The ksp of pbl2 is 7.9*10^-9. What is the expected concentration of lead ions in solution? a)1.3*10^-3 mol/L b)1.6*10-3 mol/L c)none of these d)2.0*10-3 mol/L

Question

The ksp of pbl2 is 7.9*10^-9. What is the expected concentration of lead ions in solution?

a)1.3*10^-3 mol/L

b)1.6*10-3 mol/L

c)none of these

d)2.0*10-3 mol/L

Answer 1

First, let's consider the balanced equation that describes how PbI₂ dissolves, or dissociates, in solution:

PbI₂ (s) -------> Pb²⁺ (aq) + 2 I^- (aq)

Note that according to this equation the concentration of iodide (I^-) is twice that of lead ion (Pb²⁺) due to the stoichiometric coefficients and recalling that they represent mole-to-mole ratios. Therefore if the concentration of lead ion is x, then the concentration of iodide is 2x.

K_sp is the solubility product, which is nothing more than the equilibrium constant for the above equation. Recall that the expression for an equilibrium constant is the concentration of products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. Recall that concentrations/activities of solid substances are omitted in this expression.

So, K_sp = [Pb²⁺][I^-]² = (x)(2x)² = 4x³. Note that x will represent the concentration of lead ion. So simply solve for x...

If 4x³ = 7.9 x 10^-9, then x³ = 2 x 10^-9 (when you divide both sides by 4).

Solving for x by taking the cubed root on both sides gives x = 1.25 x10^-3 M = [Pb²⁺].

So a) is the answer. Hope this helps!!