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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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