Question

Choose the correct statement(s) for the reaction.

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

• A. The equivalent weight of $H_{3}P{O}_4$ is 49
• B. The resulting solution is neutralized by 1 mole of $KOH$
• C. 1 mole of $H_{3}P{O}_4$ is completely neutralized by 1.5 moles of $Ca(OH{)}_2$
• D. 0.5 moles of $H_{3}P{O}_4$ is completely neutralized by two moles of calcium hydroxide

Answer 1

Answer: (A), (B), (C)

The correct options are
A The resulting solution is neutralized by 1 mole of $KOH$
B The equivalent weight of $H_{3}P{O}_4$ is 49
C 1 mole of $H_{3}P{O}_4$ is completely neutralized by 1.5 moles of $Ca(OH{)}_2$

$2O{H}^{-}$ ions react with $2H^{+}$ ions given by $H_{3}P{O}_4$.
Therefore, in this reaction, basicity of the acid is $2$.
$Eq.wt.=\frac{Mol.wt.}{2}$
The remaining acidic $H$ is $1$. Therefore, the solution requires $1$ mole of KOH to get completely neutralised.
The basicity of $Ca(OH{)}_2$ is $2$. Therefore, $2\times 1.5=3$equivalents.
$3$ equivalents or $1.5$ moles of $Ca(OH{)}_2$ will be neutralized by $1$ mole or $3$ equivalents of $H_{3}P{O}_4$.