Chapter 14 Acids and Bases

14.1 The Nature of Acids and Bases 14.2 Acid Strength 14.3 The pH Scale 14.4 Calculating the pH of Strong Acid Solutions 14.5 Calculating the pH of Weak Acid Solutions 14.6 Bases 14.7 Polyprotic Acids 14.8 Acid–Base Properties of Salts 14.9 The Effect of Structure on Acid–Base Properties 14.10 Acid–Base Properties of Oxides 14.11 The Lewis Acid–Base Model 14.12 Strategy for Solving Acid–Base Problems: A Summary Copyright © Cengage Learning. All rights reserved

Models of Acids and Bases Arrhenius: Acids produce H+ ions in solution, bases produce OH- ions. Brønsted–Lowry: Acids are proton (H+) donors, bases are proton acceptors. HCl + H2O  Cl- + H3O+ acid base Copyright © Cengage Learning. All rights reserved

Brønsted–Lowry Reaction Copyright © Cengage Learning. All rights reserved

HA(aq) + H2O(l) H3O+(aq) + A-(aq) Acid in Water HA(aq) + H2O(l) H3O+(aq) + A-(aq) acid base conjugate conjugate acid base Conjugate base is everything that remains of the acid molecule after a proton is lost. Conjugate acid is formed when the proton is transferred to the base. Copyright © Cengage Learning. All rights reserved

Lewis acid: electron pair acceptor Lewis base: electron pair donor Lewis Acids and Bases Lewis acid: electron pair acceptor Lewis base: electron pair donor Lewis acid Lewis base Copyright © Cengage Learning. All rights reserved

Three Models for Acids and Bases Copyright © Cengage Learning. All rights reserved

Acid Ionization Equilibrium Copyright © Cengage Learning. All rights reserved

Ionization equilibrium lies far to the right. Strong acid: Ionization equilibrium lies far to the right. Yields a weak conjugate base. See AP strong acids list! Weak acid: Ionization equilibrium lies far to the left. Weaker the acid, stronger its conjugate base. Copyright © Cengage Learning. All rights reserved

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Various Ways to Describe Acid Strength Copyright © Cengage Learning. All rights reserved

Water as an Acid and a Base Water is amphoteric: Behaves either as an acid or as a base. At 25°C: Kw = [H+][OH–] = 1.0 × 10–14 No matter what the solution contains, the product of [H+] and [OH–] must always equal 1.0 × 10–14 at 25°C. Copyright © Cengage Learning. All rights reserved

Three Possible Situations [H+] = [OH–]; neutral solution [H+] > [OH–]; acidic solution [H+] < [OH–]; basic solution Copyright © Cengage Learning. All rights reserved

Self-Ionization of Water Copyright © Cengage Learning. All rights reserved

HA(aq) + H2O(l) H3O+(aq) + A-(aq) Concept Check HA(aq) + H2O(l) H3O+(aq) + A-(aq) acid base conjugate conjugate acid base What is the equilibrium constant expression for an acid acting in water? Use the following equation to write K: HA(aq) + H2O  H3O+(aq) + A-(aq) Although you can also write this as HA(aq)  H+(aq) + A-(aq), have the students use the form above until they are used to the fact that water is acting like a base. K = [H3O+][A-] / HA Copyright © Cengage Learning. All rights reserved

A compact way to represent solution acidity. pH = –log[H+] A compact way to represent solution acidity. pH decreases as [H+] increases. Copyright © Cengage Learning. All rights reserved

Higher the pH, more basic. pH < 7; acidic pH Range pH = 7; neutral pH > 7; basic Higher the pH, more basic. pH < 7; acidic Lower the pH, more acidic. Copyright © Cengage Learning. All rights reserved

The pH Scale and pH Values of Some Common Substances Copyright © Cengage Learning. All rights reserved

Calculate the pH for each of the following solutions. Exercise Calculate the pH for each of the following solutions. 1.0 × 10–4 M H+ pH = 4.00 b) 0.040 M OH– pH = 12.60 pH = –log[H+] a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00 b) Kw = [H+][OH–] = 1.00 × 10–14 = [H+](0.040 M) = 2.5 × 10–13 M H+ pH = –log[H+] = –log(2.5 × 10–13 M) = 12.60 Copyright © Cengage Learning. All rights reserved

