1. Aqueous Equilibria Entry Task: Feb 28 th Thursday Question: Provide the K sp expression for calcium phosphate, K sp = 2.0 x 10 -29. From this expression, will there be a great deal of dissolved substance? You have 5 minutes!

2. Aqueous Equilibria I can… Define and explain the difference between:  Solubility  Solubility-product  Molar solubility Write the Ksp expression for a given soluble ionic compound. Calculate the number of ions dissolved in solution.

3. Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Sections 4

4. Aqueous Equilibria Solubility Products Define saturated solution. Is a solution in which the solution is in contact with undissolved solute. It’s the maximum amount of solute dissolved in solution at a specific temperature.

5. Aqueous Equilibria Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Equilibrium shift to the left, because not a lot of product. Slightly souble.

6. Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2- ] where the equilibrium constant, K sp, is called the solubility product.

7. Aqueous Equilibria Solubility Products Where can one find the Ksp values for a particular ionic solid? Appendix D on page 1023

8. Aqueous Equilibria Solubility Products What does a small Ksp value mean? The smaller the Ksp value only a very small amount of solid would dissolve.

9. Aqueous Equilibria Solubility Products Write the Ksp constant for CaF 2 including the # value. CaF 2 (s)  Ca +2 (aq) + 2F - (aq) 3.9 x 10 -11 = [Ca +2 ] [F - ] 2

10. Aqueous Equilibria Solubility Products Write the Ksp constant for barium carbonate include the # value. BaCO 3 (s)  Ba +2 (aq) + CO 3 -2 (aq) 5.0 x 10 -9 = [Ba +2 ] [CO 3 -2 ]

11. Aqueous Equilibria Solubility Products Write the Ksp constant for silver sulfate include the # value. Ag 2 SO 4 (s)  2Ag +1 (aq) + SO 4 -2 (aq) 1.5 x 10 -5 = [Ag +1 ] 2 [SO 4 -2 ]

12. Aqueous Equilibria Solubility and Ksp It’s important to distinguish the difference between solubility and solubility-product constant, Ksp. What is solubility? Solubility of a substance is the quantity that dissolves to form a saturated solution

13. Aqueous Equilibria Solubility and Ksp It’s important to distinguish the difference between solubility and solubility-product constant, Ksp. What is the molar solubility? The molar solubility is the number of moles of the solute that dissolves in forming a liter of saturated solution of the solute (mol/L)

14. Aqueous Equilibria Solubility and Ksp It’s important to distinguish the difference between solubility and solubility-product constant, Ksp. What is the solubility-product constant (Ksp)? The solubility-product constant (Ksp) is the equilibrium constant for the equilibrium between ionic solid and its saturated solution.

15. Aqueous Equilibria Solubility and Ksp How can solubility of a solid change? The solutions pH Concentration of other ions in solution Temperature

16. Aqueous Equilibria Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

17. Aqueous Equilibria

18. Aqueous Equilibria Calculating the Ksp

19. Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. First provide the equation: Ag 2 CrO 4 (s)  2Ag +1 (aq) + CrO 4 2- (aq)

20. Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. What do we know: Divide 1.3 x 10 -4 by 2 = 6.5 x 10 -5 We know that there is a 2 Ag 1+ ions to 1 of CrO 4 2- Ag 2 CrO 4 (s)  2Ag +1 (aq) + CrO 4 2- (aq) The Ag 1+ ions 1.3  10  4 M.

21. Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. Provide the Ksp expression: Ksp = [Ag +1 ] 2 [CrO 4 2- ]

22. Aqueous Equilibria Problem 1: Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. Plug in the numbers: Ksp = [1.3 x 10 -4 ] 2 [6.5 x 10 -5 ] Now solve: 1.1 x 10 -12

23. Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25  C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. First provide the equation: Mg(OH) 2 (s)  Mg +2 (aq) + 2OH 1- (aq)

24. Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25  C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. What we know: pH is 10.17, what can we get out of this information? We can get OH- concentration. pH = 10.17 – 14 = 3.83 is pOH 10 -3.83 = 1.47 x 10 -4 is the OH -1 concentration AND there is twice the amount OH -1 to Mg +2 ions so divide OH by 2. 7.35 x 10 -5 Mg +2 ions

25. Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25  C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. Provide the Ksp expression: Ksp = [Mg +2 ] [OH -1 ] 2

26. Aqueous Equilibria Problem 2: A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25  C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. Plug in the numbers: Ksp = [7.35 x 10 -5 ] [1.47 x 10 -4 ] 2 Ksp = 1.6 x 10 -12

27. Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. First provide the equation: CaCl 2 (s)  Ca +2 (aq) + 2Cl 1- (aq)

28. Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. What we know: CaCl 2 (s)  Ca +2 (aq) + 2Cl 1- (aq) 200.0 g / 110.98 = 1.80 mol/0.1 L or 18.0M of CaCl 2 There’s 1 Ca +2 ions to 2Cl -1 ions. There’s 18.0M Ca +2 ions to 36.0M Cl -1 ions.

29. Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. Provide the Ksp expression: CaCl 2 (s)  Ca +2 (aq) + 2Cl 1- (aq) Ksp = [Ca 2+ ][F - ] 2

30. Aqueous Equilibria Problem 3: Calculate the K sp for CaCl 2 if 200.0 g of CaCl 2 are required to saturate 100.0 mL of solution. Plug in the numbers: CaCl 2 (s)  Ca +2 (aq) + 2Cl 1- (aq) Ksp = [18.0][36.0] 2 There’s 18.0M Ca +2 ions to 36.0M Cl -1 ions. Ksp = 2.33 x10 4

31. Aqueous Equilibria Calculating Molar Solubility

32. Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: Provide equation: HgSO 4 (s)  Hg +2 (aq) + SO 4 2- (aq)

33. Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: What we know: Initial Change Equilibrium 0 0------- +x M x M HgSO 4 (s)  Hg +2 (aq) + SO 4 2- (aq)

34. Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: What we know: Notice that the stoichiometry of the equilibrium for every mole of HgSO 4 dissolves there is the same number of its ions. Provide the Ksp expression:Ksp = [Hg 2+ ][SO 4 -2 ] Plug in the equilibrium [ ]:6.3 x 10 -5 = [x][x]

35. Aqueous Equilibria Problem 4: If the solubility product of HgSO 4 is 6.4 x10 -5, then its molar solubility is: Clean up and solve: Plug in the equilibrium [ ]: x 2 = 6.3 x 10 -5 x = 7.9 x 10 -3 What does this number represent? It’s the molar solubility of HgSO 4 is 7.9 x 10 -3 6.3 x 10 -5 = [x][x]

36. Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: Provide equation: NaCl(s)  Na +1 (aq) + Cl 1- (aq)

37. Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: What we know: Initial Change Equilibrium 0 0------- +x M x M NaCl(s)  Na +1 (aq) + Cl 1- (aq)

38. Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: What we know: Notice that the stoichiometry of the equilibrium for every mole of NaCl dissolves there is the same number of its ions. Provide the Ksp expression:Ksp = [Na 1+ ][Cl -1 ] Plug in the equilibrium [ ]:1.5 x 10 -10 = [x][x]

39. Aqueous Equilibria Problem 5: The solubility product of NaCl is 1.5x10 -10 is at 298 K. its solubility in mol/L would be: Clean up and solve: Plug in the equilibrium [ ]: x 2 = 1.5 x 10 -10 x = 1.2 x 10 -5 What does this number represent? It’s the molar solubility of NaCl is 1.2 x 10 -5 1.5 x 10 -10 = [x][x]

40. Aqueous Equilibria Calculating Solubility

41. Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. First provide the equation: CaF 2 (s)  Ca +2 (aq) + 2F 1- (aq)

42. Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: **Remember solubility is the amount of solute that can dissolve in a solvent, where as the solubility-product constant is an equilibrium constant. **For this reason, we have to assume initially-that none of the salt has dissolved and allow x moles/liter of CaF 2 to dissociate completely when equilibrium is achieved

43. Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: CaF 2 (s)  Ca +2 (aq) + 2F 1- (aq) Initial Change Equilibrium 0 0------- +x M +2x M x M 2x M

44. Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: Notice that the stoichiometry of the equilibrium dictates that 2x moles/liter of F- ions are produced for each x moles/liter of CaF 2 that dissolve. Provide the Ksp expression:Ksp = [Ca 2+ ][F - ] 2 Plug in the equilibrium [ ]:3.9 x 10 -11 = [x][2x] 2

45. Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: Clean up and solve: Provide the Ksp expression:Ksp = [Ca 2+ ][F - ] 2 Plug in the equilibrium [ ]:3.9 x 10 -11 = [x][2x] 2 4x 3 = 3.9 x 10 -11 x = 3 3.9 x 10 -11 4 x = 2.1 x 10 -4 What does this number represent? It’s the molar solubility of CaF 2 is 2.1 x 10 -4

46. Aqueous Equilibria Problem 6: The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. What we know: From the molarity solubility we can get to the number of grams dissolved in 1 liter of solvent Now what: 2.1 x 10 -4 mol CaF 2 1 liter X 1.6 x 10 -2 g CaF 2 /liter solution 78.1 g CaF 2 1 mol

47. Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3  10  10 at 25  C. What is the solubility of Cu(N 3 ) 2 in grams per liter. Provide equation: Cu(N 3 ) 2 (s)  Cu +2 (aq) + 2N 3 1- (aq)

48. Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3  10  10 at 25  C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: **Remember solubility is the amount of solute that can dissolve in a solvent, where as the solubility-product constant is an equilibrium constant. **For this reason, we have to assume initially-that none of the salt has dissolved and allow x moles/liter of Cu(N 3 ) 2 to dissociate completely when equilibrium is achieved

49. Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3  10  10 at 25  C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: Initial Change Equilibrium 0 0------- +x M +2x M x M 2x M Cu(N 3 ) 2 (s)  Cu +2 (aq) + 2N 3 1- (aq)

50. Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3  10  10 at 25  C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: Notice that the stoichiometry of the equilibrium dictates that 2x moles/liter of N 3 - ions are produced for each x moles/liter of Cu(N 3 ) 2 that dissolve. Provide the Ksp expression:Ksp = [Cu 2+ ][N 3 - ] 2 Plug in the equilibrium [ ]:6.3 x 10 -10 = [x][2x] 2

51. Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3  10  10 at 25  C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: Clean up and solve: Provide the Ksp expression:Ksp = [Cu 2+ ][N 3 - ] 2 Plug in the equilibrium [ ]:6.3 x 10 -10 = [x][2x] 2 4x 3 = 6.3 x 10 -10 x = 3 6.3 x 10 -10 4 x = 5.4 x 10 -4 What does this number represent? It’s the molar solubility of Cu(N 3 ) 2 is 5.4 x 10 -4

52. Aqueous Equilibria Problem 7: The K sp for Cu(N 3 ) 2 is 6.3  10  10 at 25  C. What is the solubility of Cu(N 3 ) 2 in grams per liter. What we know: From the molarity solubility we can get to the number of grams dissolved in 1 liter of solvent Now what: 5.4 x 10 -4 mol Cu(N 3 ) 2 1 liter X 7.97 x 10 -2 g Cu(N 3 ) 2 /liter solution 147.6 g Cu(N 3 ) 2 1 mol