A 0.1 mL sample of 0.01 M HCl is added to 1000 L of water at 298ºK. What is the pH of the resulting solution?

Question

Quizplus · Accepted Answer

The concentration of HCl in the final solution is calculated by using the formula $C_1V_1 = C_2V_2$, where $C_1$ and $V_1$ are the concentration and volume of the initial solution, and $C_2$ and $V_2$ are the concentration and volume of the final solution. Given $C_1 = 0.01$ M, $V_1 = 0.1$ mL = $0.1 \times 10^{-3}$ L, and $V_2 = 1000$ L, we find $C_2 = \frac{C_1V_1}{V_2} = \frac{0.01 \times 0.1 \times 10^{-3}}{1000} = 10^{-7}$ M. The pH is calculated using the formula $pH = -\log[H^+]$, where $[H^+]$ is the concentration of hydrogen ions. Substituting $[H^+] = 10^{-7}$ M gives $pH = -\log(10^{-7}) = 7$. However, this calculation assumes no contribution from the water's self-ionization, which is not correct given the context of diluting an acid in such a large volume of water. The correct approach recognizes that the dilution makes the HCl contribution negligible, and the pH is determined primarily by the water itself, which at 25°C (298 K) has a pH of 7.0. The initial calculation misleadingly suggests a neutral pH due to an oversight in considering the significant dilution effect and the autoionization of water.

A 01 ML Sample of 0