Calculate the `pH` after `50.0mL` of this solution is treated with `25.0mL` of `0.1M HCI`
`K_(b)` for `NH_(3) = 1.77 xx 10^(-5) (pK_(b) ~~ 4.76)`.

Explanation

On addition of 25mL of 0.1 m HCI (2.5 m moles), to 50mL of 0.1M `NH_(3) (5.0 "m moles")`, 2.5 m moles of `NH_(3)` molecules are neutralised. The resulting `75mL(25mL +50mL)` solution constains the remaining unnceuralised `2.5` mmoles of `NH_(3)` molecules and 2.5 m moles of `overset(o+)NH_(4)` ions
`{:(,NH_(3),+,HCl rarr,overset(oplus)(N)H_(4)+,Cl^(Theta)),("mmoles",5.0 "mmol",,2.5 "mmol",0,0),("mmoles at",(5-25),,2.5-2.5 =02.5,2.5,),("equilibrium",=2.5,,,,):}`
`[overset(o+)NH_(4)]_(eq) = (2.5 "mmoles")/(75mL) = 0.033M`
`[NH_(3)]_(eq) ("unreacted") = (2.5 "mmol")/(75mL) = 0.033M`
This `NH_(3)` exists in the following equilibrium :
`{:(NH_(4)OH,hArr,overset(Theta)(N)H_(4),+,overset(Theta)(O)H),(0.033-y,,y,,y):}`
`:. [overset(o+)NH_(4)]_("total") = 0.33 +y ~~ 0.033` (as y is small)
`[NH_(4)OH] = 0.33 - y ~~ 0.033` (as y is small)
`:. K_(b) = ([overset(o+)NH_(4)][overset(Θ)OH])/([NH_(4)OH]) = ((0.033)[overset(Θ)OH])/((0.033)) = 1.77xx10^(-5)`
`:. [overset(Θ)Oh] = 1.77 xx 10^(-5)`
`pOH =- log (1.77 xx 10^(-5)) = 4.76`
`:. pH = 14 - 4.76 = 9.24`