Which is the correct expression for calculating the solubility (in mol L⁻¹) of lead(II) iodide in a 0.1 mol L⁻¹ solution of NaI at 25ºC? - HSC - SSCE Chemistry - Question 11 - 2024 - Paper 1

To find the solubility of lead(II) iodide (PbI₂) in that solution, we start with the solubility product expression:

$K_{sp} = [Pb^{2+}][I^{-}]^2$

In this case, we have NaI in solution which dissociates completely to provide iodide ions.

For a 0.1 mol L⁻¹ NaI solution, the concentration of iodide ions is:

$[I^{-}] = 0.1$

Thus the expression for Ksp becomes:

$K_{sp} = [Pb^{2+}](0.1)^2$

Solving for the concentration of lead ions gives us:

$[Pb^{2+}] = \frac{K_{sp}}{(0.1)^2}$

Assuming the Ksp for PbI₂ is approximately 9.8 x 10⁻⁹, the expression simplifies to:

$[Pb^{2+}] = \frac{9.8 imes 10^{-9}}{(2 imes 0.1)^2}$

Therefore, the correct option is D.