Question

The half-life of methyl mercury in our body is 70. Days. How many days' are required for the amount of methyl mercury to drop to 10% of the original value after accidental ingestion. The decomposition of methyl mercury proceeds by first order kinetics.

Answer 1

For a first order reaction, (1/2)^{number of half-lives} = the decimal amount remaining

Remaining amount of methylmercury = 0.10

(½)ⁿ = 0.10; n = number of half-lives

n log1/2 = log 0.10

n = -1/-0.3010 = 3.32

days required = 3.32 times half life = 3.32 x 70 = 233 days