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