Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 17: Acids and Bases

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 2 of 47 Contents 17-1The Arrhenius Theory: A Brief Review 17-2Brønsted-Lowry Theory of Acids and Bases 17-3The Self-Ionization of Water and the pH Scale 17-4Strong Acids and Strong Bases 17-5Weak Acids and Weak Bases 17-6Polyprotic Acids 17-7Ions as Acids and Bases 17-8Molecular Structure and Acid-Base Behavior 17-8Lewis Acids and Bases Focus On Acid Rain.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 3 of The Arrhenius Theory: A Brief Review HCl(g) → H + (aq) + Cl - (aq) NaOH(s) → Na + (aq) + OH - (aq) H2OH2O H2OH2O Na + (aq) + OH - (aq) + H + (aq) + Cl - (aq) → H 2 O(l) + Na + (aq) + Cl - (aq) H + (aq) + OH - (aq) → H 2 O(l) Arrhenius theory did not handle non OH - bases such as ammonia very well.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 4 of Brønsted-Lowry Theory of Acids and Bases An acid is a proton donor. A base is a proton acceptor. NH 3 + H 2 O  NH OH - NH OH -  NH 3 + H 2 O baseacid base acid conjugate acid conjugate base

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 5 of 47 Base Ionization Constant NH 3 + H 2 O  NH OH - Kc=Kc= [NH 3 ][H 2 O] [NH 4 + ][OH - ] K b = K c [H 2 O] = [NH 3 ] [NH 4 + ][OH - ] = 1.8  baseacid conjugate acid conjugate base

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 6 of 47 Acid Ionization Constant CH 3 CO 2 H + H 2 O  CH 3 CO H 3 O + Kc=Kc= [CH 3 CO 2 H][H 2 O] [CH 3 CO 2 - ][H 3 O + ] K a = K c [H 2 O] = = 1.8  [CH 3 CO 2 H] [CH 3 CO 2 - ][H 3 O + ] baseacid conjugate acid conjugate base

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 7 of 47 Table 17.1 Relative Strengths of Some Brønsted-Lowry Acids and Bases HClO 4 + H 2 O  ClO H 3 O + NH CO 3 2-  NH 3 + HCO 3 - HCl + OH -  Cl - + H 2 O H 2 O + I -  OH - + HI

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 8 of The Self-Ionization of Water and the pH Scale

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 9 of 47 Ion Product of Water Kc=Kc= [H 2 O][H 2 O] [H 3 O + ][OH - ] H 2 O + H 2 O  H 3 O + + OH - baseacid conjugate acid conjugate base K W = K c [H 2 O][H 2 O] = = 1.0  [H 3 O + ][OH - ]

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 10 of 47 pH and pOH The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H + ]. pH = -log[H 3 O + ]pOH = -log[OH - ] -logK W = -log[H 3 O + ]-log[OH - ]= -log(1.0  ) K W = [H 3 O + ][OH - ]= 1.0  pK W = pH + pOH= -(-14) pK W = pH + pOH = 14

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 11 of 47 pH and pOH Scales

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 12 of Strong Acids and Bases HCl CH 3 CO 2 H Thymol Blue Indicator pH < 1.2 < pH < 2.8 < pH

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 13 of Weak Acids and Bases Acetic AcidHC 2 H 3 O 2 or CH 3 CO 2 H

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 14 of 47 Weak Acids Ka=Ka= = 1.8  [CH 3 CO 2 H] [CH 3 CO 2 - ][H 3 O + ] pK a = -log(1.8  ) = 4.74 glycine H 2 NCH 2 CO 2 H lactic acid CH 3 CH(OH) CO 2 H C OH O R

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 15 of 47 Table 17.3 Ionization Constants of Weak Acids and Bases

