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 2.3 parts per million (ppm). Twenty-four hours later, Geoff tested the water again and found the amount of free chlorine to be 1.9 ppm. What will be the reading after 4 days (that is, 96 hours)? When the chlorine level reaches 1.0 ppm, Geoff must shock the pool again. How long can Geoff go before he must shock the pool again?

After 4 days the chlorine level reaches 1 ppm.

Explanation:

The formula  N(t) = N₀ e⁽⁻kt⁾  models this decay process well, where

t     = time measured in 24-hour days, so t =0, 1, 2, 3, ...

N₀   = initial measurement of HOCl at time t=0, just after first shocking the pool (given as 2.3 ppm)

N(t) = measurement of HOCl at time any time t  (N at t=1 is given as 1.9 ppm)

e = exponential function

k = rate of decay constant

First, we have to find k from the measurements at t= 0 and t=1

N(1) = N(0) e⁽⁻k(1))  so  1.9 = 2.3 e⁻k  and  e⁻k = 1.9/2.3 = 0.826

Taking the logarithm of both sides:  -k = ln (0.826) = - 0.191  

so k = + 0.191

After 4 days,

N(4) = 2.3e⁻(0.191ₓ4) = 1.07 ≈ 1 ppm

After 4 days Geoff must shock the pool again.

the first point, i am pretty sure.

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