LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide explanations of their properties based on particle views. (Sec 14.8) LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. (Sec 14.1-14.6) LO 3.7 The student is able to identify compounds as Bronsted-Lowry acids, bases, and/or conjugate acid-base pairs, using proton-transfer reactions to justify the identification. Two other learning objectives apply to multiple sections and serve as over-arching principles of acid-base equilibria. (Sec 14.1) LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec 14.1-14.12)

LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. (Sec 14.4-14.7) LO 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration. (Sec 14.2) LO. 6.14 The student can, based on the dependence of Kw on temperature, reason that neutrality requires [H+] = [OH-] as opposed to requiring pH = 7, including especially the applications to biological systems. (Sec 14.2) LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution. (Sec 14.4-14.7)

LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. (Sec 14.5-14.6)

AP Learning Objectives, Margin Notes and References LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. LO 3.7 The student is able to identify compounds as Bronsted-Lowry acids, bases, and/or conjugate acid-base pairs, using proton-transfer reactions to justify the identification. Two other learning objectives apply to multiple sections and serve as over-arching principles of acid-base equilibria. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium.

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 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Acid in Water HA(aq) + H2O(l) H3O+(aq) + A-(aq) 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

Acid Ionization Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. LO 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration. LO. 6.14 The student can, based on the dependence of Kw on temperature, reason that neutrality requires [H+] = [OH-] as opposed to requiring pH = 7, including especially the applications to biological systems.

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

Copyright © Cengage Learning. All rights reserved

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 [OH–] > [H+]; basic solution Copyright © Cengage Learning. All rights reserved

Self-Ionization of Water To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 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

CONCEPT CHECK! If the equilibrium lies to the right, the value for Ka is __________. large (or >1) If the equilibrium lies to the left, the value for Ka is ___________. small (or <1) Large (or >1); small (or < 1) The students should know this from Chapter 13, but it is worth re-emphasizing. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! HA(aq) + H2O(l) H3O+(aq) + A–(aq) If water is a better base than A–, do products or reactants dominate at equilibrium? Does this mean HA is a strong or weak acid? Is the value for Ka greater or less than 1? products; strong; greater than 1 One of the advantages in having students write out the weak acid reaction with water is that we can make comparisons between the relative strengths of the conjugate base and water acting as a base. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider a 1.0 M solution of HCl. Order the following from strongest to weakest base and explain: H2O(l) A–(aq) (from weak acid HA) Cl–(aq) The bases from strongest to weakest are: A-, H2O, Cl-. Some students will try to claim that the Kb value for water is 1.0 x 10-14 (or some will say 1.0 x 10-7) if they’ve read ahead in the chapter. Water does not have a Kb value; we must use reactions (like in the first Concept Check problem) to think about relative base strengths. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How good is Cl–(aq) as a base? Is A–(aq) a good base? The bases from strongest to weakest are: A–, H2O, Cl– Copyright © Cengage Learning. All rights reserved

Consider a solution of NaA where A– is the anion from weak acid HA: CONCEPT CHECK! Consider a solution of NaA where A– is the anion from weak acid HA: A–(aq) + H2O(l) HA(aq) + OH–(aq) base acid conjugate conjugate acid base Which way will equilibrium lie? left left Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider a solution of NaA where A– is the anion from weak acid HA: A–(aq) + H2O(l) HA(aq) + OH–(aq) base acid conjugate conjugate acid base b) Is the value for Kb greater than or less than 1? less than 1 less than 1 Copyright © Cengage Learning. All rights reserved

Consider a solution of NaA where A– is the anion from weak acid HA: CONCEPT CHECK! Consider a solution of NaA where A– is the anion from weak acid HA: A–(aq) + H2O(l) HA(aq) + OH–(aq) base acid conjugate conjugate acid base c) Does this mean A– is a strong or weak base? weak base weak base; Students often try to memorize relationships. An easy (and incorrect one) is “The conjugate base of a weak acid is a strong base.” The terms “strong” and “weak” are relative terms. We reserve the term “strong base” for hydroxides. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Acetic acid (HC2H3O2) and HCN are both weak acids. Acetic acid is a stronger acid than HCN. Arrange these bases from weakest to strongest and explain your answer: H2O Cl– CN– C2H3O2– Copyright © Cengage Learning. All rights reserved

