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Chemistry Question on Buffer Solutions

Solubility product (Ksp)(K_{sp}) of saturated PbCl2PbCl_2 in water is 1.8×10−4 mol3 dm−91.8 \times 10^{-4} \, mol^3 \, dm^{-9}. What is the concentration of Pb2+Pb^{2+} in the solution?

A

(0.45×10−4)1/3 mol dm−3(0.45 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}

B

(1.8×10−4)1/3 mol dm−3(1.8 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}

C

(0.9×10−4)1/3 mol dm−3(0.9 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}

D

(2.0×10−4)1/3 mol dm−3(2.0 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}

Answer

(0.45×10−4)1/3 mol dm−3(0.45 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}

Explanation

Solution

For the reaction of the AB2A B_{2}

i.e. (PbCl2)\left( PbCl _{2}\right)
KSP=4S3K_{SP} =4 S^{3}

or, S=[KSP4]1/3S =\left[\frac{K_{S P}}{4}\right]^{1 / 3}

Given, KSP=1.8×10−4 mol3 dm−9K_{SP}=1.8 \times 10^{-4}\, mol ^{3} \,dm ^{-9}
∴\therefore Solubility of Pb+2Pb ^{+2} ions will be

∴S=[1.8×10−44]13\therefore S=\left[\frac{1.8 \times 10^{-4}}{4}\right]^{\frac{1}{3}}
=[0.45×10−4]13mol. dm−3=\left[0.45 \times 10^{-4}\right]^{\frac{1}{3}} mol.\, dm ^{-3}