The pH of a solution is 5.85. What is the [H+] for this solution? Exercise The pH of a solution is 5.85. What is the [H+] for this solution? [H+] = 1.4 × 10–6 M [H+] = 10^–5.85 = 1.4 × 10–6 M Copyright © Cengage Learning. All rights reserved

–log Kw = –log[H+] – log[OH–] pKw = pH + pOH 14.00 = pH + pOH pH and pOH Recall: Kw = [H+][OH–] –log Kw = –log[H+] – log[OH–] pKw = pH + pOH 14.00 = pH + pOH Copyright © Cengage Learning. All rights reserved

Calculate the pOH for each of the following solutions. Exercise Calculate the pOH for each of the following solutions. 1.0 × 10–4 M H+ pOH = 10.00 b) 0.040 M OH– pOH = 1.40 pH = –log[H+] and pOH = –log[OH–] and 14.00 = pH + pOH a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00; So, 14.00 = 4.00 + pOH; pOH = 10.00 b) pOH = –log[OH–] = –log(0.040 M) = 1.40 Copyright © Cengage Learning. All rights reserved

The pH of a solution is 5.85. What is the [OH–] for this solution? Exercise The pH of a solution is 5.85. What is the [OH–] for this solution? [OH–] = 7.1 × 10–9 M pH = –log[H+] and pOH = –log[OH–] and 14.00 = pH + pOH 14.00 = 5.85 + pOH; pOH = 8.15 [OH–] = 10^–8.15 = 7.1 × 10–9 M Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 1.5 x 10–11 M solution of HCl. Concept Check Calculate the pH of a 1.5 x 10–11 M solution of HCl. pH = 7.00 pH = 7.00 Watch for student to say pH = -log(1.5 x 10-11) = 10.82. This answers neglects that water is a major species and it turns out to be the major contributor of H+. If students think about this problem and consider major species, they will answer this correctly. Major species are H+, Cl-, and H2O. Dominant reaction is: H2O + H2O  H3O+ (aq) + OH- (aq) Water is a major contributor. Thus , pH = 7.00. Copyright © Cengage Learning. All rights reserved

Equilibrium Constants for Weak Acids Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc] [H3O+] [OAc-] initial change equilib 1.00 0 0 -x +x +x 1.00-x x x

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 2. Write Ka expression This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: 10-5 or smaller is ok)

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 3. Solve Ka expression First assume x is very small because Ka is so small. Now we can more easily solve this approximate expression.

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 3. Solve Ka approximate expression x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37

Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO2H. HCO2H + H2O ↔ HCO2- + H3O+ Ka = 1.8 x 10-4 Approximate solution [H3O+] = 4.2 x 10-4 M, pH = 3.37 Exact Solution [H3O+] = [HCO2-] = 3.4 x 10-4 M [HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M pH = 3.47

Equilibrium Constants for Weak Bases Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7

Relation of Ka, Kb, [H3O+] and pH

Exercise Calculate the pH of a 0.50 M aqueous solution of the weak acid HF. (Ka = 7.2 x 10–4) Major Species: HF, H2O Possibilities for primary reaction: HF(aq) + H2O  H3O+(aq) + F-(aq) H2O + H2O  H3O+(aq) + OH-(aq) The first reaction is the primary reaction because the K value is significantly larger. Thus, it is the primary source of H3O+ (or H+). Use an ICE table, K expression, and pH = –log[H+] to solve for pH. pH = 1.72 Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH HF(aq) + H2O H3O+(aq) + F–(aq) Initial 0.50 M ~ 0 Change –x +x Equilibrium 0.50–x x See Slide 4. Ka = 7.2 x 10–4 pH = 1.72 Copyright © Cengage Learning. All rights reserved

Percent Dissociation (Ionization) For a given weak acid, the percent dissociation increases as the acid becomes more dilute. Copyright © Cengage Learning. All rights reserved

