Chapter 14 Acids and Bases AP*. AP Learning Objectives  LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide.

Chapter 14 Acids and Bases AP*

AP Learning Objectives  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)

AP Learning Objectives  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 K w 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)

AP Learning Objectives  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)

Section 14.1 The Nature of Acids and Bases AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.1 The Nature of Acids and Bases  Arrhenius  An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions.  A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.  Brønsted–Lowry  An acid is a proton donor.  A base is a proton acceptor.

Section 14.1 The Nature of Acids and Bases Brønsted–Lowry Acid and Base A Brønsted–Lowry acid must have at least one removable (acidic) proton (H + ) to donate. A Brønsted–Lowry base must have at least one nonbonding pair of electrons to accept a proton (H + ).

Section 14.1 The Nature of Acids and Bases Brønsted–Lowry Reaction Copyright © Cengage Learning. All rights reserved 8 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 HERECLICK HERE

Section 14.1 The Nature of Acids and Bases What Is Different about Water?  Water can act as a Brønsted–Lowry base and accept a proton (H + ) from an acid, as on the previous slide.  It can also donate a proton and act as an acid, as is seen below.  This makes water amphoteric.

Section 14.1 The Nature of Acids and Bases Conjugate Acids and Bases  The term conjugate means “joined together as a pair.”  Reactions between acids and bases always yield their conjugate bases and acids.

Section 14.1 The Nature of Acids and Bases Relative Strengths of Acids and Bases  Acids above the line with H 2 O as a base are strong acids; their conjugate bases do not act as acids in water.  Bases below the line with H 2 O as an acid are strong bases; their conjugate acids do not act as acids in water. The substances between the lines with H 2 O are conjugate acid–base pairs in water.

Section 14.1 The Nature of Acids and Bases

Section 14.2 Acid Strength AP Learning Objectives, Margin Notes and References  Learning Objectives  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 K w on temperature, reason that neutrality requires [H + ] = [OH - ] as opposed to requiring pH = 7, including especially the applications to biological systems.

Section 14.2 Acid Strength Acid and Base Strength  In every acid–base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.  HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl  (aq)  H 2 O is a much stronger base than Cl , so the equilibrium lies far to the right (K >> 1).  CH 3 COOH(aq) + H 2 O(l) ⇌ H 3 O + (aq) + CH 3 COO – (aq)  Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K < 1).

Section 14.2 Acid Strength  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 15

Section 14.2 Acid Strength Autoionization of Water  Water is amphoteric.  In pure water, a few molecules act as bases and a few act as acids.  This is referred to as autoionization.

Section 14.2 Acid Strength Ion Product Constant  The equilibrium expression for this process is K c = [H 3 O + ][OH  ]  This special equilibrium constant is referred to as the ion product constant for water, K w.  At 25 ° C, K w = 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.

Section 14.2 Acid Strength Aqueous Solutions Can Be Acidic, Basic, or Neutral  If a solution is neutral, [H + ] = [OH – ].  If a solution is acidic, [H + ] > [OH – ].  If a solution is basic, [H + ] < [OH – ].

Section 14.2 Acid Strength Self-Ionization of Water Copyright © Cengage Learning. All rights reserved 21 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 HERECLICK HERE

Section 14.1 The Nature of Acids and Bases Acid Ionization Equilibrium Copyright © Cengage Learning. All rights reserved 22 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 HERECLICK HERE

Section 14.2 Acid Strength

Section 14.2 Acid Strength HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid base conjugate conjugate acid base What is the equilibrium constant expression for an acid acting in water? Copyright © Cengage Learning. All rights reserved 25 CONCEPT CHECK!

Section 14.2 Acid Strength If the equilibrium lies to the right, the value for K a is __________. large (or >1) If the equilibrium lies to the left, the value for K a is ___________. small (or <1) Copyright © Cengage Learning. All rights reserved 26 CONCEPT CHECK!

Section 14.2 Acid Strength HA(aq) + H 2 O(l) H 3 O + (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 K a greater or less than 1? Copyright © Cengage Learning. All rights reserved 27 CONCEPT CHECK!

