Equal volumes of two solutions of pH = 3 and | vedclass

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(C) Let the volume of each solution be $V$. For the first solution,$[H^{+}]_1 = 10^{-3} \, M$. For the second solution,$[H^{+}]_2 = 10^{-5} \, M$. The

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$3.3$

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(C) Let the volume of each solution be $V$. For the first solution,$[H^{+}]_1 = 10^{-3} \, M$. For the second solution,$[H^{+}]_2 = 10^{-5} \, M$. The total number of moles of $H^{+}$ ions is $n = (10^{-3} 	imes V) + (10^{-5} 	imes V) = V(10^{-3} + 0.01 	imes 10^{-3}) = 1.01 	imes 10^{-3} 	imes V$. The total volume of the mixture is $2V$. The resultant concentration of $H^{+}$ ions is $[H^{+}]_{mix} = \frac{1.01 	imes 10^{-3} 	imes V}{2V} = 0.505 	imes 10^{-3} = 5.05 	imes 10^{-4} \, M$. The resultant $pH$ is calculated as $pH = -\log[H^{+}]_{mix} = -\log(5.05 	imes 10^{-4}) = 4 - \log(5.05) \approx 4 - 0.703 = 3.297 \approx 3.3$.

Equal volumes of two solutions of $pH = 3$ and $pH = 5$ are mixed. The resultant $pH$ is

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