For the reaction:

${\mathrm{H}}_3{\mathrm{PO}}_4+\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\to {\mathrm{CaHPO}}_4+2{\mathrm{H}}_2\mathrm{O}$ 

1mole         1mole

Then which of the following statements are correct?

$H_{3}P{O}_4+Ca{\left(OH\right)}_2\to CaHP{O}_4+2H_{2}O$

(A) n-factor for $H_{3}P{O}_4=2$ (Since $2H^{+}$ ions are replaced)

$eq.wt=\frac{M}{2}=\frac{98}{2}=49$            

(B) $CaHP{O}_4$ have only one replacable $H^{+}$ ion

So nf = 1   and nf of KOH = 1

 $\therefore$ eq of $CaHP{O}_4$ = eq of KOH

 (C) For complete neutralization

   No of eq of $H_{3}P{O}_4$ = no of eq of $Ca{\left(OH\right)}_2$

       Can be neutralized