[xshadow]

HI,I have to calculate the pH of a solution 0.0176M of potassium hydrogen phthalate

I have :
pk1=2.89
pk2=5.51

Now  hydrogen phthalate  can reacts in two ways:

HA^- + H₂O <----> H₂A + OH^-   1/pk1=pkb

HA^- + H₂O <----> A^2-+ H₃O⁺


SO if I have an ampholyte why some texts calculate the pH of KHA considering  HA-  only like a "weak monoprotic  acid solutiom HA "  and not an ampholyte ??
https://socratic.org/questions/how-to-calculate-the-ph-of-0-180-g-of-potassium-biphthalate-pka-5-4-in-50-0-ml-o

Isn' t the  ampholytes formula tje best way to calculate pH pf a potassium hydrogen phthalate solution?

Thanks...

[Borek]

why some texts calculate the pH of KHA considering  HA-  only like a "weak monoprotic  acid solutiom HA "  and not an ampholyte ??

Because some people have no idea what they are doing. The answer given is wrong, and it is not that difficult to prove.

He calculated:

[H⁺] = 0.000265 M

so (from the dissociation stoichiometry):

[P^2-] = 0.000265 M

[HP^-] = 0.01763 - 0.000265 M

We have

[tex]K_{a1} = \frac {[H^+][HP^-]}{[H_2P]}[/tex]

so

[tex][H_2P] = \frac {[H^+][HP^-]}{K_{a1}}[/tex]

or

[tex][H_2P] = \frac {0.000265 \times (0.01763 - 0.000265)}{10^{-2.89}} = 0.00357[/tex]

Now, total (analytical) concentration of phthalate is

[tex]C_A = [H_2P]+[HP^-]+[P^{2-}][/tex]

or (using the numbers calculated)

[tex]C_A = 0.00357+(0.01763 - 0.000265)+0.000265 = 0.0212[/tex]

So the phthalate concentration calculated from the equilibrium concentrations is 0.0212 M when we know it should be 0.01763 M. Apparently something is wrong.

Try to calculate pH from the formula for ampholyte and apply the same reasoning to see if the answer leads to contradiction (or not).

[xshadow]

why some texts calculate the pH of KHA considering  HA-  only like a "weak monoprotic  acid solutiom HA "  and not an ampholyte ??

Because some people have no idea what they are doing. The answer given is wrong, and it is not that difficult to prove.

He calculated:

[H⁺] = 0.000265 M

so (from the dissociation stoichiometry):

[P^2-] = 0.000265 M

[HP^-] = 0.01763 - 0.000265 M

We have

[tex]K_{a1} = \frac {[H^+][HP^-]}{[H_2P]}[/tex]

so

[tex][H_2P] = \frac {[H^+][HP^-]}{K_{a1}}[/tex]

or

[tex][H_2P] = \frac {0.000265 \times (0.01763 - 0.000265)}{10^{-2.89}} = 0.00357[/tex]

Now, total (analytical) concentration of phthalate is

[tex]C_A = [H_2P]+[HP^-]+[P^{2-}][/tex]

or (using the numbers calculated)

[tex]C_A = 0.00357+(0.01763 - 0.000265)+0.000265 = 0.0212[/tex]

So the phthalate concentration calculated from the equilibrium concentrations is 0.0212 M when we know it should be 0.01763 M. Apparently something is wrong.

Try to calculate pH from the formula for ampholyte and apply the same reasoning to see if the answer leads to contradiction (or not).

Hi,i I found this:

H⁺= 6.0906* 10^-5 = [ P^2-]
[HP^-] = 0.01763 - 6.0906*10^-5 = 0.017569

[H₂P] = [H⁺]  [HP^- ] / Ka1= 8.3063*10^-4


So
C_HA-= [H2A] + [HA-] + [A2-] = 0.01846M

But ,still, there is some difference witch the analytical conentration 0.0176M

Why?
Thx

EDIT
BUT if I use the expression
α_HA-* C = [HA^-] in order to calculate [HA^-] I'll get :
 [HA^-]= 0.0160M instead of 0.01846M and the sum is equal the the analytical concentration

Is it correct proceeding in thos way?

