report chem 4

8/3/2019 Report Chem 4

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Experiment 4

Topic : Solubility, Solubility Product Constant (KSP) and Ion Effect

Objective

Upon completion of this experiment, student should be able to

1. Determine the KSP for certain hydroxide salt

2. Explain the effect of common ions which influence the solubility of certain salt

3. Differentiate between solubility an KSP

Introduction

Certain ionic compound such as Ca(OH)2 are slightly soluble in water. Therefore if we take some Ca(OH)2

and dissolve it in water and then we shake the solution we observed that some of the Ca(OH)2 do not

dissolve in water. At this point we obtain a saturated solution. For a saturated solution we obtained an

equilibrium between the solid Ca(OH)2 , Ca2+

and OH-.

Ca(OH)2(S) Ca2+

(S) + 2 OH-(aq)

The solubility product constant KSP can be expressed as below

KSP = [Ca2+][ OH-]2

Common ion effect can bestudied easily when adding another salt, where one of the ion is the same

with the ion in the saturated solution. Suitable with Le Chateliar principal, if the addition of Ca2+

(in the

form of CaCI.2H2O) into a saturated solution of Ca(OH)2 , the concentration of Ca2+

ion from Ca(OH)2 will

decrease because the presence of Ca2+

ion from CaCI.2H2O, shifted the equilibrium of the Ca(OH)2

solution to the left of the equation. Thus more precipitate of Ca(OH)2 appeared in the solution. As a

result the solubility of Ca(OH)2 will be less in the presence of CaCI.2H2O. even though the cause of the

common ion will reduce the solubility of the salt, tha KSP of the salt remain unchanged. In other word the

KSP remain the same when the same ion is added to the saturated solution.

Chemicals

Saturated solution of Ca(OH)2 (Solution A)

Mixture of solution A and CaCI.2H2O salt (solution B)

Standard solution of HCL 0.0400 M

Phenolphthalein


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Apparatus

Burette

Pipette 20@25 mL

Cone flask

Filter funnel

Beaker 250 mL

Thermometer

Procedure

Experiment 4a: determination of the solubility of hydroxide calcium in water

1. 150mL of solution A was poured into a beaker

2. 20@25mL of solution A was pipette into a conical flask. The temperature of the solution was

recorded.

3. The burette was rinsed with standard solution of HCI twice. The burette was filled with standard

solution of HCI, the air bubbles was removed. The initial reading was recorded.

4. Solution A was titrated with standard solution of HCI using phenolphthalein as the indicator.

5. The titration was repeated for 3 times.

Experiment 4b: Common ion effect to molar solubility of hydroxide calcium

1. The procedure of experiment 4a was repeated using Solution B.

Result

Experiment 4a: solubility determination, molar solubility, and solubility product constant, KSP of Ca(OH)2

[HCI] = 0.04M

Temperature = 260C

Step Item Trial 1 Trial 2 Trial 3

1 Final burette reading (mL) 2.8 1.8 2.3

2 Initial burette reading (mL) 2.3 1.5 1.8

3 Volume of HCI (mL) 0.5 0.3 0.5

4 Amount of HCI (mol) 2.0x10-5 1.2x10-5 2.0x10-5

5 Amount of OH-(mol), standard solution 2.0x10-5 1.2x10-5 2.0x10-5

6 Volume of Ca(OH)2 (mL) 25 25 257 [OH-], equilibrium 8.0 x 10-4 4.8 x 10-4 8.0 x 10-4

8 [Ca2+

], equilibrium 4.0 x 10-4

2.4 x 10-4

4.0 x 10-4

9 Molar solubility of Ca(OH)2 (mol/L) 4.0 x 10-4 2.4 x 10-4 4.0 x 10-4

10 Average of molar solubility of Ca(OH)2 3.5 x 10-4

mol/L

11 KSP of Ca(OH)2 2.6 x 10-10 1.7 x 10-10 2.6 x 10-10

12 Average of KSP of Ca(OH)2 2.3 x 10-10


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Experiment 4b: Molar solubility of Ca(OH)2 and common ion effect

[HCI] = 0.04M

Temperature = 260C

Step Item Trial 1 Trial 2 Trial 3

1 Final burette reading (mL) 3.5 4.2 5.0

2 Initial burette reading (mL) 2.8 3.5 4.2

3 Volume of HCI (mL) 0.7 0.7 0.8

4 Amount of HCI (mol) 2.8 x 10-5 2.8 x 10-5 3.2 x 10-5

5 Amount of OH-(mol), standard solution 2.8 x 10

-52.8 x 10

-53.2 x 10

-5

6 Volume of Ca(OH)2 with CaCI2 (mL) 25 25 25

7 [OH-], equilibrium 1.12 x 10-3 1.12 x 10-3 1.28 x 10-3

8 Molar solubility of Ca(OH)2 (mol/L) 5.6 x 10-4

5.6 x 10-4

6.4 x 10-4

9 Average of molar solubility of Ca(OH)2 5.9 x 10-4 mol/L

Calculation for experiment 4a

Volume of HCL (mL) used = final burette reading initial burette reading

= 2.8mL - 2.3mL

= 0.5mL

No of moles of HCL = concentration(M) x volume(L)

