Answer
$$[CN^-]= 3.6 \times 10^{-3} \space M$$
Work Step by Step
$Na^+$ does not hydrolyzes.
$CN^-$ act as a weak base.
$$K_b(CN^-) = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.6 \times 10^{-5}$$
$$[H_3O^+] = 10^{-pH} = 10^{-10.38} = 4.2 \times 10^{-11} \space M$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 4.2 \times 10^{-11} } = 2.4 \times 10^{-4} \space M$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CN^- ]& [ OH^- ]& [ HCN ]\\
Initial& y & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& y -x& 0 +x& 0 +x\\
\end{vmatrix}$$
y is the initial concentration of the ion.
$x = [OH^-] = 2.4 \times 10^{-4}$
$$K_b = \frac{[OH^-][HCN]}{[CN^-]} = \frac{(x)(x)}{[CN^-]}$$
$$1.6 \times 10^{-5} = \frac{(2.4 \times 10^{-4})^2}{[CN^-]}$$
$$[CN^-]= 3.6 \times 10^{-3} \space M$$
