Free chlorine in water distributes between HOCl and OCL^- species according to the following reaction: HOCl <-->...

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Free chlorine in water distributes between HOCl and OCL^-species according to the following reaction:
HOCl <--> H^+ + OCl^-
K=[H^+][OCl^-]/[HOCl] = 10^-7.6 @ 25 degrees C
A water has a pH of 7.8, temperature of 25 degrees C, and thetotal free chlorine (HOCl + OCl^-) = 1.2*10^-4 mole/L. Calculatethe concentration of the HOCl species in mg/L.

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4.3 Ratings (578 Votes)

pH = 7.8

[H+] = 10^-7.8 = 1.58 x 10^-8 M

[OCl-] = 1.58 x 10^-8 M

HOClÂ Â Â Â Â <---------------------> 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 - xÂ 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 xÂ Â Â Â Â Â Â Â Â Â 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 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 = 9.9 x 10^-9 Mol/Lit

concentration of the HOCl species in mg/L = 9.9 x 10^-9 x 52.46

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 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 = 5.19 x 10^-7 g/L

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 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 = 5.19 x 10^-4 mg/L

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