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CHEM160 General Chemistry II Lecture Presentation Aqueous Equilibria: Acids and Bases. Chapter 16. Dr. Daniel Autrey. Why study acids and bases? . Acids and bases are common in the everyday world as well as in the lab. Some common acidic products

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chem160 general chemistry ii lecture presentation aqueous equilibria acids and bases

CHEM160 General Chemistry IILecture PresentationAqueous Equilibria: Acids and Bases

Chapter 16

Dr. Daniel Autrey

Chapter 16

why study acids and bases
Why study acids and bases?
  • Acids and bases are common in the everyday world as well as in the lab.
    • Some common acidic products
      • vinegar (dilute acetic acid), soft drinks (carbonic acid), aspirin (acetylsalicylic acid), vitamin C (ascorbic acid), lemonade (citric acid + ascorbic acid), muriatic acid (hydrochloric acid)
    • Some common basic products
      • Antacids such as milk of magnesia (magnesium hydroxide) and tums (calcium carbonate), household ammonia, oven and drain cleaners (sodium hydroxide), some bleaches (sodium hydroxide)

Chapter 16

why study acids and bases1
Why study acids and bases?
  • Many biological and geological processes involve acid-base chemistry.
    • Gastric juice contains hydrochloric acid
    • Lactic acid builds up in muscles during strenuous exercise
    • Basicity of blood must be maintained within a certain narrow range or death can result
    • Acidity/basicity of soil and water are of importance to animals and plants living there
    • Cave formation and weathering of rocks is affected by acidity of water

Chapter 16

why study acids and bases2
Why study acids and bases?
  • 15 of the top 50 chemicals produced in the largest quantities annually in the U.S. are acids or bases. (#1 = sulfuric acid)
    • Used as reactants and catalysts in manufacture of various consumer and industrial products
      • Plastics, synthetic fibers, detergents, pharmaceuticals, agricultural fertilizers, explosives, etc.

Chapter 16

why study acids and bases3
Why study acids and bases?
  • So why study this acid-base stuff?
    • Acid-base reactions constitute an important class of chemical reactions
    • Understanding acid-base chemistry is necessary for chemistry, biology, geology, and other related scientific disciplines

Chapter 16

concepts to review
Concepts to Review
  • Aqueous Reactions and Solution Stoichiometry
    • Acids and bases, neutralization, electrolytes (section 4.3)
    • Molarity and molarity calculations (section 4.5)
    • Solution stoichiometry (section 4.7)
  • Chemical equilibrium
    • Equilibrium constants, solving equilibrium problems, LeChatelier’s principle (chapter 14)

Chapter 16

arrhenius concept
Arrhenius Concept
  • Arrhenius Concept (S. Arrhenius, 1887)
    • Acids produce hydrogen ions, H+, in water
      • H+ ions attach to H2O molecules forming hydronium ions, H3O+
    • Bases produce hydroxide ions, OH-, in water
      • Base contains OH group in its formula.
    • Neutralization reaction: acid + base  salt + water

Chapter 16

arrhenius concept1
Arrhenius Concept
  • Arrhenius Concept (S. Arrhenius, 1887)
    • Acids produce hydrogen ions, H+, in water
      • H+ ions attach to H2O molecules forming hydronium ions, H3O+
    • Bases produce hydroxide ions, OH-, in water
      • Base contains OH group in its formula.
    • Neutralization reaction: acid + base  salt + water
    • Problems
      • Some basic substances do not have OH-
      • Confined to H2O solutions
      • H+ does not exist free in water
        • Forms H3O+

Chapter 16

bronsted lowry concept
Bronsted-Lowry Concept
  • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923)
    • Acid-base reaction involves proton transfer
      • HA + B  HB+ + A-
    • Acid: proton donor
    • Base: proton acceptor
      • does not have to have OH- in formula

Chapter 16

bronsted lowry concept1
Bronsted-Lowry Concept

Example:

acid

base

Chapter 16

bronsted lowry concept2
Bronsted-Lowry Concept

Example:

acid

base

Chapter 16

bronsted lowry concept3
Bronsted-Lowry Concept
  • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923)
    • Acid-base reaction involves proton transfer
    • Acid: proton donor
    • Base: proton acceptor
      • does not have to have OH- in formula
    • Water is amphoteric