Section 14.2 Acid Strength Acetic acid (HC 2 H 3 O 2 ) 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: H 2 O Cl – CN – C 2 H 3 O 2 – Copyright © Cengage Learning. All rights reserved 28 CONCEPT CHECK!

Section 14.2 Acid Strength Discuss whether the value of K for the reaction: HCN(aq) + F – (aq) CN – (aq) + HF(aq) is >1<1 =1 (K a for HCN is 6.2×10 –10 ; K a for HF is 7.2×10 –4.) Explain your answer. Copyright © Cengage Learning. All rights reserved 29 CONCEPT CHECK!

Section 14.2 Acid Strength Calculate the value for K for the reaction: HCN(aq) + F – (aq) CN – (aq) + HF(aq) (K a for HCN is 6.2×10 –10 ; K a for HF is 7.2×10 –4.) K = 8.6 × 10 –7 Copyright © Cengage Learning. All rights reserved 30 CONCEPT CHECK!

Section 14.3 The pH Scale AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.3 The pH Scale  pH is a method of reporting hydrogen ion concentration.  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 32

Section 14.3 The pH Scale Other “p” Scales  The “p” in pH tells us to take the –log of a quantity (in this case, hydrogen ions).  Some other “p” systems are  pOH: –log[OH  ]  pK w : –log K w  pK a : –log K a  pK b : –log K b

Section 14.3 The pH Scale Relating pH and pOH Because [H 3 O + ][OH  ] = K w = 1.0  10  14 we can take the –log of the equation –log[H 3 O + ] + –log[OH  ] = –log K w = 14.00 which results in pH + pOH = pK w = 14.00

Section 14.3 The pH Scale How Do We Measure pH?  Indicators, including litmus paper, are used for less accurate measurements; an indicator is one color in its acid form and another color in its basic form.  pH meters are used for accurate measurement of pH; electrodes indicate small changes in voltage to detect pH.

Section 14.3 The pH Scale Calculate the pOH for each of the following solutions. a) 1.0 × 10 –4 M H + pOH = 10.00 b) 0.040 M OH – pOH = 1.40 Copyright © Cengage Learning. All rights reserved 41 EXERCISE!

Section 14.4 Calculating the pH of Strong Acid Solutions AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.4 Calculating the pH of Strong Acid Solutions 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 44

Section 14.4 Calculating the pH of Strong Acid Solutions Strong Acids  You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4.  These are, by definition, strong electrolytes and exist totally as ions in aqueous solution; e.g., HA + H 2 O → H 3 O + + A –  So, for the monoprotic strong acids, [H 3 O + ] = [acid]  (Unless the [acid] is too low, such that the [H 3 O + ] from the acid is less than 1 x 10 -7 M.)

Section 14.4 Calculating the pH of Strong Acid Solutions Consider an aqueous solution of 2.0 × 10 –3 M HCl. What are the major species in solution? H +, Cl –, H 2 O What is the pH? pH = 2.70 Copyright © Cengage Learning. All rights reserved 46 CONCEPT CHECK!

Section 14.4 Calculating the pH of Strong Acid Solutions Let’s Think About It…  When HNO 3 is added to water, a reaction takes place immediately: HNO 3 + H 2 O H 3 O + + NO 3 – Copyright © Cengage Learning. All rights reserved 49

Section 14.4 Calculating the pH of Strong Acid Solutions Let’s Think About It…  Why is this reaction not likely? NO 3 – (aq) + H 2 O(l) HNO 3 (aq) + OH – (aq) Copyright © Cengage Learning. All rights reserved 50

Section 14.4 Calculating the pH of Strong Acid Solutions Let’s Think About It…  What reaction controls the pH?  H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq)  In aqueous solutions, this reaction is always taking place.  But is water the major contributor of H + (H 3 O + )? pH = 1.82 Copyright © Cengage Learning. All rights reserved 51

Section 14.5 Calculating the pH of Weak Acid Solutions AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.5 Calculating the pH of Weak Acid Solutions Weak Acids  For a weak acid, the equation for its dissociation is HA(aq) + H 2 O(l) ⇌ H 3 O + (aq) + A – (aq)  Since it is an equilibrium, there is an equilibrium constant related to it, called the acid-dissociation constant, K a  K a = [H 3 O + ][A – ] / [HA] The greater the value of K a, the stronger is the acid.

