Question

A₂X and MX have the same solubility product$4.0\times {10}^{-12}$. What will be the ratio of their molar solubilities$\frac{\mathrm{S}\left({\mathrm{A}}_2\mathrm{X}\right)}{\mathrm{S}\left(\mathrm{MX}\right)}$? (Round off it to the nearest integer)

Answer 1

Step 1:

The salt ${\mathrm{A}}_2\mathrm{X}$ breaks into.

• $$\begin{gathered} {\mathrm{A}}_2\mathrm{X}\rightleftharpoons 2{\mathrm{A}}^{+}+\mathrm{X} \\ \mathrm{s}2\mathrm{s}\mathrm{s} \end{gathered}$$
• ${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{A}}^{+}\right]}^2{\left[\mathrm{X}\right]}^{2-}$
• Ksp is defined as solubility product constant, it is directly proportional to solubility.
• ${\mathrm{K}}_{\mathrm{sp}}=2{\mathrm{s}}^2\times \mathrm{s}=4{\mathrm{s}}^3$-(i)

Step 2:

The salt $\mathrm{MX}$ breaks into.

• $$\begin{gathered} \mathrm{MX}\rightleftharpoons {\mathrm{M}}^{+}+{\mathrm{X}}^{-} \\ \mathrm{s}\text{'}\mathrm{s}\text{'}\mathrm{s}\text{'} \end{gathered}$$
• ${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{M}}^{+}\right]\left[{\mathrm{X}}^{-}\right]$
• $K_{sp}=\left[s\text{'}\right]\left[s\text{'}\right]={\left[s\text{'}\right]}^2$
• $K_{sp}=s{\text{'}}^2$
• $4\times {10}^{-12}=s^{\text{'}2}$-(ii)
• $2\times {10}^{-6}=s\text{'}$
• Equating (i) and (ii)
• $4\times {10}^{-12}=4s^3$
• $\mathrm{s}={10}^{-4}$

Step 3:

• The ratio of $\frac{\mathrm{S}\left({\mathrm{A}}_2\mathrm{X}\right)}{\mathrm{S}\left(\mathrm{MX}\right)}$$\frac{s}{s\text{'}}=\frac{{10}^{-4}}{2\times {10}^{-6}}$=$50$

Final answer: The ratio of their molar solubilities is $50$.