What is eq. concen. of sulfite ion, (SO3)2-, in 0.45 M H2SO3 soln? (Ka1=1.2x10-2 and Ka2=6.2x10-8)?

Question

Lecturer notes state that the answer is [(SO3)2-] = Ka2.

Why isn't the eq. concentration of (SO3)2- related to initial concentration of H2SO3?

Answer 1

There must be something wrong...

For a diprotic acid of the for H2A with concentration C you have

[A-2]= Ka1*Ka2*C / ([H+]^2+Ka1[H+] +Ka1*Ka2)

Because the difference in Ka1 and Ka2 is more than 3 orders of magnitude (6 to be exact) you can ignore the second dissociation for calculating the [H+]. Thus

Ka1=[H+][HA-]/[H2A] (1)

but [HA-]=[H+] and [H2A]=C-[H+]

so (1) becomes [H+]^2 + Ka1[H+] -Ka1C=0
which is a quadratic.
For C=0.45 and Ka1=1.2*10^-2 [H+]=0.068
so

[A-2]= (1.2*10^-2) *(6.2*10^-8) *0.45 /(0.068^2 +0.012*0.068 +0.012*6.2*10^-8) = 6.15 *10^-8 which happens to be near Ka2. If you had C=1 then you would have [H+]=0.104 and [A-2]=6.2*10^-9 (ten times lower)