Outline:Outline: 3/14/073/14/07 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 Average = 5.3

• 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 EquilibriaSolubility 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 EquilibriaSolubility Equilibria

The Common Ion Effect

• Solubility is decreased when a common ion is added (Le Châtelier again)

• 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.

Factors that Affect SolubilityFactors that Affect Solubility

CaF2(s) Ca2+(aq) + 2F-(aq)

Solubility and pH

• If F is removed, then the equilibrium shifts right and CaF2 dissolves.

• F can be removed by adding a strong acid:

• As pH decreases, [H+] increases and solubility increases.

• The effect of pH is dramatic!

CaF2(s) Ca2+(aq) + 2F-(aq)

F-(aq) + H+(aq) HF(aq)

Equilibrium Calculations:

Example: AgCl(s) Ag+ + Cl

Ksp = 1.81010

Are reactants or products favored? Question: 15.0 g of AgClare put in 100 mL of

pure water; what is the [Ag]?

Calculation of concentrations:

Keq = [Ag+][Cl] = 1.81010

x2 = 1.8 1010 [Ag] = x = 1.34 105 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+ + Cl

Ksp = 1.81010

Now add some common ions: Question: 15.0 g of AgClare put in 100 mL of

salt water ([Cl]=0.1M); what is the [Ag+]?

Calculation of concentrations:

Keq = [Ag+][Cl] = 1.81010

x (0.1 + x) = 1.8 1010 [Ag] = x = 1.8 109 M

Significant reduction of solubility by having common ion in solution (often done with pH)!

vs. 1.34 105 M

• 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 EquilibriaSolubility Equilibria

Worksheet #10Worksheet #10

Ca3(PO4)2

Ksp = [Ca2+]3[PO43]2 = 2.0 1029

[Ca2+] = 0.2 M

[PO43] = 2.0 10M

What is calcium phosphate?

Ca3(PO4)2 3 Ca2+ + 2 PO4

3

What is the Ksp expression?

Worksheet #10Worksheet #10

Ca3(PO4)2 3 Ca2+ + 2 PO4

3

Ksp = [Ca2+]3[PO43]2 = 2.0 1029

[Ca2+] = 0.2 M

[PO43] = 3.0 10M

0.2 2.0 10

2.0 103.0 10

0.197 0.0

Worksheet #10Worksheet #10

Ca3(PO4)2 3 Ca2+ + 2 PO4

3

Ksp = [Ca2+]3[PO43]2 = 2.0 1029

0.2 2.0 10

2.0 103.0 10

0.197 0.0

Ksp = [0.197+3x]3[2x ]2 = 2.0 1029

2x = 5.1 1014 M

Worksheet #10Worksheet #10

What mass of Ca3(PO4)2 precipitates?

= 2.0 103 M 2000 L = 4.0 mol

0.2 2.0 10

2.0 103.0 10

0.197 0.0

= 1.24 kg

Worksheet #10Worksheet #10

Cr(OH)3 Cr3+ + 3 OH

Ksp = [Cr3+][OH]3 = 1.6 1030

x(1.0 106 + 3x)3 = 1.6 1030

+x +3x

pH = 8.0 then [OH] = 1.0 106

x = [Cr3] = 1.6 10M

Worksheet #10Worksheet #10

Cr(OH)3 Cr3+ + 3 OH

Ksp = [Cr3+][OH]3 = 1.6 1030

x(1.0 108 + 3x)3 = 1.6 1030

+x +3x

pH = 6.0 then [OH] = 1.0 108

x = [Cr3] = 1.6 10M

x is no longer small…solve exactly!

Another way to ask KAnother way to ask Kspsp::

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 10M

Ag2SO4 2Ag+ + SO4

x + 2x + xx = 1.50 10M

Ksp = (2x)2 x = 2.70 10

Complex Formation… Stoichiometry of ComplexesStoichiometry of Complexes

A species that bonds to a metal cation to A species that bonds to a metal cation to form a complex is known as a form a complex is known as a ligandligand..

The number of ligands is called the The number of ligands is called the coordination numbercoordination number))

The stabilization of a metal complex by a The stabilization of a metal complex by a ligand with more than one donor atom is ligand with more than one donor atom is known as the known as the chelate effect.chelate effect.

The Chelate Effect Ligands that have two or more donor atoms are Ligands that have two or more donor atoms are

chelatingchelating ligands. ligands. Chelating ligands bond more tightly to the metal Chelating ligands bond more tightly to the metal

cations.cations. Ethylenediamine (HEthylenediamine (H22NCHNCH22CHCH22NHNH22) is a common ) is a common

chelating ligand - it is abbreviated as “en”chelating ligand - it is abbreviated as “en” Each nitrogen has a lone pair of electrons which Each nitrogen has a lone pair of electrons which

can be donor atoms. Thus en is said to be can be donor atoms. Thus en is said to be bidentatebidentate

Complex Formation and Solubility Complexation can enhance solubility.Complexation can enhance solubility. It removes metal cations from solution It removes metal cations from solution

causing the equilibrium to shift to the causing the equilibrium to shift to the left, and dissolve more solid.left, and dissolve more solid.

Example:

Ni(OH)Ni(OH)22 Ni Ni2+2+ + 2 OH + 2 OH KKspsp=5.5=5.510101717

NiNi2+2+ + 3 en + 3 en Ni(en) Ni(en)33 KKff= 4.1= 4.110101717

Ni(OH)Ni(OH)22 + 3 en + 3 en Ni(en) Ni(en)3 3 + 2 OH+ 2 OH

KKeqeq= 22.6= 22.6

Practice!Practice!