1. Aqueous Equilibria Entry Task: Feb 17 th Wednesday Notes on Precipitate and ions HW: Precipitate and ions ws MAYHAN

2. Aqueous Equilibria I can… Explain the factors that can affect the K sp of a substance Calculate the changes that occurs when these factors are applied to a solution MAYHAN

3. Aqueous Equilibria

4. Aqueous Equilibria Information from the Curve: There are several things you can read from the titration curve itself. Consider this titration curve. 1.What type of titration curve is above? Identify the titrant. 2. Place a dot (  ) on the curve at the equivalence point. The pH at the equivalence point is ____. Choose a good indicator for this titration from Figure 16.07 on page 604 of your textbook. Strong base into a weak acid. The titrant is the strong base 9 Phenolphthalein

5. Aqueous Equilibria Information from the Curve: There are several things you can read from the titration curve itself. Consider this titration curve. 3.What volume of base was used to titrate the acid solution? _______ mL 4.Place a box ( ) on the curve where the pH of the solution = the pK a of the acid. What is the pH at this point? _____ What is the pK a of the acid? _____ What is the K a of the acid? 25 4.8 1.6 x10 -5

6. Aqueous Equilibria Calculations knowing the Acid: 5.Hydrofluoric acid, HF, has a K a = 7.2 x 10 -4. Calculate the pH of 10.0 mL of a 0.050 M solution of HF. Plot this point on the axes. x 2 0.050M = 7.2 x 10 -4 x 2 = 3.6 x 10 -5 6.0 x 10 -3 = pH = 2.22

7. Aqueous Equilibria Calculations knowing the Acid: 6.A 0.020 M solution of NaOH is used for the titration. What volume will be needed to reach the equivalence point? (10 mls)(0.050M) = (x mls)(0.020M) = 25.0 mls

8. Aqueous Equilibria Calculations knowing the Acid: 7. Write the net reaction for the neutralization of a solution of HF with a solution of NaOH. HF + OH -  F - + H 2 O

9. Aqueous Equilibria Calculations knowing the Acid: 8.Calculate the moles of F - at the equivalence point. What is the total volume? _______ L The [F - ] at the equivalence point is _________ 10 mls (HF) + 25.0 mls (NaOH) = 35.0 mls 0.0350 (x)(35.0 ml) = (0.020M)(25.0ml) 0.0143 M

10. Aqueous Equilibria Calculations knowing the Acid: 9. Calculate the pH of the solution at the equivalence point. Use this information and the answer to question 6 to plot the equivalence point on your graph. Choose a good indicator for this titration from Figure 16.07 on page 604 of your textbook. (10.0 mls)(0.050M) = 0.50 mmol HF (25.0 mls)(0.020M) = 0.50 mmol NaOH The acid was neutralized but the c-base is left at the same amount of OH- which is 0.50 mmol. 0.50 mmol/35.0 ml = 0.0143 M X 2 0.0143 M = 1.389 x10 -11 Since its more base- change Ka to Kb 1.0 x10 -14 / 7.2 x10 -4 x 2 = 1.99 x10 -13 x= 4.46 x10 -7 6.35 - 14 = pH = 7.65

11. Aqueous Equilibria Calculations knowing the Acid: 9. Calculate the pH of the solution at the equivalence point. Use this information and the answer to question 6 to plot the equivalence point on your graph. Choose a good indicator for this titration from Figure 16.07 on page 604 of your textbook. 0.0143 M Bromthymol blue

12. Aqueous Equilibria Calculations knowing the Acid: 10. What is the pH halfway to the equivalence point? Plot this point on your graph. pH is pKa = 3.14 at 12.5 mls

13. Aqueous Equilibria Calculations knowing the Acid: 11. How many moles of HF are in the original 10.0 mL sample of HF? _______ (0.010 L)(0.050mol/L) = 0.00050 moles of HF

14. Aqueous Equilibria Calculations knowing the Acid: 12. When only 5.0 mL of 0.020 M NaOH has been added, calculate the moles of HF left and F - produced. (0.0050 L)(0.020 mol/L) = 0.00010 moles of OH- HF OH - H2OH2OF-F- i ------- c e 0.00050 0.00010 -0- -0.00010 +0.00010 0.00040 0 0.00010 0.00040 moles of HF 0.015Liters = 0.0267 M of HF

15. Aqueous Equilibria Calculations knowing the Acid: 12. When only 5.0 mL of 0.020 M NaOH has been added, calculate the moles of HF left and F - produced. (0.0050 L)(0.020 mol/L) = 0.00010 moles of OH- HF OH - H2OH2OF-F- i ------- c e 0.00050 0.00010 -0- -0.00010 +0.00010 0.00040 0 0.00010 0.00010 moles of F 0.015Liters = 0.00667 M of F -

