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Chemistry Question on Equilibrium

Solubility product of PbCl2{PbCl_2} at 298 K{298\, K} is 1×10−6{1 \times 10^{-6}}. At this temperature, solubility of PbCl2 (in mol/L){PbCl_2\, (in\, mol/L)} is

A

(1×10−6)1/2(1 \times 10^{-6} )^{1/2}

B

(1×10−6)1/3(1 \times 10^{-6} )^{1/3}

C

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

D

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

Answer

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

Explanation

Solution

PbCl2<=>Pb2++2Cl−{PbCl_2 <=> Pb^{2+} + 2Cl^-}
Ksp=[Pb2+][Cl−]2=(S)(2S)2=4S3K_{sp} = {[Pb^{2+} ][Cl^-]^2 } = (S)(2S)^2 = 4S^3
1×10−6=4S31 \times 10^{-6} = 4S^3
S=(frac1×10−64)1/3=(0.25×10−6)1/3S = \left( frac{1 \times 10^{-6}}{4} \right)^{1/3} = (0.25 \times 10^{-6} )^{1/3}