Calculate the Ka value for formic acid. Exercise A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water. Calculate the Ka value for formic acid. Ka = 1.8 x 10–4 Ka = 1.8 x 10-4 If 8.00 M of the acid is 0.47% ionized, then 0.038 M dissociates. HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) I 8.00 0 0 C -0.038 +0.038 +0.038 E 7.96 0.038 0.038 Copyright © Cengage Learning. All rights reserved

Arrhenius: bases produce OH– ions. Brønsted–Lowry: bases are proton acceptors. In a basic solution at 25°C, pH > 7. Ionic compounds containing OH- are generally considered strong bases. See strong bases list! pOH = –log[OH–] pH = 14.00 – pOH Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 1.0 x 10–3 M solution of sodium hydroxide. Concept Check Calculate the pH of a 1.0 x 10–3 M solution of sodium hydroxide. pH = 11.00 pH = 11.00 Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 1.0 x 10–3 M solution of calcium hydroxide. Concept Check Calculate the pH of a 1.0 x 10–3 M solution of calcium hydroxide. pH = 11.30 pH = 11.30 Students need to see that the concentration of the hydroxide ion is twice that of the calcium hydroxide. Copyright © Cengage Learning. All rights reserved

Equilibrium expression for weak bases uses Kb. CN–(aq) + H2O(l) HCN(aq) + OH–(aq) Copyright © Cengage Learning. All rights reserved

pH calculations for solutions of weak bases are very similar to those for weak acids. Kw = [H+][OH–] = 1.0 × 10–14 pOH = –log[OH–] pH = 14.00 – pOH Copyright © Cengage Learning. All rights reserved

Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O ↔ NH4+ + OH- Kb = 1.8 x 10-5 Step 1. Define equilibrium concs. in ICE table [NH3] [NH4+] [OH-] initial change equilib 0.010 0 0 -x +x +x 0.010 - x x x

Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 Step 2. Solve the equilibrium expression Assume x is small, so x = [OH-] = [NH4+] = 4.2 x 10-4 M and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M The approximation is valid !

Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 Step 3. Calculate pH [OH-] = 4.2 x 10-4 M so pOH = - log [OH-] = 3.37 Because pH + pOH = 14, pH = 10.63

Calculate the pH of a 2.0 M solution of ammonia (NH3). Concept Check Calculate the pH of a 2.0 M solution of ammonia (NH3). (Kb = 1.8 × 10–5) pH = 11.78 pOH = -log(0.0060); pH = 14.00 – 2.22 = 11.78 Copyright © Cengage Learning. All rights reserved

Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H+ (aq) + A- (aq) Ka A- (aq) + H2O (l) OH- (aq) + HA (aq) Kb H2O (l) H+ (aq) + OH- (aq) Kw KaKb = Kw Weak Acid and Its Conjugate Base Ka = Kw Kb Kb = Kw Ka 15.7

Acids that can furnish more than one proton. Always dissociates in a stepwise manner, one proton at a time. The conjugate base of the first dissociation equilibrium becomes the acid in the second step. For a typical weak polyprotic acid: Ka1 > Ka2 > Ka3 For a typical polyprotic acid in water, only the first dissociation step is important to pH. Copyright © Cengage Learning. All rights reserved

Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13 pH = 1.08 Exercise Calculate the pH of a 1.00 M solution of H3PO4. Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13 pH = 1.08 pH = 1.08 (if the students neglect "x", they will get 1.06) This is a good problem to discuss how to treat polyprotic acids. The students should understand why only the first dissociation reaction is important. Copyright © Cengage Learning. All rights reserved

AP always treats H2SO4 in this way: Sulfuric Acid AP always treats H2SO4 in this way: The first H+ is STRONG. Assume 100% ionization. The second H+ is WEAK. Use the Ka. Add the H+ Molarities together to find the pH. Copyright © Cengage Learning. All rights reserved

Types of Acid/Base Reactions: Summary

Salts Ionic compounds. When dissolved in water, break up into its ions (which can behave as acids or bases). Copyright © Cengage Learning. All rights reserved

The salt of a strong acid and a strong base gives a neutral solution. Salts The salt of a strong acid and a strong base gives a neutral solution. KCl, NaNO3 Copyright © Cengage Learning. All rights reserved

Use Kb when starting with base. Salts A basic solution is formed if the anion of the salt is the conjugate base of a weak acid. NaF, KC2H3O2 Kw = Ka × Kb Use Kb when starting with base. Copyright © Cengage Learning. All rights reserved

