When 10 mL of 0.1 M HCl and 40 mL of 0.2 | vedclass

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(D) $1$. Calculate the total moles of $H^+$ ions from $HCl$: $n(H^+) = M 	imes V 	imes 	ext{basicity} = 0.1 	imes 0.010 	imes 1 = 0.001 \ mol$. $

Vedclass · Accepted Answer

$0.46$

Vedclass · Answer

(D) $1$. Calculate the total moles of $H^+$ ions from $HCl$: $n(H^+) = M 	imes V 	imes 	ext{basicity} = 0.1 	imes 0.010 	imes 1 = 0.001 \ mol$. $2$. Calculate the total moles of $H^+$ ions from $H_2SO_4$: $n(H^+) = M 	imes V 	imes 	ext{basicity} = 0.2 	imes 0.040 	imes 2 = 0.016 \ mol$. $3$. Total moles of $H^+$ = $0.001 + 0.016 = 0.017 \ mol$. $4$. Total volume of the mixture = $10 \ mL + 40 \ mL = 50 \ mL = 0.050 \ L$. $5$. Concentration of $[H^+] = \frac{	ext{Total moles}}{	ext{Total volume}} = \frac{0.017}{0.050} = 0.34 \ M$. $6$. $pH = -\log[H^+] = -\log(0.34) \approx 0.468 \approx 0.47$.

When $10 \ mL$ of $0.1 \ M \ HCl$ and $40 \ mL$ of $0.2 \ M \ H_2SO_4$ solutions are mixed,the $pH$ of the resulting mixture will be:

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