Question

The solubility of PbI2 in water is s mol L-1. The concentration-solubility product is

A) s^3
B) 2s^3
C) 4s^3
D) 8s^3

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Answer 1

answer : option (c) 4s³

explanation : The solubility of PbI2 in water is s mol/L. then we have to find concentration- solubility product. or simply say, solubility product.

we know, PbI2 dissociates into Pb²+ and 2I- ions.

i.e., $PbI_2(aq)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)$

so, solubility product , $K_{sp}=[Pb^{2+}][2I^{-}]^2$

given, $[Pb^{2+}]=[I^-]=s$

so, Ksp = (s)(2s)² = 4s³

hence, solubility product of PbI2 is 4s³.

therefore, option (c) is correct choice.

Answer 2

Answer:

(c) 4s^3

Explanation:

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