What is the molar solubility of ${\mathrm{AgNO}}_3$ in a 0.1M ${\mathrm{H}}_2\mathrm{S}$ solution buffered at pH = 2 ($K_1$ and $K_2$ of $H_{2}S$ are ${10}^{–4}$ and ${10}^{–8}$ respectively) ($K_{sp}$ of ${\mathrm{Ag}}_2\mathrm{S}$ = $4\times {10}^{–3}$) (Note: No $A{g}_{2}S$ precipitate should be formed)

pH = $-log\left[H^{+}\right]$

$\left[H^{+}\right]$ = ${10}^{-2}$ =0.01 M

First dissociation of $H_{2}S$, $H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$

$K_1=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$

${10}^{–4}=\frac{[0.01]\left[H{S}^{-}\right]}{[0.1]}$

${10}^{–4}\times [0.1]$ = [0.01] [$H{S}^{-}$]

$\frac{{10}^{–5}}{0.01}$ = [$H{S}^{-}$]

[$H{S}^{-}$] = ${10}^{–3}$M

second dissociation of $H_{2}S$, $H{S}^{-}\rightleftharpoons H^{+}+S^{2-}$

$K_2=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}$

${10}^{–8}=\frac{[0.01]\left[S^{2-}\right]}{[0.001]}$

${10}^{–8}\times [0.001]$ = [0.01] [$S^{2-}$]

$\frac{{10}^{–11}}{0.01}$ = [$S^{2-}$]

[$S^{2-}$] = ${10}^{–9}$M

$A{g}_{2}S\rightleftharpoons 2A{g}^{+}+S^{2-}$

$K_{sp}=\left[A{g}^{+}{]}^2\right[S^{2-}]$

$4\times {10}^{–3}$ = $\left[A{g}^{+}{]}^2\right[{10}^{-9}]$

$\frac{4\times {10}^{–3}}{{10}^{-9}}$ = $[A{g}^{+}{]}^2$

$4\times {10}^6$ =  $[A{g}^{+}{]}^2$

$\sqrt{4\times {10}^6}$ = $\left[A{g}^{+}\right]$

$\left[A{g}^{+}\right]$ = 2000 M

Hence, the answer is 2000 M.