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` is a weak base and ionises partially whereas , ammonium chloride ionises completely in the solution. Let x be the number of moles of ammonia ionised :
`NH_3+H_2O hArr NH_4^(+)+OH^(-)`
`NH_4Cl hArr NH_4^(+) + Cl^-`
Since `NH_4Cl` is completely ionised, the concentration of `NH_4^+` ions from `NH_4Cl` =0.2 M
Total `[NH_4^+]` =0.2 +x `approx` 0.2
Total conc. of `NH_3` left unionized =0.1-x `approx` 0.1 (because x is a small)
So, at equilibrium
`{:(NH_3+H_2O hArr , NH_4^(+)+,OH^-),(0.1,0.2,x):}`
Now `pK_b=-log K_b=4.75`
`therefore log K_b=-4.75`
`K_b` =antilog (-4.75)=`1.78xx10^(-5)`
Now `K_b=([NH_4^+][OH^-])/([NH_3])`
`1.78xx10^(-5)=((0.20)x)/0.1`
`therefore x=0.89xx10^(-5)`
`therefore [OH^-]=0.89xx10^(-5)`
`[H_3O^+]=K_w/([OH^-])=(1xx10^(-14))/(0.89xx10^(-5))=1.12xx10^(-9)`
`pH=-log [H_3O^+]=-log (1.12xx10^(-9))`
=8.95