Embed Size (px)
Citation preview
1
Ch. 8: Acids and Bases
Chem 20
El Camino College
2
3
Two Acid-Base Theories
Arrhenius TheoryAn acid solution contains more H+ ions than
OH- ionsA base solution contains more OH- ions
than H+ ions
4
Two Acid-Base Theories
Note--H+ is a proton Bronsted-Lowry Theory
An acid is a proton donorA base is a proton acceptor
HBr(aq) + H2O(l) H3O+(aq) + Br- (aq)
NH3(aq) + H2O(l) NH4+(aq) + OH- (aq)
acid base
acidbase
5
6
7
Conjugate Acid-Base Pairs
Conjugate Acid-Base Pairs differ by one H+
The reactants side has the acid and the base The products side has the conjugate acid and the
conjugate base
HBr(aq) + H2O(l) H3O+(aq) + Br- (aq)conjugate
acidconjugate
baseacid base
8
Conjugate Acid-Base Pairs
Acid: H+ donor on left side Conjugate base: missing H+ on right side Base: H+ acceptor on left side Conjugate acid: received H+ on right side
9
10
Conjugate Acid-Base Pairs
Label and link conjugate acid-base pairs.
NH3(aq) + CH3CO2H(aq) NH4+(aq) + CH3CO2
- (aq)conjugate
acidconjugate
baseacidbase
H2SO4(aq) + HSO3-(aq) HSO4
-(aq) + H2SO3 (aq)
conjugateacid
conjugatebase
acid base
11
The Water Equilibrium
H2O(l) H+(aq) + OH-(aq)
Kw = (concentration H+ in M)* (conc. OH- in M)Kw = [H+][OH- ] = 1.0 x 10-14
In aq. soln, if [H+]=[OH- ], the soln is neutral.
In aq. soln, if [H+]>[OH- ], the soln is acidic.
In aq. soln, if [H+]<[OH- ], the soln is basic.
12
13
Examples
Use this equation: [H+][OH- ]=1.0 x 10-14
Ex. If the conc. of H+ is 3.5 x 10-3 M, find [OH- ]. Is the solution acidic or basic?
[OH- ] = 1 x 10-14
[H+]
= 1 x 10-14
3.5x10-3M= 2.9 x 10-12 M
acidic
14
Examples
Ex. If the conc. of H+ is 9.9 x 10-11 M, find [OH- ]. Is the solution acidic or basic?
[OH- ] = 1 x 10-14
[H+]
= 1 x 10-14
9.9x10-11M= 1.0 x 10-4 M
Ex. If [OH- ] is 1.7 x 10-10 M, find [H+]. Is the solution acidic or basic?
[H+] = 1 x 10-14
[OH- ]
= 1 x 10-14
1.7x10-10M= 5.9 x 10-5 M
basic
acidic
15
pH Values
pH = 7 is a neutral solution pH < 7 is an acidic solution pH > 7 is a basic solution
16
Table 18-2, p. 516
17
pH = -log [H+]
Ex. If [H+] = 1.5 x 10-6 M, find [OH- ] and pH
[OH- ] = 1 x 10-14
[H+]
= 1 x 10-14
1.5x10-6M= 6.7 x 10-9 M
pH = -log [H+] = -log(1.5 x 10-6) = 5.82
18
pH = -log [H+]
Ex. If [OH-] = 3.3 x 10-4 M, find [H+ ] and pH
[H+ ] = 1 x 10-14
[OH-]
= 1 x 10-14
3.3x10-4M= 3.0 x 10-11 M
pH = -log [H+] = -log(3.0 x 10-11) = 10.52
19
pH = -log [H+]
Ex. If [H+] = 8.5 x 10-1 M, find [OH- ] and pH
[OH- ] = 1 x 10-14
[H+]
= 1 x 10-14
8.5x10-1M= 1.2 x 10-14 M
pH = -log [H+] = -log(8.5 x 10-1) = 0.07
20
pH to [H+]
[H+] = 10-pH or antilog(-pH)
The minus sign goes on the pH value first
Ex. If the pH = 5.55, find [H+]
[H+] = 10-pH = 10-5.55 = 2.8 x 10-6 M
21
pH to [H+]
Ex. If the pH = 8.88, find [H+]
[H+] = 10-pH = 10-8.88 = 1.3 x 10-9 M
Ex. If the pH = 13.00, find [H+] and [OH-]
[H+] = 10-pH = 10-13.00 = 1.0 x 10-13 M
[OH- ] = 1 x 10-14
[H+]
= 1 x 10-14
1.0x10-13M= 1.0 x 10-1 M
22
Titration
In the acid-base titration we’ll do in lab, a flask contains a mixture of acid and phenolphthalein (a ccs)
Base is added by buret When the soln turns pale pink for 30 seconds,
moles acid moles base.
23
Fig. 16-8, p. 457
24
p. 459
25
p. 459
26
p. 459
27
Titration: How many mL? Ex. The flask is filled with 25.00 mL of a 0.500 M HCl soln. How many mL of 0.300 M NaOH soln will neutralize the acid?
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Start with volume of acid, convert to L
Use molarity of acid as a conversion factor
Use a mole ratio to convert mol acid to mol base
Use molarity of base as a conversion factor, convert to mL
.500 mol HCl 1 L
= 41.7 mL25.00mL
HCl1 mol HCl
1 mol NaOH .300 mol
NaOH
1 L1000 mL
1 L 1 L
1000 mL
28
Titration: How many mL? Ex. The flask is filled with 75.00 mL of a 0.200 M HCl soln.
How many mL of 0.150 M NaOH soln will neutralize the acid?
