The K _ sp of Ag _ 2 CrO _ 4 is 1 . 1 × 1 0 ^ - 1 2 at 298K . The...

$\begin{cases}{Ag}_{2}{CrO}_{4}\left(s\right) & ⇌ & 2{Ag}^{+}\left(aq\right) & + & {CrO}_{4}^{2 -} \\ \left(2\text{S}\right)\text{molar} & \left(\text{S}\right)\text{molar}\end{cases}$
$\begin{cases}{AgNO}_{3}\left(aq\right) & ⟶ & {Ag}^{+}\left(aq\right) & + & {NO}_{3}^{-}\left(aq\right) \\ \left(0\text{.}1\right)\text{molar} & \left(0\text{.}1\right)\text{molar}\end{cases}$
\left[{Ag}^{+}\right] = \left(2S + 0 . 1\right)molar$<br />\left[{CrO}_{4}^{2 -}\right] = \left(S\right)\text{molar}$
Here, $S$ is the solubility of ${Ag}_{2}{CrO}_{4}$ in aqueous~ ${AgNO}_{3}$.

$⇒{\left[{Ag}^{+}\right]}^{2 } \left[{CrO}_{4}^{2 -}\right] = {\left(2 \text{S} + 0 \text{.} 1\right)}^{2} \left(\text{S}\right) = 1 \text{.} 1 \times 1 {0}^{- 1 2}$
$\text{S} ≪ 0 \text{.} 1$
$⇒\left(0 . 1\right)2\left(S\right) = 1 . 1 \times {10}^{- 12} = {10}^{- 2} \times SS = 1 . 1 \times {10}^{- 10}M$