Let’s Think About It… H2O(l) + H2O(l) H3O+(aq) + OH–(aq) acid base conjugate conjugate acid base At 25°C, Kw = 1.0 × 10–14 The bases from weakest to strongest are: Cl–, H2O, C2H3O2–, CN– Weakest to strongest: Cl-, H2O, C2H3O2-, CN-. If HCN is weaker than HC2H3O2, the conjugate base will be stronger. Have the students see this by looking at the general equation HA(aq) + H2O  H3O+(aq) + A-(aq) They can also use this to see that water is a better base than the chloride ion. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Discuss whether the value of K for the reaction: HCN(aq) + F–(aq) CN–(aq) + HF(aq) is >1 <1 =1 (Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.) Explain your answer. The value for K is less than 1. This reaction is neither a Kb nor a Ka reaction. To think about it, we need to know which is the stronger acid, HCN or HF. This can be decided by knowing the Ka values. HF is stronger, which means that it is more likely to donate a proton. Also, F- must be a weaker base than CN-. Thus, CN- has a greater affinity for a proton than F-. Because of all of this, equilibrium lies on the left, so K is less than 1. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the value for K for the reaction: HCN(aq) + F–(aq) CN–(aq) + HF(aq) (Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.) K = 8.6 × 10–7 K = 8.6 x 10-7 K = Ka(HCN)/Ka(HF) = [6.2 x 10-10]/[7.2 x 10-4] The students should know how to manipulate the Ka expressions to derive this. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

pH changes by 1 for every power of 10 change in [H+]. pH = –log[H+] pH changes by 1 for every power of 10 change in [H+]. A compact way to represent solution acidity. pH decreases as [H+] increases. Significant figures: The number of decimal places in the log is equal to the number of significant figures in the original number. Copyright © Cengage Learning. All rights reserved

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. 1.0 × 10–4 M H+ 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

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

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. 1.0 × 10–4 M H+ 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

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

AP Learning Objectives, Margin Notes and References LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution.

Thinking About Acid–Base Problems What are the major species in solution? What is the dominant reaction that will take place? Is it an equilibrium reaction or a reaction that will go essentially to completion? React all major species until you are left with an equilibrium reaction. Solve for the pH if needed. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider an aqueous solution of 2.0 × 10–3 M HCl. What are the major species in solution? H+, Cl–, H2O What is the pH? pH = 2.70 Major Species: H+, Cl-, H2O pH = –log[H+] = –log(2.0 × 10–3 M) = 2.70 pH = 2.70 Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 1.5 × 10–11 M solution of HCl. pH = 7.00 CONCEPT CHECK! Calculate the pH of a 1.5 × 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

Calculate the pH of a 1.5 × 10–2 M solution of HNO3. CONCEPT CHECK! Calculate the pH of a 1.5 × 10–2 M solution of HNO3. Major Species: H+, NO3-, H2O The reaction controlling the pH is: H2O + H2O  H3O+(aq) + OH-(aq) This is because it is the only equilibrium reaction (NO3- is not a base in water). However, the major source for H+ is from the nitric acid, so the pH = -log(1.5 x 10-2) = 1.82. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… When HNO3 is added to water, a reaction takes place immediately: HNO3 + H2O H3O+ + NO3– Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Why is this reaction not likely? NO3–(aq) + H2O(l) HNO3(aq) + OH–(aq) Because the K value for HNO3 is very large and HNO3 virtually completely dissociates in solution. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What reaction controls the pH? H2O(l) + H2O(l) H3O+(aq) + OH–(aq) In aqueous solutions, this reaction is always taking place. But is water the major contributor of H+ (H3O+)? pH = 1.82 Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution. LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations.

