J.R. S. answered 11/02/20
Ph.D. University Professor with 10+ years Tutoring Experience
You are correct in stating that the 2nd reaction (second ionization) isn't complete because HSO4- is a weak acid. In this case, you'd have to use the Ka value for HSO4- ==> H+ + SO42-. Of course, this will add relatively little to the [H+] relative to the first ionization, but if pH = 1.3 isn't correct, perhaps they want to to include the 2nd ionization as well.
The value I have for Ka HSO4- ==> H+ + SO42- is 1.2x10-2
From the first ionization you will have 0.05 M H+ and 0.05 M HSO4-. Then you have...
HSO4- ==> H+ + SO42- ...... Ka = 1.2x10-2
0.05..........0..........0......Initial
-x...........+x..........+x.....Change
0.05-x......x...........x.......Equilibrium
Ka = [H+][SO42-]/[HSO4-] = 1.2x10-2
1.2x10-2 = (x)(x) / 0.05-x
Solve for x, and add this value to 0.05 M from the first ionization to get the TOTAL H+ concentration. Take the negative log to get the final pH.
I hope this helps.
Lily P.
I got the final pH to 1.17? I would appreciate it if you can tell me if I am correct or not? :) A million thanks!!11/02/20

J.R. S.
11/03/20
Lily P.
Its ok! Thanks anyway for your amazing help11/07/20
Lily P.
Thank you sir for your help :) Glad to know I wasn't misunderstanding a basic principle.11/02/20