Decomposing

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 his pool, Ben tested the water and found the amount of free chlorine to be 2.3 parts per million (ppm). Twenty-four hours later, Ben tested the water again and found the amount of free chlorine to be 1.9 ppm. What will be the reading after 3 days (that is, 72 hours)? When the chlorine level reaches 1.0 ppm, Ben must shock the pool again. How long can Ben go before he must shock the pool again

Answer 1

Kasi,

 

The formula  N(t) = N_o e^(-kt)  models this decay process well, where

 

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

N_o   = 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, compute k from the measurements at t= 0 and t=1

N(1) = N(0) e ^-k(1)  s0  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 3 days (t=3), the reading will be:

N(3) = 2.3 e ^-0.573  = 2.3 (0.564) = 1.297  = 1.3 ppm rounded

 

For what value of t does N(t) = 1 ppm?

1.0 = 2.3 e ^-0.191t , so  e ^-0.191t = 1/2.3 = 0.435  and -0.191 t = ln (0.435) = -0.832

So t = - 0.832/-0.191 = + 4.356 or about 4.4 days, or about 4 days and 9 hours

Then Ben must shock the pool again.