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Question

Calculate the weight (in mg) of HCl that should be added to 100 ml of 0.1 M BOH to make a buffer solution of pOH=6.
Kb(BOH)=1×108 (assume there is no change in volume on addition of HCl)

A
3.6
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B
4.6
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C
5.6
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D
6.6
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Solution

The correct option is A 3.6
Let the mass of HCl added be x mg
mmol of HCl is Mass (in mg)Molar mass=x36.5
mmol of BOH is 100×0.1=10
The mmol of each species is calculated below :
BOH + HCl BCl+H2O
t=0:(100×0.1=10) x36.5t=teq: 10x36.5 0 x36.5 =365x36.5

This constitutes basic buffer. For basic buffer, the pOH given by:

pOH=pKb+log [BCl][BOH]pOH=log[1×108]+log(x×36.536.5×(365x))6=8+log(x365x)2=log(x365x)102=(x365x)3.650.01x=x1.01x=3.65x=3.613.6

Weight of HCl required = 3.6 mg.

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