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pOH of 0.002 M HNO(3) is:...

pOH of 0.002 M `HNO_(3)` is:

A

`11+log2`

B

`11-log2`

C

`-3 +log 2`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
1
`[H^(+)]=0.002M`
`pH=log2xx10^(-3)=3-log2`
`:.pOH=11+log2`
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