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For 10^(-4) M BOH (weak base) ,K(b)=5xx...

For `10^(-4)` M BOH (weak base) ,`K_(b)=5xx10^(-5))`:

A

`alpha=0.707`

B

`[OH^(-)]=5xx10^(-5)M`

C

`pH=9.85`

D

All of these

Text Solution

Verified by Experts

`alpha=sqrt((5xx10^(-5))/(10^(-4)))gt0.1` So `(Calpha^(2))/(1-alpha)=K_(b)rArr 5xx10^(-5)=(10^(-4)alpha^(2))/(1-alpha)rArr 2alpha^(2)+alpha-1=0rArr alpha=0.5`
Hence `[OH^(-)]=Calpha=5xx10^(-5)&pH=9.7`
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