Calculate the pH after 50.0 mL of 0.1 M ammonia solution is treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia , `K_b=1.77xx10^(-5)`

Explanation

`(i) NH_(3)+H_(2)OhArrNH_(4)^(+)+OH^(-)`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])=([OH^(-)]^(2))/([NH_(3)]) (.:' [NH_(4)^(+)]=[OH^(-)])`
`:. [OH^(-)]=sqrt(K_(b)[NH_(3)])=sqrt((1.77xx10^(-5))(0.10))=1.33xx10^(-3)M`
`pOH = - log (1.33xx10^(-3))=3-0.12=2.88`
`:. pH=14-pOH = 14 - 2.88 = 11.12`
(ii) 50.0 mL of 0.10 M `NH_(3) ` : 25.0 mL of 0.10 M HCl = 2.5 mmol of HCl
2.5 mmol of HCl will neutralize 2.5 mmol of `NH_(3) ` forming 2.5 mmol of `NH_(4)Cl` and 2.5 mmol of `NH_(3)` will be left un-neutralized. Thus, now the solution contains 2.5 mmol of `NH_(3) and 2.5` mmol of `NH_(4)Cl`
Total volume of the solution = 50 + 25 = 75 mL
`:. ` In the final solution, we have `[NH_(3)] = 2.5/75 M = 0.033 M, [NH_(4)Cl ] = 2.5/75 M = 0.033M`
The dissociation of `NH_(3)` now will be less due to common ion effect. If x is the amount of `NH_(3) ` now dissociated, then
`{:(,NH_(4)Cl ,+,aq,rarr,NH_(4)^(+),+,Cl^(-)),(,,,,,0.033M,,),(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.033M,,,,,,),("In presence of "NH_(4)Cl,0.033-x,,,,0.033+x,,x),(,~= 0.033,,,,~= 0.033,,),(,,,,,,,):}`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])`
`1.77xx10^(-5)=(0.033(x))/(0.033) :. x= 1.77xx10^(-5)M ` i.e.,`[OH^(-)]=1.77xx10^(-5)`
`pOH = - log (1.77xx10^(-5))=5-0.2480 = 4.75 :. pH = 14-4.75=9.25`