Back Titration of Ethanol in Vodka with Excess Dichromate Solution

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In summary, in the reaction of Vodka (ethanol) w/ excess dichromate solution, the volume "delivered" of iron(ll) solution in order to titrate the excess dichromate is 10.35mL. The mole of iron(ll) solution is 0.0010557 mol, and the mole of dichromate in excess is 0.00017595 mol. The concentration of the Fe(II) solution is 0.03477g/.789g/ml.
  • #1
googol
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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.
 
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  • #2
googol said:
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:

googol said:
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
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
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
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
googol said:
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.
 

1. What is a back titration?

A back titration is a technique used in analytical chemistry to determine the concentration of an unknown substance by reacting it with an excess of a known reagent, and then titrating the remaining excess with a second reagent. This method is useful for substances that are difficult to titrate directly, or for reactions that are slow or have low endpoint sensitivity.

2. When should a back titration be used?

Back titrations are typically used when the substance being analyzed cannot be easily titrated directly, such as in cases where it is insoluble, volatile, or weakly acidic or basic. This method can also be used when the reaction between the unknown substance and the known reagent is slow or has a weak endpoint.

3. What are the steps involved in a back titration?

The first step in a back titration is to add an excess of a known reagent to the unknown substance. The excess is then titrated with a second reagent until the endpoint is reached. The volume of the second reagent used is then used to calculate the amount of the unknown substance present.

4. What are the advantages of using a back titration?

Back titrations offer several advantages over direct titrations. They can be used for substances that are difficult to titrate directly, and they provide a way to measure slow or weak reactions. Additionally, back titrations can be more accurate and precise than direct titrations, as they eliminate the need to measure the small volume of the unknown substance.

5. What are some potential sources of error in a back titration?

As with any analytical technique, back titrations can be subject to sources of error. Some potential sources of error include incomplete reaction between the unknown substance and the known reagent, improper technique in titrating the excess reagent, and contamination of reagents or equipment. It is important to carefully control and monitor these factors to ensure accurate results.

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