# 7.68     The solubility product constant of Ag2CrO4 and AgBr is 1.1 × 10-12 and 5.0 × 10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.

silver chromate ($Ag_2CrO_{4}$)
Ionization of silver chromate

$Ag_2CrO_{4}\rightleftharpoons 2Ag^++CrO_{4}^{2-}$
Let "$s$" be the solubility of $Ag_2CrO_{4}$
$[Ag^+] = 2s$
$[CrO_{4}^{2-}] = s$
$K_{sp}$ of $Ag_2CrO_{4}$ = $1.1\times 10^{-12}$

$\\\Rightarrow 1.1\times 10^{-12} = (2s)^2.s\\ =1.1\times 10^{-12} =2s^3$

$s = \sqrt[3]{\frac{1.1\times 10^{-12}}{4}}$
$=0.65 \times 10^{-4}$

Ionization of Silver bromide ($AgBr$)
$AgBr\rightleftharpoons Ag^++Br^-$

$K_{sp}$ of $AgBr$ = $5\times 10^{-13}$

$[Ag^+] = s'$
$[Br^-] = s'$

$\\\Rightarrow 5\times 10^{-13} = s'.s'\\ =5\times 10^{-13} =s'^2$

$s' = \sqrt{\frac{5\times 10^{-13}}{1}}=\sqrt{0.5\times 10^{-12}}$
$=7.07 \times 10^{-7}$

Now, the ratio of solubilities
$\Rightarrow \frac{s}{s'}=\frac{6.5\times 10^{-5}}{1.1\times 10^{-12}}$
$= 9.91$

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