Calculate the `pH` of the solution in which `0.2M NH_(4)Cl` and `0.1M NH_(3)` are present. The `pK_(b)` of ammonia solution is `4.75`.

Explanation

`NH_(3) +H_(2)O hArr NH_(4)^(+)+OH^(-)`
The ionization constant of `NH_(3)`
`K_(b)= "antilog" (-pK_(b))i.e.`
`K_(b)=10^(4.75)=1.77xx10^(-5)M`
`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-)`
Initial concnetration (M)
`0.10 " " 0.20 " "0`
Change in concentration to reach (M)
`-x" " +x" "+x`
At equilibrium(M)
`0.10-x " " 0.20+x " " x`
`k_(b)=[NH_(4)^(+)][OH^(-)]//[NH_(3)]`
`=(0.20xx x)(x) //(0.1-x) (x) //(0.1-x)=1.77xx10^(-5)`
As `K_(b)` is small, we can neglect x in comparison to 0.1 M and 0.1 M Thus
`[OH^(-)]=x=0.88xx10^(-5)`
Therefore, `[H^(+)]=1.12xx10^(-9)`
`pH=log[H^(+)]=8.95`