1. Chapter 15 Applications of Aqueous Equilibria

2. Catalyst Derive the Henderson Hasselbalch equation! DON’T LOOK AT YOUR NOTES

3. Solutions of Acids or Bases Containing a Common Ion We will talk about solutions that contain HA AND it’s conjugate base NaA Suppose we have a solution of HF and NaF (remember salts fully dissociate) ◦ Step 1: Identify MAJOR SPECIES  HF, Na +, F -, H 2 O (F is the common ion)

4. Solutions of Acids or Bases Containing a Common Ion Let’s Compare 2 solutions: ◦ 0.1 M HF solution ◦ 0.1 M HF solution + 0.1 M NaF ◦ Step 1: Identify MAJOR SPECIES ◦ Step II: Write out the equations How will LeChatelier’s Principle apply?

5. Solutions of Acids or Bases Containing a Common Ion Let’s Compare 2 solutions: ◦ 0.1 M HF solution ◦ 0.1 M HF solution + 0.1 M NaF Common Ion Effect: The equilibrium position of HF will shift because the F - is already in solution! …so the pH with NaF will be higher! (less acidic)

6. Example Problem The equilibrium concentration of H + in a 1.0 M HF solution is 2.7 x 10 -2 M and the % dissociation is 2.7%. Calculate the [H + ] and the % dissociation of HF in a solution containing 1.0 M HF (K a = 7.2 x 10 -4 ) and 1.0 M NaF

7. Buffered Solutions What does it mean to have a buffer?

8. Buffered Solutions A buffered solution is one that resists change in its pH when either OH - or H + ions are added. Example: Our blood – it can absorb acids and bases produced in our bodily reactions – but it must maintain a balanced pH to keep our cells alive!

9. Buffered Solutions A buffered solution is one that resists change in its pH when either OH - or H + ions are added. A buffered solution may contain a WEAK ACID and it’s SALT (HF and NaF) OR a WEAK BASE and it’s SALT (NH 3 and NH 4 Cl)

10. Buffered Solutions A buffered solution is one that resists change in its pH when either OH - or H + ions are added. By choosing the correct components, a solution can resist change at almost any pH!

11. How does a buffered solution resist changes in pH when an acid or a base is added? By solving the next set of example problems, our goal is to answer the question:

12. REMEMBER your SYSTEMATIC approach! A buffered solution contains 0.5 M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.5 M Sodium Acetate. Calculate the pH of this solution.

13. REMEMBER your SYSTEMATIC approach! Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution described in the previous example. Compare the pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

14. Example 15.4 Calculate the pH of a solution containing 0.75 M lactic acid and 0.25 M sodium lactate.

15. Example 15.5 A buffered solution contains 0.25 M ammonia and 0.40 M ammonium chloride. Calculate the pH of the solution

16. Adding strong acid to a buffered solution Calculate the pH of the solution that results when 0.10 mol gaseous HCl is aded to 1.0 L of a buffered solution that contains 0.25 M ammonia and 0.40 M ammonium chloride (15.5)

17. SO HOW DO WE KNOW WHEN WE’VE MET THE EQUIVALENCE POINT IN A TITRATION?

18. 2 ways to tell equivalence point 1. Use a pH meter 2. Use an indicator that changes color at the end point (equivalence point).

19. What is an indicator? It is a weak acid (HIn) that changes color when the H + leaves, leaving an (In - ion) Let’s try a problem to see how they function… ◦ Assume you have some hypothetical indicator HIn, Ka = 1.0 x 10 -8 ◦ Let’s write the equation… ◦ Write Ka expression… ◦ What if we add this indicator to a solution with a pH of 1.0? ◦ What color will it be? ◦ What if we add OH?...eventually…what?

20. When is the color visible for acidic solution?

21. Choosing the Appropriate indicator Ex. 15.11)

22. Use H-H equation to determine what pH will allow the indicator to change color!

23. When is the color visible for BASIC solution?

24. Figure 15.8!!!! All indicator ranges!

25. Solubility Equilibria What does it mean to be soluble? If something is NOT soluble…what will you see in the solution? Solubility product constant or solubility product = Ksp Table 15.4

26. Ex 15.12) Calculating K sp from Solubility I pg. 718 Copper (I) Bromide has a measured solubility of 2.0 x 10 -4 mol/L at 25 °C. Calculate its K sp value.

27. Ex 15.13) Calculating K sp from Solubility II pg. 719 Calculating K sp value for bismuth sulfide, which has a solubility of 1.0 x 10 -15 mol/L at 25 °C.

28. Ex 15.14) Calculating Solubility from K sp pg.720 The K sp for copper (II) iodate, Cu(IO 3 ) 2, is 1.4 x 10 -7 at 25 °C. Calculate its solubility at 25 °C.

29. Ex 15.15) Solubility and Common Ions pg. 723 Calculate the solubility of solid CaF 2 (K sp = 4.0 x 10 -11 ) in a 0.025 M NaF solution.

30. Catalyst Turn in Prelab questions Answer the following: ◦ What is the Kinetic Molecular Theory? ◦ Write the Solubility Rules ◦ Write the strong acids ◦ Write the strong bases

31. Precipitation Precipitation and Qualitative Analysis What is Q? How do we calculate it again? For precipitation predictions: ◦ Q < K : no precipitation ◦ Q > K : precipitation will occur

32. Ex 15.16) Determining Precipitation Conditions pg. 725 A solution is prepared by adding 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 to 300.00 mL of 2.00 x 10 -2 M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 x 10 -10 ) precipitated from this solution?

33. Ex 15.17 A solution is prepared by mixing 150.0 mL of 1.00 x 10 -2 M Mn(NO 3 ) 2 and 250.0 mL of 1.0 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 ).

34. Selective Precipitation Ex. 15.18) A solution contains 1.0 x 10 -4 M Cu + and 2.0 x 10 -3 M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (K sp = 1.4 x 10 -8 ) or CuI (K sp = 5.3 x 10 -12 ) precipitate first? Specify the concentration of I - necessary to begin precipitation of each salt.

35. Complete Qualitative Analysis on your own! This is how Selective Precipitation is used in the lab!

36. Complex Ions form Coordination Complex’s These are metals surrounded by ligands (Lewis Base) Common Ligands: ◦ H 2 O, NH 3, Cl -, CN - Metal Ions add ligands one at a time…in a stepwise fashion: 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

37. Complex Ion Equilibria Ex 15.19) Complex Ions Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) 2 3- in a solution prepared by mixing 150.0 mL of 1.00 x 10 -3 M AgNO 3 with 200.0 mL of 5.00 M Na 2 S 2 O 3. The stepwise formation equilibria are: 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