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To neutralize completely 20 ml of 0.1 M ...

To neutralize completely 20 ml of 0.1 M phosphorus acid, 40 mol of KOH was required. What volume of this KOH solution will be required to neutralize 0.66 g of `H_(3)PO_(2)`?

A

100 ml

B

200 ml

C

300 ml

D

66.7 ml

Text Solution

Verified by Experts

The correct Answer is:
A
Eqts of `H_(3)PO_(3)` = Eqts of KOH
`20xx0.1xx2=1xx40xxMimpliesM=0.1`
Now Eqts of KOH = Eqts of `H_(3)PO_(2)`
`0.1xx1xxV=(0.66)/(66)impliesV=0.1" lit"`
`V=100ml`
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