Question

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.

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

If you liked the answer, please rate it positively, you would help me a lot, thanks.