Question

Calculate the pH at 25 degrees Celcius of a 0.35M aqueous solution of phosphoric acid (H3PO4). (Ka1, Ka2, and Ka3 for phosphoric acid are 7.5x10^-3, 6.25x10^-8, and 4.8x10^-13, respectively.)

Answer 1

Reactions taking place are :

H₃PO₄ ----> H⁺ + H₂PO₄^- ----(I)

H₂PO₄^- ----> H⁺ + HPO₄^2- ----(II)

HPO₄^2- ----> H⁺ + PO₄^3- ---(III)

The H⁺ ions which will be obtained due to reactions (II) and (III) are extremely small, hence they can be easily neglected.

Thus, the acidity of solution is due to the H⁺ ions coming from (I) reaction

For reaction (I) :

                       H₃PO₄^- ----> H⁺ + H₂PO₄^-

Initial                  0.35            0        0

Equil.               0.35-x            x        x

From the above reaction, we get :

K_a1 = x²/(0.35-x) = 7.5*10^-3

Solving for the +ve value of x, we get :

x = 0.0474

Thus, [H⁺] = x = 0.0474 M

So, pH = -log [H⁺] = 1.32