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Question

How much water (in L) must be evaporated from 5 L of 10−3 M HCl to decrease its pH by 2 units?

A
1.50
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B
0.50
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C
2.54
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D
4.95
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Solution

The correct option is D 4.95
We know that pH=log [H+]
Th pH of a 103 M solution is 3.

pH is decreased by 2 units to the value of 1.

Hence, the concentration of the solution will be 0.1 M.
Now, M1V1=M2V2
Hence, 5×103=0.1×V2
V2=0.05 L
Hence, the change in volume =ΔV=50.05=4.95 L

Hence, the water that must be evaporated is 4.95 L.

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