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)Cl]=(0.1)/1`
`pOH =-log 1.8 xx 10^(-5)+log (["Salt"])/(["Base"])`
`=-log 1.8 xx 10^(-5)+log (0.1)/(0.1)`
pOH =4.7447
`therefore pH =9.2553`
Now 0.02 moles of HCl are added, then `HCl+NH_(4)OH to NH_(4)Cl+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) and [NH_(4)Cl]=(0.12)/(1)`
`therefore pOH_(1) =-log 1.8 xx 10^(-5) +log (0.12)/(0.08)`
`therefore pOH_(1)=4.9208`
`therefore pH_(1)=9.0792`
Change in pH `=pH -pH_(1)=9.255-9.0732=+0.1761`
Change in pH =0.1761 and pH decreases