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(B) For an aqueous solution of $HCl$ with $pH = 1.0$,the concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$. When an equal

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increase to $1.3$

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(B) For an aqueous solution of $HCl$ with $pH = 1.0$,the concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$. When an equal volume of water is added,the total volume becomes double $(2V)$. Since the number of moles of $HCl$ remains constant,the new concentration $[H^{+}]_{new}$ becomes half of the original concentration: $[H^{+}]_{new} = \frac{0.1 \ M}{2} = 0.05 \ M = 5 	imes 10^{-2} \ M$. The new $pH$ is calculated as: $pH = -\log[H^{+}]_{new} = -\log(5 	imes 10^{-2}) = -(\log 5 + \log 10^{-2}) = -(\log(10/2) - 2) = -(1 - \log 2 - 2) = -(-1 - 0.30) = 1.3$. Therefore,the $pH$ increases to $1.3$.

An aqueous solution of $HCl$ with $pH \ 1.0$ is diluted by adding an equal volume of water (ignoring the dissociation of water). The $pH$ of the $HCl$ solution would be (Given $\log 2 = 0.30$):

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