The pH of 100 mL of an aqueous solution of a | vedclass

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(C) For a strong acid,the $pH$ is given by $pH = -\log[H^+]$.Initially,$pH_1 = 1$,so $[H^+]_1 = 10^{-1} \ M = 0.1 \ M$.Initial volume $V_1 = 100 \ mL$

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$900$

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(C) For a strong acid,the $pH$ is given by $pH = -\log[H^+]$.Initially,$pH_1 = 1$,so $[H^+]_1 = 10^{-1} \ M = 0.1 \ M$.Initial volume $V_1 = 100 \ mL$.We want the final $pH_2 = 2$,so $[H^+]_2 = 10^{-2} \ M = 0.01 \ M$.Using the dilution formula $M_1V_1 = M_2V_2$:$0.1 \ M 	imes 100 \ mL = 0.01 \ M 	imes V_2$.$V_2 = (0.1 	imes 100) / 0.01 = 1000 \ mL$.The volume of water to be added is $V_2 - V_1 = 1000 \ mL - 100 \ mL = 900 \ mL$.

The $pH$ of $100 \ mL$ of an aqueous solution of a strong acid is $1$. The volume of water to be added to change its $pH$ to $2$ is .......... $mL$.

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