Use Ka when starting with acid. Salts An acidic solution is formed if the cation of the salt is the conjugate acid of a weak base. NH4Cl Kw = Ka × Kb Use Ka when starting with acid. Copyright © Cengage Learning. All rights reserved

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Qualitative Prediction of pH of Salt Solutions (from Weak Parents) Copyright © Cengage Learning. All rights reserved

Calculate the Kb values for: C2H3O2− and CN− Exercise HC2H3O2 Ka = 1.8 x 10-5 HCN Ka = 6.2 x 10-10 Calculate the Kb values for: C2H3O2− and CN− Kb (C2H3O2-) = 5.6 x 10-10 Kb (CN-) = 1.6 x 10-5 Kb (C2H3O2-) = 5.6 x 10-10 Kb (CN-) = 1.6 x 10-5 Copyright © Cengage Learning. All rights reserved

Arrange the following 1.0 M solutions from lowest to highest pH. Concept Check Arrange the following 1.0 M solutions from lowest to highest pH. HBr NaOH NH4Cl NaCN NH3 HCN NaCl HF Justify your answer. HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH The order is: HBr (strong acid), HF (Ka = 7.2 x 10-4), HCN (Ka = 6.2 x 10-10), NH4Cl (Ka = 5.6 x 10-10), NaCl (neutral), NaCN (Kb = 1.6 x 10-5), NH3 (Kb = 1.8 x 10-5), NaOH (strong base). Have the students use the Ka and Kb values to decide. They need not calculate the pH values to answer this question. Copyright © Cengage Learning. All rights reserved

Ka for HCN is 6.2 x 10–10. Exercise Calculate the pH of a 0.75 M aqueous solution of NaCN. Ka for HCN is 6.2 x 10–10. Major Species: Na+, CN-, H2O Possibilities: CN-(aq) + H2O  HCN(aq) + OH-(aq) H2O + H2O  H3O+(aq) + OH-(aq) The primary reaction is the first one, and the Kb value is 1.6 x 10-5. This is many times larger than the K value for the second reaction (Kw). pH = 11.54 Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH CN–(aq) + H2O HCN(aq) + OH–(aq) Initial 0.75 M ~ 0 Change –x +x Equilibrium 0.75–x x x can be safely neglected here. Kb = 1.6 x 10–5 pH = 11.54 Copyright © Cengage Learning. All rights reserved

Models of Acids and Bases Two factors for acidity in binary compounds: Bond Polarity (high is good) Bond Strength (low is good) Copyright © Cengage Learning. All rights reserved

Bond Strengths and Acid Strengths for Hydrogen Halides Copyright © Cengage Learning. All rights reserved

Contains the group H–O–X. Oxyacids Contains the group H–O–X. For a given series the acid strength increases with an increase in the number of oxygen atoms attached to the central atom. The greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule. Copyright © Cengage Learning. All rights reserved

Several Series of Oxyacids and Their Ka Values Copyright © Cengage Learning. All rights reserved

Acidic Oxides (Acid Anhydrides): O-X bond is strong and covalent. SO2, NO2, CO2 When H-O-X grouping is dissolved in water, the O-X bond will remain intact. It will be the polar and relatively weak H-O bond that will tend to break, releasing a proton. Copyright © Cengage Learning. All rights reserved

Basic Oxides (Basic Anhydrides): O-X bond is ionic. K2O, CaO If X has a very low electronegativity, the O-X bond will be ionic and subject to being broken in polar water, producing a basic solution. Copyright © Cengage Learning. All rights reserved

Chapter 15 Acid–Base Equilibria

15.1 Solutions of Acids or Bases Containing a Common Ion 15.2 Buffered Solutions 15.3 Buffering Capacity 15.4 Titrations and pH Curves 15.5 Acid–Base Indicators Copyright © Cengage Learning. All rights reserved

An application of Le Châtelier’s principle. Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application of Le Châtelier’s principle. Copyright © Cengage Learning. All rights reserved

