# pH at maximum concentration of a certain species (Phosphoric acid) [closed]

I am trying to find the pH at which $$[H_{2}PO_{4} ^{-}]$$ is maximum by only knowing the pKas. I have tried to derive an equation using kinetics but it failed [1]. After taking a look at the Sillén/Hägg/Bjerrum diagram, I have come to the hypothesis that the pH would be equal to 0.5*(pKa1+pKa2) - the same formula as for the isoelectric point. However, I do not know if this is true or how I'd go about proving it. My knowledge on Sillén diagrams is a tad rusty.

[1] Equations I used (I don't include [H2O] as it is basically constant):
Phosphoric acid: d[A]/dt = k1[B][H30+]
1st deprotonation: d[B]/dt = k2[A] + k3[C][H30+]
2nd deprotonation: d[C]/dt = k4[B]
Equilibrium condition (though I fear it is necessary but not sufficient):
[B]' = [A]' + [C]'

• Not sure what you are trying to achieve with kinetics, but chemistry.stackexchange.com/questions/76928 looks quite similar. Commented Apr 27 at 16:55
• Chem+Math MathJax formatting: Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. Commented Apr 27 at 18:31
• @andselisk I'm just trying to relate the concentrations of all the species with the equilibria. The acid dissociation constant comes from k2/k1 in the following example: HA + H2O -> A- + H3O+ [HA]' = k1[H3O+][A-] [A-]' = k2[HA] [HA]'=[A-] <-> Ka= k2/k1 = [H3O+][A-]/[HA] I'm trying to apply the same principle to this problem. Commented Apr 27 at 22:58
• @Poutnik I have but I'm scared that I might do an unwarranted approximation somewhere. I'm also not quite sure how I could combine both HH equations (1st deprotonation & 2nd ...) Commented Apr 27 at 23:02
• Again, forget about kinetics. Use the equilibrium equations, defining the acidity constants, in ordinary or logarithmic form. Commented Apr 28 at 3:20

The fraction of the dyhydrogenphosphate, while ignoring the third ionization, is:

$$x(\ce{H2PO4-})=\dfrac{K_\text{a1}\cdot [\ce{H+}]}{[\ce{H+}]^2+K_\text{a1}.[\ce{H+}] + K_\text{a1} \cdot K_\text{a2}}\tag{1}$$

With all three degrees of dissociation, it is:

$$x(\ce{H2PO4-})=\dfrac{K_\text{a1} \cdot [\ce{H+}]^2}{[\ce{H+}]^3+K_\text{a1} \cdot [\ce{H+}]^2 + K_\text{a1} \cdot K_\text{a2} \cdot [\ce{H+}] + K_\text{a1} \cdot K_\text{a2} \cdot K_\text{a3} },\tag{2}$$

but the difference is really negligible.(*)

For the symmetry of the case, one can use Henderson-Hasselbalch equation

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10} \left( \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right)$$

in the modified, exponential forms, for both neighbour pairs of phosphate forms:

$$10^{\mathrm{p}K_\mathrm{a1} - \mathrm{pH}}= \frac{[\ce{H3PO4}]} {[\ce{H2PO4-}]} \tag{3}$$

$$10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a2} }= \frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}\tag{4}$$

$$10^{\mathrm{p}K_\mathrm{a1} - \mathrm{pH}} + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a2} }= \frac{[\ce{H3PO4}] + [\ce{HPO4^2-}] } {[\ce{H2PO4-}]} \tag{5}$$

We then need to determine the minimum of $$10^{\mathrm{p}K_\mathrm{a1} - \mathrm{pH}} + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a2} }$$ from derivative per pH.

$$\ln{10}\cdot (10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a2} } - 10^{\mathrm{p}K_\mathrm{a1} - \mathrm{pH}} )=0 \tag{6}$$

$$\mathrm{pH} - \mathrm{p}K_\mathrm{a2} = \mathrm{p}K_\mathrm{a1} - \mathrm{pH} \tag{7}$$

$$\mathrm{pH} = \dfrac{ \mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2} \tag{8}$$

At this pH, the fraction of dihydrogenphosphate reaches it maximum.

