If solubility of sparingly soluble salt \( M_3A_2(s) \) is \(x\) gm/litre and \(y\) is the molar mass (in gm/mole) of the salt, then determine the value of
\( \dfrac{[A^{3-}]}{K_{sp}} \).
- \( \dfrac{1}{36}\dfrac{y^4}{x^4} \)
- \( \dfrac{1}{54}\dfrac{y^4}{x^4} \)
- \( \dfrac{1}{54}\dfrac{x^4}{y^4} \)
- \( \dfrac{1}{36}\dfrac{x^3}{y^3} \)
The Correct Option is B
Solution and Explanation
For sparingly soluble salts, solubility is first converted into molar solubility. Using the dissociation equation, concentrations of ions are written in terms of solubility and substituted into the solubility product expression. Step 1: Write the dissociation reaction. \[ M_3A_2(s) \rightleftharpoons 3M^{2+}(aq) + 2A^{3-}(aq) \] Step 2: Convert solubility into molarity. Given solubility \(=x\) g/L \[ \text{Molar solubility}=\frac{x}{y} \] \[ [M^{2+}] = 3\frac{x}{y}, \qquad [A^{3-}] = 2\frac{x}{y} \] Step 3: Write the \(K_{sp}\) expression. \[ K_{sp}=[M^{2+}]^3[A^{3-}]^2 \] \[ K_{sp}=\left(3\frac{x}{y}\right)^3\left(2\frac{x}{y}\right)^2 \] \[ K_{sp}=27\frac{x^3}{y^3}\times4\frac{x^2}{y^2} \] \[ K_{sp}=108\frac{x^5}{y^5} \] Step 4: Calculate required ratio. \[ \frac{[A^{3-}]}{K_{sp}} = \frac{2x/y}{108x^5/y^5} \] \[ = \frac{1}{54}\frac{y^4}{x^4} \] \[ \boxed{\frac{1}{54}\frac{y^4}{x^4}} \]
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