(C) Given,$pH_1 = 4$,so $[H^+]_1 = 10^{-4} \ M$. Initial volume $V_1 = 10 \ mL$. After dilution,$pH_2 = 5$,so $[H^+]_2 = 10^{-5} \ M$. Using the dilut

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(C) Given,$pH_1 = 4$,so $[H^+]_1 = 10^{-4} \ M$. Initial volume $V_1 = 10 \ mL$. After dilution,$pH_2 = 5$,so $[H^+]_2 = 10^{-5} \ M$. Using the dilution formula $M_1 V_1 = M_2 V_2$: $10^{-4} \times 10 = 10^{-5} \times V_2$ $V_2 = \frac{10^{-4} \times 10}{10^{-5}} = 100 \ mL$. Volume of water to be added = $V_2 - V_1 = 100 \ mL - 10 \ mL = 90 \ mL$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of strong Acids and strong Bases

Medium

The $pH$ of a $10 \ mL$ aqueous solution of $HCl$ is $4$. The amount of water to be added to this solution in order to change its $pH$ from $4$ to $5$ is $..... \ mL$

KVPY 2009 All KVPY

A
$30$
B
$60$
C
$90$
D
$120$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

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The $pH$ of a $10 \ mL$ aqueous solution of $HCl$ is $4$. The amount of water to be added to this solution in order to change its $pH$ from $4$ to $5$ is $..... \ mL$

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