Choose the correct statements (atomic masses: H 1, C 12, O 16, Na 23, P 31; equivalents = moles $\times$ n-factor).

A $H X_{3} PO X_{3}$ has n-factor 2 in its reaction with excess NaOH
B $NaH X_{2} PO X_{4}$ has n-factor 1 both when treated with NaOH (giving $Na X_{2} HPO X_{4}$ ) and with HCl (giving $H X_{3} PO X_{4}$ )
CThe equivalent mass of $Na X_{2} CO X_{3}$ (106 g/mol) is 53 g/eq when HCl converts it only to $NaHCO X_{3}$
DThe equivalent mass of $NaHCO X_{3}$ (84 g/mol) is 84 g/eq when NaOH converts it to $Na X_{2} CO X_{3}$

Answer: A, B, D

H X_{3} PO X_{3}

has two replaceable H, so n = 2 (true).

H X_{2} PO X_{4} X^{-}

gains or loses exactly one proton in both reactions, so n = 1 (true). Carbonate to bicarbonate takes one

H X^{+}

, so n = 1 and eq. mass = 106, not 53 (false).

HCO X_{3} X^{-}

loses one proton, so n = 1 and eq. mass = 84 (true).