Quantity Trial 1 Trial 2 Trial 3 1.193 M 1.193 M 1.193 M 0.20 mL 21.70 mL 36.90
ID: 517936 • Letter: Q
Question
Quantity
Trial 1
Trial 2
Trial 3
1.193 M
1.193 M
1.193 M
0.20 mL
21.70 mL
36.90 mL
21.70 mL
36.90 mL
49.90 mL
21.50 mL
15.20 mL
13.00 mL
0.0215 L
0.0152 L
0.0130 L
moles NaOH
0.0256 mol
0.0181 mol
0.0115 mol
moles AA = moles NaOH
0.0256 mol
0.0181 mol
0.0115 mol
0.0150 L
0.0150 L
0.0150 L
Actual molarity of AA
1.706 M
1.2089 M
1.0339 M
Average molarity of AA
1.3162 M
1.3162 M
1.3162 M
Using the information above, calculate
a) mass percent of acetic acid in vinegar
Also, explain why red cabbage acid-base indicator would not work as an indicator for titration.
Explain possible sources of error.
Quantity
Trial 1
Trial 2
Trial 3
M of NaOH (exact concentration)1.193 M
1.193 M
1.193 M
V (initial buret reading)0.20 mL
21.70 mL
36.90 mL
V (final buret reading)21.70 mL
36.90 mL
49.90 mL
V of NaOH added21.50 mL
15.20 mL
13.00 mL
V of NaOH in L0.0215 L
0.0152 L
0.0130 L
moles NaOH
0.0256 mol
0.0181 mol
0.0115 mol
moles AA = moles NaOH
0.0256 mol
0.0181 mol
0.0115 mol
volume of acid (vinegar)0.0150 L
0.0150 L
0.0150 L
Actual molarity of AA
1.706 M
1.2089 M
1.0339 M
Average molarity of AA
1.3162 M
1.3162 M
1.3162 M
Explanation / Answer
Molecular mass of acetic acid is 60g/mole.
Molarity is also given as 1.3162M
Mass = molarity * Volume * Molar mass
= 1.3162 * 0.015 * 60 = 1.185 grams.
We will consider density of vinegar as 1g/ml as it is not given
We have 15 ml vinegar = 15 g vinegar
% acetic acid = grams of AA/ grams of Vinegar * 100
= 1.185/15 * 100 = 7.9 %
Red cabbage indicator has vast range of colors but for quantitative measurements it is not suitable. Also the color only darkens as the pH decreases , we cannot calculate the pH accurately.