Question

Titration experiment: Determination of the acid content of vinegar.

25.00 mL of vinegar with a 5% concentration of acetic acid is combined with distilled water to prepare a dilute vinegar solution of 250.00 mL.

The burette is filled with a standardised NaOH solution with a molarity of 0.09979 M.

25.00 mL of diluted vinegar is pipetted into a conical flask, and phelphthalein indicator is added to it.

The volume of NaOH that needed to be added in order for the acetic acid to be neutralized is 21.57 mL.

The reaction equation for the titration is CH3COOH + NaOH--> CH3COONa + H2O.

Using the above data, calculate the following for this experiment (show detailed calculations):

a) From the titration results, calculate the number of moles of NaOH required to neutralize the acetic acid in the 25mL of diluted vinegar.

b) Calculate the number of moles of acetic acid in the 25mL of diluted vinegar used in the titration.

c) Find the molarity of acetic acid in the diluted vinegar.

d) Calculate the dilution factor used in preparing the diluted vinegar from the original vinegar.

e) Calculate the molarity of acetic acid in the original vinegar.

f) Express the concentration fo acetic acid in the vinegar in:

    i) g/L

    ii) g/100 mL

g) What therefore is the weight by volume percentage (w/v %) of acetic acid in the vingear.

h) Explain why it was important to determine the mass of KHP by difference.

Answer 1

a) Number of moles of NaOH required for the titration = (volume of NaOH in L)*(molarity of NaOH in mol/L) = (21.57 mL)*(1 L/1000 mL)*(0.009979 mol/L) (1 M = 1 mol/L) = 2.1525*10^-4 mole (ans).

b) As per the stoichiometric equation,

1 mole NaOH =1 mole acetic acid

Therefore, mole(s) of acetic acid in 25.00 mL vinegar solution = mole(s) of NaOH titrated = 2.1525*10^-4 mole (ans).

c) Molarity of acetic acid in the diluted sample = (moles of acetic acid)/(volume of dilute solution taken for titration in L) = (2.1525*10^-4 mole)/[(25.00 mL)*(1 L/1000 mL)]= 0.008610 mol/L = 0.008610 M (ans).

d) 25.00 mL of the original vinegar solution was taken and diluted to a final volume of 250.00 mL. The dilution factor is (volume of dilute solution)/(volume of original solution) = (250.00 mL)/(25.00 mL) = 10.00 (ans).

e) Molarity of acetic acid in the original solution = (molarity of acetic solution in the dilute solution)*(dilution factor) = (0.008610 M)*(10.00) = 0.0861 M (ans).