Summary
Consider the reaction in which the slightly soluble solid PbI2 dissolves: PbI2(s)
Pb2+(aq) + 2I–(aq)Consider the reaction quotient for this reaction: Qs(PbI2) = [Pb2+][I-]2
Consider the relationship between the solubility of PbI2 in a given solution and the equilibrium concentrations of the ions.
| S(PbI2) = | [Pb2+] | = | [I-] |
| 1 | 2 |
Note for a 1:2 or 2:1 salt the concentration of one of the ions in water is obtained by doubling the solubility.
On substitution into the Qs expression this doubled solubility is then squared.
For a saturated solution of PbI2 in water
s(PbI2) = 6.31 × 10-4 mol L–1
Ks = s(PbI2) × [2s(PbI2)]2
Ks = 1.0 × 10–9
s(PbI2) = 6.31 × 10-4 mol L–1
Ks = s(PbI2) × [2s(PbI2)]2
Ks = 1.0 × 10–9
Note that when one of the ions present in the slightly soluble compound is already a major species in solution, it is assumed that the change in concentration of that ion due to dissolving can be ignored. Also it is good practice to substitute an expression in terms of solubility for the unknown ion (I- below) rather than just calling the unknown x as then desired solubility is then calculated directly.
For a saturated solution of PbI2 in
[Pb2+] = 0.10 + s(PbI2)
[I–] = 2 × s(PbI2)
Ks = 0.10 × [2s(PbI2)]2
s(PbI2) = 5.0 × 10-5
therefore s(PbI2) = 5.0 × 10-5 mol L–1
Note that the s(PbI2) in aqueous Pb(NO3)2 is ten fold less than in water.
0.10 mol L–1 Pb(NO3)2
.[Pb2+] = 0.10 + s(PbI2)
≈ 0.10
[I–] = 2 × s(PbI2)
Ks = 0.10 × [2s(PbI2)]2
= 0.4 × [s(PbI2]2 = 1.00 × 10-9
s(PbI2) = 5.0 × 10-5
therefore s(PbI2) = 5.0 × 10-5 mol L–1
Note that the s(PbI2) in aqueous Pb(NO3)2 is ten fold less than in water.