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Chapter 9: POLYPROTIC
ACID-BASE EQUILIBRIA
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Oxalic acid,
rhubarb and
rhubarb pie
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Understanding Polyprotic Species
Diprotic acids, H2A vs. Dibasic species, A2-
Triprotic acids, H3A vs. Tribasic species, A3-
Ex. H2SO4, H2CO3 Ex. SO42-, CO3
2-
Ex. H3PO4, Ex. PO43-
H3C6H5O7 (Citric acid) C6H5O73-
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� Can donate 2 H+ � Can accept 2 H+
� Can donate 3 H+ � Can accept 3 H+
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Diprotic Acids and Bases
General formulas:
� H2A = fully acidic form
� HA- = intermediate form; amphoteric
� A2- = fully basic or fully deprotonated form
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Equilibria involved: Diprotic Acids and Bases
- +
2 2 3
a1H A + H O HA + H O
K→←First dissociation:
Second dissociation:
Diprotic Acid, H2A
Dibasic species, A2-
2- - -
2
b1A + H O HA + OH
K→←First hydrolysis:
- -
2 2
b2HA + H O H A + OH
K→←Second hydrolysis:
Q. How do we calculate Kb1 and Kb2 from Ka values?
- -2 +
2 3
a2HA + H O A + H O
K→←
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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:
2 2 3
1aH A H O HA H O
K− +→+ +←
2 2
2bHA H O H A OH
K− −→+ +←
+
2 32
wH O H O OH
K+ −→ +← Kw = Ka1 x Kb2
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Polyprotic acids: Amino Acids
NOTE: - COOH group is much more acidic (higher Ka; first to
dissociate) than the –NH3+ group.
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Diprotic AcidsExample: Leucine, H2L
Stepwise dissociation:
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Fully protonated form = fully acidic, H2A
+
Fully dissociated form = fully basic, A-
� Start with the fully acidic form, H2A+ = H2L
+
Dibasic Species
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Stepwise hydrolysis of leucine:
� Start with the fully basic form, A- = L-
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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.
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pH Calculation: Diprotic Acids and Bases
2 2 3
1aH A H O HA H O
K− +→+ +←
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]
Equil: F-x x x
2
1a
xK
F x=
−
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Fully acidic form (H2A) – Cont.
2 3 2 3 3
1aH SO H O HSO H O
K− +→+ +←
Problem (a): Find the pH and [H2SO3], [HSO3-] and [SO3
2-]
in a 0.050 M H2SO3 solution. Ka1 = 1.23 x 10-2; Ka2 = 6.6 x 10-8
Equil: 0.050-x x x
2
1
2
x1.23 10(0.050 )
aK
x
x−
= =
−
x cannot be ignored
since Ka1 isn’t too small
2 2 4x x1.23 10 6.15 10x x
− −
+ − Solve for x using
quadratic equation2
3 3x1.94 10 [ ] [ ]x M H O HSO
− + −
= = =
0.050 M - x
1.71pH = 2 3[ ] 0.031H SO M=
2
3x[ ] 1.9 10HSO M
− −
=
2
3[ ] 0SO M
−
�11
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pH Calculations: Diprotic systems – Cont.
2
2
1bA H O HA OH
K− − −→+ +←
2. The fully basic form, A2-
Approximation: 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]
Equil: F-y y y
2
1b
yK
F y=
−
pOH = -log (y)pH = 14 - pOH
Recall: Kb1 = Kw/Ka214
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Problem: Find the pH and concentrations of H2SO3, HSO3- and
SO32- in each of the following solutions:
(a) 0.050 M H2SO3 - DONE!
(b) 0.050 M NaHSO3, and
√ (c) 0.050 M Na2SO3
Ka1 = 1.23 x 10-2; Ka2 = 6.6 x 10-8
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.
2 2
2bH O H A OH
KHA −− →+ +←
3. The intermediate (amphoteric) form, HA-
Q. What is the predominant species in a solution of HA-?
Compare Ka2 and Kb2 equilibria:
� HA- can act as an acid or a base
2
2 3
2aH O A H O
KHA − +− →+ +←Dissociation:
Hydrolysis:
� HA- will dissociate/hydrolyze to form A2- and H2A
Approximation: [HA-] ≈ FHA- = FNaHA or FKHA 16
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The intermediate form, HA- (Cont.)
Calculation of pH and [species]
pH = -log [H+]
WhereK1= Ka1
K2 = Ka2
F = FHA-
Quick check: pH = ½ (pK1 + pK2)
� Solve for [H2A] and [A2-] using [H+] above and K1 & K2
equilibria
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The intermediate form, HA- (Cont.)
� Solving for [H2A] and [A2-]:
1 1
2
[ ][
[ ]
]a
H HAK
H AK
+ −
= =
2
1
[[ ]
][ ]H HATh s Au H
K
+ −
=
2 2[ ][ ]
,[ ]
K HA
ALikewise
H
−
−
+=
2 2
2
2 :[ ]
[
[ ]
]a
From nd dissociationH
K KA
HA
+ −
−= =
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Problem: Find the pH and concentrations of H2SO3, HSO3- and
SO32- in each of the following solutions:
(a) 0.050 M H2SO3 - DONE!
√(b) 0.050 M NaHSO3, and
(c) 0.050 M Na2SO3 – DONE!
Ka1 = 1.23 x 10-2; Ka2 = 6.6 x 10-8
Answer: pH =4.59; [HSO3-] ≈ 0.050 M, [H2SO3] = 1.1 x 10-4 M; [SO3
2-] = 1.3 x 10-4 M 19
Triprotic Acids and BasesExample: H3PO4; PO4
3-
3 4 2 2 4 3
1aH PO H O H PO H O
K− +→+ +←
2
2 4 2 4 3
2aH PO H O HPO H O
K− − +→+ +←
2 3
4 2 4 3
3aHPO H O PO H O
K− − +→+ +←
Successive
dissociation:
Successive
hydrolysis:3 2
4 2 4
1bPO H O HPO OH
K− − −→+ +←
2
4 2 2 4
2bHPO H O H PO OH
K− − −→+ +←
2 4 2 3 4
3bH PO H O H PO OH
K− −→+ +← 20
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Use handout on pH calculations involving triprotic systems
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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/Ka322
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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? Answer: A2-, HA-, H2A
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