[Technicalhuman]

Say I have 5M of HCl and 5M of CH3COOH in 1L of solution. And we are asked to find the pH of this solution. How are we going to find this?

I was thinking that since HCl would dissociate completely the concentration of H+ be 5M initially and then we can use the Ka value of the CH3COOH to get the number of moles of H+ produced by that reversible reaction. Then adding it to number of moles from the H+ to get the total then finally dividing it by 1L and -log it. However, I also thought since acetate ions are being produced, wouldn't there also be a reaction between the acetate ion and the H+ ions?

So I'm not sure if my method is correct. I think that we can find the concentration of H+ using my proposed method and writing an ICE table for the reaction between H+ +CH3COO-  CH3COOH to get the new concentration of H+ then using the -log to find my pH. But I'm not sure if this would work because I'm assuming that the reactions do not occur simultaneously but go through step by step. But this wouldn't affect the final answer right?

Thanks in advance for the help.

[Borek]

Your suggested approach is a correct one.

Technically you should take into account also Ka for HCl, but it won't change the result much. Much more important thing would be to remember about ionic strength of the solution, but it is too high for reasonable results anyway.

[Technicalhuman]

Your suggested approach is a correct one.

Technically you should take into account also Ka for HCl, but it won't change the result much. Much more important thing would be to remember about ionic strength of the solution, but it is too high for reasonable results anyway.

Hi Borek, thanks for the reply.

So my approach would be to first find the concentration of H⁺ ions produced by the HCl. Subsequently I would use the Ka for CH3COOH CH3COO- + H+ using ICE table and getting the concentrations of new concentrations of H+ and CH3COO-. Lastly, we use the reaction between CH3COO- +H+ CH3COOH equilibrium constant which would just be 1/Ka of the hydrolysis of CH3COOH to get the final concentration of H+ then using -log the concentration the pH would be determined?

Thank you

[Borek]

So my approach would be to first find the concentration of H⁺ ions produced by the HCl. Subsequently I would use the Ka for CH3COOH CH3COO- + H+ using ICE table and getting the concentrations of new concentrations of H+ and CH3COO-

And you are done - you don't need any further steps. Just remember that initial concentration of H⁺ is 5M.

Note that 5M/5M solution is so concentrated calculating pH of such a mixture won't yield any reliable result.

[Technicalhuman]

So my approach would be to first find the concentration of H⁺ ions produced by the HCl. Subsequently I would use the Ka for CH3COOH CH3COO- + H+ using ICE table and getting the concentrations of new concentrations of H+ and CH3COO-

And you are done - you don't need any further steps. Just remember that initial concentration of H⁺ is 5M.

Note that 5M/5M solution is so concentrated calculating pH of such a mixture won't yield any reliable result.

Hi Borek thanks for the reply.

But why don't I have to use the reaction between the H+ and acetate ion? I thought that there might be some reaction between them?

Thanks in advance.

[Borek]

You already took it into account in the dissociation equilibrium - it consist of the dissociation of the acetic acid and protonation of acetate.

[Babcock_Hall]

I am not sure that I am following this discussion.  if there is 5 M HCl, I would predict that the concentration of acetate ion would be very close to zero.  If anything I might expect that acetic acid itself would pick up a proton.  What am I missing here?

[Technicalhuman]

You already took it into account in the dissociation equilibrium - it consist of the dissociation of the acetic acid and protonation of acetate.

Hi Borek thanks again for the reply.

What do you mean by that? I don't quite understand why the reaction CH3COOH+H2O CH3COO- +H3O+ covers the reaction between the acetate ion and H3O+. I see that that reaction I refer here is just this reverse reaction. But shouldn't we also take this into account for the actual H+ concentration?

Thanks so much for the help

[Borek]

I don't quite understand why the reaction CH3COOH+H2O CH3COO- +H3O+ covers the reaction between the acetate ion and H3O+.

