CHE 106: General Chemistry - Mr. Donohue's Chemistry

CHE 106: General Chemistry - Mr. Donohue's Chemistry

CHE 116: General Chemistry 1 CHAPTER SIXTEEN Copyright Tyna L. Heise 2001-2002 All Rights Reserved Prof. T. L. Heise CHE 116 Acids and Bases: Review

2 Properties of Acids sour taste change with litmus Properties of Bases bitter taste change with litmus Prof. T. L. Heise CHE 116 Acids and Bases: Review

3 1830 - scientists have recognized that all acids contain hydrogen, but not all hydrogen bearing compounds are acids Svante Arrhenius - linked acid behavior with the presence of an H + and base behavior with the presence of an OH - Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 4

Arrhenius definition is useful, but restricts acid base reactions to aqueous conditions Johannes Bronsted and Thomas Lowry proposed a more general definition which involves the transfer of an H+ ion from one molecule to another Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 5 H+ ion is simply a proton with no surrounding valence electron.

Small particle interacts strongly with the nonbonding pairs of water molecules to form hydrated hydrogen ions chemists use H+ and H3O+ interchangeably Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases Fig 16.1 Prof. T. L. Heise 6

Demonstrates the interconnections possible between hydrogenated water CHE 116 Bronsted-Lowry Acids and Bases 7 Definitions: Acid - any compound which transfers an H+ to another molecule Base - any compound which accepts a

transfer of an H+ from another molecule * an acid and base always work together Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases Definitions: Amphoteric - some substances can be an 8 acid or a base depending on reaction

Conjugate Acid Base pairs - two compounds that differ only in the presence of an H+. The molecule with the extra H is the acid. Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 9 Fig 16.7, 16.8

Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 10 Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO3FPO43CO Prof. T. L. Heise CHE 116

Bronsted-Lowry Acids and Bases 11 Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO3FPO43CO Given a base, bases accept H+, so add an H+ to each molecule. Prof. T. L. Heise

CHE 116 Bronsted-Lowry Acids and Bases 12 Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3FPO43CO Given a base, bases accept H+, so add an H+ to each molecule.

Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 13 Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3HF FPO43CO Given a base, bases accept H+, so add an H+ to each

molecule. Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 14 Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3HF FHPO42PO43CO Given a base, bases accept

H+, so add an H+ to each molecule. Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 15 Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3HF FHPO42PO43HCO+ CO

Given a base, bases accept H+, so add an H+ to each molecule. Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 16 Sample Exercise: When lithium oxide, Li2O, is dissolved in water,

the solution turns basic from the reaction of the oxide ion, O 2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs. Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 17 Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.

O2- + H2O OH- + OH- Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 18 Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs. O2- + H2O OH- + OH-

Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 19 Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs. O2- + H2O OH- + OH- acid

Prof. T. L. Heise base base acid CHE 116 Bronsted-Lowry Acids and Bases 20 Relative Strengths of Acids and Bases:

the more readily a substance donates an H +, the less readily its conjugate base will accept one the more readily a substance accepts an H +, the less readily its conjugate acid will donate one the stronger one of the substances is, the weaker its conjugate Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 21 Relative Strengths of Acids and Bases:

strong acids completely transfer their protons to water, leaving no undissociated molecules weak acids are those that only partly dissociate in aqueous solution and therefore exist in the solution as a mixture of acid molecules and component ions Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 22 Relative Strengths of Acids and Bases:

negligible acids are those that have hydrogen but do not donate them at all, their conjugate bases would be extremely strong, reacting with water to complete their octet and leaving OH- behind. In every acid base reaction, the position of the equilibrium favors transfer of H+ from stronger side to weaker side Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 23 Fig. 16.4

Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 24 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO43-(aq)+H2O(l)HPO42-(aq)+OH-(aq) b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)

Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 25 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO43-(aq)+H2O(l)HPO42-(aq)+OH-(aq) 2 acids are: H2O and HPO422 bases are: PO43- and OH- Prof. T. L. Heise

CHE 116 Bronsted-Lowry Acids and Bases 26 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO43-(aq)+H2O(l)HPO42-(aq)+OH-(aq) 2 acids are: H2O and HPO422 bases are: PO43- and OHred indicates strength Prof. T. L. Heise CHE 116

Bronsted-Lowry Acids and Bases 27 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO43-(aq)+H2O(l)HPO42-(aq)+OH-(aq) 2 acids are: H2O and HPO422 bases are: PO43- and OHreverse reaction favored Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases

28 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO43-(aq)+H2O(l)HPO42-(aq)+OH-(aq) 2 acids are: H2O and HPO422 bases are: PO43- and OHshifts left Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 29

Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l) Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 30 Sample exercise: For each of the following reactions, use Fig. 16.4 to

predict whether the equilibrium lies predominantly to the left or right: b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l) 2 acids are: NH4+ and H2O 2 bases are: NH3 and OH- Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 31 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:

b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l) 2 acids are: NH4+ and H2O 2 bases are: NH3 and OHred indicates strength Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 32 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)

2 acids are: NH4+ and H2O 2 bases are: NH3 and OHfavors forward reaction Prof. T. L. Heise CHE 116 Bronsted-Lowry Acids and Bases 33 Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l) 2 acids are: NH4+ and H2O

2 bases are: NH3 and OHshifts right Prof. T. L. Heise CHE 116 The Autoionization of Water 34 One of the most important properties of water is its ability to ac as either a Bronsted acid or Bronsted base, depending on circumstances. One water molecule can donate a proton to another water molecule Fig 16.10

Prof. T. L. Heise CHE 116 The Autoionization of Water 35 The autoionization of water, although rapid and weak, does exist as an equilibrium, and therefore has an equilibrium constant expression: Keq = [H3O+][OH-] [H2O]2 * because water is a liquid, it can be excluded from the equation...

Prof. T. L. Heise CHE 116 The Autoionization of Water 36 Keq[H2O]2 = [H3O+][OH-] Kw = [H3O+][OH-] = 1.0 x 10-14 * this equation is not only applicable to water, but to all aqueous solutions, and it is upon this fact that the pH scale was built.

Prof. T. L. Heise CHE 116 The Autoionization of Water 37 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H+] = 2 x 10-5 b) [OH-] = 3 x 10-9 c) [OH-] = 1 x 10-7 Prof. T. L. Heise

CHE 116 The Autoionization of Water 38 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H+] = 2 x 10-5 then [OH-] must equal 1.0 x 10-14 which is 2 x10-5 [OH-] = 5.0 x 10-10 Prof. T. L. Heise

CHE 116 The Autoionization of Water 39 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H+] = 2 x 10-5 then [OH-] must equal 1.0 x 10-14 which is 2 x10-5 [OH-] = 5.0 x 10-10 [H+] > [OH-] so acidic

Prof. T. L. Heise CHE 116 The Autoionization of Water 40 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: b) [OH-] = 3 x 10-9 then [H+] must equal 1.0 x 10-14 which is 3 x10-9 [H+] = 3.3 x 10-6 Prof. T. L. Heise

CHE 116 The Autoionization of Water 41 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: b) [OH-] = 3 x 10-9 then [H+] must equal 1.0 x 10-14 which is 3 x10-9 [H+] = 3.3 x 10-6 [H+] > [OH-] so acidic Prof. T. L. Heise

CHE 116 The Autoionization of Water 42 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: c) [OH-] = 1 x 10-7 then [H+] must equal 1.0 x 10-14 which is 1 x10-7 [H+] = 1.0 x 10-7 Prof. T. L. Heise

CHE 116 The Autoionization of Water 43 Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: c) [OH-] = 1 x 10-7 then [H+] must equal 1.0 x 10-14 which is 1 x10-7 [H+] = 1.0 x 10-7 [H+] = [OH-] so neutral

Prof. T. L. Heise CHE 116 The pH Scale 44 For convenience, we can use a logarithmic version of concentration to turn the very small concentrations of [H+] and [OH-] into whole numbers. p(anything) = - log[anything] p(H) = -log[H+] p(OH) = - log[OH-] ** pH + pOH = 14

Prof. T. L. Heise CHE 116 The pH Scale 45 Prof. T. L. Heise CHE 116 The pH Scale

46 Common household products and their relative pHs. Prof. T. L. Heise CHE 116 The pH Scale 47 Sample exercise: In a sample of lemon juice

[H+] is 3.8 x 10-4 M. What is the pH? Prof. T. L. Heise CHE 116 The pH Scale 48 Sample exercise: In a sample of lemon juice [H +] is 3.8 x 10-4 M. What is the pH? pH = -log[H+] Prof. T. L. Heise

CHE 116 The pH Scale 49 Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH? pH = -log[H+] pH = -log[3.8 x 10-4] Prof. T. L. Heise CHE

