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Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #15 on: January 27, 2013, 08:45:52 AM »
I have now realized what I didn't when I typed that up yesterday at 2AM. An obvious oversight on my part:

Ka=([H+]*[salt])/[weak acid] -> [H+]=Ka*[weak acid]/[salt]. If you compare this to the exact equation for a monoprotic acid being buffered:

[weak acid]=Ca-[H+]+(Kw/[H+])
[salt]=Cs+[H+]-(Kw/[H+])

And now buffer composition becomes easy to calculate if you know the pH you desire: activity-correct back to the required [H+], plug that in along with the Ka (given) and Kw (we're taking a constant at 1*10-14 mol2dm-6 I assume) and either Ca or Cs, depending on which one you know, and you can find the other one required to get your pH.

This is ducking the point of your sum because you apparently don't know the pH for sure but it would be useful if you were a) dealing with a monoprotic acid and b) you wanted to check that you have the right amounts in your solution for the pH measured.

How do you calculate the exact pH of a potassium buffer? Known buffer info: -0.5 g/L KH2PO4, +0.5 g/L K2HPO4 in a 1 L solution at a pH of 7.00.

Experimentally, I took four pH readings. One of just the buffer and three of increased dilutions. I started with 5 mL of buffer to 45mL of water, then diluted each concentration by one tenth from there.

I tried the Henderson-Hasselbach equation. It is not giving me an accurate enough pH.
This is a very bad example for testing changes of pH during dilutions. For pH very close to 7 all diluted solutions will show practically the same pH. And, of course, an unabbreviated HH equation given by Big-Daddy with activity corrections should be used. This equation still assumes that nor acid or base undergo protolysis (1000 times dilution of phosphate buffer with pH far from 7 can change pH even by 1, mainly because of protolysis of dihydrogen phosphate)

Moreover phosphates form hydrates and usually the hydrates are sold. Check if you used anhydrous salts.

The OP has multiple salts, of polyprotic acids, together and thus that equation up there will not even come close to an exact calculation! The real deal is much more complicated.

OP if you want to proceed then first let's clarify - calculate Cs1 (analytical concentration of KH2PO4) and Cs2 (K2HPO4) in your final solution where they're combined. What do you get? (Calculate moles first by n=m/Mr then use that to calculate the concentration in the 1 L solution by C=n/V, for each salt.) We can talk about what to do next after but it's likely that if you're desperate for an exact answer you'll need to read up.

Offline Borek

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Re: Calculating exact pH of a buffer
« Reply #16 on: January 27, 2013, 11:51:38 AM »
Just a few remarks, unfortunately I have no time to delve deeper at the moment.

First of all - you need to be careful about what is the analytical concentration and what is the equilibrium concentration. HH equation is OK when you put equilibrium concentrations into it. Problems start when you forget that equilibrium concentrations differ from the analytical ones (calculated from the amount of the acid and salt/conjugate base put into solution). See discussion here:

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch

I am not convinced that the version of the equation that you listed is better. I don't understand the reasoning behind using [itex][H^+]+\frac{K_w}{H^+}[/itex] as a correcting factor.

The OP has multiple salts, of polyprotic acids, together and thus that equation up there will not even come close to an exact calculation! The real deal is much more complicated.

Actually as long as we are close enough to pKa2 (say any pH between 5 and 9) we don't have to worry - both other dissociations steps are far enough to not matter.

In the end I would use Buffer Maker ;)
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Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #17 on: January 27, 2013, 01:05:26 PM »
Just a few remarks, unfortunately I have no time to delve deeper at the moment.

First of all - you need to be careful about what is the analytical concentration and what is the equilibrium concentration. HH equation is OK when you put equilibrium concentrations into it. Problems start when you forget that equilibrium concentrations differ from the analytical ones (calculated from the amount of the acid and salt/conjugate base put into solution). See discussion here:

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch

I am not convinced that the version of the equation that you listed is better. I don't understand the reasoning behind using [itex][H^+]+\frac{K_w}{H^+}[/itex] as a correcting factor.

Have a look at the general buffer derivation of the Henderson-Hasselbalch formula from the mass balances and charge balances. The exact formula I wrote is the one you will arrive at.[itex][H^+]+\frac{K_w}{H^+}[/itex] is an extremely useful factor in this type of chemistry for exact calculation. (See "Aqueous Acid-Base Equilibria and Titrations", Robert de Levie, Oxford Chemistry Press, for some more detail.)

The OP has multiple salts, of polyprotic acids, together and thus that equation up there will not even come close to an exact calculation! The real deal is much more complicated.

Actually as long as we are close enough to pKa2 (say any pH between 5 and 9) we don't have to worry - both other dissociations steps are far enough to not matter.
[/quote]

For an exact calculation you would probably need to treat the dissociation steps separately. Obviously let's remember that there's no such thing as an exact calculation of pH because finally the best we have for ionic activity corrections is the approximate Davies equation.

Offline Borek

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Re: Calculating exact pH of a buffer
« Reply #18 on: January 27, 2013, 03:14:23 PM »
Have a look at the general buffer derivation of the Henderson-Hasselbalch formula from the mass balances and charge balances. The exact formula I wrote is the one you will arrive at.[itex][H^+]+\frac{K_w}{H^+}[/itex] is an extremely useful factor in this type of chemistry for exact calculation. (See "Aqueous Acid-Base Equilibria and Titrations", Robert de Levie, Oxford Chemistry Press, for some more detail.)

Will try if I will be able to find the book, in my position it can be rather difficult.

Besides, I have a feeling we are comparing apples and oranges here. It could be that de Levie uses the term "HH equation" in a non standard way. Typically HH equation is just the one we got by rearranging dissociation constant definition, it is not something derived using mass and charge balance.

Quote
Quote
Quote
The OP has multiple salts, of polyprotic acids, together and thus that equation up there will not even come close to an exact calculation! The real deal is much more complicated.

Actually as long as we are close enough to pKa2 (say any pH between 5 and 9) we don't have to worry - both other dissociations steps are far enough to not matter.

For an exact calculation you would probably need to treat the dissociation steps separately. Obviously let's remember that there's no such thing as an exact calculation of pH because finally the best we have for ionic activity corrections is the approximate Davies equation.

Yes, but it depends on what you mean by "exact". Exact in general means solving 5th degree polynomial for phosphoric acid alone, without adding base. That's not a thing to be done manually, numerical approach is the only viable one - and then it is better to use ready software, and to not solve each case separately.

Also note "exact" equation that you listed is for sure not exact in this context, it is obvious it relies on some simplifying assumptions as well. Easy to check that it is not self consistent, let's take an equimolar solution of acetic acid and acetate:

Ca = 5×10-4 M
Cs = 5×10-4 M
Ka = 10-4.757

Using full blown approach to the equilibrium one can find exact pH of the solution to be 4.79, or [H+] = 1.639×10-5 M (calculated using BATE, ionic strength ignored, but it is low enough to not substantially change the final result; equilibrium concentrations of [HAcetate]=4.836×10-4 M, [Acetate-]=5.164×10-4 M, you can easily check that these numbers perfectly fit initial conditions, mass & charge balances and Ka). Let's insert these values into the "exact HH" equation:

[tex]\begin{align}
[H^+] &= K_a \frac {C_a - [H^+] + \frac {K_w}{[H^+]}}{C_s + [H^+] - \frac {K_w}{[H^+]}}\\
&= 10^{-4.757} \frac {5\times 10^{-4} + 1.639\times 10^{-5} - \frac {10^{-14}}{1.639\times 10^{-5}}}{5\times 10^{-4} - 1.639\times 10^{-5} + \frac {10^{-14}}{1.639\times 10^{-5}}}\\
&=8.86\times 10^{-6}\end{align}[/tex]

So we have 1.639×10-5 = 8.86×10-6 which is obviously not true - and it means the equation is inconsistent with the numbers given. As these numbers do describe correctly the solution, equation must be wrong.

Quote
Obviously let's remember that there's no such thing as an exact calculation of pH because finally the best we have for ionic activity corrections is the approximate Davies equation.

Once again, it depends. Davies equation is an approximation used for solutions with ionic strengths above 0.1, as long as the ionic strength is below 0.1 Debye-Huckel theory (and Debye-Huckel equation, especially in the rarely used version taking ionic radii into account) gives pretty good results.
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Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #19 on: January 27, 2013, 05:00:12 PM »
Have a look at the general buffer derivation of the Henderson-Hasselbalch formula from the mass balances and charge balances. The exact formula I wrote is the one you will arrive at.[itex][H^+]+\frac{K_w}{H^+}[/itex] is an extremely useful factor in this type of chemistry for exact calculation. (See "Aqueous Acid-Base Equilibria and Titrations", Robert de Levie, Oxford Chemistry Press, for some more detail.)

Will try if I will be able to find the book, in my position it can be rather difficult.

