CORRECT

(7.b) 0.11 M

The [H+] is due primarily to the dissociation of the first H+. There is a two step dissociation. The first has a very large Ka (a strong acid) the second has a smaller Ka and must be found in tables. This turns out to be a "common ion" problem.

H2SO4 ----> H+ + HSO4-

HSO4- <----> H+ + SO4-2

 H2SO4 ---> H+ + HSO4- Initial(I) 0.10 M 0 M 0 M Change(C) -0.10 M +0.10M +0.10 M

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 Equilibrium(E) 0 M 0.10M 0.10 M

 HSO4- <---> H+ + SO4-2 Initial(I) 0.10 M 0.10 M** 0 M Change(C) -x +x +x

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 Equilibrium (E) (0.10-x)* (0.10+x)* x

*Hint : Since [H2SO4] < 100 Ka2 then x is probably not much smaller than 0.10 so (0.10 - x) and (0.10 + x) must be used in the calculation rather than assuming that (0.10 - x) ~ 0.10.

** from the first dissociation of H2SO4

Ka2 = {[H+][SO4-2]}/[ HSO4-]

1.2 x 10-2 = {(0.10 + x)(x)}/(0.10 - x)

0.10x + x2 = 1.2 x 10-2(0.10 - x)

x2 + 0.112x - 1.2 x 10-3 = 0

x = (- 0.112 + b2 - 4ac)/2a

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x = (- 0.112 + 0.0173)/2

x = 0.01 M = [SO4-2]

OR.........

x = (0.112 + 0.132)/2

x = - 0.112(not valid)

This is the contribution of H+ from the 2nd dissociation which forms SO4-2, and must be added to the 0.10 M H+ from the first dissociation of H2SO4.