Let's say I have a polyprotic acid $\ce{H2A}$ with the following properties:

$$ \begin{align} \ce{H2A &<=> HA- + H+} & K_1 &= 1\cdot 10^{-1}\\ \ce{HA- &<=> A^2-+ H+} & K_2 &= 1\cdot 10^{-4} \end{align} $$

Now a 0.5 M solution of $\ce{HA-}$ is prepared; what will be the $\mathrm{pH}$ of the solution?

My textbook (and many other sources) says it should be

$$\frac{\mathrm{p}K_1 + \mathrm{p}K_2}{2}$$

which sounds a little counterintuitive to me.

First of all, it implies that the $\mathrm{pH}$ is independent of the concentration of $\ce{HA-}$ present.

Also, if $1/\mathrm{p}K_1 > \mathrm{p}K_2$, then $\ce{HA-}$ should have the tendency to accept a $\ce{H+}$ more than it has of losing one, then why is the $\mathrm{pH}$ exactly partway between $\mathrm{p}K_1$ and $\mathrm{p}K_2$?

  • 2
    $\begingroup$ To clear up one point, what you have is a buffer solution. Indeed the pH of the buffer solution does not depend on the concentration. However a buffer solution has a buffer capacity that is its resistance to change pH when acid or base is added. The lower the concentration of the amphiprotic salt, the lower the buffer capacity. $\endgroup$
    – MaxW
    Apr 28, 2019 at 15:03
  • 4
    $\begingroup$ $\ce{pH} = \dfrac{\ce{pK_1 + pK_2}}{2}$ if only if ${\ce{[HA-] >> K_1}}$ $\endgroup$
    – MaxW
    Apr 28, 2019 at 15:33

3 Answers 3


$$\begin{align} \mathrm pH &=\mathrm pK_\mathrm{a1}+\log \frac {[\ce{HA-}]} {[\ce{H2A}]} \\ \mathrm pH &=\mathrm pK_\mathrm{a2}+\log \frac {[\ce{A^2-}]} {[\ce{HA-}]} \\ \mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + pK_\mathrm{a2} +\log \frac {[\ce{A^2-}]} {[\ce{H2A}]}} {2}\\ \end{align}$$

The last equation is the sum of the equation sides, divided by 2.

We can consider for expected acidic $\mathrm pH$, that we can neglect water dissociation and suppose $\ce{H+}$ is produced and consumed by these 2 reactions: $$\begin{align} \ce{HA- + H+ &<=> H2A} \\ \ce{HA- &<=> H+ + A^2-} \\ \end{align}$$


$$\begin{align} [\ce{H}]&={[\ce{A^2-}]}-{[\ce{H2A}]} \\ \mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \frac {[\ce{H2A}] + [\ce{H+}] } {[\ce{H2A}]}} {2}\\ \mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { [\ce{H+}] } {[\ce{H2A}]}\right)} {2}\\ K_\mathrm{1a}&=\ce{[H+][HA-] /[H2A]} \\ \ce{[H2A]}&=\ce{[H+][HA-]} /K_\mathrm{1a}\\ \mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { K_\mathrm{1a}} {[\ce{HA-}]}\right)} {2}\\ \end{align}$$

Therefore, if $\mathrm{p}K_\mathrm{a1}$ is high enough, i.e. if $\ce{H2A}$ is rather a weak acid,
the simplified formula $\mathrm{p}H=(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})/2$ is right enough.

If $\mathrm{p}K_\mathrm{a1}=\ce{[HA-]}$, the $\mathrm{p}H$ correction is cca $+0.15 = \log (2)/2$

And then we can use this equation to iterative corrections, with $\ce{[MHA]}$ being concentration of the used acidic salt. The fraction is the classical formula to calculate ion fraction.

$$\ce{[HA-]}=\ce{[MHA]} \cdot \small \frac {K_\mathrm{a1}.\ce{[H+]}} {[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}} $$

$$\frac {K_\mathrm{a1}}{\ce{[HA-]}}= \frac {[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}} {\ce{[MHA]} \cdot \small \ce{[H+]}}$$

For particular calculation, the first shoot is taking $\mathrm pH$ from the simplified formula:

$$\begin{align} \mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}=(1+4)/2=2.5 \\ \end{align}$$

I have realized, the direction of iterations must be reversed, as the formulas lead to divergence. ( It is known mathemathical problem to choose the right iteration with the small derivative $\mathrm{d}f(x)/\mathrm{d}x$.

