Transcript 14.2 Nature of Acids and Bases

Equilibrium and acids and bases
• Chapter 14 and 15 in text
1
Chemical Equilibrium
• Concentrations of all reactants and
products are constant with time.
• That does not mean [reactant] = [product]
Rather the rate of decomposition of [react]
and formation of [prod] are constant.
2
Equilibrium is a Dynamic Situation
Demonstration
B
A
3
• When a car from island A moves to island
B there is a stress placed on the
equilibrium and the reaction (the members
of the groups) must shift either left or right
to relive the stress.
• If a stress is placed on Island A then the
equilibrium will shift to the right to
compensate.
4
What Just Happened?
• When a chemical reaction is at
equilibrium, any disturbance of the system,
such as a change in temperature, or
addition or removal of one of the reaction
components, will "shift" the composition of
the reaction to a new equilibrium state.
5
A Chemical Example of a Shift in
Equilibrium
Co(H2O)62+ + 4Cl
Pink
CoCl42+ + 6H20
Blue
6
Demonstration of Equilibrium
7
Question:
Which direction would the equilibrium shift
if HCl was added to the reaction?
HCl
Co(H2O)62+ + 4Cl
Pink
CoCl42+ + 6H20
Blue
8
Answer
The addition of HCl would cause an increase in Cland thus the reaction would react by shifting the
reaction to the right (blue side) in order to
“consume” the excess Cl- and return to a state
of equilibrium.
Co(H2O)62+ + 4Cl
Pink
CoCl42+ + 6H20
Blue
9
The Law of Mass Action
For any reversible reaction:
jA + kB
lC + mD
The law of mass action is represented by the following equilibrium
expression. (Prod /React)
K=
[C]l [D]m
[A]j [B]k
K = equilibrium constant
Solids and liquids are not written in equilibrium
expressions
10
• K > 1 Favors products
• K<1 Favors reactants
11
Graph of Equilibrium
12
Question
Write the equilibrium expression for the
following equations:
PCl5 (g)
Cl2O7(g) + 8H2(g)
PCl3 (g) + Cl2 (g)
2HCl(g) + 7H20(g)
NOTE: Make sure the RXN is BALANCED!!!
13
Answer
K = [PCl3] [Cl2]
[PCl5]
K = [H2O]7 [HCl]2
[Cl2O7] [H2]8
14
Calculating Equilibrium
Example:
Calculate the equilibrium constant K, for the
following reaction at 25°C,
H2 (g) + I2 (g)
2HI (g)
if the equilibrium concentrations are
[H2] = 0.106M [I2] = 0.022M
[HI] = 1.29M
15
Answer
Equilibrium Expression:
K = [HI]2
[H2] [I2]
Equilibrium Constant:
K=
(1.29)2
= 7.1 x 102
(0.106) (0.022)
Note the units cancel out!
That’s one less thing to
remember!
16
Homework
• Pg 633 1,8,9,14,16
• Equilibrium wks
17
How do we predict which direction
equilibrium will shift to?
Le Chatelier’s Principle
If we stress a reaction out the reaction will shift
(either towards R or P) to reduce the stress.
Stress = a change in: Concentration
Pressure
Volume
Temperature
18
Concentration
• Ask yourself:
• Is what is being added a solid or liq? (No effect)
• a gas or aqueous reactant or a product?
(effect!)
• If it’s a reactant the reaction will shift toward the
products to consume the extra reactant.
• If it’s a product the reaction will shift toward the
reactants to consume the extra product.
19
Example
• N2 g + 3H2
g
2NH3 g
• Add NH3
• Remove N2
• Add H2
20
Changes in V or P
• P changes in L or aq have little effect
• P changes of gases have huge effects
(PV=nRT)
(PV=PV)
• Concentration is effected by pressure.
21
• An increase in Pressure (due to a
decrease in V) will shift to decreases the
total number of moles of gas.
• If the number of moles on R and P side
are the same then a change in P will have
no effect on equilibrium.
22
Example
• Lets do Pg 626 example 14.12 in your text
↓P shift to side of reaction with greatest
number of moles of gas
↑P shift to side of reaction with less moles of
gas
23
Temperature
• ONLY a change in TEMPERATURE can change
the value of K (equilibrium constant).