= 0.04mol/L x (0.5/1000)L

= 2.0 x 10-5 mol

Ca(OH)2 + 2HCI CaCI + 2H2O

1 mol of Ca(OH)2 react with 2 mol of HCI

Therefore, number of moles of Ca(OH)2 = x 2.0 x 10-5 mol

= 1.0 x 10-5

mol

Ca(OH)2(S) Ca2+

(S) + 2 OH-(aq)

1 mol of Ca(OH)2 produce 2 mol of OH-

Therefore, number of moles of OH-

= 2 x (1.0 x 10-5

mol)

= 2 x 10-5 mol

Concentration of [OH-] = number of moles(mol)/volume(L)

= 2.0 x 10-5

mol/(25/1000)L

= 8 x 10-4

M

Concentration of [Ca2+

] = number of moles(mol)/volume(L)

= 1.0 x 10-5 mol/(25/1000)L

= 4.0 x 10-4 M


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Molar solubility of Ca(OH)2 = number of moles of solute dissolve/volume(L)

= 1.0 x 10-5

mol/(25/1000)L

= 4.0 x 10-4 M

Average of molar solubility = (4.0 x 10-4 + 2.4 x 10-4 + 4.0 x 10-4)/3

= 3.5 x 10-4 mol/L

Solubility product constant, KSP = [Ca2+

][ OH-]

2

= (4.0 x 10-4

)( 8.0 x 10-4

)2

= 2.6 x 10-10

Average of KSP of Ca(OH)2 = (2.6 x 10-10

+ 1.7 x 10-10

+ 2.6 x 10-10

)/3

= 2.3 x 10-10


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Calculation for experiment 4b

Volume of HCL (mL) used = final burette reading initial burette reading

= 3.5mL - 2.8mL

= 0.7mL

No of moles of HCL = concentration(M) x volume(L)

= 0.04mol/L x (0.7/1000)L

= 2.8 x 10-5

mol

Ca(OH)2 + 2HCI CaCI + 2H2O

1 mol of Ca(OH)2 react with 2 mol of HCI

Therefore, number of moles of Ca(OH)2 = x 2.8 x 10-5

mol

= 1.4 x 10-5 mol

Ca(OH)2(S) Ca2+

(S) + 2 OH-(aq)

1 mol of Ca(OH)2 produce 2 mol of OH-

Therefore, number of moles of OH-

= 2 x (1.4 x 10-5

mol)

= 2.8 x 10-5

mol

Concentration of [OH-] = number of moles(mol)/volume(L)

= 2.8 x 10-5 mol/(25/1000)L

= 1.12 x 10-3M

Molar solubility of Ca(OH)2 = number of moles of solute dissolve/volume(L)

= 1.4 x 10-5

mol/(25/1000)L

= 5.6 x 10-4 mol/L

Average of molar solubility = (5.6 x 10-4 + 5.6 x 10-4 + 6.4 x 10-4)/3

= 6.4 x 10-4mol/L


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Exercises

1. Saturated solution is a solution in which no more solute will dissolve at a given temperature.

Normal solution is a solution in which more solution can dissolve at a given temperature. The

solubility of salt in certain solvent will increase as the temperature increase because more heat

is available to increase the dissolving process. The kinetic energy for each salt molecule increase

causes the bond between the molecules to break.

2. The KSP of the Ca(OH)2 remain unchanged. This is because the amount moles that dissolve does

not affect the value of KSP.

3. a) the amount of HCI used will increase because as the concentration of the Ca(OH)2 increase.

Hence, more HCI needed to neutralize it.

b) Remain unchanged because the effect of concentration does not affect the value of KSP

c) Increase because the number of moles of the compound that dissolve in 1 liter of solvent

increase.

4. a) The value of KSP is the accurate one as the phenolphthalein is used because the end point is

the point where the saturated solution obtained an equilibrium between solid Ca(OH)2, Ca+

and OH-

b) The value of KSP is not too accurate as methyl orange is used as the indicator because the end

point is the point where the saturated solution obtained are not in equilibrium between solid

Ca(OH)2 , Ca+

and OH-

5. if we use pipe water, the water contain impurities that can cause the Ca(OH)2 salt not to dissolve

completely in the water which the saturated solution obtained are not in equilibrium between

the solid Ca(OH)2, Ca+ and OH-.


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Discussion

In experiment 4a, the phenolphthalein is used as the indicator in order to indicate that both

solutions have completely neutralized. From the volume of HCI used for the neutralization, the number

of moles of Ca(OH)2 can be obtain because the number of moles of Ca(OH)2 that dissolve to give one

liter of saturated solution is the molar solubility of the compound.

In experiment 4b, CaCI2 was added in order to the explained about the effect of common ion

effect to the molar solubility of the compound. From the equation

Ca(OH)2(S) Ca2+

(S) + 2 OH-(aq), if we added another salt that providesone of the product

ions such as Ca2+

, this dissolution reaction is shifted to the lest according to the LeChateliers Principle.

Therefore, the solubility of the compound will decrease as the common ion effect take place.

Conclusion

1. The solubility constant for Ca(OH)2, KSP = 2.3 x 10-10

2. The common ion effect will decrease the molar solubility of the compound.

Methodology

1. Gathering of information

a) Chemistry the central science 11th edition

b) Whitten, Davis, Peck Stanley. Chemistry eight edition

report chem 4

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