HCN(aq) + H2O(l) H3O+(aq) + CN-(aq) Example HCN(aq) + H2O(l) H3O+(aq) + CN-(aq) Addition of NaCN will shift the equilibrium to the left because of the addition of CN-, which is already involved in the equilibrium reaction. A solution of HCN and NaCN is less acidic than a solution of HCN alone. Copyright © Cengage Learning. All rights reserved

Key Points about Buffered Solutions Buffered Solution – resists a change in pH. They are weak acids or bases containing a salt with common ion. After addition of strong acid or base, deal with stoichiometry first, then the equilibrium. Copyright © Cengage Learning. All rights reserved

Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na2CO3/NaHCO3 (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HCl is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is it conjugate acid buffer solution 16.3

Adding an Acid to a Buffer Copyright © Cengage Learning. All rights reserved

Buffers Copyright © Cengage Learning. All rights reserved

Solving Problems with Buffered Solutions Copyright © Cengage Learning. All rights reserved

Henderson–Hasselbalch Equation For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same pH. Copyright © Cengage Learning. All rights reserved

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Ka for HCOOH = 1.8 x 10 -4 x = 1.038 X 10 -4 [H+] [HCOO-] Ka = [HCOOH] pH = 3.98 16.2

Henderson-Hasselbach equation OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA. NaA (s) Na+ (aq) + A- (aq) Ka = [H+][A-] [HA] pKa = -log Ka HA (aq) H+ (aq) + A- (aq) [H+] = Ka [HA] [A-] Henderson-Hasselbach equation -log [H+] = -log Ka - log [HA] [A-] pH = pKa + log [conjugate base] [acid] -log [H+] = -log Ka + log [A-] [HA] pH = pKa + log [A-] [HA] 16.2

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x pH = pKa + log [HCOO-] [HCOOH] Common ion effect 0.30 – x  0.30 pH = 3.77 + log [0.52] [0.30] = 4.01 0.52 + x  0.52 HCOOH pKa = 3.77 16.2

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+] [OH-] [NH3] Kb = = 1.8 X 10-5 0.36 Initial 0.30 + x + x Change - x End 0.30 - x 0.36 + x x (.36 + x)(x) (.30 – x) 1.8 X 10-5 = 0.36x 0.30 1.8 X 10-5  x = 1.5 X 10-5 pOH = 4.82 pH= 9.18 16.3

final volume = 80.0 mL + 20.0 mL = 100 mL Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? final volume = 80.0 mL + 20.0 mL = 100 mL NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M start (M) 0.29 0.01 0.24 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (M) 0.28 0.0 0.25 [H+] [NH3] [NH4+] Ka= = 5.6 X 10-10 [H+] = 6.27 X 10 -10 [H+] 0.25 0.28 pH = 9.20 = 5.6 X 10-10 16.3

NH4+ (aq) H+ (aq) + NH3 (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH4+ (aq) H+ (aq) + NH3 (aq) pH = pKa + log [NH3] [NH4+] pH = 9.25 + log [0.30] [0.36] pKa = 9.25 = 9.17 final volume = 80.0 mL + 20.0 mL = 100 mL start (M) 0.29 0.01 0.24 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (M) 0.28 0.0 0.25 pH = 9.25 + log [0.25] [0.28] = 9.20 16.3

Buffered Solution Characteristics Buffers contain relatively large amounts of weak acid and corresponding conjugate base. Added H+ reacts to completion with the conjugate base. Added OH- reacts to completion with the weak acid. Copyright © Cengage Learning. All rights reserved

Buffered Solution Characteristics The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A– or B and BH+) are large compared with amounts of H+ or OH– added. Copyright © Cengage Learning. All rights reserved

Determined by the magnitudes of [HA] and [A–]. The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. Determined by the magnitudes of [HA] and [A–]. A buffer with large capacity contains large concentrations of the buffering components. Copyright © Cengage Learning. All rights reserved

Optimal buffering occurs when [HA] is equal to [A–]. It is for this condition that the ratio [A–] / [HA] is most resistant to change when H+ or OH– is added to the buffered solution. Copyright © Cengage Learning. All rights reserved

Titration Curve Plotting the pH of the solution being analyzed as a function of the amount of titrant added. Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. Copyright © Cengage Learning. All rights reserved