The reflection symmetry around $$\mathrm{pH_\mathrm{max}}$$:

$$10^{\mathrm{p}K_\mathrm{a1} - \mathrm{pH}} + 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a2} }$$ is the same for $$\mathrm{pH} = \mathrm{pH_\mathrm{max}}+x$$ and $$\mathrm{pH} = \mathrm{pH_\mathrm{max}}-x$$.

$$10^{\frac{ \mathrm{p}K_\mathrm{a1} - \mathrm{p}K_\mathrm{a2}}{2} - x} +10^{\frac{ \mathrm{p}K_\mathrm{a1} - \mathrm{p}K_\mathrm{a2}}{2} + x} \tag{10}$$

the flipping the sign of $$x$$ effectively just switched the term order.

(*) - For some reasons, I remember these formulas for about 40 years, without using them much. Probably because there is a nice pattern within them. I have derived them once in the past for myself while studying.

• Thanks for your answer. Very insightful. You have both answered half of my question (the most important part) but I still lack an intuitive understanding as to why 0.5(pKa1+pKa2) corresponds (almost perfectly) with the value that I'm looking for. Saying it's symmetrical doesn't cut it for me because you'd have to explain why it's symmetrical - they're both identical arguments. Commented Apr 28 at 17:55
• @David See the update, deriving the symmetry and verifying the assumed average formula directly from both Henderson-Hasselbalch equations. Commented Apr 29 at 6:17

Now that you have posted comment to the original question

Phosphorsäure hat die folgenden drei pKa-Werte: pKa1 : 2.15 ; pKa2 : 7.20 ; pKa3 : 12.4 i) Bei welchem pH-Wert einer wässrigen Alkaliphosphatlösung ist die Konzentration von H2PO4 ─ maximal

Nowhere it asks you to use chemical kinetics or rates of reaction expressions. This is acid-base equilibrium problem. What you are looking for are called distribution diagrams. The equations can be derived with a lot of algebraic rearrangements but it is doable. Search alpha plots or composition diagrams and you will find good information. You are right, in the German literature, they may be called Bjerrum plots but more modern name is alpha plots or distribution diagrams.

This YouTube video gives a sufficient background: Video, 11 min

Here is one example Link to the diagram. From these diagrams one can read the percentage of species as a function of pH just from the knowledge of pKa values.

• Thanks for answering but that doesn't really answer my question. I'll try to clarify my question in the comments above Commented Apr 27 at 22:55
• This is exactly what you are asking "I am trying to find the pH at which [H2PO−4] is maximum by only knowing the pKas." Did you even check alpha plots or check the expressions for alphas. It is unfortunate with anonymous downvoting that we never know if an Einstein thinks the answer is wrong or a high-school student "thinks" the approach is wrong.
– ACR
Commented Apr 28 at 3:40
• What do you mean with what an Einstein or a high school student would think? I didn't downvote the answer Commented Apr 28 at 11:53
• He meant anybody, as it is anonymous, requiring just some rep. point level. There is no way to know the knowledge level of the downvoter. Commented Apr 28 at 13:21

What people find intuitive depends on their experience, and what proofs they can follow depends on their mathematical background. Thus, this might not add anything to Poutnik's or AChem's answer.

Starting with the equation from Poutnik's answer,

$$x(\ce{H2PO4-})=\dfrac{K_\text{a1}\cdot [\ce{H+}]}{[\ce{H+}]^2+K_\text{a1}.[\ce{H+}] + K_\text{a1} \cdot K_\text{a2}},$$

we can show the symmetry by dividing by the numerator:

$$x(\ce{H2PO4-})=\dfrac{1}{\dfrac{[\ce{H+}]}{K_\text{a1}}+ 1 + \dfrac{K_\text{a2}}{[\ce{H+}]}}$$

When the pH is far away from either $$\mathrm{p}K_\text{a}$$ value (and inbetween the two), I can neglect the first and the last term of the denominator, and the fraction is almost one. As the pH approaches one or the other $$\mathrm{p}K_\text{a}$$, one of the terms will become significantly higher than zero (at pH = $$\mathrm{p}K_\text{a}$$, it will be one and the value of the entire expression will be about 1/2). Graphically, you can see this in AChem's answer.

For a proof, you can find the maximum by differentiating with respect to the hydrogen ion concentration. To make the math easier, you can also find the minimum of the reciprocal. Setting the derivative to zero, this leads to (exercise for the reader):

$$[\ce{H+}]^2 = K_\text{a1} \cdot K_\text{a2}$$

(which in turn is equivalent to the statement about the pH, if you take the negative logarithm on both sides and divide by two).

• Although the other answers were useful, this is the answer I was looking for. To be fair it's not a complicated proof but it was not immediately apparent to me. Commented Apr 28 at 20:03