Chapter 15 Lecture Outline
15.3 Definitions of Acids and Bases
The Arrhenius Definition
Bronsted-Lowry Acid-Base Reactions
- General Equation
Acid | | Base | | Conjugate Base of HA | | Congugate acid of B |
HA | + | B | --> | A- | + | HB+ |
- Strong Acid and Strong Base
Acid | | Base | | Conjugate Base of HA | | Congugate acid of B |
HCl(aq) | + | NaOH(aq) | --> | Cl- | + | Na1+ | + | H2O (aq) |
- Weak Acid and Weak Base
Acid | | Base | | Conjugate Base of HA | | Congugate acid of B |
CH3COOH(aq) |
+ |
H2O(l) |
<--> |
CH3COO-(aq) |
+ |
H3O+(aq) |
- Autoionization of water:
Acid | | Base | | Conjugate Base of HA | | Congugate acid of B |
H2O(l) |
+ |
H2O(l) |
<--> |
OH-(aq) |
+ |
H3O+(aq) |
- Polyprotic Acid
Acid | | Base | | Conjugate Base of HA | | Congugate acid of B |
H2SO4 |
+ |
H2O(l) |
--> |
HSO41- |
+ |
H3O+(aq) |
HSO41- |
+ |
H2O(l) |
<--> |
SO42-
|
+ |
H3O+(aq) |
15.4 Acid Strength and the Acid Ionization Constant (Ka)
- Binary acids X-H bond strength:
- Oxyacids, more O, stronger acid (Electron withdrawing by X in X-OH)
Acid | Formula | Acid Dissociation Constant |
hypochlorous acid | HOCl | Ka = 10-8 |
chlorous acid | HOClO | Ka = 10-2 |
chloric acid | HOClO2 | Ka = 101 |
perchloric acid | HOClO3 | Ka = 108 |
- Metal hydroxides, X is electroposative
- NaOH strong base
- Be(OH)2 amphoteric
- Carboxylic acids R-COOH
15.5 Autoionization of Water and pH
Acid | | Base | | Conjugate Base of HA | | Congugate acid of B |
H2O(l) |
+ |
H2O(l) |
--> |
OH-(aq) |
+ |
H3O+(aq) |
- The equilibrium concentrations can be described with a constant K for the autoionization of water:
- This is usually rearranged since [H2O] is essentially constant. and
given as Kw:
- At 25 C, Kw = 1.0 * 10-14
- Calculate [OH-] and [H3O+]
- 0.1 M HCl
- 0.1 M NaOH
- pH and pOH
- pH = - log[H3O+]
- pOH = - log[OH-]
- pH + pOH = 14 (Show algebra)
15.6 Finding the [H3O+] and pH of Strong and Weak Acid Solutions
- Generic acid reaction
- HA + H2O <--> A- + H3O+
- Look up Ka in table
- Acetic acid:
- For the reaction CH3COOH + H2O <--> CH3COO- + H3O1+
- Since [H2O] is constant, this is typically expressed as Ka
- Ka = 1.8 x 10-5
- Calculating pH (Given Solution concentration)
Solution | [H3O+] | pH | [OH-] | pOH |
10 M HCl | 101 | -1 | 10-15 | 15 |
0.1 M HCl | 10-1 | 1 | 10-13 | 13 |
10-8 M HCl | 10-8 | 8 | 10-6 | 6 |
10 M NaOH | 10-15 | 15 | 101 | -1 |
0.1 M NaOH | 10-13 | 13 | 10-1 | 1 |
10-8 M NaOH | 10-6 | 6 | 10-8 | 8 |
- Note: pH + pOH = 14
- Note: 10-8 M HCl and NaOH make no sense, why?
- Significant figures and logs
- Calculate [H3O+] and [OH-] given pH or pOH
- Vinegar pH=2.9
- ammonia pH = 11.5
Example Problems: Solving a Weak Acid Equlibrium System
- For a nitrous acid (Ka = 4.5 x 10-4) solution find the pH and pOH for initial nitrous acid concentrations of
- 1.0 M
- 0.001 M
- 10-8 M.
- Solutions
- Acetic Acid Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of acetic acid is added to 250.0 mL of water.
- Solution
- Nitrous Acid Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of nitrous acid is added to 250.0 mL of water.
- Solution
- Nitric Acid Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of nitric acid is added to 250.0 mL of water.
- Solution
15.7 Base Solutions
- Ka =
- Kb =
- Kw =
- Kw = Ka * Kb
15.8 The Acid-Base Properties of Ions and Salts
- Chemical System
- Reaction #1 CH3COOH + H2O <-> CH3COO- + H3O+
- Reaction #2
- step 1: NaCH3COO --> Na1+ + CH3COO1-
- step 2:
CH3COO1- + H2O <-> CH3COOH + OH-
- TOTAL rxn: 2 H2O <-> H3O+ + OH-
- Equlibrium for reaction #1 is Ka for CH3COOH
- For a 0.100 M CH3COOH at equilibrium, calculate Ka if at equlibrium [H3O] = 1.318*10-3 M:
|
CH3COOH |
H2O |
<--> |
CH3COO- |
H3O+ |
Initial (M) |
0.100 |
|
|
0 |
0 |
Change (M) |
-1.318*10-3 |
|
|
+1.318*10-3 |
+1.318*10-3 |
Final (M) |
0.0987 |
|
|
1.318*10-3 |
1.318*10-3 |
- Flow Chart
- Examples. What happens when...
- sodium acetate dissolves in water.
- potassium nitrate dissolves in water.
- ammonium chloride dissolves in water.
- sodium fluoride dissovles in water.
- potassium hydroxide dissolves in water.
Example Problems
- Sodium Acetate Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of sodium acetate is added to 250.0 mL of water.
- Solution
- Ammonium Chloride Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of ammonium chloride is added to 250.0 mL of water.
- Solution
- Sodium Nitrite Equlibrium
- Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of sodium nitrite is added to 250.0 mL of water.
- Solution
15.9 Acid Strength and Molecular Structure
15.10 Lewis Acids and Bases
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Last Updated Friday, May 25, 2001 1:59:43 PM