Find out the equivalent weight of `H_(3) PO_(4)`in the reaction:
`Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2 H_(2) O`

Answer

D

Explanation

`Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2 H_(2) O`
In the above equation, `2H` atom has reacted, so `H_(3) PO_(4)`
acts as a dibasic acid.
`( :. Mw of H_(3) PO_(4) = 3 + 31 + 16 xx 4 = 98 g)`
`:. Ew of H_(3) PO_(4) = (Mw)/(n) = (98)/(2) = 49 g`