1 , mL of 0.1 , M HCl is added to 999 , mL of | vedclass

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(B) The total volume of the resulting solution is $1 \, mL + 999 \, mL = 1000 \, mL = 1 \, L$.The number of moles of $H^+$ ions provided by $HCl$ is $

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The total volume of the resulting solution is $1 \, mL + 999 \, mL = 1000 \, mL = 1 \, L$.The number of moles of $H^+$ ions provided by $HCl$ is $n_{H^+} = M 	imes V(L) = 0.1 \, mol/L 	imes 0.001 \, L = 10^{-4} \, mol$.The concentration of $H^+$ ions in the final solution is $[H^+] = \frac{n_{H^+}}{V_{total}} = \frac{10^{-4} \, mol}{1 \, L} = 10^{-4} \, M$.The $pH$ of the solution is calculated as $pH = -\log[H^+] = -\log(10^{-4}) = 4$.

$1 \, mL$ of $0.1 \, M$ $HCl$ is added to $999 \, mL$ of $0.1 \, N$ $NaCl$ solution. The $pH$ of the resulting solution will be:

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