Solution of `0.1 N NH_(4)OH` and `0.1 N NH_(4)Cl` has `pH 9.25`, then find out `K_(b)` of `NH_(4)OH`.

Options

A. `9.25`
B. `4.75`
C. `3.75`
D. `8.25`

Answer

B

Explanation

It is a basic buffer. This,
`pOH=pK_(b)+log.(C_("salt"))/(C_("base"))`
For any aqueous solution at 298 K, we have
`pH+pOH=pK_(w)=14`
or `pOH=14-pH`
`= 14-9.25`
`=4.75`
we are given `C_("salt")=C_("base")`. Thus,
`log.(C_("salt"))/(C_("base"))=log 1= 0`
Hence, `pOH= pK_(b)=4.75`.