Post on 13-Jan-2016
description
Equilibrium constants
Equilibrium constants
K =PC
cPDd
PAaPB
b
K =[A]a[B]b
[C]c[D]d
Equilibrium constants
K =PC
cPDd
PAaPB
b
K =[A]a[B]b
[C]c[D]d
[solid] = 1
Equilibrium of solids and solutions
Equilibrium of solids and solutions
NaCl(s) + H2O(l)
Equilibrium of solids and solutions
NaCl(s) + H2O(l) Na+(aq) + Cl-
(aq)
To remove an ion from the crystal lattice, the
solvating interactions must be stronger than
the lattice interactions.
Equilibrium of solids and solutions
NaCl(s) + H2O(l) Na+(aq) + Cl-
(aq)
Solution must be saturated for this
equilibrium to take place.
Dynamic equilibrium
Equilibrium for dissolution-precipitation
reaction:
Equilibrium for dissolution-precipitation
reaction:
I2(s) + CCl4(l) I2(CCl4)
Equilibrium for dissolution-precipitation
reaction:
K [I2]CCl4
=
I2(s) + CCl4(l) I2(CCl4)
Equilibrium for dissolution-precipitation
reaction:
[I2]CCl K4=
I2(s) + CCl4(l) I2(CCl4)
Molarity of saturated solution = K
[I2]CCl K4=
K will vary when conditions are changed
based on Le Chatelier’s Principle.
[I2]CCl K4=
If solution of a material is exothermic,
increasing the temperature will
decrease K.
A(s) A(sol)
[I2]CCl K4=
If solution of a material is endothermic,
increasing the temperature will
increase K.
A(s) A(sol)
Kinetics (reaction rates) must
be considered.
The value of K is not an indicator
of how long it takes to attain
equilibrium.
A K = 5 does not guarantee a 5 M
solution in a few minutes.
Not all solutions are ‘ideal’
Not all solutions are ‘ideal’
Ideal Solution:
Widely separated species
(ions or molecules) that do not
interact.
CsCl(s) Cs+(aq) + Cl-
(aq)
CsCl(s) Cs+(aq) + Cl-
(aq)
As the concentration of ions increases,
Cs+ to Cl- distances decrease.
CsCl(s) Cs+(aq) + Cl-
(aq)
As the concentration of ions increases,
Cs+ to Cl- distances decrease.
Cs+…Cl-Cs+…Cl-
Cl-…Cs+
CsCl(s) Cs+(aq) + Cl-
(aq)
Cs+…Cl-Cs+…Cl-
Cl-…Cs+
Ion pairing may occur before equilibrium.
CsCl(s) Cs+(aq) + Cl-
(aq)
Cs+…Cl-Cs+…Cl-
Cl-…Cs+
These solutions are non-ideal.
Ion pairing may occur before equilibrium.
CsCl(s) Cs+(aq) + Cl-
(aq)
Salts of low solubilities allow the study
of solutions that are essentially ideal.
CsCl(s) Cs+(aq) + Cl-
(aq)
Salts of low solubilities allow the study
of solutions that are essentially ideal.
A saturated solution of 0.1 or less is
a sign of low solubility in a salt.
Solubility product
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
Same form as equilibrium constant
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-]
[AgCl]
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-]
1= Ksp
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-] = Ksp
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-] = Ksp
25oC Ksp = 1.6 x 10-10
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-] = Ksp
25oC Ksp = 1.6 x 10-10
[Ag+] = [Cl-] = y
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-] = Ksp
25oC Ksp = 1.6 x 10-10
[Ag+] = [Cl-] = y y2 = Ksp = 1.6 x 10-10
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-] = Ksp
y2 = Ksp = 1.6 x 10-10
= 1.26 x 10-5 M[Ag+] = [Cl-]
Solubility product
AgCl(s) Ag+(aq) + Cl-
(aq)
[Ag+][Cl-] = Ksp
y2 = Ksp = 1.6 x 10-10
= 1.26 x 10-5 M[Ag+] = [Cl-]
1.8 x 10-3 g/L
BaF2(s) Ba2+(aq) + 2 F-
(aq)
BaF2(s) Ba2+(aq) + 2 F-
(aq)
[Ba2+][F-]2 = Ksp
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
For every Fe3+ that goes into solution,
3 OH- go into solution.
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
If there is another source of OH- (NaOH)
that provides a higher [OH-] then that is
the value of [OH-] to be used.
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
27y4 = 1.1 x 10-36
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
27y4 = 1.1 x 10-36
y4 =1.1 x 10-36
27= 4.1 x 10-38
Fe(OH)3Fe3+
(aq) + 3 OH-(aq)
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
27y4 = 1.1 x 10-36
y4 =1.1 x 10-36
27= 4.1 x 10-38
y = 4.5 x 10-10
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
[Cu2+][S2-] = 2 x 10-47
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
[Cu2+][S2-] = 2 x 10-47
y2 = 2 x 10-47
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
[Cu2+][S2-] = 2 x 10-47
y2 = 2 x 10-47
y = 4.5 x 10-24 = [Cu2+] = [S2-]
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
y = 4.5 x 10-24 = [Cu2+] = [S2-]
mw CuS = 95.6
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
y = 4.5 x 10-24 = [Cu2+] = [S2-]
mw CuS = 95.6
4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
y = 4.5 x 10-24 = [Cu2+] = [S2-]
4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L
Atoms/L = moles x Ao
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
y = 4.5 x 10-24 = [Cu2+] = [S2-]
4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L
Atoms/L = moles x Ao = (4.5 x 10-24)(6 x 1023) =
Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-
(aq)
y = 4.5 x 10-24 = [Cu2+] = [S2-]
4.5 x 10-24 (95.6) = 4.3 x 10-22 g/L
Atoms/L = moles x Ao = (4.5 x 10-24)(6 x 1023) =
2.7 atoms/L
Example 9-3
Determine Ksp for a salt based
on solubility data.
