pH = 7.8
[H+] = 10^-7.8 = 1.58 x 10^-8 M
[OCl-] = 1.58 x 10^-8 M
HOCl     <--------------------->
H^+ + OCl^-
CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â 0
C -
x                                    Â
x          Â
x
[H+] = [OCl-] = x = 1.58 x 10^-8 M
K=[H^+][OCl^-]/[HOCl]
10^-7.6 = (1.58 x 10^-8)^2 / C- 1.58 x 10^-8
C = 2.57 x 10^-8 M
concentration of [HOCl] = C - x = 2.57 x 10^-8 - 1.58 x
10^-8
                                     Â
= 9.9 x 10^-9 Mol/Lit
concentration of the HOCl species in mg/L = 9.9 x 10^-9 x
52.46
                                                                   Â
= 5.19 x 10^-7 g/L
                                                                  Â
= 5.19 x 10^-4 mg/L
