Question

n-factor of ${H_3}B{O_3}$ in aqueous solution will be.

A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

Hint: We can know, 'n' factor of an acid is that the number of ions replaced by one mole of acid. Boric acid is also called acidum boricum, hydrogen borate and orthoboric acid.

Complete step by step answer: We know the formula of boric acid is which makes us believe that the n factor is three. But the particular representation of boric acid is which reacts with water to make which on further reaction with base can form. It's a tetrahedral structure with the four groups symmetrically distributed around the boron atom. During this explanation, the groups aren't acidic and thus don't contribute to a better n factor.
The n-factor for acid isn't the same as its basicity; i.e. the number of moles of usable Hydrogen atoms present in one mole of acid.
The n-factor of boric acid is,
In boric acid, albeit all three Hydrogen is linked to oxygen but not all three Hydrogen are replaceable. Here, boron acts as a Lewis acid because it is an electron-deficient species.
Boric acid may be a monobasic weak acid; it doesn't dissociate completely to offer Hydrogen ions therefore, the 'n' factor of boric acid is 1.

So, the correct answer is “Option A”.

Note: We also remember that in solution, particularly at concentrations above boric acid forms a spread of other cyclic compounds and linear polymers that have higher n factors. So even with the foremost common explanation, in most cases, the particular n factor is going to be a variety depending upon the varied compounds present and their relative concentrations.