The presence of a common ion reduces the solubility of a sparingly soluble salt. This is a direct consequence of LeChatelier's Principle.
Suppose some lead(II) chloride is dissolved in water and equilibrium is established:
PbCl2(s) → ← Pb2+(aq) + 2 Cl–(aq)
Now, add some sodium chloride:
NaCl(s) → Na+(aq) + Cl–(aq)
The added chloride ion pushes the equilibrium reaction towards reactants, i.e. creating more solid lead(II) chloride and removing some of the lead ion from solution.
This can be an effective way to remove ions from solution.
How much is the solubility of lead(II) chloride changed in the presence of 0.85 M NaCl?
Recall that lead(II) chloride has a solubility of 1.6×10–2 M in pure water.
In the presence of NaCl, we need to redo the calculation, accounting for the common ion.
NaClaq) → Na+(aq) + Cl–(aq)
PbCl2(s) → ← Pb2+(aq) + 2 Cl–(aq)
Ksp = [Pb2+]e[Cl–]e2 = 1.6×10–5
Initial00.85
Change+x+2x
Equilibriumx0.85 + 2x
x will be less than 0.016 so 0.85 + 2x ~ 0.85
1.6×10–5 = [x][0.85]2 = 0.7225x
x = 2.2×10–5
or [Pb2+]e = 2.2×10–5 M
The concentration is reduced by about 1000.
Complex ions are the result of Lewis acid/base reactions. Metal cations act as Lewis acids and many anions and species containing lone pairs can act as Lewis bases.
Cu2+(aq) + 4 NH3(aq) → ← [ H3N—Cu—NH3 ] (aq) 2+ | NH3 | NH3
Lewis acid/base complexes can react with a variety of stoichiometries that are not easily predicted. Further, the reactions are equilibria so the extent of reaction can vary considerably.
In general:
Mn+(aq) + q L(aq) → ← [MLq]n+(aq)
The mass action expression is
Kc = [MLq]n+]e [Mn+]e[L]eq = Kf
Kf is called the formation constant or stability constant for the complex ion.
Fortunately, the Table of Complex Ion Formation Constants also gives the stoichiometry of the complex ion, so we can use this information to establish stoichiometry.
Common Lewis bases that give complex ions are NH3, CN–, OH–, Cl–, Br–, and I–.
Complex ion formation can have a strong effect on solubility.
Calculate the molar solubility of zinc carbonate in pure water and in 1.0 M ammonia at 25 °C.
ZnCO3(s) → ← Zn2+(aq) + CO32–(aq)
Ksp = [Zn2+]e[CO32–]e = 1.4×10–11
Initial00
Change+x+x
Equilibriumxx
1.4×10–11 = [x][x] = x2
x = 3.7×10–6
The molar solubility of ZnCO3 in water is 3.7×10–6 M
When the zinc carbonate is dissolved in ammonia the complex ion [Zn(NH3)4]2+ can form. This complicates the solution to the problem.