vinegar solution, and mass % of acetic acid in the vinegar used to titrate 5.00 mL of the vinegar were calculated using the average of three good trials (difference no greater than 0.1 mL). Data&Observations Volume of NAOH used in the titration: Rough trial Trial Trial 2 Trial 3 Initial reading (mL) Final reading (mL) Volume 0.00 0.00 0.00 0.00 25.50 25.90 25.80 26.00 25.50 25.90 25.80 26.00 dispensed (mL) Average volume of NaOH used (Trials 1-3) 25.90 mL Molarity of NaOH = 0.200 M should ze shosn unda CalmeaOi Unknown Vinegar # 18 The actual molarity of the vinegar = 1.00 M The actual mass % of acetic acid 5.95 % A faint pink color was observed at the end point of each titration trial. Calculations& Results Calculations and results pages are attached. Below is a set of caleulation performed in this experiment: Calculation of moles HC,HaO, in 5.00 mL vinegar: L NaOHmol NaOH mol HC2H302 Average mL NaOH 0.200 mol NaOH 1 mol HC2H3 02) 1 L NaOH = 0.00518 mol HC2H302 1 L NaOH 1 mol NaOH 25.90 mL NAOH 1000 mL NAOH 5 Aion of molarity of vinegar: mol HC2H302 -molarity of vinegar 1 0.00518 mol HC2H302 = 1.04 M 0.00500 L vineg ar Calculation of mass % of acetic acid in 5.00 mL of vinegar: mass of acetic acid mass % acetic acid in vineg ar X 100 mass of vinegar 60.05 g HC2H3O2 mass of HC2H3 O2 0.00518 mol HC2H302 = = 0.311 g HC2H302 1 mol HC2H3O2 1.01 g vinegar mass of vinegar = 5.00 mL vineg ar = 5.05 g vineg ar 1.00 mL vineg ar 0.311 g HC2H3 O2 mass % acetic acid in vineg ar X 100 6.16 % 5.05 g vinegar The % error in the calculation of the molarity of the vinegar: lexperimental molarity - actual molarity actual molarity X 100 % error in mo larity of vinegar 1 1.04 M - 1.00 M| X 100 1.00 M = 4% The % error in the calculation of the mass % of acetic acid in vinegar: lexperimental mass %- actual mass % actual mass % x 100 % error in mass % of ace tic acid = 16.16%-5.95%| 5.95% - 100 = 3.5% io Ctu o Lib) elel boau HIOsVlo omu gail ein yaib

Question

How did he find the actual Molarity (1.00M)of vinegar and actual mass % of acetic acid ?( 5.95%)

vinegar solution, and mass % of acetic acid in the
vinegar used to titrate 5.00 mL of the vinegar were calculated using the average of three
good trials (difference no greater than 0.1 mL).
Data&Observations
Volume of NAOH used in the titration:
Rough trial
Trial
Trial 2
Trial 3
Initial reading
(mL)
Final reading
(mL)
Volume
0.00
0.00
0.00
0.00
25.50
25.90
25.80
26.00
25.50
25.90
25.80
26.00
dispensed (mL)
Average volume of NaOH used (Trials 1-3) 25.90 mL
Molarity of NaOH = 0.200 M
should ze shosn unda
CalmeaOi
Unknown Vinegar # 18
The actual molarity of the vinegar = 1.00 M
The actual mass % of acetic acid 5.95 %
A faint pink color was observed at the end point of each titration trial.
Calculations& Results
Calculations and results pages are attached. Below is a set of caleulation performed in this
experiment:
Calculation of moles HC,HaO, in 5.00 mL vinegar:
L NaOHmol NaOH mol HC2H302
Average mL NaOH
0.200 mol NaOH 1 mol HC2H3 02)
1 L NaOH
= 0.00518 mol HC2H302
1 L NaOH
1 mol NaOH
25.90 mL NAOH
1000 mL NAOH
Expand
Transcribed Image Text

