Chapter 15
Solubility Equilibria
AP Chemistry
The Solubility Product Constant, Ksp
• Many important ionic compounds are only slightly
soluble in water and equations are written to
represent the equilibrium between the compound
and the ions present in a saturated aqueous
solution.
• The solubility product constant, Ksp, is the
product of the concentrations of the ions involved
in a solubility equilibrium, each raised to a power
equal to the stoichiometric coefficient of that ion
in the chemical equation for the equilibrium.
The Solubility Equilibrium Equation And Ksp
CaF2 (s)  Ca2+ (aq) + 2 F- (aq)
Ksp = [Ca2+][F-]2
Ksp = 5.3x10-9
As2S3 (s)  2 As3+ (aq) + 3 S2- (aq)
Ksp = [As3+]2[S2-]3
• * Remember, solids are not in equilibrium
expressions!
Some Values For Solubility Product
Constants (Ksp) At 25 oC
Ksp And Molar Solubility
• The solubility product constant (Ksp) is related to
the solubility of an ionic solute, but Ksp and
• Molar Solubility - the molarity of a solute in a
saturated aqueous solution
• Ksp and Molar solubility are not the same thing.
• Calculating solubility equilibria fall into two
categories:
– determining a value of Ksp from experimental data
– calculating equilibrium concentrations when Ksp is
known.
Calculating Ksp From Molar Solubility
It is found that 1.2x10-3 mol of lead (II) iodide, PbI2,
dissolves in 1.0 L of aqueous solution at 25 oC.
What is the Ksp at this temperature?
• PbI2 
Pb2+ + 2I-
• Ksp = [Pb2+][I-]2
• Ksp = [1.2 x 10-3] [2(1.2 x 10-3)]2
• Ksp = 6.9 x 10-9
• 2 iodide ions
form, so you
must multiply
molarity by 2!
Calculating Molar Solubility From Ksp
Calculate the molar solubility of silver chromate,
Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.
Ag2CrO4(s)  2Ag+ + CrO42At equilibrium
2x
x
Ksp = [Ag+]2[CrO4-2]
1.1 x 10-12 = (x)(2x)2
1.1 x 10-12 = 4x3
x = 6.5 x 10-5 M = [Ag2CrO4] = [CrO42-]
[Ag+] = 2(6.5 x 10-5)
The Common Ion Effect In Solubility Equilibria
• The common ion effect also affects solubility
equilibria.
• Le Châtelier’s principle is followed for the
shift in concentration of products and
reactants upon addition of either more
products or more reactants to a solution.
• The solubility of a slightly soluble ionic
compound is lowered when a second solute
that furnishes a common ion is added to the
solution.
Solubility Equilibrium Calculation
-The Common Ion Effect
What is the solubility of Ag2CrO4 in 0.10 M K2CrO4?
Ksp = 1.1x10-12 for Ag2CrO4.
Ag2CrO4  2Ag+ + CrO42-
Comparison of solubility of Ag2CrO4
In pure water:
6.5 x 10-5 M
In 0.10 M K2CrO4:
1.7 x 10-6 M
The common ion effect!!
Adding more CrO42- ions shifts the equilibrium back
to the reactants, which is solid Ag2CrO4
Common Ion Effect Example:
What will the molar solubility of CaF2 (Ksp = 4.0 x 10-11)
in a 0.025M NaF solution
CaF2  Ca2+ + 2F-
and Ksp = [Ca2+][F-]2
Before it dissolves
[Ca2+] = 0
[F-] = 0.025
From NaF
After it dissolves
[Ca2+] = x
4.0 x
10-11
[F-] = 0.025 + 2x
= x (0.025 –
4.0 x 10-11 = x(0.025)2
Prentice-Hall ©2002
From CaF2
Assume x is very small
due to small Ksp
2x)2
x = 6.4 x 10-8 = molar
solubility
Chapter Sixteen
Slide 10 of 32
Determining Whether Precipitation Occurs
• Qsp is the ion product reaction quotient and is
based on initial conditions of the reaction.
