The reaction
`H_(3)PO_(4)+Ca(OH)_(2)rarrCa(HPO_(4))_(2)+2H_(2)O`
Which statements `(s)` is (are)true?

Options

A. the equivalent weight of `H_(3)PO_(4)` is 49
B. 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)`

Answer

A, B, C

Explanation

`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.