2. Chapter 16 Notes1 Chapter 16 Aqueous Equilibria: Applications 1. neutralization reactions 2. common ion effect; buffers 3. titrations strong acid - strong base; strong - weak; weak - weak; polyprotic acid - strong base 4. solubility equilibria review precipitation reactions and solubility rules; K sp & solubility; common ion effect; solubility & pH 5. complex ion formation K f ; solubility 6. selective precipitation; qualitative analysis - read only

3. Chapter 16 Notes2 1. neutralization reactions: In a neutralization reaction, an acid and a base react to form a salt and water. Consider a neutralization reaction. Does it proceed to completion? If so, why? What is the pH of the resulting solution? Examples: a. 0.01 M HCl(aq) mixed with 0.01 M NaOH(aq) predominant species: H 1+, Cl 1-, Na 1+, OH 1-, H 2 O direct reaction(s): H 1+ (aq)+OH 1- (aq)  H 2 O(l) K=1/K w =1x10 14 The rxn goes to completion; formation of water is driving force; “new” predominant species H 2 O; pH=7

4. Chapter 16 Notes3 Examples: b. 0.01 M NH 3 (aq) mixed with 0.01 M HClO 4 (aq) predominant species: H 1+, ClO 4 1-, NH 3, H 2 O direct reactions: NH 3 + H 2 O  NH 4 1+ +OH 1- K=K b =1.8x10 -5 H 1+ +OH 1-  H 2 OK=1/K w =1x10 14 net: NH 3 + H 1+  NH 4 1+ K=1.8x10 9 The rxn goes to completion; formation of water is driving force; new predominant species: NH 4 1+, H 2 O; pH<7

5. Chapter 16 Notes4 Examples: c. 0.01 M NH 3 (aq) mixed with 0.01 M HNO 2 (aq) predominant species: HNO 2, NH 3, H 2 O direct reactions: NH 3 + H 2 O  NH 4 1+ +OH 1- K=K b =1.8x10 -5 HNO 2  H 1+ +NO 2 1- K=K a =4.5x10 -4 H 1+ +OH 1-  H 2 OK=1/K w =1x10 14 net: NH 3 + HNO 2  NH 4 1+ + NO 2 1- K=8.1x10 5 The rxn goes mostly to completion; formation of water is driving force (compare to previous reaction); “new” predominant species: NH 4 1+, H 2 O, NO 2 1- ; pH<7 (K a of NH 4 1+ is larger than K b of NO 2 1- )

6. Chapter 16 Notes5 Does a neutralization reaction proceed to completion? yes exception: very weak acid reacts with weak base (or vice versa) What drives a neutralization reaction to completion? formation of water Is the resulting solution neutral (pH=7)? only if a strong acid and strong base react !Treat neutralization reactions as stoichiometry problems!

7. Chapter 16 Notes6 2. common ion effect The percent ionization of a 0.50 M acetic acid (K a =1.8x10 -5 ) solution is 0.60%; the pH of the solution is 2.37. Is the percent ionization of acetic acid in a solution of 0.50 M acetic acid with added 0.50 M acetate ion higher than, lower than, or equal to 0.60%? Is the pH of a solution of 0.50 M acetic acid with 0.50 M added acetate ion higher than, lower than, or equal to 2.37? Calculate the percent ionization and pH of a 0.50 M solution of acetic acid that is also 0.50 M in sodium acetate. Define common ion effect:

8. Chapter 16 Notes7 Figure 16.2

9. Chapter 16 Notes8 2. buffers Consider a solution that is 0.50 M acetic acid and 0.50 M sodium acetate; the solution is at equilibrium. a. The pH of the solution is : b. What happens when 0.010 moles HCl are added (without changing the volume)? list predominant species: H 1+, Cl 1-, HC 2 H 3 O 2, Na 1+, C 2 H 3 O 2 1-, H 2 O look for neutralization reactions (direct reactions) with large driving force list “new” predominant species: write equilibrium reaction, and calculate pH

10. Chapter 16 Notes9 2. buffers Consider a solution that is 0.50 M acetic acid and 0.50 M sodium acetate; the solution is at equilibrium. a. The pH of the solution is : b. What happens when 0.010 moles NaOH are added (without changing the volume)?

11. Chapter 16 Notes10 Consider: 0.50 M acetic acid and 0.50 M sodium acetate (the buffer) vs. water a. What is the pH change when 0.010 moles strong acid are added to the buffer? b. What is the pH change when 0.010 moles strong base are added to the buffer? c. What is the pH change when 0.010 moles strong acid are added to the water? d. What is the pH change when 0.010 moles strong base are added to the water? Define buffer

12. Chapter 16 Notes11

13. Chapter 16 Notes12 Henderson-Hasselbalch equation in the lab this equation can lead to the incorrect pH note the different forms for the acid and the base! the equation is good for helping determine the appropriate buffer - match the desired pH with pK a of available acids

14. Chapter 16 Notes13 Question: Choose the best acid to give a buffer solution with pH=4.30. a. chloroacetic acid, K a =1.35x10 -3 b. propanoic acid, K a =1.3x10 -5 c. benzoic acid, K a =6.4x10 -5 d. hypochlorous acid, K a =3.5x10 -8