• Qsp can then be compared to Ksp.
• To predict if a precipitation occurs:
- Precipitation should occur if Qsp > Ksp.
- Precipitation cannot occur if Qsp < Ksp.
- A solution is just saturated if Qsp = Ksp.
Sometimes the concentrations of the ions
are not high enough to produce a
precipitate!
Determining Whether Precipitation Occurs
– An Example
The concentration of calcium ion in blood plasma is
0.0025 M. If the concentration of oxalate ion is
1.0x10-7 M, do you expect calcium oxalate to
precipitate? Ksp = 2.3x10-9.
Three steps:
(1) Determine the initial concentrations of ions.
(2) Evaluate the reaction quotient Qip.
(3) Compare Qsp with Ksp.
Determining Whether Precipitation Occurs
The concentration of calcium ion in blood plasma is
0.0025 M. If the concentration of oxalate ion is
1.0x10-7 M, do you expect calcium oxalate,
CaC2O4, to precipitate? Ksp = 2.3x10-9.
Qsp = [Ca2+][C2O42-] = (0.0025)(1.0 x 10-7)
Qsp = 2.5 x 10-10
• Qsp < Ksp therefore, no precipitate will form!!!
Prentice-Hall ©2002
Chapter Sixteen
Slide 13 of 32
Determining Whether Precipitation Occurs
In applying the precipitation criteria, the effect of dilution when
solutions are mixed must be considered.
Example: A 250.0 mL sample of 0.0012 M Pb(NO3)2 (aq) is
mixed with 150.0 mL of 0.0640 M NaI (aq). Should
precipitation of PbI2 (s), Ksp = 7.1x10-9, occur?
Calculate new concentrations in total volume of 400mls = 0.4L
[Pb2+] = (0.250L)(0.0012M)/(0.400L) = 7.5 x 10-4 M
[I-] = (0.150L)(0.0640M)/(0.400L) = 0.024 M
Qsp = [Pb2+][I-]2 = (7.5 x 10-4)(0.024)2 = 4.32 x 10-7
Qsp > Ksp therefore a precipitate will form!
Selective Precipitation
a) The first precipitate to
form when AgNO3(aq)
is added to an aqueous
solution containing Cland I- is yellow AgI(s).
b) Essentially all the I- has
precipitated before the
precipitation of white
AgCl(s) begins.
AgCl (s)  Ag+ (aq) + Cl- (aq)
• AgI (s)  Ag+ (aq) + I- (aq)
Ksp = 1.8x10-10
Ksp = 8.5x10-17
Selective Precipitation An Example
Example: An aqueous solution that is 2.00 M in AgNO3 is
slowly added from a buret to an aqueous solution that is
0.0100 M in Cl- and also 0.0100 M in IAgCl (s)  Ag+ (aq) + Cl- (aq)
Ksp = 1.8x10-10
AgI (s)  Ag+ (aq) + I- (aq)
Ksp = 8.5x10-17
a.
Which ion. Cl- or I-, is the first to precipitate from solution?
(this depends on the Molar solubility not Ksp, but since
both compounds give equal ions, then you can use Ksp)
• AgI will precipitate first, lower Ksp
a.
Is separation of the two ions by selective precipitation
feasible?
Comparing Solubilities
•
Which salt will be the most soluble in water.
Molar solubility
1. CuS Ksp = 8.5 x 10-45
x = 9.2 x 10-23
Ksp = x2
2. Ag2S Ksp = 1.6 x 10-49
x = 3.4 x 10-17
Ksp = (2x)2x = 4x3
3. Bi2S3 Ksp = 1.1 x 10-73
x = 1.0 x 10-15
Ksp = (2x)2(3x)3 = 108x5
• Most Soluble = highest molar solubility!!!! Do not use Ksp!!
Prentice-Hall ©2002
Chapter Sixteen
Slide 17 of 32
Effect of pH on Solubility
• The solubility of an ionic solute may be greatly
affected by pH if an acid-base reaction also occurs
as the solute dissolves.
