Chapter 15 AcidBase Equilibria Section 15 1 Solutions

Chapter 15 Acid-Base Equilibria

Section 15. 1 Solutions of Acids or Bases Containing a Common Ion Effect § Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. § An application of Le Châtelier’s principle. Copyright © Cengage Learning. All rights reserved 2

Section 15. 1 Solutions of Acids or Bases Containing a Common Ion Example HCN(aq) + H 2 O(l) H 3 O+(aq) + CN-(aq) § Addition of Na. CN will shift the equilibrium to the left because of the addition of CN-, which is already involved in the equilibrium reaction. § A solution of HCN and Na. CN is less acidic than a solution of HCN alone.

Section 15. 2 Buffered Solutions Key Points about Buffered Solutions § Buffered Solution – resists a change in p. H. § They are weak acids or bases containing a common ion. § After addition of strong acid or base, deal with stoichiometry first, then the equilibrium. Copyright © Cengage Learning. All rights reserved 4

Section 15. 2 Buffered Solutions Adding an Acid to a Buffer To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 5

Section 15. 2 Buffered Solutions Buffers To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 6

Section 15. 2 Buffered Solutions Solving Problems with Buffered Solutions Copyright © Cengage Learning.

Section 15. 2 Buffered Solutions Buffering: How Does It Work? Copyright © Cengage Learning.

Section 15. 2 Buffered Solutions Henderson–Hasselbalch Equation § For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same p. H. Copyright © Cengage Learning. All rights reserved 10

Section 15. 2 Buffered Solutions EXERCISE! What is the p. H of a buffer solution that is 0. 45 M acetic acid (HC 2 H 3 O 2) and 0. 85 M sodium acetate (Na. C 2 H 3 O 2)? The Ka for acetic acid is 1. 8 × 10– 5. p. H = 5. 02 Copyright © Cengage Learning. All rights reserved 11

Section 15. 2 Buffered Solutions Copyright © Cengage Learning. All rights reserved 12

Section 15. 2 Buffered Solutions Buffered Solution Characteristics § Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base. § Added H+ reacts to completion with the weak base. § Added OH- reacts to completion with the weak acid. Copyright © Cengage Learning. All rights reserved 13

Section 15. 2 Buffered Solutions Buffered Solution Characteristics § The p. H in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the p. H will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A– or B and BH+) are large compared with amounts of H+ or OH– added. Copyright © Cengage Learning. All rights reserved 14

Section 15. 3 Buffering Capacity § The amount of protons or hydroxide ions the buffer can absorb without a significant change in p. H. § Determined by the magnitudes of [HA] and [A–]. § A buffer with large capacity contains large concentrations of the buffering components. Copyright © Cengage Learning. All rights reserved 15

Section 15. 3 Buffering Capacity § Optimal buffering occurs when [HA] is equal to [A–]. § It is for this condition that the ratio [A–] / [HA] is most resistant to change when H+ or OH– is added to the buffered solution. Copyright © Cengage Learning. All rights reserved 16

Section 15. 3 Buffering Capacity Choosing a Buffer § p. Ka of the weak acid to be used in the buffer should be as close as possible to the desired p. H. Copyright © Cengage Learning. All rights reserved 17

Section 15. 4 Titrations and p. H Curves Titration Curve § Plotting the p. H of the solution being analyzed as a function of the amount of titrant added. § Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. Copyright © Cengage Learning. All rights reserved 18

Section 15. 4 Titrations and p. H Curves Neutralization of a Strong Acid with a Strong Base To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 19

Section 15. 4 Titrations and p. H Curves The p. H Curve for the

Section 15. 4 Titrations and p. H Curves Weak Acid–Strong Base Titration Step 1: Step 2: A stoichiometry problem (reaction is assumed to run to completion) then determine concentration of acid remaining and conjugate base formed. An equilibrium problem (determine position of weak acid equilibrium and calculate p. H). Copyright © Cengage Learning. All rights reserved 22

Section 15. 4 Titrations and p. H Curves CONCEPT CHECK! Consider a solution made by mixing 0. 10 mol of HCN (Ka = 6. 2 × 10– 10) with 0. 040 mol Na. OH in 1. 0 L of aqueous solution. What are the major species immediately upon mixing (that is, before a reaction)? HCN, Na+, OH–, H 2 O Copyright © Cengage Learning. All rights reserved 23

Section 15. 4 Titrations and p. H Curves Let’s Think About It… § Why isn’t Na. OH a major species? § Why aren’t H+ and CN– major species? § List all possibilities for the dominant reaction. Copyright © Cengage Learning. All rights reserved 24

Section 15. 4 Titrations and p. H Curves Let’s Think About It… The possibilities for the dominant reaction are: 1. 2. 3. 4. 5. H 2 O(l) + H 2 O(l) H 3 O+(aq) + OH–(aq) HCN(aq) + H 2 O(l) H 3 O+(aq) + CN–(aq) HCN(aq) + OH–(aq) CN–(aq) + H 2 O(l) Na+(aq) + OH–(aq) Na. OH Na+(aq) + H 2 O(l) Na. OH + H+(aq) Copyright © Cengage Learning. All rights reserved 25