Besides, I have a feeling we are comparing apples and oranges here. It could be that de Levie uses the term "HH equation" in a non standard way. Typically HH equation is just the one we got by rearranging dissociation constant definition, it is not something derived using mass and charge balance.

Quote
Quote
Quote
The OP has multiple salts, of polyprotic acids, together and thus that equation up there will not even come close to an exact calculation! The real deal is much more complicated.

Actually as long as we are close enough to pKa2 (say any pH between 5 and 9) we don't have to worry - both other dissociations steps are far enough to not matter.

For an exact calculation you would probably need to treat the dissociation steps separately. Obviously let's remember that there's no such thing as an exact calculation of pH because finally the best we have for ionic activity corrections is the approximate Davies equation.

Yes, but it depends on what you mean by "exact". Exact in general means solving 5th degree polynomial for phosphoric acid alone, without adding base. That's not a thing to be done manually, numerical approach is the only viable one - and then it is better to use ready software, and to not solve each case separately.

Also note "exact" equation that you listed is for sure not exact in this context, it is obvious it relies on some simplifying assumptions as well. Easy to check that it is not self consistent, let's take an equimolar solution of acetic acid and acetate:

Ca = 5×10-4 M
Cs = 5×10-4 M
Ka = 10-4.757

Using full blown approach to the equilibrium one can find exact pH of the solution to be 4.79, or [H+] = 1.639×10-5 M (calculated using BATE, ionic strength ignored, but it is low enough to not substantially change the final result; equilibrium concentrations of [HAcetate]=4.836×10-4 M, [Acetate-]=5.164×10-4 M, you can easily check that these numbers perfectly fit initial conditions, mass & charge balances and Ka). Let's insert these values into the "exact HH" equation:

[tex]\begin{align}
[H^+] &= K_a \frac {C_a - [H^+] + \frac {K_w}{[H^+]}}{C_s + [H^+] - \frac {K_w}{[H^+]}}\\
&= 10^{-4.757} \frac {5\times 10^{-4} + 1.639\times 10^{-5} - \frac {10^{-14}}{1.639\times 10^{-5}}}{5\times 10^{-4} - 1.639\times 10^{-5} + \frac {10^{-14}}{1.639\times 10^{-5}}}\\
&=8.86\times 10^{-6}\end{align}[/tex]

So we have 1.639×10-5 = 8.86×10-6 which is obviously not true - and it means the equation is inconsistent with the numbers given. As these numbers do describe correctly the solution, equation must be wrong.

Quote
Obviously let's remember that there's no such thing as an exact calculation of pH because finally the best we have for ionic activity corrections is the approximate Davies equation.

Once again, it depends. Davies equation is an approximation used for solutions with ionic strengths above 0.1, as long as the ionic strength is below 0.1 Debye-Huckel theory (and Debye-Huckel equation, especially in the rarely used version taking ionic radii into account) gives pretty good results.

The equation I wrote cannot be used to solve for exact concentrations in a solution with multiple buffers or polyprotic buffers, so it will not work in this case. I already pointed that out. At the same time, it is certainly better than the Henderson-Hasselbalch equation (which, as I know it, is not a real rearrangement of the dissociation constant but rather an approximated one).

However, in the case you proposed, the equation I outlined works perfectly. Try putting the numbers in again and you'll see [H+] comes out as 1.63877*10-5 M. In fact this is without a doubt the equation used by BATE itself to find [H+]. The equation will work for all cases involving a single salt of a monoprotic acid with an acid (and if it were a single salt of a monoprotic acid with a base, one would simply have to replace all the H+ with OH-, Ka with Kb, and Ca with Cb and you could produce [OH-] and subsequently [H+]).

To do the calculations this thread asks for is a much more complicated task. First I would like the OP to find Cs1 and Cs2 for us. Then we will work on finding the exact [H+]. Beyond that there are so many different considerations - ionic strength modifications, activity corrections, etc. - that I leave it to a more energetic soul to solve this problem completely! Like I said though, there is no way to be completely exact with activity corrections - the Davies' equation may give good approximations but we still don't have an "exact" way to calculate ionic activity.

Offline Borek

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Re: Calculating exact pH of a buffer
« Reply #20 on: January 27, 2013, 06:23:44 PM »
The equation I wrote cannot be used to solve for exact concentrations in a solution with multiple buffers or polyprotic buffers, so it will not work in this case. I already pointed that out.

