(D) For the initial solution: $pH = 1$,so $[H^{+}]_1 = 10^{-1} = 0.1 \, M$. For the final solution: $pH = 2$,so $[H^{+}]_2 = 10^{-2} = 0.01 \, M$. Usi

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(D) For the initial solution: $pH = 1$,so $[H^{+}]_1 = 10^{-1} = 0.1 \, M$. For the final solution: $pH = 2$,so $[H^{+}]_2 = 10^{-2} = 0.01 \, M$. Using the dilution formula $M_1 V_1 = M_2 V_2$: $0.1 \, M \times 1 \, L = 0.01 \, M \times V_2$. $V_2 = \frac{0.1}{0.01} = 10 \, L$. The volume of water to be added is $V_2 - V_1 = 10 \, L - 1 \, L = 9 \, L$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium pH of strong Acids and strong Bases

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$L$ litres of water must be added to $1 \, L$ of an aqueous solution of $HCl$ with a $pH$ of $1$ to create an aqueous solution with $pH$ of $2$?

JEE MAIN 2013 All JEE MAIN

A
$0.1$
B
$0.9$
C
$2$
D
$9$

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2 Equilibrium: Ionic Equilibrium

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pH of strong Acids and strong Bases

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$L$ litres of water must be added to $1 \, L$ of an aqueous solution of $HCl$ with a $pH$ of $1$ to create an aqueous solution with $pH$ of $2$?

Similar Questions

Which of the following will decrease the $pH$ of a $50 \ mL$ solution of $0.01 \ M \ HCl$?

What will be the concentration of $H^{+}$ ions in a solution prepared by dissolving $4 \ g$ of $NaOH$ in $100 \ mL$ of water at $25 \ ^{\circ}C$?

Number of equivalents of $H_2SO_4$ present in $100 \ mL$ of its solution whose $pH$ is $5$.

Calculate the $pH$ of $1.0 \times 10^{-8} \ M \ NaOH$ solution.

$40 \, mg$ of pure sodium hydroxide is dissolved in $10 \, L$ of distilled water. The $pH$ of the solution is

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