SOLUBILITY AND SOLUBILITY CONSTANT

A. SOLUBILITY (s)Solubility is maximum amount of substance that can dissolve in the solvent. Solubility expressed in gram/L or mol/L. The solubility of substance affected by factors that is kind of solvents and temperature.

B. SOLUBILITY CONSTANT (Ksp)When the solid salt is first added to the water, no Ca2+ and F- ions are present. However, as dissolving occurs, the concentrations of Ca2+ and F- increase, and it becomes more and more likely that these ions will collide and re-form the solid. Thus two opposite (competing) processes are occurringthe dissolving reaction shown above and the reverse reaction to re-form the solid:

Ca2+(aq) + 2F-(aq) CaF2(s)Ultimately, equilibrium is reached. No more solid dissolves and the solution is said to be saturated.

We can write an equilibrium expression for this process according to the law of chemical equilibrium:

Ksp = [Ca2+][F-]2where [Ca2+] and [F-] are expressed in mol/L. The constant Ksp is called the solubility product constant, or simply the solubility product.C. THE RELATIONSHIP OF SOLUBILITY AND KspSee the equilibria happens in saturated Ag2CrO4Ag2CrO4(s) 2Ag+(aq) + CrO42-(aq) s mol/L 2s mol/L s mol/L

The equilibria concentration of Ag+ ion and CrO42- ion in saturated solution can be related with Ag2CrO4 solubility, that is based on its reaction coefficient. If the solubility of Ag2CrO4 is expressed with s, so the concentration of Ag+ ion in that solution is equal to 2s and concentration of CrO42- ion is equal to s. So, the value of solubility product (Ksp) is :

Ksp = [Ag+]2 [CrO42-]

= (2s)2 (s)= 4s3On certain temperature, solubility of AgCl in water is 1,435 mg L-1.

a. How many solubility of AgCl in mol L-1, if Mr AgCl = 143,5 ?

b. Determine [Ag+] and [Cl-] in that AgCl solution !

c. Determine its Ksp.

Answer :

a. s AgCl = 1,435 mg L-1= 1,435 x 10-3 g L-1= 1,435/143,5 x 10-3 mol L-1b. = 10-5 mol L-1c. AgCl(s) Ag+(aq) + Cl-(aq)10-5 mol L-1 10-5 mol L-1 10-5 mol L-1So, [Ag+] = 10-5 mol L-1[Cl-] = 10-5 mol L-1d. Ksp AgCl = [Ag+] [Cl-]

= 10-5 x 10-5

= 10-10

D. THE EFFECT OF SAME ION TO SOLUBILITYSee the equilibria reaction below :

CaSO4 Ca2+ + SO42-If we add H2SO4, so will happens shifting equilibrium (remember Le Chatelier law about shifting equilibrium)

H2SO4 2H+ dan SO42-CaSO4 Ca2+ + SO42- same ionIf equilibrium shifts to left, how about CaSO4 solubility ? The solubility of CaSO4 will decrease. If we add Ba(OH)2 so will happens shifting equilibrium.

The relationship of solubility product and solubility can describe solubility will decrease if added reagent that contain of same ion. The effect of same ion into solubility is same ion will decrease solubility. But, same ion doesnt affect the value of solubility product if its temperature doesnt change.

Example of reaction that contain of same ion :

a. The formation of salts

Example :The solubility of CaCO3(s) in the water contain of CO2 higher than water.

CaCO3(s) + H2O(l) + CO2(g) Ca(HCO3)2(aq) solubleb. Reaction between amphoteric base with strong baseExample :The solubility of Al(OH)3 in KOH is higher than solubility of Al(OH)3 in water. Al(OH)3(s) + KOH(aq) KAlO2(aq) + 2 H2O(l) soluble

c. The formation of complex ion

Example :The solubility of AgCl in NH4OH is higher than solubility of AgCl in water. AgCl(s) + NH4OH(aq) Ag(NH3)2Cl(aq) + H2O(l) soluble

E. PRECIPITATION REACTION

We can get out an ion from its solution by precipitation reaction. Such as, calcium ion (Ca2+) in hard water can be got out by add Na2CO3 solution. Ca2+ ion will gather with carbonate ion (CO32-) to form CaCO3, hard soluble salt, so precipitated.Ca2+(aq) + CO32-(aq) CaCO3(s)The process of happens AgCl precipitate if solution contaion of Cl- ion dropped with Ag+ solution. AgCl can dissolve in water, although in little amount. Its mean that Ag+ ion and Cl- ion can be on the solution until saturated solution, is until same with value of Ksp AgCl. If addition of Ag+ ion increase, so [Ag+][Cl-] > Ksp AgCl. So the excession of Ag+ ion and Cl- ion will gather to form AgCl precipitation.

Concept of Qc is :

Qc AgCl = [Ag+][Cl-]

If Qc < Ksp, so the solution is not saturated

If Qc = Ksp, so happens saturated solution If Qc > Ksp, so happens AgCl precipitateF. THE PRINCIPAL OF SOLUBILITY IN DAILY LIFEa. The formation of saltThe salt made from salt water by evaporation principal to get NaCl crystal. In salt contain of other compounds, such as MgCl2 and CaCl2. To pure the salt, done the separation of disturbance substance based on precipitation principal. Its reaction occurred :

1. CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2NaCl(aq)White precipitate of CaCO3 is separated and got pure NaCl

2. MgCl2(aq) + 2NaOH(aq) Mg(OH)2(s) + 2NaCl(aq)MgCl2 reacted with strong base sodium hydroxide and produce white precipitate of Mg(OH)2 that doesnt dissolve, so got pure NaCl b. Photograph industryc. Removing hard waterHard water disturb our life. Hard water will reduce cleaning power of detergent, because Ca2+ contain in hard water will reacts and form insoluble salt. Hard water is water contain of high Mg2+ ion and Ca2+ ion. While, contain of HCO3- anion. Because of that, added salt contain of carbonate ion (CO32-) and bicarbonate ion (HCO3-). The addition of those ion will affect Ca2+ precipitated as CaCO3.

1. If concentration of dissolved AgCl is maximum, so the equation of equilibrium constant of AgCl is . . .

2. See the equilibrium occurred in saturated solution Ag2CrO4

If concentration of Ag2CrO4 expressed by s, so the concentration of Ag+ ion in that solution is equal to 2s, and concentration of CrO42- ion is equal to s.

The right statement about relationship of solubility (s) and Ksp is . . .a. Ksp = s2

d. Ksp = 4s3

b. Ksp = s3

e. Ksp = 4s4

c. Ksp = 16s4

3. On certain temperature, solubility product (Ksp) of PbSO4 = 1 x 10-8. The equilibrium constant of reactions is.

From data above, solubility product of PbS is . . .

a. 4 x 10-28

d. 1,6 x 10-3b. 1 x 10-6

e. 1600

c. 4 x 1028

4. Solubility of PbSO4 in the water is 1,4 x 10-4 M on temperature about 30C. If dissolved in K2SO4 0,05 M, so solubility of PbSO4 is . . .a. 1,96 x 10-8 M

d. 3,92 x 10-7 M

b. 3,96 x 10-8 M

e. 3,50 x 10-6 M

c. 3,92 x 10-4 M

5. On 25C, solubility constant of raksa (II) sulfide is 60 x 10-52. The amount of Hg2+ ion contain in 1 litre of solution is . . . (Avogadro constant = 6,02 x 1023).

a. 7,746 x 10-26

d. 4,663 x 10-3b. 1,663 x 10-23

e. 3,653 x 10-1

c. 2,746 x 1036. Solubility constant of Cr(OH)2 is 1,08 x 10-19 mol3 L-3. The solubility of Cr(OH)2 is . . .a. 16,4 x 10-10 mol/litre

d. 3,28 x 10-9 mol/litre

b. 6,56 x 10-10 mol/litre

e. 3,22 x 10-9 mol/litre

c. 3,0 x 10-7 mol/litre

7. The solubility of AgCl in water on temperature 25C is 1,435 mg per litre. The solubility of AgCl in the solution contain of 0,1 M NaCl is . . .a. 1 x 10-2 M

