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

Explanation

`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-), pK_(b)=4.75`
On mixing `NH_(3) ` with `NH_(4)Cl` solution, we have
`{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.1M,,,,0.2M,,0),("At eqm.",0.1 -x,,,,0.2+x,,x):}`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])`
As `pK_(b) = 4.75 , i.e., -log K_(b) = 4.75 or log K_(b) = -4.75 = bar(5) .25 :. K_(b) = 1.77xx10^(-5)`
`:. 1.77xx10^(-5) = ((0.2+x)(x))/(0.1-x)~=((0.2)x)/(0.1)` (neglecting x in comparison to 0.1 and 0.2)
or, `x=010^(-5),i..e., [OH^(-)]=0.885xx10^(-5)=8.85xx10^(-6)M`
or `[H^(+)]=(10^(-14))/(8.85xx10^(-6))=1.13xx10^(9) :. pH = - log [H^(+)]=-log (1.13xx10^(-9))=9-0.053=8.95`
Alternatively, the given solution is a buffer solution of a weak base and is salt with a strong acid (discussed in sec. 7.31). For such a buffer,
`pOH=pK_(b)+log.(["Salt"])/(["Base"])=4.75+log. (0.2)/(0.1) = 4.75+log 2 = 4.75 + 0.30 = 5.05 `[ Eqn . (vi)]
`pH = 14 -pOH = 14 - 5.05=8.95`