# The pH of 0.10M HOCl is 4.23. Calculate the Ka and pKa for hypochlorous acid.?

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I have no idea how to do this can someone please help

• The Ka, the acid dissociation constant, is define for some acid HX as follows:

Ka = [H+][X-] / [HX]

So, for hypochlorous acid, it is as follows:

Ka = [H+][OCl-] / [HOCl]

So, how do we know any of these numbers? We know the pH, and, since pH = -log[H+], we can find [H+] concentration. Since dissociation results in equal amounts of hydrogen ions and hypochlorite, the [H+] we calculate will be equal to the value of [OCl-]

pH = – log[H+]

4.23 = -log[H+]

[H+] = 5.9*10^-5 M

So, now we know [H+] and [OCl-]. How do we know [HOCl]? Well, we started out with .1 M. Technically, 5.9*10^-5 M dissociates, so we could subtract that from 0.1 to get our term, but it won’t make an appreciable difference, so we can just say that [HOCl] = 0.1

Ka = (5.9*10^-5 M)(5.9*10^-5 M) / .1

Ka = 3.47 * 10^-8

Now, we can find pKa.

pKa = – log Ka

pKa = – log (3.47 * 10^-8)

pKa = 7.46

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