Question

The conductivity of a saturated solution of $PbC{l}_2$ is $4\times {10}^{-4}S{m}^{-1}.$ However, the equivalent conductance of this solution is $1.25\times {10}^{-5}S{m}^{2}e{q}^{-1}$. The $K_{sp}$ value of $PbC{l}_2$ will be:

• A. $2.56\times {10}^{-4}$
• B. $1.64\times {10}^{-5}$
• C. $1.24\times {10}^{-6}$
• D. $2.36\times {10}^{-8}$

Answer 1

The correct option is B $1.64\times {10}^{-5}$

${\lambda }_m^0=\frac{K\times {10}^{-3}}{S}$
$2(1.25\times {10}^{-5})=\frac{4\times {10}^{-4}\times {10}^{-3}}{S}$
$\therefore S=1.6\times {10}^{-2}mol/L.$ $\therefore K_{sp}=4S^3\approx 1.64\times {10}^{-5}$