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### Topic: Is it possible to calculate the pKa of H2SO4  (Read 545 times)

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#### Akutni

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• Mole Snacks: +0/-0 ##### Is it possible to calculate the pKa of H2SO4
« on: April 07, 2023, 12:11:09 PM »
Hello,
I know the pKa of H2SO4 is −3.

The pKa is the pH under which the acid is halfway deprotonised. The H2SO4 is a strong acid which should be able to dissociate completely if enough water is added.
The molar density of pure H2SO4 is 18,68 mol/dm³. If we had 1 litre of H2SO4 (18,67 moles) and added 37,36 moles of water, the whole H2SO4 should dissociate. If we add only 18,68 moles of water, half of the dissociable protons shall separate, leaving 18,68 moles of H3O+, 9,34 moles of SO4¯ and 9,34 unchanged H2SO4.

18,67 moles of water should be 0,336 litres. So our final solution should be 1,336 dm³.

c(H3O+) = 18,67/1,336 = 13,97 mol/dm³
c(SO4¯) = c(H2SO4) = 9,34/1,336 = 6,99 mol/dm³

Shouldn't the pKa equal −log(13,97•6,99/6,99) = −log(13,97) = −1,145

Why are my assumptions wrong?

#### Babcock_Hall

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• Mole Snacks: +307/-22 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #1 on: April 07, 2023, 02:39:47 PM »
18,67 moles of water should be 0,336 litres. So our final solution should be 1,336 dm³.
This looks wrong on the face of it.  How did you obtain the volume?

#### Akutni

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• Mole Snacks: +0/-0 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #2 on: April 07, 2023, 03:07:51 PM »
18,67 moles of water should be 0,336 litres. So our final solution should be 1,336 dm³.
This looks wrong on the face of it.  How did you obtain the volume?

V = n•M•V = 18,67•18•1 = 336 cm³ = 0,336 dm³

#### Babcock_Hall

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• Mole Snacks: +307/-22 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #3 on: April 07, 2023, 03:49:59 PM »
Yes, that part of your calculation looks OK.  One thing to bear in mind is that pKa1 of sulfuric acid is more negative than pKa2.
« Last Edit: April 07, 2023, 04:09:31 PM by Babcock_Hall »

#### Borek ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #4 on: April 07, 2023, 04:42:34 PM »
I am afraid this is wrong in so many ways it is not even clear where to start. If I were to guess it looks like you are trying to calculate equilibrium constant assuming there is no equilibrium and the reaction went to completion. It doesn't work this way.

If we had 1 litre of H2SO4 (18,67 moles) and added 37,36 moles of water, the whole H2SO4 should dissociate.

Why do you think so?

Quote
If we add only 18,68 moles of water, half of the dissociable protons shall separate, leaving 18,68 moles of H3O+, 9,34 moles of SO4¯ and 9,34 unchanged H2SO4.

Why do you think so?

Sulfuric acid has two Ka values for two dissociation steps.

It is not clear what you mean by "fully dissociated". Technically there is no such thing as "fully" when we speak about equilibrium reactions. Looks like you try to calculate Ka assuming one of the concentrations is zero. It isn't.

« Last Edit: April 08, 2023, 03:16:59 AM by Borek »
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#### Akutni

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• Mole Snacks: +0/-0 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #5 on: April 08, 2023, 01:40:16 PM »
I am afraid this is wrong in so many ways it is not even clear where to start. If I were to guess it looks like you are trying to calculate equilibrium constant assuming there is no equilibrium and the reaction went to completion. It doesn't work this way.

If we had 1 litre of H2SO4 (18,67 moles) and added 37,36 moles of water, the whole H2SO4 should dissociate.

Why do you think so?

Quote
If we add only 18,68 moles of water, half of the dissociable protons shall separate, leaving 18,68 moles of H3O+, 9,34 moles of SO4¯ and 9,34 unchanged H2SO4.

Why do you think so?

Sulfuric acid has two Ka values for two dissociation steps.

It is not clear what you mean by "fully dissociated". Technically there is no such thing as "fully" when we speak about equilibrium reactions. Looks like you try to calculate Ka assuming one of the concentrations is zero. It isn't.

I thought that strong acids dissociated completely if they had enough water. So I concluded that I could estimate the amount of water needed to make them lose half of the possible H+ and calculate the Ka. OK, so more water would have been needed to get the acid concentration equal its conjugate base concentration.

Now I realise there are other errors like not realising this Ka = –3 constant was for the equilibrium between H2SO4 and HSO4¯. But even if I counted with 9.34 moles of HSO4¯ and 9.34 moles of H2SO4 (the amount of the acid equals its conjugate base), I would only get 9.34 H+. And no matter how much water I added, I would never get the concentration of H+, the negative logarithm of which would make −3.

So my next question is how can H2SO4 have pKa of −3. That would mean that at the time the concentration of the acid equals the concentration of its conjugate base, the concentration of H+ would have to be 1000 moles/dm³. The concentration of H2SO4 is 18,67 moles/dm³.

#### Borek ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #6 on: April 08, 2023, 01:53:49 PM »
So my next question is how can H2SO4 have pKa of −3. That would mean that at the time the concentration of the acid equals the concentration of its conjugate base, the concentration of H+ would have to be 1000 moles/dm³. The concentration of H2SO4 is 18,67 moles/dm³.

No.

Please write the formula for the Ka.

Your problem stems from the initial statement of

The pKa is the pH under which the acid is halfway deprotonised.

this is a dangerous simplification, that works nicely only for well behaving weak acids in not tool large concentrations.
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#### Akutni

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• Mole Snacks: +0/-0 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #7 on: April 08, 2023, 04:26:25 PM »
So my next question is how can H2SO4 have pKa of −3. That would mean that at the time the concentration of the acid equals the concentration of its conjugate base, the concentration of H+ would have to be 1000 moles/dm³. The concentration of H2SO4 is 18,67 moles/dm³.

