Lab 8 was an experiment on acid-base titration. In this experiment you titrated
ID: 483647 • Letter: L
Question
Lab 8 was an experiment on acid-base titration. In this experiment you titrated HCI with NaOH, an example of a strong acid titrated with a strong base, and compared it to a titration of acetic acid and NaOH, an example of a weak acid titrated by a strong base. The plot of the pH of an acid solution versus the volume of strong base produces what is known as a titration curve. In this extra credit assignment you are asked to calculate the pH for a 10.0 mL sample of 0.10 M HCI titrated in 1.00 mL increments with 0.10 M NaOH. Below is a table for you to record the pH of an HCl solution throughout a titration. On the following pages are two approaches to determine the pH; the first method is typically taught and the second method is an alternative. On the last page you will be asked to answer a few survey questions to provide feedback about your impressions of the two methods. Your feedback will not be personally identifiable but will be extremely informative for we who teach acid-base titration. Please read through both methods before picking which one to use.Explanation / Answer
Initial pH when no NaOH is added
[H+] = concentration of HCl
= 0.1 M
pH = -log [H+]
= -log 0.1 = 1
pH when 1 mL NaOH is added
Total volume of the system = 11 mL
number of moles of H+ = concentration of HCl X volume of HCl
= 0.1 M X 10 X 10-3 L = 1 X 10-3 mol
number of moles of OH- = concentration of NaOH X volume of NaOH
= 0.1 M X 1 X 10-3 L = 1 X 10-4 mol
Excess number of moles of H+ = number of moles of H+ - number of moles of OH-
= 1 X 10-3 mol - 1 X 10-4 mol
= 9 X 10-4 mol
[H+] = Excess number of moles of H+ / Total volume of the system
= 9 X 10-4 mol / 11 X 10-3 L
= 0.082 M
pH = -log [H+]
= -log 0.082 = 1.09
pH when 2 mL NaOH is added
Total volume of the system = 12 mL
number of moles of H+ = concentration of HCl X volume of HCl
= 0.1 M X 10 X 10-3 L = 1 X 10-3 mol
number of moles of OH- = concentration of NaOH X volume of NaOH
= 0.1 M X 2 X 10-3 L = 2 X 10-4 mol
Excess number of moles of H+ = number of moles of H+ - number of moles of OH-
= 1 X 10-3 mol - 2 X 10-4 mol
= 8 X 10-4 mol
[H+] = Excess number of moles of H+ / Total volume of the system
= 8 X 10-4 mol / 12 X 10-3 L
= 0.067 M
pH = -log [H+]
= -log 0.067 = 1.18
We can calculate pH of the system in similar manner up until the equivalence point.
At equivalence point: pH when 10 mL NaOH is added
Total volume of the system = 20 mL
number of moles of H+ = concentration of HCl X volume of HCl
= 0.1 M X 10 X 10-3 L = 1 X 10-3 mol
number of moles of OH- = concentration of NaOH X volume of NaOH
= 0.1 M X 10 X 10-3 L = 1 X 10-3 mol
number of moles of H+ = number of moles of OH-
Therefore [H+] = [OH-]
And pH = pOH
We know, pH + pOH = 14
Therefore pH = 7
Beyond equivalence point: pH when 11 mL NaOH is added
Total volume of the system = 21 mL
number of moles of H+ = concentration of HCl X volume of HCl
= 0.1 M X 10 X 10-3 L = 1 X 10-3 mol
number of moles of OH- = concentration of NaOH X volume of NaOH
= 0.1 M X 11 X 10-3 L = 1.1 X 10-3 mol
Excess number of moles of OH- = number of moles of OH- - number of moles of H+
= 1.1 X 10-3 mol - 1 X 10-3 mol
= 1 X 10-4 mol
[OH-] = Excess number of moles of OH- / Total volume of the system
= 1 X 10-4 mol / 21 X 10-3 L
= 0.0048 M
pOH = -log [OH-]
= -log 0.0048 = 2.32
pH = 14 – pOH = 14 - 2.32 = 11.68
We can calculate pH of the system in similar manner beyond the equivalence point.
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