Calculate the change in pH of 1 litre buffer solution containing `0.1` mole each of `NH_(3)` and `NH_(4)CI` upon addition of:
(i) `0.02` mole of dissolved gasous HCI.
(ii) `0.02` mole of dissolved of NaOH.
Assume no change in volume. `K_(NH_(3))=1.8xx10^(-5)`

Explanation

Initial pH of solution, when
`[NH_(3)]=(0.1)/(1)` and `[NH_(4)CI]=(0.1)/(1)`
`pOH = log.8xx10^(-5)+log ((["Salt"])/(["Base"]))`
`= -log 1.8xx10^(-5)+ log ((0.1)/(0.1))`
`pOH= 4.7447`
`:. pH = 9.2553`
(i) Now `0.02` mole of HCI are added, then
`{:(,HCI+NH_(4)OHrarr,NH_(4)CI+,H_(2)O),("Mole before reaction", 0.02,0.1,0.1),("Mole after reaction",0,0.08,(0.1+0.02)):}`
`:. Volume = 1 litre`
`:. [NH_(4)OH]= (0.08)/(1)` na d`[NH_(4)CI]=(0.12)/(1)`
`:. pOH_(1)= -log 1.8xx10^(-5)+log ((0.12)/(0.08))`
`:. pOH_(1)= 4.9208`
`:. pH_(1)= 9.0792`
Change in `pH=pH-pH_(1)=9.2553-9.0792= +0.1761`
`:." Change in"pH= 0.1761` and pH decreases.
(ii) Now `0.02` mole of NaOH are added.
`{:(,NaOH+,NH_(4)CIrarr,NaCI+,NH_(4)OH),("Mole before reaction",0.02,0.1,0,0.1),("Mole after reaction", 0,0.08,0.02,0.012):}`
`:. pOH_(2)= -log 1.8xx10^(-5)+log(0.08)/(0.12)`
`pOH_(2)=4.5686`
`:. pH_(2)=9.4314`
Change in `pH= pH-pH_(2)`
`=9.2553-9.4314= -0.1761`
`:. "Change in pH" = 0.1761` unit i.e., pH in increases.