Solution : At equilibrium we can write equation as: PbI2 (s) < = > Pb2+(aq) + 2 I-(aq) but we know for equilibrium Ksp = [Pb2+][I-]^2 this condition should be satisfied let we suppose x = PbI2 (mol/L) which gets dissolved . It will result in => x mol/L Pb2+ and 2x mol/L I- I- is already in the solution corresponding yo=0.025 M At equilibrium: we can write: [Pb2+]= x and [I-]= 2x + 0.025 Solving: 7.9 x 10^-9 = (x)(2x + 0.025)^2 =>x =1.3 x 10^-5 M which is molar solubility Please rate.

1. Solution
: At equilibrium we can write equation as: PbI2 (s) < = > Pb2+(aq) + 2 I-(aq) but we know for
equilibrium Ksp = [Pb2+][I-]^2 this condition should be satisfied let we suppose x = PbI2
(mol/L) which gets dissolved . It will result in => x mol/L Pb2+ and 2x mol/L I- I- is already in
the solution corresponding yo=0.025 M At equilibrium: we can write: [Pb2+]= x and [I-]= 2x +
0.025 Solving: 7.9 x 10^-9 = (x)(2x + 0.025)^2 =>x =1.3 x 10^-5 M which is molar solubility
Please rate