Question

How much water (in L) must be evaporated from $5$ L of ${10}^{-3}MHCl$ to decrease its pH by $2$ units?

• A. $1.50$
• B. $0.50$
• C. $2.54$
• D. $4.95$

Answer 1

The correct option is D $4.95$

We know that $pH=-log\left[H^{+}\right]$

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 $0.1M$.

Now, $M_1{V}_1=M_2{V}_2$

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.