When solid $\ce{NaH}$ is mixed with water, the products are not $\ce{Na-}$ and $\ce{H3O+}$. The reaction yields $\ce{NaOH}$ and $\ce{H2}$ gas.

Why does solid $\ce{NaH}$ not act as a Brønsted acid in this case?

  • $\begingroup$ It's actually one of strongest bases... $\endgroup$ – Mithoron Apr 19 '16 at 23:20
  • $\begingroup$ @Mithoron Of course, since it produces the very strong base NaOH. $\endgroup$ – user3932000 Jun 6 '16 at 13:42

The answer lies in electronegativity. Sodium has a significantly lower electronegativity ($\approx 1.0$) than hydrogen ($\approx 2.1$) meaning that hydrogen pulls electron density towards itself, away from sodium to generate a sodium cation and a hydride anion.

For a compound to be a Brønsted acid, you need to draw electron density away from the hydrogen, i.e. connect it to an electronegative atom such as oxygen, fluorine, nitrogen etc. Only then can it formally be described as $\ce{H+}$ and can dissociate as such.

A hydride is much better described as $\ce{H-}$ and has a free lone pair. As such, it is a Brønsted base, not an acid. In fact, if you extend the Brønsted acid/base definition in the way Lewis did, you will arrive at the sodium ($\ce{Na+}$) being the acidic species here.

The Brønsted acid/base reaction product of the base $\ce{H-}$ and the acid $\ce{H+}$ happens to be $\ce{H2}$, and because the acidic hydrogen is abstracted directly from water (hydrogen gas can bubble away, shifting the equilibrium even if the reaction were not thermodynamically favoured), $\ce{OH-}$ ions are left which can be written with the remaining $\ce{Na+}$ cation to give sodium hydroxide.


The hydrogen is in the form of a $\ce{H^-}$ ion, negatively charged instead of the usual partial positive. Ths hydride ion strips a $\ce{H^+}$ from a $\ce{H_2O}$ molecule to become $\ce{H_2}$ gas and leaves behind $\ce{OH^-}$.

  • 3
    $\begingroup$ While it's correct, I think you should elaborate to provide better answer. $\endgroup$ – Mithoron Apr 19 '16 at 23:22

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