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 16 of 47 Example 17-5 Determining a Value of K A from the pH of a Solution of a Weak Acid. Butyric acid, HC 4 H 7 O 2 (or CH 3 CH 2 CH 2 CO 2 H) is used to make compounds employed in artificial flavorings and syrups. A M aqueous solution of HC 4 H 7 O 2 is found to have a pH of Determine K A for butyric acid. HC 4 H 7 O 2 + H 2 O  C 4 H 77 O 2 + H 3 O + K a = ? Solution: For HC 4 H 7 O 2 K A is likely to be much larger than K W. Therefore assume self-ionization of water is unimportant.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 17 of 47 Example 17-5 HC 4 H 7 O 2 + H 2 O  C 4 H 7 O 2 + H 3 O + Initial conc M00 Changes-x M+x M+x M Eqlbrm conc.(0.250-x) M x Mx M

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 18 of 47 Example 17-5 Log[H 3 O + ] = -pH = HC 4 H 7 O 2 + H 2 O  C 4 H 77 O 2 + H 3 O + [H 3 O + ] = = 1.9  = x [H 3 O + ] [C 4 H 7 O 2 - ] [HC 4 H 7 O 2 ] Ka=Ka= 1.9  · 1.9  (0.250 – 1.9  ) = K a = 1.5  Check assumption: K a >> K W.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 19 of 47 Percent Ionization HA + H 2 O  H 3 O + + A - Degree of ionization = [H 3 O + ] from HA [HA] originally Percent ionization = [H 3 O + ] from HA [HA] originally  100%

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 20 of 47 Percent Ionization K a = [H 3 O + ][A - ] [HA] K a = n H3O+H3O+ A-A- n HA n 1 V

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 21 of Polyprotic Acids H 3 PO 4 + H 2 O  H 3 O + + H 2 PO 4 - H 2 PO H 2 O  H 3 O + + HPO 4 2- HPO H 2 O  H 3 O + + PO 4 3- Phosphoric acid: A triprotic acid. K a = 7.1  K a = 6.3  K a = 4.2 

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 22 of 47 Phosphoric Acid K a1 >> K a2 All H 3 O + is formed in the first ionization step. H 2 PO 4 - essentially does not ionize further. Assume [H 2 PO 4 - ] = [H 3 O + ]. [HPO 4 2- ]  K a2 regardless of solution molarity.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 23 of 47 Table 17.4 Ionization Constants of Some Polyprotic Acids

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 24 of 47 Example 17-9 Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H 3 PO 4 solution, calculate: (a) [H 3 O + ]; (b) [H 2 PO 4 - ]; (c) [HPO 4 2- ] (d) [PO 4 3- ] H 3 PO 4 + H 2 O  H 2 PO H 3 O + Initial conc.3.0 M00 Changes-x M+x M+x M Eqlbrm conc.(3.0-x) M x Mx M

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 25 of 47 Example 17-9 H 3 PO 4 + H 2 O  H 2 PO H 3 O + [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] Ka=Ka= x · x (3.0 – x) = Assume that x << 3.0 = 7.1  x 2 = (3.0)(7.1  ) x = 0.14 M [H 2 PO 4 - ] = [H 3 O + ] = 0.14 M

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 26 of 47 Example 17-9 H 2 PO H 2 O  HPO H 3 O + [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] Ka=Ka= y · ( y) ( y) = = 6.3  Initial conc.0.14 M00.14 M Changes-y M+y M+y M Eqlbrm conc.( y) M y M(0.14 +y) M y << 0.14 M y = [HPO 4 2- ] = 6.3  10 -8

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 27 of 47 Example 17-9 HPO H 2 O  PO H 3 O + [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] Ka=Ka= (0.14)[PO 4 3- ] 6.3  = = 4.2  M [PO 4 3- ] = 1.9  M

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 28 of 47 Sulfuric Acid Sulfuric acid: A diprotic acid. H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - HSO H 2 O  H 3 O + + SO 4 2- K a = very large K a = 1.96

