Question

Determine normality and strength of HCI using (0.1N) NaOH solution conductometrically...!!!​

Answer 1

Answer:

Task:

Determine the Normality, with 0.1 n NaOH, of a HCl- and a

CH3COOH-Solution using Conductometric Titration. The solutions

should be titrated singularly, as well as simultaneously.

Basics:

The Specific Conductivity

(R = Resistance in Ω

C = Cell constant in cm-1

Experimental Procedure:

The solutions to be tested will be handed out by the technical

assistants in the lab.

The graduated flasks with the supplied substances should be filled to

100ml with distilled water.

The following solutions should be titrated:

a) 50 ml HCl-Solution

b) 50 ml CH3COOH-Solution

c) 20 ml HCl + 20 ml CH3COOH-Solution

An exact amount of solution should be pipetted into a beaker (250

ml) which has been cleaned with distilled water. Place a magnetic

stirrer and an immersion cell into the beaker. The solution must be

a minimum of 1cm over the electrodes, if necessary add distilled

water.

The digital measuring instrument will immediately show the

conductivity (what scale is being measured?)

Small amounts (ca. 0.5 ml) of 0.1 n NaOH should be added

dropwise to the solution.

After each addition the solution should be stirred well, and the

Specific Conductivity subject to the amount of titrant added

measured until the equivalence point has been significantly

exceeded.

Data Analysis:

1) The Specific Conductivity should be plotted against the

volume of added NaOH (in ml). The respective Normality of the

original solutions should be calculated from the equivalence

points. Conduct an error analysis for the HCl titration.

Reasonably estimate the possible error for each measured value,

from these calculate the error for each straight line, and

determine the error of Normality for HCl.

2) Explain the progression of the titration curve for each system.

3) Calculate the titration curve 1/R = f(V(NaOH)) for the titration

of 50ml of a 0.01 n HCl-solution with a 1 n NaOH-solution (the dilution effect can be ignored).

❤️

Answer 2

Task:

Determine the Normality, with 0.1 n NaOH, of a HCI- and a CH3COOH-Solution using Conductometric Titration. The solutions should be titrated singularly, as well as simultaneously.

Basics:

The Specific Conductivity (R = Resistance in Q C = Cell constant in cm-1

Experimental Procedure:

The solutions to be tested will be handed out by the technical assistants in the lab. The graduated flasks with the supplied substances should be filled to 100ml with distilled water.

The following solutions should be titrated:

a) 50 ml HCI-Solution

b) 50 ml CH3COOH-Solution

c) 20 ml HCI + 20 ml CH3COOH-Solution An exact amount of solution should be pipetted into a beaker (250 ml) which has been cleaned with distilled water. Place a magnetic stirrer and an immersion cell into the beaker. The solution must be a minimum of 1cm over the electrodes, if necessary add distilled water. The digital measuring instrument will immediately show the conductivity (what scale is being measured?) Small amounts (ca. 0.5 ml) of 0.1 n NaOH

The digital measuring instrument will immediately show the conductivity (what scale is being measured?) Small amounts (ca. 0.5 ml) of 0.1 n NaOH should be added dropwise to the solution. After each addition the solution should be stirred well, and the Specific Conductivity subject to the amount of titrant added measured until the equivalence point has been significantly exceeded. 1) The Specific Conductivity should be plotted against the volume of added NaOH (in ml). The respective Normality of the original solutions should be calculated from the equivalence points. Conduct an error analysis for the HCI titration. Reasonably estimate the possible error for each measured value, from these calculate the error for each straight line, and determine the error of Normality for HCI.

determine the error of Normality for HCI.

2) Explain the progression of the titration curve for each system.

3) Calculate the titration curve 1/R = f(V(NaOH)) for the titration of 50ml of a 0.01 n HCl-solution with a 1 n NaOH-solution (the dilution effect can be ignored).

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