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

$A_2X_3 ⇋ 2A + 3x$
$2S$ $3S$
$K_{sp} = (2s)^2(3s)^3$
$= 1.1 × 10^{-23}$
$S ≈ 10^{-5}$
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 $3$.