- Firstly, filter solution 1 using filter paper into a conical flask, rejecting the first 10 drops.
- Next, rinse a burette with ionized water and the filled it with 0.05M of Hydrochloric acid.
- Record the initial reading of the burette.
- Pipette 10ml of aliquot of the filtrate into a 250mL Erlenmeyer flask using a pipette.
- Drop 3 drops of methyl orange (indicator) into the Erlenmeyer flask that contains the 10mL aliquot.
- Begin titrating the aliquot using the 0.05M of Hydrochloric acid.
- When the solution turns from yellow to peachy orange color, stop the titration.
- Record down the amount of 0.05M of Hydrochloric acid used.
- Repeat Step 1 to Step 9 using solution II, III, IV.
RESULT
CALCULATION
1. From your titre values, calculate total concentration of Hydroxide ion in each of the solution I, II, III and IV.
Solutions I
[H+] = [OH-]
M1 = [H+] = 0.05M
V1 = Volume of HCl = 6.50 mL
V2 = Volume of saturated solution = 10.00 mL
M1 x V1 = M2 x V2
M2 =
= 0.05M x 6.50 mL
10.00 mL
= 0.0325M
Solution II
[H+] = [OH-]
M1 = [H+] = 0.05M
V1 = Volume of HCl = 18.25 mL
V2 = Volume of saturated solution = 10.00 mL
M1 x V1 = M2 x V2
M2 =
= 0.05M x 18.25 mL
10.00 mL
= 0.0913M
Solution III
[H+] = [OH-]
M1 = [H+] = 0.05M
V1 = Volume of HCl = 11.25 mL
V2 = Volume of saturated solution = 10.00 mL
M1 x V1 = M2 x V2
M2 =
= 0.05M x 11.25 mL
10.00 mL
= 0.0562M
Solution IV
[H+] = [OH-]
M1 = [H+] = 0.05M
V1 = Volume of HCl = 8.65 mL
V2 = Volume of saturated solution = 10.00 mL
M1 x V1 = M2 x V2
M2 =
= 0.05M x 8.65 mL
10.00 mL
= 0.0432M
2) For the solutions II, III, and IV, calculate the concentration of Hydroxide ion which is derived from the dissolved Calcium Hydroxide. This is done by subtracting the Hydroxide ion concentration derived from the Sodium Hydroxide from the total Hydroxide ion concentration.
Hence, calculate the calcium ion concentration in each of the solutions I, II, III, and IV.
Solution II
M2 = [OH-] = 0.0913M
Contribution from Noah (0.1M)
[OH-]Ca (OH) = 0.0913M – 0.1M
= -8.7 x 10-3 M
Solution III
M2 = [OH-] = 0.0562M
Contribution from NaOH (0.05M)
[OH-]Ca (OH) = 0.0562M – 0.05M
= 0.006 M
Solution IV
M2 = [OH-] = 0.0432M
Contribution from NaOH (0.025M)
[OH-]Ca (OH) = 0.0432M – 0.025M
= 0.0182 M
Calculate the calcium ion concentration in each of the solution I, II, III, IV.
Ca (OH) 2(s) Ca2+ (aq) + 2OH- (aq)
1 mol of Ca2+ = 2 mol of OH-
Solution I
[OH-] = 0.0325M
[Ca2+] = 0.0325 = 1.62 x 10-2 M
2
Solution II
[OH-] = -8.7 x 10-3 M
[Ca2+] = -8.7 x 10-3 = -4.35x 10-3 M
2
Solution III
[OH-] = 0.006 M
[Ca2+] = 0.006 = 3.00 x 10-3 M
2
Solution IV
[OH-] = 0.0182 M
[Ca2+] = 0.0182 = 9.10 x 10-3 M
2
3) Calculate the solubility product of calcium hydroxide, Ksp, from the results of each
experiment.
