Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 1 of 52 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 16: Acids and Bases

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 2 of 52 Contents 16-1The Arrhenius Theory: A Brief Review 16-2Brønsted-Lowry Theory of Acids and Bases 16-3The Self-Ionization of Water and the pH Scale 16-4Strong Acids and Strong Bases 16-5Weak Acids and Weak Bases 16-6Polyprotic Acids 16-7Ions as Acids and Bases 16-8Molecular Structure and Acid-Base Behaviour

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 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 © 2007 General Chemistry: Chapter 16 Slide 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 baseacid conjugate acid conjugate base ??

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 5 of 52 The Solvated Proton

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 6 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 7 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 8 of 52

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 9 of 52 A Weak Base

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 10 of 52 A Weak Acid

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 11 of 52 A Strong Acid

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 12 of The Self-Ionization of Water and the pH Scale

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 13 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 14 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 15 of 52 pH and pOH Scales

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 16 of Strong Acids and Bases HCl CH 3 CO 2 H Thymol Blue Indicator pH < 1.2 < pH < 2.8 < pH

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 17 of Weak Acids and Bases Lactic AcidGlycine General Chemistry: Chapter 16Prentice-Hall © 2007 Acetic Acid

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 18 of 52 Acetic Acid 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 General Chemistry: Chapter 16Prentice-Hall © 2007

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 19 of 52 Weak Bases Kb=Kb= = 4.3  [CH 3 NH 2 ] [CH 3 NH 3 + ][HO - ] pK b = -log(4.2  ) = 3.37

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 20 of 52 Table 16.3 Ionization Constants of Weak Acids and Bases

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 21 of 52 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 7 O 2 + H 3 O + K a = ? EXAMPLE 16-5

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 22 of 52 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 Equilibrium (0.250-x) M x Mx M Concentration EXAMPLE 16-5 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 © 2007 General Chemistry: Chapter 16 Slide 23 of 52 Log[H 3 O + ] = -pH = HC 4 H 7 O 2 + H 2 O C 4 H 7 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. EXAMPLE 16-5

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 24 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 25 of 52 Percent Ionization K a = [H 3 O + ][A - ] [HA] K a = n H3O+H3O+ A-A- n HA n 1 V

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 26 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 © 2007 General Chemistry: Chapter 16 Slide 27 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 28 of 52

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 29 of 52 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 Equilibrium(3.0-x) M x Mx M Concentration EXAMPLE 16-9

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 30 of 52 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 EXAMPLE 16-9

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 31 of 52 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 Equilibrium ( y) M y M(0.14 +y) M Concentration y << 0.14 M y = [HPO 4 2- ] = 6.3  EXAMPLE 16-9

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 32 of 52 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 EXAMPLE 16-9

Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 33 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 34 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 35 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 © 2007 General Chemistry: Chapter 16 Slide 36 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 37 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 © 2007 General Chemistry: Chapter 16 Slide 38 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 39 of 52 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 © 2007 General Chemistry: Chapter 16 Slide 40 of 52 Strengths of Oxoacids 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 © 2007 General Chemistry: Chapter 16 Slide 41 of 52 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