Solving Weak Acid Equilibrium Problems List the major species in the solution. Choose the species that can produce H+, and write balanced equations for the reactions producing H+. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+. Write the equilibrium expression for the dominant equilibrium. Copyright © Cengage Learning. All rights reserved

Solving Weak Acid Equilibrium Problems List the initial concentrations of the species participating in the dominant equilibrium. Define the change needed to achieve equilibrium; that is, define x. Write the equilibrium concentrations in terms of x. Substitute the equilibrium concentrations into the equilibrium expression. Copyright © Cengage Learning. All rights reserved

Solving Weak Acid Equilibrium Problems Solve for x the “easy” way, that is, by assuming that [HA]0 – x about equals [HA]0. Use the 5% rule to verify whether the approximation is valid. Calculate [H+] and pH. Copyright © Cengage Learning. All rights reserved

What are the major species in solution? HCN, H2O CONCEPT CHECK! Consider a 0.80 M aqueous solution of the weak acid HCN (Ka = 6.2 × 10–10). What are the major species in solution? HCN, H2O Major Species: HCN, H2O. This is because HCN is a weak acid. Thus, while H+ and CN- are in solution, the concentrations are very small. Primary reaction: HCN(aq) + H2O  H3O+(aq) + CN-(aq) The primary reaction is chosen because the K value is significantly larger. Thus, it is the primary source of H3O+ (or H+). Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Why aren’t H+ or CN– major species? See notes on slide 41. Copyright © Cengage Learning. All rights reserved

Consider This HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) Ka = 6.2 × 10-10 H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Kw = 1.0 × 10-14 Which reaction controls the pH? Explain. See notes on slide 1. Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 0.50 M aqueous solution of the weak acid HF. EXERCISE! Calculate the pH of a 0.50 M aqueous solution of the weak acid HF. (Ka = 7.2 × 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

Let’s Think About It… What are the major species in solution? HF, H2O Why aren’t H+ and F– major species? See Slide 4. Copyright © Cengage Learning. All rights reserved

HF(aq) + H2O(l) H3O+(aq) + F–(aq) Let’s Think About It… What are the possibilities for the dominant reaction? HF(aq) + H2O(l) H3O+(aq) + F–(aq) Ka=7.2 × 10-4 H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Kw=1.0 × 10-14 Which reaction controls the pH? Why? See Slide 4. 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 × 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

A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water. 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 × 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

EXERCISE! Calculate the pH of an 8.00 M solution of formic acid. Use the data from the previous slide to help you solve this problem. pH = 1.42 pH = –log[H+] = –log[0.038 M] = 1.42 Major species: HCHO2, H2O Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) Copyright © Cengage Learning. All rights reserved

EXERCISE! The value of Ka for a 4.00 M formic acid solution should be: higher than lower than the same as the value of Ka of an 8.00 M formic acid solution. Explain. the same; Students should understand that Ka is not a function of initial concentration. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The percent ionization of a 4.00 M formic acid solution should be: higher than lower than the same as the percent ionization of an 8.00 M formic acid solution. Explain. higher; To make 4.00 M solution from 8.00 M solution, we must add water (dilute). With a larger volume, Le Chatelier's principle tells us that equilibrium will shift to the right. Thus, percent ionization will increase. However, this increase if offset by the dilution factor, and the solution has an overall lower concentration of H+. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The pH of a 4.00 M formic acid solution should be: higher than lower than the same as the pH of an 8.00 M formic acid solution. Explain. higher; They should also understand that a lower concentration of the same weak acid will be less acidic (and thus the pH will be higher). Copyright © Cengage Learning. All rights reserved

EXERCISE! Calculate the percent ionization of a 4.00 M formic acid solution in water. % Ionization = 0.67% % dissociation = 0.67% Major species: HCHO2, H2O Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) I 4.00 0 0 C -x +x +x E 4.00-x x x Ka = 1.8 x 10-4 = (x)2/(4.00-x) x = 0.027; (0.027/4.00) x 100 = 0.67% Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 4.00 M solution of formic acid. EXERCISE! Calculate the pH of a 4.00 M solution of formic acid. pH = 1.57 pH = 1.57 Major species: HCHO2, H2O Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) From preview exercise, x = [H3O+] = [H+] = 0.027; pH = 1.57 Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution. LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations.

