Q. The solubility of $PbCl_{2}\ is$
A) $\sqrt{K_{sp}}$
B) $\left( K_{sp} \right)^{\frac{1}{3}}$
C) $\left( \frac{K_{sp}}{4} \right)^{\frac{1}{3}}$
D) $\left( 8K_{sp} \right)^{\frac{1}{2}}$
Explanation
(c)
Let the solubility of $PbCl_{2} = x$ mol/L
$PbCl_{2} \rightleftharpoons {Pb}^{2 +} + 2I^{-}$
$x$ mol/L $x$ $2x$
$K_{sp} = \left\lbrack {Pb}^{2 +} \right\rbrack\left\lbrack I^{-} \right\rbrack^{2}$
or $= (x) \times (2x)^{2}$
$\$or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4x^{3}$
$\therefore$ $x = \sqrt[3]{\frac{K_{sp}}{4}} = \left( \frac{K_{sp}}{4} \right)^{\frac{1}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.