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This experiment is done over a two-week period. There are two separate lab reports, however, and you will receive two separate grades. The first lab report is on the Standardization of NaOH (Page169 of the lab manual; see separate help file), whereas the second lab report is on the Equivalent Weight of an Unknown (Page 175 of the lab manual). To make things easier, in thediscussion to follow, we will do each of these lab reports separately.
- Click here to go to the first part of this experiment (Standardization of NaOH)
In last week's experiment, you prepared and standardized (determined the concentration of) a sodium hydroxide solution. In this week's experiment, you used your sodium hydroxide solution to titrate samples of an unknown acid sample to determine the equivalent weight of the acid.
The equivalent weight of an acid is the mass of acid required to generate one mole of hydrogen (hydronium) ion in aqueous solution. For an acid like HCl, which has only one ionizable hydrogen, the equivalent weight is the same as the molar mass. For an acid that contains more than one ionizable hydrogen, the equivalent weight is a fraction of the molar mass.
Consider the following table, which compares the molar masses of some common acids with their equivalent weights:
Acid |
Molar Mass |
Equivalent Weight |
HCl |
36.5 g |
36.5 g |
H2SO4 |
98.0 g |
98.0g/2 = 49.0 g |
H3PO4 |
98.0 g |
98.0g/3 = 32.7 g |
You should have transferred the average molarity of your NaOH solution from last week's report to this page. If you forgot to do this, you'll have to find your instructor to get the information from him or her.
Average molarity of NaOH (from last week) |
0.1104 M |
Unknown acid code # |
this was on vial you were given |
C. Mass of unknown acid samples
Mass of beaker + acid |
15.2215 g |
Mass of beaker + acid after 1st sample |
14.6712 g |
Mass of beaker + acid after 2nd sample |
14.0700 g |
Mass of beaker + acid after 3rd sample |
13.4718 g |
Mass of beaker + acid after 4th sample |
12.9099 g |
D. Buret readings
Initial Reading |
Final Reading |
Volume NaOH |
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Sample 1 |
1.27 mL |
26.34 mL |
25.07 mL |
Sample 2 |
1.58 mL |
28.99 mL |
27.41 mL |
Sample 3 |
3.75 mL |
30.59 mL |
26.84 mL |
Sample 4 |
0.73 mL |
26.33 mL |
25.60 mL |
Page 175, Part II
A. Moles of NaOH reacted with Sample #1
The number of moles of NaOH used from the buret to titrate Sample #1 is determined from the volume of NaOH used to titrate the sample and the molarity of the NaOH as determined in last week's lab. For Sample #1 above, we used 25.07 mL of 0.1104 M NaOH to reach the endpoint:
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B. Mass of unknown acid reacted, Sample #1
The mass of sample #1 is calculated by difference from the Part C data above
Mass of Sample #1 = 15.2215 g - 14.6712 g = 0.5503 g |
C. Equivalent Weight of Unknown Acid, based on Sample #1
In Part A of the calculations above, we determined that we used 0.002768 moles of NaOH to titrate Sample #1. Since OH- and H+ ions react on a 1:1 stoichiometric basis, this means that Sample #1 must have produced 0.002768 moles of H+ ion. Since the equivalent weight of an acid is the number of grams of the acid required to produced 1 mole of H+ ion, the equivalent weight of the unknown acid (based on Sample #1) is given by
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D. Summary and Results
In this table, you just summarize the calculations for Sample #1 (already done above) and the other samples. The remainder of the calculations for this lab should be a "piece of cake" for you by now! If you need review on calculation of averages, deviations, and percent deviation, see the help files for the earlier experiments you have done.
