How much of 0.3 M ammonium hydroxide should be mixed with 30 mL of 0.2 M solution of ammonium chloride to give buffer solutions of pH 8.65 and 10 (`pK_b` = 4.75) ?

Explanation

For basic buffer solution,
`pH=14-pK_b+"log""[Base]"/"[Salt]"`
pH=14-4.75 +log `"[Base]"/"[Salt]"`
or log `"[Base]"/"[Salt]"` = pH-9.25
(i)For suffer of pH=8.65
log `"[Base]"/"[Salt]"`=8.65-9.25
or `"[Base]"/"[Salt]"` =Antilog (-0.60)=0.25
or `([NH_4OH])/([NH_4Cl])=((0.3xxV)/1000)/((0.2xx30)/1000)`=0.25
`V=(0.25xx0.2xx30xx1000)/(0.3xx1000)`
=5.02 mL
Similarly for solution of pH =10
log `"[Base]"/"[Salt]"` =10-9.25
`"[Base]"/"[Salt]"` =antilog 0.75=5.62
`therefore ((0.3xxV)/1000)/((0.2xx30)/1000)=5.62`
`therefore V=(5.62xx0.2xx30xx1000)/(0.3xx1000)`
=12.94 mL