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The volume of 0.25 MH(3)PO(3) required t...

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

Text Solution

Verified by Experts

The correct Answer is:
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`
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