• You must look at the H of the reaction to see
how T will effect it
• Temperature is our stress so the reaction will
move in the direction that removes the stress.
24
↑ temperature the rxn moves in the endothermic direction (+ΔH)
ENDO  Right ( K ↑ )
↓ temperature the rxn moves in the exothermic direction (-ΔH)
EXO  Left ( K ↓ )
**K only depends on temperature, catalysts have NO EFFECT on
K**
K>>1 favors products
K<< 1 favors reactants
25
Think about heat like a R or P
• Endothermic +ΔH
Heat + A  B H = 400 kJ
Favors ↑ in T
therefore K ↑ when it is heated and K ↓ when it is
cooled.
Exothermic -ΔH
A  Heat + B H = - 400 kJ
Favors ↓ in T therefore K ↑ when it is cooled and K ↓
when it is heated
26
Example
• Examples 14.13 on pg 628-629 in text
27
Homework
• Pg 636 pg 49,51,53-55,62
28
Nature of Acids and Bases
Acids: Sour Taste
If a solution has a high
[H+] = acidic
Base: Bitter Taste /
Slippery feel
If a solution has a high
[OH-] = base
29
About the scale
• P = quantity
• So …
• pH = quantity of H
ion
• pOH = quantity of OH
ion
• pH + pOH = 14
30
Concentrations
• [H+] = [OH-] = Neutral
• [H+] > [OH-] = Acid
• [H+] < [OH-] = Base
31
Arrhenius Concept
• Focuses on what ions were formed when
acids and bases dissolved in water.
• Acids dissociate in water give hydrogen
ions (H+ or H3O+ hydronium ion)
• Bases dissociate in water give hydroxide
ions (OH- hydroxide ion ) .
32
• Arrhenius acid - Any substance that
ionizes when it dissolves in water to give
the H+ ion.
e.g.
Arrhenius base - Any substance that
ionizes when it dissolves in water to give
the OH- ion.
e.g.
33
• The theory can only classify substances when they are
dissolved in water since the definitions are based upon
the dissociation of compounds in water.
• It does not explain why some compounds containing
hydrogen such as HCl dissolve in water to give acidic
solutions and why others such as CH4 do not.
• The theory can only classify substances as bases if they
contain the OH- ion and cannot explain why some
compounds that don't contain the OH- such as Na2CO3
have base-like characteristics.
34
Bronsted Lowery Acid Base
Concept
Acid: substance that can donate a proton (+)
Base: substance that accepts a proton (+) (aka
they have a lone pair of e- to accept a
proton)
Unlike Arrhenius concept this is applicable in
both aqueous and non-aqueous states.
35
Equilibrium
• Reaction produces reactants and products
at the same rate, but not necessarily in the
same amounts
• Bathroom theory
36
Example
• NH3+ (aq) + H20 (l)  NH4 (aq) + OH- (aq)
• Equilibrium will favor the formation of the
weaker acid and the weaker base.
• In this rxn the [NH4] and [OH- ] will be low
because they are the stronger acid and
base.
37
In the above reaction, the H+ from HCl is
donated to H2O which accepts the H+ to
form H3O+, leaving a Cl- ion.
38
Conjugate Acid and Base Pairs
• The part of the acid remaining when an
acid donates a H+ ion is called the
conjugate base.
• The acid formed when a base accepts a
H+ ion is called the conjugate acid.
39
For the generic acid HA:
Formed
when a
proton Is
transferred
to the base
Everything
that is left
after a
proton to the
base.
NOTE
Strong acids have weak conjugate bases.
Strong bases have weak conjugate acids.
40
Question
If H20 is an acid what would its conjugate
base be?
• What is the conjugate acid of HPO42- ?
• What is the conjugate base of HS-
41
Answer
• H20 take away a proton OH• HPO42- add an H+ H2PO4( we added a proton and that needs to be reflected in the
molecular charge)
• HS-
S2-
42
Amphoteric
• The ability of a substance to act as an acid
or a base.
• Ex: H2PO4- and H2O can act as both acids
and bases.
43
Back to Old Faithful
In this equation the stronger base will win the competition for
H+.
If H2O is a stronger base than A-, then it will have a greater
affinity for the protons and the equilibrium will lie to the right
favoring the formation of H3O+.
