1. The Chemistry of Acids and Bases
To play the movies and simulations included, view the presentation in Slide Show Mode.

2. Learning Objectives
Acids and bases Learners be able to calculate and understand 1) Arrhenius and Bronsted-Lowry theories. 2) pH as -log10[H+], and be able to convert pH to concentration for a strong acid or base. 3) know that water is weakly dissociated 4) know the ionic product of water, Kw and that Kw = [H+] [OH-]

3. A/B Intro 5 min Fuse School

4. Strong Acids….MEMORIZE RIGHT NOW, or don’t…. HCl - hydrochloric acid
HNO3 - nitric acid
H2SO4 - sulfuric acid (HSO4- is a weak acid)
HBr - hydrobromic acid
HI - hydroiodic acid
HClO4 - perchloric acid
H2SO4 (aq) → 2H+(aq)  SO42- (aq) 
HCl(aq)  → H+ (aq) + Cl-(aq) 
HNO3 (aq) → H+ (aq) + NO3-(aq) 

5. STRONG BASES LiOH lithium hydroxide NaOH sodium hydroxide KOH
potassium hydroxide
RbOH
rubidium hydroxide
CsOH
cesium hydroxide
*Ca(OH)2
calcium hydroxide
*Sr(OH)2
strontium hydroxide
*Ba(OH)2
barium hydroxide

7. Electrolytes and Nonelectrolytes
Electrolyte: substance that dissolved in water produces a solution that conducts electricity
Contains ions
Nonelectrolyte: substance that dissolved in water produces a solution that does not conduct electricity
Does not contain ions

8. Dissociation - ionic compounds separate into constituent ions when dissolved in solution
Ionization - formation of ions by molecular compounds when dissolved

9. Strong and weak electrolytes
Strong Electrolyte: 100% dissociation
All water soluble ionic compounds, strong acids and strong bases
Weak electrolytes
Partially ionized in solution
Exist mostly as the molecular form in solution
Weak acids and weak bases

10. Method to Distinguish Types of Electrolytes
nonelectrolyte
weak electrolyte
strong electrolyte

11. Classify the following as nonelectrolyte,
weak electrolyte or strong electrolyte
NaOH
strong electrolyte
CH3OH
nonelectrolyte
H2CO3
weak electrolyte

12. Acids have a pH less than 7

13. Operational Properties of Acids
Taste sour. Ex: Vinegar is acetic acid and citrus fruits
Corrode metals and produce hydrogen gas
React with carbonates and bicarbonates to produce CO2(g)
React with bases to form a salt and water and are electrolytes
Turns blue litmus paper to red “Blue to Red A-CID”

14. Taste Sour

15. OPERATIONAL ACIDS Reacts with metals to form H2 gas.
HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

16. Acids React with Carbonates and Bicarbonates
HCl + NaHCO3
Hydrochloric acid + sodium bicarbonate
NaCl + H2O + CO2
salt + water + carbon dioxide
An old-time home remedy for relieving an upset stomach

17. Conducts electricity.

18. Neutralizes BASES

20. Acids Turn Litmus Red
Blue litmus paper turns red in contact with an acid

21. B/L Theory and Arrhenius

22. BASES

23. OPERATIONAL BASES

24. Some Properties of Bases
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes
Feel soapy, slippery
React with acids to form salts and water
pH greater than 7
Turns red litmus paper to blue “Basic Blue”

25. ALL Strong Bases produce
Some Common Bases
NaOH sodium hydroxide lye
KOH potassium hydroxide liquid soap
Ba(OH)2 barium hydroxide stabilizer for plastics
Mg(OH)2 magnesium hydroxide MOM” Milk of magnesia
Al(OH)3 aluminum hydroxide Maalox (antacid)
ALL Strong Bases produce
OH-

26. BASES OPERATE

27. BASES ARE SLIPPERY

28. BASES are Corrosive

29. Bases have a pH greater than 7

31. Bases Turn Litmus Blue
Red litmus paper turns blue in contact with a base (and blue paper stays blue).

32. Sodium hydroxide, NaOH (lye for drain cleaner; soap)
Examples of Bases
Sodium hydroxide, NaOH (lye for drain cleaner; soap)
Potassium hydroxide, KOH (alkaline batteries)
Magnesium hydroxide, Mg(OH)2 (Milk of Magnesia)

33. Review Properties of Bases (metallic hydroxides, NaOH)
React with acids forms water and salt.
Taste bitter & feel slippery
Electrolytes in aqueous
Change the color of indicators (bases go Blue, with litmus).

