27
Date:
Experiment 10: CONDUCTOMETRIC DETERMINATION OF HCl AND CH3COOH IN A
GIVEN ACID MIXTURE
Aim: Conductometric determination of HCl and CH3COOH in a given acid mixture, using
standard NaOH solution.
Principle: Measurement of conductance can be employed to determine the end point in
acid – base titration in conductometric titration, there is a sudden change in conductance
of the solution at the equivalence point. The principle underlying conductometric
titration is the substitution of ions of one mobility by ions of another mobility. Therefore,
the conductance of a solution depends on the number and mobility of ions. The
equivalence point is determined graphically by plotting conductance against titre values.
Conductometric titration may be applied for the determination of acid present in a
mixture, in the titration of a mixture of a weak acid (CH3COOH) and a strong acid (HCl)
with a strong base (NaOH), the conductance dicreases upon adding NaOH to acid mixture
owing to the substitution of highly mobile H + ion (mobility: 350ohm-1m-1) by the less
mobile ion Na+ ion.
This trend continues till all the H+ ions of HCl replaced i.e., the strong acid is
neutralized. Continued addition of NaOH raises the conductance moderately, as the weak
acid, CH3COOH is converted into its salt CH3COONa. Further addition of NaOH rises the
conductance steeply due to the presence of free OH- ions (mobility: 198ohm-1m-1). The
titration curves in the graph given depict the location of the equivalence points.
Procedure: Estimation of HCl and CH3COOH
Given acid mixture is made up to the mark by using ion exchange water. Pipette
out 50 cm3 of the given acid mixture into a clean 100cm3 beaker. Dip the conductivity cell
in the solution and note down the conductance of the solution i.e., when the volume of
NaOH added is zero.
Now add standard NaOH solution from the burette in increment of 0.5cm 3. after
each addition, stir the solution gently and note down the conductance. As the titration
proceeds, the conductance first decreases and then increases sharply. Continue titration
until the conductance is more or less the same as it was
in the beginning. Plot the graph of conductance on Y-
axis verses volume of NaOH on X-axis to get three
straight lines. The point of intersection of the first and
the second lines gives the volume of NaOH needed to
get neutralize only HCl acid. The point of intersection
of the second and third straight lines gives the volume
of NaOH required to neutralize both HCl and CH3COOH
(after drawing a perpendicular to X-axis).
Result:1) Weight of HCl/dm3 =…………………g
2) Weight of CH3COOH/dm3=………………… g
Dept of chemistry/PACE/Mangalore
Date:
Experiment 10: CONDUCTOMETRIC DETERMINATION OF HCl AND CH3COOH IN A
GIVEN ACID MIXTURE
Aim: Conductometric determination of HCl and CH3COOH in a given acid mixture, using
standard NaOH solution.
Principle: Measurement of conductance can be employed to determine the end point in
acid – base titration in conductometric titration, there is a sudden change in conductance
of the solution at the equivalence point. The principle underlying conductometric
titration is the substitution of ions of one mobility by ions of another mobility. Therefore,
the conductance of a solution depends on the number and mobility of ions. The
equivalence point is determined graphically by plotting conductance against titre values.
Conductometric titration may be applied for the determination of acid present in a
mixture, in the titration of a mixture of a weak acid (CH3COOH) and a strong acid (HCl)
with a strong base (NaOH), the conductance dicreases upon adding NaOH to acid mixture
owing to the substitution of highly mobile H + ion (mobility: 350ohm-1m-1) by the less
mobile ion Na+ ion.
This trend continues till all the H+ ions of HCl replaced i.e., the strong acid is
neutralized. Continued addition of NaOH raises the conductance moderately, as the weak
acid, CH3COOH is converted into its salt CH3COONa. Further addition of NaOH rises the
conductance steeply due to the presence of free OH- ions (mobility: 198ohm-1m-1). The
titration curves in the graph given depict the location of the equivalence points.
Procedure: Estimation of HCl and CH3COOH
Given acid mixture is made up to the mark by using ion exchange water. Pipette
out 50 cm3 of the given acid mixture into a clean 100cm3 beaker. Dip the conductivity cell
in the solution and note down the conductance of the solution i.e., when the volume of
NaOH added is zero.
Now add standard NaOH solution from the burette in increment of 0.5cm 3. after
each addition, stir the solution gently and note down the conductance. As the titration
proceeds, the conductance first decreases and then increases sharply. Continue titration
until the conductance is more or less the same as it was
in the beginning. Plot the graph of conductance on Y-
axis verses volume of NaOH on X-axis to get three
straight lines. The point of intersection of the first and
the second lines gives the volume of NaOH needed to
get neutralize only HCl acid. The point of intersection
of the second and third straight lines gives the volume
of NaOH required to neutralize both HCl and CH3COOH
(after drawing a perpendicular to X-axis).
Result:1) Weight of HCl/dm3 =…………………g
2) Weight of CH3COOH/dm3=………………… g
Dept of chemistry/PACE/Mangalore
