J.R. S. answered 10/26/17
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In order to solve this problem, one needs to know the Ka for H2SO4 (Ka1 and Ka2) and the Ka for H2SO3 (Ka1 and Ka2).
Looking up these values I have ...
H2SO4 ==> H+ + HSO4- Ka1 = large
HSO4- ==>H+ SO42- Ka2 = 1.2x10-2
H2SO3 ==> H+ + HSO3- Ka1 = 1.2x10-2
HSO3- ==> H+ + SO32- Ka2 = 6.6x10-8
0.55 M H2SO4 contains a higher concentration of H3O+ than 0.55 M H2SO3
a) Kb of conjugate base of H2SO3 is the Kb for HSO3- which is 1x10-14/Ka1 = 1x10-14/1.2x10-2 = 8.33x10-11
b) HSO3- ==> H+ + SO32-
Ka2 = 6.6x10-8 = [H+][SO32-]/[HSO3-] = (x)(x)/0.20
x2 = 1.32x10-8
x = 1.15x10-4 M = [H+]
pH = -log [H+] = -log 1.15x10-4
pH = 3.94