34. Theories Acid Base H+ H+

35. Acid-Base Theories a) Arrhenius, b) Brønsted-Lowry
c) Lewis (not covered).

36. 1. Arrhenius Definition - 1887
Acids produce (H+) in solution
(HCl → H+ + Cl-)
Bases produce OH- in water.
(NaOH → Na+ + OH-)
Reasons it SUCKS
Limited to aqueous solutions.
Only one kind of base (hydroxides)
NH3 (ammonia) could not be an Arrhenius base: no OH- produced.

37. Arrhenius: Acid: dissolve in water to ↑ H+
Base: increase ↑ hydroxide ions or OH-

38. 3 flaws with Arrhenius
1. A proton (H+) does not exist ALONE in water, it makes H3O+(aq)
2. NH3 (g) is a base when dissolved in water; but does not contain the OH-(aq)
3. Amphiprotics like HCO3-(aq) appear to contain the H+ ion BUT ARE bases

39. Bronsted-Lowry Acids and Bases
It’s all about protons (H+)
Acid: Proton donor
HCl(aq)  H+(aq) + Cl-(aq)
H2SO4(aq)  2H+(aq) + SO42-(aq)
Base: Proton acceptor
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
OH-(aq)* + H+(aq)  H2O(l)
*From any soluble hydroxide or other alkali
If we mention acid/base without mentioning the type, we generally mean a Bronsted-Lowry one.

40. Brønsted-Lowry - 1923 A broader definition than Arrhenius
Acid is a H+ ,base is H+ acceptor.
HCl is an acid.
When it dissolves in water, it gives it’s proton to water.
HCl(g) + H2O(l) ↔ H3O+(aq) + Cl-(aq)
base

41. The Bronsted-Lowry concept+ Cl H O
acid
base
conjugate acid
conjugate base
conjugate acid-base pairs

42. Acids and bases come in pairs
A “conjugate base” is the remainder of the original acid, after it donates it’s hydrogen ion
A “conjugate acid” is the particle formed when the original base gains a hydrogen ion
CONJOINED TWINS
Amphiprotic like frogs
or writing with both hands Or
Conjugate = Latin word
conjugare, like conjoined

43. DYNAMIC EQUILIBRIUM
• a reversible chemical reaction is a dynamic process
• appears stationary but the reactions are moving both ways
• the position of equilibrium can be varied by changing certain conditions
Trying to get up a “down” escalator gives an excellent idea of a non-chemical situation involving dynamic equilibrium.
Eq LAW Simply states
“If the concentrations of chemicals at equilibrium are raised to the power of the number of moles whereby, the product of the concentrations of the products divided by the product of the concentrations of the reactants, at a constant temperature”
aA bB cC Dd
a constant, (Kc) = [C]c . [D]d / [A]a . [B]b
where [ ] denotes the equilibrium concentration in mol dm-3
Kc is known as the Equilibrium Constant

44. Time to practice
Give the formula of the conjugate base for each of the following: HF
H2SO4
H3PO4
CH3COOH
H2O
NH4+
Give the formula of the conjugate acid for each of the following:
OH-
SO42-
HPO42-
(CH3)2NH
H2O

45. Some chemicals are BOTH Acid and Base…
…like these amphoterics
amphiprotic.
HCO3–
HSO4 –
H2O

47. pH of Common Substances

48. pH
pH is defined as the negative base-10 logarithm of the hydronium ion concentration.
pH = –log [H3O+]

