Welcome to ChemZipper !!!: TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

## Search This Blog

### 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>>Ka3, hence x >>y>>Z   so y and z can be neglected with respect to x   it mean total OHions count from only first step OH- ions coming from 2nd and 3rd  hydrolysis is ignored

Special Case (1):
Special Case (2)
This is the quadratic equation solve by following formulae
So approximation not valid hence follows 2nd 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: