Solubility Equilibrium, Ksp

A solubility equilibrium exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. 

The equilibrium is an example of dynamic equilibrium in that some individual molecules migrate between the solid and solution phases such that the rates of dissolution and precipitation are equal to one another. When equilibrium is established, the solution is said to be saturated.

The concentration of the solute in a saturated solution is known as the solubility. Units of solubility may be molar (mol dm−3) or expressed as mass per unit volume, such as μg ml−1. Solubility is temperature dependent. A solution containing a higher concentration of solute than the solubility is said to be supersaturated. A supersaturated solution may be induced to come to equilibrium by the addition of a "seed" which may be a tiny crystal of the solute, or a tiny solid particle, which initiates precipitation.

This equilibrium constant is dimensionless as activity is a dimensionless quantity. However, use of activities is very inconvenient, so the equilibrium constant is usually divided by the quotient of activity coefficients, to become a quotient of concentrations. 

Solubility constants defined in terms of concentrations are not only temperature dependent, but also may depend on solvent composition when the solvent contains also species other than those derived from the solute.

Temperature effect
Solubility is sensitive to changes in temperature. For example, sugar is more soluble in hot water than cool water. It occurs because solubility constants, like other types of equilibrium constant, are functions of temperature. In accordance with Le Chatelier's Principle, when the dissolution process is endothermic (heat is absorbed), solubility increases with rising temperature, but when the process is exothermic (heat is released) solubility decreases with rising temperature. The temperature effect is the basis for the process of recrystallization, which can be used to purify a chemical compound.

Sodium sulfate shows increasing solubility with temperature below about 32.4°C, but a decreasing solubility at higher temperature. This is because the solid phase is the decahydrate, Na2SO4·10H2O, below the transition temperature but a different hydrate above that temperature.

Molar Solubilities and Solubility Products 

Consider the chemical equation and equilibrium law equation for a weak electrolyte such as copper(I) chloride.Notice that since copper(I) chloride dissolves very little in water, almost any amount dropped into water will result in the formation of a saturated solution, and there will be a heterogeneous equilibrium between the solid and dissolved Cu+ (aq) and Cl- (aq) ions.

CuCl (s)    Cu2+ (aq)  + Cl- (aq)

The solubility equilibrium law equation is

K = [Cu+ (aq)][Cl- (aq)]
			[CuCl (s)]

which simplifies to

K = [Cu+ (aq)][Cl- (aq)]

as the concentration (density) of the CuCl(s) is constant and so is incorporated into the value of the equilibrium constant. The value of this equilibrium constant is:

K = 1.7 x 10-7 at 25°C.

For any solute that forms ions in solution, the solubility equilibrium constant is the product of the concentrations of the ions in solution raised to the power equal to the coefficient of each in the balanced equation, and for that reason is often called the solubility product constant of the substance, symbolized as Ksp. 

In general, for the dissociation equilibrium equation
BC(s)  b B+ (aq) + c C- (aq)
where BC(s) is a slightly soluble salt, and B+ (aq) and C- (aq) are aqueous ions,

Ksp = [B+ (aq)]b[C- (aq)]c 

Example 1:
The Ksp for AgCl is 1.8x10-10. What is the molar solubility of AgCl in pure water? 

     AgCl (s)      Ag+ (aq)  +  Cl-  (aq)
    I                           0                0
   C                          +x             + x
   E                            x                x

             Ksp = [Ag+][Cl-]			solubility of AgCl = [Ag+] = [Cl-]
1.8 x 10-10 =  (x)(x)				           = 1.3 x 10-5 mol / L 
               x2= 1.8 x 10-10
               x  = 1.3 x 10-5

Example 2
The Ksp for Ag2CrO4 is 9 x 10-12. What is the molar solubility of Ag2CrO4 in pure water? 

Let x be the molar solubility of Ag2CrO4, then 
      Ag2CrO4 (s)      2 Ag+ (aq) +  CrO42- (aq)
 I                                       0                  0
C                                  + 2 x              + x
E                                     2 x                 x

              Ksp = [Ag+]2[CrO42-]				solubility Ag2CrO4  = ½ [Ag+] = [CrO42-]
9 x 10-12 = (2 x)2 (x)						             = 1.3 x 10-4 mol / L
            4x3 = 9 x 10-12
                x = 1.3 x10-4

Example 3:
Very careful experiment indicates that the molar solubility of Bi2S3 is 1.8x10-15 mol/L, what value of Ksp does this compound have? 

    Bi2S3 (s)   2 Bi3+(aq)  +     3 S2- (aq)

Solubility of Bi2S3 = ½ [Bi3+] =  [S2-]

[Bi3+]eqm = 2 x solubility Bi2S3 = 2 (1.8x10-15 mol/L) = 3.6x10-15  
 [S2-]eqm =  3 x solubility Bi2S3 = 3 (1.8x10-15 mol/L) = 5.4x10-15

    Bi2S3 (s)   2 Bi3+(aq)  +     3 S2- (aq)
E                      3.6x10-15       5.4x10-15

     Ksp  =  [Bi3+]2[S2-]3
    Ksp  =  (3.6 x 10-15)2 (5.4 x 10-15)3
         =  2.0 x10-72

Discussion
You should be able to calculate the Ksp if you know the molar solubility. 

Review of skills
1.	Be able to write the Ksp expression for the ionization of any salts. and calculate Ksp from molar solubility 
2.	Calculate Ksp from molar solubility and vice versa. Pay attention to the stoichiometry of the salt. You may have to understand what the salt really is in order to know how they ionize in the solution. You have learned the method to calculate molar solubility from Ksp 
3.	Perform calculations and be able to tell if a precipitate will form. The calculation is an important part of chemistry. 

Common Ion Effect

Solubility effects

The solubility of a sparingly soluble salt is reduced in a solution that contains an ion in common with that salt. For instance, the solubility of silver chloride in water is reduced if a solution of sodium chloride is added to a suspension of silver chloride in water.

A practical example used very widely in areas drawing drinking water from chalk or limestone aquifers is the addition of sodium carbonate to the raw water to reduce the hardness of the water. In the water treatment process, highly soluble sodium carbonate salt is added to precipitate out sparingly soluble calcium carbonate. The very pure and finely divided precipitate of calcium carbonate that is generated is a valuable by-product used in the manufacture of toothpaste.
Another example is the salting out process used in the manufacture of soaps. Soaps are sodium salts of fatty acids. Addition of sodium chloride reduces the solubility of the soap salts.

Sea, brackish and other waters that contain appreciable amount of Na+ interfere with the normal behavior of soap because of common ion effect. In the presence of excess Sodium ions the solubility of soap salts is reduced, making the soap less effective.

Example:
Determine the solubility of lead (II) bromide in (a) a solution of pure water and (b) a 0.100 mol/L solution of NaBr
Ksp = 6.6 x 10-6

(a)			PbBr2 (s)    Pb2+ (aq) + 2 Br- (aq)
		            I	 	           0                0				Ksp = [Pb2+][Br-]2
		            C                           +x             +2x			6.6 x 10-6 = (x)(2x)2 = 4x3
		            E                             x               2x			   x = 0.0118

solubility PbBr2 = [Pb2+] = ½ [Br-] = 0.012 mol / L

(b) 			PbBr2 (s)    Pb2+ (aq) + 2 Br- (aq)
		            I	 	           0              0.100			
		            C                           +x             +2x			
		            E                             x              0.100 + 2x			  

		check
		   [Br-]intital    =   0.100          >  100    therefore, [Br-]eq’m ~ 0.100
		      Ksp               6.6 x 10-6

		           Ksp = [Pb2+][Br-]2
6.6 x 10-6 = (x)(0.100)2
               x =  6.6 x 10-4

solubility PbBr2 = [Pb2+] = 6.6 x 10-4 mol/L

Qsp, Ksp and Saturation 

For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated. These substances have a tendency to form oversaturated solutions. For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids. For oversatureated solutions, Qsp is greater than Ksp. When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring Qsp to approach Ksp. 

