Question

1 mole each of $H_{3}P{O}_2,H_{3}P{O}_3$ and $H_{3}P{O}_4$ were given in three test tubes. In a volumetric titration, standard $1MCa(OH{)}_2$ solution was provided. Volume of $Ca(OH{)}_2$ used by :

• A. $490mL$
• B. $750mL$
• C. $980mL$
• D. $1500mL$

Answer 1

The correct option is D $1500mL$

$\underset{2mol}{2H_{3}P{O}_2}+\underset{1mol}{Ca\left(O{H}_2\right)}⟶Ca(H_{2}P{O}_2{)}_2+2H_{2}O$
$1mol0.5mol\equiv 500mLof1MCa(OH{)}_2$
$\underset{1mol}{H_{3}P{O}_3}+\underset{1mol}{Ca(OH{)}_2}⟶CaHP{O}_3+2H_{2}O$
$\equiv 1000mLof1MCa(OH{)}_2$ solution
$\underset{2mol}{2H_{3}P{O}_4}+\underset{3mol}{3Ca(OH{)}_2}⟶C{a}_3(P{O}_4{)}_2+3H_{2}O$
$1mol1.5mol\equiv 1500mLof1MCa(OH{)}_2$
Hence option D is correct.