Respuesta :

saadhussain514

PbI(ii) ionization in the solution of PBI(ii) into water is:

PbI₂(solution) <==> Pb₂⁺ + 2I⁻

If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.

Therefore,

Ksp=[Pb²⁺][I-]²

Plugging the values:

1.4×10⁻⁸ = x ⋅ (2x)²

1.4×10⁻⁸ = 4x³

x³ = {1.4×10⁻⁸}÷4

x³ = 0.35 x 10⁻⁸

or 

x³ = 3.5 x 10⁻⁹

x = 1.51 x 10⁻³

Hence,

Concentration of iodide ions in the solution:

2x = 3.02 x 10⁻³ 

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