What is the equivalent mass of :
(a) `H_(3)PO_(4)` when neutralised to `HPO_(4)^(2-)`
(b) `HClO_(4)`
(c ) `NaIO_(3)` when reduced to `I^(-)`
(d) `NaIO_(3)` when reduced to `I_(2)`
(e ) `Al(OH)_(3)`.

Explanation

(a) Molecular mass of `H_(3)PO_(4)=(3+31+64)=98g. H_(3)PO_(4)` when neutralised to `HPO_(4)^(2-)`, two `H^(+)` ions have been replaced.
Thus, eq. `"mass"=("Mol. Mass")/("No. of replaceable hydrogen atoms")`
`=(98)/(2)=49.0 g`
(b)` HClO_(4)` molecule contains one replaceable hydrogen atom. Thus, eq. mass`=("Mol. mass")/(1)=100.5`
(c ) `NaIO_(3) to I^(-)`
Oxidation no. `+5 " " -1`
Change in oxidation number=6
Mol. mass of `NaIO_(3)=(23+127+48)=198 g`
Eq. mass of `NaIO_(3) = ("Mol. mass")/("Change in O.N.")=(198)/(6)=33.0`
(d) `NaIO_(3) to I_(2)`
Oxidation no. ` +5 " " 0`
Change in oxidation number =5
Eq. mass of `NaIO_(3) = (198)/(5)=39.6`
(e ) The acidity of `Al(OH)_(3)` is 3.
Eq. mass of `Al(OH)_(3) = ("Mol. mass" )/("Acidity")=(78)/(3)=26.0 g`