Chemistry pH problem?

Ok, calculate the [SO4^2-] in .20M H2SO4

I assumed that since I have H2, I would multiply .2M by 2 then get the pH. But its not working out very well. Can anyone tell me what Im doing wrong?

Answer:
We assume 100 % dissociation of H2SO4 >> H+ + HSO4-.
This means that the 0.2 M H2SO4 can be regarded as 0.2 M H+ and 0.2 M HSO4-.
Let x = moles/L HSO4- that dissociate.This leaves 0.2-x moles/L HSO4- undissociated and produce x moles/L SO42- while ading x moles/L H+ to yhe 0.2 M arleady there from the first dissociation
HSO4- <> H+ + SO42-
initial concentration
0.2
at equilibrium
0.2-x . . . . .0.2+x . .. . x
K = 1.26 x 10^-2 = ( 0.2+x)x/ 0.2 -x
x = 0.0113
[HSO4-] = 0.2 - 0.0113 = 0.189 M
[H+] = 0.2 + 0.0113 = 0.211 M >> pH = 0.675
[SO42-] = 0.0113 M

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