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
B
0.50
C
2.54
D
4.95

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, $$5 \times 10^{-3}=0.1 \times V_2$$$$V_2=0.05$$ LHence, the change in volume $$= \Delta V = 5-0.05=4.95$$ LHence, the water that must be evaporated is $$4.95$$ L.Chemistry

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