Question

What is the equivalent mass of the following?
(a) $H_{3}P{O}_4$ when neutralised to ${HP{O}_4}^{2-}$
(b) $HCl{O}_4$ neutralization by a base
(c) $NaI{O}_3$ when reduced to $I^{-}$
(d) $NaIO$ when reduced to $I_2$
(e) $Al(OH{)}_3$ neutralization by an acid

Answer 1

(a)Molecular mass of $H_{3}P{O}_4=(3+31+64)=98g.H_{3}P{O}_4$ when neutralised to $HP{O}_4^{2-}$, two $H^{+}$ ions have been replaced.
Thus, eq. mass $=\frac{\text{Mol. mass}}{\text{No. of replaceable hydrogen atoms}}$
$=\frac{98}{2}=49.0g$
(b) $HCl{O}_4$ molecule contains one replaceable hydrogen atom.
Thus, eq. mass $=\frac{Mol.mass}{1}=\frac{100.5}{1}=100.5$
(c) Oxidation no. $\underset{+5}{NaI{O}_3}\to \underset{-1}{I^{-}}$
Change in oxidation number $=6$
Mol. mass of $NaI{O}_3=(23+127+48)=198g$
Eq. mass of $NaI{O}_3=\frac{Mol.mass}{ChangeinO.N.}=\frac{198}{6}=33.0$
(d) $\underset{+5}{NaI{O}_3}\to \underset{0}{I_2}$
Change in oxidation number $=5$
Eq. mass of $NaI{O}_3=\frac{198}{5}=39.6$
(e) The acidity of $Al(OH{)}_3$ is $3$.
Eq. mass of $Al(OH{)}_3=\frac{Mol.mass}{Acidity}=\frac{78}{3}=26.0g$.