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TITRATION OF WEAK ACID WITH STRONG BASE:


TITRATION OF ACETIC ACID VS SODIUM HYDROXIDE: 
ILLUSTRATIVE EXAMPLE: Give the answers of following questions when 20 ml of acetic acid (CH3COOH) is titrated with 0.10 M NaOH the (Given that Ka=2×10-5).
(A) Write out the reactions and equilibrium expression associated with Ka.
(B) Calculate the PH when:
(1) 20 ml of 0.10M CH3COOH + 0.0 ml of 0.10M NaOH 
(2) 20 ml of 0.10M CH3COOH + 5.0 ml of 0.10M NaOH
(3) 20 ml of 0.10M CH3COOH + 10 ml of 0.10M NaOH
(4) 20 ml of 0.10M CH3COOH + 15 ml of 0.10M NaOH
(5) 20 ml of 0.10M CH3COOH + 19 ml of 0.10M NaOH
(6) 20 ml of 0.10M CH3COOH + 20 ml of 0.10M NaOH
(7) 20 ml of 0.10M CH3COOH + 21 ml of 0.10M NaOH
(8) 20 ml of 0.10M CH3COOH + 25 ml of 0.10M NaOH
(9) 20 ml of 0.10M CH3COOH + 20 ml of 0.10M NaOH

SOLUTION:

(A) Write out the reactions and equilibrium expression associated with Ka.
(B) Calculate the PH when:

S.N.
Given condition
comments
PH
1
20 ml of 0.10M CH3COOH + 0.0 ml of 0.10 ml NaOH 
WA, PH=1/2(Pka-logC)
2.85
2
20 ml of 0.10M CH3COOH + 5.0 ml of 0.10 ml NaOH 
ABS, PH=PKa+ log[S]\[A]
4.22
3
20 ml of 0.10M CH3COOH + 10 ml of 0.10 ml NaOH 
BB , PH=PKa
Half of equivalent point
4.70
4
20 ml of 0.10M CH3COOH + 15 ml of 0.10 ml NaOH 
ABS, PH=PKa+ log[S]\[A]
5.17
5
20 ml of 0.10M CH3COOH + 19 ml of 0.10 ml NaOH 
ABS, PH=PKa+ log[S]\[A]
5.98
6
20 ml of 0.10M CH3COOH + 20 ml of 0.10 ml NaOH 
SH, PH=7+ 1/2(Pka-logC)
Equivalent point
8.7
7
20 ml of 0.10M CH3COOH + 21 ml of 0.10 ml NaOH 
Strong Base
11.39
8
20 ml of 0.10M CH3COOH + 25 ml of 0.10 ml NaOH 
Strong Base
12.04
9
20 ml of 0.10M CH3COOH + 30 ml of 0.10 ml NaOH 
Strong Base
12.30


(C) Sketch the titration curve for this titration.

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