116 The pH Scale 50 Sample exercise: In a sample of lemon juice [H +] is 3.8 x 10-4 M. What is the pH? pH = -log[H+] pH = -log[3.8 x 10-4] pH = 3.42 Prof. T. L. Heise CHE 116

The pH Scale 51 Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH? Prof. T. L. Heise CHE 116 The pH Scale 52

Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH? pH = -log[H+] Prof. T. L. Heise CHE 116 The pH Scale 53 Sample exercise: A commonly available window-cleaning solution has a [H +] is

5.3 x 10-9 M. What is the pH? pH = -log[H+] pH = -log[5.3 x 10-9] Prof. T. L. Heise CHE 116 The pH Scale 54 Sample exercise: A commonly available window-cleaning solution has a [H +] is 5.3 x 10-9 M. What is the pH? pH = -log[H+]

pH = -log[5.3 x 10-9] pH = 8.28 Prof. T. L. Heise CHE 116 The pH Scale 55 Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].

Prof. T. L. Heise CHE 116 The pH Scale 56 Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+]. pH = -log[H+] Prof. T. L. Heise CHE

116 The pH Scale 57 Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+]. pH = -log[H+] 9.18 = -log[H+] Prof. T. L. Heise CHE 116

The pH Scale 58 Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+]. pH = -log[H+] 9.18 = -log[H+] 10-9.18 = [H+] Prof. T. L. Heise CHE 116 The pH Scale

59 Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+]. pH = -log[H+] 9.18 = -log[H+] 10-9.18 = [H+] 6.61 x 10-10 = [H+] Prof. T. L. Heise CHE 116 The pH Scale

60 Measuring pH: a pH can be measured quickly and accurately using a pH meter. A pair of electrodes connected to a meter capable of measuring small voltages a voltage which varies with pH is generated when the electrodes are placed in a solution calibrated to give pH Prof. T. L. Heise CHE 116 The pH Scale

61 Measuring pH: a pH can be measured quickly and accurately using a pH meter. Electrodes come in a variety of shapes and sizes a set of electrodes exists that can be placed inside a human cell acid base indicators can be used, but are much less precise Prof. T. L. Heise CHE 116 The pH Scale

62 Fig 16.7 Prof. T. L. Heise CHE 116 Strong Acids and Bases 63 Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. Strong Acids

HCl HBr HI monoprotic HNO3 HClO3 HClO4 H2SO4 diprotic Prof. T. L. Heise CHE 116 Strong Acids and Bases

64 Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. Calculating the pH of a solution made up entirely of ions means the [H+] is proportional to [acid] Prof. T. L. Heise CHE 116 Strong Acids and Bases 65

An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? Prof. T. L. Heise CHE 116 Strong Acids and Bases 66 An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H+]

Prof. T. L. Heise CHE 116 Strong Acids and Bases 67 An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H+] 2.66 = -log[H+] Prof. T. L. Heise CHE

116 Strong Acids and Bases 68 An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+] 2.2 x 10-3 = [H+] Prof. T. L. Heise CHE 116

Strong Acids and Bases 69 An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+] 2.2 x 10-3 = [H+] 2.2 x 10-3 M H+ 1 mole HNO3 1 mole H+ Prof. T. L. Heise

CHE 116 Strong Acids and Bases 70 An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+] 2.2 x 10-3 = [H+] 2.2 x 10-3 M H+ 1 mole HNO3 1 mole H+ 2.2 x 10-3 HNO3

Prof. T. L. Heise CHE 116 Strong Acids and Bases 71 The most soluble common bases are the ionic hydroxides of the alkali and alkaline earth metals. Due to the complete dissociation of the base into its ion components makes the pH calculation equally straightforward

Prof. T. L. Heise CHE 116 Strong Acids and Bases 72 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? Prof. T. L. Heise CHE

116 Strong Acids and Bases 73 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+] Prof. T. L. Heise CHE 116 Strong Acids and Bases

74 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+]11.89 = -log[H+] Prof. T. L. Heise CHE 116 Strong Acids and Bases 75

Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+]11.89 = -log[H+]10-11.89 = [H+] Prof. T. L. Heise CHE 116 Strong Acids and Bases 76 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+]

11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+] Prof. T. L. Heise CHE 116 Strong Acids and Bases 77 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+]11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+] 1.0 x 10-14 = [OH-]