d. 1 x 10-5 M

b. 1 x 10-4 M

e. 1 x 10-9 M

c. 1 x 10-11 M

8. Solubility of CaCO3 in the water on certain temperature = 10-3 mol/litre. On the same temperature solubility of CaCO3 in Ca(NO3)2 solution 0,05 M is . . .a. 2 x 10-4 mol/L

d. 2 x 10-8 mol/L

b. 2 x 10-5 mol/L

e. 2 x 10-9 mol/L

c. 5 x 10-9 mol/L

9. If Ksp of Ag2CrO4 = 1,1 x 10-12 mol/L. Solubility of Ag2CrO4 in AgNO3 solution 0,1 M is . . .a. 1,1 x 10-12 mol/L

d. 1,1 x 10-17 mol/L

b. 1,1 x 10-21 mol/Le. 1,1 x 10-13 mol/L

c. 1,1 x 10-10 mol/L

10. pH of Mg(OH)2 if Ksp = 1,2 x 10-11 is . . .a. 10,447

d. 10,474

b. 10,477

e. 11,447

c. 11,477

11. Saturated base L(OH)3 has pH = 10. Ksp of that base is . . .a. 3,3 x 10-5

d. 3,3 x 10-17

b. 4 x 10-12

e. 4 x 10-18

c. 2,7 x 10-2012. Solubility of L(OH)2 in the water is 5 x 10-4 mol/litre. pH of saturated solution L(OH)2 in the water is . . .a. 10,3

d. 9,7

b. 11,0

e. 3,7

c. 2,0

13. Solubility of Mg(OH)2 (Ksp = 3 x 10-11) in the solution that has pH = 12 is . . .a. 1 x 10 -7

d. 3 x 10-7

b. 2 x 10-7

e. 4 x 10-7

c. 5 x 10-714. In 50 mL PbCl2 solution contain of 0,2207 g dissolved PbCl2. If Ksp of PbCl2 = 2,4 x 10-4 mol3/L3, so Qc is . . .a. Qc < Ksp , so doesnt produce precipitateb. Qc = Ksp, so produce precipitatec. Qc > Ksp, so doesnt produce precipitated. Qc < Ksp, so produce precipitatee. Qc > Ksp, so produce precipitate15. A solution contain of Pb(NO3)2, Mn(NO3)2, dan Zn(NO3)2 salts has concentration 0,01 M. On this solution, solid NaOH dissolved until pH of solution is 8. Based on Ksp data are :

Pb(OH)2= 2,8 x 10-16Mn(OH)2= 4,5 x 10-14Zn(OH)2= 4,5 x 10-17Hydroxide precipitated is . . .

a. none

d. only Zn(OH)2b. Zn(OH)2 and Pb(OH)2

e. only Mn(OH)2c. all of three

16. On every solution of 500 mL CaCl2 0,001 M, 500 mL SrCl2 0,001 M, and 500 mL BaCl2 0,001 M added 500 mL K2SO4 0,001 M. If Ksp of CaSO4 = 2 x 10-5; SrSO4 = 3 x 10-7; BaSO4 = 1 x 10-10, so the precipitate occurred is . . .a. BaSO4

d. BaSO4 and SrSO4b. SrSO4

e. CaSO4 and BaSO4c. CaSO4, SrSO4 and BaSO4by study this chapter, students will can :

Explain the equilibrium in saturated solution

Explain the solubility and solubility constant principal, and

Determine pH of solution and predict precipitate from its solubility constant (Ksp)

Example :

Conclusion :

Solubility is maximum amount of substance that can dissolve in the solvent. Solubility expressed in gram/L or mol/L. The solubility of substance affected by factors that is kind of solvents and temperature.

Generally, the formula of solubility constant (Ksp) is Ksp AmBn = [An+]m[Bm-]n

The relationship between solubility (s) and solubility constant (Ksp) is :

If in the equilibrium system added same ion will causes compound solubility is decreased

Higher Ksp of substance, easier to soluble of that solution

Competencies Test