How many litres of water must be added to 1 | vedclass

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(D) Using the formula $pH = -\log [H^{+}]$.Initial $pH = 1$,so $[H^{+}]_{1} = 10^{-1} = 0.1 \ M$.Final $pH = 2$,so $[H^{+}]_{2} = 10^{-2} = 0.01 \ M$.

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$9$

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(D) Using the formula $pH = -\log [H^{+}]$.Initial $pH = 1$,so $[H^{+}]_{1} = 10^{-1} = 0.1 \ M$.Final $pH = 2$,so $[H^{+}]_{2} = 10^{-2} = 0.01 \ M$.For dilution,the number of moles of $HCl$ remains constant: $M_{1}V_{1} = M_{2}V_{2}$.Substituting the values: $0.1 \ M 	imes 1 \ L = 0.01 \ M 	imes V_{2}$.$V_{2} = \frac{0.1}{0.01} = 10 \ L$.The volume of water to be added is $V_{2} - V_{1} = 10 \ L - 1 \ L = 9 \ L$.

How many 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 $2$?

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