The simplified equation at the bottom of the post expects not too acidic or alkaline $\mathrm pH$. It supposes the equilibrium reactions affecting $[\ce{H}]$ do not significantly affect ratio of concentrations $$\frac {[\ce{A^2-}]}{[\ce{H2A}]}$$

As the $\mathrm pH$ is near the both $\mathrm pK_\mathrm a$, it is quite a fair assumption for most cases.

$$\begin{align} \ce{2 HA- &<=> H2A + A^2-} \\ \frac {[\ce{H2A}]} {[\ce{HA-}]} &=\frac {[\ce{A^2-}]} {[\ce{HA-}]}\\ [\ce{H2A}] &= [\ce{A^2-}] \\ \mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2} \\ \end{align}$$

  • $\begingroup$ In step three how did you took the average of both pH values to get the final pH? $\endgroup$ Apr 28, 2019 at 14:42
  • 1
    $\begingroup$ It is formally an average, but of the same values. It was pure mathematical trick of equivalent operations over equations. You can add or multiply both sides by the same nonzero value, even if the expressions with the same value are not the same. I summed left sides, right sides and divided by 2. $\endgroup$
    – Poutnik
    Apr 28, 2019 at 14:50
  • $\begingroup$ But you assumed that pH is independently defined by the first two equations. The real pH is dependent on both of these pK values. Hence, the first and second equation is true if only the first and second ionizations respectively are considered. $\endgroup$ Apr 28, 2019 at 14:57
  • $\begingroup$ The beginning is exact. pH must obey simultaneously both first equations. pH from both equations are equal. $\endgroup$
    – Poutnik
    Apr 28, 2019 at 14:59
  • $\begingroup$ I think there is a factor two missing in line 4. $\endgroup$
    – Karsten
    Apr 29, 2019 at 2:58

Brilliant question, and brilliant comment by MaxW, that deserves a non-textbook answer.

First of all, it implies that the pH is independent of the concentration of HA− present.

Yes, like a buffer.

Also, if 1/pK1>pK2, then HA− should have the tendency to accept a H+ more than it has of losing one, then why is the pH exactly partway between pK1 and pK2?

The tendency to accept an H+ (as expressed in the ratio of conjugate acid and conjugate base) depends both on the pH and the pKa. If the pH were neutral, we could not have both reactions be at equilibrium. We need a pH that is midway between the pKa values.

Let's treat this as a buffer:

$$\ce{H2A <=> 2 H+ + A^2-}$$

with $K_{a12} = K_{a1} * K_{a2}.$

Let's check out what happens when $\ce{[H2A] = [A^2-]}$. Inserting the concentrations and taking the negative logarithm, we get:

$$\ce{[H2A]} \times K_{a12} = \ce{[H+]^2 \times [A^2-]}$$

$$ \mathrm{pK_{a12} = pK_{a1} + pK_{a2} = 2 pH}$$

$$ \mathrm{pH = \frac{pK_{a1} + pK_{a2}}{2}}$$

So why is $\ce{[H2A] = [A^2-]}$?

Well, it works. If you calculate $\ce{[H2A]}$ from pH and $\ce{[HA-]}$, and then calculate $\ce{[A^2-]}$ from pH and $\ce{[HA-]}$, you get the same answer.

If we estimate the pH as: $$ \ce{[H+]} \overset{!}{=} \sqrt{\mathrm K_{a1} \times K_{a2}} $$

For $\ce{[H2A]}$, we get: $$\ce{[H2A]} \times K_{a1} = \ce{[H+] \times [AH^-]}$$

$$\ce{[H2A]} = \frac{\ce{[H+]} \times \ce{[AH^-]}}{\mathrm K_{a1}} = \ce{[AH^-]} \times \sqrt{ \frac{\mathrm K_{a2}}{\mathrm K_{a1}}}$$

And for $\ce{[A^2-]}$, we get: $$\ce{[AH^-]} \times \mathrm{K_{a2}} = \ce{[H+] \times \ce{[A^2-]}}$$

$$\ce{[A^2-]} = \frac{\mathrm{K_{a2}} \times \ce{[AH^-]}}{\ce{[H+]}} = \ce{[AH^-]} \times \sqrt{ \frac{\mathrm K_{a2}}{\mathrm K_{a1}}}$$

This is a bit of a circular argument, but it is fine because there is always just one equilibrium state for a system. So if you happen to find it and can show everything is at equilibrium, it must be correct.

