Macro

Solubility and Ksp 
It is important to distinguish carefully between solubility and solubility product.  The solubility of a substance is the quantity that dissolves to form a saturated solution.  Solubility is often expressed as grams of solute per liter of solution.  The molar solubility is the number of moles of the solute that dissolve in forming a liter of saturated solution of thesolute.  The solubility product constant is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution.  The solubility of a substance changes as the concentration of other solutes change.  In contrast the solubility product for a given solute is constant at a specific temperature.
In studying solubility equilibria, it is important to be able to interconvertsolubility and solubility product.  The following examples will illustrate this.
 Example 1 
The Ksp for CaF2 is 3.9 x 10-11 at 25oC.  What is the (a) molar solubility of CaF2 in water?  (b) What is the solubility in grams per liter?
First set up an equilibrium table and let x = [Ca2+] (and the molar solubility of CaF2 )
 
Concentrations(M) | CaF2(s)  | <==> | Ca2+(aq) | + | 2 F-(aq) |Starting | | | 0 | | 0 |
Change | | | +x | | +2x |
Equilibrium | | | x | | 2x |
 Ksp = [Ca2+][F-]2 = (x)(2x)2 = 4x3 = 3.9 x 10-11 
(a) x = (3.9 x 10-11 /4)1/3 = 2.1 x 10-4 M
(b) Convert moles per liter of CaF2 to grams per liter.
2.1 x 10-4 mol CaF2 78.1 g CaF2
------------------- x ------------ = 1.6 x 10-2 g CaF2 /L soln
1 Lsoln 1 mol CaF2
 Example 2 
Analysis of a saturated solution of silver chromate, Ag2CrO4, indicates that the concentration of silver ion is 1.3 x 10-4 M.  What is the Ksp of Ag2CrO4?  
The equilibrium equation and the solubility product expression are
Ag2CrO4(s)   <==>    2 Ag+(aq) + CrO42-    Ksp = [Ag+]2[CrO4]
From the equation we can see that at equilibrium, the concentrationof CrO42- is going to be half that of Ag+. 
[ CrO42-] = 1/2  [Ag+] = 1/2 (1.3 x 10-4 M) = 6.5 x 10-5 
 Ksp = [Ag+]2[CrO4] =   (1.3 x 10-4)2( 6.5 x 10-5) = 1.1 x 10-12 


nitrate ion: NO31-


Equilibrium :  Solubility Equilibrium |
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Solubility equilibrium is base on the assumption that solids dissolve in water to give the basic particles from which they are formed.
 Molecular solidsdissolve to give individual aqueous molecules.

 Ionic solids dissociate to give their respective positive and negative ions:

The ions in formed from the dissociation of ionic solids can carry an electrical current.  Salt solutions, therefore, are good conductors of electricity.  Molecular solids, however, do not dissociate in water to give ions, so no electrical current can be carried.Solubility
1. The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve.
1. Generally expressed in two ways:
1. grams of solute per 100 g of water
2. moles of solute per Liter of solution
Soluble: Dissolve - Do NOT form a solid precipitate. 1. **alkali metal ions and ammonium ion: Li+, Na+, K+, NH4+  2. acetate ion:C2H3O21-  3. nitrate ion: NO31-  4. halide ions (X): Cl-, Br-, I-  (Exceptions:  AgX, HgX, and PbX2 are insoluble)  5. sulfate ion: SO42-  (Exceptions:  SrSO4, BaSO4, and PbSO4 are insoluble;
                     AgSO4, CaSO4, and Hg2SO4 are slightly soluble)  |

Insoluble: Do NOT Dissolve - Do form a solid precipitate. 1. carbonate ion: CO32-  2. chromate ion: CrO42-  3. phosphate ion:PO43-  4. sulfide ion: S2-  (Exceptions:  CaS, SrS, and BaS are soluble)  5. hydroxide ion: OH-  (Exceptions:  Sr(OH)2 and Ba(OH)2 are soluble; 
                     Ca(OH)2 is slightly soluble)  |
* A salt is considered soluble if it dissolves in water to give a solution with a concentration of at least 0.1 M at room temperature.
* A salt is considered insoluble if the...