Aim of Experiment: To determine  amount of the each component of the ternary mixture of HCl,CH₃COOH and CuSO₄ by conductometric titration.

Theory : During the titration HCl, being the strong acid will be neutralized first and conductance will be fall of rapidly. Then the neutralization of acetic acid will take place resulting in slight increase in conductance. Finally precipitation takes place.

                CuSO₄   +  2 NaOH -------------------- Cu (OH) ₂ + Na₂SO₄

where Cu²⁺  ions are replaced by slightly less mobile Na ⁺ ions. Consequently the conductance decreases very slowly until the precipitation is complete . There after the conductance rises due to the excess of NaOH added. Thus the titration curve will be marked by three breaks as shown below.

If V₁, V₂ and V₃ are the volumes of NaOH added as indicated by first, second and third breaks in the titration curve then

                V ₁ => HCl

                V₂ => CH₃COOH

                V₃ => CuSO₄

Requirements: 1)Conductometric bridge

                          2) Conductometric cell

                          3) Burette

                          4) Standard Na OH solution

Procedure: 5 ml HCl (0.1 M); 5 ml CH₃COOH (0.1 M) and 5 ml CuSO₄ (0.1 M)  are taken in a beaker and 85 ml of distilled water is added to it. The whole mixture is then titrated with 0.1 M NaOH conductometrically and the conductance readings are recorded.

Observation:

Temperature of the experiment=..........

Table:

Calculation:

                    Total volume of mixture =V

                    Strength of NaOH        = 0.1 M

             From the titration curve ,we have:

                    V₁ ml of Na OH => Amount of HCl in V ml of mixture

                    So............................................................................

                    (V₂ - V₁ ) ml of NaOH => Amount of CH₃COOH in V ml of mixture

                    So..................................................................................................

                    ( V₃ - V₂ ) ml of NaOH => Amount of CuSO₄ in V ml of mixture

                    So...........................................................................................

Therefore 1) Strength of HCl in V ml of mixture = [V₁ (0.1)] / V = .......N

Hence amount of HCl in 1000 ml = .......................g

      2) Strength of CH₃COOH in V ml of mixture = [(V₂ - V₁ )(0.1)] / V = .......N

Hence amount of CH₃COOH in 1000 ml = .......................g

      3) Strength of CuSO₄ in V ml of mixture = [( V₃ - V₂ ) (0.1)] / V = .......N

Hence amount of CuSO₄  in 1000 ml = ....................... g