Question

The volume of 0.25 `MH_(3)PO_(3)` required to neutralise 25 ml of 0.03 M `Ca(OH)_(2)` is
A. (a)1.32 mL
B. (b)3 mL
C. (c )26.4 mL
D. (d)2.0 mL

Answer 1

Correct Answer - b
Meq. Of `H_(3)PO_(4)=Meq. Of Ca(OH)_(2)`
`implies Vxx0.25xx2=25xx0.03xx2(H_(3)PO_(3)` is dibasic acd)
`:. V=(25xx3xx2)/(25xx2)=3 mL`