Applications of Aqueous Equilibria Chapter 15 Solubility Equilbriua Sections 6-8
Solubility Product CaF 2 (s) Ca 2+ (aq) + 2F - (aq) then Ca 2+ (aq) + 2F - (aq) CaF 2 (s) CaF 2 (s) Ca 2+ (aq) + 2F - (aq) the solution is saturated K sp = [Ca 2+ ] [F - ] 2 solubility product (constant) The amount of solid doesn t affect the equilibrium. Increasing solid - increases the surface area to dissolve, however because the ions reform on the surface of the solid it also increases the undissolving Solubility vs Solubility Product Solubility product is an equilibrium constant with one value at a certain temp Solubility is an equilibrium position. Salts has a specific solubility at a certain temp but a common ion could change the solubility.
Solubility Product Table 15.4 page 718
K sp Sample Exercise 15.13 page 719 Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0 x 10-15 mol/l at 25 C.
K sp Sample Exercise 15.14 page 720 The K sp value for copper II iodate, Cu(IO 3 ) 2, is 1.4 x 10-7 at 25 C. Calculate its solubility at 25 C.
Relative Solubilities Be careful using K sp to compare solubility salts must prodeuce the same number of ions then largest K sp = most soluble if salts don t produce the same number of ions then calculate the solubility from the K sp
Comparing Solubilities of Salts Which is more soluble? BaSO 4 PbI 2 Bi 2 S 3 K sp = 1.5 x 10-9 1.4 x 10-8 1.1 x 10-73 Look at the mole ratio They are not all 1:1 If they were all 1:1 then PbI 2 is more soluble But you can t do it that way. Find solubility BaSO 4 [Ba 2+ ] [SO 4 2- ] = 1.5 x 10-9 = [x] [x] = x 2 sol. = 1.5 x 10-9 = 3.9 x 10-5 PbI 2 [Pb 2+ ] [I 1- ] 2 = 1.4 x 10-8 = [x] [2x] 2 = 4x 3 sol. = 3 (1.4 x 10-8 / 1 1 2 2 ) = 1.5 x 10-3 Bi 2 S 3 [Bi 3+ ] 2 [S 2- ] 3 = 1.1 x 10-73 = [2x] 2 [3x] 3 = 108x 5 sol. = 5 (1.1 x 10-73 / 2 2 3 3 ) = 1.0 x 10-15
Solubility K sp and Q Ion product, Q just like K sp except initial concentrations are used. Q > K sp then then there is too much product and it won t all dissolve Q < K sp then then more will dissolve Q = K sp then equilibrium then it is a saturated solution
Solubility K sp and Q If 5.0 g BaSO 4 is dissolved in 100 g H 2 O, will all the salt dissolve? If not, how much will dissolve?
Common Ion Effect Sample Exercise 15.15 page 723 Calculate the solubility of solid CaF 2 (K sp = 4.0 x 10-11 ) in a 0.025 M NaF solution. Try #89, #92
ph and Solubility Mg(OH) 2 Mg 2+ + 2OH - In basic solution, solubility decreases In acidic solution, solubility increases AgPO 4 3Ag 1+ + PO 4 3- PO 4 3- is basic, reacting with H + to produce HPO 4 2- In acidic solution, solubility increases In a salt, MX, if X - is a base then solubility is increased in an acidic solution Examples: OH -, S 2-, CO 3 2-, C 2 O 4 2-, CrO 4 2-
Common Ion Effect Sample Exercise 15.16 page 725 A solution is prepared by adding 750.0 ml of 4.00 x 10-3 M Ce(NO 3 ) 3 to 300.0 ml of 2.00 x 10-2 M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 x 10-10 ) precipitate from this solution? (Peek at next slide.)
Determining Precipitation Conditions
Common Ion Effect Sample Exercise 15.17 page 726 A solution is prepared by mixing 150.0 ml of 1.00 x 10-2 M Mg(NO 3 ) 2 and 250.0 ml of 1.00 x 10-1 M NaF. Calculate the concentrations of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp = 6.4 x 10-9 ). (Peek at next slide.) Try #97, #98, #99, #100
Precipitation
Selective Precipitation The separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S Figure 15.11
Selective Precipitation Classic method for separating the common cations by selective precipitation. Figure 15.12
Selective Precipitation The Separation of the Group I Ions in the Classic Scheme of Qualitative Analysis Figure 15.13
H N H H : Complex Ion Equilibria AgCl (s) + NH 3 Ag(NH 3 ) 2 + + Cl - Ag + lost e- will accept e- acid Complex Ion H N H H e- donor = base must have a high concentration to break up the AgCl A metal surrounded by ligands Ligands Lewis bases, e- pair donor The metal is a Lewis acid Coordination number The number of ligands attached to a metal. Common numbers 6, 4, 2 H H N H : Ag : Coordinate covalent bond Chapter 21 Coordination Chemistry
Complex Ion Equilibria
Complex Ion Equilibria
Complex Ion Equilibria
Figure 21.7 The Coordination of EDTA with a 2 + Metal Ion Complex Ion Equilibria
Figure 21.29 The Heme Complex Complex Ion Equilibria
Tannic Acid C 76 H 52 O 46 Complex Ion Equilibria
Complex Ion Equilibria Figure 21.30 Chlorophyll is a Porphyrin Complex of Mg 2+
Complex Ion Equilibria Metals add ligands one at a time Equilibrium constant describes the stability Formation constant or stability constants Ligands have lots of lone pairs to donate H 2 O, NH 3, CN -, Cl - The more ligands the more AgCl (s) can dissolve AgCl (s) Ag + + Cl - K sp = 1.6 x 10-10 Ag + + NH 3 Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + + NH 3 Ag(NH 3 ) + 2 K 2 = 8.2 x 10 3 AgCl (s) + 2NH 3 Ag(NH 3 ) 2+ + Cl - K = K sp x K 1 x K 2 K = [Ag(NH 3) 2+ ] [Cl - ] [NH 3 ] 2 = K sp x K 1 x K 2 If you add HNO 3 (also high concentrations) then it reacts with the NH 3 and the Ag + and Cl - make a precipitate Try #113 (from homework) or #111 or #115, 107 (conceptual) or #103 (step wise formations)