These are TWO reactions, forward and backward. You even used the arrow that shows that.

Do you know kinetic approach to equilibrium? It says that reaction goes in both directions at the same time, and equilibrium is when forward and backward speeds are identical - while reactions still go, concentrations don't change.

[Technicalhuman]

I don't quite understand why the reaction CH3COOH+H2O CH3COO- +H3O+ covers the reaction between the acetate ion and H3O+.

These are TWO reactions, forward and backward. You even used the arrow that shows that.

Do you know kinetic approach to equilibrium? It says that reaction goes in both directions at the same time, and equilibrium is when forward and backward speeds are identical - while reactions still go, concentrations don't change.

Oh yeah! I forgot about that. So since the back reaction has been taken cared off that it the final pH got it.

Thanks Borek.

[Big-Daddy]

I am not sure that I am following this discussion.  if there is 5 M HCl, I would predict that the concentration of acetate ion would be very close to zero.  If anything I might expect that acetic acid itself would pick up a proton.  What am I missing here?

I think it's ok. The concentration of the acetate ion will certainly turn out to be nearly 0. Since we don't have any Kb for the pick-up by ethanoic acid of hydrogen, we can't say. It's just an equilibrium yield calculation:

[tex]K_a = \frac{c_{H+} \cdot c_{CH3COO-}}{CH3COOH}[/tex]
or
[tex]K_a = \frac{(c_{0,H+} + \Delta_{H+}) \cdot (c_{0,CH3COO-} + \Delta_{H+})}{c_{0,CH3COOH} - \Delta_{H+}}[/tex]

And now we know that initial concentration of acetate is 0 so you just plug that in and solve for change in [H+]. It cannot be very big and thus acetate cannot be very big.

I am not entirely convinced this is an exact method but anyway. I know that you cannot use ICE tables for an exact calculation with multiple equilibrium constants, but maybe it is ok here because we only have 1.

[magician4]

in my opinion, in this case we safely could assume that the contribution of acetic acid to total pH is negligible for all practical purposes

I much more would worry about the "5 mole per liter" part ref. HCl , as in this case we'd be way beyond "activity [itex] \approx [/itex] concentration" , and hence all simple approaches become debateable

taking into account the "contribution of acetic acid " part here is, compared to this , looking at the false problem


regards

Ingo

[Big-Daddy]

We're talking about a 5M solution of HCl. The absolute maximum limit where Debye-Huckel's extended equation could be useful is with an ionic strength of 1. What approach relates activity to concentration at such high ionic strength? The only equation I know of is the Robinson-Stokes equation and it's listed in little detail in my book, and virtually no detail anywhere I can find online.

[magician4]

to the best of my knowledge, there are no reliable "simple" approaches for such high concentrations - let alone that the meaning of "pH" becomes kind of debateable here

remember: "5 M HCl + 5 M AcOH" would mean that there would be like 5 M H⁺ , 5 M Cl^- and 5 M (next to undissociated) AcOH competing for approx. 55 moles of water to hydrolyse them

if "Cl^-_(aq.)" and "H⁺_(aq.)" would be expressed as " [Cl (H₂O)₆)]^- " and " [H (H₂O)₄]⁺ " (which is the least of hydrate hull we should consider), it is very obvious that with approx. 50 moles of water required here there is next to no "free" water left  ...

... which would mean that in a certain sense we're not talking a "waterbased" system any more , and not only would "conc. [itex] \neq [/itex]  activity " become a problem, but that the waterbased defined pH - value would become meaningless (as we're more or less dealing with a "solution" of named hydrate - complexes in acetic acid instead )


regards

Ingo

[Big-Daddy]

I see what you're saying. The method used would have to be general enough to consider the activity of the solvent. In that case we could (in principle - we probably can't actually?) calculate the activities of all the species in the mixture. At that point we then get the opportunity to redefine pH on the basis that water's activity is no longer considered 1.