first, we can neglect the Ka2 value and use Ka1:
according to this equation and by using the ICE table:
H2Se ⇄ H+ +HSe-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X
so,
Ka1 = [H+][HSe-] / [H2Se]
so by substitution:
1.3 x 10^-4 = X*X / (0.4 -X) by solving for X
∴X = 0.00715
∴[H+] = 0.00715 m
∴PH = -㏒[H+]
= -㏒ 0.00715
= 2.15
