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 HNO₃ 

 _{ANSWERS KEY:
(A): PH=12.6989  (B): PH=2.6}

Basicity of Guanidine :

Guanidine is strongest organic nitrogenous compound with the formula HNC(NH₂)₂. Guanidine is analogue of carbonic acid. That is, the C=O group in carbonic acid is replaced by a C=NH group, and each OH is replaced by a NH₂ group.

A guanidine group also appears in larger organic molecules, including on the side chain of arginine (a basic amino acid).

It is a colourless solid that dissolves in polar solvents. It is a strong base that is used in the production of plastics and explosives. It is found in urine as a normal product of protein metabolism.

Basicity of Guanidine:

Guanidine is the strongest base among neutral compounds:

The remarkable basicity of guanidine is attributed to the fact that the positive charge on

the guanidinium ion is delocalized equally over the three nitrogen atoms, as shown by

these three equivalent resonating  structures:

Basicity of nitrogen can be increased by attachment to pi-donors (NH₂) group. These two pi-donating NH₂ groups donate electron density to the (pi-accepting) C=NH. Hence, the guanidinium ion is a highly stable cation.
IIT UPDATE:
QUESTION:

SOLUTION: (B)

Related Questions:

1. How is base strength related to the availabihty of the electron-pair?
2. Amines are more basic than ammonia why?
3. What is relative basic strength order 1° amines , 2°amines and 3° amines ? Explain:
4. Give an explanation for the fact that Guanidine NH=C(CH3)2 is a stronger base than most of amines?
5. Arrange in correct order of basic Character of aniline, pyrrol, pyridine and piperidine?
6. What is correct basicity order of pyridine, pyridazine, pyrimidine and pyrazine ?
7. Give an explanation for the fact that Guanidine NH=C(CH3)2 is a stronger base than most of amines?
8. Arrange in correct order of basic Character of aniline, pyrrol, pyridine and piperidine?
9. What is correct basicity order of pyridine, pyridazine, pyrimidine and pyrazine ?
10. Why pyridine is more basic than Pyrrole?
11. Why pyrimidine is less basic than pyridine?
12. Imidazole is more basic than pyridine? Why?
13. Biological Important of Imidazole and structure:
14. Pyridine is almost 1 million times less basic than piperidine? Why?
15. Cyclohexylamine amine is the stronger base than Aniline? Why?
16. Tetrahydroquinoline amine is the stronger base than Tetrahydroisoquinoline? Why?
17. Arrange the following in the order of increasing basicity : p-Toluidine, N, N-Dimethyl-p-toluidine, p-Nitroaniline, Aniline. (I.I.T.1986)
18. Arrange the following in the increasing order of their acid strength : Methyl amine, Dimethyl amine, Aniline, N-methyl aniline (I.I.T, 1988).

Nucleophilic Aromatic Substitution Reactions : (SN-Ar):

(1) SN-Ar-(Substitution of hydrogen of benzene):

(2) Substitution of group other than hydrogen:

(A) SN1 (Aromatic) Type:

(B) SN2 (Aromatic) Type:

      SN2-Ar Mechanism: Meisenheimer complex Intermediate:

 (3) Substitution of unactivated hydrogen:

      SN-Ar Mechanism: Benzyne Intermediate:

(3) Substitution of unactivated hydrogen: Benzyne Intermediate Mechanism:

SN-Ar-Elimination-Addition:  Benzyne Intermediate Mechanism:

An aryl halide can undergo a nucleophilic substitutoin reaction in the presence of a very strongbase such as NH2 When chlorobenzene – that has the carbon to which chlorine is attached isotopically labeled with Cabon-14 –is treated with amide ion in liquid ammonia,  aniline is obtained as a product. Half of the product has the amino group attached to the isotopically labelled carbon (14) as expected, but the other half has the amino group attached to the carbon adjacent to the labelled carbon.

The mechanisms that accounts for the experimental observations involves formation of a benzene intermediate which has two equivalent carbon atoms to which amino group can be attached. Benzyne has an extra (Pi) bond between two adjacent carbon atoms of benzene and can be formed as

Step-(1): Strong base NH₂- removes a proton from the position ortho to halogen:

Step-(2): Anion formed in step (1) eliminates the halide ion, thereby forming Benzyne:

The incoming nucleophile can attack either of the carbons of the “triple bond” of benzyne. Protonation of the resulting anion form the substitution product. The overall reaction is an elimination-addition reaction; benzyne is formed in an elimination reaction and immediately undergoes an addition reaction.

Substitution at the carbon (C-14) that was attached to the leaving group is called direct substitution product (DSP). Substitution at the adjacent labeled carbon (C-14) of is called cine substitution product (CSP).

Characteristic of Benzyne reaction:

(1) It is a S_N-E_A reaction  proceed via elimination –Addition Nucleophilic Substitution reaction.

(2) Aryl halide + strong base required

(3) Presence of beta hydrogen must be needed.

(4) NaNH₂ and KNH₂ can be considered to be essentially the same for our purposes

(5) For symmetrical intermediate which is attacked equally on either side. so that the roughly   50:50 ratio of products are obtained .

Benzyne Intermediate Mechanism: SN-Ar Reaction:

The mechanisms that accounts for the experimental observations involves formation of a benzene intermediate which has two equivalent carbon atoms to which amino group can be attached. Benzyne has an extra (Pi) bond between two adjacent carbon atoms of benzene and can be formed as

Step-(1): Strong base NH₂- removes a proton from the position ortho to halogen:

Step-(2): Anion formed in step (1) eliminates the halide ion, thereby forming Benzyne:

The incoming nucleophile can attack either of the carbons of the “triple bond” of benzyne. Protonation of the resulting anion form the substitution product. The overall reaction is an elimination-addition reaction; benzyne is formed in an elimination reaction and immediately undergoes an addition reaction.

Substitution at the carbon (C-14) that was attached to the leaving group is called direct substitution product (DSP). Substitution at the adjacent labeled carbon (C-14) of is called cine substitution product (CSP).