200 mL of an aqueous solution of HCl (pH=2) | vedclass

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(B) For $HCl$ solution: $pH = 2$, so $[H^+] = 10^{-2} \ M$. Moles of $H^+ = 10^{-2} \ mol/L 	imes 0.2 \ L = 2 	imes 10^{-3} \ mol$. For $NaOH$ solut

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$11$

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(B) For $HCl$ solution: $pH = 2$, so $[H^+] = 10^{-2} \ M$. Moles of $H^+ = 10^{-2} \ mol/L 	imes 0.2 \ L = 2 	imes 10^{-3} \ mol$. For $NaOH$ solution: $pH = 12$, so $pOH = 14 - 12 = 2$. $[OH^-] = 10^{-2} \ M$. Moles of $OH^- = 10^{-2} \ mol/L 	imes 0.3 \ L = 3 	imes 10^{-3} \ mol$. Since $n(OH^-) > n(H^+)$, the reaction $H^+ + OH^- ightarrow H_2O$ occurs. Remaining moles of $OH^- = 3 	imes 10^{-3} - 2 	imes 10^{-3} = 1 	imes 10^{-3} \ mol$. The total volume of the resulting solution is $1.0 \ L$. Concentration of $OH^- = \frac{1 	imes 10^{-3} \ mol}{1.0 \ L} = 10^{-3} \ M$. $pOH = -\log(10^{-3}) = 3$. $pH = 14 - pOH = 14 - 3 = 11$.

$200 \ mL$ of an aqueous solution of $HCl$ $(pH=2)$ is mixed with $300 \ mL$ of aqueous solution of $NaOH$ $(pH=12)$ and is diluted to $1.0 \ L$. The $pH$ of the resulting solution is

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