Chapter 16

bronsted lowry concept4
Bronsted-Lowry Concept

acid

base

base

acid

Chapter 16

bronsted lowry concept5
Bronsted-Lowry Concept
  • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923)
    • Acid-base reaction involves proton transfer
    • Acid: proton donor
    • Base: proton acceptor
      • does not have to have OH- in formula
    • Water is amphoteric
    • Acids & bases can be molecules or ions

Chapter 16

bronsted lowry concept6
Bronsted-Lowry Concept

acid

base

Chapter 16

bronsted lowry concept7
Bronsted-Lowry Concept
  • Bronsted-Lowry Concept (J. Bronsted, T. Lowry, 1923)
    • Acid-base reaction involves proton transfer
    • Acid: proton donor
    • Base: proton acceptor
      • does not have to have OH- in formula
    • Water is amphoteric
    • Acids & bases can be molecules or ions
    • Back reaction is also a proton transfer.

Chapter 16

bronsted lowry concept8
Bronsted-Lowry Concept

acid

base

base

acid

Chapter 16

bronsted lowry concept9
Bronsted-Lowry Concept

acid

base

base

acid

Acid-base conjugate pairs

Chapter 16

bronsted lowry concept10
Bronsted-Lowry Concept
  • Conjugate Acid-Base Pair (HA/A-, HB+/B)
    • Two chemical species whose formulas differ by only one proton

acid  conj. base + H+

base + H+ conj. acid

Chapter 16

acid base strength
Acid-Base Strength
  • Strong acids
    • Ionize completely in water

HX + H2O  H3O+ + X-

  • Weak acids
    • Ionize only partially in water

HX + H2O  H3O+ + X-

    • Common weak acids: Numerous molecules and certain cations (we’ll learn more about these later)

Chapter 16

strong acids

HX

H3O+

X-

Strong Acids

Before Equilibrium

At Equilibrium

100 %

Chapter 16

acid base strength1
Acid-Base Strength
  • Common Strong Acids (Know these!)

HCl

HBr

HI

HNO3

HClO4

H2SO4

Chapter 16

weak acids

HX

Weak Acids

Before Equilibrium

At Equilibrium

HX

< 100 %

H3O+

X-

Chapter 16

acid base strength2
Acid-Base Strength
  • Strong bases
    • Ionize completely in water

NaOH  Na+ + OH-

    • Common strong bases: MOH and M(OH)2, M2O and MO (M = Grp IA and IIA metals)
  • Weak bases
    • Ionize only partially in water

B + H2O  BH+ + OH-

    • Common weak bases: NR3 (R = H or other chemical species) and certain anions

Chapter 16

relative acid base strength
Relative Acid-Base Strength

Strongest HClO4 ClO4- Weakest

Acids H2SO4 HSO4- Bases

HCl Cl-

HNO3 NO3-

H3O+ H2O

HSO4- SO42-

HNO2 NO2-

HClO ClO-

Weakest NH4+ NH3 Strongest

Acids H2O OH- Bases

OH- O2-

H2 H-

Strongest acid that can exist in H2O

Strongest base that can exist in H2O

Chapter 16

more on neutralization reactions
More on Neutralization Reactions
  • Acids react with bases:
    • acid + base  salt + water

or

    • acid + base  salt

Chapter 16

autoionization of water
Autoionization of Water
  • In pure water, the following equilibrium occurs: H2O + H2O  H3O+ + OH-
  • The equilibrium constant for this reaction is the ion product, Kw:

Kw = [H3O+][OH-]

    • [H3O+] = [OH-] = 1 x 10-7 M (25 °C)

so

Kw = 1 x 10-14 (25 °C)

Chapter 16

example problem 1 1a and 1b on example problem handout
Example Problem 1(1a and 1b on Example Problem Handout)
  • Calculate the [OH-] in a solution with [H3O+] = 0.00028 M (25 °C). (ans.: [OH-] = 3.6 x 10-11 M)
  • Calculate the [H3O+] in a solution with [OH-] = 0.056 M (25 °C). (ans.: [H3O+] = 1.8 x 10-13 M)

Chapter 16

example problem 2 5a and 5b on example problem handout
Example Problem 2(5a and 5b on Example Problem Handout)
  • Calculate the concentration of [H3O+] and [OH-] for the following aqueous solutions at 25 °C:
  • 0.0035 M HCl

