[AllTheGoodNamesAreTaken94]

I've been working on this problem for several hours now, and I don't know if I am doing this right at all. I honestly do not understand back titrations.

The problem is:

The ethyl acetate (CH3COOC2H5; fw 88.10) concentration in a solution of alcohol was determined using a back titration procedure. A 10.00 mL sample of the alcohol was diluted to 100.0 mL and a 20.0 mL portion of the diluted solution added to 40.00 mL of 0.04672 M KOH. The equation for the reaction is:
CH3COOC2H5 + OH- → CH3COO- + C2H5OH
The excess KOH was titrated by 3.41 mL of 0.05042 M H2SO4. Calculate the ppm of the ethyl acetate in the original solution.

I've gotten to the point where I know the moles of ethyl acetate reacted with the OH- is 1.52x10^-3. Where do I go from there? I keep getting conflicting answers. Do I then multiply 1.52x10^-3 mol by (100ml/ 20ml)? If so, I get:

1.52x10^-3 (100ml/20ml) = 7.6x10^-3 mol
7.6x10^-3/(10ml(L/1000ml))=0.76M
7.6x10^-3 mol (88.11g/mol) = 0.67 g

ppm= mg/L = 6.7x10-4 mg/ 0.01L = 0.067ppm 

I am almost certain that number is wrong, but I have no idea what to do to solve this problem. I could just turn it in like this, my professor really wouldn't care so long as I made an attempt, but it is really frustrating me that I can't figure this out. Please help. Thank you!

[Borek]

You did the "back titration" part correctly, you got lost in dilutions and conversions.

I've gotten to the point where I know the moles of ethyl acetate reacted with the OH- is 1.52x10^-3.

OK. From here I have to guess what you did, as you have not described the numbers.

1.52x10^-3 (100ml/20ml) = 7.6x10^-3 mol

Number of moles in the original 10 mL sample, OK.

7.6x10^-3/(10ml(L/1000ml))=0.76M

Concentration in the original sample, OK.

7.6x10^-3 mol (88.11g/mol) = 0.67 g

Mass of the ethyl acetate in the original 10 mL sample, OK.

ppm= mg/L = 6.7x10-4 mg/ 0.01L = 0.067ppm 

No, 0.67 g is not 6.7×10^-4 mg. Besides, ppm is typically not weight/volume, rather weight per weight. That means you can't use the 1 L = 1 kg approximation, as the solution you started with was not based on water, but on ethanol.

[AllTheGoodNamesAreTaken94]

ppm= mg/L = 6.7x10-4 mg/ 0.01L = 0.067ppm 

No, 0.67 g is not 6.7×10^-4 mg. Besides, ppm is typically not weight/volume, rather weight per weight. That means you can't use the 1 L = 1 kg approximation, as the solution you started with was not based on water, but on ethanol.

Wow, sorry about that. Grams to mg would be 670mg.
Okay, but where do I go from there? The density of Ethanol is 897.00 kg/m³.
ppm is mg/kg. So,

density ethyl acetate = 897.00 kg/m³
897.00 kg/m³(m3/1000L) = 0.897 kg/L (0.01L) = 8.97x10-3 Kg
ppm = mg/kg = 670mg/8.97x10-3 kg = 74693.42 ppm?

Am I even coming close to doing this right? 

[Borek]

ppm = mg/kg = 670mg/8.97x10-3 kg = 74693.42 ppm?

Looks OK to me (just don't abuse significant figures).