Question

under certain water​ conditions, the free chlorine​ (hypochlorous acid,​ hocl) in a swimming pool decomposes according to the law of uninhibited decay. after shocking a​ pool, the pool​ boy, geoff, tested the water and found the amount of free chlorine to be 2.6 parts per million​ (ppm). twenty-four hours​ later, geoff tested the water again and found the amount of free chlorine to be 2.1 ppm. what will be the reading after 2 days​ (that is, 48 ​hours)

Answer 1

1.7 ppm

Step-by-step explanation:

Original amount N' = 2.6 ppm

time to testing t = 24 hr

final amount N = 2.1 ppm

Using exponential inhibited decay, we have

N = N'e^(-kt)

Where

N is the new reading

N' is the original reading

t is the decay time

k is the decay constant

Substituting, we have

2.1 = 2.6 x e^(-k x 24)

2.1 = 2.6 x e^(-24k)

0.808 = e^(-24k)

We take the natural log of both sides of the equation

Ln 0.808 = Ln (e^(-24k))

-0.213 = - 24k

K = 0.213/24 = 0.00886

After 48 hrs, the reading of free chlorine will be

N = 2.6 x e^(-0.00886 x 48)

N = 2.6 x e^(-0.425)

N = 2.6 x 0.654

N = 1.7 ppm