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AllTheGoodNamesAreTaken94

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Back Titration Problem
« on: December 07, 2014, 03:17:23 am » 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 mol7.6x10^-3/(10ml(L/1000ml))=0.76M7.6x10^-3 mol (88.11g/mol) = 0.67 gppm= 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! Logged

Offline Borek

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Re: Back Titration Problem
« Reply #1 on: December 07, 2014, 04:02:41 am » You did the "back titration" part correctly, you got lost in dilutions and conversions.Quote from: AllTheGoodNamesAreTaken94 on December 07, 2014, 03:17:23 am
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.Quote
1.52x10^-3 (100ml/20ml) = 7.6x10^-3 mol
Number of moles in the original 10 mL sample, OK.Quote
7.6x10^-3/(10ml(L/1000ml))=0.76M
Concentration in the original sample, OK.Quote
7.6x10^-3 mol (88.11g/mol) = 0.67 g
Mass of the ethyl acetate in the original 10 mL sample, OK.Quote
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. Logged Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

AllTheGoodNamesAreTaken94

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Re: Back Titration Problem
« Reply #2 on: December 07, 2014, 01:06:04 pm » Quote
Quote
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/m³897.00 kg/m³(m3/1000L) = 0.897 kg/L (0.01L) = 8.97x10-3 Kgppm = mg/kg = 670mg/8.97x10-3 kg = 74693.42 ppm? Am I even coming close to doing this right? ??? Logged

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Re: Back Titration Problem
« Reply #3 on: December 07, 2014, 02:00:53 pm » Quote from: AllTheGoodNamesAreTaken94 on December 07, 2014, 01:06:04 pm
ppm = mg/kg = 670mg/8.97x10-3 kg = 74693.42 ppm?
Looks OK to me (just don't abuse significant figures). Logged Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info
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