To determine the concentration of acetic acid in an unknown sample of vinegar by titrating a known volume ofthe acid solution against a standardized basic solution.

Part B: Deterimination of acetic acid content of vinegar

1. A pipette was used to place exactly 25.0 cm3 of vinegar into a 250 cm3 volumetric flask.
2. The volumetric flask was filled to the graduation mark with deionized water.
3. The volumetric flask was stopped. The flask was turned upside down in order to mix the solution well.

4. 25cm of diluted vinegar solution from the volumetric flask was transferred into a conical flask using a pipette and pipette filler.
5. A few drop of phenolphthalein were added.
6. A burette with the sodium hydroxide solution was filled.
7. The sodium hydroxide solution was run from the burette into the conicial flask until the colour of the indicator changes from colourless to pink.
8. The titration was carried out 2-3 times to obtain an average value.

Result:

The solution will change from colourless to pink when it is neutralized.

Solution NaOH versus solution KHC8H4O4

Solution NaOH versus diluted vinegar

Calculation:

The volumes of NaOH solution required for neutralization are: 25.55 cm3, 25.30 cm3, 25.50 cm3.

Therefore the average volume of 0.0500 M sodium carbonate required for neutralization:

= 25.55 + 25.30 +25.50

3.       cm3

=25.45 cm3

No. of moles of 250 cm3 KHC8H4O4 in volumetric flask        

=   Mass  

 Molar Mass

=  5.010  

   204.23  mol

=0.025 mol

No of mole of 25 cm3 of KHC8H4O4

=0.025 ×  25  

          250

=0.0025mol

According to the equation,

1 mole of NaOH required 1 mole of KHC8H4O4 for complete neutralization.

∴No of moles of NaOH= 0.0025mol

Molarity of NaOH solution

= No. of moles of NaOH    

   Volume of solution

= 0.0025mol

 0.02545dm3

=0.098M

The volumes of 0.098M NaOH solution required for neutralization are: 24.50 cm3, 22.65 cm3.

Therefore the average volume of 0.098M sodium hydroxide required for neutralization:

= 24.50 + 22.65

2.    cm3

= 23.58 cm3

NaOH (aq) + CH3COOH(aq)CH3COONa (aq) + H2O(l)

 0.098M        ?M

 23.58 cm3      25 cm3

No. of moles of NaOH in 23.58 cm3

=  Molarity of solution ×Volume of solution

=  0.098 mol dm3 ×0.02358 dm3 

=  0.0023mol

According to the equation, 1 mole of NaOH required 1 mole of CH3COOH for complete neutralization.

∴No. of moles of CH3COOH in 25cm3 solution =0.0023mol

∴No. of moles of CH3COOH in 250cm3 solution =0.0024 × 250

                                                   25                                      

                                       =0.024mol

Concentration of CH3COOH solution

=Number of moles of CH3COOH  

       Volume of solution

= 0.024mol  

  0.025dm3

=0.96mol dm3

Question:

1. Why is the burette usually not allowed to be filled with alkalis?

Ans: It is because some of the alkali e.g. NaOH will absorb carbon dioxide in the air to form carbonate. The carbonate will block the mouth of the burette and deposited on the stop cork, which cannot be washed away easily. This will affect the reading of the burette. Also, all alkali are corrosive, it will attack the glass of the burette, as a result the volume cannot be measured accurately.

2. What is a primary standard? Why can potassium hydrogen phthalate be used as a primary standard in this experiment?

Ans: A primary standard is a compound that serves as a reference material in a titrimetric analysis. It is usually a reagent that has known concentration and it is also extremely pure, stable, has no waters of hydration, and has a high molecular weight to prevent weighing error.

Since solid sodium hydroxide is hygroscopic. Pellets of NaOH exposed to air will increase in mass as they become hydrated so the actual mass of pure NaOH is not accurately known. Sodium hydroxide in solution reacts with carbonic acid and its concentration decreases over time. The acid is formed when small amounts of carbon dioxide present in the air dissolves in solution. Therefore, if we want to know the exact concentration of NaOH, we should standardize it. Potassium hydrogen phthalate can be act as a primary standard to standardize NaOH. It is because it is a monoprotic acid whose formula is KHC8H4O4 and has a high molecular weight (204.22 g/mol). Also, the exact mass can be easily determi ned by weighing the dried acid on an analytical balance. It is stable and not hydroscopic, and it fits all the propeties of a primary standard.

3. Is methyl orange a suitable indicator in this experiment? Why?

Ans: Methyl orange is not a suitable indicator in this experiment. Because for strong base (NaOH) with weak acid (KHC8H4O4), the pH at the equivalence point is always greater than 7 because of the conjugate base is basic. And from the titration curve for strong acid with weak base, we know that pH at the equivalence point is approximately 8.8, and the dramatic change in pH cannot cover the pH range of methyl orange which is (3.1-4.4). But it can cover the pH range of phenolpthalein (8.3-10.0), therefore it is more suitable for us to use phenolpthalein in this experiment.

Titration for weak acid with strong base