Two salts $A_2 X $ and MX have the same value of solubility product of $4.0 \times 10^ {-12} $. The ratio of their molar solubilities i.e. $\frac{{S({A_2}X)}}{{S(MX)}} =$ _ _ _ _ . (Round off to the nearest integer)

Solution

We have, $4.0 \times {10^{ – 12}} = 4[S{({A_2}X)]^3}$

$ \Rightarrow S({A_2}X) = {10^{ – 4}}$

Also, $4.0 \times {10^{ – 12}} = [S{(MX)]^2}$

$ \Rightarrow S(MX) = 2.0 \times {10^{ – 6}}$

The ratio, $\frac{{S({A_2}X)}}{{S(MX)}} = \frac{{{{10}^{ – 4}}}}{{2.0 \times {{10}^{ – 6}}}} = 50$