pH of a saturated solution of silver salt of monobasic acid HA is found to be 9. Now Find the ${\mathrm{K}}_{\mathrm{sp}}$ of sparingly soluble salt Ag A(s). Given ${\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HA}\right)={10}^{-10}.$

$$\begin{gathered} \mathrm{Ag}\left(1\right)\rightleftharpoons {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{A}}^{-}\left(\mathrm{aq}\right) \\ \mathrm{x} \mathrm{x} - \mathrm{y} \end{gathered}$$

$$\begin{gathered} \mathrm{Ag}\left(1\right) + {\mathrm{H}}_2\mathrm{O}\left(\mathcal{l}\right) \rightleftharpoons \mathrm{HA} \left(\mathrm{aq}\right) + {\mathrm{OH}}^{-} \left(\mathrm{aq}\right) \\ \mathrm{x}-\mathrm{y} \mathrm{y} \mathrm{y} \end{gathered}$$

Now, ${\mathrm{K}}_{\mathrm{sp}}=\mathrm{x}(\mathrm{x}-\mathrm{y})$

$x-y={10}^{-6}$

$x={10}^{-5}+{10}^{-6}$

${\mathrm{K}}_{\mathrm{sp}}=1.1\times {10}^{-5}\times {10}^{-6}$

${\mathrm{K}}_{\mathrm{sp}}=1.1\times {10}^{-11}$