For the diprotic weak acid H2A, Ka1 = 3.5 × 10-6 and Ka2 = 6.2 × 10-9. What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
apply
H2A <-> H+ +HA-
HA- <-> H+ + A-2
Ka1 = [H+][HA-]/[H2A]
3.5*10^-6 = (x*x)/(0.08-x)
x = 5.27*10^-4
then
Ka2 = [H+][A-2] / [HA-]
Ka2 = (y*y)/(x-y)
6.2*10^-9 = y*y/(5.27*10^-4 -y)
y = 1.8*10^-6
then
Htotal = x+y = 5.27*10^-4+1.8*10^-6 = 0.0005288
pH= -log(0.0005288) = 3.276
H2A = M-x = 0.0800 -5.27*10^-4 = 0.079473 M
A-2 = y = 1.8*10^-6 M
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