After 4 ​days, or 96 ​hours, the amount of free chlorine in the pool will be____ppm.

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

2.52.5

parts per million​ (ppm). ​ Twenty-four hours​ later, Geoff tested the water again and found the amount of free chlorine to be

2.12.1

ppm. What will be the reading after

44

days​ (that is,

9696

​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?

Answer 1

Uninhibited decay: b1 = bo * exp(k*t0)

Let b0 be the initial concentration (here listed as 2.52.5ppm (whatever that value is supposed to be).

Let b1 be the concentration after 1 day (listed as 2.12.1ppm (see above).

Let t0 = 24 hours.

Then k = ln(b1/bo) / t (use ln, natural log!)

Then b2 = b0 * exp(k*t2)

Then t3 = ln(b3/b0) / k