Question

The solubility product constant of ${Ag}_2{CrO}_4$ and $AgBr$ is $1.1\times {10}^{-12}$ and $5.0\times {10}^{-13}$ respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer 1

Silver chromate: $A{g}_{2}Cr{O}_4\rightleftharpoons 2A{g}^{+}+Cr{O}_4^{2-}$
$\left[A{g}^{+}\right]=2s_1,Cr{O}_4^{2-}=s_1$
$K_{sp}=\left(2s_1{)}^2\right(s_1)=4s^3=1.1\times {10}^{-12}$
$s_1=6.5\times {10}^{-5}$.....(1)
Silver bromide: $AgBr\rightleftharpoons A{g}^{+}+B{r}^{-}$
$\left[A{g}^{+}\right]=\left[B{r}^{-}\right]=s_2$
$K_{sp}=\left(s_2\right)\times \left(s_2\right)=s_2^2=5.0\times {10}^{-13}$
$s_2=7.07\times {10}^{-7}$......(2)
Divide equation (1) by equation (2) to obtain the ratio of the molarities of saturated solutions:
$\frac{s_1}{s_2}=\frac{6.50\times {10}^{-5}}{7.07\times {10}^{-7}}=91.9$