Outline:3/14/07 è Chem. Dept. Seminar 4pm è Pick up Quiz #7 – from me è last lecture before Exam 2… è No class Friday = Spring Break! Today: è Finish Chapter 18 Solubility Product (K sp )

Quiz # 7 Average = 5.3 

The solubility product is another example of equilibrium calculations Solubility product calcs depend on the common ion effect (LeChâtelier). They have particular applications with metal ions and pH calculations (environmental applications). Solubility Equilibria

The solubility product is another example of equilibrium calculations Solubility product calcs depend on the common ion effect (LeChâtelier). They have particular applications with metal ions and pH calculations (environmental applications). Solubility Equilibria

The Common Ion Effect Solubility is decreased when a common ion is added (Le Châtelier again) as F - (from NaF, say) is added, the equilibrium shifts left, therefore CaF 2 (s) is formed (precipitation occurs). As NaF is added to the system, the solubility of CaF 2 decreases. Factors that Affect Solubility

Solubility and pH If F  is removed, then the equilibrium shifts right and CaF 2 dissolves. F  can be removed by adding a strong acid: As pH decreases, [H + ] increases and solubility increases. The effect of pH is dramatic!

Equilibrium Calculations:  Example: AgCl (s)  Ag + + Cl   K sp = 1.8  10  10 Are reactants or products favored? Question:  15.0 g of AgCl  are put in 100 mL of pure water; what is the [Ag  ]?

Calculation of concentrations: K eq = [Ag + ][Cl  ] = 1.8  10  10  x 2 = 1.8  10  10  [Ag  ] = x = 1.34  10  5 M Don’t need to know how much solid is there….it is a solely a function of the K eq of the ions in solution!

Equilibrium Calculations:  Example: AgCl (s)  Ag + + Cl   K sp = 1.8  10  10 Now add some common ions: Question:  15.0 g of AgCl  are put in 100 mL of salt water ([Cl  ]=0.1M); what is the [Ag + ]?

Calculation of concentrations: K eq = [Ag + ][Cl  ] = 1.8  10  10  x (0.1 + x) = 1.8  10  10  [Ag  ] = x = 1.8  10  9 M Significant reduction of solubility by having common ion in solution (often done with pH)!  vs  10  5 M

The solubility product is another example of equilibrium calculations Solubility product calcs depend on the common ion effect (LeChâtelier). They have particular applications with metal ions and pH calculations (environmental applications). Solubility Equilibria

Worksheet #10 Ca 3 (PO 4 ) 2 K sp = [Ca 2+ ] 3 [PO 4 3  ] 2 = 2.0  10  29 [Ca 2+ ] = 0.2 M [PO 4 3  ] = 2.0  10  M What is calcium phosphate? Ca 3 (PO 4 ) 2  3 Ca PO 4 3  What is the K sp expression?

Worksheet #10 Ca 3 (PO 4 ) 2  3 Ca PO 4 3  K sp = [Ca 2+ ] 3 [PO 4 3  ] 2 = 2.0  10  29 [Ca 2+ ] = 0.2 M [PO 4 3  ] = 3.0  10  M  10   2.0  10   3.0  10 

Worksheet #10 Ca 3 (PO 4 ) 2  3 Ca PO 4 3  K sp = [Ca 2+ ] 3 [PO 4 3  ] 2 = 2.0  10   10   2.0  10   3.0  10  K sp = [ x] 3 [2x ] 2 = 2.0  10  29 2x = 5.1  10  14 M

Worksheet #10 What mass of Ca 3 (PO 4 ) 2 precipitates? = 2.0  10  3 M  2000 L = 4.0 mol  10   2.0  10   3.0  10  = 1.24 kg

Worksheet #10 Cr(OH) 3  Cr OH  K sp = [Cr 3+ ][OH  ] 3 = 1.6  10  30 x(1.0  10  6 + 3x) 3 = 1.6  10  30 +x +3x pH = 8.0 then [OH  ] = 1.0  10  6 x = [Cr 3  ] = 1.6  10  M

Worksheet #10 Cr(OH) 3  Cr OH  K sp = [Cr 3+ ][OH  ] 3 = 1.6  10  30 x(1.0  10  8 + 3x) 3 = 1.6  10  30 +x +3x pH = 6.0 then [OH  ] = 1.0  10  8 x = [Cr 3  ] = 1.6  10  M x is no longer small…solve exactly!

Another way to ask K sp : Exactly 4.68 g of silver sulfate will dissolve in 1.00 L of water. What is the K sp of silver sulfate at this temp? Ag 2 SO 4 : 312 g/mol 4.68 g/L = 1.50  10  M Ag 2 SO 4  2Ag + + SO 4   x + 2x + x x = 1.50  10  M K sp = (2x) 2 x = 2.70  10 

Complex Formation… n Stoichiometry of Complexes u A species that bonds to a metal cation to form a complex is known as a ligand. u The number of ligands is called the coordination number) n The stabilization of a metal complex by a ligand with more than one donor atom is known as the chelate effect.

The Chelate Effect n Ligands that have two or more donor atoms are chelating ligands. n Chelating ligands bond more tightly to the metal cations. n Ethylenediamine (H 2 NCH 2 CH 2 NH 2 ) is a common chelating ligand - it is abbreviated as “en” n Each nitrogen has a lone pair of electrons which can be donor atoms. Thus en is said to be bidentate

Complex Formation and Solubility n Complexation can enhance solubility. n It removes metal cations from solution causing the equilibrium to shift to the left, and dissolve more solid.

Example: Ni(OH) 2  Ni OH  K sp =5.5  10  17 Ni en  Ni(en) 3 K f = 4.1  10  17 Ni(OH) en  Ni(en) OH  K eq = 22.6 Practice!