3/28/2014
Chapter 9: POLYPROTIC
ACID-BASE EQUILIBRIA
Oxalic acid,
rhubarb and
rhubarb pie
Images available at http://www.800mainstreet.com/acid_base/0000-01a-oxyacids-.html and1
http://www.sciencephoto.com/images/
Understanding Polyprotic Species
Diprotic acids, H2A
vs.
Can donate 2 H+
Ex. H2SO4, H2CO3
Triprotic acids, H3A
Dibasic species, A2-
Can accept 2 H+
Ex. SO42-, CO32-
vs.
Tribasic species, A3-
Can donate 3 H+
Can accept 3 H+
Ex. H3PO4,
H3C6H5O7 (Citric acid)
Ex. PO43C6H5O73-
2
1
3/28/2014
Diprotic Acids and Bases
General formulas:
H2A = fully acidic form
HA- = intermediate form; amphoteric
A2- = fully basic or fully deprotonated form
3
Equilibria involved: Diprotic Acids and Bases
Diprotic Acid, H2A
K
First dissociation:
a1→
+

H 2 A + H 2O ←
 HA + H3O
K
a2 → -2
+

HA + H 2O ←
 A + H3O
Second dissociation:
-
Dibasic species, A2-
K
First hydrolysis:
b1→

A 2- + H 2O ←
 HA + OH
Second hydrolysis:
b2 →

HA - + H 2O ←
 H 2 A + OH
K
Q. How do we calculate Kb1 and Kb2 from Ka values?
4
2
3/28/2014
Note that H2A and HA- species in the Ka1 expression both appear
in the Kb2 expression. Similarly, the conjugates HA- and A2- in the
Ka2 expression both appear in the Kb1 expression.
Thus,
Proof:
K
a1→
−
+

H 2 A + H 2O ←
 HA + H 3O
+
K
b2 →
−

HA− + H 2O ←
 H 2 A + OH
K
w→
+
−

2 H 2O ←
 H 3O + OH
Kw = Ka1 x Kb2
5
Polyprotic acids: Amino Acids
NOTE: - COOH group is much more acidic (higher Ka; first to
dissociate) than the –NH3+ group.
6
3
3/28/2014
Diprotic Acids
Example: Leucine, H2L
Fully protonated form =
fully acidic, H2A+
Fully dissociated
form = fully basic, A-
Stepwise dissociation:
Start with the fully acidic form, H2A+ = H2L+
7
Dibasic Species
Stepwise hydrolysis of leucine:
Start with the fully basic form, A- = L-
8
4
3/28/2014
pH Calculations: Diprotic Acids and Bases
Problem: Find the pH and concentrations of H2SO3, HSO3- and
SO32- in each of the following solutions:
(a) 0.050 M H2SO3
(b) 0.050 M NaHSO3, and
(c) 0.050 M Na2SO3
Note that for diprotic acids and bases, there are 3 species in
solution (i.e. 3 unknowns: H2A, HA- and A2-) so we need 3
independent equations to solve the problem.
9
pH Calculation: Diprotic Acids and Bases
1. The fully acidic form, H2A
Approximation: In a solution of H2A (Ex. 0.050 M H2SO3),
the 2nd dissociation is usually negligible that H2A behaves
as a monoprotic acid. Also, [A2-] ≈ 0 M.
Calculation of pH and [species]
K
a1→
−
+

H 2 A + H 2O ←
 HA + H 3O
Equil:
F-x
x
x
x2
K a1 =
F−x
10
5
3/28/2014
Fully acidic form (H2A) – Cont.
Problem (a): Find the pH and [H2SO3], [HSO3-] and [SO32-]
in a 0.050 M H2SO3 solution. Ka1 = 1.23 x 10-2; Ka2 = 6.6 x 10-8
K
a1→
−
+

H 2 SO3 + H 2O ←
 HSO3 + H 3O
Equil:
0.050-x
K a1 = 1.23 x 10−2 =
x
x
x2
x cannot be ignored
since Ka1 isn’t too small
(0.050 − x)
x 2 + 1.23 x 10−2 x − 6.15x10−4
Solve for x using
quadratic equation
x = 1.94 x 10−2 M = [ H 3O + ] = [ HSO3− ]
pH = 1.71 [ H 2 SO3 ] = 0.031M
[ HSO3− ] = 1.9 x 10−2 M
0.050 M - x
[ SO3−2 ] 0 M
11
12
6
3/28/2014
13
pH Calculations: Diprotic systems – Cont.
2. The fully basic form, A2Approximation: In a solution of A2- (Ex. 0.050 M Na2SO3),
the 2nd hydrolysis is usually negligible that A2- behaves
as a monobasic species. Also, [H2A] ≈ 0 M.
Calculation of pH and [species]
K
b1→
−
−

