Transcript • 18.1 – Acids and Bases in Water

Chapter 18 – Acid-Base Equilibria
• 18.1 – Acids and Bases in Water
• 18.2 – Autoionization of Water and the pH Scale
• 18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base
Definition
• 18.4 – Solving Problems Involving Weak-Acid Equilibria
• 18.5 – Weak Bases and Their Relation to Weak Acids
• 18.6 – Molecular Properties and Acid Strength
• 18.7 – Acid-Base Properties of Salt Solutions
• 18.8 – Generalizing the Brønsted-Lowry Concept: The
Leveling Effect
• 18.9 – Electron-Pair Donation and the Lewis Acid-Base
Definition
1
18.1 – Acids and Bases in Water
2
18.1 – Acids and Bases in Water
Strong acid: HA(g or l) + H2O(l)
H3O+(aq) + A-(aq)
Zn with 1M HCl(aq)
vs. 1M CH3COOH(aq)
Weak acid: HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
3
ACID STRENGTH
18.1 – Acids and Bases in Water
4
18.1 – Acids and Bases in Water
SAMPLE PROBLEM 18.1:
PROBLEM:
Classify each of the following compounds as a strong acid,
weak acid, strong base, or weak base.
(a) H2SeO4
PLAN:
Classifying Acid and Base Strength from the
Chemical Formula
(b) (CH3)2CHCOOH
(c) KOH
(d) (CH3)2CHNH2
Pay attention to the text definitions of acids and bases. Look at O for
acids as well as the -COOH group; watch for amine groups and
cations in bases.
SOLUTION: (a) Strong acid - H2SeO4 - the number of O atoms exceeds
the number of ionizable protons by 2.
(b) Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group.
(c) Strong base - KOH is a Group 1A(1) hydroxide.
(d) Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and
is an amine.
5
18.2 – Autoionization of Water and the pH Scale
+
H2O(l)
+
H2O(l)
H3O+(aq)
OH-(aq)
6
18.2 – Autoionization of Water and the pH Scale
H2O(l) + H2O(l)
Kc =
H3O+(aq) + OH-(aq)
[H3O+][OH-]
[H2O]2
The Ion-Product Constant for Water, Kw:
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 at 25 C
A change in [H3O+] causes an inverse change in [OH-], and vice
versa.
in an acidic solution, [H3O+] > [OH-]
in a basic solution, [H3O+] < [OH-]
in a neutral solution, [H3O+] = [OH-]
7
18.2 – Autoionization of Water and the pH Scale
[H3O+]
Divide into Kw
[OH-]
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
ACIDIC
SOLUTION
NEUTRAL
SOLUTION
BASIC
SOLUTION
8
18.2 – Autoionization of Water and the pH Scale
SAMPLE PROBLEM 18.2:
PROBLEM:
PLAN:
Calculating [H3O+] and [OH-] in an Aqueous
Solution
A research chemist adds a measured amount of HCl gas to pure
water at 25 °C and obtains a solution with [H3O+] = 3.0x10-4M.
Calculate [OH-]. Is the solution neutral, acidic, or basic?
Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].
SOLUTION: K = 1.0x10-14 = [H O+] [OH-] so
w
3
[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 = 3.3x10-11M
[H3O+] is > [OH-] and the solution is acidic.
9
18.2 – Autoionization of Water and the pH Scale
•
•
•
•
•
•
•
•
pH = -log [H3O+]
# of sig figs
pH of a neutral soln = 7.00
pH of an acidic soln < 7.00
pH of a basic soln > 7.00
1 pH unit = 10x change
[H3O+] = 10-pH
p-scales
– pOH = -log [OH-]
– pK = -log K
10
18.2 – Autoionization of Water and the pH Scale
Table 18.3 The Relationship Between Ka and pKa
Acid Name (Formula)
Ka at 25 C
pKa
Hydrogen sulfate ion (HSO4-)
1.02x10-2
1.991
Nitrous acid (HNO2)
7.1x10-4
3.15
Acetic acid (CH3COOH)
1.8x10-5
4.74
Hypobromous acid (HBrO)
2.3x10-9
8.64
Phenol (C6H5OH)
1.0x10-10
10.00
11
18.2 – Autoionization of Water and the pH Scale
12
18.2 – Autoionization of Water and the pH Scale
SAMPLE PROBLEM 18.3:
PROBLEM:
PLAN:
Calculating [H3O+], pH, [OH-], and pOH
In an art restoration project, a conservator prepares copperplate etching solutions by diluting concentrated HNO3 to 2.0M,
0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and
pOH of the three solutions at 25 °C.
HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-]
and then convert to pH and pOH.
