2. Solubility Allows us to flavor foods -- salt & sugar. Solubility of tooth enamel in acids. Allows use of toxic barium sulfate for intestinal x-rays.

3. Solubility Product See Table 15.4 on page 759 for common solubility products. Relative solubilities can be predicted by comparing K sp values only for salts that produce the same total number of ions. AgI (s) K sp = 1.5 x 10 -16 CuI (s) K sp = 5.0 x 10 -12 CaSO 4(s) K sp = 6.1 x 10 -5 CaSO 4(s) > CuI (s) > AgI (s)

4. Solubility Product CuS (s) K sp = 8.5 x 10 -45 Ag 2 S (s) K sp = 1.6 x 10 -49 Bi 2 S 3(s) K sp = 1.1 x 10 -73 Bi 2 S 3(s) > Ag 2 S (s) > CuS (s) Why does this order from most to least soluble appear to be out of order?

5. Solubility Product For solids dissolving to form aqueous solutions. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2  (aq) K sp = solubility product constant and K sp = [Bi 3+ ] 2 [S 2  ] 3 Why is Bi 2 S 3(s) not included in the solubilty product expression?

6. Solubility Product “Solubility” = s = concentration of Bi 2 S 3 that dissolves. The [Bi 3+ ] is 2s and the [S 2  ] is 3s. Note:K sp is constant (at a given temperature) s is variable (especially with a common ion present) Solubility product is an equilibrium constant and has only one value for a given solid at a given temperature. Solubility is an equilibrium position.

7. Solubility Product Calculations Cupric iodate has a measured solubility of 3.3 x 10 -3 mol/L. What is its solubility product? Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) 3.3 x 10 -3 M ---> 3.3 x 10 -3 M + 6.6 x 10 -3 M K sp = [Cu 2+ ][IO 3 - ] 2 K sp = [3.3 x 10 -3 ][6.6 x 10 -3 ] 2 K sp = 1.4 x 10 -7

8. Solubility Product Calculations If a 0.010 M solution of sodium iodate is mixed with a 0.0010 M cupric nitrate, will a precipitate form? 2 NaIO 3(aq) + Cu(NO 3 ) 2(aq) ---> Cu(IO 3 ) 2(s) + 2 NaNO 3(aq) Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) Q sp = [Cu 2+ ][IO 3 - ] 2 Q sp = [1.0 x 10 -3 ][1.0 x 10 -2 ] 2 Q sp = 1.0 x 10 -7 Q sp < K sp  no precipitate forms.

9. Solubility Product Calculations Cu(IO 3 ) 2(s) Cu 2+ (aq) + 2 IO 3 - (aq) K sp = [Cu 2+ ][IO 3 - ] 2 If solid cupric iodate is dissolved in HOH; double & square the iodate concentration. If mixing two solutions, one containing Cu 2+ and the other IO 3 -, then use the concentration of iodate and only square it.

10. Common Ion Effect CaF 2(s) Ca 2+ (aq) + 2F - (aq) What will be the effect on this equilibrium if solid sodium fluoride is added? Explain. Equilibrium will shift to the left, due to Le Chatelier’s Principle. Solubility product must stay constant, so the amount of Ca 2+ & F - must decrease by forming solid CaF 2. See Sample Exercise 15.15 on pages 764-765.

11. pH & Solubility If a solid precipitate has an anion X - that is an effective base (HX is a weak acid), then the salt MX will show increased solubility in an acidic solution. Salts containing OH -, S 2-, CO 3 2-, C 2 O 4 2-, & CrO 4 2- are all soluble in acidic solution. Limestone caves are made up of insoluble CaCO 3, but dissolve in acidic rain water (H 2 CO 3 ).

12. Ion Product, Q sp If 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 is added to 300.0 mL of 2.00 x 10 -2 M KIO 3, will Ce(IO 3 ) 3 precipitate? [Ce 3+ ] = (750.0 mL)(4.00 x 10 -3 mol/mL) (750.0 mL + 300.0 mL) [Ce 3+ ] = 2.86 x 10 -3 M [IO 3 - ] = (300.0 mL)(2.00 x 10 -2 mmol/mL) (750.0 mL + 300.0 mL) [IO 3 - ] = 5.71 x 10 -3 M

13. Ion Product, Q sp Continued Q sp = [Ce 3+ ] 0 [IO 3 - ] o 3 Q sp = [2.86 x 10 -3 ][5.72 x 10 -3 ] 3 Q sp = 5.32 x 10 -10 Q sp > K sp  Ce(IO 3 ) 3 will precipitate. K sp = 1.9 x 10 -10

14. Progressive Precipitation A Solubility Experience

15. An experiment to show the effect of solubility on an equilibrium system!

16. Solutions of: AgNO 3 Na 2 SO 4 K 2 CrO 4 NaCl (NH 4 ) 2 S

17. If AgNO 3 is mixed with Na 2 SO 4 what ions are most abundant in the solution? AgNO 3 Na 2 SO 4 K 2 CrO 4 NaCl (NH 4 ) 2 S With what ions is the solution saturated?