Section 14.5 Calculating the pH of Weak Acid Solutions Solving Weak Acid Equilibrium Problems 1.List the major species in the solution. 2.Choose the species that can produce H +, and write balanced equations for the reactions producing H +. 3.Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H +. 4.Write the equilibrium expression for the dominant equilibrium. Copyright © Cengage Learning. All rights reserved 54

Section 14.5 Calculating the pH of Weak Acid Solutions Solving Weak Acid Equilibrium Problems 5.List the initial concentrations of the species participating in the dominant equilibrium. 6.Define the change needed to achieve equilibrium; that is, define x. 7.Write the equilibrium concentrations in terms of x. 8.Substitute the equilibrium concentrations into the equilibrium expression. Copyright © Cengage Learning. All rights reserved 55

Section 14.5 Calculating the pH of Weak Acid Solutions Solving Weak Acid Equilibrium Problems 9.Solve for x the “easy” way, that is, by assuming that [HA] 0 – x about equals [HA] 0. 10.Use the 5% rule to verify whether the approximation is valid. 11.Calculate [H + ] and pH. Copyright © Cengage Learning. All rights reserved 56

Section 14.5 Calculating the pH of Weak Acid Solutions Consider a 0.80 M aqueous solution of the weak acid HCN (K a = 6.2 × 10 –10 ). What are the major species in solution? HCN, H 2 O Copyright © Cengage Learning. All rights reserved 57 CONCEPT CHECK!

Section 14.5 Calculating the pH of Weak Acid Solutions Consider This HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) K a = 6.2 × 10 -10 H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) K w = 1.0 × 10 -14  Which reaction controls the pH? Explain. Copyright © Cengage Learning. All rights reserved 59

Section 14.5 Calculating the pH of Weak Acid Solutions Calculate the pH of a 0.50 M aqueous solution of the weak acid HF. (K a = 7.2 × 10 –4 ) Copyright © Cengage Learning. All rights reserved 60 EXERCISE!

Section 14.5 Calculating the pH of Weak Acid Solutions Let’s Think About It…  What are the major species in solution? HF, H 2 O  Why aren’t H + and F – major species? Copyright © Cengage Learning. All rights reserved 61

Section 14.5 Calculating the pH of Weak Acid Solutions Let’s Think About It…  What are the possibilities for the dominant reaction? HF(aq) + H 2 O(l) H 3 O + (aq) + F – (aq) K a =7.2 × 10 -4 H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) K w =1.0 × 10 -14  Which reaction controls the pH? Why? Copyright © Cengage Learning. All rights reserved 62

Section 14.5 Calculating the pH of Weak Acid Solutions HF(aq) +H2OH2OH 3 O + (aq) + F – (aq) Initial0.50 M~ 0 Change–x+x Equilibrium0.50–xxx Steps Toward Solving for pH K a = 7.2 × 10 –4 pH = 1.72 Copyright © Cengage Learning. All rights reserved 63

Section 14.5 Calculating the pH of Weak Acid Solutions Calculating K a from the pH  The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is 2.38. Calculate K a for formic acid at this temperature.  We know that [H 3 O + ][HCOO – ] [HCOOH] K a =  To calculate K a, we need the equilibrium concentrations of all three things.  We can find [H 3 O + ], which is the same as [HCOO – ], from the pH.  [H 3 O + ] = [HCOO – ] = 10 –2.38 = 4.2 × 10 –3

Section 14.5 Calculating the pH of Weak Acid Solutions Calculating K a from pH Now we can set up a table for equilibrium concentrations. We know initial HCOOH (0.10 M) and ion concentrations (0 M); we found equilibrium ion concentrations (4.2 × 10 –3 M); so we calculate the change, then the equilibrium HCOOH concentration. [HCOOH], M[H 3 O + ], M[HCOO  ], M Initially0.1000 Change  4.2  10  3 +4.2  10  3 At equilibrium 0.10  4.2  10  3 = 0.0958 = 0.10 4.2  10  3

Section 14.5 Calculating the pH of Weak Acid Solutions Calculating K a from pH [4.2  10  3 ][4.2  10  3 ] [0.10] K a = = 1.8  10  4 This allows us to calculate K a by putting in the equilibrium concentrations.

