The pH of 1 L of an aqueous solution of HCl | vedclass

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(C) For the initial solution: $pH_1 = 1$,so $[H^+]_1 = 10^{-pH_1} = 10^{-1} \ M$. Volume $V_1 = 1 \ L$. Number of moles of $H^+ = M_1 	imes V_1 = 10^

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$9$

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(C) For the initial solution: $pH_1 = 1$,so $[H^+]_1 = 10^{-pH_1} = 10^{-1} \ M$. Volume $V_1 = 1 \ L$. Number of moles of $H^+ = M_1 	imes V_1 = 10^{-1} 	imes 1 = 0.1 \ mol$. For the final solution: $pH_2 = 2$,so $[H^+]_2 = 10^{-pH_2} = 10^{-2} \ M$. Let the final volume be $V_2$. Since the number of moles of $H^+$ remains constant: $M_1 	imes V_1 = M_2 	imes V_2$. $0.1 	imes 1 = 0.01 	imes V_2$. $V_2 = 10 \ L$. Volume of water to be added = $V_2 - V_1 = 10 \ L - 1 \ L = 9 \ L$.

The $pH$ of $1 \ L$ of an aqueous solution of $HCl$ is $1$. How many liters of water must be added to make the $pH$ of this solution $2$ (in $L$)?

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