We kept adding pbi2 into water until no more will dissolve. what is the concentration of lead (ii) iodide in the solution

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,

2x = 3.02 x 10⁻³