• In other words, some salts will not dissolve well in
pure water, but will dissolve in an acid or a base.
• If the anion (A-) of the salt/precipitate is that of a
weak acid, the salt/precipitate will dissolve more
when in a strong acid (H+ ions will form HA with
A-)
• However, if the anion of the precipitate is that of a
strong acid, adding a strong acid will have no
effect on the precipitate dissolving more.
Effect of pH on Solubility
• How would the addition of HCl affect the solubility of
PbCl2?
– Cl- is the conjugate base of a strong acid, thus it is a weak
base.
– It will not react with H+ ions, so there is no effect.
• How would the addition of HCl affect the solubility of
FeS?
– S2- is a strong base (conjugate base of weak acid)
– Thus, it will react with H+ ions to form H2S,
– This will shift the equilibrium to make more FeS dissolve!
Prentice-Hall ©2002
Chapter Sixteen
Slide 19 of 32
Summary
• The solubility product constant, Ksp,
represents equilibrium between a slightly
soluble ionic compound and its ions in a
saturated aqueous solution.
• The common ion effect is responsible for the
reduction in solubility of a slightly soluble
ionic compound.
• The solubilities of some slightly soluble
compounds depends strongly on pH.
Prentice-Hall ©2002
Chapter Sixteen
Slide 20 of 32
Qualitative Inorganic Analysis
• Acid-base chemistry, precipitation reactions,
oxidation-reduction, and complex-ion formation all
come into sharp focus in an area of analytical
chemistry called classical qualitative inorganic
analysis.
• “Qualitative” signifies that the interest is in
determining what is present, not how much is
present.
• Although classical qualitative analysis is not as
widely used today as instrumental methods, it is
still a good vehicle for applying all the basic
concepts of equilibria in aqueous solutions.
Cations of Group 1
• If aqueous HCl is added to an unknown solution of
cations, and a precipitate forms, then the unknown
contains one or more of these cations: Pb2+, Hg22+,
or Ag+.
• These are the only ions to form insoluble chlorides.
• If there is no precipitate, then these ions must be
absent from the mixture.
• If there is a precipitate, it is filtered off and saved for
further analysis.
• The supernatant liquid is also saved for further
analysis.
Cation Group 1 (continued)
Analyzing For Pb2+
• Of the three possible ions in solution, PbCl2 is the
most soluble in water.
• The precipitate is washed with hot water and the
washings then treated with aqueous K2CrO4.
• If Pb2+ is present, chromate ion combines with lead
ion to form a precipitate of yellow lead chromate,
which is less soluble than PbCl2.
• If Pb2+ is absent, then the washings just become
tinged yellow but no precipitate is in evidence.
Cation Group 1 (continued)
Analyzing For Ag+
• Next, the undissolved precipitate is treated with
aqueous ammonia.
• If AgCl is present, it will dissolve in this solution.
• If there is any remaining precipitate, it is separated
from the supernatant liquid and saved for further
analysis.
• The supernatant liquid (which contains the Ag+, if
present) is then treated with aqueous nitric acid.
• If a precipitate reforms, then Ag+ was present in the
solution, if no precipitate forms, then Ag+ was not
present in the solution.
Cation Group 1 (continued)
Analyzing For Hg22+
• When precipitate was treated with aqueous
ammonia in the previous step, any Hg22+
underwent an oxidation-reduction reaction to form
a dark gray mixture of elemental mercury and
HgNH2Cl that precipitates from the solution.
• If this dark gray precipitate was observed, then
mercury was present in the original unknown
sample.
• If this dark gray precipitate was not observed,
then mercury must have been absent from the
original unknown sample.
Group 1 Cation Precipitates
left: cation goup 1 ppt: PbCl2,
PbCl2, AgCl (all white)
middle: product from test for
Hg22+: mix of Hg (black) and
HgNH2Cl (white)
right: product from test for Pb2+:
PbCrO4 (yellow) when
K2CrO4(aq) is reacted with
saturated PbCl2