In the following reaction, the salt $\ce{CH3NH3Br}$ is formed:

$$\ce{CH3NH2 + HBr -> CH3NH3Br} $$

In water this dissociates completely into $\ce{CH3NH3+}$ and $\ce{Br-}$

In my chemistry book, it says that the $\ce{CH3NH3+}$ dissociates further in water into $\ce{CH3NH2}$ and $\ce{H3O+}$ (meaning that $\ce{CH3NH3Br}$ an acidic salt) according to the following:

$$\ce{CH3NH3+ + H2O <=> CH3NH2 + H3O+}$$

Fair enough, but why does the $\ce{Br-}$ not do the same and dissociate in water?

$$\ce{Br- + H2O <=> HBr + OH-}$$

In this way, the dissociation of the $\ce{CH3NH3Br}$ salt would not have so much of an effect on the pH, because the quantity $\ce{OH-}$ ions has also increased?

  • $\begingroup$ Somehow related: chemistry.stackexchange.com/questions/42696/… $\endgroup$ Commented Nov 20, 2016 at 7:32
  • $\begingroup$ Please have a look at the acid dissociation constant of $\ce{HBr}$. Is hydrobromic acid a strong acid? If so, what does that mean for bromide, its corresponding base? $\endgroup$ Commented Nov 20, 2016 at 7:50
  • 1
    $\begingroup$ @KlausWarzecha I see now; because $\ce{HBr}$ is a strong acid, the $\ce{Br-}$ won't dissociate in water because that would be effectively undoing the dissociation of HBr. Thanks for that. $\endgroup$
    – Adrian
    Commented Nov 20, 2016 at 8:34

1 Answer 1


The key is comparing $\mathrm{p}K_\mathrm{a}$ values. A good resource for this is the Evans $\mathrm{p}K_\mathrm{a}$ table. Unfortunately, the table does not contain $\ce{CH3NH3+}$, but in the section protonated nitrogen there is a value quoted for $\ce{C2H5NH3+}$ — which is chemically similar enough to methyl ammonium to assume the $\mathrm{p}K_\mathrm{a}$ values to be essentially identical.

\begin{array}{lr}\hline \text{compound} & \mathrm{p}K_\mathrm{a}\\ \hline \ce{EtNH3+} & 10.6\phantom{0}\\ \ce{HBr} & -9.00 \\ \ce{H3O+} & -1.7\phantom{0} \\ \hline\end{array}

Notice how the $\mathrm{p}K_\mathrm{a}$ values differ by almost $20$ logarithmic units. $\ce{HBr}$’s $\mathrm{p}K_\mathrm{a}$ value is even lower than water’s by $7$ to $9$ logarithmic units depending on the source used. Thus, while the amine exhibits extensive acid/base chemistry in water, bromide will not be protonated to any noticeable extent. Or you could say that your final equation will be strongly favouring the reactant side.

I’ll also point you to a recent answer of mine wherein I calculated the concentration of $\ce{HF}\ (\mathrm{p}K_\mathrm{a} \approx 3.17$, so notably weaker than $\ce{HBr}$) in a 1 molar solution of $\ce{NaF}$. The case is very comparable to your’s.

  • $\begingroup$ So, when we say that a strong acid dissociates completely, and we say that it is not an equilibrium reaction, it actually is protonated as well, but the equilibrium lies so far to the right that the reverse reaction is negligible? $\endgroup$
    – Adrian
    Commented Nov 23, 2016 at 20:41
  • $\begingroup$ @Adrian For all practical intents and purposes, a strong acid is fully dissociated. Taking a fully correct look at it, the equilibrium is only shifted very far to one side. $\endgroup$
    – Jan
    Commented Nov 23, 2016 at 21:06

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