Calculate the `pH` of `0.10M` ammonia solution. Calcualte the `pH` after `50.0mL` of this solution is treated with `25.0mL` of `0.10M HCl`. The dissociation constant of ammonia, `K_(b)=1.77xx10^(-5)`.

Explanation

`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-)`
`K_(b)=[NH_(4)^(+)][OH^(-)]//[NH_(3)]=1.77xx10^(-5)`
Before neutralization,
`NH_(4)^(+)][OH^(-)]=x`
`[NH_(3)]=0.10-x=0.10, x^(2)//0.10=1.77xx10^(-5)`
Thus, `x=1.33xx10^(-3)=[OH^(-)]`
Therefore, `[H^(+)]=k_(w)//[OH^(-)]=10^(-14)//(1.33xx10^(-3))=7.51xx10^(-12)`
`pH=-log(7.5xx10^(-12))=11.12`
On additio of 25 mL of 0.1 M HCl solution (i.e., 2.5 m mol of HCl) to 50 mol of `NH_(3))`. 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of `NH_(3)` molecules and 2.5 mmol of `NH_(4)^(+)`
`{:(NH_(3),+HCl, to NH_(4)^(+), Cl^(-)),(2.5,2.5,0,0),("At equilibrium",,,),(0,0,2.5,2.5):}`
The resulting 75 mL of solution contains 2.5 mmole of `NH_(4)^(+)` ions(i.e., 0.33M) and 2.5 mmol (i.e., `0.033M)` of uneutralised `NH_(3)` molecules. this `NH_(3)` exists as `NH_(4) OH` in the following equilibrium :
`{:(NH_(4)OH hArr NH_(4)^(+),+,OH^(-)),(0.033M-y,y,y):}`
where `y=[OH]=[NH_(4)^(+)]`
The final 75 mL solution after neutralisation alredad contains 2.5 m mol `[NH_(4)^(+)]` ions (i.e., thus total concentration fo `[NH_(4)^(+)]` ions is given as `[NH_(4)^(+)]=0.03+y`
As y is small, `[NH_(4)OH]~~ 0.033 M and [NH_(4)^(+)]~~0.033M`
We know, ,
`K_(b)=[NH_(4)^(+)][OH^(-)]//[NH_(4)OH]`
`=y(0.033)//(0.033)1.77xx10^(-5)M`
Thus, `y=1.77xx10^(-5)=[OH^(-)]`
`[H^(+)]=10^(-14)//1.77xx10^(-5)=0.56xx10^(-9)`
Hence `pH=9.24`.