CHEM 2400/2480
Lecture 5
Review of Chemical Equilibrium
Reactions used in analytical chemistry are never complete. Instead,
they proceed to a state of chemical equilibrium in which the ratio of
concentrations of reactants and products is constant.
i.e. For the reaction:
aA + bB
º cC + dD
the concentration equilibrium constant, K, is
[C]c [D]d
K =
[A]a [B]b
[ ] is the concentration of species relative to its standard state.
Standard state for solute = 1 M, gases = 1 atms
and solids and liquids = 1
Activity coefficient, γ , (chapter 8) will be omitted from calculations in
this course.
Week 3-1
CHEM 2400/2480
Basic Information required to follow Equilibrium
•
Aqueous reactions involve H2O.
•
Amphiprotic species behaves as an acid or a base depending on the
properties of the solute
e.g.
CH3COOH
Acid 1
+
H2O º CH3COO- + H3O+
Base 2
Base 1
Acid 2
and
NH3 +
Base 1
H2O
º
Acid 2
NH4+ +
Acid 1
+ ve ion (cation) / acid
•
;
OHBase 2
- ve ion (anion) / base
Autoprotolysis
- amphiprotic solvents undergo self ionization e.g.
CH3OH + CH3OH º CH3O- + CH3OH2+
2 H2O º H3O+ + OH-
[H 3 O + ] [OH −]
+ ] [OH −]
K =
=
[H
O
3
[H 2 O]2
Note: in dilute solutions [H2O] = 55.6 M
Bronsted - Lowry Acids and Bases
Week 3-2
CHEM 2400/2480
Acids - proton donors
Bases - proton acceptors
pH = - log [H+]
pKw = pH + pOH = 14
neutral
pH = 7
acid
pH < 7
base
pH > 7
Conjugate Acids and Bases
HA +
Acid 1
A- +
Base 1
B
º
Base 2
BH+
Acid 2
Acid Dissociation Constant
HA
º
H
+
+
A
-
Ka
[H +] [A −]
=
[HA]
Base Hydrolysis Constant
B
+ H2 O
º
BH+ + OH-
Week 3-3
[BH +] [OH −]
Kb =
[B]
CHEM 2400/2480
Relation Between Ka and Kb for a Conjugate pair
Ka Kb = Kw
+
e.g. NH3 + H2O º NH4 + OH
-
NH4+ + H2O º NH3 + H3O+
[NH 4+] [OH −]
Kb =
[NH 3 ]
[NH 3 ] [H 3 O +]
Ka =
[NH 4+]
______________________________________________
2 H2O º H3O+ + OH-
add:
K w = [H 3 O +] [OH −]
______________________________________________
[NH 4+] [OH −]
[NH 3 ] [H 3 O]
=
x
= [H 3 O +] [OH −]
+
[NH 3 ]
[NH 4 ]
K aK b
∴
K
a
Kb
= K
w
Week 3-4
CHEM 2400/2480
SYSTEMATIC TREATMENT OF EQUILIBRIUM
Before we proceed, you must be comfortable with two new
concepts: ‘Mass Balance’ and ‘Charge Balance’
1.
Mass Balance relates to the conservation of matter i.e.
•
the relation of the equilibrium concentrations, [ ], of various
species in a solution to one another and to the analytical
concentration, C, of the various solute
can have more than one mass balance equation
misleading title because concentrations rather than masses are used
•
•
e.g. write the mass balance equations for a solution that is 0.050 M in
acetic acid (HAc)
In solution we have:
HAc + H2O º H3O+ + Ac2H2O º H3O+ + OH1st mass balance equation:
CHAc = 0.050 M = [HAc] + [Ac-]
2nd mass balance equation:
[H3O+] = [Ac-] + [OH-]
Week 3-5
CHEM 2400/2480
Write the mass balance equations for 0.10 M Na2CO3 solution
Na2CO3 º 2Na+ + CO32-
[Na +]
[CO 23−]
=
2
1
Week 3-6
CHEM 2400/2480
Write the mass balance for a saturated solution of Ag3PO4
Ag3PO4 º 3 Ag+ + PO43-
ratio:
[Ag+]
3
=
1
[PO 43−]
Week 3-7
CHEM 2400/2480
Lecture 6
CHARGE BALANCE
An electrolyte solution is NEUTRAL and
the sum of the molar concentration of the +ve charges in solution
= the sum of the molar concentration of the -ve charges in
solution.
Note: There can only be ONE charge balance equation for a given
solution.
Contribution of charge by an ion
= molar concentration of that ion times its charge
e.g. Mg2+
mol +ve charge/L = mol Mg2+
L
x
2 mol +ve charge
mol Mg2+
= 2 x [Mg2+]
Write the charge balance equation for MgCl2 with C MgCl2 = 0.100 M
2 [Mg2+]
2 (0.100)
+
[H3O+]
1 x 10-7
=
[Cl-]
+
[OH-]
2 (0.100)
For neutral solution [H3O+] and [OH-] are very small
ˆ
2 [Mg2+] = [Cl-] = 0.200 M
Week 3-8
1 x 10-7
CHEM 2400/2480
ACID - BASE EQUILIBRIA
Let’s consider weak acids and bases. We will deal with weak acids in
our lectures.
