When 200 mL solution of HCl of pH=2 is mixed | vedclass

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(B) For $HCl$ solution: $pH = 2$,so $[H^+] = 10^{-2} \ M = 0.01 \ M$. Number of moles of $H^+ = M 	imes V(L) = 0.01 	imes 0.2 = 0.002 \ mol$. For $N

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$11.3$

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(B) For $HCl$ solution: $pH = 2$,so $[H^+] = 10^{-2} \ M = 0.01 \ M$. Number of moles of $H^+ = M 	imes V(L) = 0.01 	imes 0.2 = 0.002 \ mol$. For $NaOH$ solution: $pH = 12$,so $pOH = 14 - 12 = 2$. $[OH^-] = 10^{-2} \ M = 0.01 \ M$. Number of moles of $OH^- = M 	imes V(L) = 0.01 	imes 0.3 = 0.003 \ mol$. Since $n_{OH^-} > n_{H^+}$,the resulting solution is basic. Remaining moles of $OH^- = 0.003 - 0.002 = 0.001 \ mol$. Total volume = $200 \ mL + 300 \ mL = 500 \ mL = 0.5 \ L$. $[OH^-]_{final} = \frac{0.001 \ mol}{0.5 \ L} = 0.002 \ M = 2 	imes 10^{-3} \ M$. $pOH = -\log(2 	imes 10^{-3}) = 3 - \log 2 = 3 - 0.3 = 2.7$. $pH = 14 - pOH = 14 - 2.7 = 11.3$.

When $200 \ mL$ solution of $HCl$ of $pH=2$ is mixed with $300 \ mL$ solution of $NaOH$ of $pH=12$,the $pH$ of resulting solution is $(\log 2=0.3)$

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