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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.

TITRATION OF STRONG ACID WITH STRONG BASE:


The titration of HCl (aq) with a standardized NaOH solution illustrated the titration of strong acid by a strong base.
The molecular and net ionic equation is.
Case (1): At the start point before any titrant has been added the receiving flask contains only 0.10 M HCl and 50 ml. Because it is strong acid so
Case (2): After starting but before equivalent point.
Case (3): At equivalent point
Case (4): Before equivalent point
TITRATION SUMMARY TABLE:

S.N.
Volume of
HCl Taken
Volume
of NaOH
PH

1
50.0 ml (In ml)
And 0.10 M
0.0 (In ml)
And 0.10M
1.0

2

10
1.17

3

20
1.36

4

30
1.60
NAVA>NBVB
5

40
1.95

6
(Vertical Over)
45
2.27

7

49
2.99

8
50.0 ml (In ml)
And 0.10 M
50
7.0
NAVA=NBVB
9

51
11
NAVA<NBVB
10

60
11.95























GRAPHICAL REPRESENTATION:
ILLUSTRATIVE EXAMPLE: Find the pH of following titrations:
(A) 500 ml, 0.10 M HCl + 500 ml 0.10 M Ca(OH)2
(B) 400 ml, M/200 Ca(OH)2 + 400 ml M/50 HNO
 ANSWERS KEY:
(A): PH=12.6989  (B): PH=2.6

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