[Borek]

H⁺= 6.0906* 10^-5 = [ P^2-]

1. Not sure where you got the H⁺ concentration from, that's not the value I got (not that you are very off, but definitely not correct either).

2. You can't assume [P^2-] = [H⁺], as if HP^- is an ampholyte some will get converted to H₂P. I could calculate it this way as it was a direct conclusion of the (erroneous) assumption that HP^- dissociates only - so this approach was consistent with the assumptions made in calculations.

Can't think of an easy way of calculating the equilibrium concentrations to check the mass balance, sorry about that. For now let's agree that it is easy to show the calculations you linked to are self contradictory, perhaps I will have some better idea later.

For now see the attachment. These are results of equilibrium calculation done with specialized software (Buffer Maker in case you wonder), you can check that the pH equals that given by the ampholyte approximation and that values calculated by the program  are in agreement with every mass/charge balance and equilibrium equation describing the solution.

[xshadow]

Sorry so I can't use  the general ampholytes formula?

I used this formula and I get pH=4.215

[H⁺]²= (Ka₁Ka₂C + ka₁k_w ) / Ka₁ +C

Using this formula pH=4.215 . .

[Borek]

I would need to see the derivation to judge, but at first sight it can't be entirely correct, as it doesn't depend on Ka1 (note it cancels out).

The most commonly used one (and the the one I used) is derived here: http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

It is much simpler and a bit limited (won't work for diluted solutions) but is also much easier to remember.

[xshadow]

http://

I would need to see the derivation to judge, but at first sight it can't be entirely correct, as it doesn't depend on Ka1 (note it cancels out).

The most commonly used one (and the the one I used) is derived here: http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

It is much simpler and a bit limited (won't work for diluted solutions) but is also much easier to remember.

The formula I used is the general expression from which I can  get the following 1) and 2) expressions

1) [H⁺]= Sqrt [(C_HA Ka₁ka₂) / C_HA + ka₁

With this expression pH= 4.2153


Using a more approximated formula like:
2) [H⁺]= sqrt (ka₁ka₂) I have pH= 4.2



 2) is a more approximated formula than 1) (so the pH value of 1) is always less " right")

(And the  1) expression is also more approximated than the general one I used  (obtained from the  Charge balance and  the mass  one)


But none of these gives the  pH value of 4.19
!?!?! So strange....

[AWK]

Both formula you uses are still an approximation. They do not take into account the autoprotolysis of water (but this is unimportant in this case).
Much more important is the correction for ionic strength of solution.
For this correction you should use the extended Debye–Hückel equation (also the approximation). For your data this correction will be approximately about -0.02 pH unit.
 Use correction to your more exact result (equation 1)

[xshadow]

Both formula you uses are still an approximation. They do not take into account the autoprotolysis of water (but this is unimportant in this case).
Much more important is the correction for ionic strength of solution.
For this correction you should use the extended Debye–Hückel equation (also the approximation). For your data this correction will be approximately about -0.02 pH unit.
 Use correction to your more exact result (equation 1)

I know that:
Log f_i= -A z_i²* sqrt (I)

For the ionic strenght "I"  which components should I have to consider?
I =  1/2 Σ_i z_i²C_i

Now the charged species are HA- / A2- / OH- /H+ /K+

BUT if I don't know pH I can't calculate the equilibrium concentration of  HA-/ A2-/ H+/OH-   in order to evalutate the ionic strenght

I only know the K⁺ concentration?!
Or I have to calculare the ionic strange witch the concentrations obtained per pH= 4.2,then recalculate the acid contstant K1 and k2 and at last recalculating the pH with the new  constants?

Thanks.