18. 2AgNO 3(aq) + Na 2 SO 4(aq)  2NaNO 3 (aq) + Ag 2 SO 4(s) K sp = 1.2  10 -5 Silver Sulfate Precipitate 2Ag + (aq) + 2NO 3 - (aq) + 2Na + (aq) + SO 4 2- (aq)  2Na + (aq) + 2NO 3 - (aq) + Ag 2 SO 4(s) 2Ag + (aq) + SO 4 2- (aq)  Ag 2 SO 4(s) First Precipitation Molecular Equation Overall Ionic Equation Net Ionic Equation

19. Silver Sulfate Precipitate Ag 2 SO 4(s)  2Ag + (aq) + SO 4 2- (aq) K sp = [Ag + ] 2 [SO 4 2- ] = 1.2  10 -5 K sp = [2x] 2  [x] = 1.2  10 -5 Molar Solubility = 1.4  10 -2 mol/liter

20. Silver Sulfate Precipitate Potassium Chromate solution What ions will be most abundant in the solution when these are mixed? With what ions will the solution be saturated?

21. Silver Sulfate Precipitate Second Reaction Potassium Chromate solution

22. Ag 2 SO 4(s) + K 2 CrO 4(aq)  K 2 SO 4(aq) + Ag 2 CrO 4(s) K sp = 9.0  10 -12 Ag 2 SO 4(s) + 2K + (aq) + CrO 4 2- (aq)  2K + (aq) + SO 4 2- (aq) + Ag 2 CrO 4(s) Ag 2 SO 4(s) + CrO 4 2- (aq)  Ag 2 CrO 4(s) + SO 4 2- (aq) Molecular Equation Overall Ionic Equation Net Ionic Equation Silver Chromate precipitate

23. K sp = 9.0  10 -12 Silver Chromate precipitate Ag 2 CrO 4(s)  2Ag + (aq) + CrO 4 2- (aq) K sp = [Ag + ] 2 [CrO 4 2- ] = 9.0  10 -12 K sp = [2x] 2  [x] = 9.0  10 -12 Molar Solubility = 1.3  10 -4 mol/liter

24. What ions will be most abundant in the solution when NaCl solution is added to Ag 2 CrO 4 precipitate? With what ions will the solution be saturated?

25. Third Reaction

26. Ag 2 CrO 4(s) + 2NaCl (aq)  Na 2 CrO 4(aq) + 2AgCl (s) K sp = 9.0  10 -12 Ag 2 CrO 4(s) + 2Na + (aq) + 2Cl - (aq)  2Na + (aq) + CrO 4 2- (aq) + 2AgCl (s) Ag 2 CrO 4(s) + 2Cl - (aq)  2AgCl (s) + CrO 4 2- Molecular Equation Overall Ionic Equation Net Ionic Equation Silver Chloride precipitate

27. K sp = 1.6  10 -10 Silver Chloride precipitate AgCl (s)  Ag + (aq) + Cl - (aq) K sp = [Ag + ] [Cl - ] = 1.6  10 -10 K sp = [x]  [x] = 1.6  10 -10 Molar Solubility = 1.3  10 -5 mol/liter

28. What ions will be most abundant in the solution when (NH 4 ) 2 S solution is added to AgCl precipitate? With what ions will the solution be saturated?

29. Fourth Reaction

30. 2AgCl (s) + (NH 4 ) 2 S (aq)  2NH 4 Cl (aq) + Ag 2 S (s) K sp = 9.0  10 -12 2AgCl (s) + 2NH 4 + (aq) + S 2 - (aq)  2NH 4 + (aq) + 2Cl - (aq) + Ag 2 S (s) 2AgCl (s) + S 2 - (aq)  Ag 2 S (s) + 2Cl - (aq) Molecular Equation Overall Ionic Equation Net Ionic Equation Silver Chloride precipitate

31. K sp = 1.6  10 -49 Silver Chloride precipitate Ag 2 S (s)  2Ag + (aq) + S 2 - (aq) K sp = [Ag + ] 2 [S 2- ] = 1.6  10 -49 K sp = [2x] 2  [x] = 1.6  10 -49 Molar Solubility = 3.4  10 -17 mol/liter

32. Of the four “insoluble” compounds, which one is the most insoluble? Molar Solubility of Ag 2 S = 3.4  10 -17 mol/liter Molar Solubility of AgCl = 1.3  10 -5 mol/liter Molar Solubility of Ag 2 CrO 4 = 1.3  10 -4 mol/liter Molar Solubility of Ag 2 SO 4 = 1.4  10 -2 mol/liter Molar Solubility of Ag 2 S = 3.4  10 -17 mol/liter

33. Of the four “insoluble” compounds, which one is the least insoluble? Molar Solubility of Ag 2 S = 3.4  10 -17 mol/liter Molar Solubility of AgCl = 1.3  10 -5 mol/liter Molar Solubility of Ag 2 CrO 4 = 1.3  10 -4 mol/liter Molar Solubility of Ag 2 SO 4 = 1.4  10 -2 mol/liter

34. Qualitative Analysis The separation of ions by selective precipitation. Much descriptive chemistry can be learned from qualitative analysis. Qualitative analysis can be done for both cations and anions.

35. Qualitative Analysis Group I -- Insoluble chlorides -- Ag +, Pb 2+, & Hg 2 2+ Group II -- Sulfides insoluble in acid solution -- Hg 2+, Cd 2+, Bi 3+, Cu 2+, & Sn 4+ Group III -- Sulfides insoluble in basic solution -- Co 2+, Zn 2+, Mn 2+, Ni 2+, & Fe 3+ Group IV -- Insoluble carbonates --Ba 2+, Ca 2+, & Mg 2+ Group V -- alkali metal and ammonium ions -- soluble so must be identified by flame tests, etc.