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)`

Options

A. degree of dissociation of `NH_(3)` approaches to zero.
B. change in `pH` would be `1.85`
C. concentration of `[Na^(+)]=0.1 M, [NH_(3)]=0.1 M`, `[OH^(-)]=0.2 M`
D. on addition of `OH^(-)`,`K_(b)` of `NH_(3)` does not changes

Answer

C

Explanation

Initial `pOH=(1)/(2)(pK_(b)-log C)`
`= (1)/(2)(4.7-log 0.1)=2.85`
Final `pOH=1`
Change in `pOH=`Change in `pH=1.85`