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When 0.1 m "mole" of solid NaOH is added...

When `0.1 m "mole"` of solid `NaOH` is added in `1 L` of `0.1 M NH_(3) (aq)` then which statement is going to be wrong?
`(K_(b)=2xx10^(-5), log 2=0.3)`

A

degree of dissociation of `NH_(3)` approaches to zero.

B

change in `pH` by adding `NaOH` would be `1.85`.

C

in solution `[Na^(+)]=0.1 M ,[NH_(3)]=0.1M, [OH^(-)]=0.2M`

D

on addition of `OH^(-),K_(b)` of `NH_(3)` does not changes.

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