Under certain water conditions, the free chlorine (hypochlorous acid, HOCI) in a

Question

Under certain water conditions, the free chlorine (hypochlorous acid, HOCI) 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 2.2 parts per million (ppm). Twenty-four hours later, Geoff tested the water again and found the amount of free chlorine to be 1.8 ppm. What will be the reading after 3 days (that is, 72 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 3 days, or 72 hours, the amount of free chlorine in the pool will be ppm

Explanation / Answer

Decay will be at a constant rate

1.8 = 2.2* e ^( - k * 24)

1.8/ 2.2 = e^(-k* 24)

So e^(-k * 72) = (e^(-k* 24))^3.

So level will be (1.8/2.2)^3

Ppm will be after 72 hours = (1.8/2.2)^3

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