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 to 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 addition of `25mL` of `0.1MHCl` solution (i.e., `2.5` mmol of `HCl`) to `50mL` of `0.1M` ammonia solution (i.e., `5`mmol of `NH_(3)`), `2.5` mmol of ammonia molecules are neturalized. The resulting `75mL` 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 `75mL` of solution contains `2.5`mmol of `NH_(4)^(+)` ions (i.e., `0.03M`) of uneutralised `NH_(3)` molecules . This `NH_(3)` exists in the following equilibrium :
`{:(NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-)),(0.033M-y,,y,,y):}`
where `y=[OH^(-)]=[NH_(4)^(+)]`
The final `75mL` solutions after neutralisation already contains `2.5m` mol `NH_(4)^(+)` ions (i.e, `0.033M`), thus total concentration of `NH_(4)^(+)` ions is given as :
`[NH_(4)^(+)]=0.033+y`
As `y` is small , `[NH_(4)OH]=0.033M` 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`