200 mL of 0.01 M HCl is mixed with 400 mL | vedclass

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(B) The total moles of $H^{+}$ ions from $HCl$ is $n_{1} = M 	imes V = 0.01 	imes 0.2 = 0.002 \ mol$.The total moles of $H^{+}$ ions from $H_{2}SO_{

Vedclass · Accepted Answer

$1.78$

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(B) The total moles of $H^{+}$ ions from $HCl$ is $n_{1} = M 	imes V = 0.01 	imes 0.2 = 0.002 \ mol$.The total moles of $H^{+}$ ions from $H_{2}SO_{4}$ is $n_{2} = M 	imes V 	imes 2 = 0.01 	imes 0.4 	imes 2 = 0.008 \ mol$.Total moles of $H^{+} = 0.002 + 0.008 = 0.01 \ mol$.Total volume of the mixture = $200 \ mL + 400 \ mL = 600 \ mL = 0.6 \ L$.Concentration of $[H^{+}] = \frac{0.01 \ mol}{0.6 \ L} = \frac{1}{60} \ M$.$pH = -\log[H^{+}] = -\log(\frac{1}{60}) = \log(60) = \log(6 	imes 10) = \log(6) + \log(10) \approx 0.778 + 1 = 1.778 \approx 1.78$.

$200 \ mL$ of $0.01 \ M \ HCl$ is mixed with $400 \ mL$ of $0.01 \ M \ H_{2}SO_{4}$. The $pH$ of the mixture is

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