Answer
Since the $K_b$ for $CN^-$ is greater than the $K_a$ for $N{H_4}^+$, the solution of this salt is basic.
Work Step by Step
$$K_a(NH_4^+) = \frac{K_w}{K_b(NH_3)}= \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$
$$K_b (CN^-) = \frac{K_w}{K_a(HCN)} = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.6 \times 10^{-5}$$
$$K_b \gt K_a$$
Therefore, the solution of $NH_4CN$ is basic.
