The first ionization is 100%; therefore,
..........H2SO4 ==> H^+ + HSO4^-
I.........0.150......0.....0
C........-0.l50...0.150..0.150
The second H is weak.
............HSO4^- ==> H^+ + SO4^2-
I.........0.150......0.150..0
C...........-x.........+x...x
E........0.150-x....0.150+x..x
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Substitute the E line of the second ICE chart and solve for x, then
total (H^+) = 0.150+x
(SO4^2-) = x
(HSO4^-) = 0.150-x
Of course (H2SO4) = 0
1- Calculate the equilibrium concentration for H+, HSO4- and SO4^2- , given a .150M H2SO4 solution. the Ka for HSO4- is 1.20*10^-2
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