1.29 x 10-12 Prof. T. L. Heise CHE 116 Strong Acids and Bases 78 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+]11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+] 1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12

Prof. T. L. Heise CHE 116 Strong Acids and Bases 79 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? 1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12 7.8 x 10-3 M OH1 mol KOH

1 mol OH- Prof. T. L. Heise CHE 116 Strong Acids and Bases 80 Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? 1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12

7.8 x 10-3 M OH1 mol KOH 1 mol OH7.8 x 10-3 M KOH Prof. T. L. Heise CHE 116 Weak Acids 81 Most acids are weak acids and only partially dissociate in aqueous solution. The extent to which a weak acid dissociates can be expressed using an equilibrium constant for the ionization reaction HX + H2O H3O+

+ XKa = [H3O+][X-] [HX] *the larger the Ka the stronger the acid Prof. T. L. Heise CHE 116 Weak Acids 82 Calculating Ka from pH: a much more complicated calculation is required for the determinations of weak acids, in many cases, due to the extremely small

magnitude of the values, some simpler approximations can be made. Prof. T. L. Heise CHE 116 Weak Acids 83 Sample exercise: Niacin, one of the B vitamins, has the following molecular structure: C O

H O N A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? b) What is the acid-dissociation constant? Prof. T. L. Heise CHE 116 Weak Acids 84

Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? pH = -log[H+] 3.26 = -log[H+] e-3.26 = [H+] 5.5 x 10-4 = [H+] Prof. T. L. Heise CHE 116 Weak Acids Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? C6H4NOOH H+ initial

0.020 0 0 change -5.5 x 10-4 +5.5 x 10-4 +5.5 x 10-4 equil 0.01945 5.5 x 10-4 5.5 x 10-4 Prof. T. L. Heise 85 C6H4NOO- + CHE 116

Weak Acids 86 Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 a) What % of the acid is ionized in this solution? % = part/total x 100 = 5.5 x 10-4/0.01945 x 100 = 2.8% Prof. T. L. Heise CHE 116

Weak Acids 87 Sample exercise: A 0.020 M solution of niacin has a pH of 3.26 b) What is the acid-dissociation constant? C6H4NOOH C6H4NOO- + H+ Ka = [C6H4NOO- ][H+] 1.55 x10-5 Prof. T. L. Heise [C6H4NOOH] = (5.5 x 10-4 )(5.5 x 10-4 ) 0.01945

= CHE 116 Weak Acids 88 Using Ka to calculate pH: Using the value of Ka and knowing the initial concentration of the weak acid, we can calculate the concentration of H +(aq). Example: Calculate the pH of a 0.30 M solution of acetic acid. HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

Prof. T. L. Heise CHE 116 Weak Acids 89 Example: Calculate the pH of a 0.30 M solution of acetic acid. HC2H3O2(aq) H+(aq) + C2H3O2-(aq) From Table 16.2, Ka = 1.8 x 10-5 Ka = 1.8 x 10-5 = [H+][C2H3O2-] [HC2H3O2] set up data table of concentrations involved...

Prof. T. L. Heise CHE 116 Weak Acids 90 Ka = 1.8 x 10-5 = [H+][C2H3O2-] [HC2H3O2] set up data table of concentrations involved [HC2H3O2] [H+] [C2H3O2-] Initial 0.30 M 0

0 Change -x +x +x Equilibrium 0.30 - x x x Prof. T. L. Heise CHE 116 Weak Acids

91 Input concentrations in formula Ka = 1.8 x 10-5 = [x][x] [0.30 -x] This will lead to a quadratic equation, but we can simplify the problem a bit. All weak acids dissociate so little that we can assume the initial concentration of the acid remains essentially the same, that means we can rewrite the formula to read... Prof. T. L. Heise CHE 116 Weak Acids

Ka = 1.8 x 10-5 = [x][x] 1.8 x 10-5 (0.30) = x2 1.8 x 10-5 (0.30) = x 0.0023 = x 0.0023 = [H+] pH = -log [H+] = -log[0.0023] = 2.64 Prof. T. L. Heise 92 [0.30] CHE 116

Weak Acids 93 Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin? Prof. T. L. Heise CHE 116 Weak Acids 94

Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin? C6H4NOOH C6H4NOO- + H+ Ka = 1.5 x10-5 = [C6H4NOO-][H+] [C6H4NOOH] Prof. T. L. Heise CHE 116 Weak Acids 95

Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin? C6H4NOOH C6H4NOO- H+ initial 0 change +x equilibrium x Prof. T. L. Heise 0.010 -x 0.010 -x 0

+x x CHE 116 Weak Acids 96 Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin? C6H4NOOH C6H4NOO- + H+ Ka = 1.5 x10-5 = [x][x] [0.010]

Prof. T. L. Heise CHE 116 Weak Acids 97 Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin? 1.5 x10-5 [0.010] = x2 1.5 x10-5 [0.010] = x 3.87 x 10-4 = x Prof. T. L. Heise

CHE 116 Weak Acids 98 Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin? 3.87 x 10-4 = x 3.87 x 10-4 = [H+] Prof. T. L. Heise

pH = -log[H+] pH = -log[3.87 x 10-4] pH = 3.41 CHE 116 Weak Acids 99 The results seen in the two previous examples are typical of weak acids. The lower number of ions produced during the partial dissociation causes less electrical conductivity and a slower reaction rate with metals.

Percent ionization is a good way to discover the actual conductivity, however... Prof. T. L. Heise CHE 116 Weak Acids 100 As the concentration of a weak acid increases, the % ionized decreases. Prof. T. L. Heise

CHE 116 Weak Acids 101 As the concentration of a weak acid increases, the % ionized decreases. Prof. T. L. Heise CHE 116 Weak Acids

102 Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise b) a 1.0 x 10-3 M solution Prof. T. L. Heise CHE 116 Weak Acids 103

Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise Our approximation was good so % = part x 100 total = 3.87 x 10-4 x 100 = 3.9% 0.010 Prof. T. L. Heise CHE 116 Weak Acids

104 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3] 1.5 x10-5 (1.0 x 10-3) = x2 1.2 x 10-4 Prof. T. L. Heise CHE 116 Weak Acids

105 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3] but, 1.2 x 10-3 x 100 1.5 x10-5 (1.0 x 10-3) = x2 1.0 x 10-3 1.2 x 10-3 Prof. T. L. Heise is greater than 5% so use quadratic... CHE 116

Weak Acids 106 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3 - x] -1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0 Prof. T. L. Heise CHE 116

Weak Acids 107 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution -1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0 x = -b b2 - 4ac 2a x= Prof. T. L. Heise CHE 116

Weak Acids 108 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution -1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0 x = -b b2 - 4ac 2a x = 1.1 x 10-4 or -1.3 x 10-4 Prof. T. L. Heise CHE 116

Weak Acids 109 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution x = 1.1 x 10-4 1.1 x 10-4 x 100 1.0 x 10-3 Prof. T. L. Heise CHE 116

Weak Acids 110 Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution x = 1.1 x 10-4 1.1 x 10-4 x 100 = 11% 1.0 x 10-3 Prof. T. L. Heise CHE 116

Weak Acids 111 Polyprotic Acids: many acids have more than one ionizable H atom. H2SO3(aq) H+(aq) + HSO3-(aq) Ka1 HSO3-(aq) H+(aq) + SO3-2(aq) Ka2 The Ka are labeled according to which proton is dissociating. -it is always easier to remove the first proton than the second

Prof. T. L. Heise CHE 116 Weak Acids 112 Prof. T. L. Heise CHE 116 Weak Acids

113 If Ka values differ by 103 or more, only use Ka1 to determine calculations. Prof. T. L. Heise CHE 116 Weak Acids 114 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020

M solution of oxalic acid. Prof. T. L. Heise CHE 116 Weak Acids 115 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. H2C2O4 HC2O4- + H+ Ka = 5.9 x 10-2 = [HC2O4- ][H+]

Prof. T. L. Heise [H2C2O4] CHE 116 Weak Acids 116 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. H2C2O4 HC2O4- + H+ strong acid so 100% dissociation 0.020 M H+

Prof. T. L. Heise CHE 116 Weak Acids 117 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. HC2O4- C2O4-2 + H+ I 0.020 0 0.020

E 0.020 - x x 0.020 + x Prof. T. L. Heise -x +x +x CHE 116

Weak Acids 118 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. Ka = 6.4 x 10-5 = [0.020 + x ][x] [0.020 - x] use your assumption Prof. T. L. Heise CHE 116 Weak Acids

119 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. Ka = 6.4 x 10-5 = [0.020 ][x] [0.020] 6.4 x 10-5 = x = [C2O42-] Prof. T. L. Heise CHE 116 Weak Acids

