Question

The solubility product of a sparingly soluble salt \(A_2X_3\) is \(1.1 × 10^{–23}\). If the specific conductance of the solution is \(3×10^{–5}\) S m^–1, the limiting molar conductivity of the solution is \(x×10^{–3}\) S m² mol^–1. The value of x is _______.

Answer 1

Correct Answer: 3

            \(A_2X_3 ⇋ 2A + 3x\)
                             \(2S\)     \(3S\)
\(K_{sp} = (2s)^2(3s)^3\)
\(= 1.1 × 10^{-23}\)
\(S ≈ 10^{-5}\)
For sparingly soluble salts
\(∧m = ∧°m\)
\(∧m = \frac {k}{S × 10^3}\)

\(=\frac { 3 × 10^{-5}}{10-5  × 10^{-3}}\)
\(= 3 × 10^{-3} Sm^2 mol^{-1}\)

So, the answer is \(3\).