Chem 1011 – Intersession 2011
Class #32
13-Jun-11
1
Class 32: Solubility Equilibria /
Solubility Product (Ksp)
•
Sec 16.5 – Solubility Equilibria and Solubility
Product Constant
▫
▫
▫
▫
•
Ksp and Molar Solubility
Ksp and Relative Solubility
The Effect of a Common Ion on Solubility
The Effect of pH on Solubility
Sec 16.5 – Precipitation
2
Solubility
•
•
•
•
▫
▫
▫
Soluble
a compound dissolves in water
Insoluble
a compound does not dissolve in water.
Note:
All ionic compounds dissolve in water to some
degree, however, many compounds have such low
solubility in water that we classify them as insoluble
There is a continuum of solubility, and we are now
going to examine degrees of solubility
3
Solubility
• In Chem 1010 you learned solubility rules. Please review
these; it is important to know what compounds are
always soluble. (See Tro, Table 14.1)
Solubility Equilibria / Solubility Product
Constant (Ksp)
1
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
4
Solubility Equilibria and Solubility
Product Constant (Ksp)
•
▫
•
Solubility Product constant (Ksp)
equilibrium expression for the dissolution of an ionic
compound.
Example
CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
Ksp = 1.46 x 10-10
Ksp = [Ca2+][F–]2
•
Just like any other equilibrium, solids do not appear in K
expressions – just the aqueous ions!
•
The value of Ksp is a measure of the solubility of a
compound
5
Solubility Product Constant (Ksp)
6
Ksp and Molar Solubility
•
•
▫
▫
Molar solubility
the solubility in units of moles per litre (mol/L)
Note
Ksp is NOT molar solubility; Ksp can be used to determine
molar solubility.
•
Ksp has only one value for every temperature.
•
Molar solubility can have different values, depending
on the solution conditions (common ion effect, etc…).
•
Molar solubility is normally given the symbol “s” (to
represent solubility) when working with Ksp ICE
tables.
Solubility Equilibria / Solubility Product
Constant (Ksp)
2
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
7
Ksp and Molar Solubility
• Example: Calculate the molar solubility of PbCl2 in pure
water at 25 C. (Ksp = 1.17 x 10-5)
Write the
dissociation
reaction and
Ksp expression.
Create an ICE
table defining
the change in
terms of the
solubility of the
solid.
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl−(aq)
[I]
0
0
[C]
+s
+2s
[E]
s
2s
8
Ksp and Molar Solubility
Substitute into
the Ksp
expression.
Ksp = [Pb2+][Cl−]2
Ksp = (s)(2s)2
Substitute the
value of Ksp
into the
equation, and
solve for s.
9
Problems
• Calculate the molar solubility of Fe(OH)2 in pure
water. Ksp (Fe(OH)2) = 4.87 x 10-17
• The molar solubility of AgBr in pure water is 7.3
x 10-7 M. Calculate Ksp.
Solubility Equilibria / Solubility Product
Constant (Ksp)
3
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
10
Ksp and Relative Solubility
•
You cannot always use Ksp values to directly compare the
molar solubility of two compounds.
•
This is because the relationship between Ksp and molar
solubility also depends on the stoichiometry of the
dissociation process.
If two reactions have the same dissociation stoichiometry,
their molar solubilities can be directly compared.
•
11
The Common Ion Effect
•
The solubility of an ionic compound is lower in a
solution containing a common ion than in pure water.
•
This makes sense, based upon Le Chatelier’s principle.
•
Example: Consider a solution of PbCl2, to which 0.010
mol of NaCl is added.
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl−(aq)
Increase the
concentration
of Cl–
Equilibrium
shifts to the
left.
12
The Common Ion Effect
• Example: Calculate the molar solubility of CaF2 in 0.100
M NaF at 25 C. (Ksp = 1.46 x 10-10)
Write the
dissociation
reaction and
Ksp expression.
Create an ICE
table defining
the change in
terms of the
solubility of the
solid.
CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq)
[I]
0
0.100 M
[C]
+s
+2s
[E]
s
0.100 + 2s
Solubility Equilibria / Solubility Product
Constant (Ksp)
4
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
13
The Common Ion Effect
Substitute into
the Ksp
expression.
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Since Ksp is small, assume s is small…
So, 0.100 + 2s ≈ 0.100
Ksp = (S)(0.100)2
Substitute the
value of Ksp
into the
equation, and
solve for s.
