The solubility product constants of $Ag_{2}CrO_{4}$ and $AgBr$ are $1.1 \times 10^{-12}$ and $5.0 \times 10^{-13}$ respectively. Calculate the ratio of the molarities of their saturated solutions.

For $Ag_{2}CrO_{4}$:
$Ag_{2}CrO_{4} \longleftrightarrow 2Ag^{+} + CrO_{4}^{2-}$
$K_{sp} = (2s)^{2} \cdot s = 4s^{3} = 1.1 \times 10^{-12}$
$s^{3} = 0.275 \times 10^{-12} = 275 \times 10^{-15}$
$s = (275)^{1/3} \times 10^{-5} \approx 6.5 \times 10^{-5} \, M$
For $AgBr$:
$AgBr \longleftrightarrow Ag^{+} + Br^{-}$
$K_{sp} = (s^{\prime})^{2} = 5.0 \times 10^{-13} = 50 \times 10^{-14}$
$s^{\prime} = \sqrt{50} \times 10^{-7} \approx 7.07 \times 10^{-7} \, M$
Ratio of molarities:
$\frac{s}{s^{\prime}} = \frac{6.5 \times 10^{-5}}{7.07 \times 10^{-7}} \approx 91.9$