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?
.
2 answers
You've omitted k2 for H2SO4. You must look up k2 and add it to the equation. And no, the 1+x and 1-x don't necessarily cancel.
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.
3 answers