(ans.: [H3O+] = 0.0035 M, [OH-] = 2.9 x 10-12 M)

(b) 0.22 M NaOH

(ans.: [OH-] = 0.22 M, [H3O+] = 4.5 x 10-14 M)

Chapter 16

problem solving strategy
Problem Solving Strategy
  • Determine whether the acid or base is strong or weak. HCl and NaOH are strong electrolytes.
  • If autoionization is considered to be negligible (Cacid > 10-6 M, Cbase > 10-6 M), then:
    • For strong acids, [H3O+] = CAcid
    • For strong bases, [OH-] = CBase for MOH or [OH-] = 2CBase for M(OH)2
  • [OH-] or [H3O+] are then calculated using the expression for Kw:
    • Kw = 1 x 10-14 = [H3O+] [OH-]

Chapter 16

ph and poh scales convenient ways to express acidity and basicity
pH and pOH Scales: Convenient ways to express acidity and basicity
  • We can represent the concentration of [H3O+] as
    • pH = -log [H3O+]
  • We can represent the concentration of [OH-] as
    • pOH = -log [OH-]
  • The p function simply means -log. For a number X:
    • pX = -log X

Chapter 16

ph and poh scales
pH and pOH Scales
  • A relationship between pH and pOH may be derived for all aqueous solutions:
    • Kw = [H3O+][OH-]
    • Take the -log of each side:

-logKw = -log([H3O+][OH-])

-logKw = -log[H3O+] - log[OH-]

pKw = pH + pOH

14.00 = pH + pOH ( at 25 °C)

Chapter 16

example 3 2a on example problem handout
Example 3(2a on Example Problem Handout)
  • Calculate the pH and pOH of a solution in which [H3O+] = 0.00028 M at 25 °C.

(ans.: pH = 3.55, pOH = 10.45)

Chapter 16

ph calculations with calculators

2.5567839

log

inv

ln

yx

cos

sin

tan

Exp

8

7

9

x

5

4

6

2

1

3

+

+/-

0

·

-

pH Calculations with Calculators
  • To calculate pH from [H3O+]:
    • Enter [H3O+]
    • Press “log” key
    • Change sign by pressing “+/-” key
  • pOH is calculated in an analogous manner.

Chapter 16

example 4 3a on example problem handout
Example 4(3a on Example Problem Handout)
  • Calculate the [H3O+] and [OH-] of a solution if the pH is 4.88. (ans.: [H3O+] = 1.3 x 10-5 M, [OH-] = 7.6 x 10-10 M)

Chapter 16

calculating h 3 o from ph

0.00055783198

log

inv

ln

yx

cos

sin

tan

Exp

8

7

9

x

5

4

6

2

1

3

+

+/-

0

·

-

Calculating [H3O+] from pH
  • To calculate [H3O+] from pH:
    • Enter pH
    • Change sign by pressing “+/-” key
    • Take the antilog by pressing the “2nd/inv/shift” key, followed by the “log” key
  • [OH-] is calculated in an analogous manner.

Chapter 16

classifying aqueous solutions
Classifying Aqueous Solutions

[H3O+] [OH-] pH pOH

neutral = 10-7 M = 10-7 M = 7.0 = 7.0

acidic > 10-7 M < 10-7 M < 7.0 > 7.0

basic < 10-7 M > 10-7 M > 7.0 < 7.0

Chapter 16

example 5 6a on example problem handout
Example 5(6a on Example Problem Handout)
  • Calculate the [H3O+], [OH-], pH, and pOH for a solution prepared by dissolving 0.100 moles HCl(g) into enough water to give a volume of 0.350 L.

(ans.: [H3O+] = 0.29 M, [OH-] = 3.5 x 10-14 M, pH = 0.54, pOH = 13.46)

Chapter 16

problem solving strategy1
Problem Solving Strategy
  • Identify the substance as an acid or base
  • Determine whether the acid or base is strong or weak
    • If strong acid, [H3O+] = CAcid
    • If strong base, [OH-] = CBase for MOH or [OH-] = 2CBase for M(OH)2
  • Calculate CHA (moles/L) (= [H3O+])
  • Calculate pH = -log[H3O+]

Chapter 16

problem solving strategy2
Problem Solving Strategy
  • Calculate pOH from pOH = 14.00 - pH
  • Calculate [OH-] from either [OH-] = 10-pOH or [OH-] = Kw/ [H3O+]

Chapter 16

example 6 7a on example problem handout
Example 6(7a on Example Problem Handout)
  • Calculate the [H3O+], [OH-], pH, and pOH for a solution prepared by dissolving 3.1 g KOH(s) (FW = 56 g/mol) into enough water to give a volume of 0.500 L.

