Chapter 17: Additional Aspects of Acid-Base Equilibria

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Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring.
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Presentation transcript:

Chapter 17: Additional Aspects of Acid-Base Equilibria Chemistry 140 Fall 2002 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci • Harwood • Herring • Madura Chapter 17: Additional Aspects of Acid-Base Equilibria Juana Mendenhall, Ph.D. Assistant Professor Morehouse College Feb 22 Lecture 2 General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Chemistry 140 Fall 2002 Objectives Explain how an acid-base indicator works to determine the equivalence point in a titration Calculate the pH values and plot the titration curve of a strong acid with a strong base, or strong base with a strong acid Calculate pH values and plot the titration curve of a weak acid with a s strong base or of a weak base with a strong acid. General Chemistry: Chapter 17 Prentice-Hall © 2007

Acid-Base Indicators Color of some substances depends on the pH. Chemistry 140 Fall 2002 Acid-Base Indicators Color of some substances depends on the pH. HIn + H2O In- + H3O+ In the acid form the color appears to be the acid color. In the base form the color appears to be the base color. Intermediate color is seen in between these two states. The complete color change occurs over about 2 pH units. General Chemistry: Chapter 17 Prentice-Hall © 2007

Indicator Colors and Ranges Chemistry 140 Fall 2002 Indicator Colors and Ranges Slide 27 of 45 General Chemistry: Chapter 17 Prentice-Hall © 2007 General Chemistry: Chapter 17 Prentice-Hall © 2007

Neutralization Reactions and Titration Curves Chemistry 140 Fall 2002 Neutralization Reactions and Titration Curves Equivalence point: The point in the reaction at which both acid and base have been consumed. Neither acid nor base is present in excess. End point: The point at which the indicator changes color. Titrant: The known solution added to the solution of unknown concentration. Titration Curve: The plot of pH vs. volume. General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Chemistry 140 Fall 2002 The millimole Typically: Volume of titrant added is less than 50 mL. Concentration of titrant is less than 1 mol/L. Titration uses less than 1/1000 mole of acid and base. L/1000 mol/1000 = M = L mol mL mmol http://michele.usc.edu/java/acidbase/acidbase.html General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Strong Acid with a Strong Base Chemistry 140 Fall 2002 Titration of a Strong Acid with a Strong Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Strong Acid with a Strong Base Chemistry 140 Fall 2002 Titration of a Strong Acid with a Strong Base The pH has a low value at the beginning. The pH changes slowly: until just before the equivalence point. The pH rises sharply: perhaps 6 units per 0.1 mL addition of titrant. The pH rises slowly again. Any Acid-Base Indicator will do. As long as color change occurs between pH 4 and 10. General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Chemistry 140 Fall 2002 Example (calc pH values & plot titration curve of a weak acid & strong base) Calculate the pH in the titration of 25.0 mL of 0.100 M CH3COOH by NaOH after the addition of the acid solution of a) 10.0mL of 0.100 M NaOH b) 25.0mL of 0.100 M NaOH, and c) 35.0mL of 0.100 M NaOH. General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Strong Base with a Strong Acid Chemistry 140 Fall 2002 Titration of a Strong Base with a Strong Acid General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Weak Acid with a Strong Base Chemistry 140 Fall 2002 Titration of a Weak Acid with a Strong Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Weak Acid with a Strong Base Chemistry 140 Fall 2002 Titration of a Weak Acid with a Strong Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Weak Polyprotic Acid Chemistry 140 Fall 2002 Titration of a Weak Polyprotic Acid NaOH NaOH H3PO4 H2PO4- HPO42-  PO43- General Chemistry: Chapter 17 Prentice-Hall © 2007

17-5 Solutions of Salts of Polyprotic Acids Chemistry 140 Fall 2002 17-5 Solutions of Salts of Polyprotic Acids The third equivalence point of phosphoric acid can only be reached in a strongly basic solution. The pH of this third equivalence point is not difficult to calculate. It corresponds to that of Na3PO4 (aq) and PO43- can ionize only as a base. PO43- + H2O → OH- + HPO42- Kb = Kw/Ka = 2.410-2 General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Chemistry 140 Fall 2002 EXAMPLE 17-9 Determining the pH of a Solution Containing the Anion (An-) of a Polyprotic Acid. Sodium phosphate, Na3PO4, is an ingredient of some preparations used to clean painted walls before they are repainted. What is the pH of 1.0 M Na3PO4? Kb = 2.410-2 PO43- + H2O → OH- + HPO42- Initial concs. 1.0 M 0 M 0 M Changes -x M +x M +x M Equilibrium (1.00 - x) M x M x M Concentration General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Chemistry 140 Fall 2002 EXAMPLE 17-9 [OH-] [HPO42-] [PO43-] Kb= x · x (1.00 - x) = = 2.410-2 x2 + 0.024x – 0.024 = 0 x = 0.14 M pOH = +0.85 pH = 13.15 It is more difficult to calculate the pH values of NaH2PO4 and Na2HPO4 because two equilibria must be considered simultaneously. General Chemistry: Chapter 17 Prentice-Hall © 2007

Concentrated Solutions of Polyprotic Acids Chemistry 140 Fall 2002 Concentrated Solutions of Polyprotic Acids For solutions that are reasonably concentrated (> 0.1 M) the pH values prove to be independent of solution concentrations. for H2PO4- pH = 0.5 (pKa1 + pKa2) = 0.5 (2.15 + 7.20) = 4.68 for HPO42- pH = 0.5 (pKa1 + pKa2) = 0.5 (7.20 + 12.38) = 9.79 General Chemistry: Chapter 17 Prentice-Hall © 2007

17-6 Acid-Base Equilibrium Calculations: A Summary Chemistry 140 Fall 2002 17-6 Acid-Base Equilibrium Calculations: A Summary Determine which species are potentially present in solution, and how large their concentrations are likely to be. Identify possible reactions between components and determine their stoichiometry. Identify which equilibrium equations apply to the particular situation and which are most significant. General Chemistry: Chapter 17 Prentice-Hall © 2007