pH and Dilutions

Calculate the pH of the following solution:

100cm3 of 2M H2SO4 is made up to a 500cm3 solution.

So would you do 0.1/0.5 and then times by 2 (original [H+]) and then do -log[H+] which is 0.397?

Converting the volumes to dm^3 first isn’t necessary, but if this is an A level question, you are most of the way there.

Before you use pH = -log[H^+], have you taken into account that each H2SO4 molecule can donate as many as 2 H^+ ions?

I thought the 2M concentration would have already taken into account the fact it is diprotic?

Unfortunately it doesn’t- it means the “formal” concentration of H2SO4 (i.e the what the concentration of H2SO4 molecules would be if there was no dissociation whatsoever), would be 2.0 M.

Assuming full dissociation, what would [H^+] be before and after dilution?

Ohhh ok so it will yield 2H+ in the concentration so double the concentration to 4M?

Yes, before dilution. After dilution, however, what would it be?

0.8? (4 x 100/500)

Correct. Now all you have to do is find the pH.

Bear in mind that normally the pH should only be given to 2 decimal places.

Nice, so 0.097?

Nice, so 0.10?

That is correct.