Chapter 6 Problems

6.6, 6.9, 6.15, 6.16,

6.19, 6.21, 6.24
Chapter 6
Chemical
Equilibrium
Chemical Equilibrium

Equilibrium Constant




Solubility product (Ksp)
Common Ion Effect
Separation by precipitation
Complex formation
Solubility Product
Introduction to Ksp
Solubility Product
solubility-product
the product of the solubilities
solubility-product constant => Ksp
constant that is equal to the solubilities
of the ions produced when a substance
dissolves
Solubility Product
For silver sulfate
Ag2SO4 (s)
2 Ag+(aq) + SO4-2(aq)
Ksp = [Ag+]2[SO4-2]
Solubility of a Precipitate
in Pure Water
EXAMPLE: How many grams of AgCl (fw
= 143.32) can be dissolved in 100. mL
of water at 25oC?
AgCl <=> Ag+ + ClKsp = [Ag+][Cl-] = 1.82 X 10-10
(Appen. F)
EXAMPLE: How many grams of AgCl (fw
= 143.32) can be dissolved in 100. mL of
water at 25oC?
AgCl(s)
Initial
Change
Equilibrium
Ag+ (aq) + Cl- (aq)
Some
-
-
-x
-x
+x
+x
+x
+x
(x)(x) = Ksp = [Ag+][Cl-] = 1.82 X 10-10
x = 1.35 X 10-5M
EXAMPLE: How many grams of AgCl (fw
= 143.32) can be dissolved in 100. mL of
water at 25oC?
x = 1.35 X 10-5M

How many grams is that in 100 ml?
# grams = (M.W.) (Volume) (Molarity)
= 143.32 g mol-1 (.100 L) (1.35 x 10-5 mol L-1)
= 1.93X10-4 g = 0.193 mg
Solubility Product
For silver sulfate
Ag2SO4 (s)
2 Ag+(aq) + SO4-2(aq)
Ksp = [Ag+]2[SO4-2]
The Common Ion Effect
Ag2CO3
2 Ag+ + CO3-2
What is the effect on solubility of Ag2CO3 if I add CO3-2?
The Common Ion Effect
Add common ion effect
 a salt will be less soluble if one of its
constituent ions is already present in
the solution
The Common Ion Effect
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M
in Na2CO3.
Ag2CO3
2 Ag+ + CO3-2
Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M in
Na2CO3.
Ag2CO3
2 Ag+ + CO3-2
Initial
Solid
-
0.0200M
Change
-x
Solid
+2x
+2x
+x
0.0200+x
Equilibrium
Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12
Ksp=(2x)2(0.0200M + x) = 8.1 X 10-12
4x2(0.0200M + x) = 8.1 X 10-12
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M in
Na2CO3.
4x2(0.0200M + x) = 8.1 X 10-12
no exact solution to a 3rd order equation,
need to make some approximation
first, assume the X is very small compared to
0.0200 M
4X2(0.0200M) = 8.1 X 10-12
4X2(0.0200M) = 8.1 X 10-12
X= 1.0 X 10-5
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M
in Na2CO3.
X = 1.0 X 10-5 M
(1.3 X 10-4 M in pure water)
Second, check assumption
[CO3-2] = 0.0200 M + X
~ 0.0200 M
0.0200 M + 0.00001M ~ 0.0200M
Assumption is ok!
Separation by
Precipitation
Separation by
Precipitation
Complete separation can mean a lot …
we should define complete.
Complete means that the concentration of
the less soluble material has decreased
to 1 X 10-6M or lower before the more
soluble material begins to precipitate
Separation by
Precipitation
EXAMPLE:
Can Fe+3 and Mg+2 be separated
quantitatively as hydroxides from a solution that is
0.10 M in each cation? If the separation is possible,
what range of OH- concentrations is permissible.
Add OH-
Mg2+
Mg2+
Fe3+
Fe3+
Fe3+
3+
Fe
2+
Mg2+
Mg
Mg2+
Mg2+
Fe3+
3+
Fe
3+
Mg2+
2+
Fe
Mg
Fe3+
Fe3+
Fe3+
Mg2+
2+
Mg2+ Mg
Fe3+ Fe3+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+ Mg
2+
Mg2+
Mg2+
Fe3+
@ equilibrium
What is the [OH-] ^when
this happens
Is this [OH-] (that is in
solution) great enough
to start precipitating
Mg2+?
Fe(OH)3(s)
Separation by
Precipitation
EXAMPLE:
Can Fe+3 and Mg+2 be separated
quantitatively as hydroxides from a solution that is
0.10 M in each cation? If the separation is possible,
what range of OH- concentrations is permissible.
Two competing reactions
Fe(OH)3(s)
Fe3+ + 3OH-
Mg(OH)2(s)
Mg2+ + 2OH-
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume quantitative separation requires
that the concentration of the less
soluble material to have decreased to <
1 X 10-6M before the more soluble
material begins to precipitate.
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume [Fe+3] = 1.0 X 10-6M
What will be the [OH-] @ equilibrium required
to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
(1.0 X 10-6M)*[OH-]3 = 2 X 10-39
[OH  ]3  2 1033
[OH  ]  1.31011
EXAMPLE: Separate Iron and
Magnesium?
What [OH-] is required to begin the
precipitation of Mg(OH)2?
[Mg+2] = 0.10 M
Ksp = (0.10 M)[OH-]2 = 7.1 X 10-12
[OH-] = 8.4 X 10-6M
EXAMPLE: Separate Iron and
Magnesium?
@ equilibrium
[OH-] to ‘completely’ remove Fe3+
^ -11
= 1.3 X 10 M
[OH-] to start removing Mg2+
= 8.4 X 10-6M
“All” of the Iron will be precipitated b/f any
of the magnesium starts to precipitate!!
Complex Ion Formation
Complex Formation

