# Having trouble understanding pH = pKa + log([A-]/[HA])

In summary: Ah, I think I finally got what your problem is. Textbooks is a bit wrong. What Henderson-Hasselbalch equation says is that at pH=pKa the ratio of concentrations of the acid and conjugate base equals 1. For monoprotic acid that's equivalent to saying they are 50:50. For multiprotic acid that's no longer the case - while the ratio of concentrations in the conjugate pair is still 1:1, concentrations of other forms are not zero, so the 50:50 doesn't have to hold (although - depending on the acid - it can be quite close to that).In summary, textbooks state that when [H+] = Ka, there will be 50%
So I have a problem with the equation: pH = pKa + log([A-]/[HA])

The textbook states that when [H+] = Ka, there will be 50% A- and 50% HA.

However, this won't work when the coefficients of the reactants to the products are not the same. The equation only works for reactions where it is a 1:1 mole ratio of reactants to products.

So, my question is, why is it generalized and said that pH = pKa at half-equivalence?

What if we have H3PO4?
Wouldn't the overall equation be: H3PO4 + H2O <-> 3 [H3O+] +[PO4 3-]
and then that equation won't work?

It is about a single step of dissociation, and Ka used in the Henderson-Hasselbalch equation is a stepwise contant, not the overall one.

Borek said:
It is about a single step of dissociation, and Ka used in the Henderson-Hasselbalch equation is a stepwise contant, not the overall one.
That makes a lot of sense. Then for all acids (and bases) it will be a 1:1 <-> 1:1 mole ratio, right? Otherwise this falls apart?

Well yes, just look at the definition of an acid dissociation constant Ka. The ratio of concentration of acid and its conjugate base is 1 when [H+] = Ka. Or pH = pKa, same thing.

This then is true for each of the three equilibria you can write for the successive dissociations.

What is not true is that this gives 50% of total phosphate to each of the two forms - that is true only when there are only two forms. (Though actually when, as with phosphate and often, when the pK's are well separated then it is still a pretty good approximation, e.g. when making phosphate buffers you only need to consider one dissociation.)

That makes a lot of sense. Then for all acids (and bases) it will be a 1:1 <-> 1:1 mole ratio, right? Otherwise this falls apart?

Depends on what you mean by "all".

It works for each conjugate acid/base pair separately. Each conjugate pair has its own Ka value, and for pH=pKa concentrations of these particular conjugate acid and base will be identical.

epenguin said:
Well yes, just look at the definition of an acid dissociation constant Ka. The ratio of concentration of acid and its conjugate base is 1 when [H+] = Ka. Or pH = pKa, same thing.

This then is true for each of the three equilibria you can write for the successive dissociations.

What is not true is that this gives 50% of total phosphate to each of the two forms - that is true only when there are only two forms. (Though actually when, as with phosphate and often, when the pK's are well separated then it is still a pretty good approximation, e.g. when making phosphate buffers you only need to consider one dissociation.)
I see.
So in reality, when you have 3 successive deprotonations of H3PO4 in the same solution, such as:
H3PO4 <-> [H2PO4-] + [H+]
[H2PO4-] <-> [HPO4 2-] + [H+]
[HPO4 2-] <-> [PO4 3-] + [H+]

At the last dissociation, when pH = pKa, we actually have some of the H3PO4 and [H2PO4]- still present right? So the it is not exactly 50% to 50%, but close enough due to the difference in the pKa values (which makes sense, but I want to make sure I have this right).

Ah, I think I finally got what your problem is. Textbooks is a bit wrong. What Henderson-Hasselbalch equation says is that at pH=pKa the ratio of concentrations of the acid and conjugate base equals 1. For monoprotic acid that's equivalent to saying they are 50:50. For multiprotic acid that's no longer the case - while the ratio of concentrations in the conjugate pair is still 1:1, concentrations of other forms are not zero, so the 50:50 doesn't have to hold (although - depending on the acid - it can be quite close to that).

For example, for 0.1 M phosphoric acid partially neutralized so that pH=pKa2=7.2

 H3PO4 4.46×10-7 M 0.0% H2PO4- 0.05 M 50.0% HPO42- 0.05 M 50.0% PO43- 3.53×10-7 M 0.0%

but for citric acid with much closer pKa values (again, 0.1 M and pH=pKa2=4.76)

 C6H8O7 1.13×10-3 M 1.1% C6H7O7- 0.0488 M 48.8% C6H6O72- 0.0488 M 48.8% C6H5O73- 1.13×10-3 M 1.1%

In the case of polyprotic acids confusion is usually avoided by naming the successive dissociation constants with number: Ka1, Ka2, Ka3 and pKa1, pKa2, pKa3 &c.

In the case of 0.1 M phosphate when pH = pKa2 then [H3PO4] = 4.3×10-7 and [PO43-] = 3.6 × 10-7, while [H2PO4-] is 0.0499992, not exactly 0.05 but exactly equal to [HPO42-].

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For buffer systems using polyprotic weak acids, a simplification to this conundrum is to assume all H+ come from the 1st ionization step. Except for some of the lower pKa value weak acids, subsequent ionization steps make very little contribution to the total H+ ion concentration. Even so, the common ion effect would limit the effect of further ionizations beyond the 1st ionization step giving such small contributions, one would see almost no change in pH even if higher ionization steps are considered. Typically, when using the Henderson-Hasselbalch equation one should choose a pKa value close to the desired pH value. Such, as noted earlier, is only a one step ionization consideration. Subsequent ionization would not affect the pH generated by the 1st ionization step and is - again, for simplification - still a 1 to 1 ionization ratio and therefore applicable to the Henderson - Hasselbalch equation.

## 1. What is the significance of the equation pH = pKa + log([A-]/[HA]) in chemistry?

The equation pH = pKa + log([A-]/[HA]) is known as the Henderson-Hasselbalch equation and is used to calculate the pH of a solution containing a weak acid and its conjugate base. It helps in understanding the relationship between the pH, pKa, and the ratio of the concentrations of the acid and its conjugate base.

## 2. How is the Henderson-Hasselbalch equation derived?

The Henderson-Hasselbalch equation is derived from the equilibrium constant expression for a weak acid, Ka = [H+][A-]/[HA]. By taking the negative logarithm of both sides and rearranging the terms, we get pH = pKa + log([A-]/[HA]).

## 3. What does the pKa value represent in the Henderson-Hasselbalch equation?

The pKa value is the negative logarithm of the acid dissociation constant (Ka) and represents the strength of an acid. A lower pKa value indicates a stronger acid, while a higher pKa value indicates a weaker acid.

## 4. How does the Henderson-Hasselbalch equation help in understanding acid-base titrations?

The Henderson-Hasselbalch equation is used to calculate the pH of a solution at different points during an acid-base titration. It helps in determining the pH at the equivalence point, where the moles of acid and base are equal, and also the buffering capacity of the solution at different points.

## 5. Can the Henderson-Hasselbalch equation be applied to all acid-base systems?

No, the Henderson-Hasselbalch equation is only applicable to weak acid-strong base or weak base-strong acid systems. It cannot be used for strong acid-strong base or weak acid-weak base systems as they do not have a well-defined pKa value.

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