`50.0` mL of `0.10` M ammonia solution is treated with `25.0` mL of `0.10M HCI`. If `K_(b)(NH_(3))=1.77xx10^(-5)`, the pH of the resulting solution will be

Explanation

`NH_(3) + H_(2)O rarr NH_(4)^(+) + OH`
`K_(b) = [NH_(4^(+))][OH^(-)]//[NH_(3)] = 1.77 xx 10^(-6)`
Before neutralization,
`[NH_(4)^(+)] = [OH^(-)] = x`
`[NH_(3)]= 0.10 - x = 0.10`
`x^(2)//0.10 = 1.77 xx 10^(-5)`
Thus, `x = 1.33 xx 10^(-3) = [OH^(-)]`
Therefore, `[H^(+)] = K_(w)//[OH]=10^(-14)//(1.33 xx 10^(-3)) = 7.51 xx 10^(-12)`
`pH = -log(7.5 xx 10^(-12)) = 11.12`
On addition of `25 mL` of `0.1 M HCl` to `50 mL` of `0.1 M` ammonia solution (i.e., `5` mmol 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,rarr,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` mmol of `NH_(4)^(+)` ions (i.e., `0.033 M`) and `2.5` mmol (i.e., `0.033 M`) of uneutralised `NH_(3)` molecules. This `NH_(3)` exists in the following equilibrium :
`{:(NH_(4)OH,hArr,NH_(4)^(+),+,OH^(-)),(O.033M-y,,y,,y):}`
where, `u = [OH^(-)] = [NH_(4)^(+)]`
The final `75` mL solution after neutralisation already contains `2.5 m` mol `NH_(4)^(+)` ions (i.e., `0.0333 M`) thus total concentration of `NH_(4)^(+)` ions is given as :
`[NH_(4)^(+)] = 0.033+ y`
As y is small. `[NH_(4)OH] = 0.33` M and
`[NH_(4)^(+)] = 0.033 +y`
As y is small`[NH_(4)OH] = 0.033 M` and
`[NH_(4)^(+)] = 0.033 M`
We know,
`K_(b) = [NH_(4)^(+)][OH^(-)]//[NH_(4)OH]`
`= y (0.033)//(0.033) = 1.77 xx 10^(-5) M`
Thus, `y = 1.77 xx 10^(-5) = [OH^(-)]`
`[H^(+)] = 10^(-4) // 1.77 xx 10^(-5) = 0.56 xx 10^(-9)`
Hence, `pH = 9.24`