How many moles of `NaOH` can be added to `0.1L` of solution of `0.1M NH_(3)` and `0.1M NH_(4)CI` without changing `pOH` by more than pne unit `(pK_(a) "of" NH_(3) = 4.75)` ?

Explanation

It is a basic buffer
`pOH = pK_(b) + log [("Salt")/("base")]`
`poH = 4.75 + log [(0.1)/(0.1)]`
`pOH = pK_(b) - 4.75`.
Let `x` mol of `NaOH` ids added (rule `BBB)`. In a basic buffer `(B)`, on adding strong base `(B)`, the concentration of weak base `(B)` increases and that of salt decreases. on adding the base, `pH` will increase and `pOH` will decrease.
Nwe `pOH = 4.75 - 1 = 3.75`
`:. 3.75 = 4.75 + log [(0.1 -x)/(0.2 +x)]`
Solve, `x = 0.082 mol = 0.082 M (V = 1L)`