I'm not help in chem, but...

You start out trying to figure out how much H2SO4 you need...and then all of a sudden you start figuring out HCl. And then you get the answer 10mL of HCl. But the first thing you wrote said that 10mL HCl was required.

So, frankly, I'm not sure what you did, but I THINK you have the right idea...you just figured out something you already know. Do it again for H2SO4

As Dewayne said, The answer you give in brackets is wrong. Either you made a mistake in copying it over or the back of the book didn't print it right because what you are looking for is the mL of H2SO4, not HCl. The answer is 5.0 mL of H2SO4.

It doesn't require math, so you're doing way too much work than you need to be. The logic is this, I think: HCl liberates 1 H+ (well, in the form of H3O+, but the point is that it releases one positive, acidic group) while H2SO4 liberates 2 H+. Therefore, H2SO4 neutralises a base twice as efficiently as HCl. Therefore, one would require half as much H2SO4 in order to neutralise a base.

There's one IMPORTANT premise though: this only works for strong acids. Both HCl and H2SO4 are strong acids. This means that it ionises COMPLETELY in water. Theoretically, this should mean that, just as 1M HCl → 1M (H+) and 1M (Cl-), 1M H2SO4 → 2M (H+) and 1M (SO4--). But that's not entirely true. You see, the pKa — the dissociation constant — changes for H2SO4 after the first proton is liberated and it becomes a weaker acid. It is still relatively strong, but it is not classified as a strong acid after it has lost its first proton. But I suppose for the sake of this question, it can be assumed that the reaction progresses to completion and so you get exactly 2M of H+, when in an actual lab, you wouldn't. So basically what I'm saying is that in a lab setting, you'd probably need to use a bit more than 5 mL to neutralise the OH- liberated from NaOH, but you don't really need to worry about that to do this question.

Doesn't come out too clearly on gt. I'll just retype those two equations:

• 1M HCl → 1M (H+) and 1M (Cl-)

• 1M H2SO4 → 2M (H+) and 1M (SO4- -)

And of course with my reservations about the second equation because that's not quite true. Diprotic strong acids cannot be treated as if they're necessarily strong acids for both protons because it's a stepwise reaction, so the dissociation constant changes, and for the second reaction, it's harder to yank the proton away, so the second time around, it's not as good an acid as it was to begin with.