Question

Assuming the water sample has a pH of 7.2, determine the amount of HOCl and OCl- in mg/L contained in the free chlorine residual of the treated sample.

Answer 1

pH= -log[H+]
7.2 = - log [H+]
[H+] = 6.31*10^-8 mol/L

HOCL   ----> H+   +   OCl-

[H+] = [OCL-]
so,
[OCl-] = 6.31*10^-8 mol/L

Molar mass of OCL- = 51.5 g/mol
AMount of OCl - = 6.31*10^-8 mol/L * 51.5 g/mol
                                   = 3.25*10^-6 g/L
                                    = 3.25*10^-3 mg/L

Ka of HOCl =3.5*10^-8
HOCL   ----> H+   +   OCl-

use:
Ka = [H+] [OCL-] / [HOCl]
3.5*10^-8 = (6.31*10^-8)*(6.31*10^-8) / [HOCl]
[HOCl] = 1.14*10^-7 mol/L

Molar mass of HOCL = 52.5 g/mol

AMount of HOCl =1.14*10^-7 mol/L * 52.5 g/mol
                                   = 5.97*10^-6 g/L
                                    = 5.97*10^-3 mg/L