No.

Please write the formula for the Ka.

Your problem stems from the initial statement of

The pKa is the pH under which the acid is halfway deprotonised.

this is a dangerous simplification, that works nicely only for well behaving weak acids in not tool large concentrations.

Ka = [H3O+][A-]/[HA]

The more water we add to pure HA, the more it dissociates. There is an amount of water which causes [HA] = [A-] and then Ka = [H3O+]. For H2SO4 [H3O+] = 1000 mol/dm³. I'd like to understand where the H+ came from.
(thank you for helping me)

#### mjc123

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• Mole Snacks: +294/-12 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #8 on: April 08, 2023, 06:13:15 PM »
Remember that water is involved in the reaction, and that properly
Ka = [H3O+][A-]/[HA][H2O]
Normally we're considering fairly dilute solutions, where water is in large excess and [H2O] can be regarded as unchanging, so we ignore it. But that is far from being the case in your scenario.
Also, at these very high concentrations, the equilibrium "constant" in terms of concentrations will not be constant, as the activity coefficients will be very different from 1.
So this is not a promising way of determining pKa.

#### Akutni

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• Mole Snacks: +0/-0 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #9 on: April 09, 2023, 02:06:43 AM »
Remember that water is involved in the reaction, and that properly
Ka = [H3O+][A-]/[HA][H2O]
Normally we're considering fairly dilute solutions, where water is in large excess and [H2O] can be regarded as unchanging, so we ignore it. But that is far from being the case in your scenario.
Also, at these very high concentrations, the equilibrium "constant" in terms of concentrations will not be constant, as the activity coefficients will be very different from 1.
So this is not a promising way of determining pKa.

So a better formula would be Ka = 55.56 · [H3O+][A-]/[HA][H2O]. For well diluted solutions – [H2O] = 55.56 mol/dm³.
So the Henderson–Hasselbalch equation only works for well diluted solutions and cannot be used for determining Ka just by mixing the acid with water to the half-deprotonation (because then the solution doesn't have [H2O] = 55.56 mol/dm³, the amount of water is limited).

But I found a similar method on Wikipedia. We dilute the acid and then titrate it by a strong base of a half amount (so that the half-neutralisation occurs). Here the water is in large excess, then that should mean pKa = pH.
This method should be correct, but still I can't imagine there would be 1000 mol/dm³ of H+.
« Last Edit: April 09, 2023, 02:20:19 AM by Akutni »

#### Borek ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #10 on: April 09, 2023, 03:53:42 AM »
I can't imagine there would be 1000 mol/dm³ of H+.

That's not what pH -3 means.

pH is not -log of concentration, it is -log of actitivity.

For very diluted solution (like 0.001 M) the difference between activity and concentration is negligible. The more concentrated solution, the higher the difference. Initially activity is lower than the concentration (google for Debye-Huckle theory), but when you get to really concentrated solutions things get really strange.

Besides, I am not convinced pH -3 solutions were ever observed in practice. That doesn't mean we can't determine pKa=-3, we just can't go the half-neutralization path to do so.
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#### Akutni

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• Mole Snacks: +0/-0 ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #11 on: April 09, 2023, 12:38:24 PM »
I can't imagine there would be 1000 mol/dm³ of H+.

That's not what pH -3 means.

pH is not -log of concentration, it is -log of actitivity.

For very diluted solution (like 0.001 M) the difference between activity and concentration is negligible. The more concentrated solution, the higher the difference. Initially activity is lower than the concentration (google for Debye-Huckle theory), but when you get to really concentrated solutions things get really strange.

Besides, I am not convinced pH -3 solutions were ever observed in practice. That doesn't mean we can't determine pKa=-3, we just can't go the half-neutralization path to do so.

One last comment (then I will give up). I would like to understand the method I read about on Wikipedia. It says:

If we add half an equivalent of a base to one equivalent of an acid, the measured pH is equal to the pK of the acid.

I must've been wrong about diluting the acid  unlimitedly in this method, because for different amounts of water different pH would be measured and the equation pH = pK wouldn't work (there would be more A¯ ions with more water – [HA] ≠ [A¯]).
So how much water is used?

I guess not a big amount of water, because that only would solve my problem with the concentration of H3O+ by making the equation 1000 = 55.56 · [H3O+]/[H2O].

Also how does this method work. I thought the acid would've dissociated completely and by eliminating half of the acid by the base (BOH), we would get the concentration of H3O2+ with [A¯] = [BA], but the dissociation is never total (and not at all in weak acids).

#### Borek ##### Re: Is it possible to calculate the pKa of H2SO4
« Reply #12 on: April 09, 2023, 02:14:36 PM »
I would like to understand the method I read about on Wikipedia. It says:

If we add half an equivalent of a base to one equivalent of an acid, the measured pH is equal to the pK of the acid.

This is simply wrong as a general statement.

For reasonably weak acids - like acetic acid - it works. For very weak acids, or strong acids - it doesn't.

Dichloroacetic acid has pKa of 1.5. If you take 0.1M solution and add identical volume of 0.05M NaOH solution - which technically 50% neutralized, has pH not of 1.48 as expected, but 1.93, and the ration of HA/A- is not 1 as expected, but around 1/3 (that's because the free acid dissociated further, so some of HA was converted to H+ and A-).

the equation pH = pK wouldn't work

Because it is in general faulty and an approximation at best, holding only for a specific range of conditions. As long you try to make it a starting point for a further discussion we won't move ahead.