What is the ph of a 0.40 m h2se solution that has the stepwise dissociation constants k a1 = 1.3 × 10-4 and k a2 = 1.0 × 10-11?

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