October 29, 2020, 03:53:52 PM
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### Topic: Three-protonic acid, dissociation, formula question  (Read 327 times)

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#### schnorch

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• Mole Snacks: +0/-0 ##### Three-protonic acid, dissociation, formula question
« on: October 07, 2020, 03:01:53 PM »
Hello!

There is this formula for dissociation of an acid:

Ka = H+ x A- / HA

And if it is the three-proton-acid citric acid:

Ka1 = H+ x H2Cit- / H3Cit
Ka2 = H+ x HCit-2 / H2Cit-
Ka3 = H+ x Cit-3 / HCit-2

Now my question:
Is the H+ the total amount of H+ or only the amount of the dissociation step. I mean like this:

Ka1 = H+1 x H2Cit- / H3Cit
Ka2 = H+2 x HCit-2 / H2Cit-
Ka3 = H+3 x Cit-3 / HCit-2

And H+a + H+b + H+c = total H+
And H+a > H+b > H+c

For me it sounds logical, if the formula doesn't mean the total amount of H+.
But I calculated a bit and the results look better, if it does mean the total amount.

I am no native English speaker.

#### Borek ##### Re: Three-protonic acid, dissociation, formula question
« Reply #1 on: October 07, 2020, 03:04:00 PM »
Total. No way to say where did this particular H+ came from, they are all identical.
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#### chenbeier

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« Reply #2 on: October 07, 2020, 03:30:15 PM »
Additional 1 mol citric acid contain 3 mol H+, which dissociate in steps.
There are  no partical H+.

#### schnorch

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• Mole Snacks: +0/-0 ##### Re: Three-protonic acid, dissociation, formula question
« Reply #3 on: October 07, 2020, 03:47:32 PM »
So if I have a pH of 4 (H = 0.0001), this is correct:

log10(3.13) = 0.0001 x H2Cit- / H3Cit
log10(4.76) = 0.0001 x HCit-2 / H2Cit-
log10(6.40) = 0.0001 x Cit-3 / HCit-2

And so I have the correct ratios of the citric acid and the anions.
Right?

« Last Edit: October 07, 2020, 04:12:09 PM by schnorch »

#### chenbeier

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« Reply #4 on: October 07, 2020, 05:10:04 PM »

#### Borek ##### Re: Three-protonic acid, dissociation, formula question
« Reply #5 on: October 07, 2020, 05:46:58 PM »
So if I have a pH of 4 (H = 0.0001), this is correct:

log10(3.13) = 0.0001 x H2Cit- / H3Cit
log10(4.76) = 0.0001 x HCit-2 / H2Cit-
log10(6.40) = 0.0001 x Cit-3 / HCit-2

And so I have the correct ratios of the citric acid and the anions.
Right?

Yes.
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#### schnorch

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• Mole Snacks: +0/-0 ##### Re: Three-protonic acid, dissociation, formula question
« Reply #6 on: October 08, 2020, 08:39:10 AM »
Thanks.

I calculated with 0.0001 in every line further and yes, the most often anion is the di hydrogen citrate: I calculate 20% tri hydrogen citrate, 74% di hydrogen citrate, 6% mono hydrogen citrate and almost nothing of the zero hydrogen citrate.

This means the ratios of the anions depend on the pH and only indirectly on the citric acid concentration. So I could change the ratios of the anions both by adding more or less citric acid but also by adding a strong acid or a strong base. So, if I have a pH 4 citric acid solution (with mainly di hydrogen citrate) and add HCl to it until the pH is 3, I will have new ratios of anions (I calculate ca. 73% three hydrogen citrate, 27% di hydrogen citrate and almost nothing of the others).
Is this right generally? (You don't need to check the exact numbers of course)

#### Borek ##### Re: Three-protonic acid, dissociation, formula question
« Reply #7 on: October 08, 2020, 09:05:25 AM »
This means the ratios of the anions depend on the pH and only indirectly on the citric acid concentration.

Actually as long as we ignore the ionic strength of the solution ratios depend only on pH (and Ka values). Note that there is nothing else in the Ka definition. pH _can_ depend on the citric acid concentration, but can be controlled by other means as well.

Quote
So I could change the ratios of the anions both by adding more or less citric acid but also by adding a strong acid or a strong base

in other words: by changing the pH, by any means.

Quote
So, if I have a pH 4 citric acid solution (with mainly di hydrogen citrate) and add HCl to it until the pH is 3, I will have new ratios of anions (I calculate ca. 73% three hydrogen citrate, 27% di hydrogen citrate and almost nothing of the others).
Is this right generally? (You don't need to check the exact numbers of course)

General; idea sounds OK, but taking into account that pKa1≈3.1 at pH=3 I would expect the ratio of H3Citrate to H2Ctitrate- to be closer to one.
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#### schnorch

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• Mole Snacks: +0/-0 ##### Re: Three-protonic acid, dissociation, formula question
« Reply #8 on: October 08, 2020, 09:20:58 AM »
Quote
Actually as long as we ignore the ionic strength of the solution ratios depend only on pH (and Ka values). Note that there is nothing else in the Ka definition. pH _can_ depend on the citric acid concentration, but can be controlled by other means as wel.
Thanks. I needed a few words more than only "Yes".

Quote
General; idea sounds OK, but taking into account that pKa1≈3.1 at pH=3 I would expect the ratio of H3Citrate to H2Ctitrate- to be closer to one.

This is right unfortunately. I calculated from the pKa1=3.13 the ratio at pH=3 must be 1.35 but I calculated 2.7. Strange. Pretty exact twice the number. I think I will find the mistake soon.

#### schnorch

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• Mole Snacks: +0/-0 ##### Re: Three-protonic acid, dissociation, formula question
« Reply #9 on: October 08, 2020, 09:28:49 AM »
And I found the mistake.

At pH=3 I have 57% tri, 42% di 1% mono and 0% zero hydrogen citrate.
At pH=4 I have 10% tri, 76% di, 13% mono and 1% zero hydrogen citrate.

Thanks a lot!