Let V mL of `NH_(4)OH` be mixed with `NH_(4)CI` to have a buffer of `pH 8.65`. The total volume after mixing becomes `(V+30)mL`.

m mole of `NH_(4)OH= 0.3xxV`

`:. [NH_(4)OH]=(0.3xxV)/((V+30))`

m mole of `NH_(4)CI= 0.2xx30`

`:. [NH_(4)CI]=(0.2xx30)/((V+30))`

Also pOH of buffer mixture is given by:

`pOH=pK_(b)`=log `(["Salt"])/(["Base"])`ltbgt or 14-8.65=4.74+log`((0.2xx30)//(V+30))/((0.3xxV)//(V+30)) , (pOH=n 14-pH)`

0.61=log`(6)/(0.3xxV)`

`:. V= 4.91 mL`

Similary calculate,

`(14-10=4.74+log((0.2xx30//(V_(1)+30))/(0.3xxV_(1)//(V_(1)+30))))`

For `pH= 10, V= 109.9 mL`