And I already explained that for all practical purposes pKa1 and pKa3 don't matter here, as for pHs involved neither of these equilibria change the situation. It can be easily shown that at pH 5 [H3PO4] is less than 10-3 and at pH 9 less than 10-9 of the total amount of phosphoric acid present in the solution. Situation is similar for PO43- produced in the third dissociation step. As long as pH is between 5 and 9 ignoring H3PO4 and PO43- we don't make errors higher that 0.1%, these are negligible.

Quote
At the same time, it is certainly better than the Henderson-Hasselbalch equation (which, as I know it, is not a real rearrangement of the dissociation constant but rather an approximated one).

All sources I know list Henderson-Hasselbalch equation as just rearranged form of the dissociation constant and that's the usual way it is derived. Looks like it can be historically not true, but that's the way it is usually shown.

From what I have learned outside of the forum (thanks to AWK) equation you listed is not called a Henderson-Hasselbalch equation, but Charlot equation. It is not the equation used by BATE, as BATE uses much more advanced version describing mixture of fourprotic acid and foruprotic base.

There is some mistake in the calculations I presented. It is not the typo I did while entering LaTeX (switched signs), as calculations were done with the correct signs. But it is already past midnight here and I need to wake up early tomorrow, so I can't investigate further.
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Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #21 on: January 28, 2013, 03:36:09 PM »
The equation I wrote cannot be used to solve for exact concentrations in a solution with multiple buffers or polyprotic buffers, so it will not work in this case. I already pointed that out.

And I already explained that for all practical purposes pKa1 and pKa3 don't matter here, as for pHs involved neither of these equilibria change the situation. It can be easily shown that at pH 5 [H3PO4] is less than 10-3 and at pH 9 less than 10-9 of the total amount of phosphoric acid present in the solution. Situation is similar for PO43- produced in the third dissociation step. As long as pH is between 5 and 9 ignoring H3PO4 and PO43- we don't make errors higher that 0.1%, these are negligible.

Any factor you consider negligible takes you further away from an "exact" solution, which is what the OP wanted.


Quote
At the same time, it is certainly better than the Henderson-Hasselbalch equation (which, as I know it, is not a real rearrangement of the dissociation constant but rather an approximated one).

All sources I know list Henderson-Hasselbalch equation as just rearranged form of the dissociation constant and that's the usual way it is derived. Looks like it can be historically not true, but that's the way it is usually shown.

From what I have learned outside of the forum (thanks to AWK) equation you listed is not called a Henderson-Hasselbalch equation, but Charlot equation. It is not the equation used by BATE, as BATE uses much more advanced version describing mixture of fourprotic acid and foruprotic base.

There is some mistake in the calculations I presented. It is not the typo I did while entering LaTeX (switched signs), as calculations were done with the correct signs. But it is already past midnight here and I need to wake up early tomorrow, so I can't investigate further.
[/quote]

BATE does not generally deal with salts at all does it? That is more the realm of the Buffer Maker. And yes, there is no doubt that the equation will boil down to the same thing, except that the Buffer Maker is kitted out to handle multiple polyprotic acids and this equation applies only to a single monoprotic acid's salt.

There is no "more advanced" way of doing this, for a single salt of monoprotic acid in solution with monoprotic acid (I said this already). This equation has not undergone approximations; it is derived straight from the mass and charge balances. For that reason it is absolutely exact when it comes to calculating concentrations (assuming there are no interfering equilibria).

I have seen your link. Yes it is the Charlot equation. But if we have this exact version then for what purposes is Henderson-Hasselbalch useful?

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Re: Calculating exact pH of a buffer
« Reply #22 on: January 29, 2013, 01:50:42 AM »
I have seen your link. Yes it is the Charlot equation. But if we have this exact version then for what purposes is Henderson-Hasselbalch useful?

Even using an exact equation we still do inaccurate calculations because all our data are not exact.

HH equation is simple and sufficient for most of our purposes.
As I stated before the problem is wrongly defined. Dilution of buffer solution with pH 7 should not cause distinct change of pH. And of course, HH equation in this case is completely unsuitable. Chariot equation is also unsuitable  for exact pH calculations in this case since is devoted only for buffer solutions containing monoprotic acids or bases. Change of pH caused by changing activities is also negligible (~0.01  pH unit for "monoprotic buffers", and in this case may be, say ~0.03 pH unit). Much more important may be change in protolysis of H2PO4- anion during 100 times dilution though at pH close to 7  Kw will brake this change.
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Offline curiouscat

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Re: Calculating exact pH of a buffer
« Reply #23 on: January 29, 2013, 09:23:09 AM »
With the amount of computing power and software so easily available these days, shouldn't we soon move to adopting a full simultaneous equilibrium approach?

