What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?

When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

          6.86 = -㏒ Ka 

∴Ka = 1.38 x 10^-7

by using ICE table:

               H2PO4- →  H+  + HPO4
initial      0.4 m            0         0

change     -X                +X       +X

Equ       (0.4-X)               X          X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X)   by solving for X

∴X = 2.3x 10^-4 

∴[H+] = X = 2.3 x 10^-4

∴PH = -㏒[H+]

        = -㏒ (2.3 x 10^-4)
 ∴PH  =  3.6