When 50 mL of 0.1 M HCl and 50 mL of 0.2 | vedclass

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(C) $1$. Calculate the millimoles of $HCl$ and $NaOH$: $n(HCl) = 50 \ mL 	imes 0.1 \ M = 5 \ mmol$ $n(NaOH) = 50 \ mL 	imes 0.2 \ M = 10 \ mmol$ $2$

Vedclass · Accepted Answer

$12.70$

Vedclass · Answer

(C) $1$. Calculate the millimoles of $HCl$ and $NaOH$: $n(HCl) = 50 \ mL 	imes 0.1 \ M = 5 \ mmol$ $n(NaOH) = 50 \ mL 	imes 0.2 \ M = 10 \ mmol$ $2$. The reaction is $HCl + NaOH ightarrow NaCl + H_2O$. Since $NaOH$ is in excess,the remaining $OH^-$ ions are: $10 \ mmol - 5 \ mmol = 5 \ mmol$. $3$. Total volume of the mixture = $50 \ mL + 50 \ mL = 100 \ mL$. $4$. Concentration of $[OH^-] = \frac{5 \ mmol}{100 \ mL} = 0.05 \ M = 5 	imes 10^{-2} \ M$. $5$. Calculate $pOH$: $pOH = -\log[OH^-] = -\log(5 	imes 10^{-2}) = 2 - \log(5) = 2 - 0.699 = 1.301$. $6$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699 \approx 12.70$.

When $50 \ mL$ of $0.1 \ M$ $HCl$ and $50 \ mL$ of $0.2 \ M$ $NaOH$ solutions are mixed,the $pH$ of the resulting solution will be:

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