Question

A certain volume of $0.001$ N NaOH solution is diluted with $900$ ml of water. The decrease in pH of the solution is one unit. The original volume (in ml) of the solution was:

• A. $1$
• B. $10$
• C. $100$
• D. $1000$

Answer 1

Answer: (C)

$100$

Change in $pH$ is given by $\Delta pH=p{H}_2-p{H}_1=-log\frac{[H^{+}{]}_2}{[H^{+}{]}_1}=1$ (given)

Thus, the ratio of hydrogen ion concentration $=\frac{[H^{+}{]}_2}{[H^{+}{]}_1}={10}^{-1}$

The hydrogen ion concentration increases by $10$ times.

Let $x$ ml be the original volume of the solution.
Total final volume $=(x+900)$ ml

Thus, the ratio of hydrogen ion concentration $=\frac{[H^{+}{]}_2}{[H^{+}{]}_1}=\frac{V_1}{V_2}$

Substituting values in the above expression, we get,

$\frac{1}{10}=\frac{x}{900+x}⟹x=100$ (positive value)

Hence, the original volume was $100$ ml.

Option C is correct.