Additional Problems 1. Given that for HOCl, Ka = 2.7 × 10-8, deduce the fraction

Question

Additional Problems 1. Given that for HOCl, Ka = 2.7 × 10-8, deduce the fraction of a sample of the acid in water that exists in the molecular form at phH values (predetermined by the presence of other species) of 7.0, 7.5, 8.0, and 8.5. [Hint: Derive an expression that relates the fraction of HOCl that is ionized to the concentration of hydrogen ions.] Would it be a good idea to allow the pH of swimming pool water to rise to 8.5? 2. For efficient disinfection to occur, treated drinking water should contain about 0.5 mg L-1 of Cl2 that remains after most of the chlorine has

Explanation / Answer

HOCl(aq) <--> H+(aq) + OCl-(aq)

The acid constant equation is:

Ka = [H+(aq)][OCl-(aq)] / [HOCl(aq)]

[HOCl(aq)] = [H+(aq)][OCl-(aq)] / Ka

from the above equation it is clear that [H+] = [OCl-]

at pH 7

Concentration of [H+] = 10-7

therefore [HOCl] = (10-7 X 10-7) /(2.7 X 10-8)= 3.7 X 10-7

similarly at pH 7.5, [HOCl] = (3.16X10-8 X 3.16X10-8) /(2.7 X 10-8)= 3.703 X 10-9

at pH 8, [HOCl] = (10-8 X 10-8) /(2.7 X 10-8)= 3.703 X 10-9

at pH 8.5, [HOCl] = (3.1622X10-9 X 3.1622X10-9 ) /(2.7 X 10-8)= 3.703 X 10-10

No it isnot advisable to leave swimming pool water at pH 8.5

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