Precipitation Equilibrium

The solubility of compounds can be affected by the presence of ions in the solution. For example, the solubility of PbCl2

PbCl2(s) < = > Pb+2(aq) + 2Cl-(aq)
will be affected if there are lead or chloride ions already present in the solution before the lead chloride is added. This is a reflection of LeChatlier's Principle- if you add lead or chloride ions to the mixture, you will push the equilibrium to the left in the above equation. This will reduce the solubility of lead chloride. This is known as the common ion effect.

To compute the solubility of a compound in a solution with a common ion, simply work a solubility from Ksp problem as usual, except you should make sure to include the concentration of the common ion(s) as well


Example: What is the solubility of PbCl2 in 0.10 M NaCl? Ksp for PbCl2 is 1.7*10-5.

Solution: Set up the problem as a solubility problem: first, write down the balanced chemical equation and the Ksp expression

PbCl2(s) < = > Pb+2(aq) + 2Cl-(aq)
Ksp = [Pb+2][Cl-]2 = 1.7*10-5
For each mole of PbCl2 that dissolves, 1 mole of Pb+2 and two moles of Cl- are formed. If we define the solubility of PbCl2 in moles/L as s, then [Pb+2] = s at equilibrium. For [Cl-], we form 2s from the equilibrium, but we already have 0.10 M Cl- from the salt NaCl, so we have 2s+ 0.10 M concentration at equilibrium. Thus, our equilibrium expression is
Ksp = [Pb+2][Cl-]2
1.8*10-5 = (s)(2s + .1)2
This is an ugly equation, so we'll try to simplify. If s is small, as we expect, 2s + 0.1 ~ 0.1, so we can try to substitute this in
1.8*10-5 = (s)(0.1)2
s = 1.8*10-3
This is indeed much smaller than 0.1, so the approximation 2s+0.1 ~ 0.1 is valid. Thus, the molar solubility of PbCl2 in 0.1M NaCl is 1.8*10-3 mol/L. (For comparision, it is 1.6*10-2 mol/L in pure water- it's much less soluble in salt water.)


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