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:

ACID-BASE TITRATION:

THEORY OF INDICATOR'S:

It is method to determine concentration of any Solution with the help of a solution of known concentration.
EQUIVALENT POINT:
When the complete reaction occurs between the solutions, it is called equivalent point of titration.
END POINT: 
When sudden change in colour of solution occurs, it is called end point of titration.
INDICATORS:
(1) They are very weak organic acids or bases that show different colours at different pH value.
(2) They exist in ionized and unionized form which have different colours .
(3) The point at which Number of equivalent of acid and base become equal is known as equivalent point / Stoichiometric point / neutralization point.
(4) We assure that indicators do affect the pH of Solution.
(5) Around the equivalent point pH change the drastic change.
(6) For the successful titration end point should be as close as possible to the equivalent point.
TYPE OF INDICATOR'S:
(1) ACIDIC INDICATOR'S
(2) BASIC INDICATOR'S

(1) ACIDIC INDICATOR'S:

Phenophthalene (HPh): It is weak acid and can represented as [HPh] If it ionised gives [H⁺ ] and [Ph^- ] 

Addition of [H⁺]: In addition of an acid the ionization of HPh is practically negligible and as the equilibrium shift left hand side due to high concentration of [H⁺] ion thus solution would remain colourless.

Addition of [OH^- ]: In the presence of a base H⁺]^[ ions  are removed by [OH^-] ions in the in the form of water molecules and the above equilibrium shift to right hand side .Thus the concentration of Ph- ions increases in solution and they impart a pink colours to the solution.

INDICATOR THEORY: Let consider acidic indicators [HPh] Phenophthalene. An indicator has two colouring parts.

(1) Unionized part of indicators 

(2) Ionized part of indicators

The relative concentration of these species will depend upon PH of medium .

Take negative log both sides

CASE (1): It is observed that 

Then colour of indicator decided by concentration of [Ph^-)]

CASE (2): It is observed that 

Then colour of indicator decided by concentration of [HPh]

PH RANGE OF INDICATORS:

The given indicators work between a pH range i.e.

Then working of indicators is best;

ILLUSTRATIVE EXAMPLE (1):  For an acidic indicator, dissociation constant, K_in is 2×10^-6 Calculate pH range of indicator.

SOLUTION:  

ILLUSTRATIVE EXAMPLE (2): The pH range of a basic indicator is 4 to 6.5 Calculate the dissociation constant of indicator?.

SOLUTION:  pK_In must be midpoint of pH range for acidic indicators and pOH range for basic indicators

The pH range = 4 to 6.5 so pOH range is 10 to 7.5

Hence Pk_In =   (10+7.5) /2 = 8.75 

ILLUSTRATIVE EXAMPLE (3):  For an indicator pKa is 6 Calculate pH of Solution having this indicator such that 40% indicator molecules remain in ionised form.

SOLUTION:  We know that  

(2) BASIC INDICATOR'S

Methyl orange (MeOH): It is weak base and can represented an MeOH If it ionised gives [ Me⁺] and[ OH^- ]  and methyl orange is an intensely coloured indicator that is red below pH 3.1 and orange-yellow above pH 4.4

The red (acid) form has an [H+] attached to one of the N atoms and the yellow (basic) form has lost the [H+]

Addition of [H⁺ ] ^: In the presence of an acid, OH^- ions are removed in the form of water molecules and the above equilibrium shift to right hand side .This effect Me⁺ ions are produced which impart red coloure to the Solution.

Addition of [OH^- ]^: In addition of alkali the concentration of OH^- increases in the solution and equilibrium shift left hand side i.e. the ionization of MeOH is practically negligible .thus solution acquired the colour of unionised methyl orange molecules i.e. yellow

DEGREE OF DISSOCISTION OF INDICATOR'S (DOD): Consider a general indicator dissociation;

ILLUSTRATIVE EXAMPLE (4):  For acidic indicator, pH range is 3 to 4.6 calculate the ratio of[ In^- ] and H⁺ for the appearance of solution in a single colour.

SOLUTION:

Given pH range 3.0 to 4.6 so pK_In= (3.0+ 4.6)/2 =3.6

ILLUSTRATIVE EXAMPLE (5): An indicator with Ka = 10^-5 is solution with pH = 6 Calculate % of indicator in ionised form?

SOLUTION:

ILLUSTRATIVE EXAMPLE (6): The pH of at which an acid indicator with Ka is 10^-15 changes colour when indicator concentration 1×10^-5 M is? 

SOLUTION:

TYPE OF TITRATION:

(A) ACID-BASE TITRATION: 

(A-1): Strong acid Vs Strong base:

(A-2): Weak acid Vs Strong base:
(A-3): Weak base Vs Strong base:
(A-4): Weak acid Vs Weak base: weak acid and weak base titration can not be carried out because due to very low PH change , their is no suitable indicator for this titration.
(A-5); Salt of SB and WA Vs Strong acid:

(A-6); Double Indicators Titration:

(A-7): Back Titration:
(B):TITRATION OF POLYPROTIC ACIDS:
(1) Titration of diprotic acid with strong base:
(2) Titration of triprotic acid with strong base:

(C) REDOX TITRATION:

TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

TRIPROTIC Vs NaOH: [H₃PO₄ Vs NaOH]
ILLUSTRATIVE EXAMPLE: Phosphoric acid (H₃PO₄), is a Triprotic acid with Ka₁= 10^-3 , Ka₂=10^-8 and Ka₃= 10^-12 .Consider the titration of 0.10M 100 ml H₃PO4 with 0.1M NaOH Solution and answers the following questions.

