Chapter 16: Acids and Bases Chemistry 140 Fall 2002 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci • Harwood • Herring • Madura Chapter 16: Acids and Bases Juana Mendenhall, Ph.D. Assistant Professor Morehouse College General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Objectives Calculate one of Ka, [H+] or [H3O+], or the molarity of a weak acid, given the other two (doing the same for a weak base). Become familiar with percent ionization Calculate the strength of acids and bases General Chemistry: Chapter 16 Prentice-Hall © 2007

General Approach to Solution Equilibrium Calculations Chemistry 140 Fall 2002 General Approach to Solution Equilibrium Calculations Identify species present in any significant amounts in solution (excluding H2O). Write equations that include these species. Write the Ionization Equation Arrange in tabular form and emphasize:. Initial Conditions Changes that occur, & Equilibrium concentrations Substitute equilibrium concentrations into the ionization constant expression. Solve for the quantity of interest. General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Example A 0.101 M solution of HN3 has pH = 2.86. What is Ka for this acid? [H3O+][N3-] HN3(aq) + H2O(aq) H3O+(aq) + N3-(aq) Ka = [HN3] 1. Next, [H3O+] = 10-pH = 10-2.86 = 1.4 x 10-3 M 2. Make chart HN3(aq) + H2O(aq) H3O+(aq) + N3-(aq) I: 0.101 M +0.0014 M +0.0014 M C: -0.0014 M 0.0014 M 0.0014 M E: 0.100 M [H3O+][N3-] (0.0014)(0.0014) Ka = = = 2.0 x 10-5 M [HN3] 0.100 General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 EXAMPLE 16-9 Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H3PO4 solution, calculate: (a) [H3O+]; (b) [H2PO4-]; (c) [HPO42-] (d) [PO43-] H3PO4 + H2O H2PO4- + H3O+ Initial conc. 3.0 M 0 0 Changes -x M +x M +x M Equilibrium (3.0-x) M x M x M Concentration General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 EXAMPLE 16-9 H3PO4 + H2O H2PO4- + H3O+ [H3O+] [H2PO4-] x · x Ka= = = 7.110-3 [H3PO4] (3.0 – x) Assume that x << 3.0 x2 = (3.0)(7.110-3) x = 0.14 M [H2PO4-] = [H3O+] = 0.14 M General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 EXAMPLE 16-9 H2PO4- + H2O HPO42- + H3O+ Initial conc. 0.14 M 0 0.14 M Changes -y M +y M +y M Equilibrium (0.14 - y) M y M (0.14 +y) M Concentration [H3O+] [HPO42-] [H2PO4-] Ka= y · (0.14 + y) (0.14 - y) = = 6.310-8 y << 0.14 M y = [HPO42-] = 6.310-8 General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 EXAMPLE 16-9 HPO4- + H2O PO43- + H3O+ [H3O+] [HPO42-] (0.14)[PO43-] = 4.210-13 M Ka= = [H2PO4-] 6.310-8 [PO43-] = 1.910-19 M General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Percent Ionization HA + H2O H3O+ + A- Degree of ionization = [H3O+] from HA [HA] originally Percent ionization = [H3O+] from HA [HA] originally  100% General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 % Ionization Example A student prepared a 0.10 M solution of formic acid, HCHO2, and measured its pH using a pH meter. The pH at 25 °C was found to be 2.38. Calculate the Ka for formic acid at this temperature. What percentage of the acid is ionized in this 0.10 M solution? Answer HCHO2(aq) H+(aq) + CHO2-(aq) General Chemistry: Chapter 16 Prentice-Hall © 2007

16-7 Ions as Acids and Bases Chemistry 140 Fall 2002 16-7 Ions as Acids and Bases CH3CO2- + H2O CH3CO2H + OH- base acid [NH3] [H3O+] Ka= [NH4+] = ? NH4+ + H2O NH3 + H3O+ acid base [NH3] [H3O+] [OH-] Ka= [NH4+] [OH-] = KW Kb = 1.010-14 1.810-5 = 5.610-10 Ka Kb = Kw General Chemistry: Chapter 16 Prentice-Hall © 2007

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Hydrolysis Water (hydro) causing cleavage (lysis) of a bond. Na+ + H2O → Na+ + H2O No reaction Cl- + H2O → Cl- + H2O No reaction NH4+ + H2O → NH3 + H3O+ Hydrolysis General Chemistry: Chapter 16 Prentice-Hall © 2007

16-8 Molecular Structure and Acid-Base Behavior Chemistry 140 Fall 2002 16-8 Molecular Structure and Acid-Base Behavior Why is HCl a strong acid, but HF is a weak one? Why is CH3CO2H a stronger acid than CH3CH2OH? There is a relationship between molecular structure and acid strength. Bond dissociation energies are measured in the gas phase and not in solution. Strength depends on properties of the solvent and temperature. General Chemistry: Chapter 16 Prentice-Hall © 2007

Strengths of Binary Acids Chemistry 140 Fall 2002 Strengths of Binary Acids HI HBr HCl HF Bond length 160.9 > 141.4 > 127.4 > 91.7 pm Bond energy 297 < 368 < 431 < 569 kJ/mol Acid strength 109 > 108 > 1.3106 >> 6.610-4 More energy is required to break the H—F bone, HF is the weak acid. HI has the lowest bond energy, so HI is the strongest acid of the group. The polarity of the bond actually decreases from HF to HI b/c F is the most electonegative of the halogens. HF + H2O → [F-·····H3O+] F- + H3O+ ion pair H-bonding free ions General Chemistry: Chapter 16 Prentice-Hall © 2007