Aim of Experiment: To determine  amount of the each component of the ternary mixture of HCl, NH₄Cl and NaCl  by conductometric titration.

Theory: The mixture of HCl, NH₄Cl and NaCl need two titrants for their determination of amounts in the mixture viz. NaOH and AgNO₃ .

              During the titration with NaOH, HCl being the strong acid will be neutralized first and the conductance will decrease rapidly due to the replacement of H⁺ ion by Na⁺ ion  up to the equivalent point of HCl . Then there will be a little change in the conductance due to  the formation of weak base NH₄OH (resulting in replacement of the NH₄⁺ ions in solution by the somewhat less mobile) up to the end  point of NH₄Cl. The alkali has no effect on the salt NaCl and finally conductance will increase rapidly due to excess of free NaOH added . Thus the titration by NaOH  for the mixture of HCl, NH₄Cl and NaCl will consist of a titration curve marked by two breaks, first corresponding to the equivalent point of the HCl and the second to that of NH₄Cl. 

              Again a direct titration of the mixture is possible with AgNO₃ which yield the total halide ion concentration giving a single end point corresponding to HCl, NH₄Cl and Na Cl in the titration curve.

              If  V₁ and V₂ are the volumes of NaOH added as indicated by first and second breaks in the titration curve (1), as shown in the following then:

                    Titration Curve (1):

                                                                                       V ₁ a [HCl]

                                                                                        V₂ a [HCl] + [NH₄Cl]   

            Again if V₃ is the volume of AgNO₃ solution added in the direct titration of the mixture as indicated in the titration curve (2), as shown below, then 

                                                                                        V₃ a [HCl] + [NH₄Cl] + [NaCl]

So amount of NaCl = ( amount of HCl + NH₄Cl + NaCl ) – (amount of HCl + amount of  NH₄Cl) .

Requirements: 1) Conductometric bridge

                         2) Conductometric cell

                         3) Burette

                        4) Standard NaOH solution

                        5) Standard AgNO₃ solution

Procedure: 5 ml HCl (~0.1 M)  5 ml NH₄Cl (~0.1 M) and 5 ml NaCl (~0.1 M) are taken in two beakers (or a 15 mL solution of the mixture of the three was supplied by the teacher) and 85 ml of distill water is added to both the beakers. The whole mixture are then titrated against two titrants viz. 0.1 M NaOH and 0.1 M AgNO₃ solution conductometrically and the conductance readings are then noted.

Observation:

Temperature of the experiment:........

Titration curve (1) reading:

Similarly Titration curve (2) reading;

(similar table)

Calculation:

Total volume of the mixture V ml.

Strength of NaOH:......................

Strength of AgNO₃.....................

Using titration curve (1) we have,

V₁ ml of NaOH = Amount of HCl in V ml of mixture

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

V₂ – V₁ ml of NaOH = Amount of NH₄Cl in V ml of mixture

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

V₃ ml of AgNO₃ = Amount of total Cl^–  ion in V ml of mixture

or, Amount of NaCl = Amount of total Cl^–  ion – Amount of (HCl + NH₄Cl)

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

Therefore

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

Hence amount of Cl^– in HCl 1000 ml = .......................

 2) Strength of NH₄ Cl in V ml of mixture = [(V₂ - V₁ )(0.1)] / V = .......

Hence amount of Cl^- in NH₄Cl 1000 ml = .......................

 3) Strength of ( HCl + NH₄Cl + Na Cl ) in V ml of mixture = [V₃ (0.1)] / V = .......

Hence amount of Cl^-   in 1000 ml = .......................

So amount of NaCl :......................................