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(B) For $HCl$ solution: $pH = 2$,so $[H^{+}] = 10^{-2} \ M$. Milli-moles of $H^{+} = 100 \ mL 	imes 10^{-2} \ M = 1 \ mmol$.For $NaOH$ solution: $pH

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$11.48$

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(B) For $HCl$ solution: $pH = 2$,so $[H^{+}] = 10^{-2} \ M$. Milli-moles of $H^{+} = 100 \ mL 	imes 10^{-2} \ M = 1 \ mmol$.For $NaOH$ solution: $pH = 12$,so $pOH = 14 - 12 = 2$. Thus,$[OH^{-}] = 10^{-2} \ M$. Milli-moles of $OH^{-} = 200 \ mL 	imes 10^{-2} \ M = 2 \ mmol$.Reaction: $H^{+} + OH^{-} \longrightarrow H_{2}O$.Remaining milli-moles of $OH^{-} = 2 - 1 = 1 \ mmol$.Total volume of the mixture = $100 \ mL + 200 \ mL = 300 \ mL$.Concentration of $[OH^{-}] = \frac{1 \ mmol}{300 \ mL} = 3.33 	imes 10^{-3} \ M$.$pOH = -\log(3.33 	imes 10^{-3}) = 3 - \log(3.33) \approx 3 - 0.52 = 2.48$.$pH = 14 - pOH = 14 - 2.48 = 11.52$.

What will be the resultant $pH$ when $100 \ mL$ of an aqueous solution of $HCl$ $(pH = 2)$ is mixed with $200 \ mL$ of an aqueous solution of $NaOH$ $(pH = 12)$?

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