Alkaline buffers

THe answer is supposed to be (B) but I'm not understanding properly

I initially chose A - to form an alkaline buffer, the # mol of acid should be half of this. But this doesn't seem to be the case

I don't understand why the mol of base in this case should be 5x the mole of acid

your help is greatly appreciated, as always!

Hint: sulfuric acid is diprotic.

Ok that does help - so we should be looking at bisulfate rather than sulfuric acid
but wouldn't the concentration of bisulfate be the same i.e. the second part of the henderson hasselbach would still be

for a. log (0.05/0.1) = log(0.5)

and for
b. log(0.1/0.5) = log(0.2)

I feel like i'm missing something obvious ....

stunner5000pt said:

a. 0.1 mol ammonia & 0.05 mol sulfuric acid

stunner5000pt said:

bisulfate

Nope, not bisulfate. Just look at the reaction stoichiometry and amounts of substances involved.