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Chemistry Question on Equilibrium

Solubility product of PbCl2PbCl _{2} at 298 K298 \,K is 1×10−61 \times 10^{-6}. At this temperature solubility of PbCl2PbCl _{2} in mol/Lmol / L is

A

(1×10−6)1/2\left(1 \times 10^{-6}\right)^{1 / 2}

B

(1×10−6)1/3\left(1 \times 10^{-6}\right)^{1 / 3}

C

(0.25×10−6)1/3\left(0.25 \times 10^{-6}\right)^{1 / 3}

D

(2.5×10−6)1/2\left(2.5 \times 10^{-6}\right)^{1 / 2}

Answer

(0.25×10−6)1/3\left(0.25 \times 10^{-6}\right)^{1 / 3}

Explanation

Solution

PbCl2⇌Pb2+S+KSP2sPbCl _{2} \rightleftharpoons \underset{S}{ Pb ^{2+}}+\underset{2 s}{K_{S P}} KSP=(S)(2S)2K_{SP}=(S)(2 S)^{2} 1×10−6=4S31 \times 10^{-6} =4 S^{3} S=(14×10−6)1/3{S} {=\left(\frac{1}{4} \times 10^{-6}\right)^{1 / 3}} =(0.25×10−6)1/3=\left(0.25 \times 10^{-6}\right)^{1 / 3}