Wikipedia gives two pKa values for "Bis-tris propane".

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What about bis-tris-propane gives this compound two pKa values where other compounds have only one? How am I to choose which pKa value to use in a specific scenario?

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    $\begingroup$ It has 2 secondary amino groups, which, both in protonized forms, formally form biprotic acid due 2 respective conjugate acids $\ce{R- NH2^{+}-R.}$ $\endgroup$
    – Poutnik
    Dec 16, 2022 at 15:32
  • 1
    $\begingroup$ And the two identical amino groups are sufficiently far apart to act almost independently - hence very similar pKa's. If both are involved in ionisation, you must include both in your equilibrium calculations. $\endgroup$
    – Ian Bush
    Dec 16, 2022 at 15:39
  • $\begingroup$ @IanBush hm, similar enough to consider both e.g. for pH buffer calculations, but otherwise different enough. It has 6 bonds H-to-H. Not having data for alkylendiamins, but for analogical dicarboxylic acids, HOOC-CH2-COOH has pKa 2.83, 5.69 with the same count 6 bonds HtoH . HOOC-(CH2)2-COOH has pKa 4.2, 5.6 with 7 bonds HtoH . HOOC-(CH2)4-COOH has pKa 4.43, 5.41 with 9 bonds HtoH . $\endgroup$
    – Poutnik
    Dec 16, 2022 at 15:59
  • 1
    $\begingroup$ In the protonated forms, I would guess the slight difference in acidity comes from there not being a destabilizing +/+ interaction and perhaps to some degree, inductive effect. $\endgroup$
    – xasthor
    Dec 17, 2022 at 5:38
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    $\begingroup$ @Karsten, I'm more interested in the latter question: Why, with symmetric functional groups, are there two equilibrium constants? $\endgroup$ Dec 31, 2022 at 20:38

2 Answers 2


The compound "bis-tris-propane"(1,3-bis(tris(hydroxymethyl)methylamino)propane, also known as BTP) has 2 secondary amino groups and can be written as


which, both in protonized forms, formally forms a biprotic acid:


dissociating in the schema:

\begin{align} \ce{R-NH2^{+}-R-NH2^{+}-R(aq) &<=>[\mathrm{p}K_\mathrm{a1}] H+(aq) + R-NH-R-NH2^{+}-R(aq)}\\ \ce{R-NH-R-NH2^{+}-R(aq) &<=>[\mathrm{p}K_\mathrm{a2}] H+(aq) + R-NH-R-NH-R(aq)} \end{align}

The acidity of the respective substituted ammonium depends on if the other nitrogen is protonated or not, therefore 2 different acidity constants:

$\mathrm{p}K_\mathrm{a1}=6.8$ and $\mathrm{p}K_\mathrm{a2}=9.0$.

The particular usage of one of $\mathrm{p}K_\mathrm{a}$, or both, depends on the context. Note that the $\mathrm{p}K_\mathrm{a}$ difference 2.2 is almost ideal for a broad pH buffering scope

$$\mathrm{pH} \approx 5.8 - 10.0.$$

(ported from my prior comments)
The compound has in its bi-protonated form 6 bonds between the respective acidic hydrogens.

Not having data for alkylendiamines, but for analogical dicarboxylic acids:

Compound $\ce{H}$ to $\ce{H}$ distance $\mathrm{p}K_\mathrm{a1}$ $\mathrm{p}K_\mathrm{a2}$ $\mathrm{p}K_\mathrm{a}$ difference
$\ce{HOOC-CH2-COOH}$ 6 2.83 5.69 2.86
$\ce{HOOC-(CH2)2-COOH}$ 7 4.2 5.6 1.4
$\ce{HOOC-(CH2)4-COOH}$ 9 4.43 5.41 0.98
"Bis-tris-propane" $\\ \ce{R-NH-(CH2)3-NH-R} $ 6 6.8 9.0 2.2
$\ce{NH2-CH2-CH2-NH2}$ 5 6.99 10.08 3.09

Why, with symmetric functional groups, are there two equilibrium constants?

If the functional groups are close together, there will be electrostatic interactions (coupling). Adding the first proton (to the neutral molecule) is easier than adding the second proton (to the +1 cation). So adding the first proton (no matter on which side) will break the symmetry.

If the functional groups are spatially and electronically distant (uncoupled) from one another, you still expect distinct apparent association constants. Take the case where the pH equals the (microscopic) $\mathrm{p}K_\mathrm{a}$. Each site will be exactly 50% protonated. There are 4 microstates (not protonated, two cases of singly protonated, doubly protonated), and 3 macrostates (neutral, +1, +2). The +1 macrostate will be present at twice the concentration compared to the other two states, so the apparent equilibrium constant for the first protonation step and the second protonation step will be different by a factor of 4. As a consequence, the apparent $\mathrm{p}K_\mathrm{a}$ values will be different by $\log_{10} 4 = 0.602$.

For the related example of dicarboxylic acids mentioned in Poutnik's answer, you can see that the difference of $\mathrm{p}K_\mathrm{a}$ values does not approach zero, reflecting this statistical reason for a difference in the apparent equilibrium constants.

How am I to choose which pKa value to use in a specific scenario?

You first figure out what the major species are. At pH = 6.8, you have the +2 and +1 cations as major species, at pH = 9.0, the +1 cation and the neutral species. The compound is useful as buffer in the pH range of 5.8 to 10.0, but does not buffer well at around 7.9 (major species is the +1 cation, with very little, i.e. less than 10% each, of the other two species).

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