Chapter 15 Acid–Base Equilibria. Chapter 15 Table of Contents Copyright © Cengage Learning. All rights reserved 2 15.1Solutions of Acids or Bases Containing.

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Presentation transcript:

Chapter 15 Acid–Base Equilibria

Chapter 15 Table of Contents Copyright © Cengage Learning. All rights reserved Solutions of Acids or Bases Containing a Common Ion 15.2 Buffered Solutions 15.3 Buffering Capacity 15.4Titrations and pH Curves 15.5 Acid–Base Indicators

Section 15.1 Solutions of Acids or Bases Containing a Common Ion Return to TOC Copyright © Cengage Learning. All rights reserved 3 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.

Section 15.1 Solutions of Acids or Bases Containing a Common Ion Return to TOC Copyright © Cengage Learning. All rights reserved 4 Example HCN(aq) + H 2 O(l) H 3 O + (aq) + CN - (aq) Addition of NaCN 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 NaCN is less acidic than a solution of HCN alone.

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 5 Key Points about Buffered Solutions Buffered Solution – resists a change in pH. They are weak acids or bases containing a common ion. After addition of strong acid or base, deal with stoichiometry first, then the equilibrium.

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 6 Adding an Acid to a Buffer

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 7 Buffers

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 8 Solving Problems with Buffered Solutions

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 9 Buffering: How Does It Work?

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 10 Buffering: How Does It Work?

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 11 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 pH.

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 12 Exercise What is the pH of a buffer solution that is 0.45 M acetic acid (HC 2 H 3 O 2 ) and 0.85 M sodium acetate (NaC 2 H 3 O 2 )? The K a for acetic acid is 1.8 × 10 –5. pH = 5.02

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 13

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 14 Buffered Solution Characteristics Buffers contain relatively large amounts of weak acid and corresponding conjugate base. Added H + reacts to completion with the conjugate base. Added OH  reacts to completion with the weak acid.

Section 15.2 Atomic MassesBuffered Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 15 Buffered Solution Characteristics The pH 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 pH 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.

Section 15.3 The MoleBuffering Capacity Return to TOC Copyright © Cengage Learning. All rights reserved 16 The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. Determined by the magnitudes of [HA] and [A – ]. A buffer with large capacity contains large concentrations of the buffering components.

Section 15.3 The MoleBuffering Capacity Return to TOC Copyright © Cengage Learning. All rights reserved 17 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.

Section 15.3 The MoleBuffering Capacity Return to TOC Copyright © Cengage Learning. All rights reserved 18 Choosing a Buffer pK a of the weak acid to be used in the buffer should be as close as possible to the desired pH.

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 19 Titration Curve Plotting the pH 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.

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 20 Neutralization of a Strong Acid with a Strong Base

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 21 The pH Curve for the Titration of 50.0 mL of M HNO 3 with M NaOH

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 22 The pH Curve for the Titration of mL of 0.50 M NaOH with 1.0 M HCI

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 23 Weak Acid–Strong Base Titration Step 1:A stoichiometry problem (reaction is assumed to run to completion) then determine remaining species. Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate pH).

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check Consider a solution made by mixing 0.10 mol of HCN (K a = 6.2 x 10 –10 ) with mol NaOH 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

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 25 Let’s Think About It… Why isn’t NaOH a major species? Why aren’t H + and CN – major species? List all possibilities for the dominant reaction.

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 26 Let’s Think About It… The possibilities for the dominant reaction are: 1.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 2.HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) 3.HCN(aq) + OH – (aq) CN – (aq) + H 2 O(l) 4.Na + (aq) + OH – (aq) NaOH 5.Na + (aq) + H 2 O(l) NaOH + H + (aq)

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 27 Let’s Think About It… How do we decide which reaction controls the pH? 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 pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 28 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 +

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 29 Let’s Think About It… Now you can treat this situation as before. List the possibilities for the dominant reaction. Determine which controls the pH.

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 30 Concept Check Calculate the pH of a solution made by mixing 0.20 mol HC 2 H 3 O 2 (K a = 1.8 x 10 –5 ) with mol NaOH in 1.0 L of aqueous solution.

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 31 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 NaOH a major species? Why aren’t H + and C 2 H 3 O 2 – major species?

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 32 Let’s Think About It… What are the possibilities for the dominant reaction? 1.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 2.HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 – (aq) 3.HC 2 H 3 O 2 (aq) + OH – (aq) C 2 H 3 O 2 – (aq) + H 2 O(l) 4.Na + (aq) + OH – (aq) NaOH 5.Na + (aq) + H 2 O(l) NaOH + H + (aq) Which of these reactions really occur?

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 33 Let’s Think About It… Which reaction controls the pH? 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?

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 34 Let’s Think About It… K = 1.8 x 10 9 HC 2 H 3 O 2 (aq) +OH –  C 2 H 3 O 2 – (aq)+ H 2 O Before0.20 mol mol0 Change–0.030 mol mol After0.17 mol mol

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 35 Steps Toward Solving for pH K a = 1.8 x 10 –5 pH = 3.99 HC 2 H 3 O 2 (aq) +H2OH2OH3O+H3O+ + C 2 H 3 O 2 - (aq) Initial M~ M Change –x+x Equilibrium – xx x

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 36 Exercise Calculate the pH of a mL solution of M acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 x 10 –5. pH = 2.87

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 37 Concept Check Calculate the pH of a solution made by mixing mL of a M solution of acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 x 10 –5, and 50.0 mL of a 0.10 M NaOH solution. pH = 4.74

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 38 Concept Check Calculate the pH of a solution at the equivalence point when mL of a M solution of acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 x 10 –5, is titrated with a 0.10 M NaOH solution. pH = 8.72

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 39 The pH Curve for the Titration of 50.0 mL of M HC 2 H 3 O 2 with M NaOH

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 40 The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various K a Values with 0.10 M NaOH

Section 15.4 Titrations and pH Curves Return to TOC Copyright © Cengage Learning. All rights reserved 41 The pH Curve for the Titration of 100.0mL of M NH 3 with 0.10 M HCl

Section 15.5 Acid–Base Indicators Return to TOC Copyright © Cengage Learning. All rights reserved 42 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).

Section 15.5 Acid–Base Indicators Return to TOC Copyright © Cengage Learning. All rights reserved 43 The Acid and Base Forms of the Indicator Phenolphthalein

Section 15.5 Acid–Base Indicators Return to TOC Copyright © Cengage Learning. All rights reserved 44 The Methyl Orange Indicator is Yellow in Basic Solution and Red in Acidic Solution

Section 15.5 Acid–Base Indicators Return to TOC Copyright © Cengage Learning. All rights reserved 45 Useful pH Ranges for Several Common Indicators