If A- is stronger then equilibrium will fall to the left and acid in
the form HA will form.
44
Strong Acids
HCl
HBr
HI
HNO3
HClO4
HClO3
H2SO4
Conj. Bases
ClBr INO3ClO4ClO3HSO4-
You must memorize all of these
45
Common Strong Bases
Formula
Name
NaOH
sodium hydroxide
LiOH
lithium hydroxide
KOH
Mg(OH)2
potassium hydroxide
Magnesium Hydroxide
Ca(OH)2
Ba(OH)2
Sr(OH)2
Calcium hydroxide
Barium Hydroxide
Strontium Hydroxide
46
Example
• Identify the CA and CB for each reaction
• HNO3 + H2O ↔
• NH3 + H2O ↔
47
Acid-dissociation equilibrium
constant (Ka)
• The relative strength of an acid is
described as an acid-dissociation
equilibrium constant.
• The acid-dissociation equilibrium constant
is the mathematical product of the
equilibrium concentrations of the products
of this reaction divided by the equilibrium
concentration of the original acid
48
Think Products over reactants
49
Question
• Write an ionization equation for the
following and then write the acid
dissociation constant for both. (all occur in
water)
• Hydrochloric acid
• Acetic acid HC2H3O2
50
Answer
• HCl ↔ H+ + ClKa = [H+] [Cl-] / [HCl]
HC2H3O2 ↔ H+ + C2H3O2Ka = [H+] [C2H3O2-] / [HC2H3O2 ]
51
54
Temperature increases as Kw increases
55
Question
• Calculate the H+ in aqueous household
ammonia if the OH- concentration is
0.0025 M. Then tell me if the solution is an
acid or a base.
This solution is in water so we can use Kw
56
Answer
Kw = [H] [OH] = 1.0 x 10 -14
[H] = Kw /[OH] = 1.0 x 10 -14
[H] = Kw /[0.0025M] = 1.0 x 10 -14
[H] = 4.0 x 10-12 M
0.0025M OH > 4.0 x 10-12 M H solution is a
base
57
Question
• Calculate the values of [H+] [OH-] in a neutral
solution at 25ºC
• If we know Kw = [H][OH] and [H] = [OH] in a
neutral solution
• Kw = 1.0 E-14
• then 1.0 E-14 = [x][x]
• 1.0 E-14 = X2 (QUADRATIC TIME!!!!)
• X = 1.0E-7 M [H] and [OH]
58
Example
Calculate the concentrations using
Kw= [H+] [OH-] =10-14
For the following and state if it is an acid,
base, or neutral
• [OH-] = 10-5 M
• [H+] = 10.0 M
59
Answer
• Kw = 10-14= [H+] [10-5] = 1.0 E-9 M
• [H+] < [OH-] = Base
• Kw = 10-14=[10] [OH-] = 1.0E-15 M
• [H+] >[OH-] = Acid
60
pH and pOH
• Tell us the concentration of H and OH in solution.
• pH = -log [H+]
pOH = -log [OH-]
• [H] = 10-pH
[OH] = 10-pOH
• pH + pOH =14
• (really strong acids may have a negative pH)
61
Find pH
pH = -log [H+]
Calculate the pH of a neutral solution that
has a [H+] of 1.0 x 10-7 M
pH = - log 1.0 x 10-7 M
pH = 7.00
62
pOH
• pOH = -log [OH-]
• or
• pH + pOH = 14
• So we can use the pH (7) from the last
problem to find the pOH of that solution.
• 14-7 = 7 or we could have been given the
[OH-]
63
Given pH find [H]
• pH = -log [H]
• pH = 4 find [H]
• 4 = -log X
• 10 -4 = X
64
Strong Acids
• Strong acids dissociate almost completely
in water and therefore have relatively large
Ka values.
• Equilibrium lies far to the right
• HA dissociates almost completely
• Yield weak conjugate bases
• Strong electrolytes (100% conductivity)
• Ka Large > 1
66
Weak Acids
• Weak acids dissociate only slightly in
water and therefore have relatively small
Ka values.
• Equilibrium lies far to the left (HA does not
dissociate)
• Yields strong conjugate base
• Ka small < 1
67
Solving Strong Acid Equations
(and Bases too)
• 1. Strong acids dissociate 100% and are
the main source of H+ in a solution.