(ans.: [OH-] = 0.11 M, [H3O+] = 9.0 x 10-14 M, pOH = 0.96, pH = 13.04)

Chapter 16

weak acid and base ionizations
Weak Acid and Base Ionizations
  • How do we calculate the [H3O+] and pH of weak acid or base solutions if these do not ionize completely?
    • Consider the specific weak acid or base dissociation equilibrium involved.
    • Perform equilibrium calculations to determine concentrations of species in the reaction mixture at equilibrium.

Chapter 16

weak acid ionization
Weak Acid Ionization
  • Consider the ionization of a weak acid:

HA + H2O <=> H3O+ + A-

  • The equilibrium constant for this reaction is the acid dissociation constant, Ka:

Ka = [H3O+][A-]/[HA]

(Note that these are all equilibrium concentrations.)

Chapter 16

k a s for some weak acids
Ka’s for Some Weak Acids

Increasing acid strength

H+ ion that ionizes shown in red.

Chapter 16

calculations with k a
Calculations with Ka
  • Ka can be used to calculate:
    • equilibrium concentrations
    • pH
    • percent ionization

(Any equilibrium constant can be used to calculate equilibrium concentrations for solution species.)

Chapter 16

calculations using k a
Calculations Using Ka
  • Basic Steps for Weak Acid Calculations Using Ka
    • Write balanced chemical equation and the expression for Ka
      • Look up value for Ka
    • For each chemical species involved in the equilibrium (except H2O), write:
      • Initial concentration
      • Equilibrium concentration
        • Let the change in the [H3O+] be the variable “x”
    • Substitute the equilibrium concentrations into Ka and solve for x using either
      • quadratic approach
      • simplified approach
    • Calculate pH, equilibrium concentrations, % ionization, etc., as specified in the problem.

Chapter 16

example 7 8a on the example problem handout
Example 7(8a on the Example Problem Handout)
  • Calculate the equilibrium concentration of each species, the pH, and the percent ionization for a 0.500 M solution of acetic acid, represented as either HOAc or CH3COOH (Ka = 1.8 x 10-5).

(ans.: [HOAc] = 0.497 M, [OAc-] = [H3O+]

= 0.0030 M, [OH-] = 3.3 x 10-12 M, 0.60%)

Chapter 16

weak acid ionization1
Weak Acid Ionization
  • Some acids have more than one ionizable proton.
    • Called polyprotic acids
    • Ionize in steps
  • Consider the ionization of a weak diprotic acid:

H2A + H2O <=> H3O+ + HA- Ka1 = [H3O+][HA-]/[H2A]

HA- + H2O <=> H3O+ + A2- Ka2 = [H3O+][A2-]/[HA-]

  • The 1st ionization effectively controls solution pH for most polyprotic acids:
    • Ka1 >> Ka2 > Ka3, etc.
    • When calculating pH, ignore any ionization after the first ionization if Ka1 >> Ka2.

Chapter 16

example 8 example 9a on example problem handout
Example 8(Example 9a on Example Problem Handout)
  • Calculate the equilibrium concentration of each species and the pH of a 0.250 M solution of H2CO3 (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11).

(ans.: [H2CO3] = 0.250 M, [HCO3-] = [H3O+] = 0.00033 M,

[CO32-] = 5.6 x 10-11 M], [OH-] = 3.0 x 10-11 M, pH = 3.48)

Chapter 16

weak base ionization
Weak Base Ionization
  • Consider the ionization of a weak base:

B + H2O <=> BH+ + OH-

  • The equilibrium constant for this reaction is the base ionization constant:

Kb = [BH+][OH-]/[B]

(Note that these are all equilibrium concentrations.)

  • The same principles that apply to weak acids apply to weak bases.

