I completed a practice (not for marks) assignment for an organic chemistry course, and one of the questions asked for the pKa of methylamine.

I looked on PubChem, and it gave a result of $10.6$. Marking myself, I found that the answer key states the $\mathrm{p}K_\mathrm{a}$ of methylamine is $33$. Especially on a log scale, that is not a trivial difference.

I talked to my professor about it; he was confident the true $\mathrm{p}K_\mathrm{a}$ is in the low thirties and stated that I need to be skeptical of material online. His point in part was that there was no way methylamine was more acidic than water.

I am not personally convinced on either number yet.

What is the $\mathrm{p}K_\mathrm{a}$ of methylamine, and what evidence supports that number?

  • 6
    $\begingroup$ The value 10.6 is the $\mathrm pK_\mathrm a$ of the conjugate acid of methylamine. $\endgroup$
    – user7951
    Commented Sep 13, 2016 at 22:36

1 Answer 1


Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values.

$$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$

The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.

Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.

  • $\begingroup$ Nice answer. I know what you mean for the equation with Ka,2 but that is a really weird way to write the reaction. I really think you should have written the reactions as two separate equations. $$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ }$$ $$\ce{H3C-NH2 <=>[$K_\mathrm{a2}$] H3C-NH- + H+}$$ $\endgroup$
    – MaxW
    Commented Sep 14, 2016 at 3:47
  • $\begingroup$ @MaxW That was actually my first thought as well, and I would have done it your way, but mathematically it cancels out so I suppose it is fine. $\endgroup$ Commented Sep 14, 2016 at 4:03
  • $\begingroup$ Also @Jan i.imgur.com/ZtWjUOY.jpg the Ka's above the arrows display a bit weird on mobile. (+1 anyway) $\endgroup$ Commented Sep 14, 2016 at 4:03
  • 1
    $\begingroup$ Of course it isn't Jan's fault that the interpretation of the formatting is different on mobile phones. That absolutely isn't his problem. // Stupid !@#$%^&* mobile phones... $\endgroup$
    – MaxW
    Commented Sep 14, 2016 at 4:11
  • $\begingroup$ @orthocresol They don’t look perfect on PC either. Especially the equilibrium arrows are way to short. I think I should file a bug report to mhchem ;) $\endgroup$
    – Jan
    Commented Sep 14, 2016 at 15:31

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