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Question

How much water from 5 litre of 103M HCl should be evaporated to change its pH by 2 units?

A
3 litre
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B
0.5 litre
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C
2.54 litre
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D
4.95 litre
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Solution

The correct option is C 3 litre
Current concentration is [H+]=103M
no. of moles of HCl is 103×5=5×103
Desired concentration is when pH=2. pH=log[H+]
2=log[H+]
[H+]=102 is desired concentration.
The concentration should increase by 10-fold.
so n=CnewVnew=102Vnew
5×103=102Vnew
Vnew=1025×103=2liter
So 52=3 liter of water should be evaporated to change concentration from 103 to pH=2

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