CHE 116: General Chemistry CHAPTER SIXTEEN Copyright © Tyna L. Heise 2001-2002 All Rights Reserved

Acids and Bases: Review Properties of Acids sour taste change with litmus Properties of Bases bitter taste change with litmus

Acids and Bases: Review 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-

Bronsted-Lowry Acids and Bases 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

Bronsted-Lowry Acids and Bases 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

Bronsted-Lowry Acids and Bases Demonstrates the interconnections possible between hydrogenated water Fig 16.1

Bronsted-Lowry Acids and Bases 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

Bronsted-Lowry Acids and Bases Definitions: Amphoteric - some substances can be an 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.

Bronsted-Lowry Acids and Bases Fig 16.7, 16.8

Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO3- F- PO43- CO

Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO3- F- PO43- CO Given a base, bases accept H+, so add an H+ to each molecule.

Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3- F- PO43- CO Given a base, bases accept H+, so add an H+ to each molecule.

Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3- HF F- PO43- CO Given a base, bases accept H+, so add an H+ to each molecule.

Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3- HF F- HPO42- PO43- CO Given a base, bases accept H+, so add an H+ to each molecule.

Bronsted-Lowry Acids and Bases Sample Exercise: Write the formula for the conjugate acid of each of the following: H2SO3 HSO3- HF F- HPO42- PO43- HCO+ CO Given a base, bases accept H+, so add an H+ to each molecule.

Bronsted-Lowry Acids and Bases 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.

Bronsted-Lowry Acids and Bases 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-

Bronsted-Lowry Acids and Bases 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-

Bronsted-Lowry Acids and Bases 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 base base acid

Bronsted-Lowry Acids and Bases Relative Strengths of Acids and Bases: the more readily a substance donates an H+, the less readily it’s conjugate base will accept one the more readily a substance accepts an H+, the less readily it’s conjugate acid will donate one the stronger one of the substances is, the weaker it’s conjugate

Bronsted-Lowry Acids and Bases 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

Bronsted-Lowry Acids and Bases 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

Bronsted-Lowry Acids and Bases Fig. 16.4

Bronsted-Lowry Acids and Bases 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)

Bronsted-Lowry Acids and Bases 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 HPO42- 2 bases are: PO43- and OH-

Bronsted-Lowry Acids and Bases 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 HPO42- 2 bases are: PO43- and OH- red indicates strength

Bronsted-Lowry Acids and Bases 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 HPO42- 2 bases are: PO43- and OH- reverse reaction favored

Bronsted-Lowry Acids and Bases 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 HPO42- 2 bases are: PO43- and OH- shifts left

Bronsted-Lowry Acids and Bases 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)

Bronsted-Lowry Acids and Bases 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-

Bronsted-Lowry Acids and Bases 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- red indicates strength

Bronsted-Lowry Acids and Bases 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- favors forward reaction

Bronsted-Lowry Acids and Bases 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- shifts right

The Autoionization of Water 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

The Autoionization of Water 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...

The Autoionization of Water 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.

The Autoionization of Water 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

The Autoionization of Water 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

The Autoionization of Water 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

The Autoionization of Water 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

The Autoionization of Water 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

The Autoionization of Water 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

The Autoionization of Water 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

The pH Scale 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

The pH Scale

The pH Scale Common household products and their relative pH’s.

The pH Scale Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?

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

The pH Scale 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]

The pH Scale 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

The pH Scale Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH?

The pH Scale 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+]

The pH Scale 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]

The pH Scale 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

The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].

The pH Scale Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+]. pH = -log[H+]

The pH Scale 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+]

The pH Scale 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+]

The pH Scale 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+]

The pH Scale 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

The pH Scale 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

The pH Scale Fig 16.7

Strong Acids and Bases 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

Strong Acids and Bases 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]

Strong Acids and Bases An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?

Strong Acids and Bases An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid? pH = -log[H+]

Strong Acids and Bases 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+]

Strong Acids and Bases 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+]

Strong Acids and Bases 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+

Strong Acids and Bases 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

Strong Acids and Bases 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

Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?

Strong Acids and Bases Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89? pH = -log[H+]

Strong Acids and Bases 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+]

Strong Acids and Bases 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+]

Strong Acids and Bases 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+]

Strong Acids and Bases 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

Strong Acids and Bases 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

Strong Acids and Bases 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 OH- 1 mol KOH 1 mol OH-

Strong Acids and Bases 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 OH- 1 mol KOH 1 mol OH- 7.8 x 10-3 M KOH

Weak Acids 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+ + X- Ka = [H3O+][X-] [HX] *the larger the Ka the stronger the acid

Weak Acids 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.

