# Why does bromide not react with water when methyl ammonium bromide is dissolved in water?

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?

• Somehow related: chemistry.stackexchange.com/questions/42696/… – Nilay Ghosh Nov 20 '16 at 7:32
• 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? – Klaus-Dieter Warzecha Nov 20 '16 at 7:50
• @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. – Adrian Nov 20 '16 at 8:34

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.
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.