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Question

Calculate the weight in mg of HCl added to 100 ml of 0.1 N BOH to have its pH = 6.6 and Kb=6.25×108; Antilog (7.4)=3.98×108. (Assume there is no change in volume on addition of HCl)

A
300
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B
100
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C
425
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D
223
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Solution

The correct option is D 223
Given, [BOH]=0.1N, pH=6.6, Kb=6.25×108,V=100 ml
Let the mass of HCl added be x mg
BOH+HClBCl+H2OMillimoles 10 x36.5 0 0at eqb 10x36.5 0 x36.5
This constitutes basic buffer.
For basic buffer, the [OH] is given by Henderson equation
pOH=log Kb+log[BCl][BOH]
[OH]=Kb[BOH][BCl]..........(1)
we know, pH+pOH=14
6.6+pOH=14
pOH=7.4
​​​​​​​log[OH]=7.4
​​​​​​​log[OH]=7.4
​​​​​​​[OH]=3.98×108
now substitue values in equation (1),
3.98×108=6.25×108×(365x36.5)(36.5x)
(365x36.5)(36.5x)=0.6368
(365xx)=0.6368
On solving, x = 223.92
Weight of HCl required = 223 mg.

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