Section 14.5 Calculating the pH of Weak Acid Solutions Copyright © Cengage Learning. All rights reserved 67 Calculating Percent Ionization  Percent ionization =  100  In this example, [H 3 O + ] eq = 4.2  10  3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent ionization =  100 4.2  10  3 0.10 = 4.2%

Section 14.5 Calculating the pH of Weak Acid Solutions Strong vs. Weak Acids—Another Comparison  Strong Acid: [H + ] eq = [HA] init  Weak Acid: [H + ] eq < [HA] init  This creates a difference in conductivity and in rates of chemical reactions.

Section 14.5 Calculating the pH of Weak Acid Solutions Strong vs. Weak Acids  For a given strong acid, the properties of the acid solution are directly related to the acid concentration.  For a given weak acid, this is not the case. As the concentration of a weak acid is increased in solution, the percent dissociation decreases as the acid becomes more concentrated.

Section 14.5 Calculating the pH of Weak Acid Solutions A solution of 8.00 M formic acid (HCHO 2 ) is 0.47% ionized in water. Calculate the K a value for formic acid. K a = 1.8 × 10 –4 Copyright © Cengage Learning. All rights reserved 70 EXERCISE!

Section 14.5 Calculating the pH of Weak Acid Solutions 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 Copyright © Cengage Learning. All rights reserved 71 EXERCISE!

Section 14.5 Calculating the pH of Weak Acid Solutions The value of K a for a 4.00 M formic acid solution should be: higher than lower thanthe same as the value of K a of an 8.00 M formic acid solution. Explain. Copyright © Cengage Learning. All rights reserved 72 EXERCISE!

Section 14.5 Calculating the pH of Weak Acid Solutions The percent ionization of a 4.00 M formic acid solution should be: higher than lower thanthe same as the percent ionization of an 8.00 M formic acid solution. Explain. Copyright © Cengage Learning. All rights reserved 73 CONCEPT CHECK!

Section 14.5 Calculating the pH of Weak Acid Solutions The pH of a 4.00 M formic acid solution should be: higher than lower thanthe same as the pH of an 8.00 M formic acid solution. Explain. Copyright © Cengage Learning. All rights reserved 74 CONCEPT CHECK!

Section 14.5 Calculating the pH of Weak Acid Solutions Calculate the percent ionization of a 4.00 M formic acid solution in water. % Ionization = 0.67% Copyright © Cengage Learning. All rights reserved 75 EXERCISE!

Section 14.5 Calculating the pH of Weak Acid Solutions

Section 14.6 Bases AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.6 Bases  Arrhenius: bases produce OH – ions.  Brønsted–Lowry: bases are proton acceptors.  In a basic solution at 25°C, pH > 7.  pOH = –log[OH – ]  pH = 14.00 – pOH Copyright © Cengage Learning. All rights reserved 79

Section 14.6 Bases Copyright © Cengage Learning. All rights reserved 80 Strong Bases  Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ).  Again, these substances dissociate completely in aqueous solution; e.g., MOH(aq) → M + (aq) + OH – (aq) or M(OH) 2 (aq) → M 2+ (aq) + 2 OH – (aq)

Section 14.6 Bases Weak Bases  Ammonia, NH 3, is a weak base.  Like weak acids, weak bases have an equilibrium constant called the base dissociation constant.  Equilibrium calculations work the same as for acids, using the base dissociation constant instead.

Section 14.6 Bases Base Dissociation Constants

Section 14.6 Bases Example  What is the pH of 0.15 M NH 3 ? 1) NH 3 + H 2 O ⇌ NH 4 + + OH – 2) K b = [NH 4 + ][OH – ] / [NH 3 ] = 1.8 × 10 –5 3) NH 3 (M)NH 4 + (M)OH – (M) Initial Concentration (M) 0.1500 Change in Concentration (M) –x–x+x+x+x+x Equilibrium Concentration (M) 0.15 – xxx

Section 14.6 Bases Example (completed) 4)1.8 × 10 – 5 = x 2 / (0.15 – x) If we assume that x << 0.15, 0.15 – x = 0.15. Then: 1.8 × 10 –5 = x 2 / 0.15 and:x = 1.6 × 10 –3 Note: x is the molarity of OH –, so –log(x) will be the pOH (pOH = 2.80) and [14.00 – pOH] is pH (pH = 11.20).

Section 14.6 Bases Types of Weak Bases  Two main categories 1) Neutral substances with an Atom that has a nonbonding pair of electrons that can accept H + (like ammonia and the amines) 2) Anions of weak acids ClO - (aq) +H 2 O (l) ↔ HClO(aq) + OH – (aq)

Section 14.6 Bases A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Using the equation below, calculate the number of moles of NaClO added to the water. ClO - (aq) +H 2 O (l) ↔ HClO(aq) + OH – (aq) Using pH to Determine the Concentration of a Salt

Section 14.6 Bases Relationship between K a and K b For a conjugate acid–base pair, K a and K b are related in this way: K a × K b = K w Therefore, if you know one of them, you can calculate the other. © 2015 Pearson Education, Inc. Calculate K b for the fluoride ion.

Section 14.8 Acid-Base Properties of Salts HC 2 H 3 O 2 K a = 1.8 × 10 -5 HCN K a = 6.2 × 10 -10 Calculate the K b values for: C 2 H 3 O 2 − and CN − K b (C 2 H 3 O 2 - ) = 5.6 × 10 -10 K b (CN - ) = 1.6 × 10 -5 Copyright © Cengage Learning. All rights reserved EXERCISE!

Section 14.7 Polyprotic Acids AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.7 Polyprotic Acids  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: K a1 > K a2 > K a3  If the factor in the K a values for the first and second dissociation has a difference of 3 or greater, the pH generally depends only on the first dissociation. Copyright © Cengage Learning. All rights reserved 93

Section 14.7 Polyprotic Acids

Section 14.7 Polyprotic Acids Calculate the pH of a 1.00 M solution of H 3 PO 4. K a1 = 7.5 × 10 -3 K a2 = 6.2 × 10 -8 K a3 = 4.8 × 10 -13 pH = 1.08 (if use quadratic) Copyright © Cengage Learning. All rights reserved 95 EXERCISE!

Section 14.7 Polyprotic Acids Calculate the equilibrium concentration of PO 4 3- in a 1.00 M solution of H 3 PO 4. K a1 = 7.5 × 10 -3 K a2 = 6.2 × 10 -8 K a3 = 4.8 × 10 -13 [PO 4 3- ] = 3.6 × 10 -19 M Copyright © Cengage Learning. All rights reserved 96 CONCEPT CHECK!

Section 14.7 Polyprotic Acids  Be sure to read page 683-685 to learn about problems concerning sulfuric acid where the acid is strong in the first ionization step and weak in the second.

Section 14.8 Acid-Base Properties of Salts AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.8 Acid-Base Properties of Salts 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 99  The salt of a strong acid and a strong base gives a neutral solution.  KCl, NaNO 3

Section 14.8 Acid-Base Properties of Salts Acid–Base Properties of Salts  Many ions react with water to create H + or OH –. The reaction with water is often called hydrolysis.  To determine whether a salt is an acid or a base, you need to look at the cation and anion separately.  The cation can be acidic or neutral.  The anion can be acidic, basic, or neutral.

Section 14.8 Acid-Base Properties of Salts Anions  Anions of strong acids are neutral. For example, Cl – will not react with water, so OH – can’t be formed.  Anions of weak acids are conjugate bases, so they create OH – in water; e.g., C 2 H 3 O 2 – + H 2 O ⇌ HC 2 H 3 O 2 + OH –  Protonated anions from polyprotic acids (HSO 3 - ) can be acids or bases: If K a > K b, the anion will be acidic; if K b > K a, the anion will be basic.

