1. Phenolphthalein indicator was chosen in the neutralisation titration involving acetic acid and
sodium hydroxide. Explain why?
Evidence of the reaction between acetic acid sodium hydroxide isn’t visible, so we will add an
indicator to determine the equivalence point for the titration. A common acid-base indicator is
phenolphthalein. Phenolphthalein is colorless in acidic solution, but changes to pink when the
solution becomes basic. At the equivalence point, the moles of acetic acid are equal to the moles of
sodium hydroxide, and the solution will turn pink as more sodium hydroxide is added.
Phenolphthalein is itself a weak acid. So when enough sodium hydroxide has been added to the
acetic acid, such that there is no longer any unreacted acetic acid, the sodium hydroxide will then
begin reacting with phenolphthalein. The reaction of phenolphthalein with sodium hydroxide results
in a pink solution.
2. Write the chemical equation for the neutralisation of acetic acid with sodium hydroxide?
Stoichiometrically, states the mention chemical reaction in sentences.
Only one of the hydrogen atoms of the acetic acid molecule is acidic:
Equation 1. HC2H3O2 (aq)  H + (aq) + C2H3O2 – (aq)
The hydrogen attached to the oxygen atom is acidic while the other hydrogen atoms are not.
Another way of representing the acidity of acetic acid is to show its reaction with water:
Equation 2. HC2H3O2(aq) + H2O(l)  H3O + (aq) + C2H3O2 – (aq)
Here the acetic acid protonates (transfers a proton to) the water molecule. In fact, the hydrogen ion
(H+ ) is very reactive and doesn't exist in water. However, there is evidence that the hydronium ion
does exist. Sometimes equation 1 is used because of its simplicity. To determine the amount of
acetic acid in vinegar (typically 4-5% by mass) we will use an acid-base titration (neutralization
reaction). In this experiment we titrate acetic acid with sodium hydroxide (a strong base).
The reaction of acetic acid with sodium hydroxide is shown below:
Equation 3. HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2(aq) + H2O (l)
3. How to calculate the molarity of the commercial acetic acid in vinegar, if only the concentration of
sodium hydroxide solution and its volume are known?
Moles of base can be calculated from molarity times volume
(molesbase = Mbase × Vbase).
As a reminder on concentration units, molarity is defined as the number of moles of solute in a liter of solution
(M = mol/L). This is numerically equal to the number of millimoles of solute in a milliliter of solution (M =
mmol/mL). It is often convenient to use this second definition of molarity in titrations and other work where small
quantities are involved. There are 1000 mmol in 1 mol and 1000 mL in 1 liter.For example, 10.2 mL of
0.100 M NaOH solution contains 1.02 mmol of NaOH.
4. What is the present mass acetic aid in vinegar? Does it comply with the concentration stated in
the Malaysia Food Act and Regulations?
v/v % =volume of acetic acid x 100%
volume of vinegar
In this experiment, a carefully measured volume of vinegar (Vacid) is placed into a flask and the mass
determined. The sample of vinegar is then titrated with a NaOH solution of known concentration (Mbase), and
the volume of NaOH solution required to reach the endpoint (Vbase) is determined. Vbase, Mbase, and Vacid are all
known, so the concentration of the acid (Macid) can be determined as described above. In addition, the mass of
acetic acid in the sample can be determined from the number of moles present and the molar mass of acetic
acid (gacid = MWacid × molesacid). Finally, the mass percent of acetic acid in the vinegar can be determined from
the mass of the acetic acid in the sample and mass of the vinegar solution that was titrated and it comply with
the concentration stated in the Malaysia Food Act and Regulations.
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2(aq) + H2O (l)
4% = (mass of acetic acid / 20 ml ) x 100%
= 0.8 g
CH3COOH = 12 + 3 + 12 + 16 + 16 + 1 =60
Molarity = (mass / molar mass)
= (0.8/60)
m1v1 = m2v2
(20)(0.665) = (0.1)v2
v2 = 131 ml
= 0.01333 g/mol
n = (MV/1000)
0.01333 x (20/1000) = 0.665 mol
The titration process in Question 1 needs a lot of NaOH to neutralise acetic acid.
Hence, we should use strong acid such as HCl to conduct this experiment. This is
because weak acid needs a lot of NaOH to neutralise compare to strong acid.
(32.87)(0.114) =(5)(m2)
M2 = 0.74943636 mol / dm 3
n = MV
= 0.75 X ( 5 / 1000) = 0.00375 g/mol
CH3COOH = 12 + 3 + 12 + 16 + 16 + 1 =60
60 x 0.00375 = 0.225g
Assume ml equal to g
(0.225 / 5) x 100% = 4.5 %
(a) molarity of acetic acid in commercial vinegar
M1V1 = M2V2
M1(20) = (0.144) (26.30)
3
M1 = 0.18936 mol/ dm
(b) the percent mass of acetic acid in the commercial vinegar
CH3COOH = 12 + 3 + 12 + 16 + 16 + 1 =60
0.18936 x (20/1000) = 0.0037872
60 x 0.0037872 = 0.227232 g
Assume ml equal to g
(0.227232 / 20) X 100 % = 1.13616%
Evidence of the reaction between acetic acid sodium hydroxide isn’t visible, so we will add an
indicator to determine the equivalence point for the titration. A common acid-base indicator is
phenolphthalein. Phenolphthalein is colorless in acidic solution, but changes to pink when the
solution becomes basic. At the equivalence point, the moles of acetic acid are equal to the moles of
sodium hydroxide, and the solution will turn pink as more sodium hydroxide is added.
Phenolphthalein is itself a weak acid. So when enough sodium hydroxide has been added to the
acetic acid, such that there is no longer any unreacted acetic acid, the sodium hydroxide will then
begin reacting with phenolphthalein. The reaction of phenolphthalein with sodium hydroxide results
in a pink solution.