Solve the following

Question:

Two salts $A_{2} X$ and $\mathrm{MX}$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S\left(A_{2} X\right)}{S(M X)}=$ ______________/(Round off to the Nearest Integer).

Solution:

(50)

For $A_{2} X$

$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}_{1}^{3}=4 \times 10^{-12}$

$\mathrm{S}_{1}=10^{-4}$

for $M X$

$\mathrm{MX} \rightarrow \mathrm{M}^{+}+\mathrm{X}^{-}$

$\mathrm{S}_{2} \quad \mathrm{~S}_{2}$

$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}_{2}^{2}=4 \times 10^{-12}$

$\mathrm{S}_{2}=2 \times 10^{-6}$

So $\frac{\mathrm{S}_{\mathrm{A}_{2} \mathrm{X}}}{\mathrm{S}_{\mathrm{MX}}}=\frac{10^{-4}}{2 \times 10^{-6}}=50$

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