1990 II 3
The pH value of a 0.02 M ethanedioic acid solution is 1.8. Calculate the concentration of each species present in the solution at 25 degree Celsius.
(Dissociation constants of ethanedioic acid at 25 degree Celsius are :
K1 = 6.5 x 10 ^(-2)
K2 = 6.1 x 10 ^(-5)
In Suggested solution guide, it says that
H2A -> H+ + HA-
at eqm a 1x 10 ^(-4) b
HA- > H+ + A2-
at eqm b 1x 10 ^(-4) c
then [H2A]eqm+[HA-]eqm+[A2-]eqm = 0.02 M
1 x 10 ^(-4)/6.5 x 10 ^(-2)b +b + 6.1 x 10 ^(-5)/1 x 10 ^(-4)b = 0.02M
Why [H+]eqm = 1x 10 ^(-4)?
Why [H2A]eqm+[HA-]eqm+[A2-]eqm = 0.02 M?
Please help me! Please!