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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

20ml

B

25 ml

C

40 ml

D

50ml

Text Solution

Verified by Experts

The correct Answer is:
A
`0.025 MH_(3)PO_(4) -= 0.075N H_(3)PO_(4)`
`0.03 M Ca( OH)_(2) -= 0.06 N Ca( OH)_(2)`
`H_(3) PO_(4)` `Ca(OH)_(2)`
`N_(1)V_(1) =N_(2)V_(2)`
`V_(1)=(0.06 xx 25 )/( 0.075)`
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