Ca (OH) 2 (s) → Ca2+ (aq) + 2OH- (aq)
Solution I
Ksp = [Ca2+] [OH-] 2
Ksp =1.62 x 10-2 x (0.0335)2
= 1.81 x 10-5
Solution II
Ksp = [Ca2+] [OH-]2
Ksp = -4.35 x 10-3 x (-8.7 x 10-3)2
= -3.29 x 10-4
Solution III
Ksp = [Ca2+] [OH-]2
Ksp = 3.00 x 10-3 x (0.006)2
= 1.08 x 10-7
Solution IV
Ksp = [Ca2+] [OH-]2
Ksp = 9.10 x 10-3 x (0.0182)2
= 3.014 x 10-6
4) Compare the Ksp values obtained. What effect, if any, has the sodium hydroxide on the Ksp value on the Calcium Hydroxide?
Ans: By comparing the Ksp values of solution II, III, IV, we can clearly see that the Ksp increases when the concentration of the NaOH decreases. A decrease in the solubility of the salt (NaOH) solution is the effect of adding a common ion into the solution. The concentration will influence the reaction of equilibrium.
5) What effect, if any, does the Hydroxide ion concentration have on the solubility of calcium hydroxide in water?
Ans: When Ksp increases, the hydroxides are more soluble, the [OH-] becomes greater, the pH increases while the pOH decreases.
DISCUSSION
Since a saturated solution of Calcium Hydroxide is a sparingly salt that obeys the Law of Chemical Equilibrium. Therefore we can say that the solubility equilibrium of Calcium Hydroxide CaOH is :
Ca (OH) 2 (s) ↔ Ca2+ (aq) + 2OH-(aq)
And the Ksp is = [Ca2+] [OH-]2
As we know that Ksp is the solubility product constant, generally the molar concentrations of the constituent ions is the solubility product of a compound. Besides that, each is raised to the power of its stoichiometric coefficient in the equilibrium equation. The solubility of an ionic compound is indicated by the Ksp value of Ca(OH)2, the smaller the value, the less solubility the compound is in water. When the Ksp increases, the pOH decreases, pH increases, the Hydroxides become more soluble, and the [OH-] becomes greater. Furthermore, the concentration of NaOH decreases when the Ksp increases. A decrease in the solubility of NaOH is the effect of adding a common ion. As predicted in the Le Chatelier’s Principle, while re establishing the solution, some precipitate may form until the Ksp is equal with the ion product.
As we all know, every experiment is not error free and can never be so. Therefore, some error may have occurred during this experiment. The reason I am stating this is because there seems to be an error in my calculation for the Ksp of solution II. I received a negative value for the Ksp. This maybe due to the carelessness of the lab assistant while preparing the concentration of the Hydrochloric acid. The concentration of the Hydrochloric acid may not be exactly 0.05M.
Besides that, there may also be the possibility that, while conducting the experiment, some human error may have occurred or mistakes made by me. Firstly, the parallax error, while reading the burette or pipette, the meniscus of the solution may not be exactly on the line of the measurement. Thus, the measurement may be slightly out
Furthermore, the apparatus that we used, maybe clean properly thus this will also affect the final results. As the solutions are left in open air, there is a high possibility that the solutions are contaminated with some impurities and affect the concentration of the solution.
While conducting this experiment, there are a few pre-caution that we must take to rest assured the experiment goes and runs smoothly. Firstly, we must always wear our lab coat as we are dealing with harmful acid and bases. We would not wan the solutions to still on our clothes.
Second, we should listen carefully and follow exactly the instructions given by the lecturer for the experiment. We must pay full attention in the lab and must not joke around to avoid any mistake or accidents.
CONCLUSION
The Ksp value for:
-
Solution 1 = 1.81 x 10-5
-
Solution 2 = -3.29 x 10-4
-
Solution 3 = 1.08 x 10-7
-
Solution 4 = 3.014 x 10-6
REFERENCE
Books:
Chemistry 7th edition – Raymond Chang
Websites:
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