16. Aqueous Equilibria

17. Aqueous Equilibria

18. Aqueous Equilibria

19. Aqueous Equilibria

20. Aqueous Equilibria

21. Aqueous Equilibria Factors Affecting Solubility pH  If a substance has a basic anion, it is more soluble in an acidic solution.  Substances with acidic cations are more soluble in basic solutions. MAYHAN

22. Aqueous Equilibria Factors Affecting Solubility Complex Ions  The formation of these complex ions increases the solubility of these salts. MAYHAN

23. Aqueous Equilibria Factors Affecting Solubility Amphoterism  Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.  Examples of such cations are Al 3+, Zn 2+, and Sn 2+. MAYHAN

24. Aqueous Equilibria I can… Explain the ion product and how its value will affect the equilibrium. Predict whether a precipitate will form when two ions are mixed in solution MAYHAN

25. Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Sections 6 MAYHAN

26. Aqueous Equilibria 17.6- Precipitation and Separation of Ions Up till now we have placed ionic substances in water so it can dissociate until its saturated and reaches equilibrium. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Equilibrium can also be reached by mixing two solutions containing the ions to create a precipitate MAYHAN BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq)

27. Aqueous Equilibria 17.6- Precipitation and Separation of Ions By mixing BaCl 2 and Na 2 SO 4 together will create a precipitate of BaSO 4 if the product of the ion concentrations, Q = [Ba 2+ ][SO 4 2- ], greater than K sp. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Q is referred to simply as the ion product MAYHAN

28. Aqueous Equilibria Will a Precipitate Form? In a solution,  If Q = K sp, the system is at equilibrium and the solution is saturated.  If Q < K sp, more solid will dissolve until Q = K sp.  If Q > K sp, the salt will precipitate until Q = K sp. MAYHAN

29. Aqueous Equilibria 17.6- Precipitation and Separation of Ions SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? What precipitate COULD be made?_______________ Look up the Ksp in Appendix D:_________________ To determine if PbSO 4 will precipitate, we have to calculate the ion product, Q=[Pb +2 ][SO 4 -2 ] and compare it with Ksp. PbSO 4 6.3 x10 -7 MAYHAN

30. Aqueous Equilibria SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? In 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 there are: (0.10 L) (8.0  10  3 ) = 8.0 x 10 -4 moles of Pb +2 ions MAYHAN

31. Aqueous Equilibria SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? In 0.40 L of 5.0  10  3 M Na 2 SO 4 there are: (0.40 L) (5.0  10  3 ) = 2.0 x 10 -3 moles of SO 4 -2 ions MAYHAN

32. Aqueous Equilibria SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? We have to convert the moles in to molarity but use the combined volume. 2.0 x 10 -3 moles/0.50L= 8.0 x 10 -4 moles/0.50L = 1.6 x10 -3 of Pb +2 ions 4.0 x10 -3 of SO 4 -2 ions Substitute the values into the Ksp expression and solve for Q MAYHAN

33. Aqueous Equilibria SAMPLE EXERCISE 17.15 Will a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? (1.6 x10 -3 )(4.0 x10 -3 ) = Q = [Pb +2 ][SO 4 2  ] 6.4  10  6 Q= 6.4  10  6 Ksp= 6.3 x10 -7 Q is larger than Ksp that means A precipitate will occur MAYHAN

34. Aqueous Equilibria Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 What precipitate COULD be made?___________ Look up the K sp in Appendix D:_______________ CaF 2 3.9 x10 -11 MAYHAN

35. Aqueous Equilibria Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 In 0.050 L of 2.0  10  2 M NaF there are: (0.050 L) (2.0  10  2 ) = 1.0 x 10 -3 moles of F -1 ions MAYHAN

36. Aqueous Equilibria Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 ? In 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 there are: (0.010 L) (1.0  10  2 ) = 1.0 x 10 -4 moles of Ca +2 ions MAYHAN

37. Aqueous Equilibria Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 We have to convert the moles in to molarity but use the combined volume. 1.0 x 10 -3 moles/0.060L= 1.0 x 10 -4 moles/0.060L = 1.67 x10 -3 of Ca +2 ions 1.67 x10 -2 of F -1 ions Substitute the values into the Ksp expression and solve for Q MAYHAN

38. Aqueous Equilibria Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 (1.67 x10 -3 )(1.67 x10 -2 ) 2 = Q = [Ca +2 ][F 1  ] 2 4.7  10  7 Q= 4.7  10  7 Ksp= 3.9 x10 -11 Q is larger than Ksp that means A precipitate will occur MAYHAN