• 2. Therefore [H+] equals the original
concentration of the acid
ex: 0.2 M HCl = [H] = 0.2 M
[Cl] = 0.2 M
68
Calculating pH for a strong acid
example
• What is the pH of a 0.04 M solution of
HClO4?
• HClO4 = strong acid equilibrium lies to the
right and completely dissociates.
• [H+] = [ClO4-]= 0.04 M
• pH = -log [0.04] = 1.40
69
Calculating Ka
1.
Write the ionization equation for the reaction
2.
Write the equilibrium expression for the reaction (Ka)
3.
Calculate the [H+] using a given pH
4.
ICE box ([initial], [change], [equilibrium] to determine the
values for concentration that you will substitute into your
equilibrium equation.
5.
Insert concentration into equilibrium expression and
solve for Ka
70
Solving for Ka example
• A student prepared a 0.1M solution of
formic acid (HCHO2) and measured its pH
to be 2.38. Calculate Ka.
• How will you attack this problem?
71
Answer
• Write the ionization rxn.
HCHO2 ↔H+ + CHO2• Write Ka equation
Ka = [H+] [CHO2-]
[HCHO2]
• Solve for [H] using pH given
pH = -log [H+] =
2.38 = -log [H+]
[H+] = 10-2.38 =0.004168 = 4.2 X 10-3
72
Not done yet
• ICE Box
HCHO2 ↔
H+ +
CHO2-
0 .10
0
0
[Change]
-4.2 x10-3 +4.2 x10-3 +4.2 x10-3
[Equilibrium] 0.1- 4.2 x10-3 M 4.2 x10-3 M 4.2 x10-3M
[Initial]
73
Still not yet
• Plug values into Ka equation
HCHO2 ↔
H+ +
CHO2[Equilibrium] 0.1- 4.2 x10-3 4.2 x10-3 4.2 x10-3
Ka = [H+][CHO2] = [4.2 x10-3] [4.2 x10-3]
[HCHO2]
[0.1- 4.2 x10-3 ]
Solve for Ka = 1.8 x 10-4
74
Lets do another one
• A student calculates the pH of HOCl to be 3.5
find Ka. The initial concentration of the acid is
0.01M
•
•
•
•
•
Write equation for reaction
Write Ka equation
Use pH to solve for [H+]
ICE Box
Plug values into Ka equation
75
Using Ka and [acid] to calculate pH
• Write the dissociation equation for the
reaction.
• Write the equation for Ka
• ICE Box but now we know our initial
concentration of our acid so we can use it
to solve for the [H+]
• Use found [H+] to calculate pH
76
Weak acid example
• Calculate the pH of a 0.1 M aqueous
solution of HOCl (Ka= 3.4x10-8 weak acid).
• HOCl has a higher Ka value than water
and will dissociate to produce the most H+
ions, so we use it to write our equation for
Ka
77
Write the rxn:
H+ +
HOCl
Write the equation for:
OCl-
Ka = 3.4x10-8 = [H+][OCl-]/HOCL
Plug into ICE Box using X for unknown concentrations:
HOCl
H+ + OClI
C
E
0.1
-X
0.1-X
0
X
X
0
X
X
X = the amount
of HOCl that
dissociates
Substitute E values into Ka equation
3.4x10-8 = (x) (x)
We can ignore this x since it
0.1-X
is so small since stronger
acid
78
Plug into Ka and solve for X = [H+]:
3.4x10-8 =
X2
0.1
X = 5.9 x 10-5 M
Use pH = -log [H+] to solve for pH:
pH = -log 5.9 x 10-5
pH = 4.23
79
Percent Dissociation (aka %
ionization)
% dissociation = [ H+]final X 100
[ H+]initial
This equation allows us to identify for the
exact concentration of [ H+] that must
dissociate for the equation to reach
equilibrium.
The stronger the acid the greater the
ionization.
81
Example
Calculate the % dissociation for 1.0 M
HC2H3O2
Ka = 1.8 X 10-5
HC2H3O2
I
C
E
1.0
-X
1.0-X
H + + C 2H 3O 2
0
X
X
0
X
X
82
Ka = 1.8 X 10-5 = X2/ 1.0
[X] = 4.2 x 10-3
% diss = [H+ ]/ HC2H3O2 * 100
= 4.2 x 10-3/ 1.0
= 0.42%
83
Bases
• When strong bases are dissolved in
aqueous solutions they dissociate 100% in
OH-, so we can treat strong base
equations like we treat strong acid
equations.
