An analyst wants to convert 1 L of HCl solut | vedclass

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(B) For $pH = 1$,the concentration of $HCl$ is $[H^+]_1 = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$.For $pH = 2$,the concentration of $HCl$ is $[H^+]_2 = 10^{

Vedclass · Accepted Answer

$9000$

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(B) For $pH = 1$,the concentration of $HCl$ is $[H^+]_1 = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$.For $pH = 2$,the concentration of $HCl$ is $[H^+]_2 = 10^{-pH} = 10^{-2} \ M = 0.01 \ M$.Using the dilution formula $M_1 	imes V_1 = M_2 	imes V_2$:$0.1 \ M 	imes 1 \ L = 0.01 \ M 	imes V_2$.$V_2 = \frac{0.1}{0.01} \ L = 10 \ L$.The volume of water to be added is $V_{added} = V_2 - V_1 = 10 \ L - 1 \ L = 9 \ L$.Since $1 \ L = 1000 \ mL$,the volume of water added is $9 	imes 1000 \ mL = 9000 \ mL$.

An analyst wants to convert $1 \ L$ of $HCl$ solution with $pH = 1$ to a solution of $HCl$ with $pH = 2$. The volume of water needed to do this dilution is $...... \ mL$. (Nearest Integer)

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