Download Lecture 26

Outline:
3/14/07
Chem. Dept. Seminar today @ 4pm
 Pick up Quiz #7 – from me
 last lecture before Exam 2…
 No class Friday = Spring Break!

Today:

Finish Chapter 18
Solubility Product (Ksp)
Quiz # 7
Average = 5.3

Solubility Equilibria
• The solubility product is another
example of equilibrium calculations
• Solubility product calcs depend on
the common ion effect (LeChâtelier).
• They have particular applications
with metal ions and pH calculations
(environmental applications).
Solubility Equilibria
• The solubility product is another
example of equilibrium calculations
• Solubility product calcs depend on
the common ion effect (LeChâtelier).
• They have particular applications
with metal ions and pH calculations
(environmental applications).
Factors that Affect Solubility
The Common Ion Effect
• Solubility is decreased when a common
ion is added (Le Châtelier again)
CaF2(s)
Ca2+(aq) + 2F-(aq)
• as F- (from NaF, say) is added, the
equilibrium shifts left, therefore CaF2(s)
is formed (precipitation occurs).
• As NaF is added to the system, the
solubility of CaF2 decreases.
Solubility and pH
• If F- is removed, then the equilibrium
shifts right and CaF2 dissolves.
CaF2(s)
Ca2+(aq) + 2F-(aq)
• F- can be removed by adding a strong
acid:
F-(aq) + H+(aq)
HF(aq)
• As pH decreases, [H+] increases and
solubility increases.
• The effect of pH is dramatic!
Equilibrium Calculations:
Example: AgCl(s)  Ag+ + ClKsp = 1.810-10
Are reactants or products favored?
Question:
15.0 g of AgCl are put in 100 mL of
pure water; what is the [Ag+]?
Calculation of concentrations:

Keq = [Ag+][Cl-] = 1.810-10
x2 = 1.8  10-10
[Ag+] = x = 1.34  10-5 M
Don’t need to know how much solid
is there….it is a solely a function of
the Keq of the ions in solution!
Equilibrium Calculations:
Example: AgCl(s)  Ag+ + ClKsp = 1.810-10
Now add some common ions:
Question:
15.0 g of AgCl are put in 100 mL of salt
water ([Cl-]=0.1M); what is the [Ag+]?
Calculation of concentrations:

Keq = [Ag+][Cl-] = 1.810-10
x (0.1 + x) = 1.8  10-10
[Ag+] = x = 1.8  10-9 M
vs. 1.34  10-5 M
Significant reduction of solubility
by having common ion in solution
(often done with pH)!
Solubility Equilibria
• The solubility product is another
example of equilibrium calculations
• Solubility product calcs depend on
the common ion effect (LeChâtelier).
• They have particular applications
with metal ions and pH calculations
(environmental applications).
Worksheet #10
What is calcium phosphate?
Ca3(PO4)2
What is the Ksp expression?
Ca3(PO4)2  3 Ca2+ + 2 PO43Ksp = [Ca2+]3[PO43-]2 = 2.0  10-29
[Ca2+] = 0.2 M
[PO43-] = 2.0  10-3 M
Worksheet #10
Ca3(PO4)2  3 Ca2+ + 2 PO430.2
2.0  10-3
-3.0  10-3 -2.0  10-3
0.197
0.0
Ksp = [Ca2+]3[PO43-]2 = 2.0  10-29
[Ca2+] = 0.2 M
[PO43-] = 3.0  10-3 M
Worksheet #10
Ca3(PO4)2  3 Ca2+ + 2 PO430.2
2.0  10-3
-3.0  10-3 -2.0  10-3
0.197
0.0
Ksp = [Ca2+]3[PO43-]2 = 2.0  10-29
Ksp = [0.197+3x]3[2x ]2 = 2.0  10-29
2x = 5.1  10-14 M
Worksheet #10
What mass of Ca3(PO4)2 precipitates?
0.2
2.0  10-3
-3.0  10-3 -2.0  10-3
0.197
0.0
= 2.0  10-3 M  2000 L = 4.0 mol
= 1.24 kg
Worksheet #10
Cr(OH)3  Cr3+ + 3 OH+x
+3x
Ksp = [Cr3+][OH-]3 = 1.6  10-30
pH = 8.0 then [OH-] = 1.0  10-6
x(1.0  10-6 + 3x)3 = 1.6  10-30
x = [Cr3+] = 1.6  10-12 M
Worksheet #10
Cr(OH)3  Cr3+ + 3 OH+x
+3x
Ksp = [Cr3+][OH-]3 = 1.6  10-30
pH = 6.0 then [OH-] = 1.0  10-8
x(1.0  10-8 + 3x)3 = 1.6  10-30
x = [Cr3+] = 1.6  10-6 M
x is no longer small…solve exactly!
Another way to ask Ksp:
Exactly 4.68 g of silver sulfate will
dissolve in 1.00 L of water. What is
the Ksp of silver sulfate at this temp?
Ag2SO4 : 312 g/mol
4.68 g/L = 1.50  10-2 M
Ag2SO4  2Ag+ + SO42-x
+ 2x + x
x = 1.50  10-2 M
Ksp = (2x)2 x = 2.70  10-5
Complex Formation…
Stoichiometry of Complexes
A species that bonds to a metal cation to
form a complex is known as a ligand.
The number of ligands is called the
coordination number)
 The stabilization of a metal complex by a
ligand with more than one donor atom is
known as the chelate effect.

The Chelate Effect




Ligands that have two or more donor atoms are
chelating ligands.
Chelating ligands bond more tightly to the metal
cations.
Ethylenediamine (H2NCH2CH2NH2) is a
common chelating ligand - it is abbreviated as
“en”
Each nitrogen has a lone pair of electrons which
can be donor atoms. Thus en is said to be
bidentate
Complex Formation and Solubility
Complexation can enhance solubility.
 It removes metal cations from solution
causing the equilibrium to shift to the
left, and dissolve more solid.

Example:
Ni(OH)2  Ni2+ + 2 OH- Ksp=5.510-17
Ni2+ + 3 en  Ni(en)3
Kf= 4.110+17
Ni(OH)2 + 3 en  Ni(en)3 + 2 OHKeq= 22.6
Practice!