How many moles of NaOH can be added to `1.0L` solution of `0.10MNH_(3)` and `0.10MN_(4)CI` without changing in volume. `(K_(b)for NH_(3)=1.8xx10^(-5))`

Explanation

For buffer of `NH_(3)` and `NH_(4)CI`,
`pOH= -log 1.8xx10^(-5)+log((1)/(1))`
`pOH= 4.75`
Original `pOH=4.75`, on addition of NaOH, it will change only 1 unit (decrease) i.e., new `pOH= 3.75` or pH will increase from `9.25` to `10.25`.
`OH^(-)NH_(4)^(+)rarrNH_(3)+H_(2)O`
`:. 3.75= -log 1.8xx10^(-5)+log (([NH_(4)^(+)])/(NH_(3)])`
`([NH_(4)^(+)])/([NH_(3)])=0.1`
Thus NaOH addition must not change the concentration ration by more than `0.1`
i.e., `[NH_(4)^(+)]=0.1[NH_(3)]` ......(1)
Initially
`[NH_(4)^(+)]+[NH_(3)]=0.1+0.1=0.2` ......(2)
By eqs. (1) and (2),
`0.1xx[NH_(3)]+[NH_(3)]=0.2`
`:. [NH_(3)]=0.182`
`:. [NH_(4)^(+)]=0.018`
Thus new solution must have `[NH_(4)^(+)]=0.018` in place of `0.10`. Therefore additio of `0.1-0.82` mole of NaOH be made.