vinegar solution, and mass % of acetic acid in the vinegar used to titrate 5.00 mL of the vinegar were calculated using the average of three good trials (difference no greater than 0.1 mL). Data&Observations Volume of NAOH used in the titration: Rough trial Trial Trial 2 Trial 3 Initial reading (mL) Final reading (mL) Volume 0.00 0.00 0.00 0.00 25.50 25.90 25.80 26.00 25.50 25.90 25.80 26.00 dispensed (mL) Average volume of NaOH used (Trials 1-3) 25.90 mL Molarity of NaOH = 0.200 M should ze shosn unda CalmeaOi Unknown Vinegar # 18 The actual molarity of the vinegar = 1.00 M The actual mass % of acetic acid 5.95 % A faint pink color was observed at the end point of each titration trial. Calculations& Results Calculations and results pages are attached. Below is a set of caleulation performed in this experiment: Calculation of moles HC,HaO, in 5.00 mL vinegar: L NaOHmol NaOH mol HC2H302 Average mL NaOH 0.200 mol NaOH 1 mol HC2H3 02) 1 L NaOH = 0.00518 mol HC2H302 1 L NaOH 1 mol NaOH 25.90 mL NAOH 1000 mL NAOH

5
Aion of molarity of vinegar:
mol HC2H302
-molarity of vinegar
1
0.00518 mol HC2H302
= 1.04 M
0.00500 L vineg ar
Calculation of mass % of acetic acid in 5.00 mL of vinegar:
mass of acetic acid
mass % acetic acid in vineg ar
X 100
mass of vinegar
60.05 g HC2H3O2
mass of HC2H3 O2
0.00518 mol HC2H302
=
= 0.311 g HC2H302
1 mol HC2H3O2
1.01 g vinegar
mass of vinegar
= 5.00 mL vineg ar
= 5.05 g vineg ar
1.00 mL vineg ar
0.311 g HC2H3 O2
mass % acetic acid in vineg ar
X 100 6.16 %
5.05 g vinegar
The % error in the calculation of the molarity of the vinegar:
lexperimental molarity - actual molarity
actual molarity
X 100
% error in mo larity of vinegar
1
1.04 M - 1.00 M|
X 100
1.00 M
= 4%
The % error in the calculation of the mass % of acetic acid in vinegar:
lexperimental mass %- actual mass %
actual mass %
x 100
% error in mass % of ace tic acid =
16.16%-5.95%|
5.95%
- 100
= 3.5%
io
Ctu o
Lib) elel
boau HIOsVlo omu
gail ein
yaib
Expand
Transcribed Image Text

5 Aion of molarity of vinegar: mol HC2H302 -molarity of vinegar 1 0.00518 mol HC2H302 = 1.04 M 0.00500 L vineg ar Calculation of mass % of acetic acid in 5.00 mL of vinegar: mass of acetic acid mass % acetic acid in vineg ar X 100 mass of vinegar 60.05 g HC2H3O2 mass of HC2H3 O2 0.00518 mol HC2H302 = = 0.311 g HC2H302 1 mol HC2H3O2 1.01 g vinegar mass of vinegar = 5.00 mL vineg ar = 5.05 g vineg ar 1.00 mL vineg ar 0.311 g HC2H3 O2 mass % acetic acid in vineg ar X 100 6.16 % 5.05 g vinegar The % error in the calculation of the molarity of the vinegar: lexperimental molarity - actual molarity actual molarity X 100 % error in mo larity of vinegar 1 1.04 M - 1.00 M| X 100 1.00 M = 4% The % error in the calculation of the mass % of acetic acid in vinegar: lexperimental mass %- actual mass % actual mass % x 100 % error in mass % of ace tic acid = 16.16%-5.95%| 5.95% - 100 = 3.5% io Ctu o Lib) elel boau HIOsVlo omu gail ein yaib

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