The molar solubility of PbI2in 0.2M Pb(NO2)2 solution in terms of solubility product, Ksp

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Ksp/0.21.2

Ksp/0.41/2

Ksp/0.81/2

Ksp/0.81/3

answer is C.

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Detailed Solution

Molar solubility PbI2 will get suppressed in presence of Pb(NO2)2 due to commonion effect, and the following equilibria gets established.PbI2(s)⇌PbI2(aq)→Pb+2(aq)+2I-(aq)PbNO22→Pb+2(aq)+2NO20.2+S'-(aq)Ksp = (S' + 0.2) x (2S')2⇒ Ksp = 0.2 x 4(S')2⇒Ksp0.8=S'2⇒S'=Ksp0.812

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