Question

At AJAX Chemical, INC. a critical reaction requires the solubility of PbCl2 to be minimal in a critical patented reaction. As one of the staff chemists, it is your job to find the molar solubility of this salt in a solution that is also 0.1180 M in cesium chloride, CsCl. The Ksp of lead(II) chloride is 1.70e-5. The correct answer is 0.00122. Please explain how to get to this question.

Answer 1

Solution: Solubility product (Ksp) is related to Solubility (S) for PbCl₂ as,

Ksp=[Pb²⁺] [Cl⁻]²

In pure water, Ksp = (S) (2S)²=4S³, where S is the solubility of PbCl2.

However, here the concentration of Cl^- ion has been increased by the presence of cesium chloride (CsCl) . So the new concentration of Cl ^- ion will be , [Cl⁻] = (0.1180 + 2S).

hence, Now, Ksp=(S) (0.1180+2S}² = 1.70 x 10^-5 ( given Ksp)

S x [ (0.118)² + 4S² + 2 x 2S x 0.118] =1.70 x 10^-5

On solving, S= 0.00122