Example 9-3
Determine Ksp for a salt based
on solubility data.
PbCl2 8.67 g/L for a saturated solution
Example 9-3
Determine Ksp for a salt based
on solubility data.
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
Example 9-3
Determine Ksp for a salt based
on solubility data.
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
(y)(2y)2 = Ksp
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
(y)(2y)2 = Ksp
4y2 = Ksp
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
4y2 = Ksp 8.67 g PbCl2
278.1 g/mol=
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp 8.67 g PbCl2
278.1 g/mol= 0.031 mol
< 0.1 mol
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp 8.67 g PbCl2
278.1 g/mol= 0.031 mol
Ksp = 4(0.031)3
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp 8.67 g PbCl
278.1 g/mol= 0.031 mol
Ksp = 4(0.031)3 =1.2 x 10-4
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp 8.67 g PbCl
278.1 g/mol= 0.031 mol
Ksp = 4(0.031)3 =1.2 x 10-4 Table = 1.6 x 10-5
Example 9-3
PbCl2 8.67 g/L for a saturated solution
PbCl2 Pb2+(aq) + 2 Cl-
(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp 8.67 g PbCl
278.1 g/mol= 0.031 mol
Ksp = 4(0.031)3 =1.2 x 10-4 Table = 1.6 x 10-5
Non-ideal solution
Precipitation from solution
Precipitation from solution
Start with solution that is not saturated.
Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Remove solvent by evaporation.
Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Remove solvent by evaporation.
Change nature of solvent.
Change nature of solvent.
The polarity of a solvent
system can be adjusted.
Change nature of solvent.
The polarity of a solvent
system can be adjusted.
NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)
Not saturated
Change nature of solvent.
The polarity of a solvent
system can be adjusted.
NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)
Not saturatedAdd EtOH to solution.
Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Remove solvent by evaporation.
Change nature of solvent.
Cool or warm system.
Precipitation of a product from
a reaction.
Will it precipitate?
Determine reaction quotient.
AgNO3 (solution) mixed with
NaCl (solution)
Will AgCl precipitate?
Determine reaction quotient.
AgNO3 (solution) mixed with
NaCl (solution)
Will AgCl precipitate?
Q = [Ag+][Cl-]
Q = reaction quotient
AgNO3 (solution) mixed with
NaCl (solution)
Q = reaction quotient
Qinit = conditions just as solutions are mixed
Q = [Ag+][Cl-]
AgNO3 (solution) mixed with
NaCl (solution)
Q = reaction quotient
Qinit = conditions just as solutions are mixed
If Qinit < Ksp then no AgCl willprecipitate.
Q = [Ag+][Cl-]
[Ag+][Cl-] too low
AgNO3 (solution) mixed with
NaCl (solution)
Q = reaction quotient
Qinit = conditions just as solutions are mixed
If Qinit > Ksp then AgCl will precipitate.
Q = [Ag+][Cl-]
Ag+ Cl- too high
AgNO3 (solution) mixed with
NaCl (solution)
Q = reaction quotient
Qinit = conditions just as solutions are mixed
If Qinit > Ksp then AgCl will precipitate.
As the reaction continues, precipitation willstop when Q = Ksp
Q = [Ag+][Cl-]
precipitate
No precipitate
Exercise page 384
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Will TlIO3 precipitate atequilibrium?
Exercise page 384
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Will TlIO3 precipitate atequilibrium?
Ksp = 3.1 x 10-6
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
[Tl3+] = .555(0.0022) 1.000
= 0.0012 M
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
[Tl+] = 0.0012 M
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
[Tl+] = 0.0012 M .555 L x 0.0022 M = 0.0012 moles
[IO3-] = .00099 M + 0.0012 M = 0.0022
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-] = (0.0012)(.00099) = 1.2 x 10-6
[Tl+] = 0.0012 M .555 L x 0.0022 M = 0.0012 moles
[IO3-] = .0022 M
For this solution
TlIO3 555 ml 0.0022 M
NaIO3 445 ml 0.0022 M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-] = (0.0012)(.0022) = 2.6 x 10-6
Q < Ksp
No precipitate
For mixedsolutions
Common ion effect
Common ion effect
Adding more of a cation or anion
that is already in solution will
cause Q to change.
Common ion effect
Adding more of a cation or anion
that is already in solution will
cause Q to change.
Q = [A+][B-]
AgCl(s)Ag+
(aq) + Cl-(aq)
Q = [Ag+][Cl-]
Add more Cl- (NaCl)
Q < Ksp
Q > Ksp
AgCl(s)Ag+
(aq) + Cl-(aq)
Q = [Ag+][Cl-]
Add more Cl- (NaCl)
This raises Q above Ksp
Common ion effect
If a solution and a solid salt to be
dissolved in it have an ion in common,
the solubility of the salt is depressed.
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
What is the solubility of TlIO3
in this solution?
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y [IO3-] = y + 0.050
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y [IO3-] = y + 0.050 0.050
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y [IO3-] = y + 0.050 0.050
3.1 x 10-6 = (y)(0.050) = 0.050y
Ksp = 3.1 x 10-6
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y [IO3-] = y + 0.050 0.050
3.1 x 10-6 = (y)(0.050) = 0.050y
y = 6.2 x 10-5 mol/L
Ksp = 3.1 x 10-6
Exercise page 387
TlIO3(s)Tl+
(aq) + IO3-(aq)
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y [IO3-] = y + 0.050 0.050
3.1 x 10-6 = (y)(0.050) = 0.050y
y = 6.2 x 10-5 mol/L [Tl+] = 1.8 x 10-3
No common ion