Weak Acid
HA + H2O º H3O+
2 H 2 O º H3 O+
+
A-
+ OH-
Mass balance:
CHA = [HA] + [A-]
Charge balance:
[H3O+] = [A-] + [OH-]
Week 3-9
Ka
Kw
CHEM 2400/2480
[H 3 O +] [A −]
=
[HA]
Ka
but from massbalance
[HA] = C HA − [A −]
∴
Ka
=
[H 3 O + ] [A −]
C HA − [A −]
also from charge balance
[H 3 O +]
and
[A −]
=
[A −]
+
=
[H 3 O +] − [OH - ]
=
[H 3 O +] −
Ka =
Kw
[H 3 O +]
⎧
⎪
+
⎨ [H 3 O ] −
⎪
⎩
⎧
⎪
C HA − ⎨ [H 3 O +] −
⎪
⎩
[H 3 O +]
∴
[OH - ]
This is a cubic equation.
Week 3-10
Kw
[H 3 O +]
K w ⎫⎪
⎬
[H 3 O +] ⎪⎭
⎫
⎪
⎬
⎪
⎭
CHEM 2400/2480
⎧
⎪
⎨
⎪
⎩
⎧
⎪
⎨
⎪
⎩
[H O +] [
3
Ka
=
C HA −
] [
Kw
H 3O + −
H 3O +
[H O +] − [HKO +]
w
3
3
− K H 3 O +⎤⎥ + K a K w =
⎡
⎦
+⎤
⎢H 3 O ⎥
⎡
a⎢
⎣
K a C HA
⎣
[H O +]
2
3
[
]
[
]
H 3O + =
K
2
a
2
⎤
+
H 3O ⎥ − K w
⎡
⎢⎣
⎦
⎦
⎛
⎜
⎜K w
⎜
⎝
+ K a H 3O + −
−K a +
]
⎫
⎪
⎬
⎪
⎭
⎫
⎪
⎬
⎪
⎭
+
+ K a C HA
⎛
⎜
4⎜
⎜
⎝
K
w
2
Week 3-11
⎞
K aK w ⎟
+
⎟ =0
+
H 3 O ⎟⎠
+ K
[
a
C HA
]
K aK w
+
H 3O +
[
]
⎞
⎟
⎟
⎟
⎠
CHEM 2400/2480
Use spreadsheet to solve this equation.
Excel uses ‘circular defintion’ Quattro uses ‘solve for’
Lecture 9
1. CONTRIBUTION from WATER
In real systems, one rarely has to worry about the contribution
of H+ or OH- from water. If you had a 10-8 M KOH solution,
you would have to protect it from the atmosphere to keep the
pH at the calculated value of 7.02.
Contributions from water are normally negligible. In this case:
[H 3 O +] >> [OH −] ≈
∴
[H 3 O +]
Kw
[H 3 O +]
= [A −]
and
Ka
[H 3 O +] 2
=
C HA − [H 3 O +]
This equation holds for:
- small and large concentration
- small and large Ka’s
Week 3-12
CHEM 2400/2480
But NOT extremes i.e.
- very small Ka’s and very small CHA
Week 3-13
CHEM 2400/2480
2.
For the Extreme case of very little dissociation
C HA >>
⎧
⎪
+
⎨ [H 3 O ]
⎪⎩
K w ⎫⎪
−
⎬
[H 3 O +] ⎪⎭
and
⎧
⎪
[H 3 O +] ⎨ [H 3 O +] −
Ka =
∴
⎪⎩
[H 3 O +] =
K w ⎫⎪
⎬
[H 3 O +] ⎪⎭
C HA
K w + K a C HA
This equation holds for very small Ka’s
Week 3-14
CHEM 2400/2480
Relationship between pH, pKa and CHA
Analogy Between Weak Acid and Weak Base Equations
Week 3-15
CHEM 2400/2480
Weak acid
Weak base
(1)
[H+][A-] = Ka[HA]
(1)
[OH-][BH+] = Kb[B]
(2)
[H+][OH-] = Kw
(2)
[OH-][H+] = Kw
(3)
[H+] = [OH-] + [A-]
(3)
[OH-] = [H+] + [BH+]
(4)
[HA] + [A-] = C
(4)
[B] + [BH+] = C
[H+]
replaced by
[OH-]
[OH-]
replaced by
[H+]
[HA]
replaced by
[B]
[A-] replaced by
Ka replaced by
[BH+]
Kb
Kw = Ka Kb
Week 3-16
CHEM 2400/2480
Also, salts of SB and WA: Species, Na+, AKb: A- + H2O º OH- + HA
charge balance:
mass balance:
[H+] + [Na+] = [A-] + [OH-]
CNaA = [Na+]
CNaA = [A-] + [HA]
ˆ [H+] = [OH-] - [HA]
and, salts of WB and SA: species, BH+, ClKa: BH+ + H2O º B + H3O+
charge balance:
[BH+] + [H+] = [OH-] + [Cl-]
mass balance:
CBHCl = [Cl-]
CBHCl = [B] + [BH+]
ˆ [H+] = [OH-] + [B]
Week 3-17