Titration of Weak acid Vs Strong base:

 (II):  Titration of Weak acid Vs Strong base:

Titration of a weak acid by strong base is a bit more complex than both acid and base are strong this is because we have consideration the equilibrium involving the weak acid and its conjugate base.

Titration Table and curve:

S.N	Acid (0.1M)	Base (0.1M )	pH	Nature	Case
1		0  ml	2.85	WA	pH=1/2(Pka –logC)
2	Volume	5 ml	4.22	ABS	pH=(Pka +log [S]/[A]
3	Taken	10 ml	4.7	ABS	pH=Pka Best Buffer
4	20.0 ml	15 ml	5.17	ABS	pH=(Pka +log [S]/[A]
5	Ka for	19 ml	5.98	ABS	pH=(Pka +log [S]/[A]
**6	CH3COOH	20 ml	8.7	SH	pH=7+1/2(Pka+logC)
7	2* 10^-10	21 ml	11.39	SB	NbVb - NaVa=NrVr
8		25 ml	12.04	SB	[OH-]= Nr
9		30 ml	12.3	SB	Strong base
10					Remains present

 

In titration of any weak acid with strong base, The pH at equivalent point will be greater than “7”

TITRATIONS OF DIPROTIC ACID WITH STRONG BASE:

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.

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 2^nd 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- 1^st 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 2^nd 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 2^nd 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

Graphical representation: 

TITRATION OF TRIPROTIC ACID WITH STRONG BASE: