Dissolution of CaCl2, CaCl2 <==> Ca2+ + 2Cl- Dissolution of Ca(IO3)2 in solution
ID: 1008746 • Letter: D
Question
Dissolution of CaCl2,
CaCl2 <==> Ca2+ + 2Cl-
Dissolution of Ca(IO3)2 in solution,
Ca(IO3)2 <==> Ca2+ + 2IO3-
So we have Ca2+ and IO3- in solution when dissolving calcium iodate. When we add CaCl2 to it. More of Ca2+ would be present in solution. The increased concentration of Ca2+ would force the recombination of some of the Ca2+ with IO3- back to form undissociated Ca(IO3)2. This is due to common ion effect, wherein, presence of a common ion on one end of a reaction forces reestablishment of equilibrium by recombination of products back to reactants.
IO3-(aq) + 5 I-(aq) + 6 H3O+(aq)-----> 3 I2(aq) + 9 H2O (l)
3 I2 (aq) + 6 S2O3^2-(aq)---> 6 I-(aq) + 3 S4O6^2-(aq)
IO3-(aq) + 6 S2O3^2-(aq) + 6 H3O+(aq)-----> I- (aq) + 3 S4O6^2- (aq) + 9 H2O(l)
Solubility Product of Ca(IO3)2 Procedures:
1. To a 250 mL Erlenmeyer flask, add approximately 2.0 g of KI and 50 mL of water.
2. Then pipet 10.00 mL of the saturated calcium iodate solution into the flask and add 10.0 mL of 1.0 M HCl.
3. Insert a piece of white paper under the flask.
4. Add 40 drops of a 0.2% starch solution to the flask.
5. Add to the buret and record the initial volume of sodium thiosulfate.
6. Titrate the purple solution until it turns colorless.
7. Record the volume of sodium thiosulfate it took to reach the endpoint.
8. Record the temperature of the solution.
9. Repeat the procedure again and record your data for the second trial.
Dissolution of CaCl2,
CaCl2 <==> Ca2+ + 2Cl-
Dissolution of Ca(IO3)2 in solution,
Ca(IO3)2 <==> Ca2+ + 2IO3-
So we have Ca2+ and IO3- in solution when dissolving calcium iodate. When we add CaCl2 to it. More of Ca2+ would be present in solution. The increased concentration of Ca2+ would force the recombination of some of the Ca2+ with IO3- back to form undissociated Ca(IO3)2. This is due to common ion effect, wherein, presence of a common ion on one end of a reaction forces reestablishment of equilibrium by recombination of products back to reactants.
IO3-(aq) + 5 I-(aq) + 6 H3O+(aq)-----> 3 I2(aq) + 9 H2O (l)
3 I2 (aq) + 6 S2O3^2-(aq)---> 6 I-(aq) + 3 S4O6^2-(aq)
IO3-(aq) + 6 S2O3^2-(aq) + 6 H3O+(aq)-----> I- (aq) + 3 S4O6^2- (aq) + 9 H2O(l)
Solubility Product of Ca(IO3)2 Procedures:
1. To a 250 mL Erlenmeyer flask, add approximately 2.0 g of KI and 50 mL of water.
2. Then pipet 10.00 mL of the saturated calcium iodate solution into the flask and add 10.0 mL of 1.0 M HCl.
3. Insert a piece of white paper under the flask.
4. Add 40 drops of a 0.2% starch solution to the flask.
5. Add to the buret and record the initial volume of sodium thiosulfate.
6. Titrate the purple solution until it turns colorless.
7. Record the volume of sodium thiosulfate it took to reach the endpoint.
8. Record the temperature of the solution.
9. Repeat the procedure again and record your data for the second trial.
questions
A.
Initial reading mL Na2S2O3
Final reading mL Na2S2O3
Volumeof Na2S2O3
tritation 1
2.7 mL
17.1
14.4
tritation 2
0 mL
15
15
B. Molarity of Na2S2O3 solution used________________
C. Moles of Na2S2O3 used in trial 1 and 2
D. Moles of IO3 present initially in trial 1 and 2
E. Molarity of IO3 in the 10 ml of saturated solution in trial 1 and 2
F. Molar solubility of Ca(IO3)2 in the saturated solution in trial 1 and 2
G. Ksp of Ca(IO3)2 in trial 1 and 2
Initial reading mL Na2S2O3
Final reading mL Na2S2O3
Volumeof Na2S2O3
tritation 1
2.7 mL
17.1
14.4
tritation 2
0 mL
15
15
Explanation / Answer
IO3-(aq) + 5 I-(aq) + 6 H3O+(aq)-----> 3 I2(aq) + 9 H2O (l)
3 I2 (aq) + 6 S2O3^2-(aq)---> 6 I-(aq) + 3 S4O6^2-(aq)
IO3-(aq) + 6 S2O3^2-(aq) + 6 H3O+(aq)-----> I- (aq) + 3 S4O6^2- (aq) + 9 H2O(l)
number of moles of KI = 2/166*50 = 0.6024 mmol
1 mole of IO3- requires 5 moles of I-
number of moles of IO3- = 5 * 0.6024 = 3.012 mmol
1mole of IO3- requires 6 moles of S2O32- then
number of moles of S2O32- = 6*3.012 = 18.072 mmol
molarity of Na2S2O3 = 18.072/1000 = 0.01807 mol/L
number of moles of Na2S2O3 in trail 1 = 0.01807 * 14.4 /1000 = 2.6 * 10^-4 moles
number of moles of Na2S2O3 in trail 2 = 0.01807 * 15 /1000 = 2.71*10^-4 moles
number of moles of IO3- in trail 1 = 2.6*10^-4/6 = 4.33 * 10^-5 moles
number of moles of IO3- in trail 2 = 2.71*10^-4/6 = 4.5167 * 10^-5 moles
molarity of IO3- in trail 1 = 4.33*10^-5*1000/10 = 0.00433 mol/L
molarity of IO3- in trail 2 = 4.5167*10^-5*1000/10 = 0.00452 mol/L