CHEM 122
Expt #9
Analysis of % Acetic Acid in
Vinegar
CCBC-Catonsville
Outline
Part A: Standardization of NaOH soln:
HCl + NaOH  H2O + NaCl
To determine the molarity of the NaOH solution to
be used in Part B.
(3 acceptable trial with very pale pink endpoint and
vol HCl: vol NaOH within 0.04)
Part B: Analysis of Vinegar
HC2H3O2 + NaOH  H2O + NaC2H3O2
Using the average molarity of NaOH from Part A,
determine the percent acetic acid in vinegar
(prefer 3 acceptable trials).
NaOH
Initial
Reading
1.58 mL
Final
Reading
19.89 mL
Part A:
Standardization of NaOH soln:
18.31 A 25.00-mL sample of standard
mL
0.2000 M HCl is titrated with a
NaOH(aq )solution. The total volume
of NaOH solution used to reach the
end point was 18.31 mL. What is the
molarity of the NaOH solution?
HCl (aq) + NaOH (aq) H2O (l) +NaCl
HCl
25.00 mL
0.2000M
(aq)
25.00 mL 18.31 mL
0.2000 M ? M
DO NOT USE M1V1 = M2V2 because
this is NOT a dilution problem.
NaOH
Initial
Reading
1.58 mL
18.31
mL
Final
Reading
19.89 mL
Repeat to get 3 trials with
acceptable endpoints. Find
average molarity of NaOH to be
used in Part B.
 # mol NaOH   ? mol NaOH 
x M NaOH  
=

 # L soln   0.01831 L soln 
 0.005000 mol NaOH 
HCl
=
25.00 mL
 = 0.2731 M NaOH
 0.01831 L soln 
0.2000M
 0.2000 mol HCl   1 mol NaOH 
x mol NaOH = 0.02500 L HCl soln 


 1 L HCl soln   1 mol HCl 
V
x
M
= 0.005000 mol NaOH
NaOH
Part B:
Analysis of Vinegar:
16.87
mL A 5.00-mL sample of vinegar is titrated using
the NaOH solution described in question #2.
Final
Reading It requires 16.87 mL of NaOH to react with all
of the acetic acid (HC2H3O2) in the vinegar
sample. What is the percent by weight of
acetic acid in the vinegar? Assume the density
vinegar of vinegar to be 1.02 g/mL)
5.00 mL
?%
HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2
5.00 mL
16.87 mL
?%
0.2731 M (Av of 3 trials)
Initial
Reading
NaOH
Initial
Reading
HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2
16.87 mL
0.2731M NaOH
Final
Reading
? g aa
% aa in vin by mass 
x 100
? g vin
% aa in vin by mass  =
vinegar
5.00 mL
?%
XXX g aa
x 100 = XXX % aa
5.10 g vin
 1.02 g vin 
x g vin = 5.00 mL vin 
 = 5.10 g vin
 1 mL vin 
 0.2731 mol NaOH  1 mol aa  MM g aa 
x g aa = 0.01687 LNaOH



1 L NaOH

 1 mol NaOH  1 mol aa 
= XXX g aa