Ksp: The Solubility Product Constant Learning Goals: I will understand the solubility product constant, Ksp, and use it to compare to Q, the trial ion product, to determine if a precipitate will form I will understand how the common ion effect affects solubility

Ksp: The Solubility Product Constant. An equilibrium can exist between a partially soluble substance and its solution (i.e. not all ionic compounds are fully soluble)                                                                             

For example: BaSO4 (s)  Ba2+ (aq) + SO42- (aq) When writing the equilibrium constant expression for the dissolution of BaSO4, we remember that the concentration of a solid is constant. The equilibrium expression is therefore: K = [Ba2+][SO42-] K = Ksp, the solubility-product constant. Ksp = [Ba2+][SO42-]

The Solubility Expression AaBb(s)  aAb+ (aq) + bBa- (aq) Ksp = [Ab+]a [Ba-]b Example: PbI2 (s)  Pb2+ + 2 I- Ksp = [Pb2+] [I-]2 The greater the Ksp the more soluble the solid is in H2O.

Solubility and Ksp Three important definitions: Solubility: quantity of a substance that dissolves to form a saturated solution Molar solubility: the number of moles of the solute that dissolves to form a liter of saturated solution Ksp (solubility product): the equilibrium constant for the equilibrium between an ionic solid and its saturated solution

Comparing Q and Ksp to Determine Precipitate Formation -The Reaction Quotient, Q, is used to determine if a system is at equilibrium (but is called trial ion product when comparing to Ksp) -If Q=Ksp, then the solution is perfectly saturated and a precipitate will not form -If Q<Ksp, more solid can still dissolve, therefore, no precipitate forms -If Q>Ksp, a precipitate forms

Example #1: Calculating Molar Solubility when given Ksp Calculate the molar solubility of Ag2SO4 in one liter of water. Ksp = 1.4 x 10-5 Write the balanced equation Write the Ksp equation Use an ICE table to find [ions]

Calculate the molar solubility of Ag2SO4 in one liter of water. Ksp = 1.4 x 10-5 Ag2SO4  2Ag1+ + SO42- Ksp = [Ag]2[SO4] ICE table 1.4 x 10-5 = (2x)2(x) 1.4 x 10-5 = 4x3 …… = x Ag2SO4 2Ag1+ SO42- I - C +2x +x E 2x x

Example 2 What is the Ksp of a saturated solution of zinc hydroxide if the molar solubility is 2.5 x 10-2 mol/L? Zn(OH)2  Zn + 2OH Ksp = [Zn][OH]2 Ratio = 1:1:2 Ksp = [2.5 x 10-2 ] 2[5 x 10-2 ] Ksp = 1.25 x 10-3

Predicting if Precipitation Occurs Step 1: Write the balanced equation Step 2: Write the Ksp equation Step 3: Use the concentration of ions to Calculate Q (trial ion product) Step 4: Compare Q to Ksp

Example 1 The concentration of nickel (II) ion in a solution is 1.5 x 10-6 mol/L. If enough Na2CO3 is added to make the solution 6.0 x 10-4 mol/L in the carbonate ion, CO3-2, will precipitation of nickel (II) carbonate occur? The Ksp of NiCO3 is 6.6 x 10-9.

Factors Affecting Solubility Some of the main factors that affect solubility of a solute are:    (1) Temperature    (2) Nature of solute or solvent    (3) Pressure

Temperature Generally solubility increases with the rise in temperature and decreases with the fall of temperature but not in all cases. In endothermic process solubility increases with the increase in temperature and vice versa.    In exothermic process solubility decrease with the increase in temperature.  Gases are more soluble in cold solvent than in hot solvent.

Nature of Solute and Solvent A polar solute dissolved in polar solvent. A polar solute has low solubility or is insoluble in a non-polar solvent. In general like dissolves like

Pressure The effect of pressure is observed only in the case of gases. An increase in pressure increases of solubility of a gas in a liquid. For example carbon dioxide is filled in cold drink bottles (such as coca cola, Pepsi 7up etc.) under pressure.

Homework Pg 471 Q4, 6, 8 (don’t do)