# What reaction in a titration?

1. Jan 24, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
Cointreau is a French liqueur sometimes used to enhance the flavour of strawberry desserts. Its label states that the concentration of the alcohol is 40%alc/v. To test this claim, a 5.00ml sample of the liqueur was pipetted into a 100ml standard flask and the volume was made up to the 100.00ml mark. Successive 20.00ml aliquots of the diluted solution were then withdrawn and titrated against acidified potassium dichromate. If the K2Cr2O7 was at a concentration of 0.2154M and the mean titre was 21.15ml, what was the actual concentration of the alcohol in the liqueur? Hence comment on the claim on the label. (The density of ethanol is 0.785g/ml)

To get this question going, I have trouble with the reaction at the titration stage which the question did not specify.

2. Relevant equations

3. The attempt at a solution
Alc/v => ml of alcohol / 100ml of solution

Alcohol is ethanol which is CH3CH2OH.

How does the titration reaction occur between potassium dichromate and ethanol? Is it K2Cr2O7 + 2CH3CH2OH -> H2Cr2O7 + 2CH3CH2O- + K2+

2. Feb 2, 2007

### pivoxa15

I found out that the reaction of the tritrate (from another thread I created) was
3C2H5OH + 2(Cr2O7)2- + 16H+ --> 3CH3COOH + 4Cr3+ + 11H2O

The mean titrate of (Cr2O7)2- recquired .0046 moles so .0068 moles of ethanol contained in the 20ml aliquot. There are 5 times this many moles in the 100ml standard flask. Hence .034moles in 5ml of liqueur. So there are .683 moles of ethanol in 100ml of the liqueur. This translates to 31.49g/100ml. by %(w/v). To convert to %(alc/v) we use the data and find that 31.49g/(.785g/ml)=40.1ml of alcohol in 100ml of the liqueur solution. This is close to the value on the label 40%(alc/v) and same as the answer at the back of the book.