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(D) Let the volume of each solution be $V \ L$. Total volume of the mixture $= 2V \ L$. Moles of $NaOH = 0.1 	imes V = 0.1V$. Moles of $HCl = 0.01

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$12.65$

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(D) Let the volume of each solution be $V \ L$. Total volume of the mixture $= 2V \ L$. Moles of $NaOH = 0.1 	imes V = 0.1V$. Moles of $HCl = 0.01 	imes V = 0.01V$. Since $NaOH$ is a strong base and $HCl$ is a strong acid,they react as: $NaOH + HCl ightarrow NaCl + H_2O$. Moles of $NaOH$ left unneutralized $= 0.1V - 0.01V = 0.09V$. Concentration of $[OH^-] = \frac{0.09V}{2V} = 0.045 \ M$. $pOH = -\log(0.045) = 1.3468 \approx 1.35$. $pH = 14 - pOH = 14 - 1.35 = 12.65$.

What is the $pH$ of the resulting solution when equal volumes of $0.1 \ M \ NaOH$ and $0.01 \ M \ HCl$ are mixed?

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