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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» Correct Answer - D `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` |
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