In the early1950s, people, cats, and sea birds around Minamata Bayin Japan suffe
ID: 814090 • Letter: I
Question
In the early1950s, people, cats, and sea birds around Minamata Bayin Japan suffered brain damage, which in some cases led to death, from eating fish contaminated with mercury from a local factory. Elimination of mercury from the body is first order with a half-life of 60 days (zero not significant). If a quarter- pound serving of fish containing 5.7x10-3 g of mercury (the concentration of mercury in the Minamata Bay fish) is eaten by a 150-lb (68-kg) person and the maximum normal level of mercury in people is 25 ppb by mass, how many days will be required for the mercury level to return to normal?
Explanation / Answer
Ans:
Given t1/2 = 60 days
t1/2= 0.693/k
k= 0.693/ t1/2= 0.693/60
= 0.01155 per day
Given normal level or mercury in humanbeings = 25 ppb
= 25/109
= 25*10-9
Lets take 68*103 grams as 1 part.
5.7*10-3 grams will be = (5.*10-3)/68*103
= 84*10-9 = 84 ppb
It was the initial conentration present in = 84 ppb
According to rate equation:
k = [(2.303)*log(a/a-x)]/t
t = [(2.303)*log(a/a-x)]/k
a = initial concentration= 84 ppb
(a-x) = 25 ppb = 25*10-9
t = [(2.303)*log(84*10-9/25*10-9)]/0.01155
= [(2.303)*log(3.36)]/0.01155
= (2.303*0.52)/0.01155
= 1.2121/0.01155
= 104.94 days
So it will take 104.94 days to reach normal range of 25ppb for mercury.