Question

In: Chemistry

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

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 the total free chlorine (HOCl + OCl^-) = 1.2*10^-4 mole/L. Calculate the concentration of the HOCl species in mg/L.

Expert Solution

pH = 7.8

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

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

HOCl <---------------------> H^+ + OCl^-

C 0 0

C - x x 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

= 9.9 x 10^-9 Mol/Lit

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

= 5.19 x 10^-7 g/L

= 5.19 x 10^-4 mg/L

queen_honey_blossom answered 2 years ago

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