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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 $$10^{-3}\ M$$ 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, $$M_1V_1=M_2V_2$$
Hence, $$5 \times 10^{-3}=0.1 \times V_2$$
$$V_2=0.05$$ L
Hence, the change in volume $$= \Delta V = 5-0.05=4.95$$ L

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

Chemistry

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