Question

What will be the n-factor $\left(n_f\right)$ for $H_{3}P{O}_2$ and $H_{3}B{O}_3$ respectively?

• A. $1,3$
• B. $2,3$
• C. $3,3$
• D. $1,1$

Answer 1

The correct option is D $1,1$

In case of acid-base reaction, the $n_f$ is the actual number of $H^{+}$ or $O{H}^{-}$ replaced in the reaction. The acid or base may contain more number of replaceable $H^{+}$ or $O{H}^{-}$ than actually replaced in reaction.
$n_f$ for acid is the number of $H^{+}$ ion replaced per molecule of the acid.

For $H_{3}P{O}_2$,

The number of replaceable $H^{+}$ ions are 1 .
So, the n-factor is 1
For, $H_{3}B{O}_3$


$H_{3}B{O}_3$ has 3 hydrogen linked to oxygen yet all are not replacable
$H_{3}B{O}_3\left(s\right)+H_{2}O\left(l\right)\to \left[B\right(OH{)}_4{]}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Thus 1 mole of boric acid can give only one $H^{+}$ ion.
Hence valency factor is 1.