Question

If $1$ mole of $H_{3}P{O}_4$ is reacted with $1$ mole of $X(OH{)}_2$ as:
$H_{3}P{O}_4+X(OH{)}_2\to XHP{O}_4+2H_{2}O$, then:

• A. the equivalent weight of base is $\frac{Mol.wt}{2}$
• B. the equivalent weight of $H_{3}P{O}_4$ is $\frac{98}{3}$
• C. the resulting solution is required $1$ mole $NaOH$ for complete neutralization
• D. minimum $1$ mole of $X(OH{)}_2$ more required for complete neutralization of $XHP{O}_4$

Answer 1

Answer: (A), (C)

the equivalent weight of base is $\frac{Mol.wt}{2}$
the resulting solution is required $1$ mole $NaOH$ for complete neutralization

In the reaction, every mole of base is giving $2$ moles of $O{H}^{-}$ ions so, equivalent weight of base $=\frac{Mol.wt.}{2}$.
Also, every mole of $H_{3}P{O}_4$ gives $3$ moles of $H^{+}$ ions so, resulting solution required $1$ mole NaOH for complete neutralization.