Chapter 16

example 9 10a on example problem handout
Example 9(10a on Example Problem Handout)
  • Calculate the pOH, pH, and the percent ionization for a 0.25 M solution of ammonia, NH3 (Kb = 1.8 x 10-5). (ans.: pOH = 2.67, pH = 11.32, 0.85%)

Chapter 16

acid base properties of salts
Acid-Base Properties of Salts
  • Acids and bases can be either neutral molecules (CH3COOH, NH3) or ions (NH4+, F-).
    • Ionic acids are typically conjugate acids of weak molecular bases
    • Ionic bases are conjugate bases of weak molecular acids
  • Examples of ionic acid/basehydrolysis reaction

NH4+ + H2O <=> NH3 + H3O+

weak acid

F- + H2O <=> HF + OH-

weak base

Chapter 16

acid base properties of salts1
Acid-Base Properties of Salts
  • What is the relationship between the strength of an acid and its conjugate base?
    • The stronger the acid, the weaker the conjugate base.
  • We can quantify this mathematically in terms of Ka and Kb.

Kw = KaKb

    • For ionic acids and bases, Ka and Kb must often be calculated using this equation.

Chapter 16

example 10 11a on example problem handout
Example 10(11a on Example Problem Handout)
  • Calculate the pH and % hydrolysis of a 0.22 M solution of CH3COONa (Ka for CH3COOH, the conjugate acid of CH3COO- is 1.8 x 10-5) (ans.: pH = 9.05, 0.0050%)

Chapter 16

acid base properties of salts predicting ph of salt solutions
Acid-Base Properties of SaltsPredicting pH of Salt Solutions
  • Salt from reaction of:
    • Strong base MOH and strong acid HX is neutral
    • Strong base MOH and weak acid HX is basic
    • Weak base B and strong acid HX is acidic
    • Weak base B and weak acid HX is:
      • Acidic if Ka for cation > Kb for anion
      • Basic if Ka for cation < Kb for anion
      • Neutral if Ka for cation = Kb for anion

Chapter 16

lewis concept
Lewis Concept
  • Lewis concept (1923)
    • Emphasis is on shared electron pair between the acid and base
      • Acid-base reaction forms a coordinate covalent bond
    • Acid
      • electron pair acceptor
    • Base
      • electron pair donor

Chapter 16

lewis concept1
Lewis Concept
  • Example:

Lewis acid

Lewis base

Chapter 16

lewis concept2
Lewis Concept
  • Example:

Lewis acid

Lewis base

Chapter 16

lewis concept3
Lewis Concept
  • Arrhenius acid-base reactions are also Bronsted-Lowry acid-base reactions.
  • Bronsted-Lowry acid-base reactions are also Lewis acid-base reactions.
    • Reverse is not necessarily true!

Chapter 16

lewis acid base concept hydrolysis of metal ions
Lewis Acid-Base ConceptHydrolysis of Metal Ions
  • Metal cations from the central region of the periodic table tend to be acidic in water.
    • Involves both Lewis and Bronsted-Lowry acid-base chemistry
  • Example: Al3+. In H2O, Al3+ reacts with 6 H2O molecules, forming hydrated Al3+ ions, Al(H2O)63+. All metal ions undergo hydration in water.

Al3+ + 6H2O Al(H2O)63+

Lewis acid Lewis base

Chapter 16

slide63

Al(H2O)63+ = Al3+(aq)

Oxygen

Al3+

Lone pair

Hydrogen

Chapter 16

lewis acid base concept hydrolysis of metal ions1
Lewis Acid-Base ConceptHydrolysis of Metal Ions
  • High + charge/small size of Al3+ attracts e- density from bound H2O
    • Al3+ has high + charge density
  • O-H bond becomes more polar, making bound H2O more acidic than free H2O

e- shift

Chapter 16

lewis acid base concept hydrolysis of metal ions2
Lewis Acid-Base ConceptHydrolysis of Metal Ions
  • Hydrated metals act as weak Bronsted-Lowry acids:

Al(H2O)63+ + H2O <=> Al(H2O)5(OH)2+ + H3O+

    • Ka for metal hydrolysis increases with:
      • Increasing + charge
      • Decreasing ionic radius
  • Other example metals that form acidic solutions:

Zn2+, Fe3+, Co2+, Cr3+, Ni2+, Cu2+

  • Group IA and IIA metals generally do not undergo hydrolysis due to large ionic radius (low + charge density).

Chapter 16