Section 14.8 Acid-Base Properties of Salts Cations  Group I or Group II (Ca 2+, Sr 2+, or Ba 2+ ) metal cations are neutral.  Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH 4 +.  Transition and post-transition metal cations are acidic. Why? (There are no H atoms in these cations!)

Section 14.8 Acid-Base Properties of Salts Hydrated Cations  Transition and post-transition metals form hydrated cations.  The water attached to the metal is more acidic than free water molecules, making the hydrated ions acidic.

Section 14.8 Acid-Base Properties of Salts

Section 14.8 Acid-Base Properties of Salts Salt Solutions—Acidic, Basic, or Neutral? 1)Group I/II metal cation with anion of a strong acid: neutral (NaCl, RbClO 4 ) 2)Group I/II metal cation with anion of a weak acid: basic (like the anion) (NaClO, BaSO 3 ) 3)Transition/Post-transition metal cation or polyatomic cation with anion of a strong acid: acidic (like the cation) (NH 4 NO 3, AlCl 3, Fe(NO 3 ) 3 4)Transition/Post-transition metal cation or polyatomic cation with anion of a weak acid: compare K a and K b ; whichever is greater dictates what the salt is. (NH 4 ClO, Al(CH 3 COO) 3, CrF 3

Section 14.8 Acid-Base Properties of Salts Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Solution Analyze We are asked to predict whether a solution of Na 2 HPO 4 is acidic or basic. This substance is an ionic compound composed of Na + and HPO 4 2− ions. Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na + is a cation of group 1A, it has no influence on pH. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO 4 2− ion. We need to consider that HPO 4 2− can act as either an acid or a base: As acid HPO 4 2– (aq) H + (aq) + PO 4 3– (aq) [16.46] As base HPO 4 2– (aq) + H 2 O H 2 PO 4 – (aq) + OH – (aq) [16.47] Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic. Predict whether the salt Na 2 HPO 4 forms an acidic solution or a basic solution when dissolved in water.

Section 14.8 Acid-Base Properties of Salts Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Solve The value of K a for Equation 16.46 is K a3 for H 3 PO 4 : 4.2 × 10 −13 (Table 16.3). For Equation 16.47, we must calculate K b for the base HPO 4 2− from the value of K a for its conjugate acid, H 2 PO 4 −, and the relationship K a × K b = K w (Equation 16.40). The relevant value of K a for H 2 PO 4 − is Ka 2 for H 3 PO 4 : 6.2 × 10 –8 (from Table 16.3). We therefore have This K b value is more than 10 5 times larger than K a for HPO 4 2− ; thus, the reaction in Equation 16.47 predominates over that in Equation 16.46, and the solution is basic.

Section 14.8 Acid-Base Properties of Salts Arrange the following 1.0 M solutions from lowest to highest pH. HBr NaOH NH 4 Cl NaCN NH 3 HCN NaCl HF Justify your answer. HBr, HF, HCN, NH 4 Cl, NaCl, NaCN, NH 3, NaOH Copyright © Cengage Learning. All rights reserved 109 CONCEPT CHECK!

Section 14.8 Acid-Base Properties of Salts Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba(CH 3 COO) 2, (b) NH 4 Cl, (c) CH 3 NH 3 Br, (d) KNO 3, (e) Al(ClO 4 ) 3.

Section 14.8 Acid-Base Properties of Salts Consider a 0.30 M solution of NaF. The K a for HF is 7.2 × 10 -4. What are the major species? Na +, F -, H 2 O Copyright © Cengage Learning. All rights reserved 111 CONCEPT CHECK!

