Answer
a.
$$K_a=\frac{[H^+][CN^-]}{[HCN]}$$
b.
$$K_a=\frac{[H^+][OC_6H_5^-]}{[HOC_6H_5]}$$
c.
$$K_a=\frac{[H^+][C_6H_5NH_2]}{[C_6H_5NH_3]}$$
Work Step by Step
The general formula for the $K_a$ equilibrium constant for the reaction:
$$HA\Rightarrow H^++A^-$$
is $K_a=\frac{[H^+][A^-]}{[HA]}$.
Therefore, the dissociation reactions are as follows:
$$HCN_{(aq)}\Rightarrow H^+_{(aq)}+CN^-_{(aq)}$$
$$HOC_6H_{5(aq)}\Rightarrow H^+_{(aq)}+OC_6H_{5(aq)}^-$$
$$C_6H_5NH_{3(aq)}\Rightarrow H^+_{(aq)}+C_6H_5NH_{2(aq)}$$
and the dissociation constants are:
$$K_a=\frac{[H^+][CN^-]}{[HCN]}$$
$$K_a=\frac{[H^+][OC_6H_5^-]}{[HOC_6H_5]}$$
$$K_a=\frac{[H^+][C_6H_5NH_2]}{[C_6H_5NH_3]}$$
