Slide 1

Ethanol-dichromate titration


Slide 2

Acidified potassium dichromate can be used to oxidise
ethanol to ethanoic acid.
These days instruments are used to quickly determine the
concentration of ethanol in blood, but at one time blood
alcohol levels (for drink-driving convictions) were
determined by titration.
The ethanol was separated from the blood sample and
reacted with the dichromate solution.

Calculate the concentration of ethanol contained in a
10.0 mL sample of blood that reacted with 13.7 mL of
0.0209 mol L–1 solution of potassium dichromate, and hence
determine whether the blood exceeded the legal limit of
80 mg of ethanol per 100 mL of blood.


Slide 3

Step 1: Determine the amount of the known reagent (the
dichromate)

V(Cr2O72–) = 13.7 mL

c(Cr2O72–) = 0.0209 mol L–1

= 13.7  10–3 L

n(Cr2O72–)

= cV
=
= 2.863  10–4
mol

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Slide 4

Step 2: Use the equation for the chemical reaction to
determine the amount of ethanol present
3
2 2O72– + 16H+ → 3CH3COOH + 4Cr3+ + 11H2O
3CH3CH2OH + 2Cr
Unknown on top
Known on
the bottom

n(CH3CH2OH)
=
2–
n(Cr2O7 )
n(CH3CH2OH)
3
=
2–
n(Cr2O7 )

Rearrange to get the
unknown on its own.

2
3  2.863  10–4
=
mol 2
= 4.295  10–4
mol


Slide 5

Step 3: Calculate the concentration of ethanol in the
sample

n(CH3CH2OH) = 4.295  10–4 mol
V(CH3CH2OH) = 10.0 mL

Don’t forget to
convert mL to L.

= 10.0  10–3 L

n
c(CH3CH3OH) =
V
=

=

4.295  10–2 mol L–1


Slide 6

Step 4: Convert the concentration of ethanol in the
blood sample into mg per 100 mL of blood

c(CH3CH2OH) = 4.295  10–2 mol L–1

M(CH3CH2OH) = 46.0 g mol–1
To convert concentration in mol L–1 into g L–1, multiply by
the molar mass (in g mol–1).

c(CH3CH2OH) =
= 1.976 g L–1
= 0.198 g per 100 mL
= 198 mg per 100 mL

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