Question

How many liters of water has evaporated on concentrating 10 liters of $H_{2}S{O}_4$ such that its pH decreases from 6 to 5?

• A. 9
• B. 7
• C. 5
• D. 10

Answer 1

The correct option is A 9

Concentration of $\left[H^{+}\right]$ a pH 6 : $\left[H^{+}\right]$ = antilog (-pH) = antilog (-6) = $1\times {10}^{-6}M$

Concentration of $\left[H^{+}\right]$ a pH 5 : $\left[H^{+}\right]$ = antilog (-pH) = antilog (-5) = $1\times {10}^{-5}M$

Since the pH 6 is at the initial condition at which the volume was 10 L, then the volume at which the pH is 5 will be:

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

$1\times {10}^{-6}M\times 10L=1\times {10}^{-5}M\times V_2$

​$V_2=1L$

Hence, the volume of water evaporated = 10 - 1 = 9 L

Hence, option A is correct.