Section 15. 4 Titrations and p. H Curves Let’s Think About It… § How do we decide which reaction controls the p. H? H 2 O(l) + H 2 O(l) H 3 O+(aq) + OH–(aq) HCN(aq) + H 2 O(l) H 3 O+(aq) + CN–(aq) HCN(aq) + OH–(aq) CN–(aq) + H 2 O(l)

Section 15. 4 Titrations and p. H Curves Let’s Think About It… HCN(aq) + OH–(aq) CN–(aq) + H 2 O(l) § What are the major species after this reaction occurs? HCN, CN–, H 2 O, Na+ Copyright © Cengage Learning. All rights reserved 27

Section 15. 4 Titrations and p. H Curves Let’s Think About It… § Now you can treat this situation as before. § List the possibilities for the dominant reaction. § Determine which controls the p. H. Copyright © Cengage Learning. All rights reserved 28

Section 15. 4 Titrations and p. H Curves CONCEPT CHECK! Calculate the p. H of a solution made by mixing 0. 20 mol HC 2 H 3 O 2 (Ka = 1. 8 × 10– 5) with 0. 030 mol Na. OH in 1. 0 L of aqueous solution. Copyright © Cengage Learning. All rights reserved 29

Section 15. 4 Titrations and p. H Curves Let’s Think About It… § What are the major species in solution? Na+, OH–, HC 2 H 3 O 2, H 2 O § Why isn’t Na. OH a major species? § Why aren’t H+ and C 2 H 3 O 2– major species? Copyright © Cengage Learning. All rights reserved 30

Section 15. 4 Titrations and p. H Curves Let’s Think About It… § What are the possibilities for the dominant reaction? 1. 2. 3. 4. 5. H 2 O(l) + H 2 O(l) H 3 O+(aq) + OH–(aq) HC 2 H 3 O 2(aq) + H 2 O(l) H 3 O+(aq) + C 2 H 3 O 2–(aq) HC 2 H 3 O 2(aq) + OH–(aq) C 2 H 3 O 2–(aq) + H 2 O(l) Na+(aq) + OH–(aq) Na. OH(aq) Na+(aq) + H 2 O(l) Na. OH + H+(aq) § Which of these reactions really occur? Copyright © Cengage Learning. All rights reserved 31

Section 15. 4 Titrations and p. H Curves Let’s Think About It… § Which reaction controls the p. H? H 2 O(l) + H 2 O(l) H 3 O+(aq) + OH–(aq) HC 2 H 3 O 2(aq) + H 2 O(l) H 3 O+(aq) + C 2 H 3 O 2–(aq) HC 2 H 3 O 2(aq) + OH–(aq) C 2 H 3 O 2–(aq) + H 2 O(l) § How do you know? Copyright © Cengage Learning. All rights reserved 32

Section 15. 4 Titrations and p. H Curves Let’s Think About It… HC 2 H 3 O 2(aq) + OH– Before Change 0. 20 mol C 2 H 3 O 2–(aq) + H 2 O 0. 030 mol – 0. 030 mol After 0. 17 mol 0 0 +0. 030 mol K = 1. 8 × 109 Copyright © Cengage Learning. All rights reserved 33

Section 15. 4 Titrations and p. H Curves Steps Toward Solving for p. H HC 2 H 3 O 2(aq) + H 2 O Initial Change Equilibrium H 3 O+ + C 2 H 3 O 2 -(aq) 0. 170 M ~0 0. 030 M –x +x +x 0. 170 – x x 0. 030 + x Ka = 1. 8 × 10– 5 p. H = 3. 99 Copyright © Cengage Learning. All rights reserved 34

Section 15. 4 Titrations and p. H Curves EXERCISE! Calculate the p. H of a 100. 0 m. L solution of 0. 100 M acetic acid (HC 2 H 3 O 2), which has a Ka value of 1. 8 × 10– 5. p. H = 2. 87 Copyright © Cengage Learning. All rights reserved 35

Section 15. 4 Titrations and p. H Curves CONCEPT CHECK! Calculate the p. H of a solution made by mixing 100. 0 m. L of a 0. 100 M solution of acetic acid (HC 2 H 3 O 2), which has a Ka value of 1. 8 × 10– 5, and 50. 0 m. L of a 0. 10 M Na. OH solution. p. H = 4. 74 Copyright © Cengage Learning. All rights reserved 36

Section 15. 4 Titrations and p. H Curves CONCEPT CHECK! Calculate the p. H of a solution at the equivalence point when 100. 0 m. L of a 0. 100 M solution of acetic acid (HC 2 H 3 O 2), which has a Ka value of 1. 8 × 10– 5, is titrated with a 0. 10 M Na. OH solution. p. H = 8. 72 Copyright © Cengage Learning. All rights reserved 37

Section 15. 4 Titrations and p. H Curves The p. H Curves for the Titrations of 50. 0 -m. L Samples of 0. 10 M Acids with Various Ka Values with 0. 10 M Na. OH Copyright © Cengage Learning. All rights reserved 39

Section 15. 5 Acid-Base Indicators § Marks the end point of a titration by changing color. § The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible). Copyright © Cengage Learning. All rights reserved 41