Neutralization of a Strong Acid with a Strong Base Copyright © Cengage Learning. All rights reserved

The pH Curve for the Titration of 50. 0 mL of 0. 200 M HNO3 with 0 The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH Copyright © Cengage Learning. All rights reserved

The pH Curve for the Titration of 100. 0 mL of 0. 50 M NaOH with 1 The pH Curve for the Titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCI Copyright © Cengage Learning. All rights reserved

Weak Acid–Strong Base Titration Step 1: A stoichiometry problem (reaction is assumed to run to completion) then determine remaining species. Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate pH). Copyright © Cengage Learning. All rights reserved

Concept Check Calculate the pH of a solution made by mixing 0.20 mol HC2H3O2 (Ka = 1.8 x 10–5) with 0.030 mol NaOH in 1.0 L of aqueous solution. Major Species: HC2H3O2, Na+, OH-, H2O Possibilities for reactions: 1) HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) 2) H2O + H2O  H3O+(aq) + OH-(aq) 3) HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) Reaction 3 is most likely. In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka/ Kw = 1.8 x 109. We can assume the reaction goes to completion. In this problem, OH- is limiting. After the reaction takes place, the major species are: HC2H3O2, C2H3O2-, H2O, Na+ The primary reaction could be: 3) C2H3O2-(aq) + H2O  HC2H3O2(aq) + OH-(aq) We will see that reactions 1 and 3 will give us the same answer (as long as "x" is negligible in calculations, or solved for exactly). In this case, though, Ka (reaction 1) is greater than Kb (reaction 3) and should be used. When students solve this, make sure they include the initial concentration of the conjugate base (they tend to forget and call it zero). pH = 3.99 Copyright © Cengage Learning. All rights reserved

K = 1.8 x 109 Let’s Think About It… HC2H3O2(aq) + OH–  C2H3O2–(aq) Before 0.20 mol 0.030 mol Change –0.030 mol –0.030 mol +0.030 mol After 0.17 mol 0 0.030 mol K = 1.8 x 109 Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH HC2H3O2(aq) + H2O H3O+ + C2H3O2-(aq) Initial 0.170 M ~0 0.030 M Change –x +x Equilibrium 0.170 – x x 0.030 + x Ka = 1.8 x 10–5 pH = 3.99 Copyright © Cengage Learning. All rights reserved

Exercise Calculate the pH of a 100.0 mL solution of 0.100 M acetic acid (HC2H3O2), which has a Ka value of 1.8 x 10–5. pH = 2.87 pH = 2.87 Major Species: HC2H3O2, H2O Dominant reaction: HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) The 100.0 mL is not necessary to know for this problem. Copyright © Cengage Learning. All rights reserved

Concept Check Calculate the pH of a solution made by mixing 100.0 mL of a 0.100 M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 x 10–5, and 50.0 mL of a 0.10 M NaOH solution. pH = 4.74 pH = 4.74 Major Species: HC2H3O2, Na+, OH-, H2O Dominant reaction: HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka/Kw = 1.8 x 109. We can assume the reaction goes to completion. In this problem, OH- is limiting. After the reaction takes place, the major species are: HC2H3O2, C2H3O2-, H2O Primary reaction is: HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) When students solve this, make sure they include the initial concentration of the conjugate base (they tend to forget and call it zero). Copyright © Cengage Learning. All rights reserved

Finding the Equivalence Point (calculation method) Strong Acid vs. Strong Base 100 % ionized! pH = 7 No equilibrium! Weak Acid vs. Strong Base Acid is neutralized; Need Kb for conjugate base equilibrium Strong Acid vs. Weak Base Base is neutralized; Need Ka for conjugate acid equilibrium Weak Acid vs. Weak Base Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

Exactly 100 mL of 0. 10 M HNO2 are titrated with 100 mL of a 0 Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? start (moles) 0.01 0.01 HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l) end (moles) 0.0 0.0 0.01 [NO2-] = 0.01 0.200 = 0.05 M Final volume = 200 mL NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq) Initial (M) Change (M) Equilibrium (M) 0.05 0.00 0.00 -x +x +x 0.05 - x x x Kb = [OH-][HNO2] [NO2-] = x2 0.05-x pOH = 5.98 = 2.2 x 10-11 pH = 14 – pOH = 8.02 0.05 – x  0.05 x  1.05 x 10-6 = [OH-]

