Why is NaH a strong base?

1 Answer

Because it produces #NaOH# and #H_2# when placed in water.

Explanation:

In the Bronsted-Lowry definition, bases are proton acceptors.
To be a strong base, the substance needs to basically completely dissociate in an aqueous solution to give high #"pH"#.

This is the balanced equation of what happens when #NaH# solid is placed into water:

#NaH(aq) + H_2O(l) -> NaOH(aq) + H_2(g)#

#NaOH#, as you may already know, is another very strong base that basically completely dissociates in an aqueous solution to form #Na^+# and #OH^-# ions.

So, another way to write our equation is this:

#NaH(aq) + H_2O(l) -> Na^+(aq) + OH^(-)(aq) + H_2(g)#

The #H^(-)# in #NaH# accepts an #H^+# ion from water to form #H_2# gas, making it a Bronsted-Lowry base.

If we were going by the Arrhenius definition of acids and bases, #NaH# would be a base not because it dissociates to give #OH^-# directly from its chemical structure, but because it results in #[OH^-]# increasing upon dissociation.

This reaction happens with a large equilibrium constant, so we can say that #NaH# almost completely dissociates when placed into an aqueous solution. This makes it a strong base.

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You might be wondering why this reaction doesn't happen instead, which would make #NaH# an acid:

#NaH(aq) + H_2O(l) -> Na^(-)(aq) + H_3O^+(aq)#

This reaction doesn't happen because sodium has a lower electronegativity than hydrogen.

For example, #HCl# can form #H_3O^+# and #Cl^-# ions in an aqueous solution.

#HCl# can do this because hydrogen is less electronegative than chlorine. Electrons is drawn toward chlorine. So, #H^+# is easily pulled off of #HCl# to form #H_3O^+#.

But #NaH# has again, hydrogen more electronegative than hydrogen, so we more-or-less have #Na^+# cation and a #H^-# anion, a consequence of electrons being drawn toward hydrogen.

So, instead of an #H^+# adding onto water to form #H_3O^+#, the electrons go with #H# to form #H^-# ion and form #H_2# gas by stealing an #H^+# from water.