Question #02c04

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Answer 1 · 2016-10-28T08:48:03

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

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

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

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

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

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

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

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

$pH=14-pOH=14-5.74=10.26$