49. Hydrogen Ions from Water
Water ionizes, or falls apart into ions:
H2O ↔ H+ + OH-
Called the “self ionization” of water
Occurs to a very small extent:
[ H+ ] = [OH-] = 1 x 10-7 M
Since they are equal a neutral solution
Kw = [H+ ] x [OH-] = 1 x 10-14
14 = pH + pOH

50. The ionic product of water, Kw
Water can be both an acid and a base, this leads to the following equilibrium:
H2O + H2O  H3O+ + OH- ..or more simply:
H2O  H+ + OH-
The equilibrium for this reaction is called the ionic product of water and has the symbol

51. Calculating [H+] and [OH-] of pure water:
Kw = [H+][OH-] = 1.00x mol2 dm-6
Since when pure, [H+] = [OH-]
[H+]2 = 1.00x10-14
[H+] = √1.00x10-14 = 1.00x10-7 mol dm-3
Kw varies with temperature:
At 273 K, Kw = 1.14x10-15, calculate [H+]
At 373 K, Kw = 5.13x10-13, calculate [OH-]
Do these changes mean the self-dissociation of water is endothermic or exothermic? Justify your answer.

53. If you know one, you know them all:
[OH-]
1.0 x 10-14
[OH-]
10-pOH
1.0 x 10-14
[H+]
-Log[OH-]
[H+]
pOH
10-pH
14 - pOH
-Log[H+]
14 - pH pH

54. pH and Significant Figures
[H+] = M = 1.0 x 10-3 M
has 2 sig figs
the pH = 3.00, with the two numbers to the right of the decimal corresponding to the 2 sig figs

56. (Remember that the [ ] mean Molarity, or mol/L)
Calculating the pH
pH = - log [H+]
(Remember that the [ ] mean Molarity, or mol/L)
Example: If [H+] = 1 X pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.0 X 10-5 pH = - log 1.0 X 10-5
pH = - (- 5)
pH = 5

57. A pH Number line Number lines. [H+] =10-15 [H+] < [OH -] pH = 15
basic
pH = 15
[H+] =10-12
[OH -] = 10-2
basic
pH = 12
[H+] < [OH -]
basic
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 7
acidic
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
pH = 2

58. Try These! Find the pH of these:
A 0.15 M solution of Hydrochloric acid
Answer: pH (- 2 sig figs)
2) A 3.00 X 10-7 M solution of Nitric acid
Answer: pH (- 3 sig figs) remember leading zero do NOT count

59. pH calculations – Solving for H+
If the pH of Coke is 3.12, then [H+] = ???
Because pH = - log [H+] then
- pH = log [H+]
Take antilog (10x) of both sides and get
[H+] = = 7.6 x 10-4 M
*** to find antilog on your calculator,
look for “Shift” or “2nd function” and
then the log button

60. [H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution?
[OH-] = (or 1.0 X 10-3 M)
pOH = - log
pOH = 3
pH = 14 – 3 = 11
OR Kw = [H3O+] [OH-]
[H3O+] = 1.0 x M
pH = - log (1.0 x 10-11) = 11.00

61. Calculating [H3O+], pH, [OH-], and pOH
Problem 1:
A chemist dilutes concentrated HCl to M.
Calculate the [H3O+], pH, [OH-], and pOH
ANSWERS: pH=2.62, pOH = 11.38, 4.2x10 -12
Problem 2:
What is the [H3O+], [OH-], and pOH of a solution with a
pH = 3.67? Is this an acid, base, or neutral?
ANSWERS : An acid. [H3O+], M, [OH-], 4.7x10-11
pOH = 10.33

63. ANSWERS to Previous Slide

64. Refresh
100 cm3 of a NaOH solution of pH 12 is mixed with 900 cm3 of water. What is the pH of the resulting solution? 1 3 11 13

65. STRONG V WEAK

66. STRONG VS WEAK

67. Strength
In any acid-base reaction, the equilibrium favors the reaction that moves the proton to the stronger base.
HCl(aq) + H2O(l)  H3O+(aq) + Cl–(aq)
Cl– equilibrium lies so far to the right K is not measured (K>>1), so is not a base, it is neutral

68. Strength Strong acids completely dissociate in water.
Their conjugate bases are actually so weak as not to be able to act as a base at all.

