Question

The solubility product constant of Ag2CrO4 and AgBr are 1.1×10–12 and 5.0×10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer 1

Let s be the solubility of $Ag_2CrO_4$
Then, $Ag_2CrO_4 ↔ Ag^{2+} + 2CrO_4^-$
$K_{sp} = (2s)^2.s = 4s^3$
$1.1×10^{-12} = 4s^3$
$s = 6.5×10^{-5}M$
Let s ´ be the solubility of $AgBr.$
$AgBr(s) ↔ Ag^+ + Br^-$
$K_{sp} = s'^2 = 5.0×10^{-13}$
∴ $s' = 7.07×10^{-7} M$
Therefore, the ratio of the molarities of their saturated solution is:
$\frac {s}{s'}= \frac {6.5×10^{-5} M}{7.07×10^{-7}M} = 91.9$