Select the correct statement(s) about n-factors in the stated acid-base reactions.

A $H X_{3} PO X_{3}$ (a dibasic acid) has $n = 2$ with excess $NaOH$
B $H X_{3} PO X_{2}$ (a monobasic acid) has $n = 1$ with $NaOH$
C $H X_{3} BO X_{3}$ has $n = 3$ with $NaOH$ because it has three $H$ atoms
D $NaH X_{2} PO X_{4}$ has $n = 2$ when it is converted to $Na X_{3} PO X_{4}$ by $NaOH$

Answer: A, B, D

The n-factor counts replaceable (ionisable)

H X^{+}

actually neutralised, not the total number of H atoms.

H X_{3} PO X_{3}

has 2 acidic H,

H X_{3} PO X_{2}

has 1, and

NaH X_{2} PO X_{4} Na X_{3} PO X_{4}

replaces 2 H, so options 1, 2 and 4 are correct.

H X_{3} BO X_{3}

is a Lewis acid that accepts one

OH X^{-}

, so

n = 1

and option 3 is wrong.