How is this different from a buffer?

The buffer capacity (compared to a conventional 1:1 buffer with a pKa equal to the desired pH) is miserable because the main species is $\ce{HA-}$. You would be able to change the concentration ratio of diprotic acid and the dianion by adding minimal amount of strong acids or base, with a large effect on the pH.

  • 1
    $\begingroup$ The exercise you left for the readers was essentially my question, as I also prepared a solution with only $HA^-$ and then questioned why [$H_2A$] is the same as [$A^2-$] as their reactions have a different equilibrium constant. $\endgroup$ Apr 28, 2019 at 15:31
  • $\begingroup$ Ok I’ll add that bit. $\endgroup$
    – Karsten
    Apr 28, 2019 at 16:00
  • $\begingroup$ Huh? Since the main species is $\ce{HA-}$ the buffer capacity is at a maximum. See this figure for oxalic acid which is from this paper. $\endgroup$
    – MaxW
    Apr 28, 2019 at 16:50
  • 1
    $\begingroup$ Maybe I'm thinking of a different reference point. A 1:1 buffer of H2A and HA- would have better buffer capacity, and a 1:1 buffer of HA- and A(2-) would have better buffer capacity. Of course, they would buffer at a different pH. Also, finding a buffer with a pKa at (pKa1 + pKa2)/2 would be a better buffer as 1:1 buffer at the pH in question. $\endgroup$
    – Karsten
    Apr 28, 2019 at 17:07
  • 1
    $\begingroup$ @AdnanAL-Amleh I added the calculation. If you like numbers better than symbols, you can try the example Poutnik gave in another answer. $\endgroup$
    – Karsten
    Jun 20, 2021 at 9:31

I first found the derivation here https://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt but the hydrolysis was missing, i tried to derive with hydrolysis and got similar result just with $K_w$ added.

enter image description here

$k_{a1}$, $k_{a2}$ are first and second dissociation constants of the weak acid $\ce{H2A}$, $k_w$ is dissociation constant of water
I have followed equation $\ce{3->1->2}$ (second has the final concentrations boxed)
the upper concentrations are at $t=0$ and lowers are at $t=t_{eq}$

enter image description here

assuming $x, y << C$, $C=[\ce{HA-}]^\circ$ (initial concentration of $\ce{HA-}$)

enter image description here

the two are almost similar and $k_w$ can be neglected in many cases (example of $\ce{H3PO3}$ is at the site above, it can be proved by putting $k_a$ values of $\ce{H3PO3}$, for easy calculation, take
$pk_{a1}=2$ and $pk_{a2}=7$
and [$\ce{HA-}$] (0.001M will be good for a significant difference in pH) in either of the equations, in second $k_w$ has to be neglected),

the pH of 0.001 M $\ce{H2PO2-}$ ion must come around 5.2 (assuming third dissociation step is negligible) .
At $[\ce{HA-}] < 0.01M$ , difference in pH calculated by the above and calculated by
$pH=\frac{pK_{a1}+pK_{a2}}{2}$, increases significantly as $k_{a1}$ cannot be neglected.

$pH=\frac{pK_{a1}+pK_{a2}}{2}$ this can be derived by assuming $k_{a1} <<[\ce{HA-}]$ in the above equation. Everything is mentioned at the site

  • $\begingroup$ Please see the site's guidelines for typesetting. Your post is not very readable in its current state. $\endgroup$ Jun 17, 2021 at 23:06
  • $\begingroup$ You can refer this link for how to write equations and chemical formulae on your regular keyboard using MathJax: chemistry.stackexchange.com/help/notation $\endgroup$
    – TRC
    Jun 18, 2021 at 3:36
  • $\begingroup$ thanks for pointing out and thank you for the edit $\endgroup$ Jun 18, 2021 at 7:17

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