Weak Acids 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?

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? pH = -log[H+] 3.26 = -log[H+] e-3.26 = [H+] 5.5 x 10-4 = [H+]

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  C6H4NOO- + 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

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? % = part/total x 100 = 5.5 x 10-4/0.01945 x 100 = 2.8%

Weak Acids 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+] [C6H4NOOH] = (5.5 x 10-4 )(5.5 x 10-4 ) 0.01945 = 1.55 x10-5

Weak Acids 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)

Weak Acids 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...

Weak Acids 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

Weak Acids 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...

Weak Acids Ka = 1.8 x 10-5 = [x][x] [0.30] 1.8 x 10-5 (0.30) = x2 0.0023 = [H+] pH = -log [H+] = -log[0.0023] = 2.64

Weak Acids Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?

Weak Acids 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]

Weak Acids 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 0.010 0 0 change -x +x +x equilibrium 0.010 -x x x

Weak Acids 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]

Weak Acids 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

Weak Acids 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+] pH = -log[H+] pH = -log[3.87 x 10-4] pH = 3.41

Weak Acids 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...

Weak Acids As the concentration of a weak acid increases, the % ionized decreases.

Weak Acids As the concentration of a weak acid increases, the % ionized decreases.

Weak Acids Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise b) a 1.0 x 10-3 M solution

Weak Acids 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

Weak Acids 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

Weak Acids 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 is greater than 5% so use quadratic...

Weak Acids 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

Weak Acids 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 =

Weak Acids 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

Weak Acids 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

Weak Acids 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

Weak Acids 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

Weak Acids

Weak Acids If Ka values differ by 103 or more, only use Ka1 to determine calculations.

Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid.

Weak Acids 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+] [H2C2O4]

Weak Acids 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+

Weak Acids 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 D -x +x +x E 0.020 - x x 0.020 + x

Weak Acids 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

Weak Acids 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-]

Weak Acids Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O42-] in a 0.020 M solution of oxalic acid. [H+] = 0.020

Weak Acids 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

Weak Bases 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+ + OH- Kb = [NH4+][OH-] [NH3] * Kb is the base dissociation constant utilizing the [OH-]

Weak Bases * 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

Weak Bases 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

Weak Bases 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

Weak Bases 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

Weak Bases 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

Weak Bases 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

Weak Bases 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

Weak Bases 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

Weak Bases Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?

Weak Bases 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

Weak Bases 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

Weak Bases Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? NH3 + H20  NH4+ + OH- x 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

Weak Bases Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? NH3 + H20  NH4+ + OH- Kb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4] [x - 3.16 x 10-4]

Weak Bases Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? NH3 + H20  NH4+ + OH- Kb = 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

Relationship Between Ka and Kb 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)

Relationship Between Ka and Kb 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

Relationship Between Ka and Kb Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2- B) PO43- C) N3-

Relationship Between Ka and Kb Sample exercise: Which of the following anions has the largest base-dissociation 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

Relationship Between Ka and Kb Sample exercise: Which of the following anions has the largest base-dissociation constant? B) PO43- 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

Relationship Between Ka and Kb Sample exercise: Which of the following anions has the largest base-dissociation constant? C) N3- Ka = 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

Relationship Between Ka and Kb 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

Relationship Between Ka and Kb Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline?

Relationship Between Ka and Kb 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

Relationship Between Ka and Kb 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] [Kb] = 7.9 x 10-10

Acid Base Properties of Salt Soln’s 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

Acid Base Properties of Salt Soln’s 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.

Acid Base Properties of Salt Soln’s If a weak acid and weak base are produced, the resultant solution will be dependent on the Ka values.

Acid Base Properties of Salt Soln’s 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

Acid Base Properties of Salt Soln’s 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 SA

Acid Base Properties of Salt Soln’s 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 SA

Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO

Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO KBr + H2O  KOH + HBr SB SA

Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO KBrO + H2O  KOH + HBrO SB WA

Acid Base Properties of Salt Soln’s Sample exercise: In each of following, which salt will form the more acidic solution. B) KBr, KBrO

Acid-Base Behavior & Chem Structure 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

Acid-Base Behavior & Chem Structure 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 OH’s attached are equal in number to the O’s present, acid strength increases with electronegativity

Acid-Base Behavior & Chem Structure 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 O’s present compared to the OH’s, the more polarized the OH bond becomes and the stronger the acid is

Acid-Base Behavior & Chem Structure 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