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.