[Borek]

The formula I used is the general expression from which I can  get the following 1) and 2) expressions

1) [H⁺]= Sqrt [(C_HA Ka₁ka₂) / C_HA + ka₁

[H⁺]²= (Ka₁Ka₂C + ka₁k_w ) / Ka₁ +C

This is inconsistent, hard to decipher what the formula really is, can you please elaborate?

Please note the second formula (the one you posted earlier) has no Ka1 (and is inconsistent when it comes to units):

[tex][H^+]^2= \frac {K_{a1}K_{a2}C + K_{a1}K_w}{ K_{a1}} +C = \frac{K_{a1}\times(K_{a2}C + K_w)}{ K_{a1}} +C= K_{a2}C + K_w +C[/tex]

4.19 is calculated by solving full system of equations describing the solution. No wonder the result it produces is a bit different from the result given by approximated methods.



Don't worry about thermodynamic corrections, they are difficult to calculate by hand and will only confuse you at this stage. Let's get the basics right first. Correction is much larger than AWK suggested and has to be calculated iteratively (see the attachment for results corrected according to DH theory).

[xshadow]

τΛ

The formula I used is the general expression from which I can  get the following 1) and 2) expressions

1) [H⁺]= Sqrt [(C_HA Ka₁ka₂) / C_HA + ka₁

[H⁺]²= (Ka₁Ka₂C + ka₁k_w ) / Ka₁ +C

This is inconsistent, hard to decipher what the formula really is, can you please elaborate?

Please note the second formula (the one you posted earlier) has no Ka1 (and is inconsistent when it comes to units):

[tex][H^+]^2= \frac {K_{a1}K_{a2}C + K_{a1}K_w}{ K_{a1}} +C = \frac{K_{a1}\times(K_{a2}C + K_w)}{ K_{a1}} +C= K_{a2}C + K_w +C[/tex]

4.19 is calculated by solving full system of equations describing the solution. No wonder the result it produces is a bit different from the result given by approximated methods.



Don't worry about thermodynamic corrections, they are difficult to calculate by hand and will only confuse you at this stage. Let's get the basics right first. Correction is much larger than AWK suggested and has to be calculated iteratively (see the attachment for results corrected according to DH theory).

That formula came from my slide:





From this expression I can get the other expressions seen before (much approximated)

In this expression there is only  one approximation...the one made  in one of the final steps ,in the denominator


Edit: sorry in the denominator is .. / (Ka1 + C)

[Borek]

OK, I see where it comes from (and now, with Ka1+C being in the denominator, it makes perfect sense, also when it comes to units).

Still, I feel like it is quite impractical - it is much more difficult to remember and for most cases the [itex]pH = \frac 1 2 (pK_{a1}+pK_{a2})[/itex] approximation is perfectly enough. On the page with the derivation (the one I linked earlier) there is a table showing the formula applicability - it is surprisingly wide.

[xshadow]

Π

OK, I see where it comes from (and now, with Ka1+C being in the denominator, it makes perfect sense, also when it comes to units).

Still, I feel like it is quite impractical - it is much more difficult to remember and for most cases the [itex]pH = \frac 1 2 (pK_{a1}+pK_{a2})[/itex] approximation is perfectly enough. On the page with the derivation (the one I linked earlier) there is a table showing the formula applicability - it is surprisingly wide.

Understand!

So this case of pH calculation is an ""exception" ...where these expression don't lead to a correct pH value!?

pH= 4.2 vs 4.19 calculated using that software...
Thx

[Borek]

So this case of pH calculation is an ""exception" ...where these expression don't lead to a correct pH value!?

Whenever we use approximations we end with a value that is a bit off.

As we rarely know Ka values with a high accuracy, as we rarely know concentrations of compounds present with a high accuracy, as we rarely know exact composition of the solution (think about purity of the reagents), as we are never able to use really exact calculations of the activity coefficients, second digit after coma (hundredths of the unit) is almost never exact. For most practical application it is simply not worth wasting time to get it better than "good enough"