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.