Question

\(M_3A_2\) is a sparingly soluble salt of molar mass \(y \text{ g mol}^{-1}\) and solubility \(x \text{ g L}^{-1}\). The ratio of the molar concentration of the anion (\(A^{3-}\)) to the solubility product (\(K_{sp}\)) of the salt is:

• A. \(\frac{1}{54} \frac{y^4}{x^4}\)
• B. \(\frac{y^5}{108x^4}\)
• C. \(108 \frac{x^5}{y^5}\)
• D. \(\frac{1}{108}\frac{y^4}{x^4}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The topic is Ionic Equilibrium, specifically dealing with the solubility product (\(K_{sp}\)).
We need to derive an expression for the ratio of the anion concentration to the \(K_{sp}\) in terms of the salt's solubility in g/L (\(x\)) and its molar mass (\(y\)).
Step 2: Key Formula or Approach:
1. Convert solubility from g/L to molar solubility (\(s\)) in mol/L.
2. Write the dissociation equilibrium for the salt \(M_3A_2\).
3. Express the ion concentrations and \(K_{sp}\) in terms of \(s\).
4. Form the required ratio and substitute the expression for \(s\).
Step 3: Detailed Explanation:
1. Molar Solubility (\(s\)):
Molar solubility \(s (\text{mol/L}) = \frac{\text{Solubility (g/L)}}{\text{Molar Mass (g/mol)}} = \frac{x}{y}\).
2. Dissociation and Ion Concentrations:
The salt dissociates as: \(M_3A_2(s) \rightleftharpoons 3M^{2+}(aq) + 2A^{3-}(aq)\).
In a saturated solution, the concentrations are:
\([M^{2+}] = 3s\)
\([A^{3-}] = 2s\)
3. Solubility Product (\(K_{sp}\)):
\(K_{sp} = [M^{2+}]^3 [A^{3-}]^2 = (3s)^3 (2s)^2 = (27s^3)(4s^2) = 108s^5\).
4. Calculate the Required Ratio:
We need to find \(\frac{[A^{3-}]}{K_{sp}}\).
\[ \frac{[A^{3-}]}{K_{sp}} = \frac{2s}{108s^5} = \frac{1}{54s^4} \]
5. Substitute \(s = x/y\):
\[ \text{Ratio} = \frac{1}{54\left(\frac{x}{y}\right)^4} = \frac{1}{54\frac{x^4}{y^4}} = \frac{1}{54} \frac{y^4}{x^4} \]
Step 4: Final Answer:
The ratio is \(\frac{1}{54} \frac{y^4}{x^4}\).