Example: What are the concentrations of each of the ions in a saturated solution of PbBr 2 , given
that the Ksp of PbBr 2 is 2.1 × 10−6? If 5 g of PbBr 2 are dissolved in water to make 1 L of
solution at 25°C, would the solution be saturated, unsaturated, or supersaturated?
Solution: The first step is to write out the dissociation reaction:
Let x be the concentration of Pb2+. Note that this is also the maximum molarity of
PbBr 2 that can dissolve (i.e., its solubility). Since every mole of PbBr 2 that dissolves
yields one mole of Pb2+ and two moles of Br−, [Br−] is two times [Pb2+]; i.e., the
concentration of Br− in the saturated solution at equilibrium = 2x.(x)(2x)^2 = 4x^3 = Ksp = 2.1 × 10−6Solving for x, the concentration of Pb2+ in a saturated solution is 8.07 × 10−3 M and the
concentration of Br− is 2x = 1.61 × 10–2 M.
For the second part of the problem, we convert 5 g of PbBr 2 into moles:1.36 × 10−2 mol of PbBr 2 is dissolved in 1 L of solution, so the concentration of the solution is 1.36 ×
10 −2 M. Since this is higher than the concentration of a saturated solution, this solution would be
supersaturated. Any slight disturbance would cause the salt to precipitate out of solution, reducing
the ion concentration until the solution is just saturated.