Chapter 14: Aqueous Equilibria: Acids and Bases

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Presentation transcript:

Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 Copyright © 2008 Pearson Prentice Hall, Inc.

14.11 Polyprotic Acids Acids that contains more than one dissociable proton Dissociate in a stepwise manner Each dissociation step has its own Ka Stepwise dissociation constants decreases in the order Ka1 > Ka2 > Ka3 More difficult to remove a positively charge proton from negative ion

14.11 Polyprotic Acids Diprotic acid solutions contain a mixture of acids: H2A, HA, H2O Strongest acid – HA Principle reaction – dissociation of H2A All of H3O+ come from the first ionization H2CO3(aq) + H2O(l) H3O+( aq) + HCO3-(aq) Ka1 = 4.3 x 10-7 HCO3-(aq) + H2O(l) H3O+( aq) + CO32-(aq) Ka2 = 4.8 x 10-11

Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 Polyprotic Acids Copyright © 2008 Pearson Prentice Hall, Inc.

14.12 Equilibria in Solutions of Weak Bases Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 14.12 Equilibria in Solutions of Weak Bases BH1+(aq) + OH1-(aq) B(aq) + H2O(l) Base Acid Acid Base [BH1+][OH1-] [B] Kb = Base-Dissociation Constant: NH41+(aq) + OH1-(aq) NH3(aq) + H2O(l) [NH41+][OH1-] [NH3] Kb = Copyright © 2008 Pearson Prentice Hall, Inc.

Equilibria in Solutions of Weak Bases Chemistry: McMurry and Fay, 6th Edition Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 4:44:26 AM Equilibria in Solutions of Weak Bases Copyright © 2011 Pearson Prentice Hall, Inc.

Equilibria in Solutions of Weak Bases Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 Equilibria in Solutions of Weak Bases Calculate the [-OH] and pH of a 0.40 M NH3 solution. At 25 °C, Kb = 1.8 x 10-5. While the initial hydroxide ion concentration is not 0, it’s insignificant when compared to that which comes from the dissociation of NH3. Copyright © 2008 Pearson Prentice Hall, Inc.

Example Morphine (C17H19NO3), a narcotic used in painkillers, is a weak organic base. If the pH of a 7.0 x 10-4 M solution of morphine is 9.5, what is the value of Kb?

14.13 Relation Between Ka and Kb Chapter 14: Aqueous Equilibria: Acids and Bases 14.13 Relation Between Ka and Kb 4/14/2018 H3O1+(aq) + NH3(aq) NH41+(aq) + H2O(l) Ka NH41+(aq) + OH1-(aq) NH3(aq) + H2O(l) Kb H3O1+(aq) + OH1-(aq) 2H2O(l) Kw [H3O1+][NH3] [NH41+][OH1-] Ka x Kb = x = [H3O1+][OH1-] = Kw [NH41+] [NH3] = (5.6 x 10-10)(1.8 x 10-5) = 1.0 x 10-14 Copyright © 2008 Pearson Prentice Hall, Inc.

Relation Between Ka and Kb Chapter 14: Aqueous Equilibria: Acids and Bases Relation Between Ka and Kb 4/14/2018 Ka x Kb = Kw conjugate acid-base pair Kb Kw Ka Kw Ka = Kb = pKa + pKb = pKw = 14.00 Copyright © 2008 Pearson Prentice Hall, Inc.

Examples Calculate Kb for CN- (Ka = 4.9 x 10-10) Ka for HOCl = 3.5 x 10-8, find Kb and pKb

14.14 Acid-Base Properties of Salts pH of a salt solution is determined by the acid-base properties of the consistuent cations and anions In an acid-base reaction, the influence of the stronger partner is predominant Strong acid + Strong Base  Neutral solution Strong acid + Weak Base  Basis solution Weak acid + Strong Base  Acidic solution

Neutral Salt A salt of a strong base and a strong acid. E.g NaCl Neutral cation + neutral anion  neutral salt Na+ Cl- NaCl Na+(aq) + H2O(l)  NR Cl –(aq) + H2O(l)  NR

Basic Salts A salt of a strong base and a weak acid. E.g NaCN Neutral cation + basic anion  basic salt Na+ CN- NaCN Na+(aq) + H2O(l)  NR CN-(aq) + H2O(l)  HCN(aq) + -OH(aq)

Acidic Salts A salt of a weak base and a strong acid. E.g NH4Cl Acidic cation + neutral anion  Acidic salt NH4+ + Cl- NH4Cl NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) Cl –(aq) + H2O(l)  NR

Acid-Base Properties of Salts A salt of a weak base and a weak acid Acidic cation + basic anion  (50 :50 mixture) must compare Ka and Kb Ka > Kb: The solution will contain an excess of H3O1+ ions (pH < 7). Ka < Kb: The solution will contain an excess of OH1- ions (pH > 7). Ka = Kb: The solution will contain approximately equal concentrations of H3O1+ and OH1- ions (pH ≈ 7).

Examples Calculate Ka for the cation, and Kb for the anion in an aqueous NH4CN solution. Is the solution acidic, basic or neutral? Write the hydrolysis reaction of the salt (Kb for NH3 = 1.8 x 10-5, Ka for HCN = 4.9 x 10-10)

Acid-Base Properties of Salts Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 Acid-Base Properties of Salts Salts That Yield Acidic Solutions Hydrated cations of small, highly charged metal ions, such as Al3+. The cations listed previously are slightly acidic even though we ignore them at this level. Acidity increases as the charge increases and also as the ion size decreases. Copyright © 2008 Pearson Prentice Hall, Inc.

Acid-Base Properties of Salts Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 Acid-Base Properties of Salts Copyright © 2008 Pearson Prentice Hall, Inc.

Example Predict the acidity of this salt solution and calculate the pH of a 0.10M solution of sodium fluoride (NaF) at 25oC. Ka = 7.1 x 10-4

Example Find the pH of a 0.100 M NaCHO2 solution. The salt completely dissociate into Na+(aq) and CHO2-(aq) and Na+ ion has no acid or base properties. Ka (HCHO2)= 1.8 x 10-4

Example What is the pH of a 0.30M solution for methylammonium chloride, CH3NH3Cl? Kb = 4.4 x 10-4

Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 14.16 Lewis Acids and Bases Lewis Acid: An electron-pair acceptor. Include cations and neutral molecule having vacant valence orbitals that can accept a share in a pair of electrons from a Lewis Base Lewis Base: An electron-pair donor. All Lewis bases are Bronsted-Lowry bases Copyright © 2008 Pearson Prentice Hall, Inc.

Lewis Acids and Bases Lewis Base Lewis Acid Chemistry: McMurry and Fay, 6th Edition Chapter 14: Aqueous Equilibria: Acids and Bases Lewis Acids and Bases 4/14/2018 4:44:26 AM Lewis Base Lewis Acid Copyright © 2011 Pearson Prentice Hall, Inc.

Chapter 14: Aqueous Equilibria: Acids and Bases 4/14/2018 Lewis Acids and Bases Copyright © 2008 Pearson Prentice Hall, Inc.

Examples For each of the following reactions, identify the Lewis acid and the Lewis base CO2(g) + -OH(aq) HCO3-(aq) B(OH)3(aq) + H2O(l) B(OH)4-(aq) + H+ (aq)