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:
We need to find the ratio \(\frac{[A^{3-}]}{K_{sp}}\). First, we must convert the solubility from \(\text{g/L}\) to \(\text{mol/L}\) and then express concentrations in terms of this molar solubility.
Step 2: Key Formula or Approach:
1. Molar solubility \(s = \frac{\text{Solubility in g/L}}{\text{Molar mass}} = \frac{x}{y} \text{ mol/L}\).
2. Dissociation reaction: \(M_3A_2 (s) \rightleftharpoons 3M^{2+} (aq) + 2A^{3-} (aq)\).
3. Concentrations: \([M^{2+}] = 3s\) and \([A^{3-}] = 2s\).
4. Solubility product: \(K_{sp} = [M^{2+}]^3 [A^{3-}]^2 = (3s)^3 \times (2s)^2 = 108s^5\).
Step 3: Detailed Explanation:
The required ratio is \(\frac{[A^{3-}]}{K_{sp}}\):
\[ \text{Ratio} = \frac{2s}{108s^5} = \frac{1}{54s^4} \]
Now substitute \(s = \frac{x}{y}\) into the expression:
\[ \text{Ratio} = \frac{1}{54 \times (\frac{x}{y})^4} = \frac{1}{54 \times \frac{x^4}{y^4}} = \frac{1}{54} \frac{y^4}{x^4} \]
This matches Option (A).
Step 4: Final Answer:
The ratio is \(\frac{1}{54} \frac{y^4}{x^4}\).