Question

please find the following and show work so i have a better understanding: - volume of...

please find the following and show work so i have a better understanding:

- volume of NaOH used

- moles of NaOH used

- moles of HC2H3O2

- mass of HC2H3O2

- mass percent of vinegar solution

- molarity of HC2H3O2

- deviation

- percent error

- average mass percent of vinegar

- average molarity of vinegar

Table 2: Titration of standard NaOH with a HC2H3O2 Solution

Trail 1

Trial 2

Trial 3

M of NaOH

0.4936 M

0.4936 M

0.4936 M

Volume of Vinegar Solution

25.00 mL

25.00 mL

25.00 mL

Mass of Erlenmeyer flask:

139.347 g

139.347 g

139.347 g

Mass of Erlenmeyer flask plus vinegar solution:

159.428 g

159.428 g

159.428 g

Mass of Vinegar solution

20.08 g

20.08 g

20.08 g

Initial Burette Reading:

50.00 mL

50.00 mL

50.00 mL

Final Burette Reading:

44.25 mL

44.40 mL

44.30 mL

Volume of NaOH used:

Moles of NaOH added:

Moles of HC2H3O2

Mass of HC2H3O2

Mass percent of Vinegar solution

Molarity of HC2H3O2

Deviation

Percent error

- how did the mass percent you found compare to the manufacturer claim. What can we conclude about the manufacturers claim? Explain by supporting with experimental evidence.

Homework Answers

Answer #1

The volume of NaOH used is calculated:

V NaOH = Vf - Vi = 50 - 44.25 = 5.75 mL

Trial 2: 5.60 mL

Trial 3: 5.70 mL

The moles of NaOH are calculated:

n NaOH = M * V = 0.4936 M * 0.00575 L = 2.84e-3 mol

Trial 2: 2.76e-3 mol

Trial 3: 2.81e-3 mol

The moles of acetic acid and those of NaOH are the same.

The mass of acetic acid is calculated:

m AA = n * MM = 2.84e-3 * 60.05 = 0.0171 g

Trial 2: 0.0166 g

Trial 3: 0.0169 g

The percentage is calculated:

% AA = g AA * 100 / g solution = 0.0171 g * 100 / 20.08 g = 0.085%

Trial 2: 0.083%

Trial 3: 0.084%

Average = 0.084%

The molarity is calculated:

M = n / V = ​​2.84e-3 mol / 0.025 L = 0.1136 M

Trial 2: 0.1104 M

Trial 3: 0.1124 M

Average = 0.1121 M

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