Electrical conductivity of solutions and titration curve question

[Paulo2014]

hi
The following question I am stuck on:

Some automatic titrators used in modern analytical laboratories measure the electrical conductivity of the solution in the flask as the titration proceeds. The equivalence point is determined by monitoring changes in the conductivity of the solution. These changes depend on the concentration and the relative conductivities of the ions in solution.

The relative conductivities of some ionic species are shown in the table below.

Ionic species
Relative conductivity
Na+, Cl–, CH3COO– 1

OH– 3

H3O+ 5



(a) Sketch the plot of conductivity of the titration mixture against the volume of NaOH added for the titration of 1.00 mol L–1 HCl with 1.00 mol L–1 NaOH. (See hint below.)

Justify your answer by showing how significant points on the graph were determined OR by discussing the reasons for the changing conductivity.

(b) Show on the graph drawn in (a) how the shape will change if 1.00 mol L–1 CH3COOH is used in the titration instead of HCl. Justify your answer.

Hint: It may be useful to construct a table to identify the ions present and their relative concentrations and conductivities at different stages in the titration, including a point after the equivalence point. It can be assumed that the total conductivity in a solution is the sum of conductivities of the ions present.

How would I do this?

 

[Nino MMP]

Answer: a) The plot of conductivity of the titration mixture against the volume of NaOH added for the titration of 1.00 mol L–1 HCl with 1.00 mol L–1 NaOH would look like a curve with an initial low conductivity, increasing as the titration proceeds until the equivalence point is reached. At the equivalence point, the conductivity will be the sum of the conductivities of Na+ and Cl- ions, which is equal to 1. Beyond the equivalence point, the conductivity of the solution will increase due to the presence of OH– ions, which have a relative conductivity of 3. b) If 1.00 mol L–1 CH3COOH is used in the titration instead of HCl, the shape of the graph will remain the same, but the initial conductivity will be higher due to the presence of CH3COO- ions with a relative conductivity of 1. At the equivalence point, the conductivity will be the sum of the conductivities of Na+, Cl- and CH3COO- ions, which is equal to 2. Beyond the equivalence point, the conductivity of the solution will increase due to the presence of OH– ions, which have a relative conductivity of 3.

 

[Blueshift5]

For part (a), you can use the information in the table to determine the relative conductivities of the ions present at different stages in the titration. As NaOH is added, the concentration of OH- ions will increase, while the concentration of H3O+ ions will decrease. This will result in a decrease in conductivity, as OH- ions have a higher relative conductivity than H3O+ ions. The equivalence point can be determined by looking for a sudden change in conductivity, as this indicates the point where the concentration of H3O+ and OH- ions are equal.

To sketch the plot, you can plot the volume of NaOH added on the x-axis and the conductivity on the y-axis. As NaOH is added, the conductivity will decrease until it reaches the equivalence point, where it will suddenly increase. After the equivalence point, the conductivity will continue to increase as the concentration of OH- ions increases.

For part (b), if 1.00 mol L–1 CH3COOH is used in the titration instead of HCl, the shape of the graph will be different. CH3COOH is a weak acid, so it will not fully dissociate in solution. This means that the concentration of H3O+ ions will not decrease as much, and the concentration of CH3COO- ions will increase. This will result in a smaller decrease in conductivity compared to the titration with HCl. Additionally, the equivalence point will occur at a higher volume of NaOH, as more NaOH is needed to neutralize the weaker acid. This will result in a more gradual increase in conductivity after the equivalence point. You can show this on the graph by plotting the titration curve for CH3COOH as a smoother curve compared to the sharp curve for HCl.