Experiment 29: Choice I
Acid-Base Titration
Goals: (1) To standardize a newly prepared solution of NaOH. Standardization is the process of
determining the exact concentration of the solution. This is the case where concentrations are
reported in at least 3 significant figures.
(2) To determine the concentration of commercial vinegar.
Calculations: (1) Standardization
Equation 1: (MV)acid
=
(MV)base
Since:
Mole of acid =
mole of base, therefore:
Equation 2: moles of acid = Molarity of NaOH x Vol. of NaOH in L
Example: If 0.600 grams of KHP were used, then moles of KHP would be:
mole KHP =
0.600 grams
=
0.00294 mole
204.2 gms/mole
Then: 0.0029 mole = (MV)base
So:
Molarity of NaOH = 0.00294 mole / volume of NaOH used in L
(2) Analysis of Unknown Acid: Commercial Vinegar
Given: CH3COOH + NaOH
Acetic Acid

CH3COO- Na+
+ H2O
To calculate for % Mass of Acetic Acid in Vinegar, do the following:
Since ratio of acid to base is 1:1, then
(MV)acid = (MV)base therefore,
A. To find moles of acetic acid in vinegar:
molesAcetic Acid = Mean Molarity of vinegar x Vol. of vinegar used in L
B. To Convert Molarity of Vinegar to grams
If: Molarity = [grams / molar mass] / Liter of solution
Then:
x grams Acetic Acid = MolarityAcetic Acid x Liter of soln. x Molar MassCH3COOH
Note: Molar Mass of CH3COOH (Acetic Acid) = 60.05 grams /mole
C. To find mass of vinegar
Given that the density of vinegar is 1.01 gm/mL
If Density = mass/volume, then
Mass of 5.00 grams of vinegar = Density x 5.00 mL = 5.05 grams
D. Calculate % Mass of Acetic Acid in vinegar
% by Mass of CH3COOH = Mass of CH3COOH
mass of vinegar
x 100