1 L of 0.02 M aqueous HCl is mixed with 1 | vedclass

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(A) Moles of $H^{+}$ from $HCl = 1 \ L 	imes 0.02 \ M = 0.02 \ mol$. Moles of $H^{+}$ from $H_2SO_4 = 1 \ L 	imes 0.01 \ M 	imes 2 = 0.02 \ mol$. T

Vedclass · Accepted Answer

$1.7$

Vedclass · Answer

(A) Moles of $H^{+}$ from $HCl = 1 \ L 	imes 0.02 \ M = 0.02 \ mol$. Moles of $H^{+}$ from $H_2SO_4 = 1 \ L 	imes 0.01 \ M 	imes 2 = 0.02 \ mol$. Total moles of $H^{+} = 0.02 + 0.02 = 0.04 \ mol$. Total volume of the mixture $= 1 \ L + 1 \ L = 2 \ L$. Concentration of $[H^{+}] = \frac{0.04 \ mol}{2 \ L} = 0.02 \ M$. $pH = -\log_{10}[H^{+}] = -\log_{10}(0.02) = -\log_{10}(2 	imes 10^{-2})$. $pH = -(\log_{10} 2 + \log_{10} 10^{-2}) = -(0.3 - 2) = 1.7$.

$1 \ L$ of $0.02 \ M$ aqueous $HCl$ is mixed with $1 \ L$ of $0.01 \ M$ aqueous $H_2SO_4$ solution. Assuming complete dissociation and no change in the volume upon mixing,the $pH$ of the resultant solution is $(\log_{10} 2 = 0.3)$.

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