Forget the specific simplifications and their restrictions always set up the multiple equilibria and let the solvers crunch it out.

Kinda like what Borek's software does.

Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #24 on: January 29, 2013, 12:51:13 PM »
I have seen your link. Yes it is the Charlot equation. But if we have this exact version then for what purposes is Henderson-Hasselbalch useful?

Even using an exact equation we still do inaccurate calculations because all our data are not exact.

HH equation is simple and sufficient for most of our purposes.
As I stated before the problem is wrongly defined. Dilution of buffer solution with pH 7 should not cause distinct change of pH. And of course, HH equation in this case is completely unsuitable. Chariot equation is also unsuitable  for exact pH calculations in this case since is devoted only for buffer solutions containing monoprotic acids or bases. Change of pH caused by changing activities is also negligible (~0.01  pH unit for "monoprotic buffers", and in this case may be, say ~0.03 pH unit). Much more important may be change in protolysis of H2PO4- anion during 100 times dilution though at pH close to 7  Kw will brake this change.

It is very possible to solve this exactly as curiouscat pointed out but I for one do not have the energy to type up the necessary equations into mathematical format here.

The use of HH becomes somewhat superfluous except for quick, easy ideas of the "region" of the pH, since if you have even a few more seconds available you only need to plug in six more numbers (two times [H+] and one times Kw, on both the top and bottom) to get whatever result you want. Literally, if HH is Ka=([H+]*[salt])/[weak acid], then

[weak acid]=Ca-[H+]+(Kw/[H+])
[salt]=Cs+[H+]-(Kw/[H+])

From the exact Charlot equation. Which means that for anything you'd use the HH for, you need only to plug in 3 more numbers and you'll get the exact result! Thus the HH is ok for a very very approximate guess, e.g. an order of magnitude estimate, but shouldn't even be discussed when you're talking about exact calculations - because it's so easy to just use the more exact version! Of course if you were calculating H+ it would be perfectly excusable just to solve the HH equation because a cubic is not great.

Offline Borek

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Re: Calculating exact pH of a buffer
« Reply #25 on: January 29, 2013, 02:39:04 PM »
The use of HH becomes somewhat superfluous except for quick, easy ideas of the "region" of the pH, since if you have even a few more seconds available you only need to plug in six more numbers (two times [H+] and one times Kw, on both the top and bottom) to get whatever result you want. Literally, if HH is Ka=([H+]*[salt])/[weak acid], then

[weak acid]=Ca-[H+]+(Kw/[H+])
[salt]=Cs+[H+]-(Kw/[H+])

From the exact Charlot equation. Which means that for anything you'd use the HH for, you need only to plug in 3 more numbers and you'll get the exact result! Thus the HH is ok for a very very approximate guess, e.g. an order of magnitude estimate, but shouldn't even be discussed when you're talking about exact calculations - because it's so easy to just use the more exact version! Of course if you were calculating H+ it would be perfectly excusable just to solve the HH equation because a cubic is not great.

Please explain - in more details - the procedure you are proposing. Give some example and solve it.
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Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #26 on: January 29, 2013, 07:44:00 PM »
The use of HH becomes somewhat superfluous except for quick, easy ideas of the "region" of the pH, since if you have even a few more seconds available you only need to plug in six more numbers (two times [H+] and one times Kw, on both the top and bottom) to get whatever result you want. Literally, if HH is Ka=([H+]*[salt])/[weak acid], then

[weak acid]=Ca-[H+]+(Kw/[H+])
[salt]=Cs+[H+]-(Kw/[H+])

From the exact Charlot equation. Which means that for anything you'd use the HH for, you need only to plug in 3 more numbers and you'll get the exact result! Thus the HH is ok for a very very approximate guess, e.g. an order of magnitude estimate, but shouldn't even be discussed when you're talking about exact calculations - because it's so easy to just use the more exact version! Of course if you were calculating H+ it would be perfectly excusable just to solve the HH equation because a cubic is not great.

Please explain - in more details - the procedure you are proposing. Give some example and solve it.

OK, let's take a high school problem (to which one would usually apply the Henderson-Hasselbalch). In this particular example please note that a very similar value could perhaps have been reached simply using the Henderson-Hasselbalch approximation, but as you will see if you look at the Wikipedia page, the Charlot equation does not rely on the same assumptions (the reason this problem probably yields the same value both ways is because it is a high school problem and thus designed to be solved by pupils with no idea of any solution beyond the HH).