Ex. The flask is filled with 55.00 mL of a 1.5 M HCl soln. How many mL of 0.90 M NaOH soln will neutralize the acid?
.200 mol HCl 1 L
= 100. mL75.00mL
HCl1 mol HCl
1 mol NaOH .150 mol
NaOH
1 L1000 mL
1 L 1 L
1000 mL
1.5 mol HCl 1 L
= 91.7 mL55.00mL
HCl1 mol HCl
1 mol NaOH 0.90 mol
NaOH
1 L1000 mL
1 L 1 L
1000 mL
29
Titration: Find Molarity Ex. The flask is filled with 30.00 mL of a 0.100 M HCl soln. What is the molarity of the NaOH soln if it takes 23.45 mL to neutralize the acid?
Start with volume of acid, convert to L
Use molarity of acid as a conversion factor
Use a mole ratio to convert mol acid to mol base
Solve for mol base
***Divide mol base by mL base to find molarity. Convert to L.
.100 mol HCl 1 L
30.00mL
HCl1 mol HCl
1 mol NaOH1000 mL
1 L= 0.00300 mol NaOH
= 0.128 M 1 L 1000 mL0.00300 mol NaOH
23.45 mL
30
Titration: Find Molarity Ex. The flask is filled with 45.00 mL of a 0.996 M HCl soln.
What is the molarity of the NaOH soln if it takes 52.33 mL to neutralize the acid?
.996 mol HCl 1 L
45.00mL
HCl1 mol HCl
1 mol NaOH1000 mL
1 L= 0.0448 mol NaOH
= .856 M 1 L 1000 mL0.0448 mol NaOH
52.33 mL
31
Titration: Find Molarity Ex. The flask is filled with 80.00 mL of a 2.30 M
H2SO4 soln. What is the molarity of the NaOH soln if it takes 70.00 mL to neutralize the acid?
H2SO4(aq) + 2 NaOH(aq) Na2SO4 (aq) + 2 H2O(l)
2.30 mol H2SO4
1 L
80.00mL
H2SO4 1 mol H2SO4
2 mol NaOH1000 mL
1 L= 0.184 mol NaOH
= 2.63 M 1 L 1000 mL0.184 mol NaOH
70.00 mL
32
Strong and Weak Acids
Strong acids completely break down into ions when dissolved in water Weak acids only break down into a few ions in water. Most of the
weak acid molecules stay together in water.
33
34
Strong and Weak Acids These are the only strong acids
HCl hydrochloric acidHBr hydrobromic acidHI hydroiodic acidHNO3 nitric acid
H2SO4 sulfuric acid
HClO3 chloric acid
HClO4 perchloric acid.
35
Strong and Weak Acids
All other acids are weak acids. Here are some examplesHF hydrofluoric acidH3PO4 phosphoric acid
CH3CO2H acetic acid
H2CO3 carbonic acid.
36
Buffers
When a small amount of acid or base is added to pure water, pH changes dramatically A buffer resists change in pH when small amounts of acids or bases are added Blood is buffered in the body to a pH of 7.4 A buffer is a combination of a weak acid and its conjugate base (found in an ionic cmpd) A strong acid cannot make a buffer.
37
38
39
How Buffers Work
This is a buffer made of CH3COOH and CH3COONa
When acid is added, the extra H+ reacts with CH3COO- to form more CH3COOH
When base is added, the extra OH- reacts with CH3COOH to form more CH3COO-
40
Buffers
Which of the following represents a buffer system?HCl and NaClHF and NaFHNO3 and NaNO3
CH3COOH and CH3COONa
noyes
noyes
41
Practice: Conjugate Acid-Base Pairs
Label and link conjugate acid-base pairs.
HNO3(aq) + CH3CO2- (aq) CH3CO2H (aq) + NO3
- (aq)conjugate
acidconjugate
basebaseacid
HCl(aq) + SO42-(aq) Cl-(aq) + HSO4
- (aq)conjugate
acidconjugate
baseacid base
42
Practice: pH to [H+]
Ex. If the pH = 3.68, find [H+] and [OH-] in scientific notation.
[H+] = 10-pH = 10-3.68 = 2.1 x 10-4 M
[OH- ] = 1 x 10-14
[H+]
= 1 x 10-14
2.1x10-4M= 4.8 x 10-11 M
43
Practice: Titration Ex. The flask is filled with 20.50 mL of a 0.0996 M HCl soln.
How many mL of 0.194 M NaOH soln will neutralize the acid?
.0996 mol HCl 1 L
= 10.5 mL20.50mL
HCl1 mol HCl
1 mol NaOH .194 mol
NaOH
1 L1000 mL
1 L 1 L
1000 mL
44
Practice: Titration
Ex. The flask is filled with 25.00 mL of a 2.5 M HCl soln. How many mL of 0.50 M Ca(OH)2 soln will neutralize the acid?
2 HCl(aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2 H2O(l)
= 62.5 mL2.5 mol HCl 1 L
25.00mL
HCl2 mol HCl
1 mol Ca(OH)2
0.50 mol
Ca(OH)2
1 L1000 mL
1 L 1 L
1000 mL
45
Practice: Titration Ex. The flask is filled with 20.60 mL of a 0.09662 M HCl soln.
What is the molarity of the NaOH soln if it takes 20.92 mL to neutralize the acid?
.09662 mol HCl 1 L
20.60mL
HCl1 mol HCl
1 mol NaOH1000 mL
1 L= 0.001990 mol NaOH
= 0.09512 M 1 L
1000 mL0.001990 mol NaOH20.92 mL
