Posted by Sarah on Sunday, April 6, 2008 at 10:53pm.
Calculate the pH of the following aqueous solution:
1.00 mol/L sulfuric acid, H2SO4(aq)
H2SO4 ==> H^+ + HSO4^ 100%
HSO4^ ==> H^+ + SO4^= not 100%
(H^+) = 1.00 M for the first ionization.
The second one is guided by k2.
k2 = (H^+)(HSO4^)/(HSO4^)
Plug into k2 as follows:
(H^+) = 1.00 + x
(SO4^=) = x
(HSO4^) = 1.00  x
Solve for x.
k2 = (H^+)(HSO4^)/(HSO4^)
= (1.00 + x)(1.00  x)/(1.00 x)
=Im guessing the 1.00 cancels out?
.

chemistry  DrBob222, Sunday, April 6, 2008 at 11:27pm
You've omitted k2 for H2SO4. You must look up k2 and add it to the equation. And no, the 1+x and 1x don't necessarily cancel.

chemistry  DrBob222, Sunday, April 6, 2008 at 11:34pm
If you work the quadratic (it can be done with successive approximations), the answer comes out to be (H^+) = 1.1072 which I would round to 1.11 M and take  log that for pH of about 1.97.
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