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The reaction `H_(3)PO_(4)+Ca(OH)_(2)rarrCa(HPO_(4))_(2)+2H_(2)O` Which statements `(s)` is (are)true?A. the equivalent weight of `H_(3)PO_(4)` is 49B. the resulting solution is neutralized by 1 mole of `KOH`C. 1 mole of `H_(3)PO_(4)` is completely neutralized by `1.5` mole of `Ca(OH)_(2)`D. 1 mole of `H_(3)PO_(4)` can be neutralised completely by 1 mole of `Ca(OH)_(2)` |
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Answer» Correct Answer - A::B::C `H_(3)PO_(4) + Ca(OH)_(2) rarr CaHPO_(4)+2H_(2)O` n-factor for `H_(3)PO_(4) =2` (since `2H^(+)` ions are replaced) eq. wt `=(M)/(2) = (98)/(2) = 49` resulting solution of `CaHPO_(4)` have only are replaceable `H^(+)` so nf =1 hence no of eq = 1 for more so can be neutralized by 1 mole of KOH eq of `CaHPO_(4) =` eq. of KOH For complete neutralization-no. of eq. of `H_(3)PO_(4) =` no. of eq. of `Ca(OH)_(2)` `1 xx 3 = 1.5 xx 2` can be neutralized. |
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