120 Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. [H+] = 0.020 Prof. T. L. Heise CHE 116 Weak Acids 121

Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. pH = -log[H+] pH = - log[0.020] pH = 1.70 Prof. T. L. Heise CHE 116 Weak Bases 122

Many substances behave as weak bases in water. Such substances react with water, removing protons from water, leaving the OH- ion behind. NH3 + H2O NH4+ + OHKb = [NH4+][OH-] [NH3] * Kb is the base dissociation constant utilizing the [OH-] Prof. T. L. Heise CHE 116 Weak Bases 123

* Kb is the base dissociation constant utilizing the [OH -] bases must contain one or more lone pair to bond with the H + from water. as before, the larger the Kb the stronger the base stronger base have low pOH, but high pH Prof. T. L. Heise CHE 116 Weak Bases 124

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine B) methylamine C) nitrous acid Prof. T. L. Heise CHE 116 Weak Bases 125 Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine

Kb = 1.7 x 10-9 B) methylamine Kb = 4.4 x 10-4 C) nitrous acid Kb = 2.2 x 10-11 Prof. T. L. Heise CHE 116 Weak Bases

126 Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine Kb = 1.7 x 10-9 Kb = 1.7 x 10-9 = [x][x] [0.05] x = [OH-] = 9.2 x 10-6 pOH = 5.0 so pH = 9.0 Prof. T. L. Heise CHE 116 Weak Bases

127 Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? B) methylamine Kb = 4.4 x 10-4 Kb = 4.4 x 10-4 = [x][x] [0.05] x = [OH-] = 4.6 x 10-3 pOH = 2.32 so pH = 11.68 Prof. T. L. Heise CHE 116 Weak Bases

128 Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? C) nitrous acid Kb = 2.2 x 10-11 Kb = 2.2 x 10-11 = [x][x] [0.05] x = [OH-] = 1.0 x 10-6 pOH = 5.97 so pH = 8.02 Prof. T. L. Heise CHE 116

Weak Bases 129 Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine pH = 9.0 B) methylamine pH = 11.68 C) nitrous acid pH = 8.02 Prof. T. L. Heise CHE 116

Weak Bases 130 Identifying a Weak Base Neutral substances that have an atom with a nonbonding pair of electrons that can serve as a proton acceptor. Most of these are nitrogen atoms Anions of weak acids Prof. T. L. Heise CHE

116 Weak Bases 131 Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? Prof. T. L. Heise CHE 116 Weak Bases

132 Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? pH = 10.50 so pOH = 3.50 Prof. T. L. Heise CHE 116 Weak Bases 133

Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? pOH = 3.50 NH3 + H20 NH4+ + OH[OH-] = 3.16 x 10-4 Prof. T. L. Heise CHE 116 Weak Bases 134 Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?

NH3 + H20 NH4+ + OHx 0 0 -3.16 x 10 -4 +3.16 x 10-4 +3.16 x 10-4 x - 3.16 x 10-4 3.16 x 10-4 3.16 x 10-4 Prof. T. L. Heise CHE 116 Weak Bases 135 Sample exercise: A solution of NH3 in water has a pH of

10.50. What is the molarity of the solution? NH3 + H20 NH4+ + OHKb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4][x - 3.16 x 10-4] Prof. T. L. Heise CHE 116 Weak Bases 136 Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? NH3 + H20 NH4+ + OHKb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4]

[x - 3.16 x 10-4] x = 0.0058 M Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb 137 When two reactions are added to give a third reaction, the equilibrium constant for the third reaction is equal to the product of the equilibrium constants for the two added reactants.

NH4+(aq) NH3(aq) + H+(aq) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) H2O(l) H+(aq) + OH-(aq) Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb 138 NH4+(aq) NH3(aq) + H+(aq) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) H2O(l) H+(aq) + OH-(aq)

Ka x Kb = Kw pKa + pKb = pKw = 14 Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb 139 Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2B) PO43C) N3- Prof. T. L. Heise

CHE 116 Relationship Between Ka and Kb 140 Sample exercise: Which of the following anions has the largest basedissociation constant? A) NO2- Ka = 4.5 x 10-4 Ka x Kb = 1.0 x 10-14 Kb = 1.0 x 10-14 = 2.2 x 10-11 4.5 x 10-4 Prof. T. L. Heise CHE

