ok, this question i think i did use of formulas good but something my answer is wrong:

How much grams of CH3COOH is required to add to make 2 liter pH=3.5 solution a pH=3 solution?

Ka(CH3COOH)=1.8*20^-5.

Answer: 6gr.

i found out that the moles of H+ in the solution is 10^-3.5*2 (n=c*V). we need that there will be 2*10^-3 moles of H+ (because pH need to be 3) so there must be add 1.3675*10^-3 moles of H+ (i call it n).

we know the Ka of this acid. we also know that the number of moles in equillibrium of the acid is x-n so that the number of moles of H+ will be n.

after some algebra (it's tough to show it here) i found that x=0.0533317.

Mw of the acid is 60 so the m we need to add is only ~3gr and not 6!!!

can you please tell me where were i wrong???

thx.