Solubility product constant and common ion effect lab answers

Chemistry Lab Report

“Determination of a Solubility Product Constant”

Purpose of Experiment:

According to the textbook, the purpose of this experiment is to determine the solubility product constant Ksp for Ca(IO3)2.xH2O by investigating the dissolved iodate ion concentration in saturated solutions of calcium iodate.

Procedure:

First we added 50 mL of 0.240 M KI, 10 mL of Ca(IO3)2 in water solution 1 and 10 mL of 1 M HCl to a 250 mL Erlenmeyer flask for each trail. Adding the components will make the solution turn brown. Moreover, we had to set up a buret and titrate Na2S2O3 until the solution turns into a yellow. Then finally add 10 drops of 1% starch indicator which will turn the solution to blue-black and we kept titrating until the solution just turns colorless. Then we recorded all buret readings. To sum, we repeated the procedure about 4 to 6 times until we obtained the accurate values.

Then we will make titrations of saturated calcium iodate in KIO3. Add 50 mL of 0.240 M KI, 10 mL of Ca(IO3)2 in KIO3 (solution 2) and 10 mL of 1 M HCl to a 250 mL erlenmeyer flask. The solution will turn to brown. Then we set up a buret and titrate Na2S2O3 until the solution turns to yellow. Then finally add 10 drops of 1% starch indicator which will change the solution to dark blue and keep titrating until the solution just turns colorless. And we recorded all the buret readings.

Observation:

After we added all the components to the Erlenmeyer flask, the solution turned to brown and by titrating Na2S2O3, the solution turned to yellow and finally we add the starch indicator and we also will continue titrating until the solution turns colorless.

Data for the saturated Calcium Iodate titration in water (Solution 1):

1

2

3

4

Final Buret Reading

10.5mL

25 mL

30.5

X

Initial Buret Reading

0 mL

10.5 mL

25.5

X

Vol. of Na2S2O3 used

10.5 mL

14.5 mL

5.5

X

[Na2S2O3]

0.05 L

0.05 L

0.05

X

1

2

3

4

mol S2O32-

5.22*10^-4L

7.25*10^-4L

2.75*10^-4L

X

mol IO3-

8.75*10^-5L

1.208*10^-4L

4.583*10^-5L

X

vol. of Ca(IO3)2

0.01L

0.01L

0.01L

X

[IO3-]

8.75*10^-3L

1.208*10^-2L

4.583*10^-3L

X

[Ca2+]

4.375*10^-3L

6.04*10^-3L

2.2915*10^-3L

X

Ksp

3.35*10^-3L

8.814*10^-7L

4.813*10^-8L

X

Data for saturated Calcium Iodate in KIO3 (Solution 2):

1

2

3

4

Final Buret Reading

28L

38.5

48

X

Initial Buret Reading

0

28L

38.5

X

Vol. of Na2S2O3 used

28L

10.5 L

9.5

X

[Na2S2O3]

0.05

0.05

0.05

X

1

2

3

4

Mol S203- reacted (L)

0.0014L

5.25*10^-4L

4.75*10^-4L

X

Mol IO3- reacted

0.00023(mol)

8.75*10^-5

7.92*10^-5

X

Vol. of Ca(IO3-)2 (with KIO3) used

0.01L

0.01L

0.01L

X

Mol IO3- from KIO3

.0000056(mol/L)

.0000056(mol/L)

.0000056(mol/L)

X

Mol IO3- from Ca(IO3)2

2.44*10^-4

8.19*10^-5

7.3*10^-5

X

[IO3-] from Ca(IO3)2

0.02244

8.19*10^-3

7.36*10^-3

X

[Ca2+]

1.122*10^-2

4.095*10^-3

3.68*10^-3

X

Ksp

5.94*10^-6

3.135*10^-7

2.31*10^-7

X

Total

1

2

3

4

[IO3-]

2.3*10^-2

8.75*10^-3

7.92*10^-3

X

Conclusion:

Overall, our data indicated that Ca(IO3)2.H2O is actually more soluble in water. Since we didn’t get the accurate values from the beginning, we had to redo the experiment to get more accurate results and values. However, the Ksp Data remained constant and similar. Also, we could have used sodium iodate as another salt option. We were able to determine and investigate the [IO3-] and the [Ca2+] and refer to the values to obtain Ksp.

Questions:

Q3: It will be different because Ksp of Ca(IO3)(IO-3) and if Ca(Io2)2 in H2O then the concentration of Ca2+ and (IO-3) is higher so the value of Ksp is high.

Q4: Another salt solution that could have been used to provide a common ion effect is Cacl2.

Q5: From Le Chatelier’s principle we would predict that the molar solubility of calcium idotae would be smaller so adding CaCl2 Ca(No3)2 the equilibrium shifts towards left. Also, dirty glassware.

Q6: Saturated Solution – Chemical solution containing maximum concentration of solute dissolved in a solvent so, additional solute will not dissolve in the saturated solution.

Calculations: