The pH of a solution obtained by mixing 10 m | vedclass

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(C) Millimoles of $H^{+}$ from $HCl = 0.1 	imes 10 = 1 \ mmol$. Millimoles of $H^{+}$ from $H_2SO_4 = 0.2 	imes 40 	imes 2 = 16 \ mmol$. Total mill

Vedclass · Accepted Answer

$0.470$

Vedclass · Answer

(C) Millimoles of $H^{+}$ from $HCl = 0.1 	imes 10 = 1 \ mmol$. Millimoles of $H^{+}$ from $H_2SO_4 = 0.2 	imes 40 	imes 2 = 16 \ mmol$. Total millimoles of $H^{+} = 1 + 16 = 17 \ mmol$. Total volume of solution = $10 \ mL + 40 \ mL = 50 \ mL$. Concentration of $H^{+} = \frac{17 \ mmol}{50 \ mL} = 0.34 \ M$. $pH = -\log [H^{+}] = -\log (0.34) = -\log (\frac{34}{100}) = -\log (\frac{17}{50}) = -(\log 17 - \log 50) = -(\log 17 - (\log 5 + \log 10)) = -(1.23 - (0.7 + 1)) = -(1.23 - 1.7) = 0.47$.

The $pH$ of a solution obtained by mixing $10 \ mL$ of $0.1 \ M$ $HCl$ and $40 \ mL$ of $0.2 \ M$ $H_2SO_4$ is (Given $\log 17 = 1.23, \log 5 = 0.7, \log 10 = 1$)

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