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At 25^(@)C K(b) for BOH=1.0xx10^(-12).0....

At `25^(@)C K_(b)` for `BOH=1.0xx10^(-12).0.01M` solution of `BOH` has `[OH^(-)]`:

A

`1.0xx10^(-6)M`

B

`1.0xx10^(-7)M`

C

`1.0xx10^(-5)M`

D

`2.0xx10^(-6)M`

Text Solution

Verified by Experts

The correct Answer is:
B
`[OH^(-)]=Calpha=Csqrt((K_(b))/(C ))`
`= sqrt((K_(b).C))= sqrt(1.0xx10^(-12)xx0.01)`
`=1.0xx10^(-7)M`
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