Question

Calculate the pH of a 5.0 M H₃PO₄ solution and the equilibrium concentrations of the species H₃PO₄, H₂PO₄^-, HPO₄^2- and PO₄^3- (ka1 = 7.5 x 10^-3, ka2 = 6.2 x 10^-8, ka3 = 4.8 x 10^-13)

Answer 1

Sol.

Major contribution of [H+] is from the first ionization of acid and contributions from second and third ionizations of acid are negligible . So , pH is calculated only from the first ionization of acid .

Ist ionization reaction :

H3PO4 <---> H+ + H2PO4-

initial 5.0 0 0

change - x + x + x

equilibrium 5.0 - x x x

So , Ka1 = [H+] [H2PO4-] / [H3PO4]  

7.5 × 10^-3 = 0.0075 = x² / ( 5.0 - x )   

x² + 0.0075x - 0.0375 = 0

Solving this ,  

x = ( - 0.0075 +- ( 0.0075 × 0.0075 - 4 × 1 × - 0.0375 )^1/2 ) / 2

x = ( - 0.0075 +- 0.3874 ) / 2  

x = ( - 0.0075 + 0.3874 ) / 2 = 0.1899

or , x = ( - 0.0075 - 0.3874 ) / 2 = - 0.1974

But x cannot be negative  

So , [H+] = x = 0.1899 M

And , pH = - log[H+] = - log(0.1899) =   0.72  

Also , Equilibrium Concentration of H2PO4- = [H2PO4-] = x = 0.1899 M

and , Equilibrium Concentration of H3PO4 = [H3PO4] = 5.0 - x = 5.0 - 0.1899 =   4.8101 M  

Now , 2nd ionization :

H2PO4- <---> H+ + HPO42-

initial 0.1899 0.1899 0

change - x + x + x

equilibrium 0.1899 - x 0.1899 + x x

So , Ka2 = [H+] [HPO42-] / [H2PO4-]   

6.2 × 10^-8 = (0.1899 + x ) x / ( 0.1899 - x )

As Ka2 is very small , so , 0.1899 + x = 0.1899 - x = 0.1899

and , x = 6.2 × 10^-8  

Therefore , Equilibrium Concentration of HPO42- = [HPO42-] = 6.2 × 10^-8 M  

Now , 3rd ionization :     

HPO42- <---> H+ + PO43-

initial 6.2 × 10^-8 0.1899 0

change - x + x + x

equilibrium 6.2 × 10^-8  - x 0.1899 + x x

So , Ka3 = [H+] [PO43-] / [HPO42-]   

4.8 × 10^-13 = (0.1899 + x ) x / ( 6.2 × 10^-8 - x )

As Ka3 is very small , so , 0.1899 + x = 0.1899   and 6.2 × 10^-8 - x = 6.2 × 10^-8

So , x = 4.8 × 10^-13 × 6.2 × 10^-8 / 0.1899 = 1.6 × 10^-19

Therefore , Equilibrium Concentration of PO43- = [PO43-] = 1.6 × 10^-19    M