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 = largeHSO4

^{-}==>H^{+}SO4^{2-}Ka2 = 1.2x10^{-2}H2SO3 ==> H

^{+}+ HSO3^{-}Ka1 = 1.2x10^{-2}HSO3

^{-}==> H^{+}+ SO3^{2-}Ka2 = 6.6x10^{-8}0.55 M H2SO4 contains a higher concentration of H3O

^{+}than 0.55 M H2SO3a) K

_{b}of conjugate base of H2SO3 is the K_{b}for HSO3^{-}which is 1x10^{-14}/Ka1 = 1x10^{-14}/1.2x10^{-2}= 8.33x10^{-11}b) HSO3

^{-}==> H^{+}+ SO3^{2-}Ka

_{2}= 6.6x10^{-8 }= [H^{+}][SO3^{2-}]/[HSO3^{-}] = (x)(x)/0.20x

^{2}= 1.32x10^{-8}x = 1.15x10

^{-4}M = [H^{+}]pH = -log [H

^{+}] = -log 1.15x10^{-4}pH = 3.94