JEE Main Chemistry question

Chemistry Equilibrium JEE Main

If the solubility product of lead iodide (PbI₂) is 3.2×10⁻⁸, its solubility will be:

(A)2×10⁻³ MCorrect
(B)4×10⁻⁴ M
(C)1.6×10⁻⁵ M
(D)1.8×10⁻⁵ M

Answer: (A)

Solution

4s³=3.2×10⁻⁸ → s³=8×10⁻⁹ → s=2×10⁻³ M.

Question ID VK-136509