69. Strength Strength & Concentration Night V Day
Acids & Bases are classified according to the degree they ionize in water:
Strong are completely ionized, 100%
Weak ionize only slightly , like 1%
Strength & Concentration
Night V Day

70. Compared with strong acids of the same concentration, weak acids:
Have lower electrical conductivity
React more slowly
pH is higher (less acid)
Change pH more slowly when diluted
However, they neutralise the same volume of alkali
Weak bases follow a similar pattern

71. Strong Acid Dissociation (makes 100 % ions)

72. Weak Acid Dissociation (only partially ionizes)

73. STRONG V WEAK ACIDS

74. LOL

75. Measuring strength Ionization is reversible for WEAK acids:
HA + H2O ↔ H+ + A-
This makes an equilibrium
Ka = [H+ ][A- ] [HA]
Strong acid = more products & large Ka
(Note that the arrow goes both directions.)
(Note that water is NOT shown,

76. Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, C2H3O2H, at 25°C.
Ka for acetic acid at 25°C is 1.8  10-5.

77. Calculating pH from Ka Use the ICE table: [C2H3O2], M [H3O+], M
Initial
0.30
Change –x +x
Equilibrium
0.30 – x x

78. Calculating pH from Ka Use the ICE table: [C2H3O2], M [H3O+], M
Initial
0.30
Change –x +x
Equilibrium
0.30 – x ≈ 0.30 x
Simplify: how big is x relative to 0.30?

79. Calculating pH from Ka (1.8  10-5) (0.30) = x2 5.4  10-6 = x2
Now,
(1.8  10-5) (0.30) = x2
5.4  10-6 = x2
2.3  10-3 = x
pH = = 2.64

80. Another Ka problem with % ionization found at the end

81. Calculating Ka from the pH
The pH is 2.38 for a 0.10 M solution of formic acid, HCOOH,. Calculate Ka
– 2.38 = log [H3O+]
= 10log [H3O+] = [H3O+]
4.2  10-3 = [H3O+] = [HCOO–] = X

82. Calculating Ka from pH All values In Mol/L [HCOOH] [H3O+] [HCOO−]
Initially
0.10
Change
–4.2  10-3
+4.2  10-3
Equilibrium
0.10 – 4.2  10-3
= = 0.10
4.2  10-3
4.2 

83. Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10]
= 1.8  10-4

84. Calculating Percent Ionization
[A-]eq = [H3O+]eq = 4.2  10-3 M
[A-]eq + [HCOOH]eq = [HCOOH]initial= 0.10 M
%rx = [4.2  10-3 / 0.10] (x 100)
= 4.2%

85. What about bases?
Strong bases dissociate completely.
MOH + H2O ↔ M+ + OH- (M = a metal)
Base dissociation constant = Kb
Kb = [M+ ][OH-] [MOH]
Stronger base = more dissociated ions are produced, thus a
larger Kb and a small Ka.

86. Weak Bases
The Kb for this reaction is

87. pH of Basic Solutions What is the pH of a 0.15M solution of NH3?
[NH4+] [OH−]
[NH3]
Kb =
= 1.8  10-5

88. pH of Basic Solutions [NH3] [NH4+] [OH−] Initial 0.15 Equil
Equil
x  0.15 x

89. pH of Basic Solutions (x)2 (0.15-X) 1.8  10-5 = Ignore X
pOH = –log (1.6  10-3)
pOH = 2.80
pH = 11.20