• Strong Base = Large Kb = high pOH
84
• All hydroxides of group 1A and 2A are
strong bases.
• Exception Be(OH)2
• Strong bases will have large Kb values.
85
Weak Bases
B (aq) + H2O
Base Acid
BH+ (aq) + OH-(aq)
Conj. Acid Conj Base
React with water to form conjugate acid of the
base and OH- ion.
Small Kb values (low pOH)
86
Example
Calculate the pH of 0.05 M solution of Base
X, Kb = 1.7 x 10-9
**** Kb is small and tells us Base x is a
weak base
87
Answer
B (aq) + H2O
Base Acid
BH+ (aq) + OH-(aq)
Conj. Acid Conj Base
( H2O is and acid so we can ignore it since we are given kb)
I 0.05
C -x
E 0.05-x
0
x
x
0
x
x
Plug equilibrium values into Kb = [BH+][OH-]
[B]
88
1.7 x 10-9 =
x2
0.05
(Ignore the X)
[x] = [OH-] = 9.2 x 10-6
pOH = - log (9.2 x 10-6) = 5.04
pH + pOH = 14
pH + 5.04 = 14
pH = 8.96
89
Lewis Acids
• Lewis Acid: a substance that accepts an electron
pair
• Lewis Base: a substance that donates and
electron pair.
91
Keeping it straight
Definition of
Acid
Definition of
Base
Arrehnius
H+ producer
In water
OHProducer
In water
BronstedLowery
H+
donor
H+
acceptor
Lewis
Electron pair
acceptor
Model
Electron pair
92
donor
Lewis Acid Base Example
Lewis
Acid
Lewis
Base
Lewis
complex
• Brønsted-Lowry acid-base reaction =
donation and acceptance of a proton
• Lewis base (OH -) = hydroxide ion donates a
pair of electrons for covalent bond formation,
Lewis acid (H+ ) = accepts the pair of
electrons.
93
Lewis Example
For each rxn identify the Lewis acid and
base.
H + + H2O ↔ H 3O +
94
Answer
The proton (H+) is the Lewis acid and the
water (H2O) is the Lewis base.
95
Note
• Note:
• Every Bronsted-Lowery base is a Lewis
base because they all have a lone pair of
e- to accept the protons.
• BUT not every Lewis base is a BronstedLowery base because not all LB can
accept protons.
96
• Chemicals which have no hydrogen to
donate (aka the Bronsted-Lowry scheme)
can still be acids according to the lewis
scheme.
example, BF3 . If we determine Lewis
structure of BF3 , we find that B is octet
deficient and can accept a lone pair. Thus
it can act as a Lewis acid. Thus, when
reacting with ammonia, the reaction would
look like:
97
98
Titration
• Demo
• https://www.youtube.com/watch?v=3P8FP
sbseqM
• Titration Curves
• http://www.ausetute.com.au/titrcurv.html
99
Label the Diagram
100
Selecting an Indicator
101
102
Which indicator would you
choose to use in the lab?
A
B
C
D
103
Sample problem
30 mL of 0.10M NaOH neutralized 25.0mL of
hydrochloric acid. Determine the concentration
of the acid
• Write the balanced chemical equation for the reaction
• Calculate moles NaOH
(known volume and concentration)
• From the balanced chemical equation find the mole ratio
• Find moles HCl
• Calculate concentration of HCl: M = n ÷ V
104
•
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
• Extract the relevant information from
• the question:
NaOH
V = 30mL , M = 0.10M HCl
V = 25.0mL, M = ?
• Calculate moles NaOH
• NaOH
V = 30mL , M = 0.10M
3 x 10-3 moles
• From the balanced chemical equation find the mole ratio
• Find moles HCl
NaOH: HCl is 1:1
So [NaOH] = [HCl] = 3 x 10-3 moles at the equivalence
point
• Calculate concentration of HCl: M = n ÷ V
n = 3 x 10-3 mol,
V = 25.0 x 10-3L
0.12 M HCl
105