1. CHEM160 General Chemistry II Lecture Presentation Aqueous Equilibria: Acids and Bases
Chapter 16
Dr. Daniel Autrey
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2. 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)
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3. 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
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4. 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.
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5. 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
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6. 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)
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7. 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
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8. 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+
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9. The Hydrated Proton: Hydronium Ion
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10. 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
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11. Bronsted-Lowry Concept
Example:
acid
base
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12. Bronsted-Lowry Concept
Example:
acid
base
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13. 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
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14. Bronsted-Lowry Concept
acid
base
base
acid
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15. 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
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16. Bronsted-Lowry Concept
acid
base
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17. 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.
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18. Bronsted-Lowry Concept
acid
base
base
acid
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19. Bronsted-Lowry Concept
acid
base
base
acid
Acid-base conjugate pairs
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20. 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
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21. Acid-Base Strength Strong acids Weak 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)
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22. Strong Acids 100 % Before Equilibrium At Equilibrium HX H3O+ X-
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23. Acid-Base Strength Common Strong Acids (Know these!) HCl HBr HI HNO3
HClO4
H2SO4
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24. Weak Acids < 100 % Before Equilibrium At Equilibrium HX HX H3O+ X-
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25. Acid-Base Strength Strong bases Weak 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
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26. Relative Acid-Base Strength
Strongest HClO ClO Weakest
Acids H2SO HSO Bases
HCl Cl-
HNO NO3-
H3O H2O
HSO SO42-
HNO NO2-
HClO ClO-
Weakest NH NH Strongest
Acids H2O OH Bases
OH O2-
H2 H-
Strongest acid that can exist in H2O
Strongest base that can exist in H2O
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27. More on Neutralization Reactions
Acids react with bases:
acid + base  salt + water or
acid + base  salt
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28. 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 (25 °C)
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29. Example Problem 1 (1a and 1b on Example Problem Handout)
Calculate the [OH-] in a solution with [H3O+] = M (25 °C). (ans.: [OH-] = 3.6 x M)
Calculate the [H3O+] in a solution with [OH-] = M (25 °C). (ans.: [H3O+] = 1.8 x M)
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30. 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:
M HCl
(ans.: [H3O+] = M, [OH-] = 2.9 x M)
(b) 0.22 M NaOH
(ans.: [OH-] = 0.22 M, [H3O+] = 4.5 x M)
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31. 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 = [H3O+] [OH-]
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32. 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
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33. 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)
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34. Example 3 (2a on Example Problem Handout)
Calculate the pH and pOH of a solution in which [H3O+] = M at 25 °C.
(ans.: pH = 3.55, pOH = 10.45)
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35. 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.
inv
log ln yx
sin
cos
tan 7 8 9 x 4 5 6 1 2 3 +
+/- - 
Exp
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36. Example 4 (3a on Example Problem Handout)
Calculate the [H3O+] and [OH-] of a solution if the pH is (ans.: [H3O+] = 1.3 x 10-5 M, [OH-] = 7.6 x M)
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37. 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.
inv
log ln yx
sin
cos
tan 7 8 9 x 4 5 6 1 2 3 +
+/- -
Exp 
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38. Classifying Aqueous Solutions
[H3O+] [OH-] pH pOH
neutral = 10-7 M = 10-7 M = = 7.0
acidic > 10-7 M < 10-7 M < > 7.0
basic < 10-7 M > 10-7 M > < 7.0
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39. Example 5 (6a on Example Problem Handout)
Calculate the [H3O+], [OH-], pH, and pOH for a solution prepared by dissolving moles HCl(g) into enough water to give a volume of L.
(ans.: [H3O+] = 0.29 M, [OH-] = 3.5 x M, pH = 0.54, pOH = 13.46)
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40. 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+]
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41. Problem Solving Strategy
Calculate pOH from pOH = pH
Calculate [OH-] from either [OH-] = pOH or [OH-] = Kw/ [H3O+]
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42. 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 L.
(ans.: [OH-] = 0.11 M, [H3O+] = 9.0 x M, pOH = 0.96, pH = 13.04)
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43. 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.
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44. 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.)
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45. Ka’s for Some Weak Acids
Iodic, HIO3
0.17
Chloroacetic, CH2ClCOOH
1.4 x 10-3
Hydrofluoric, HF
6.8 x 10-4
Nitrous, HNO2
4.5 x 10-4
Formic, HCOOH
1.8 x 10-4
Acetic, CH3COOH
1.8 x 10-5
Hypochlorous, HOCl
3.0 x 10-8
Hypoiodous, HOI
2.3 x 10-11
Increasing acid strength
H+ ion that ionizes shown in red.
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46. 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.)
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47. 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.
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48. Example 7 (8a on the Example Problem Handout)
Calculate the equilibrium concentration of each species, the pH, and the percent ionization for a M solution of acetic acid, represented as either HOAc or CH3COOH (Ka = 1.8 x 10-5).
(ans.: [HOAc] = M, [OAc-] = [H3O+]
= M, [OH-] = 3.3 x M, 0.60%)
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49. 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+ + A 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.
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50. Example 8 (Example 9a on Example Problem Handout)
Calculate the equilibrium concentration of each species and the pH of a M solution of H2CO3 (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11).
(ans.: [H2CO3 ] = M, [HCO3-] = [H3O+] = M,
[CO32-] = 5.6 x M], [OH-] = 3.0 x M, pH = 3.48)
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51. (Note that these are all equilibrium concentrations.)
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.
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52. 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%)
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53. 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/base hydrolysis reaction
NH4+ + H2O <=> NH3 + H3O+
weak acid
F- + H2O <=> HF + OH-
weak base
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54. 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.
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55. 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, %)
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56. 11/14/2018
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57. Acid-Base Properties of Salts Predicting 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
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58. 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
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59. Lewis Concept
Example:
Lewis acid
Lewis base
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60. Lewis Concept
Example:
Lewis acid
Lewis base
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61. 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!
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62. Lewis Acid-Base Concept Hydrolysis 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
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63. Al(H2O)63+ = Al3+(aq) Oxygen Al3+ Lone pair Hydrogen 11/14/2018
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64. Lewis Acid-Base Concept Hydrolysis 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
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65. Lewis Acid-Base Concept Hydrolysis 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).
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66. End of Presentation
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