Section 14.8 Acid-Base Properties of Salts 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 112

Section 14.8 Acid-Base Properties of Salts Let’s Think About It… The possibilities for the dominant reactions are: 1.F – (aq) + H 2 O(l) HF(aq) + OH – (aq) 2.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 3.Na + (aq) + H 2 O(l) NaOH + H + (aq) 4.Na + (aq) + F – (aq) NaF Copyright © Cengage Learning. All rights reserved 113

Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  How do we decide which reaction controls the pH? F – (aq) + H 2 O(l) HF(aq) + OH – (aq) H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq)  Determine the equilibrium constant for each reaction. Copyright © Cengage Learning. All rights reserved 114

Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  What are the major species in solution? Na +, CN –, H 2 O  Why isn’t NaCN considered a major species? Copyright © Cengage Learning. All rights reserved 116

Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  What are all possibilities for the dominant reaction?  The possibilities for the dominant reaction are: 1.CN – (aq) + H 2 O(l) HCN(aq) + OH – (aq) 2.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 3.Na + (aq) + H 2 O(l) NaOH + H + (aq) 4.Na + (aq) + CN – (aq) NaCN  Which of these reactions really occur? Copyright © Cengage Learning. All rights reserved 117

Section 14.8 Acid-Base Properties of Salts Let’s Think About It…  How do we decide which reaction controls the pH? CN – (aq) + H 2 O(l) HCN(aq) + OH – (aq) H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) Copyright © Cengage Learning. All rights reserved 118

Section 14.8 Acid-Base Properties of Salts Steps Toward Solving for pH K b = 1.6 × 10 –5 pH = 11.54 CN – (aq) +H2OH2OHCN(aq) +OH – (aq) Initial0.75 M0~ 0 Change–x+x Equilibrium0.75–xxx

Section 14.9 The Effect of Structure on Acid-Base Properties AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.9 The Effect of Structure on Acid-Base Properties Factors that Affect Acid Strength 1)H—A bond must be polarized with δ + on the H atom and δ – on the A atom 2)Bond strength: Weaker bonds can be broken more easily, making the acid stronger. 3)Stability of A – : More stable anion means stronger acid.

Section 14.9 The Effect of Structure on Acid-Base Properties Binary Acids  Binary acids consist of H and one other element.  Within a group, H—A bond strength is generally the most important factor.  Within a period, bond polarity is the most important factor to determine acid strength.

Section 14.9 The Effect of Structure on Acid-Base Properties 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 123

Section 14.9 The Effect of Structure on Acid-Base Properties Arrange the compounds in each series in order of increasing acid strength: (a) AsH 3, HBr, KH, H 2 Se; (b) H 2 SO 4, H 2 SeO 3, H 2 SeO 4. Sample Exercise 16.20 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic

Section 14.9 The Effect of Structure on Acid-Base Properties Carboxylic Acids  Carboxylic acids are organic acids containing the —COOH group.  Factors contributing to their acidic behavior:  Other O attached to C draws electron density from O—H bond, increasing polarity.  Its conjugate base (carboxylate anion) has resonance forms to stabilize the anion.

Section 14.10 Acid-Base Properties of Oxides AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.10 Acid-Base Properties of Oxides Oxides  Acidic Oxides (Acid Anhydrides):  O—X bond is strong and covalent. SO 2, NO 2, CO 2  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 129

Section 14.10 Acid-Base Properties of Oxides Oxides  Basic Oxides (Basic Anhydrides):  O—X bond is ionic. K 2 O, 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 130

Section 14.11 The Lewis Acid-Base Model AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.11 The Lewis Acid-Base Model Lewis Acids and Bases  Lewis acid: electron pair acceptor  Lewis base: electron pair donor Lewis acid Lewis base Copyright © Cengage Learning. All rights reserved 132

Section 14.11 The Lewis Acid-Base Model

Section 14.12 Strategy for Solving Acid-Base Problems: A Summary AP Learning Objectives, Margin Notes and References  Learning Objectives  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.

Section 14.12 Strategy for Solving Acid-Base Problems: A Summary 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 136

Chapter 14 Acids and Bases AP*. AP Learning Objectives  LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide.

Similar presentations

Presentation on theme: "Chapter 14 Acids and Bases AP*. AP Learning Objectives  LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chapter 14 Acids and Bases AP*. AP Learning Objectives  LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide.

Similar presentations

Presentation on theme: "Chapter 14 Acids and Bases AP*. AP Learning Objectives  LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide."— Presentation transcript:

Similar presentations

About project

Feedback