A2− + H 2O ←
 HA + OH
Equil:
F-y
K b1 =
y
y2
F−y
Recall: Kb1 = Kw/Ka2
y
pOH = -log (y)
pH = 14 - pOH
14
7
3/28/2014
Problem: Find the pH and concentrations of H2SO3, HSO3- and
SO32- in each of the following solutions:
Ka1 = 1.23 x 10-2;
(a) 0.050 M H2SO3 - DONE!
Ka2 = 6.6 x 10-8
(b) 0.050 M NaHSO3, and
√ (c) 0.050 M Na2SO3
Answer: pH = 9.94; [H2SO3] ≈ 0 M; [SO32-] = 0.04991 M ≈ 0.050 M and [HSO3-] =
[OH-] = 8.7 x 10-5 M 15
pH Calculations: Diprotic systems – Cont.
3. The intermediate (amphoteric) form, HA-
HA- can act as an acid or a base
Q. What is the predominant species in a solution of HA-?
Compare Ka2 and Kb2 equilibria:
Dissociation:
Ka 2

→ A2− + H 3O +
HA− + H 2O ←

Hydrolysis:
Kb2

→ H 2 A + OH −
HA− + H 2O ←

HA- will dissociate/hydrolyze to form A2- and H2A
Approximation: [HA-] ≈ FHA- = FNaHA or FKHA
16
8
3/28/2014
The intermediate form, HA- (Cont.)
Calculation of pH and [species]
Where
K1= Ka1
K2 = Ka2
F = FHA-
pH = -log [H+]
Quick check: pH = ½ (pK1 + pK2)
Solve for [H2A] and [A2-] using [H+] above and K1 & K2
equilibria
17
The intermediate form, HA- (Cont.)
Solving for [H2A] and [A2-]:
[ H + ][ HA− ]
K1 = K a1 =
[ H 2 A]
[ H + ][ HA− ]
Thus [ H 2 A] =
K1
K 2 [ HA− ]
Likewise, [ A ] =
[H + ]
2−
From 2 nd dissociation :
K2 = Ka2
[ H + ][ A 2 − ]
=
[ HA− ]
18
9
3/28/2014
Problem: Find the pH and concentrations of H2SO3, HSO3- and
SO32- in each of the following solutions:
Ka1 = 1.23 x 10-2;
Ka2 = 6.6 x 10-8
(a) 0.050 M H2SO3 - DONE!
√(b) 0.050 M NaHSO3, and
(c) 0.050 M Na2SO3 – DONE!
Answer: pH =4.59; [HSO3-] ≈ 0.050 M, [H2SO3] = 1.1 x 10-4 M; [SO32-] = 1.3 x 10-4 M
19
Triprotic Acids and Bases
Example: H3PO4; PO43Successive
dissociation:
K
a1→
−
+

H 3 PO4 + H 2O ←
 H 2 PO4 + H 3O
K
a2 →
−2
+

H 2 PO4 − + H 2O ←
 HPO4 + H 3O
K
a3 →
3−
+

HPO4 2− + H 2O ←
 PO4 + H 3O
Successive
hydrolysis:
K
b1→
−2
−

PO43− + H 2O ←
 HPO4 + OH
K
b2 →
−
−

HPO4 −2 + H 2O ←
 H 2 PO4 + OH
K
b3 →
−

H 2 PO4 − + H 2O ←
 H 3 PO4 + OH
20
10
3/28/2014
Use handout on pH calculations involving
triprotic systems
21
Treatment of triprotic systems
1. H3A is treated as monoprotic weak acid. Ka1 = K1.
2. H2A- is treated as the intermediate form of a diprotic acid.
3. HA2- is also treated as the intermediate form of a diprotic
acid. However, HA2- is “surrounded” by H2A- and A3-, so the
equil. constants to use are Ka2 (= K2) and Ka3 (= K3)
4. A3- is treated as monobasic. Kb1 = Kw/Ka3
22
11
3/28/2014
What is the major species at a given pH?
1. When pH < pK1, H2A predominates
2. When pH = pK1, [H2A] = [HA-]
3. When pK1 > pH < pK2, [HA-] predominates
4. When pH = pK2, [HA-] = [A2-]
5. When pH > pK2, [A2-] predominates
Q. Which of the species above predominate at pH 6.50?
pH 4.00? pH 2.00?
2-
23
Answer: A , HA , H2A
24
12