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TITRATIONS OF DIPROTIC ACID WITH STRONG BASE:

DIPROTIC ACID  Vs NaOH:  [H2CO3 and Oxalic acid H2A]
ILLUSTRATIVE EXAMPLE: Give the answers of following questions when 100 ml of Malonic acid is titrated with 0.10 M NaOH the (Given that Ka1=1.5×10-3 and Ka2=2.0×10-6 for Malonic acid HOOC-CH2-COOH represented as H2A ).
(A) Write out the reactions and equilibrium expression associated with Ka1 and Ka2.
(B) Calculate the PH when:
(1)100 ml of 0.10M H2A + 0.0 ml of 0.10 ml NaOH 
(2) 100 ml of 0.10M H2A+ 50 ml of 0.10 ml NaOH
(3) 100 ml of 0.10M H2A+ 100 ml of 0.10 ml NaOH
(4) 100 ml of 0.10M H2A+ 150 ml of 0.10 ml NaOH
(5) 100 ml of 0.10M H2A + 200 ml of 0.10 ml NaOH
(C) Sketch the titration curve for this titration.

(A) Write out the reactions and equilibrium expression associated with Ka1 and Ka2
SOLUTION:

(B) Calculate the PH when:

SOLUTION:
 (1) At Point-(1):100 ml of 0.10 M H2A + 0.0 ml of 0.10 ml NaOH 
Note: Ignore the amount of [H+] coming from 2nd dissociation of acid 

(2) At Point-(2): 100 ml of 0.10 M H2A + 50 ml of 0.10 ml NaOH 
At Point -2- is the half of the first equivalent point   where [H2A=HA-]   hence pH of solution is due to best buffer
(3) At Point-(3):100 ml of 0.10M H2+ 100 ml of 0.10 ml NaOH]
At point-3- 1st equivalent point occurs and here major species is HA- which is act as both acid and base (Amphoteric species).
(4) At Point-(4): 100 ml of 0.10 M H2A + 150 ml of 0.10 ml NaOH
At Point -4- is the half of the 2nd equivalent point   where [HA- = A-2]   hence pH of solution is due to best buffer .
(5) At Point-(5):100 ml of 0.10 M H2A + 200 ml of 0.10 ml NaOH
At point-5- It is 2nd equivalent point there is only A-2 present which act as weak base and undergo polyvalent anion  hydrolysis;
(C) Sketch the titration curve for the titration point of question (B) titration.


S.N.
Given condition
comments
PH
1
100 ml of 0.1M H2A + 0.0 ml of 0.1 ml NaOH 
H2A Diprotic acid
1.94
2
100 ml of 0.1M H2A + 50 ml of 0.1 ml NaOH 
Acidic Best buffer
PH= Pka1 (1st half E-Point)
2.82
3
100 ml of 0.1M H2A + 100 ml of 0.1 ml NaOH 
Amphoteric salt Hydrolysis
PH=1/2(Pka1+Pka2)
4.26
4
100 ml of 0.1M H2A + 150 ml of 0.1 ml NaOH
Acidic Best buffer
PH= Pka2 (2st half E-Point)
5.70
5
100 ml of 0.1M H2A + 200 ml of 0.1 ml NaOH
Anionic salt Hydrolysis
9.11

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