The pH of a 0.40 m h2se solution that has the stepwise dissociation constants is 2.15.
First, we can neglect the Ka2 value and use Ka1:
so,
Ka1 = [H+][HSe-] / [H2Se]
so by substitution:
1.3 x 1/10⁴ = X*X / (0.4 -X) by solving for X
∴X = 0.00715
∴[H+] = 0.00715
∴pH = -㏒[H+]
= -㏒ 0.00715
= 2.15
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