Physical Chemistry Laboratory - Conductometric Titration

Physical Chemistry Laboratory

Conductometric Titration

Letitia Sarah

061308256

Laboratory partner: Akane Hattori

Experiment period: 17 November 2015- 20 November 2015

Submission date: 27 November 2015

Day I

Objective:

• to determine temperature coefficient of NaCl and  molar conductivity of HCl at infinite dilution
• to observe titration of strong acid (HCl) with strong base (NaOH)

Procedure:

Experiment 1 Determination of temperature coefficient

1. Prepare 0.5 N NaCl solution in beaker
2. Heat the solution using hot plate
3. Measure the temperature and temperature dependence conductivity at 4 different times
4. Plot the relationship between the temperature and the conductivity on a graph
5. Determine α from the measurement data

Data

Theoretical mass of NaCl needed to prepare 50 mL 0.5 N solution is:

[pic 1]

[pic 2]

[pic 3]

[pic 4]

[pic 5]

Actual mass used: 1.4591 g

The measurement data of temperature, conductivity and the resulting α of NaCl

[pic 6]

From the line function obtained from graph above, conductivity of solution at 25oC can be determined

[pic 7]

[pic 8]

[pic 9]

α of NaCl can be determined by rearranging linear temperature correction formula as follows

[pic 10]

[pic 11]

[pic 12]

Using the formula above and by substituting the values of  and T obtained in experiment, we can determine α when T are 20.6oC, 26.7oC, 26 oC, 31 oC, 32.9 oC. The resulting α for those temperatures are shown in the table above, while calculations are omitted. After obtaining the value of α at 5 different temperature, we can take the average of α which gives us the value of 0.023908.[pic 13]

Discussion

This experiment has purpose to determine temperature coefficient (α) for NaCl which can be used in further experiments to convert the conductivity of solution to standard conductivity of solution (at 25oC). In reality, determining α for all the ionic species is complex so for simplicity α for NaCl salt can be used instead. After doing the calculation, it is concluded that α value of NaCl is 0.023908.

Exercise 1:

In the normal measurement of electric resistance, DC resistance is often measured. On the other hand, in the measurement of ionic conductivity of a solution, the AC current in the solution is measured by applying a sufficiently small AC voltage that does not cause an electrode reaction. Explain the reason for this.

A simple electrical conductivity meter is subject to errors resulting from solution polarization and redox reactions that can lead to the formation of precipitate on the electrodes if DC current is used. However, the formation of precipitates on electrodes can be largely controlled by selecting inert electrode materials and using alternating current (AC).

Experiment 2 Determination of hydrochloric acid concentration by conductometric titration

Procedure:

• Prepare 20 mL of 0.5 N HCl solution and titrate it with 0.5 N NaOH
• During titration, measure the electrical conductivity while adding NaOH dropwise to the HCl solution, and plot on a graph the change of electrical conductivity with respect to the amount of titrant.
• Determine the concentration of HCl solution

Data

[pic 14]

Concentration of HCl solution:

Volume of 12 M HCl needed to make 100 mL ±0.5 M HCl solution =

[pic 15]

Amount of NaOH titrant used at equivalence point: 16.9 mL

Conductivity of titrated HCl solution at equivalence point: 42.14972843 (mS/c m)

Mol of NaOH = Mol of HCl = [pic 16]

[pic 17]

[pic 18]

[pic 19]

Discussion

In this part of experiment, strong acid (HCl) is titrated with strong base (NaCl) while the conductivity of the solution is repeatedly measured. In order to create a more accurate measurement, the conductivity data obtained from the electrical conductivity meter must be altered by considering the difference in temperature and volume. This could be done by first changing the conductivity measured to the conductivity at temperature 25oC using this equation

[pic 20]

After getting the conductivity value at 25oC, the now obtained  needs to be converted again. We need to consider the fact that the volume of HCl solution changes during the titration. Hence, we need to alter conductivity by fixing volume to 20 mL as if the conductivity now only depends on concentration, not temperature or volume. This could be done by changing  using this follow equation[pic 21][pic 22]

[pic 23]

Before the titration, HCl solution has conductivity of 133.2259829. This high conductivity value is observed due to the abundance of highly mobile hydrogen ions. When the base (NaOH) is added, the conductivity falls due to the replacement of hydrogen ions by the added cation (Na+) as H+ ions react with OH- ions to form undissociated water. This would decrease the conductivity up to the equivalence point. At the equivalence point, the solution contains only NaCl so this would be the lowest conductivity point compared to other points on the graph. However, the conductivity would increase again if NaOH is kept added to the titrated solution. This is because the concentration of OH- which has large conductivity is increasing.

Exercise 2

Derive equation (1) (clarify the physical meaning of the mobility u)

[pic 24]

Consider a solution of fully dissociated strong electrolyte with molar concentration of C. The number density for the ions in the solution can further be noted as CNA.

[pic 25]

The number of ions passing through an imaginary window of area A during interval Δt is equal to the number within distance s Δt and therefore the number of passing ions in the volume for the whole solution will be s ΔtA CNA.

The flux through one window is:

[pic 26]

When considering the fact that each ion carries a charge ze, the flux of charge will be

[pic 27]

*because s = uE (s= drift speed, u=ion mobility, E= potential difference)

In addition, the charge flux times area is

[pic 28]

Because the electric field is the potential gradient (, we can write[pic 29]

[pic 30]

Furthermore, current and potential difference are related by Ohm’s law, [pic 31]

[pic 32]

By comparing (1) and (2), we know that

[pic 33]

Also, Ion mobility (u) is the ability of free ions to move in an electric field. It is the velocity that ion attains per unit of electric field. We know that from the equation [pic 34]

Exercise 3

Explain the change of the electrical conductivity obtained in the measurement by referring to these equations

[pic 35]

[pic 36]

At first there is only HCl solution without NaOH added, this solution has abundance of H+ and will give rise to the ion mobility (u). Then, the conductivity  will increase as well resulting in high conductivity. When NaOH is starting to be added, concentration of OH- increases. OH- will then react with H+ to give H2O (liquid). H2O has a negligible mobility so it does not add to the overall ion mobility of the solution. As a consequence, the conductivity of the solution decreases. This conductivity decrease is not only caused by the ion mobility decrease, but also the concentration decrease. As more NaOH is added, the solution concentration drops resulting in additional decrease of conductivity. Furthermore at equivalence point, all H+ has reacted with OH- to form water. This results in neutral pH and lowest conductivity point. Later, more OH- is added, but there is no more H+ to react with OH-. If so, further addition of NaOH will only increase Na+ and OH- such that the ion mobility (u) increases rapidly making the conductivity trend also increases. [pic 37]