[dorask]

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.

[AWK]

But you need 2 liters of solution!

[dorask]

i didn't understand what you said.. i did regard the fact that there are 2 liters!! (i multiplied by 2 the concentration etc. can you be more speciefic?? thx

[Borek]

It is not possible to answer this question not knowing what was the source of initial pH. Was is a strong acid? A weak acid? A buffer?

[dorask]

just assume that theres 2*10^-3.5 moles of h+ in the first place.

[Borek]

It is obvious how much H⁺ there is in the solution - but it doesn't change the situation. It is presence of other, unknown ions, that is a problem. Simplest case would be to assume initially only a strong acid is present - in a way it is equivalent to assuming "just 10^-3.5 M of H⁺".

As far as I can tell with this assumption neither 3g nor 6g is a correct answer, I got something between 4 and 5.

Do the calculations using concentrations, use volume to convert to moles and mass at the end. You can show your calculations using LaTeX (as described here: http://www.chemicalforums.com/index.php?topic=59314.0).