The pH Curve for the Titration of 50. 0 mL of 0. 100 M HC2H3O2 with 0 The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH Copyright © Cengage Learning. All rights reserved

The pH Curve for the Titration of 100. 0mL of 0. 050 M NH3 with 0 The pH Curve for the Titration of 100.0mL of 0.050 M NH3 with 0.10 M HCl Copyright © Cengage Learning. All rights reserved

Marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible). Copyright © Cengage Learning. All rights reserved

The Acid and Base Forms of the Indicator Phenolphthalein Copyright © Cengage Learning. All rights reserved

The Methyl Orange Indicator is Yellow in Basic Solution and Red in Acidic Solution Copyright © Cengage Learning. All rights reserved

Useful pH Ranges for Several Common Indicators Copyright © Cengage Learning. All rights reserved

Solubility and Complex Ion Equilibria Chapter 16 Solubility and Complex Ion Equilibria

16.3 Equilibria Involving Complex Ions Copyright © Cengage Learning. All rights reserved

Complex Ion Equilibria Charged species consisting of a metal ion surrounded by ligands. Ligand: Lewis base Formation (stability) constant. Equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution. Copyright © Cengage Learning. All rights reserved

Transition Metals and Coordination Chemistry Chapter 21 Transition Metals and Coordination Chemistry

21.1 The Transition Metals: A Survey 21.2 The First-Row Transition Metals 21.3 Coordination Compounds Copyright © Cengage Learning. All rights reserved

Forming Ionic Compounds More than one oxidation state is often found. Cations are often complex ions – species where the transition metal ion is surrounded by a certain number of ligands (Lewis bases). Copyright © Cengage Learning. All rights reserved

The Complex Ion Co(NH3)63+ Copyright © Cengage Learning. All rights reserved

Ionic Compounds with Transition Metals Most compounds are colored because the transition metal ion in the complex ion can absorb visible light of specific wavelengths. Many compounds are paramagnetic. Copyright © Cengage Learning. All rights reserved

Electron Configurations First-row transition metal ions do not have 4s electrons. Energy of the 3d orbitals is less than that of the 4s orbital. Ti: [Ar]4s23d2 Ti3+: [Ar]3d1 Copyright © Cengage Learning. All rights reserved

Coordination Number Number of bonds formed between the metal ion and the ligands in the complex ion. 6 and 4 (most common) 2 and 8 (least common) Copyright © Cengage Learning. All rights reserved

Monodentate ligand – one bond to a metal ion Ligands Neutral molecule or ion having a lone electron pair that can be used to form a bond to a metal ion. Monodentate ligand – one bond to a metal ion Bidentate ligand (chelate) – two bonds to a metal ion Polydentate ligand – more than two bonds to a metal ion Copyright © Cengage Learning. All rights reserved

Complex Ion Formation These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)42+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2+. Memorize the common ligands.

Common Ligands H2O aqua NH3 ammine OH- hydroxy Cl- chloro Br- bromo Names used in the ion H2O aqua NH3 ammine OH- hydroxy Cl- chloro Br- bromo CN- cyano SCN- thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

Names Names: ligand first, then cation Examples: tetraamminecopper(II) ion: Cu(NH3)42+ diamminesilver(I) ion: Ag(NH3)2+. tetrahydroxyzinc(II) ion: Zn(OH)4 2- The charge is the sum of the parts (2+) + 4(-1)= -2.

When Complexes Form The odd complex ion, FeSCN2+, shows up once in a while Acid-base reactions may change NH3 into NH4+ (or vice versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)42+, forms. Keywords such as "excess" and "concentrated" of any solution may indicate complex ions in REACTIONS! AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.

Some Coordination Complexes molecular formula Lewis base/ligand Lewis acid donor atom coordination number Ag(NH3)2+ NH3 Ag+ N 2 [Zn(CN)4]2- CN- Zn2+ C 4 [Ni(CN)4]2- Ni2+ [PtCl6] 2- Cl- Pt4+ Cl 6 [Ni(NH3)6]2+