Question

What is the pH of a 2:1 mixture of H2PO4– to HPO42– at 0.050 M ionic strength? (See the table for the activity coefficients.)

Answer 1

The equation of Ka with ionic strength is given by

Ka = a[R^-].a[H⁺]/a[RH] where a[X] is effective concentration, i.e activity of X.
a[X] = activity coefficient of X * [X]

Ionic strength is given by the table

H2PO4- is got from the equation

H3PO4 H+ + H2PO4-

H2PO4- H+ + HPO42-

HPO42- = 1/2 H2PO4-

Ionic strength =1/2( H2PO4– + 1/2 x H2PO42– x (Charge of HPO42-)² + H+ equal to H2PO42- + 1/2 x H+ equal to H2PO42-)

Ionic strength =1/2( H2PO4– + 1/2 x H2PO42– x (2)² + H+ equal to H2PO42- + 1/2 x H+ equal to H2PO42-)

0.1 M = 4.5 x H2PO4-

H2PO4- = 0.0222M

HPO42- = 1/2 H2PO4-

HPO42- = 1/2 x 0.0222

HPO42- = 1/2 H2PO4-

HPO42- = 0.0111M

From this we can calculate the concentration of [H+]

[H+] = 0.0111M + 0.0222M = 0.0333 M

pH = -log [H+]

pH = - log (0.0333M)

pH = 1.477