116 Relationship Between Ka and Kb Sample exercise: Which of the following anions has the largest base-dissociation constant? B) PO43- 141 Ka = 4.2 x 10-13 Ka x Kb = 1.0 x 10-14 Kb = 1.0 x 10-14 = 2.4 x 10-2 4.2 x 10-13 Prof. T. L. Heise CHE 116

Relationship Between Ka and Kb 142 Sample exercise: Which of the following anions has the largest basedissociation constant? C) N3Ka = 1.9 x 10-5 Ka x Kb = 1.0 x 10-14 Kb = 1.0 x 10-14 = 5.2 x 10-10 1.9 x 10-5 Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb

143 Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2- Kb = 2.2 x 10-11 B) PO43- Kb = 2.4 x 10-2 C) N3- Kb = 5.2 x 10-10 Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb

144 Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline? Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb 145

Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline? pKa + pKb = 14 4.90 + x = 14 x = 9.1 Prof. T. L. Heise CHE 116 Relationship Between Ka and Kb 146 Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation

constant for quinoline? pKb = 9.1 pKb = -log[Kb] 9.1 = -log[Kb] Prof. T. L. Heise [Kb] = 7.9 x 10-10 CHE 116 Acid Base Properties of Salt Solns 147

Salt solutions have the potential to be acidic or basic. Hydrolysis of a salt acid base properties are due to the behavior of their cations and anions perform the necessary double replacement reaction and examine the products using your strength rules Prof. T. L. Heise CHE 116 Acid Base Properties of Salt Solns 148 If a strong acid and strong base are produced, the resultant solution will

be neutral. If a strong acid and weak base are produced, the resultant solution will be acidic. If a strong base and a weak acid are produced, the resultant solution will be basic. Prof. T. L. Heise CHE 116 Acid Base Properties of Salt Solns 149 If a weak acid and weak base are produced, the

resultant solution will be dependent on the K a values. Prof. T. L. Heise CHE 116 Acid Base Properties of Salt Solns 150 Sample exercise: In each of following, which salt will form the more acidic solution. A) NaNO3, Fe(NO3)3 NaNO3 + H2O NaOH + HNO3 SB

SA Prof. T. L. Heise CHE 116 Acid Base Properties of Salt Solns 151 Sample exercise: In each of following, which salt will form the more acidic solution. A) NaNO3, Fe(NO3)3 Fe(NO3)3 + 3H2O Fe(OH)3 + 3HNO3 WB

Prof. T. L. Heise SA CHE 116 Acid Base Properties of Salt Solns 152 Sample exercise: In each of following, which salt will form the more acidic solution. A) NaNO3, Fe(NO3)3 Fe(NO3)3 + 3H2O Fe(OH)3 + 3HNO3 WB

Prof. T. L. Heise SA CHE 116 Acid Base Properties of Salt Solns 153 Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO

Prof. T. L. Heise CHE 116 Acid Base Properties of Salt Solns 154 Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO KBr + H2O KOH + HBr SB Prof. T. L. Heise

SA CHE 116 Acid Base Properties of Salt Solns 155 Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO KBrO + H2O KOH + HBrO SB Prof. T. L. Heise

WA CHE 116 Acid Base Properties of Salt Solns 156 Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO Prof. T. L. Heise CHE

116 Acid-Base Behavior & Chem Structure 157 How does the chemical structure determine which of the behaviors will be exhibited? Acidic a molecule containing H will transfer a proton only if the H-X bond is polarized like H -- X the stronger the bond the weaker the acid and vice versa

Prof. T. L. Heise CHE 116 Acid-Base Behavior & Chem Structure 158 How does the chemical structure determine which of the behaviors will be exhibited? Acidic oxyacids exist when the H is attached to an oxygen bonded to a central atom if the OHs attached are equal in number to the Os present, acid strength increases with electronegativity

Prof. T. L. Heise CHE 116 Acid-Base Behavior & Chem Structure 159 How does the chemical structure determine which of the behaviors will be exhibited? Acidic oxyacids exist when the H is attached to an oxygen bonded to a central atom the more Os present compared to the OHs, the more polarized the OH bond becomes and the stronger the acid is

Prof. T. L. Heise CHE 116 Acid-Base Behavior & Chem Structure 160 How does the chemical structure determine which of the behaviors will be exhibited? Acidic carboxylic acids exist when the functional group COOH is present the strength

also increases as the number of electronegative atoms in the molecule increase Prof. T. L. Heise CHE 116

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