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. LiOH, NaOH, KOH, Ca(OH)2 pOH = –log[OH–] pH = 14.00 – pOH Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a 1.0 × 10–3 M solution of sodium hydroxide. pH = 11.00 pH = 11.00 Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a 1.0 × 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

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

AP Learning Objectives, Margin Notes and References LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution.

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

Calculate the pH of a 1.00 M solution of H3PO4. EXERCISE! Calculate the pH of a 1.00 M solution of H3PO4. Ka1 = 7.5 × 10-3 Ka2 = 6.2 × 10-8 Ka3 = 4.8 × 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

CONCEPT CHECK! Calculate the equilibrium concentration of PO43- in a 1.00 M solution of H3PO4. Ka1 = 7.5 × 10-3 Ka2 = 6.2 × 10-8 Ka3 = 4.8 × 10-13 [PO43-] = 3.6 × 10-19 M [PO43-] = 3.6 x 10-19 M From the previous exercise, the students should see that [H+] = [H2PO4-] = 0.0829 M (from the first dissociation reaction). Using the Ka2 expression, students can solve for [HPO42-] = 6.2 x 10-8 M. H2PO4-(aq) + H2O  H3O+(aq) + HPO42-(aq) I 0.0829 0.0829 0 C -x +x +x E 0.0829 - x 0.0829 + x x Many students will use “0” as the initial value of [H3O+]. Note that [HPO42-] is equal to Ka2, which should make sense since [H+] = [H2PO4-] (“x” is negligible). Using the Ka3 expression, students can solve for the phosphate ion concentration. HPO42-(aq) + H2O  H3O+(aq) + PO43-(aq) I 6.2 x 10-8 0.0829 0 C -x +x +x E 6.2 x 10-8 - x 0.0829 + x x Again, students may forget to include the initial concentrations from the previous Ka table. Since “x” is negligible, the values are the same. Also, make sure the students understand that [H3O+] can only have one value; that is, there are not different [H3O+]’s for the different dissociations. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide explanations of their properties based on particle views. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

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

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

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

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

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

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

CONCEPT CHECK! Consider a 0.30 M solution of NaF. The Ka for HF is 7.2 × 10-4. What are the major species? Na+, F-, H2O Major Species: Na+, F-, H2O Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Why isn’t NaF considered a major species? What are the possibilities for the dominant reactions? Copyright © Cengage Learning. All rights reserved

Let’s Think About It… The possibilities for the dominant reactions are: F–(aq) + H2O(l) HF(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Na+(aq) + H2O(l) NaOH + H+(aq) Na+(aq) + F–(aq) NaF Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How do we decide which reaction controls the pH? F–(aq) + H2O(l) HF(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Determine the equilibrium constant for each reaction. The primary reaction is the first one, and the Kb value is 1.39 x 10-11. This is over 1000 times larger than the K value for the second reaction (Kw). Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 0.75 M aqueous solution of NaCN. EXERCISE! Calculate the pH of a 0.75 M aqueous solution of NaCN. Ka for HCN is 6.2 × 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

Let’s Think About It… What are the major species in solution? Na+, CN–, H2O Why isn’t NaCN considered a major species? Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are all possibilities for the dominant reaction? The possibilities for the dominant reaction are: CN–(aq) + H2O(l) HCN(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Na+(aq) + H2O(l) NaOH + H+(aq) Na+(aq) + CN–(aq) NaCN Which of these reactions really occur? Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How do we decide which reaction controls the pH? CN–(aq) + H2O(l) HCN(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) 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 × 10–5 pH = 11.54

AP Learning Objectives, Margin Notes and References LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide explanations of their properties based on particle views. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

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

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

Comparison of Electronegativity of X and Ka Value Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

SO2, NO2, CO2 Oxides 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

Oxides 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

AP Learning Objectives, Margin Notes and References LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

Lewis acid Lewis base 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

AP Learning Objectives, Margin Notes and References LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

When analyzing an acid-base equilibrium problem: Ask this question: What are the major species in the solution and what is their chemical behavior? What major species are present? Does a reaction occur that can be assumed to go to completion? What equilibrium dominates the solution? Let the problem guide you. Be patient. Copyright © Cengage Learning. All rights reserved