Question

Calculate the solubility of $AgCN$ in a buffer solution of $pH$$=3.0$. $K_{sp}\left(AgCN\right)=$ $1.2\times {10}^{-16},K_a\left(HCN\right)=4.8\times {10}^{-10}$. There is no $C{N}^{-}$ or $A{g}^{+}$ ion in the buffer previously.

Answer 1

$AgCN⟶A{g}^{+}+C{N}^{-}...............Ksp$

$H^{+}+C{N}^{-}⟶HCN........................\frac{1}{K_a}$

$AgCN+H^{+}⟶A{g}^{+}+HCN$ $K_0=\left(Ksp\right)\left(\frac{1}{K_a}\right)$

$K_0=\frac{(6\times {10}^{-17})}{4.8\times {10}^{-10}}{K}_0=1.22\times {10}^{-7}{K}_0=\frac{\left[{Ag}^{+}\right]\left[HCN\right]}{\left[H^{+}\right]}$

$\left[H^{+}\right]⟹-log\left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-3}=0.0010$ (solution buffered to $3$)

For every $x-$moles of $AgCN$ which dissolves in a liter of buffer, $X$ moles of $HCN$ are in solution.

$1.22\times {10}^{-7}=\frac{x^2}{0.001}$

$x=1.10\times {10}^{-5}M$