Calcium Hydroxide, Ca (OH) 2 is an ionic solid that is slightly soluble in water

September 19, 2017 September 1st, 2019 Free Essays Online for College Students

Calcium Hydroxide, Ca (OH) 2 is an ionic solid that is slightly soluble in water. A saturated solution of a sparingly soluble salt obeys the Law of Chemical Equilibrium.

Therefore,

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Ca (OH) 2 (s) – Ca2+ (aq) + 2OH-(aq)

Keq = [Ca2+ (aq)] [OH-(aq)] 2

[Ca (OH) 2(s)]

Keq is the Equilibrium Constant of the reaction. Whenever you see the symbol Ksp you know that it is referring to a solubility equation, written with the solid to the left of the equilibrium sign, and the dissolved products to the right. Since the concentration of the solid Ca (OH) 2 is a constant, it maybe included in the Keq for the reaction, and a new constant Ksp, the Solubility Product, is obtained. Thus for Calcium Hydroxide:

Ksp = [Ca2+ (aq)] [OH-(aq)] 2

Every substance that forms a saturated solution will have a Ksp. However, for very soluble substances like NaCl, the value is so large that the concept is rarely used. In slight and low solubility substances, the value of Ksp is a useful quantity that lets us predict and calculate solubility of substances in solution.

Ksp is also known as the course of the equilibrium constant, and is constant at constant temperature. As predicted in the Le Chatelier’s Principle, precipitate of solid Calcium Hydroxide will form when Ca2+ or OH- ions is added into a saturated solution of Calcium Hydroxide. Therefore, from the concentration of the Ca (OH) 2 in the saturated solution, we will be able to determine both the Ca+ and OH- concentration in the saturated solution.

OBJECTIVE

1) To determine the solubility product of calcium hydroxide

APPARATUS

Burette, Pipette, Erlenmeyer flask, Beaker, Retort Stand, Filter paper, Funnel

PROCEDURE

Solution I : A saturated solution of Calcium Hydroxide in deionized water.

Solution II : A saturated solution of Calcium Hydroxide in 0.1M Sodium Hydroxide.

Solution III : A saturated solution of Calcium Hydroxide in 0.05M Sodium Hydroxide.

Solution IV : A saturated solution of Calcium Hydroxide in 0.025M Sodium Hydroxide.

1. Firstly, filter solution 1 using filter paper into a conical flask, rejecting the first 10 drops.

2. Next, rinse a burette with ionized water and the filled it with 0.05M of Hydrochloric acid.

3. Record the initial reading of the burette.

4. Pipette 10ml of aliquot of the filtrate into a 250mL Erlenmeyer flask using a pipette.

5. Drop 3 drops of methyl orange (indicator) into the Erlenmeyer flask that contains the 10mL aliquot.

6. Begin titrating the aliquot using the 0.05M of Hydrochloric acid.

7. When the solution turns from yellow to peachy orange color, stop the titration.

8. Record down the amount of 0.05M of Hydrochloric acid used.

9. Repeat Step 1 to Step 9 using solution II, III, IV.

RESULT

Solution

I

II

III

IV

Final reading (cm3)

43.50

6.50

18.50

37.00

38.00

49.5

17.50

26.80

First reading (cm3)

37.00

0.00

0.00

19.0

27.00

38.00

9.00

18.00

Volume of HCl used (cm3)

6.50

6.50

18.50

18.00

11.00

11.50

8.50

8.80

Average (cm3)

6.50

18.25

11.25

8.65

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.

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