Problem: Please answer this question with a brief explanation.The solubility of lead (II) chloride (PbCl2) is 1.6 x 10-2 M. What is the Ksp of PbCl2? A) 4.1 x 10-6 B) 1.6 x 10-5 C) 5.0 X 10-4 D) 3.1 x 10-7 E) 1.6 x 10-2

FREE Expert Solution

Dissociation of PbCl₂:

PbCl₂(s) ⇌  Pb²⁺(aq) + 2Cl^–(aq)

ICE table. 

The K_sp expression then is:

$\boxed{{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\frac{\mathbf{products}}{\cancel{\mathbf{reactants}}}\mathbf{=}\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{[}{\mathbf{Cl}}^{\mathbf{-}}{\mathbf{]}}^{\mathbf{2}}}$