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Topic: pH of polyprotic acid  (Read 2853 times)

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Offline UG

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pH of polyprotic acid
« on: April 24, 2010, 07:49:52 AM »
The question I am stuck on simply goes: The dissociation constants of acid A are 10–3.46 an 10–5.10. Calculate the pH of a 0.100 M solution of A.
So I've got [H2A] + [HA-] + [A2-] = 0.100
[HA-] + 2[A2-] = [H+] I am ignoring [OH-] here because the solution is going to be around 2-3 pH and the two Ka expressions. After a couple of pages of working, I ended up with

[HA-] = 0.10 x Ka1[H+] / ([H+]2 + Ka1[H+] + Ka2)

and [A2-] = Ka2 x 0.10 / ([H+]2 + Ka1[H+] + Ka2)

Substituting these into the charge balance gives
[0.10 x Ka1[H+] / ([H+]2 + Ka1[H+] + Ka2)] + [2(Ka2 x 0.10 / ([H+]2 + Ka1[H+] + Ka2))] = [H+]

I don't want to get to far ahead of myself, is my working correct up to this stage?

Offline Borek

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Re: pH of polyprotic acid
« Reply #1 on: April 24, 2010, 12:27:08 PM »
Most likely OK.
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Offline UG

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Re: pH of polyprotic acid
« Reply #2 on: April 24, 2010, 06:27:46 PM »
Thanks, but I see where I went wrong.

Offline Borek

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Re: pH of polyprotic acid
« Reply #3 on: April 25, 2010, 04:19:16 AM »
So you are better than me.
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