Titration Analysis

Topics: Ethanol, Laboratory glassware, Titration Pages: 3 (505 words) Published: January 21, 2011
Investigation 8.4- Titration Analysis of Vinegar

Prediction: We predict that the amount concentration of acetic acid in a sample of vinegar is 0.83 mol/L.

VCH3COOH (aq) = 10.00 mL
VNaOH (aq) = ?
CCH3COOH = 0.83 mol/L
CNaOH (aq) = 0.145 mol/L

Materials:

Sodium hydroxide 100 mL
Acetic acid 50 mL
250 mL beaker x 2
Distilled water x 1 bottle
Erlenmeyer flask x 1
Burette x 1
Pipette x 1
Pipette bulb x 1
Volumetric pipette x 1
Phenolphalein

Procedure:

Step 1- Gather materials, put on eyeglasses and apron.
Step 2- Pour 50 mL of acetic acid into one 250 beaker and 100 mL of sodium hydroxide into another 250 mL beaker. Step 3- Clean burette, volumetric pipette and pipette bulb with distilled water. Step 4- Use a funnel or just the beaker to pour the sodium hydroxide (not above head) into the volumetric pipette. Use the pipette and the pipette bulb to gather 10 mL of acetic acid and empty into Erlenmeyer flask. Add a few drops of phenolphalein to the flask. Let 9 mL of the sodium hydroxide drain into the flask, placed underneath the volumetric pipette until it turns pink. Step 5- Record results. Repeat step four to determine more results. Step 6- Clean up workspace, put away materials and safety tools.

Evidence:



Qualitative/Quantitative: The acetic acid + phenolphalein changed pink quicker, and turned a darker shade of pink when more sodium hydroxide was added to it. For each trial, we added at least 9 mL of sodium hydroxide to the acetic acid + phenolphalein.

Analysis:

CH3COOH (aq) + NaOH (aq) à H2O (l) + NaCH3COO (aq)
V = 10.00 mL V = 11.00 mL
C =? C = 0.145 mol/L

According to the evidence obtained, we concluded that the amount concentration of acetic acid in a sample of vinegar is 0.1595 mol/L.

0.1595 x 5 = 0.80

Evaluation:

% difference: 0.80-0.83 = 0.036 x 100 = 3.61 %
0.83

% yield: 0.80 = 0.9638 x 100 = 96.4 %
0.83

The...
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