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)OhArrNH_(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 concentration (M)",,,,,,),(0.10,,,,0.20,,0),("Change to reach",,,,,,),("equilibrium (M)",,,,,,),(-x,,,,+x,,+x),("At equilibrium(M)",,,,,,),(0.10-x,,,,0.20+x,,x):}`
`K_(b)=[NH_(4)^(+)][OH^(-)]//[NH_(2)]`
`=(0.20+x)(x)//(0.1-x)=1.77xx10^(-5)`
As `K_(b)` is small, we can neglect x in comparison to `0.1M` and `0.2M`. Thus,
`[OH^(-)]=x=0.88xx10^(-5)`
Therefore, `[H^(+)]=1.12xx10^(-9)`
`pH=-log[H^(+)]=8.95`