14
Problem
• Calculate the molar solubility of CaF2 in a
solution containing 0.250 M Ca(NO3)2
▫ (Ksp (CaF2) = 1.46 x 10-10)
15
The Effect of pH on Solubility
•
The solubility of an ionic compound with a strongly or
weakly basic anion increases with increasing acidity
(decreasing pH).
•
This makes sense, based upon Le Chatelier’s principle.
•
Example: Consider a solution of Mg(OH)2, in basic
media (high pH; high [OH–].)
Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH−(aq)
Equilibrium
shifts to the
left.
High [OH–]
Solubility Equilibria / Solubility Product
Constant (Ksp)
5
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
16
The Effect of pH on Solubility
•
The solubility of an ionic compound with a strongly or
weakly basic anion increases with increasing acidity
(decreasing pH).
•
Again, this makes sense based on LeChatelier’s
principle.
•
Example: Consider a solution of Mg(OH)2, in acidic
media (low pH; OH– being neutralized by H3O+.)
Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH−(aq)
Equilibrium
shifts to the
right.
H3O+ reacts with OH–;
effectively, OH– is being
removed from the system.
17
Problem
• Which compound, FeCO3 or PbBr2, is more
soluble in acid than in base? Why?
18
Precipitation
•
Precipitation will occur upon
the mixing of two solutions
containing ionic compounds if
the cross product
(combination of the cation
from one solution and anion
from the other) is “insoluble”.
•
We need to quantify
“insoluble” in order to
determine if precipitation will
take place.
Solubility Equilibria / Solubility Product
Constant (Ksp)
6
Chem 1011 – Intersession 2011
Class #32
Chem 1011
13-Jun-11
Winter 201019
Dr. L. Dawe
Precipitation
(Soluble)
(“Insoluble”)
20
Precipitation
•
If we compare the reaction quotient, Q, for the
current solution concentrations to the value of
Ksp, we can determine whether precipitation
will occur.
▫
Q = Ksp: the solution is saturated; no precipitation
▫
Q < Ksp: the solution is unsaturated; no precipitation
▫
Q > Ksp: the solution is above saturation; the salt
above saturation will precipitate
21
Precipitation
• Will a precipitate form when we mix Pb(NO3)2(aq) with
NaBr(aq) if the concentrations after mixing are 0.0150 M
and 0.0350 M, respectively?
1. Write an equation for the reaction
Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)
2. Determine the ion concentrations of the original salts
Pb(NO3)2 = 0.0150 M
NaBr = 0.0350 M
Pb2+ = 0.0150 M
 Na+ = 0.0350 M
NO3− = 2(0.0150 M)
Br– = 0.0350 M
Solubility Equilibria / Solubility Product
Constant (Ksp)
7
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
22
Precepitation
3. Determine the Ksp for any insoluble products
▫
▫
In the equation, PbBr2 is a solid product
In a test or exam, Ksp values are provided – when
practicing, there is a list in your textbook
Ksp (PbBr2) = 4.67 x 10–6
4. Write the dissociation equation for the solid:
PbBr2(s) ⇌ Pb2+(aq) + 2 Br−(aq)
23
Precipitation
5. Write the expression for Qsp, and substitute the ion
concentrations to solve
6. Compare Qsp to Ksp and determine the direction of the
shift
Qsp = 1.84 x 10–7 < Ksp = 4.67 x 10–6
▫
Since Qsp < Ksp, the equilibrium shifts right (towards
products; away from the solid) and so no precipitate
forms
24
Problem
• A solution containing 0.0600M Pb(NO3)2 and a
solution containing 0.0158 M NaBr are mixed. Will
a precipitate form? (Ksp (PbBr2) = 4.67 x 10-6)
Solubility Equilibria / Solubility Product
Constant (Ksp)
8
Chem 1011 – Intersession 2011
Class #32
13-Jun-11
25
Last Chapter!
•
Sec 18.3 – Voltaic (or Galvanic) Cells:
Generating Electricity from Spontaneous
Chemical Reactions
▫
•
Electrochemical Cell Notation
Sec 18.4 – Standard Electrode Potentials
▫
▫
Predicting the Spontaneous Direction of an
Oxidation-Reduction Reaction
Predicting Whether a Metal will Dissolve in Acid
Solubility Equilibria / Solubility Product
Constant (Ksp)
9