0.1 (M) HCl and 0.1 (M) H_2SO_4 each of v | vedclass

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(B) Step $1$: Calculate the total millimoles of $H^{+}$ ions. Millimoles of $H^{+}$ from $HCl = 0.1 \ (M) 	imes 2 \ mL 	imes 1 = 0.2 \ mmol$. Millim

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(B) Step $1$: Calculate the total millimoles of $H^{+}$ ions. Millimoles of $H^{+}$ from $HCl = 0.1 \ (M) 	imes 2 \ mL 	imes 1 = 0.2 \ mmol$. Millimoles of $H^{+}$ from $H_2SO_4 = 0.1 \ (M) 	imes 2 \ mL 	imes 2 = 0.4 \ mmol$. Total millimoles of $H^{+} = 0.2 + 0.4 = 0.6 \ mmol$. Step $2$: Calculate the concentration of $H^{+}$ in the final mixture. Total volume $= 2 \ mL + 2 \ mL + 2 \ mL = 6 \ mL$. $[H^{+}] = \frac{	ext{Total millimoles}}{	ext{Total volume in } mL} = \frac{0.6}{6} = 0.1 \ (M)$. Step $3$: Calculate the $pH$. $pH = -\log_{10} [H^{+}] = -\log_{10} (0.1) = 1$.

$0.1 \ (M) \ HCl$ and $0.1 \ (M) \ H_2SO_4$ each of volume $2 \ mL$ are mixed and the volume is made up to $6 \ mL$ by adding $2 \ mL$ of $0.01 \ (N) \ NaCl$ solution. The $pH$ of the resulting mixture is

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