1. Will it all dissolve, and if not, how much?

2. Looking at dissolving of a salt as an equilibrium. If there is not much solid it will all dissolve. As more solid is added the solution will become saturated. Soliddissolved The solid will precipitate as fast as it dissolves.

3. MNm  aM + + b Nm - K sp = [M + ] a [Nm - ] b Example: CaF2(s)  Ca 2+ (aq) + F - (aq) K sp = [Ca 2+ ][F - ] Called the solubility product constant (solubility product) for each compound.

4. Speed up the dissolving… You can increase the surface area by grinding up the solid You can stir the solution Will this change the amount of solid dissolved at equilibrium? Neither the amount of excess solid nor the size of the particles present will shift the position of the solubility equilibrium

5. Solubility is not the same as solubility product. Solubility product is an equilibrium constant. - it doesn’t change except with temperature. Solubility is an equilibrium position for how much can dissolve. -A common ion can change this.

6. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution. 16.6

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Copper(I) bromide has a measured solubility of 2.0 x 10 -4 mol/L at 25C. Calculate the K sp. 1.Write the equation. 2.What is the K sp equation? 3.What are the initial and final concentrations? CuBr(s)  Cu + (aq) + Br - (aq) K sp = [Cu + ][Br - ] Initial [Cu+] 0 = 0 [Br-] 0 = 0

8. 8 Copyright©2000 by Houghton Mifflin Company. All rights reserved. CuBr(s)  Cu + (aq) + Br - (aq) I 0 0 C + 2.0 x 10 -4 + 2.0 x 10 -4 E 2.0 x 10 -4 2.0 x 10 -4 K sp = (2.0 x 10 -4 )(2.0 x 10 -4 ) 4.0 x 10 -8

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Write the equilibrium expression. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2  (aq) K sp = [Bi 3+ ] 2 [S 2  ] 3

10. 16.6

11. The solubility is determined by equilibrium. Its an equilibrium problem. Watch the coefficients Calculate the solubility of SrSO 4, with a Ksp of 3.2 x 10 -7 in mol/L and g/L. Calculate the solubility of Ag 2 CrO 4, with a Ksp of 9.0 x 10 -12 in mol/L and g/L.

12. What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.6 x 10 -10 16.6

13. Ksp will only allow us to compare the solubility of solids that fall apart into the same number of ions. The bigger the Ksp of those the more soluble. If they fall apart into different number of pieces you have to do the math.

14. If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.

15. Where as the solubility in water is 1.3 x 10 -4 mol/L 15 Calculate the solubility of solid silver chromate (Ag 2 CrO 4, K sp 9.0 x 10 -12 ) in a 0.1 M AgNO 3 solution. Before dissolving: Ag +, NO 3 -, and H 2 O Initial (M) Change (M) Equilibrium (M) Ag 2 CrO 4 (s ) 2 Ag + (aq) + CrO 4 2 - (aq) K sp = 9.0 x 10 -12 K sp = [Ag + ] 2 [CrO 4 2 - ].100 mol/L0.00 +2x+x o.00+x.100+2x x mol/lAg 2 CrO 4 (s ) 2 x mol/L Ag + (aq) + x mol/LCrO 4 2 - (aq) K sp = 9.0 x 10 -12 = [.1oo + 2x] 2 [x] ≈ [0.100] 2 [x] = 9.0 x 10 -10 mol/L

16. Mg(OH)2 (s) Mg2+(aq) +2OH-(aq) If you increase the pH: OH - can be a common ion, becomes less soluble. If you decrease the pH: add H +, more soluble in acid.

17. Common anions are: OH -, S 2-, CO3 2-, C2O4 2-,and CrO4 2- CaC 2 O 4 Ca +2 + C 2 O 4 -2 H + + C 2 O 4 -2 HC 2 O 4 - Reduces [C 2 O 4 -2 ] in acidic solution.

18. Use Ion Product (initial concentrations) Q =[M + ] a [Nm - ] b to predict precip formation. If Q>Ksp supersaturated Precipitate If Q<Ksp unsaturated No precipitate If Q = Ksp saturated Equilibrium.

19. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? 16.6 The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). NaOH + CaCl 2  Ca(OH)2 + NaCl Is Q > K sp for Ca(OH) 2 ? Ca(OH) 2  Ca 2+ + 2OH- [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form Need to calculate the additive molarities.oo2L + 1.00L = 1.002L moles OH- = 2.00 x10^-3 L x 0.200 M= 0.000400 mol moles Ca2+ = 1.00 L x 0.100 M= 0.100 mol 0.000400 mol/1.002L = 4.0 x 10 -4 M 0.100 mol / 1.002L = 0.100M

20. 1. Determine whether solid PbI 2 (K sp = 1.4 x 10 -8 ) forms [Pb 2+ ] 0 = 1.67 x 10 -2 M [I - ] 0 = 6.67 x 10 -2 M Q>K sp precip forms 2. Do the stoichiometry (run it to completion) Pb 2+ + 2I -  PbI 2 Before: (100.omL)(0.0500M) (200.0mL)(o.100M) solid does 5.00 mmol 20.o mmol not effect After: 0 10.0 mmol equilibrium Q = [Pb 2+ ] 0 [I - ] 0 2 = [1.67 x 10 -2 M][6.67 x 10 -2 M] 2 = 7.43 x 10 -5

21. PbI2 is formed, very small amounts re-dissolve to reach equilibrium. I - is in excess, the PbI2 is dissolving into a solution that contains 10.0 mmol of I - in 300.0 mL of solution. = 3.33 x 10 -2 M I - PbI2  Pb +2 + 2I - I 0 3.33 x 10 -2 C +x +2x E x 3.33 x 10 -2 +2x Ksp = 1.4 x 10 -8 = [x][3.33 x 10 -2 +2x] 2 ≈ [x][3.33 x 10 -2 ] 2 [Pb2+] = 1.3 x 10 -5 M [I-] = 3.33 x 10 -2 M

22. Used to separate mixtures of metal ions in solutions. Add anions that will only precipitate certain metals at a time. Used to purify mixtures.

23. The sulfide ion is often used to separate metal ions since the salts differ dramatically in their solubilities S 2- is basic and its concentration can be controlled by pH H 2 S  H + + HS - K= 1.0 x 10-7 In acidic solutions: lies far to the left ; [H+] high In basic solutions: lies far to the right; [H+] low Therefore most insoluble sulfide salts, can precip in an acidic solution (Hg 2+, Cd 2+, Bi 3+ Cu + and Hg 2+ )

24. Then add OH - solution [S -2 ] will increase so more soluble sulfides will precipitate. Co +2, Zn +2, Mn +2, Ni +2, Fe +2, Cr(OH) 3, Al(OH) 3

25. Follow the steps First with insoluble chlorides (Ag, Pb, Ba) Then sulfides in Acid. Then sulfides in base. Then insoluble carbonate (Ca, Ba, Mg) Alkali metals and NH 4 + remain in solution.

26. Qualitative Analysis of Cations 16.11 Separate the following metal ions from an aqueous sample containing ions of Ag +, Pb 2+, Cd 2, and Ni 2+ What is the order of precip? Ag + Cd 2+ Pb 2+ Ni 2+ HOT WATER

27. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27 Complex Ion: A charged species consisting of a metal ion surrounded by ligands (Lewis bases). Coordination Number: Number of ligands attached to a metal ion. (Most common are 6 and 4.) Formation (Stability) Constants: The equilibrium constants characterizing the stepwise addition of ligands to metal ions.

28. A charged ion surrounded by ligands. Ligands are Lewis bases using their lone pair to stabilize the charged metal ions. Common ligands are NH 3, H 2 O, Cl -,CN - Coordination number is the number of attached ligands. Cu(NH 3 ) 4 2+ has a coordination # of 4

29. Usually the ligand is in large excess. And the individual K’s will be large so we can treat them as if they go to completion. The complex ion will be the biggest ion in solution.

30. Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) -3 in a solution made by mixing 150.0 mL of 0.010 M AgNO 3 with 200.0 mL of 5.00 M Na 2 S 2 O 3 Ag + + S 2 O 3 -2 Ag(S 2 O 3 ) - K 1 =7.4 x 10 8 Ag(S 2 O 3 ) - + S 2 O 3 -2 Ag(S 2 O 3 ) 2 -3 K 2 =3.9 x 10 4