Question

Is NaH a stronger base than NaOH?

Answer 1

To find the solution for the question we should know the ions formed by the dissociation of these compounds. And we know that according to the Bronsted-Lowry concept, we define bases as those species which can accept electrons.

Complete step-by-step answer: Here in the question, we are asked to comment on whether NaH is a stronger base than NaOH. To get a solution for the question, first, we should discuss what a base means and the species of ions formed when the compounds given in the question undergo dissociation.
We are very familiar that, according to the Bronsted-Lowry concept of acids and bases, bases are those species that can accept the protons and acids are those species that can furnish protons in the reaction vessel.
And if we define a compound as a strong acid or base then that species can undergo complete dissociation.
Here NaH and NaOH are the compounds of our concern, both the compounds are strong bases i.e. they undergo complete dissociation reaction when dissolved in water.
When NaH undergoes dissociation then hydride ion (${{H}^{-}}$) ion is formed when NaOH dissociates $O{{H}^{-}}$is formed.
When NaH reacts with water then,
$NaH+{{H}_{2}}O\to NaOH+{{H}_{2}}$
From the equation, we know that NaOH is formed and it is also a strong base, so it also will undergo complete dissociation and we can write the equation as :
$NaH+{{H}_{2}}O\to N{{a}^{+}}+O{{H}^{-}}+{{H}_{2}}$
And the reaction happens with a very large equilibrium constant, which means the NaH is undergoing complete dissociation in an aqueous solution and making it a strong base.
Hence we can say that NaH is a stronger base than NaOH.

Note: While writing the chemical equation of the reaction involving NaH and${{H}_{2}}O$, students often make mistake by writing equation as follows:
$NaH+{{H}_{2}}O\to N{{a}^{+}}+{{H}_{3}}O$
But this reaction does not happen since the electronegativity value of sodium is less than that of the hydrogen atom.