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Determination of Solubility Product

Autor:   •  December 11, 2017  •  1,238 Words (5 Pages)  •  545 Views

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[2] Repeat the same procedure to each test tube.

C) Analysis of Equilibrium Mixture

[1] To each test tube obtained from procedure B, add one milliliters of one molar KI, one milliliters of one molar HCL and three milliliters of distilled water. Make an observation.

[2] Stir each solution well using a stirring rod. Make sure to wipe the rod before proceeding to the next solution. Using a spectrophotometer, measure the absorbance of each test tube, Set the wavelength at five hundred nm.

RESULTS AND DISCUSSION

- Data and Calculations

Test tube No.

1

2

3

4

5

Vol of 0.0350M KlO3 (mL)

1.00

2.00

3.00

4.00

5.00

Vol of 0.0200M Ba(NO3)2 (mL)

5.00

5.00

5.00

5.00

5.00

Total volume (mL)

6.00

7.00

8.00

9.00

10.00

Absorbance of solution

0.472

0.474

0.660

0.740

1.170

(IO3-) in solution

430.54

432.36

602.03

675.00

1067.24

Processing the data (all data x 10-5 except for Ksp values)

Test tube No.

1

2

3

4

5

Initial moles of Ba2+

1 x 10-4

1 x 10-4

1 x 10-4

1 x 10-4

1 x 10-4

Initial moles of IO3-

3.5 x 10-5

7.0 x 10-5

1.1 x 10-4

1.4 x 10-4

1.75 x 10-4

No. of moles of IO3- in equilibrium solution (12mL)

4.2 x 10-4

4.2 x 10-4

4.2 x 10-4

4.2 x 10-4

4.2 x 10-4

No. of moles of IO3- precipitated

No. of moles of Ba2+ precipitated

(Ba2+) in equilibrium solution

Ksp Ba(IO3-)2

To find the Ksp for this experiment inserted the concentrations of Ba2+ and OH- in to the Ksp equation for this chemical reaction. The accepted value of the Ksp for barium hydroxide and compare it with your value. There are discrepancy and possible sources of experimental error.

Accepted Ksp = 6.5x10-6

The Ksp for this experiment was calculated to be 3.47x10-5. The calculated value for the Ksp was significantly different than he accepted value of Ksp for Ba(OH)2. This can be caused by error occurring in our experiment. Some places where error could have occurred are during the recording of our data. The Measurements for the pH change during the titration in our experiment were very rough values and could have been more specific for each trial. This discrepancy could have greatly swayed the calculated Ksp for our experiment.

CONCLUSION

In conclusion, both of our results for the experiment ended up having very different concentrations. This is most likely due to the error in the design of both experiments, as well as the miniscule concentration required to use for both experiments. First, the spectrophotometer experiment gave more promising data, When the solution contained some solids, it was known that it is saturated. Since this solution was saturated it was implied that it was at equilibrium. Being at equilibrium the forward reaction had to equal the reverse reaction. This implied that the dissolving and crystallization reactions happened at an equal rate and simultaneously. For these reactions to be equal, some precipitate would have been present, indicating the saturated solution was at equilibrium.

It was necessary to remove all of the solid Ba(OH)2 before titrating because the it would have affected the rate of reaction. When the HCl was titrated with Ba(OH)2, some of the solid would go down with the liquid because of a minuscule amount of filtering..

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