Back titrations PLEASE HELP!

  • Thread starter googol
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  • #1
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In the reaction of Vodka (ethanol) w/ excess dichromate solution
volumes:
Diluted vodka = 5ml
6.0M H2SO4 = 3ml
K2Cr2O7 = 5ml

In the back titration of excess dichromate with iron(ll) solution, the volume "delivered" of iron(ll) solution in order to titrate the excess dichromate is 10.35mL.

There were also three stock solutions that were made initially.
For the K2CrO7 solution, 4 grams of K2CrO7 was used and the concentration is 0.136mol/L.
For the Iron solution, 4 grams were used and the concentration is 0.102 mol/L.
And for the vodka solution (2 ml was used!!)

So here's what I've done and I know its wrong T_T:
First I found mole of iron(ll) solution
n = MxL = (.102)(0.01035L) = 0.0010557 mol Fe

From the chemical equation, there's a 6:1 ratio when it reacts w/ dichromate. So 0.0010557 mol/6 = 0.00017595 mol dichromate (left in excess)

Initial dichromate reacted w/ ethanol
n = (0.005)(.136) = 0.00068 mol
0.00068 mol - 0.00017595 = 0.0005 mol dichromate that reacted w/ the ethanol

From the chemical equation with the reaction b/w ethanol and dichromate, there is a 3:2 ratio
so ... 0.0005 mol x 1.5 = 0.000756 mol ethanol

0.000756 mol x 46g/mol(molar mass) = 0.03477grams ethanol
volume = m/P = 0.03477g/.789g/ml = 0.04408 mL

.... So basically in this experiment, it says "the quality of a shipment of vodka has been questioned, so quality contro ltests have been requested. The label on the vodka indicated that ethanol content is 40% by volume, where:
%ethanol content by volume = (volume ethanol)/(total volume)*100%"

When I plug the volume ethanol = 0.04408 and divide by 2ml (the total volume) and multiply it by a 100 .. I don't get 40% or above that or less that... i get a small percent like 2% which is obviously wrong. What am I doing wrong??? I'm so confused!!!

I would APPRECIATE any sort of help.
 

Answers and Replies

  • #2
Borek
Mentor
28,628
3,102
Diluted vodka = 5ml
Diluted? So it is not 2 mL of the original 40% vodka?

In general you need to be more precise in what you do, as I spent 20 minutes trying to decipher what you did. For example:

For the K2CrO7 solution, 4 grams of K2CrO7 was used and the concentration is 0.136mol/L.
4 grams in what volume? How can anyone know if the concentration is calculated correctly?

For the Iron solution, 4 grams were used and the concentration is 0.102 mol/L.
4 grams of what, and in what volume?
 
  • #3
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Oh sorry, the volume of the solutions for both was a 100mL .. because we put 4 grams of each (iron and dichromate) into separate volumetric flasks and then filled the volumetric flasks to the mark (to the 100ml mark) ...

For the vodka, I initially took 2ml (undiluted) and added it to a volumetric flask, and then filled it to the mark. So the volume of the solution became 100mL. Then, when I was preparing for the reaction of the vodka(ethanol) with the dichromate. I took 5mL of vodka (now diluted) from the 100mL solution, and I took 5ml of dichromate from the 100ml solution that we initially made with the 4grams of dichromate.

Does that make more sense?? Sorry I skipped those things, I just assumed by concentrations were calculated correctly. I was just confused more about the later steps.
 
  • #4
Borek
Mentor
28,628
3,102
It doesn't matter much, but you have still not explained correctly how you prepared Fe(II) solution - I guess I know what salt was dissolved, but I have to guess.

It was already stated several times - vodka solution was diluted.
 
  • #5
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no first i took 2ml of pure vodka, and added it to a volumetric flask & filled it to the mark.

The fe(11) solution, I took 4grams Fe, put it into a volumetric flask and filled it to the mark
 
  • #6
Borek
Mentor
28,628
3,102
no first i took 2ml of pure vodka, and added it to a volumetric flask & filled it to the mark.
And this procedure is called dilution.

The fe(11) solution, I took 4grams Fe, put it into a volumetric flask and filled it to the mark
What is fe eleven? Fe(II) means divalent iron, fe(11) doesn't mean anything.

Are you sure you know what you did? 4 g Fe dissolved in 100 mL of water would yield 0.716M solution, not 0.102M. That is, assuming Fe dissolves in water. It doesn't.
 

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