Question

If at infinite dilution, molar conductivities of BaCl\(_2\), HCl and H\(_2\)SO\(_4\) are \( x_1 \), \( x_2 \), \( x_3 \) S cm\(^2\)mol\(^{-1}\) respectively, find solubility product of BaSO\(_4\). Given \( K_{\text{BaSO}_4} \) (Specific conductance) = \( x \) S cm\(^{-1}\)

• A. \(\left( \frac{x}{x_1 + x_3 - 2x_2} \right)^2 \times 10^6 \)
• B. \(\left( \frac{x_1 + x_3 - 2x_2}{x} \right)^2 \times 10^6 \)
• C. \(\left( \frac{x}{x_1 + x_3 - x_2} \right)^2 \times 10^3 \)
• D. \(\left( \frac{x_1 + x_3 - x_2}{x} \right)^2 \times 10^6 \)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires finding the solubility product (\(K_{sp}\)) for \(BaSO_4\). This involves calculating its molar conductivity at infinite dilution using Kohlrausch's Law and then relating it to its solubility.
Step 2: Key Formula or Approach:
1. Kohlrausch's Law: \(\Lambda_m^\circ (BaSO_4) = \lambda^\circ (Ba^{2+}) + \lambda^\circ (SO_4^{2-})\).
2. Molar Conductivity: \(\Lambda_m = \frac{\kappa \times 1000}{S}\) (where \(S\) is solubility in mol/L).
3. Solubility Product: \(K_{sp} = S^2\) (for a 1:1 salt like \(BaSO_4\)).
Step 3: Detailed Explanation:
Using Kohlrausch's Law to find \(\Lambda_m^\circ (BaSO_4)\):
\[ \Lambda_m^\circ (BaSO_4) = \Lambda_m^\circ (BaCl_2) + \Lambda_m^\circ (H_2SO_4) - 2\Lambda_m^\circ (HCl) \]
\[ \Lambda_m^\circ (BaSO_4) = x_1 + x_3 - 2x_2 \]
Since \(BaSO_4\) is a sparingly soluble salt, its molar conductivity at saturation is approximately equal to its molar conductivity at infinite dilution (\(\Lambda_m \approx \Lambda_m^\circ\)):
\[ \Lambda_m^\circ = \frac{\kappa \times 1000}{S} \]
\[ x_1 + x_3 - 2x_2 = \frac{x \times 1000}{S} \]
Rearranging for solubility \(S\):
\[ S = \frac{x \times 10^3}{x_1 + x_3 - 2x_2} \]
Now, calculate \(K_{sp}\):
\[ K_{sp} = S^2 = \left( \frac{x \times 10^3}{x_1 + x_3 - 2x_2} \right)^2 \]
\[ K_{sp} = \left( \frac{x}{x_1 + x_3 - 2x_2} \right)^2 \times 10^6 \]
Step 4: Final Answer:
The solubility product is \( \left( \frac{x}{x_1 + x_3 - 2x_2} \right)^2 \times 10^6 \).