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 29 of 47 General Approach to Solution Equilibrium Calculations Identify species present in any significant amounts in solution (excluding H 2 O). Write equations that include these species. –Number of equations = number of unknowns. Equilibrium constant expressions. Material balance equations. Electroneutrality condition. Solve the system of equations for the unknowns.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 30 of Ions as Acids and Bases NH H 2 O  NH 3 + H 3 O + base acid CH 3 CO H 2 O  CH 3 CO 2 H + OH - base acid [NH 3 ] [H 3 O + ] [OH - ] Ka=Ka= [NH 4 + ] [OH - ] [NH 3 ] [H 3 O + ] Ka=Ka= [NH 4 + ] = ? = KWKW KbKb = 1.0   = 5.6  K a K b = K w

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 31 of 47 Hydrolysis Water (hydro) causing cleavage (lysis) of a bond. Na + + H 2 O → Na + + H 2 O NH H 2 O → NH 3 + H 3 O + Cl - + H 2 O → Cl - + H 2 O No reaction Hydrolysis

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 32 of Molecular Structure and Acid-Base Behavior Why is HCl a strong acid, but HF is a weak one? Why is CH 3 CO 2 H a stronger acid than CH 3 CH 2 OH? There is a relationship between molecular structure and acid strength. Bond dissociation energies are measured in the gas phase and not in solution.

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 33 of 47 Strengths of Binary Acids HIHBrHClHF 160.9>141.4>127.4>91.7 pm 297<368<431<569 kJ/mol Bond length Bond energy 10 9 >10 8 >1.3  10 6 >> 6.6  Acid strength HF + H 2 O → [F - ·····H 3 O + ]  F - + H 3 O + ion pair H-bonding free ions

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 34 of 47 Strengths of Oxoacids Factors promoting electron withdrawal from the OH bond to the oxygen atom: –High electronegativity (EN) of the central atom. –A large number of terminal O atoms in the molecule. H-O-ClH-O-Br EN Cl = 3.0EN Br = 2.8 K a = 2.9  K a = 2.1  10 -9

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 35 of 47 S O O O O HH ·· - 2+ ·· - S O O O HH - + S O O O O HH S O O O HH K a  10 3 K a =1.3  10 -2

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 36 of 47 Strengths of Organic Acids C O C O HH ·· H H O C HH H H C H H K a = 1.8  K a =1.3  acetic acidethanol

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 37 of 47 Focus on the Anions Formed O C H ·· H H C H H C O C O H - H H C O C O H - H H -

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 38 of 47 Structural Effects C H H C O C O H - ·· H H C H H H C O O - C H H C H H C H H C H H C H H K a = 1.8  K a = 1.3  10 -5

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 39 of 47 Structural Effects C O C O H - ·· H H K a = 1.8  K a = 1.4  C O C O H - ·· H Cl ··

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 40 of 47 Strengths of Amines as Bases N H H H ·· N Br H H ·· pK b = 4.74pK a = 7.61 ammoniabromamine

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 41 of 47 Strengths of Amines as Bases C H H H C H H C H H H C H H C H H H C H H pK b = 4.74pK a = 3.38pK b = 3.37 methylamineethylaminepropylamine NH 2

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 42 of 47 Resonance Effects

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 43 of 47 Inductive Effects

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 44 of Lewis Acids and Bases Lewis Acid –A species (atom, ion or molecule) that is an electron pair acceptor. Lewis Base –A species that is an electron pair donor. baseacid adduct

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 45 of 47 Showing Electron Movement

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 46 of 47 Focus On Acid Rain CO 2 + H 2 O  H 2 CO 3 H 2 CO 3 + H 2 O  HCO H 3 O + 3 NO 2 + H 2 O  2 HNO 3 + NO

Prentice-Hall © 2002General Chemistry: Chapter 17Slide 47 of 47 Chapter 17 Questions Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding. Choose a variety of problems from the text as examples. Practice good techniques and get coaching from people who have been here before.