How much of `0.3M` ammonium hydroxide should be mixed with 30 mL of `0.2M` solution of ammonium chloride to give buffer solutions of `pH 8.65` and 10 ? (Give: `pK_(b) of NH_(4)OH=4.75`)

Explanation

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`