Consider Lead Iodide
PbI2 (s)
Pb2+ + 2I-
Ksp = 7.9 x 10-9
What should happen if I- is added to a
solution?
Should the solubility go up or down?
Complex Formation
complex ions (also called coordination ions)
Lewis Acids and Bases
acid => electron pair acceptor (metal)
base => electron pair donor (ligand)
Effects of Complex Ion
Formation on Solubility
Consider the addition of I- to a solution
of Pb+2 ions
Pb2+ + I- <=> PbI+
[ PbI  ]
2
K1 

1
.
0
x
10
[ Pb2 ][I  ]
PbI+ + I- <=> PbI2
K2 = 1.4 x 101
PbI2 + I- <=> PbI3-
K3 =5.9
PbI3+ I- <=> PbI42-
K4 = 3.6
Effects of Complex Ion
Formation on Solubility
Consider the addition of I- to a solution
of Pb+2 ions
Pb2+
+
I-
<=>
PbI+
[ PbI  ]
2
K1 

1
.
0
x
10
[ Pb2 ][I  ]
PbI+ + I- <=> PbI2
K2 = 1.4 x 101
Pb2+ + 2I- <=> PbI2
K’ =?
Overall constants are designated with b
This one is b2
Protic Acids and Bases
Section 6-7
Question


Can you think of a salt that when dissolved
in water is not an acid nor a base?
Can you think of a salt that when dissolved
in water IS an acid or base?
Protic Acids and Bases Salts

Consider Ammonium chloride

Can ‘generally be thought of as the product
of an acid-base reaction.
NH4+Cl- (s)
NH4+ + Cl-
From general chemistry – single positive and single negative
charges are STRONG ELECTROLYTES – they dissolve completely
into ions in dilute aqueous solution
Protic Acids and Bases
Conjugate Acids and Bases in the B-L concept
CH3COOH + H2O  CH3COO- + H3O+
acid
+
base
<=> conjugate base + conjugate acid
conjugate base => what remains after a B-L acid donates its proton
conjugate acid => what is formed when a B-L base accepts a proton
Question:
Calculate the Concentration of
H+ and OH- in Pure water at 250C.
EXAMPLE: Calculate the Concentration
of H+ and OH- in Pure water at 250C.
Initial
Change
Equilibrium
H2O
H+ + OH-
liquid
-
-
-x
+x
+x
+x
+x
Liquid-x
Kw = [H+][OH-] = 1.01 X 10-14
KW=(X)(X) = 1.01 X 10-14
(X) = 1.00 X 10-7
Example
Concentration of OHif [H+] is 1.0 x 10-3 M @ 25 oC?

“From now on,
Kw =
assume the
to
1 x 10-14 = [1 x 10-3][OH-] temperature
be 25oC unless
1 x 10-11 = [OH-]
otherwise
stated.”
[H+][OH-]
pH
~ -3 -----> ~ +16
pH + pOH = - log Kw = pKw = 14.00
Is there such a thing as Pure
Water?


In most labs the answer is NO
Why?
CO2 + H2O

HCO3- + H+
A century ago, Kohlrausch and his students
found it required to 42 consecutive
distillations to reduce the conductivity to a
limiting value.
6-9 Strengths of Acids and Bases
Strong Bronsted-Lowry Acid

A strong Bronsted-Lowry Acid is one
that donates all of its acidic protons to
water molecules in aqueous solution.
(Water is base – electron donor or the
proton acceptor).

HCl as example
Strong Bronsted-Lowry Base


Accepts protons from water molecules
to form an amount of hydroxide ion,
OH-, equivalent to the amount of base
added.
Example: NH2- (the amide ion)
Weak Bronsted-Lowry acid

One that DOES not donate all of its
acidic protons to water molecules in
aqueous solution.


Example?
Use of double arrows! Said to reach
equilibrium.
Weak Bronsted-Lowry base


Does NOT accept an amount of protons
equivalent to the amount of base
added, so the hydroxide ion in a weak
base solution is not equivalent to the
concentration of base added.
example:
NH3
Common Classes of Weak
Acids and Bases
Weak Acids


carboxylic acids
ammonium ions
Weak Bases


amines
carboxylate anion
Weak Acids and Bases
HA
Ka
H+ + A-
[ H  ][ A ]
Ka 
[ HA]
Ka’s ARE THE SAME
HA + H2O(l)
H3O+ + A-
[ H 3O  ][ A ]
Ka 
[ HA]
Weak Acids and Bases
B + H2O
Kb
BH+ + OH

[ BH ][OH ]
Kb 
[ B]
Relation Between Ka and Kb
Relation between Ka and Kb

Consider Ammonia and its conjugate
base.
NH3 + H2O
NH4+ + H2O
H2O + H2O
Ka
Kb

NH4+ + OH-
[ NH 4 ][OH  ]
Ka 
[ NH 3 ]
NH3 + H3O+
[ NH 3 ][H 3O ]
Kb 

[ NH 4 ]
OH- + H3O+
[ NH 3 ][H 3O ] [ NH 4 ][OH ]
K


[ NH 3 ]
[ NH34 ]
w
K  [H O ][OH ]
Example
The Ka for acetic acid is 1.75 x 10-5. Find Kb
for its conjugate base.
Kw = Ka x Kb
Kw
Kb 
Ka
1.0 1014
10
Kb 

5
.
7

10
1.75105
1st Insurance Problem
Challenge on page 107