Question

Question #02c04

Answer 1

`"volume of "NH_4Cl "solution"=400mL=0.4L`

`"Strength of "NH_4Cl "solution"=1.5M`

`"volume of "NH_3 "solution"=600mL=0.6L`

`"Strength of "NH_3 "solution"=0.1M`

`"Total volume of buffer soln"=0.4+0.6=1L`

`"Strength of "NH_4Cl "in buffer soln"=(1.5xx0.4)/1M=0.6M`

`"Strength of "NH_3 "in buffer soln"=(0.1xx0.6)/1M=0.06M`

For `NH_3` the

`K_b=1.8xx10^-5`

`=>pK_b=-log(1.8xx10^-5)=5-log1.8=4.74`

Now by Henderson Hasselblach equation `pOH` of the buffer

`pOH=pK_b+log([[NH_4Cl]]/[[NH_3]])`

`pOH=4.74+log((0.6M)/(0.06M))=4.74+1=5.74`

So for the buffer solution

`pH=14-pOH=14-5.74=10.26`