We have developed relationships showing how solubility is related to change in concentration of the ions on dissolving. These relationships are true irrespective of the solution in which the substance is dissolved. However it is important to realise that solubility is directly related to equilibrium concentration of only those ions present in solution exclusively due to the dissolving reaction.
Compare dissolving PbCl2 in water with dissolving PbCl2 in 0.1 mol L-1 NaCl.
In any solution:
PbCl2(s) 
Pb2+(aq) + 2Cl–(aq)
Ks = [Pb2+][Cl–]2 = 1.6 × 10–5
Compare dissolving PbCl2 in water with dissolving PbCl2 in 0.1 mol L-1 NaCl.
In any solution:
| s(PbCl2) = | change in [Pb2+] | = | change in [Cl–] |
| 1 | 2 |
Pb2+(aq) + 2Cl–(aq)Ks = [Pb2+][Cl–]2 = 1.6 × 10–5
| In water | [Pb2+] | [Cl–] |
| before PbCl2 dissolves | 0 | 0 |
| change on dissolving | +s | +2s |
| equilibrium | s | 2s |
| In 0.1 mol L-1 NaCl | [Pb2+] | [Cl–] |
| before PbCl2 dissolves | 0 | 0.1 |
| change on dissolving | +s | +2s |
| equilibrium | s | 0.1 + 2s |
| In 0.1 mol L-1 Pb(NO3)2 | [Pb2+] | [Cl–] |
| before PbCl2 dissolves | 0.1 | 0 |
| change on dissolving | +s | +2s |
| equilibrium | 0.1 + s | 2s |
In water:
The initial concentrations of Pb2+ and Cl– are zero,and the equilibrium concentrations of both ions are related directly to solubility through the stoicheiometry of the dissolving reaction.
Substituting the expressions for the ion concentrations into the Ks expression shows that s(PbCl2) in water is 0.016.
In 0.1 mol L-1 NaCl:
Pb2+ arises exclusively from the dissolving reaction and is related to solubility as shown above. At equilibrium the [Cl–] will be higher than in water and the [Pb2+] will be lower than in water because Ks does not depend on the composition of the solution.
Because [Pb2+] is lower than in water [Cl–] at equilibrium in this solution can be approximated by 0.10. s(PbCl2) calculated using this approximation (0.0016) shows that the approximation is valid.
In 0.1 mol L-1 Pb(NO3)2:
At equilibrium [Pb2+] will be higher than in water and the [Cl–] will be lower than in water.
The solubility will be equal to one half of the [Cl–] as shown above.
The initial concentrations of Pb2+ and Cl– are zero,and the equilibrium concentrations of both ions are related directly to solubility through the stoicheiometry of the dissolving reaction.
Substituting the expressions for the ion concentrations into the Ks expression shows that s(PbCl2) in water is 0.016.
In 0.1 mol L-1 NaCl:
Pb2+ arises exclusively from the dissolving reaction and is related to solubility as shown above. At equilibrium the [Cl–] will be higher than in water and the [Pb2+] will be lower than in water because Ks does not depend on the composition of the solution.
Because [Pb2+] is lower than in water [Cl–] at equilibrium in this solution can be approximated by 0.10. s(PbCl2) calculated using this approximation (0.0016) shows that the approximation is valid.
In 0.1 mol L-1 Pb(NO3)2:
At equilibrium [Pb2+] will be higher than in water and the [Cl–] will be lower than in water.
The solubility will be equal to one half of the [Cl–] as shown above.