Find the equivalent weight of $ ext{H}_3 ext{PO}_4$ in the following reaction:
$ ext{Ca(OH)}_2 +

Question

Find the equivalent weight of $	ext{H}_3	ext{PO}_4$ in the following reaction:
$	ext{Ca(OH)}_2 + 	ext{H}_3	ext{PO}_4 ightarrow 	ext{CaHPO}_4 + 2	ext{H}_2	ext{O}$
(Given molar mass of P = 31 g/mol, Ca = 40 g/mol, O = 16 g/mol, H = 1 g/mol)

Professor PC Thomas Classes · Answer

In this specific neutralization reaction, two acidic hydrogens of $	ext{H}_3	ext{PO}_4$ are replaced by calcium to form the $	ext{HPO}_4^{2-}$ ion. Therefore, its n-factor (basicity) in this reaction is 2. The molar mass of $	ext{H}_3	ext{PO}_4 = 3(1) + 31 + 4(16) = 98 	ext{ g/mol}$. Equivalent weight = Molar mass / n-factor = $98 / 2 = 49$.