Because of my inability to manipulate the mathematical font on this forum everything will be expressed in plain text but the brackets will make it clear what goes where.

"Calculate the mass of CH3COONa that has to be added to 100 cm3 of a 1.00 mol/dm3 solution of CH3COOH to make a buffer solution of pH=4.38. Ka[CH3COOH]=1.8*10-5 mol/dm3."

(Assume that the volume of the solution does not change on addition of the salt; that Kw=1.00*10-14; etc., etc.)

Now first we'll describe what we know, in the terms of our Charlot equation:

Ca=1.00 mol/dm3
[H+]=10-4.38 mol/dm3
Ka=1.8*10-5 mol/dm3
Kw=1.00*10-14

We appear only to lack a Cs from our Charlot equation, and that's what we want to find anyway. Rearrange the Charlot equation for Cs (can be done in your head):

Cs=((Ka*(Ca-[H+]+(Kw/[H+])))/[H+])-[H+]+(Kw/[H+])

The volume of the solution does not change so ns=V*Cs. ms=ns*Mr[CH3COONa]. And now our solution is ready:

ms=V*Cs*Mr[CH3COONa]
ms=V*Mr[CH3COONa]*((Ka*(Ca-[H+]+(Kw/[H+])))/[H+])-[H+]+(Kw/[H+])

Let's substitute in the numbers (V=0.1, Mr=82, and the rest as written above):

ms=3.54019 g

Which is, of course, correct. Now in this case using the Henderson-Hasselbalch would have worked fine (to my calculation, it results in an answer of 3.54068 g), but that's not as accurate as it could be. We've increased the accuracy and limited our approximations (to an extent - many are still being made but at least none in the concentration-solving stage; now our approximations are in the conversion from pH to [H+], which is vastly more complicated to even attempt properly than just [H+]=10-pH), and there is virtually no added work - just three more added numbers!

But why are we discussing this again? Neither the Charlot equation nor the Henderson-Hasselbalch, as far as I can see, can come anywhere near to the task at hand. We need a more serious polyprotic method (don't ask me to type one up).

Offline Big-Daddy

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Re: Calculating exact pH of a buffer
« Reply #27 on: January 29, 2013, 07:58:22 PM »
With the amount of computing power and software so easily available these days, shouldn't we soon move to adopting a full simultaneous equilibrium approach?

Forget the specific simplifications and their restrictions always set up the multiple equilibria and let the solvers crunch it out.

Kinda like what Borek's software does.

Any proper method would do this, but who wants to conceive, and then plug in the known numbers for, this many equilibria? I would prefer to search for general formulae to deal with solutions of salts.

Certain other facts pique my curiousity about this question. Is there a difference in exact concentration calculations between an acid whose species range from H4A to A4- and one whose species range from H4A2+ to A2-? (Or a similar salt)

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Re: Calculating exact pH of a buffer
« Reply #28 on: January 30, 2013, 02:46:17 AM »

"Calculate the mass of CH3COONa that has to be added to 100 cm3 of a 1.00 mol/dm3 solution of CH3COOH to make a buffer solution of pH=4.38. Ka[CH3COOH]=1.8*10-5 mol/dm3."

Let's substitute in the numbers (V=0.1, Mr=82, and the rest as written above):


For exact calculations you should use an exact molecular mass and activities.
In above example neglecting activities gives a few tens percent error.

Quote
Certain other facts pique my curiousity about this question. Is there a difference in exact concentration calculations between an acid whose species range from H4A to A4- and one whose species range from H4A2+ to A2-? (Or a similar salt)
Math is the same, values of activity coefficients - different.
AWK

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Re: Calculating exact pH of a buffer
« Reply #29 on: January 30, 2013, 04:49:48 AM »
ms=3.54019 g

Which is, of course, correct. Now in this case using the Henderson-Hasselbalch would have worked fine (to my calculation, it results in an answer of 3.54068 g), but that's not as accurate as it could be.

3.54068 g vs 3.54019 g. You have just proven it is almost perfect. If you forgot - whole discussion started with your claim:

If you're using the Henderson-Hasselbalch approximation then activity coefficients are unlikely to be relevant - using the approximation will already take you 0.1-0.4 pH points away from a true value.

Yes, HH equations has its limitations, but using more complicated method when it is not needed is rarely a good thing. HH equation works pretty nicely for reasonably concentrated solutions and pH somewhere between 3-11 - and these are parameters of most of the buffers used in the lab practice.
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