90. Example 1: Calculation of [OH-]
What is the concentration of OH- ions in a mol dm-3 solution of ammonia (Kb = 1.8x10-5)? What % of the NH3 molecules have dissociated?
Since it is a weak base and the equilibrium is to the right we assume that at equilibrium, [NH3] is the same as stated in the question. So:
Kb = [NH4+][OH-]/[NH3] Sub all known values into equation
1.8x10-5 = [NH4+][OH-]/ Looks like there are 2 unknowns
However, since [NH4+] = [OH-]:
1.8x10-5 = [OH-]2 / Rearrange to make [OH-] the subject
[OH-] = √(1.8x10-5 x 0.500) Perform calculation
[OH-] = mol dm-3
% Dissociation = / x 100 = 0.60%

91. Example 2: Calculating Kb from pH
A mol dm-3 solution of methylamine (CH3NH2) has a pH of Determine Kb of methylamine and Ka of the methylammonium ion (CH3NH3+).
Since pH = 11.59
pOH = 14 – = 2.41
[OH-] = = 3.92x10-3 mol dm-3
Remember:
[BOH] at equilibrium is same as stated in question
[OH-] = [B+]
Kb = [B+][OH-]/[BOH]
Known values subbed in
Kb = (3.92x10-3).(3.92x10-3)/0.0350
Kb = 4.39x10-4
To calculate Ka of the conjugate acid use:
Ka x Kb = Kw
Ka = Kw / Kb
= 1.00x10-14 / 4.39x10-4
= 2.78x10-11

92. Example 3: Calculating Ka, pKa, Kb and pKb
The pKa of benzoic acid is Calculate Ka, Kb and pKb
Ka = 10-pKa = = 6.31x10-5
Kb x Ka = Kw
Kb = Kw / Ka = 1.00x10-14 / 6.31x10-5 = 1.58x10-10
pKb = -log10(Kb) = -log10(1.58x10-10) = 9.80 OR
Since pKa + pKb = pKw
pKb = 14 – pKa = 14 – 4.20 = 9.80

93. Acid and Base Strength
Acetate is a stronger base than H2O, LOWER on table,
The stronger base “wins” the proton.
HC2H3O2(aq) + H2O
H3O+(aq) + C2H3O2–(aq)

95. TITRATIONS

96. Titration
Titration is the process of adding a known amount of solution of known concentration to determine the concentration of another solution
The equivalence point is when the moles of hydrogen ions is equal to the moles of hydroxide ions (= neutralized!)

97. Titration Lab Demo

98. Titration
The concentration of acid (or base) in solution can be determined by performing a neutralization reaction
An indicator is used to show when neutralization has occurred
Often we use phenolphthalein- because it is colorless in neutral and acid; turns pink in base

99. Indicators Universal (actually a mixture of indicators) Litmus:
Acid is Red / Base is Blue
Phenolpthalein:
Acid is Colourless / Alkali is Pink

100. The pH Scale Runs from 1 for most acid up to 14 for most alkali?
Nonsense, can go below 0 for very strong acids and above 14 for very strong alkalis.

102. Steps - Neutralization reaction
#1. A measured volume of acid of unknown concentration is added to a flask
#2. Several drops of indicator added
#3. A base of known concentration is slowly added, until the indicator changes color; measure the volume

103. Neutralization
The solution of known concentration is called the standard solution
added by using a buret
Continue adding until the indicator changes color
called the “end point” of the titration

104. strong acid (HCl) v. strong base (NaOH)

105. strong acid (HCl) v. strong base (NaOH)
Very sharp change in pH over the addition of less than half a drop of NaOH
Very little pH change during the initial 20cm3
pH 1 as 0.1M HCl (strong acid)

106. Refresh
If 20 cm3 samples of 0.10 mol dm–3 solutions of the acids below are taken, which acid would require a different volume of 0.10 mol dm–3 sodium hydroxide for complete neutralization?
Nitric acid
Sulfuric acid
Ethanoic acid
Hydrochloric acid

107. w. acid (CH3COOH) v s.base (NaOH)
Curve levels off at pH 13 due to excess 0.1M NaOH
(a strong alkali)
Sharp change in pH over the addition of less than half a drop of NaOH
Steady pH change
pH 4 due to 0.1M CH3COOH (weak monoprotic acid)

