\ce{OH-}] = [\ce{HCN}]$ and $C$ – concentration of sodium cyanide) as follows:

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{C-[\ce{OH-}]} \label{eq:2}\tag{2}$$

Equating \eqref{eq:1} and \eqref{eq:2}: 

$$\frac{10^{-14}}{10^{-9.21}} = \frac{[\ce{OH-}]^2}{0.25-[\ce{OH-}]}\, ,\label{eq:3}\tag{3}$$

and solving the quadratic equation \eqref{eq:3} for $[\ce{OH-}]$, one can find that $[\ce{OH-}] \approx \pu{2e-3 M}$, $\mathrm{pOH} = 2.70$, hence

$$\mathrm{pH} = 14.00 - \mathrm{pOH} = 14.00 - 2.70 = 11.30$$