In Part C of the experiment, a student begins with a beaker containing 10.0 mL o

Question

In Part C of the experiment, a student begins with a beaker containing 10.0 mL of 0.10 M HCl solution and adds different volumes of 0.10 M NaOH.

mL NaOH mmol OH ? added pH concentration 0.0 1.00 [H3O+?] >> [OH??][H3O+?] ? [OH??]    [H3O+?] << [OH??] 3.0 1.27 [H3O+?] >> [OH??][H3O+?] ? [OH??]    [H3O+?] << [OH??] 6.0 1.60 [H3O+?] >> [OH??][H3O+?] ? [OH??]    [H3O+?] << [OH??] 12.0 11.96 [H3O+?] >> [OH??][H3O+?] ? [OH??]    [H3O+?] << [OH??] In Part C of the experiment, a student begins with a beaker containing 10.0 mL of 0.10 M HCl solution and adds different volumes of 0.10 M NaOH. How many millimoles of acid are initially in the beaker? (If in doubt, look back at your calculation from experiment 9). Fill in the table below with mmol of NaOH added and the relative concentrations of H3O+ and OH ? in each solution. What would the approximate pH of the solution be after the addition of 10.0 mL of 0.10 M NaOH?

Explanation / Answer

1. mmoles is just the concentration (0.10 M) multiplied by the amount of solution (10 mL). Multiply these together and you should get 1.

2. This one is easier than it seems,
Multiply each amount of added ( 0mL, 3 mL, 6 mL etc..) by the concentration of NaOH (0.1 M).
ex. (5 mL NaOH) x (0.1 M) = 0.5 mmoles of OH-. The pH isn't factored into the calculation of mmoles.

(c) It is based on the definition of pH which is defined as -log(10) [H+].

The concentration of hydrogen ions (protons) x the concentration of hydroxide ions is always 10^-14 M^2. In this particular example, we have an equal concentration of H+ and OH-. So the concentration of each of these must be 10^-7M. So the approximate pH is -log [10^-7]. The logarithm of 10^-7 is -7, the minus on the front converts it back to a positive number. So the expected pH is 7.

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