39. Aqueous Equilibria Consider a solution containing Ag + ions and Cu +2 ions then HCl is added, AgCl precipitate will form and the CuCl 2 is soluble and remain as ions in solution. Separation of ions in an aquesous solution by using a reagent that forms a precipatate with one or few of the ions is called selective precipatation MAYHAN

40. Aqueous Equilibria A solution contains 1.0  10  2 M Ag + and 2.0  10  2 M Pb 2+. When Cl  is added, both AgCl (K sp = 1.8  10  10 ) and PbCl 2 (K sp = 1.7  10  5 M) can precipitate. What concentration of Cl  is necessary to begin the precipitation of each salt? Which salt precipitates first? We know that both Ag + and Pb +2 would form a precipitate with Cl- but which will form first? MAYHAN

41. Aqueous Equilibria A solution contains 1.0  10  2 M Ag + and 2.0  10  2 M Pb 2+. When Cl  is added, both AgCl (K sp = 1.8  10  10 ) and PbCl 2 (K sp = 1.7  10  5 M) can precipitate. What concentration of Cl  is necessary to begin the precipitation of each salt? Which salt precipitates first? Lets look at Ag + with CI - : Ksp = [Ag + ][Cl - ] 1.8 x10 -10 = (1.0 x10 -2 )(x) = Cl- ions 1.8 x10 -10 = 1.0 x10 -2 1.8 x 10 -8 Cl- ions MAYHAN

42. Aqueous Equilibria A solution contains 1.0  10  2 M Ag + and 2.0  10  2 M Pb 2+. When Cl  is added, both AgCl (K sp = 1.8  10  10 ) and PbCl 2 (K sp = 1.7  10  5 M) can precipitate. What concentration of Cl  is necessary to begin the precipitation of each salt? Which salt precipitates first? Lets look at Pb +2 with CI - : Ksp = [Pb + ][Cl - ] 2 1.7 x10 -5 = (2.0 x 10 -2 )(x) 2 = Cl- ions 1.7 x10 -5 = 2.0 x 10 -2 x 2 =8.5 x 10 -4 Cl- ions x=2.9 x 10 -2 Cl- ions MAYHAN

43. Aqueous Equilibria A solution contains 1.0  10  2 M Ag + and 2.0  10  2 M Pb 2+. When Cl  is added, both AgCl (K sp = 1.8  10  10 ) and PbCl 2 (K sp = 1.7  10  5 M) can precipitate. What concentration of Cl  is necessary to begin the precipitation of each salt? Which salt precipitates first? Which concentration is smaller? 2.9 x 10 -2 Cl- ions with Pb +2 1.8 x 10 -8 Cl- ions with Ag+ This means that it will AgCl precipitate as such a small concentration verses PbCl 2. MAYHAN

44. Aqueous Equilibria A solution consists of 0.050 M Mg 2+ and Cu 2+. Which ion precipitates first as OH  is added? What concentration of OH  is necessary to begin the precipitation of each cation? [K sp = 1.8  10  11 for Mg(OH) 2, and K sp = 4.8  10  20 for Cu(OH) 2.] Lets look at Mg +2 with OH - : Ksp = [Mg +2 ][OH - ] 2 1.8 x 10 -11 = (0.050)(x) 2 = OH- ions 1.8 x 10 -11 = 0.050 x 2 = 3.6 x 10 -10 x= 1.9 x 10 -5 OH- ions

45. Aqueous Equilibria A solution consists of 0.050 M Mg 2+ and Cu 2+. Which ion precipitates first as OH  is added? What concentration of OH  is necessary to begin the precipitation of each cation? [K sp = 1.8  10  11 for Mg(OH) 2, and K sp = 4.8  10  20 for Cu(OH) 2.] Lets look at Cu +2 with OH - : Ksp = [Cu +2 ][OH - ] 2 4.8 x 10 -20 = (0.050)(x) 2 = OH- ions 4.8 x 10 -20 = 0.050 x 2 = 9.6 x 10 -19 x= 9.8 x 10 -10 OH- ions

46. Aqueous Equilibria A solution consists of 0.050 M Mg 2+ and Cu 2+. Which ion precipitates first as OH  is added? What concentration of OH  is necessary to begin the precipitation of each cation? [K sp = 1.8  10  11 for Mg(OH) 2, and K sp = 4.8  10  20 for Cu(OH) 2. 9.8 x 10 -10 OH- ions with Cu +2 1.9 x 10 -5 OH- ions with Mg +2 Which concentration is smaller? This means that it will Cu(OH) 2 precipitate as such a small concentration verses Mg(OH) 2. MAYHAN

47. Aqueous Equilibria MAYHAN