This document describes an experiment to determine the percentage of ethanoic acid in a vinegar sample. Vinegar is made from the oxidation of ethanol and contains acetic acid as its main component. The experiment involves titrating a vinegar sample with a standardized NaOH solution using phenolphthalein as an indicator. Multiple titrations are performed and the data is used to calculate the molarity and concentration of acetic acid in the original vinegar sample. The percentage of acetic acid is determined to be 3.63% w/v.

1. Introduction
Vinegar is an acidic liquid, which is made both naturally and synthetically, from the
oxidation of ethanol, CH3CH2OH, in an alcohol-containing liquid such as wine,
fermented fruit juice (e.g. cider) or beer. It also has taste sour. It has been used since
ancient times as an important cooking ingredient, e.g. in salad dressings and on fish and
chips. The key chemical component of vinegar is acetic acid, CH3COOH (systematic
name: ethanoic acid). In this experiment,there are NaOH, potassium hydrogen phthalate
and ethanoic acid. In this experiment also have two reaction, such as below :
Reaction between NaOH and potassium hydrogen phthalate :
NaOH + KHC8H4O4 → KNaC8H4O4 + H2O
Reaction between NaOH and ethanoic acid :
NaOH + CH3COOH → NaCH3COO + H2O
Objective
To determine the percent of ethanoic acid in a vinegar sample.

2. Apparatus
25 ml pipette
Burette
Conical flask
Retort stand
250 ml volumetric flask
Chemical reagents
Standard potassium hydrogen phthalate solution (0.1 M)
Sodium hydroxide
Phenolphthalein indicator
Vinegar
Procedure
(a) Standardization of NaOH solution
1. A burette was filled with the NaOH solution. The burette reading was recorded.
2. 25 ml of the potassium hydrogen phthalate, KHC8H4O4 solution was pipetted into a
conical flask. 2-3 drops of phenolphthalein indicator was added.
3. The KHC8H4O4 was titrated with the NaOH solution until thr colour of the indicator

3. turn to light pink. The burette reading was recorded.
4. The tirtration was repeated 3 times.
(b) Titration of the vinegar
1. 25 ml of the given vinegar was pipetted and dilute it with distilled water in a 250 ml
volumetric flask.
2. 25 ml of the diluted vinegar was pipetted into a conical flask. 2-3 drops of
phenolphthalein was added.
3. The diluted vinegar was titrated with the NaOH solution.
4. The titration was repeated 3 times.
Results
Data
(a) Standardization of NaOH solution

4. (b) Titration of vinegar
Rough 1 2 3
Initial burette
reading (mL)
0.2 14.3 0.8 14.5
Final burette
reading (mL)
14.3 28.2 14.5 28.8
Volume of
NaOH solution
used (mL)
14.1 13.9 13.7 14.3
Rough 1 2 3
Initial burette
reading (mL)
0.0 8.4 16.9 25.4
Final burette
reading (mL)
8.4 16.9 25.4 33.9
Volume of
NaOH solution
used (mL)
8.4 8.5 8.5 8.5

5. Calculations :
1) Calculate the molarity of the NaOH solution.
Average volume of NaOH used
mL97.13
3
3.149.139.13



NaOH + KHC8H4O4 → KNaC8H4O4 + H2O
From the equation ;
1 mol of NaOH Ξ 1 mol of KHC8H4O4
MMb
Mb
Mb
Mb
b
a
MbVb
MaVa
179.0
)(97.135.2
1
1
)97.13(
5.2
1
1
)97.13)((
)25)(1.0(





2) Calculate the molarity of ethanoic acid in the diluted vinegar.
Average volume of NaOH used

6. mL5.8
3
5.85.85.8



NaOH + CH3COOH → NaCH3COO + H2O
From the equation;
1 mol of NaOH Ξ 1 mol of CH3COOH
MMa
Ma
Ma
b
a
MbVb
MaVa
0608.0
52.1)25(
1
1
)5.8)(179.0(
)25)((




3) Calculate the molarity of ethanoic acid in the vinegar simple
M1V1 = M2V2
M1 (25) = (0.0608)(250)
M1 (25) = 15.2
M1 = 0.608 M
4) Calculate the concentration ofethanoic acid (g/L) in the vinegar.

7. )(
L
m o l
M → concentration )(
L
g
Mol → g
Mass = molarity × molar mass
= 0.608 × ( 12+3+12+16+16+1)
= 0.608 × 60
= 36.48 g
Concentration = g
L
= 36.48
1
= 36.48 g/L
5) Calculate the percent (w/v) of ethanoic acid in the vinegar.

8.  
  %6 4 8.3/%
1 0 0
1 0 0 0
48.36
/%
100
1000
)/%(



vw
vw
mass
vw