(A) Write out the reactions associated with Ka₁, Ka₂ and Ka₃.

(B) Calculate the pH after the following titration points:

(1) 100 ml of 0.1M H₃PO₄ + 0.0 ml of 0.1 ml NaOH
(2) 100 ml of 0.1M H₃PO₄ + 50 ml of 0.1 ml NaOH
(3) 100 ml of 0.1M H₃PO₄ + 100 ml of 0.1 ml NaOH
(4) 100 ml of 0.1M H₃PO₄ + 150 ml of 0.1 ml NaOH
(5) 100 ml of 0.1M H₃PO₄ + 200 ml of 0.1 ml NaOH
(6) 100 ml of 0.1M H₃PO₄ + 250 ml of 0.1 ml NaOH
(7) 100 ml of 0.1M H₃PO₄ + 300 ml of 0.1 ml NaOH 

(C) Sketch the titration curve for this titration.

(D) What weak acid and what conjugate base make the best phosphate Buffer at pH ~7.0?

(A) Write out the reactions associated with Ka₁, Ka₂ and Ka₃.

 SOLUTION:

Step-1    

Step-2  

 Step-3    

(B)  Calculate the pH after the following titration points;

(1) 100 ml of 0.1M H₃PO₄ + 0.0 ml of 0.1 ml NaOH 

SOLUTION:

By approximation we know that if

(2) 100 ml of 0.1M H₃PO₄ + 50 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100= 10 and base 0.1x50 = 5.0

Here H₃PO₄/NaH₂PO₄ remain same in solution and which are act acidic buffer

(3) 100 ml of 0.1M H₃PO₄ + 100 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x100 = 10

Here only NaH₂PO₄ remain same in solution and which are under go amphoteric hydrolysis. PH of amphoteric salts is independent of concentration of salt.

(4) 100 ml of 0.1M H₃PO₄ + 150 ml of 0.1 ml NaOH

SOLUTION:

Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x150 = 15

Here only NaOH (5 millimoles) and NaH₂PO₄ both are remain present in solution hence which are further go titration.

Here NaH₂PO₄ and Na₂HPO₄ remain same in solution and which are act acidic buffer

(5) 100 ml of 0.1M H₃PO₄ + 200 ml of 0.1 ml NaOH

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 200 = 20

Here only NaOH (10 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here only Na₂HPO₄ remain same in solution and which are under go amphoteric hydrolysis. pH of amphoteric salts is independent of concentration of salt.

(6) 100 ml of 0.1M H₃PO₄ + 250 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 250 = 25

Here NaOH (15 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here Na₂HPO₄ (10 millimoles) and (5.0millimoles) both remain same in solution and which are further go titration.

Here Na₂HPO₄ and Na₃PO₄ remain same in solution and which are act acidic buffer

(7) 100 ml of 0.1M H₃PO₄ + 300 ml of 0.1 ml NaOH 

SOLUTION:
Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x 300 = 30

Here NaOH (20 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here NaOH (10 millimoles) and Na₂HPO₄ (10 millimoles) both remain same in solution and which are further go titration.

Finally 10 millimoles (Molarity=10/400) N₃PO₄ is formed which undergo polyvalent salt hydrolysis 

HYDROLYSIS OF ANION (PO₄^-3):

Step wise illustration of hydrolysis of poly basic acids and polyacidic base is given as -

Na+ ion do not under goes hydrolysis while PO₄^-3 undergoes step wise hydrolysis

Experimentally we know that   Ka₁>>Ka₂>>Ka_3, hence x >>y>>Z   so y and z can be neglected with respect to x   it mean total OH^-  ions count from only first step OH- ions coming from 2^nd and 3^rd  hydrolysis is ignored 

Special Case (1):  

Special Case (2) 

This is the quadratic equation solve by following formulae

So approximation not valid hence follows 2^nd case

(C) Sketch the titration curve for this titration:

SOLUTION:    

S.N.	Given condition	comments	PH
1	100 ml of 0.1M H3PO4 + 0.0 ml of 0.1 ml NaOH	H3PO4 Polyprotic acid	02.02
2	100 ml of 0.1M H3PO4 + 50 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka1 (1st half E-Point)	03.00
3	100 ml of 0.1M H3PO4 + 100 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka1+Pka2)	05.50
4	100 ml of 0.1M H3PO4 + 150 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka2 (2st half E-Point)	08.00
5	100 ml of 0.1M H3PO4 + 200 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka2+Pka3)	10.00
6	100 ml of 0.1M H3PO4 + 250 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka3 (3st half E-Point)	12.00
7	100 ml of 0.1M H3PO4 + 300 ml of 0.1 ml NaOH	Poly anionic salt Hydrolysis 3rd Equivalent point	12.07

 Graphical representation of titration: