The blood alcohol (C2H5OH) level can bedetermined by titrating a sample of blood plasma with an acidicpotassium dichromate solution, resulting in the production ofCr3+(aq) and carbon dioxide. The reaction canbe monitored because the dichromate ion(Cr2O72-) is orange in solution, and the Cr3+ ion is green. The unbalanced redox equationis shown below.
Cr2O72-(aq) +C2H5OH(aq) → Cr3+(aq) +CO2(g)
If 31.05 mL of 0.0600 M potassium dichromate solution isrequired to titrate 30.0 g of blood plasma, determine the masspercent of alcohol in the blood.

Step-by-step explanation:

In acidic reaction , equivalent weight of potassium dichromate is

mol weight / 6

= 294 / 6

49 g

equivalent weight of alcohol = 46 g

31.05 cc of .06M of potassium dichromate

= 31.05 x .06 cc of M potassium dichromate

= 1.863 cc of M potassium dichromate

= 1.863 x 6 cc of N potassium dichromate

= 11.178 cc of N potassium dichromate

== 11.178 cc of N alcohol

= (11.178 / 1000 ) x equivalent weight of alcohol

= (11.178 / 1000 ) x 46 g

= .514 g

30 g of blood contains .514 g of alcohol

100 g of blood contains (0.514 / 30) x 100

= 1.71 % of alcohol.

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