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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.) |