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.
SOLUTION:
(B) Calculate the PH when:
SOLUTION:
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 H2A + 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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