[zindarg]

"An unknown amount of acid can often be determined by adding an excess of base and then "back-titrating" the excess. A 0.3471 g sample of a mixture of oxalic acid, which has two ionizable protons and benzoic, which has one, is treated with 100. mL of 0.1000 M NaOH. The excess NaOH is titrated with 20.0 mL of .2 M HCl. Find the mass % of benzoic acid."
Well I started by finding number of moles of OH^- actually used by finding total minus excess, which I got to be .01 moles -.004 moles = .006 moles of actual used OH^- in the neutralization reaction. After that I tried dividing by three since the mixture contains a total of 3 H⁺ ions but benzoic acid only has one of the three, and I got .002 moles of benzoic acid times its molecular mass (122.12g/mol) and got .24424 g divided by the total (.3471g) multiplied by 100 and got 70.4%, however when I try to apply the same principles to oxalic acid, to try and check myself, I get a too large value and end up with .002 moles of oxalic acid times molecular mass (90.03g/mol) and get .18g divided by the total and multiplied by 100 which yields 51%... needless to say, these don't add up to 100% and I don't have any clue what to do now, any help would be appreciated 

[Borek]

You can't divide by three - that's equivalent to assuming the mixture contains exactly one mole of oxalic acid per each mole of benzoic acid. If you would know they are mixed in 1:1 ratio there will be no need to determine the mixture composition.

Assume you have x moles of benzoic and y moles of oxalic - try to express given information (mass of the sample, amount of NaOH required for neutralization) using these variables. That will give you two equations in two unknowns - just solve.

[zindarg]

Ok, so using the X moles of benzoic acid and Y moles of oxalic acid and I get the mass of the mixture = .3471 g = ((x moles benzoic acid)*(122.12 g/mole))+((y moles oxalic acid)*(90.03g/mole)) and for moles of OH^- needed for neutralization I get .006 mole OH^- = x moles benzoic acid + y moles of oxalic acid. Are these right? and from there I am lost, I tried solving for one variable and substituting but I never get a plausible answer, I think either my logic is flawed or my equations are wrong, and thanks again for all the help

[Borek]

.006 mole OH^- = x moles benzoic acid + y moles of oxalic acid

Close, but wrong. How many acidic protons in each acid?

[zindarg]

So is it .006 moles OH^- = .006 H⁺ = (1) x moles H⁺ (from benzoic acid) + (2) Y moles H⁺ (from oxalic acid) ? And then do I solve for X moles benzoic and substitute it into mass %?

[zindarg]

I think I got it! Thanks so much!