The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is: [Assume: No cyano complex is formed; $$K_{sp}(AgCN) = 2.2 \times 10^{-16}$$ and $$K_a(HCN) = 6.2 \times 10^{-10}$$]

We need to find the solubility of AgCN in a buffer solution of pH = 3. Given: $$K_{sp}(\text{AgCN}) = 2.2 \times 10^{-16}$$ and $$K_a(\text{HCN}) = 6.2 \times 10^{-10}$$.

Let the solubility of AgCN be $$s$$. When AgCN dissolves: $$\text{AgCN} \rightleftharpoons \text{Ag}^+ + \text{CN}^-$$. So $$[\text{Ag}^+] = s$$.

In the acidic buffer (pH = 3), the $$\text{CN}^-$$ ions react with $$\text{H}^+$$: $$\text{CN}^- + \text{H}^+ \rightleftharpoons \text{HCN}$$. Since $$[\text{H}^+] = 10^{-3}\,\text{M}$$ is much larger than $$K_a = 6.2 \times 10^{-10}$$, most of the $$\text{CN}^-$$ is converted to HCN.

The total cyanide concentration is $$[\text{CN}^-] + [\text{HCN}] = s$$. From the acid dissociation: $$K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}$$, so $$[\text{CN}^-] = \frac{K_a \cdot [\text{HCN}]}{[\text{H}^+]}$$. Since $$[\text{H}^+] \gg K_a$$, essentially all cyanide is as HCN, so $$[\text{HCN}] \approx s$$ and $$[\text{CN}^-] = \frac{K_a \cdot s}{[\text{H}^+]} = \frac{6.2 \times 10^{-10} \times s}{10^{-3}} = 6.2 \times 10^{-7} \cdot s$$.

Applying the solubility product: $$K_{sp} = [\text{Ag}^+][\text{CN}^-] = s \times 6.2 \times 10^{-7} \cdot s = 6.2 \times 10^{-7} \cdot s^2$$.

Solving: $$s^2 = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-7}} = \frac{2.2}{6.2} \times 10^{-9} \approx 0.3548 \times 10^{-9} = 3.548 \times 10^{-10}$$.

$$s = \sqrt{3.548 \times 10^{-10}} \approx 1.88 \times 10^{-5} \approx 1.9 \times 10^{-5}$$.

Therefore, the solubility of AgCN is $$1.9 \times 10^{-5}$$, which corresponds to option (1).