How much ml of `0.3 M NH_(4)OH` should be mixed with `30mL` of `0.2M` solution of `NH_(4)Cl` to given buffer solution of `pH 8.65?` (pKb =4.74)

Explanation

It is basic buffer.
i. `pH = 8.65, pOH = 14 - 8.65 = 5.35`
`pOH =pK_(b) +log .([overset(o+)NH_(4)])/([NH_(4)OH])`
`5.35 = 4.75 +log .([overset(o+)NH_(4)])/([NH_(4)OH])`
`([overset(o+)NH_(4)])/([NH_(4)OH]) - "antilog" (5.35 - 4.75)`
`"antilog" (0.6)~~ 4.0`
`:. ([overset(o+)NH_(4)])/([NH_(4)OH]) = (30xx0.2)/(0.3xxV)~~ 4.0`
`:. V = 5mL`