108. strong acid (HCl) v. strong base (NaOH)
PHENOLPHTHALEIN
LITMUS
METHYL ORANGE
Any of the indicators listed will be suitable - they all change in the ‘vertical’ portion

109. weak (CH3COOH) v. weak (NH3)
Curve levels off at pH 10 due to excess 0.1M NH3
(a weak alkali)
NO SHARP
CHANGE IN pH
Steady pH change
pH 4 due to 0.1M CH3COOH (weak monoprotic acid)

110. Polyprotic Acids?
Some compounds have more than one ionizable hydrogen to release
HNO3 nitric acid - monoprotic
H2SO4 sulfuric acid - diprotic - 2 H+
H3PO4 - triprotic - 3 H+
IMPT for TITRATING BUT NOT FOR pH calculating…why?

111. Polyprotic acids (H3PO4) There are 3 sharp pH changes
Phosphoric acid is triprotic; it reacts with sodium hydroxide in three steps...
Step 1 H3PO NaOH ——> NaH2PO H2O
Step 2 NaH2PO4 + NaOH ——> Na2HPO H2O
Step 3 Na2HPO4 + NaOH ——> Na3PO H2O
There are 3 sharp pH changes
Each successive addition of NaOH is the same as equal number of moles are involved.

112. Polyprotic Acids
The first dissociation constants for phosphoric acid is greater than the second, 100,000 times greater
This means nearly all the H+ ions in the solution comes from the first step of dissociation.
H3PO4(s)   + H2O(l)    H3O+(aq) + H2PO4−(aq)       Ka1= 7.25×10−3
H2PO4−(aq)+ H2O(l)    H3O+(aq) + HPO42−(aq)       Ka2= 6.31×10−8
HPO42−(aq)+ H2O(l)    H3O+(aq) +  PO43−(aq)        Ka3= 3.98×10−13

113. Salt Hydrolysis A salt is an ionic compound that:
formed in a neutralization reaction
some neutral; others acidic or basic

114. Salt Hydrolysis
To see if the resulting salt is acidic or basic, check the by dissociating the salt
NaCl, a neutral salt
(NH4)2SO4, acidic salt
CH3COOK, basic salt

115. Adding a “salt” to water
Neutral salt the pH is 7, for example KNO3.
acid salt, example NH4Cl, pH =acidic.
A basic salt, for example NaCH3COO, and the pH will be basic.

116. Salt solutions: hydrolysis
NaCl (aq) NH4Cl (aq) NaClO (aq)

117. Buffers
Buffers are solutions in which the pH remains relatively constant, even when small amounts of acid or base are added
made from a pair of chemicals: a weak acid and one of it’s salts; or a weak base and one of it’s salts

119. Buffers
A buffer system is better able to resist changes in pH than pure water
It is a pair of chemicals (conjugates)
one chemical neutralizes any acid added, while the other chemical would neutralize any base

120. Buffers
Example: Ethanoic (acetic) acid and sodium ethanoate (also called sodium acetate)
The buffer capacity is the amount of acid or base that can be added before a significant change in pH

121. Buffers
The two buffers that are crucial to maintain the pH of human blood are:
1. carbonic acid (H2CO3) & hydrogen carbonate (HCO3-)
2. dihydrogen phosphate (H2PO41-) & monohydrogen phoshate (HPO42-)

122. Aspirin (which is a type of acid) sometimes causes stomach upset; thus by adding a “buffer”, it does not cause the acid irritation.
Bufferin is one brand of a buffered aspirin that is sold in stores.

123. Indicators
1. Moisten the pH indicator paper strip with a few drops of solution, by using a stirring rod.
2.Compare the color to the chart on the vial – then read the pH value.

124. Some of the many pH Indicators and their pH range

125. How Do We Measure pH? An indicator
Compound that changes color in solution.

126. Effects of Acid Rain on Marble (marble is calcium carbonate CaCO3)
George Washington:
BEFORE acid rain
AFTER acid rain

127. End