Chapter 16: Aqueous Ionic Equilibria III
Solubility equilibria
Chem 102
Dr. Curtis
Solubility equilibria
• Solubility: amount of a solid (usually an ionic compound
- salt) that dissolves in solution
• molar solubility uses mol/L
• Equilibrium between the solid and the solute
• Solubility product: Ksp describes the equilibrium
CaF2 (s) )* Ca2+ (aq) + 2 F (aq)
Ksp = [Ca2+ ][F ]2
2
Solubility
• Even “insoluble” compounds establish a dynamic
equilibrium with the ions in solution
3
Calculating molar solubility
• Set up an ICE chart as usual
• Example: Calculate the molar solubility of AgCl(s) in a
saturated solution
I
C
E
AgCl(s) )* Ag+ (aq) + Cl (aq)
0
0
+S
+S
S
S
Ksp = [Ag + ][Cl ] = S 2 = 1.77 ⇥ 10
S = 1.33 ⇥ 10
5
10
M
4
The common ion effect
• If one of the ions from a solid is already in the solution, it
will drive the equilibrium left, reducing the solubility
• Example: What is the solubility of AgCl in a 0.10 M
solution of NaCl?
Equilibrium shifts left
AgCl(s) )* Ag+ (aq) + Cl (aq)
I
0
0.10
C
+S
+S
E
S
0.10 + S
Ksp = [Ag + ][Cl ] = S(0.10 + S) ⇡ 0.10S
S = 1.77 ⇥ 10
9
M
decreased solubility compared to previous example
5
Effect of pH on solubility
• Solubility of Mg(OH)2 would decrease in a base,
• but increase in an acid
6
Precipitation
• Saturated solution: If concentration of ions in solution is
equal to the concentration expected from Ksp
• Q = Ksp
• An equilibrium should be established between the solid and
the ions and the solid should precipitate from the solution
• If Q < Ksp, the solution is unsaturated, and more ions could
be added to the solution
• Sometimes, the ion concentration is greater than expected
from Ksp, in this case the solution is supersaturated, and a
precipitate would be expected to form
7
Precipitation summary
• Q < Ksp
• unsaturated, more solute could go into solution
• Q = Ksp
• saturated, at equilibrium with the solid
• Q > Ksp
• supersaturated, a precipitate should form
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Supersaturation
• Note: Sometimes a solution can become
supersaturated and a precipitate will not form because
there is a kinetic barrier to precipitate forming
• A seed crystal or a
scratch on the side of
the beaker can cause
precipitation to occur
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Precipitation reactions
• Precipitation can be used to remove ions from
solution
• Example:
• Na2CrO4 (aq) = 2 Na+(aq) + CrO42-(aq)
• AgNO3(aq) = Ag+(aq) + NO3-(aq)
• If Q > K,
• Ag2CrO4(s) ⇌ 2 Ag+(aq) + CrO42-(aq)
• then an equilibrium is established
Q = [Ag + ]2 [CrO42 ]
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Example
• A solution containing potassium bromide is mixed with one
containing lead (II) acetate to form a solution that is 0.013 M
KBr and 0.0035 M in Pb(C2H3O2)2. Does a precipitate form in
the mixed solution? If so, identify the precipitate.
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Selective precipitation
• If a solution contains more than one ion, you might be
able to remove one by selective precipitation
• Add a reagent that precipitates one ion, but not the
others
• Example: Sea water contains both Mg2+ and Ca2+.
Could they be separated by adding KOH?
• Ca(OH)2 Ksp = 4.68 x 10-6
• Mg(OH)2 Ksp = 2.06 x 10-13
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Qualitative Analysis
• Non-quantitative determination of the ions in an
unknown solution
• Add regents to see if a precipitate forms, then determine
what the precipitate could be
13
14
15
Complex ion equilibria
• Transition metals can generally act as Lewis acids to
form a “complex ion”
• Central metal ion surrounded by a neutral molecule or
ion that acts as a Lewis base
• Example:
Ag+ (aq) + 2 H2 O(`) )* Ag(H2 O)2 + (aq)
• Even though we write the ion as Ag+, it is understood
that it might be hydrated in aqueous solution
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Complex ion equilibria
• The water can be displaced by something else in the
solution
• Example Ag+ (aq) + 2 NH3 (aq) )* Ag(NH3 )2 + (aq)
really Ag(H2 O)2 + (aq)
• This complex has a “formation constant”,
[Ag(N H3 )+
7
2]
Kf =
=
1.7
⇥
10
[Ag + ][N H3 ]2
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Kf values are usually large
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Effect of complex ions on solubility
• Formation of complex ions can increase the solubility of
a salt that normally is not very soluble
AgCl(s) )* Ag+ (aq) + Cl (aq) Ksp = 1.77 ⇥ 10
10
Ag+ (aq) + NH3 (aq) )* Ag(NH3 )2 + (aq) Kf = 1.7 ⇥ 107
AgCl(s) + 2 NH3 (aq) )* Ag(NH3 )2 + (aq) + Cl (aq)
K = Ksp ⇥ Kf = 3.0 ⇥ 10
3
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Tips for this section
• Be able to write a solubility expression for Ksp
• Write the balanced reactions
• Calculate molar solubility (or mass solubility) from a
given Ksp
• Calculate Ksp from solubility
• Common ion effect
• Identify a potential precipitate in a reaction
• Identify whether or not a precipitate forms
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