Ratio of Anion Concentration to Solubility Product for Sparingly Soluble Salt

To find the ratio of the molar concentration of the anion $(\text{A}^{3-})$ to the solubility product ($K_{sp}$) of the salt $\text{M}_3\text{A}_2$, we follow these steps:

1. Calculate the molar solubility ($S$): Given:

   • Solubility of $\text{M}_3\text{A}_2 = x \text{ g L}^{-1}$
   • Molar mass of $\text{M}_3\text{A}_2 = y \text{ g mol}^{-1}$

   The molar solubility $S$ (in $\text{mol L}^{-1}$) is given by: $S = \frac{x}{y} \text{ mol L}^{-1}$

2. Determine the equilibrium concentrations of the ions: The dissociation equation for the sparingly soluble salt $\text{M}_3\text{A}_2$ in water is: $\text{M}_3\text{A}_2(s) \rightleftharpoons 3\text{M}^{2+}(aq) + 2\text{A}^{3-}(aq)$

   From the stoichiometry of the reaction:

   • Molar concentration of cation, $[\text{M}^{2+}] = 3S$
   • Molar concentration of anion, $[\text{A}^{3-}] = 2S$

3. Calculate the solubility product ($K_{sp}$): $K_{sp} = [\text{M}^{2+}]^3 [\text{A}^{3-}]^2$ $K_{sp} = (3S)^3 \cdot (2S)^2 = (27S^3) \cdot (4S^2) = 108S^5$

4. Find the required ratio: We are required to find the ratio of the molar concentration of the anion $[\text{A}^{3-}]$ to the solubility product $K_{sp}$: $\text{Ratio} = \frac{[\text{A}^{3-}]}{K_{sp}} = \frac{2S}{108S^5} = \frac{1}{54S^4}$

   Substituting $S = \frac{x}{y}$ into the expression: $\text{Ratio} = \frac{1}{54\left(\frac{x}{y}\right)^4} = \frac{1}{54} \cdot \frac{y^4}{x^4}$

Thus, the correct option is A.