Sodium acetate trihydrate, NaCH3COO.3H2O, when heated to 370 K will become a liquid. The sodium acetate is said to be dissolved in its own water of crystallization. The substance stays as a liquid when cooled to room temperature or even below 273 K. As soon as a seed crystal is present, crystallization occur rapidly. In such a process, heat is released, and the liquid feels warm. Thus, the relationship among Qsp, Ksp and saturation is given below: 

Qsp < Ksp 	Unsaturated solution 	
Qsp = Ksp 	Saturate solution 	
Qsp > Ksp 	Oversaturate solution 	

Does a precipitate form when form if 50.0 mL of a 0.00150 mol / L Na2SO4 solution is combined with 100.0 mL of a 0.00220 mol / L BaCl2 solution if BaSO4 has a Ksp of 1.1 x 10-10?
Molecular Equation: 	                              Na2SO4 (aq) + BaCl2 (aq) →  2 NaCl (aq) + BaSO4 (s)
Total Ionic Equation:     2 Na+ (aq)  +  SO42- (aq) + Ba2+ (aq) + 2 Cl- (aq) →  2 Na+ (aq) + 2 Cl- (aq) + BaSO4 (s)
Net Ionic Equation:  			        SO42- (aq) + Ba2+ (aq) → BaSO4 (s)

Equilibrium: 	BaSO4 (s)    SO42- (aq) + Ba2+ (aq) 

Since the two solutions are mixed together, they dilute each other prior to their reacting. Therefore, the initial concentrations have to be calculated using the dilution equation: C1V1  = C2V2

[SO42-]initial = [Na2SO4] x mol ration x VNa2SO4 / Vtotal
         
                   =  0.00150 mol Na2SO4  x     1 mol SO42-       x    50.0 mL   =   5.00 x 10-4  mol SO42-       
                                              L                   1 mol Na2SO4          150.0 mL                                L

[Ba2+]initial = [BaCl2] x mol ratio x VBaCl2 / Vtotal

                 = 0.00220 mol BaCl2    x      1 mol Ba2+     x   100.0 mL = 1.47 x 10-3  mol Ba2+
                                          L                   1 mol BaCl2          150.0 mL                               L

Example:
Does a precipitate form when 250.0 mL of a 0.0250 mol/L solution of sodium acetate (NaC2H3O2) is combined with 350.0 mL of a 0.0340 mol/L solution of a silver nitrate (AgNO3) solution?

	NaC2H3O2 (aq) + AgNO3 (aq) → AgC2H3O2 (s)  + NaNO3 (aq)
		C2H3O2- (aq) + Ag+ (aq) → AgC2H3O2 (s)

		AgC2H3O2 (s)  C2H3O2- (aq) + Ag+ (aq)

	[C2H3O2-]initial = [NaC2H3O2] x mol ratio x VNaC2H3O2 / Vtotal
		        = 0.0250 mol NaC2H3O2   x    1 mol C2H3O2-       x 250.0 mL = 0.0104 mol C2H3O2-
                                                                        L                1 mol NaC2H3O2       600.0 mL                           L

	[Ag+]initial = [AgNO3] x mol ratio x VAgNO3 / Vtotal
                                = 0.0340 mol AgNO3  x     1 mol Ag+      x   350.0 mL =  0.0198 mol Ag+
                                                        L                 1 mol AgNO3       600.0 mL                       L

	Qsp = [Ag+][C2H3O2-] = (0.0104)(0.0198) = 2.1 x 10-4

		Qsp     >    Ksp  
                         2.1 x 10-4 > 6.6 x 10-6

Therefore, a precipitate will form