SOLUTION:
For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30
For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48
For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80
13
18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition
Lone pair
binds H+
+
+
HCl
H2O
(acid, H+ donor)
Cl-
H3O+
(base, H+ acceptor)
Lone pair
binds H+
+
+
NH3
(base, H+ acceptor)
H2O
NH4+
OH-
(acid, H+ donor)
14
18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition
Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions
Conjugate Pair
Acid
+
Base
Base
+
Acid
Conjugate Pair
Reaction 1
HF
+
H2O
F-
+
H3O+
Reaction 2
HCOOH
+
CN-
HCOO-
+
HCN
Reaction 3
NH4+
+
CO32-
NH3
+
HCO3-
Reaction 4
H2PO4-
+
OH-
HPO42-
+
H2O
Reaction 5
H2SO4
+
N2H5+
HSO4-
+
N2H62+
Reaction 6
HPO42-
+
SO32-
PO43-
+
HSO315
18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition
SAMPLE PROBLEM 18.4:
PROBLEM:
Identifying Conjugate Acid-Base Pairs
The following reactions are important environmental processes.
Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO32-(aq)
(b) H2O(l) + SO32-(aq)
PLAN:
HPO42-(aq) + HCO3-(aq)
OH-(aq) + HSO3-(aq)
Identify proton donors (acids) and proton acceptors (bases).
conjugate pair2
conjugate pair
1
SOLUTION:
(a) H2PO4-(aq) + CO32-(aq)
proton
donor
proton
acceptor
HPO42-(aq) + HCO3-(aq)
proton
acceptor
conjugate pair2
conjugate pair1
(b) H2O(l) + SO32-(aq)
proton proton
donor acceptor
proton
donor
OH-(aq) + HSO3-(aq)
proton
acceptor
proton
donor
16
18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition
17
18.3 – Proton Transfer and the Brønsted-Lowry Acid-Base Definition
SAMPLE PROBLEM 18.5:
PROBLEM:
Predicting the Net Direction of an Acid-Base
Reaction
Predict the net direction and whether Ka is greater or less than 1
for each of the following reactions (assume equal initial
concentrations of all species):
(a) H2PO4-(aq) + NH3(aq)
HPO42-(aq) + NH4+(aq)
(b) H2O(l) + HS-(aq)
PLAN:
OH-(aq) + H2S(aq)
Identify the conjugate acid-base pairs and then consult Figure 18.10
to determine the relative strength of each. The stronger the species,
the more preponderant its conjugate.
SOLUTION:
(a) H2PO4-(aq) + NH3(aq)
stronger acid
HPO42-(aq) + NH4+(aq)
stronger base
weaker base weaker acid
Net direction is to the right with Kc > 1.
(b) H2O(l) + HS-(aq)
OH-(aq) + H2S(aq)
weaker acid weaker base
stronger basestronger
acid
Net direction is to the left with Kc < 1.
18
18.4 – Solving Problems Involving Weak-Acid Equilibria
SAMPLE PROBLEM 18.6:
PROBLEM:
PLAN:
Finding the Ka of a Weak Acid from the pH of
Its Solution
Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc)
builds up in the blood of persons with phenylketonuria, an
inherited disorder that, if untreated, causes mental retardation
and death. A study of the acid shows that the pH of 0.13M
HPAc is 2.62. What is the Ka of phenylacetic acid?
Write out the dissociation equation. Use pH and solution concentration
to find the Ka.
Assumptions:
SOLUTION:
With a pH of 2.62, the [H3O+]HPAc >> [H3O+]water.
[PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial [HPAc]dissociation
HPAc(aq) + H2O(l)
Ka =
H3O+(aq) + PAc-(aq)
[H3O+][PAc-]
[HPAc]
19
18.4 – Solving Problems Involving Weak-Acid Equilibria
SAMPLE PROBLEM 18.6:
Finding the Ka of a Weak Acid from the pH of
Its Solution
continued
Concentration(M)
Initial
Change
Equilibrium
H2O(l)
H3O+(aq) +
0.12
-
1x10-7
0
-x
-
+x
+x
0.12-x
-
x +(<1x10-7)
x
HPAc(aq) +
PAc-(aq)
[H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water)
x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-]
So Ka =
(2.4x10-3) (2.4x10-3)
[HPAc]equilibrium = 0.12-x ≈ 0.12 M
= 4.8 x 10-5
0.12
Be sure to check for % error.
[H3
O+]
from water;
[HPAc]dissn;
1x10-7M
x100 = 4x10-3 %
2.4x10-3M
2.4x10-3M
x100 = 2.0 %
0.12M
20
18.4 – Solving Problems Involving Weak-Acid Equilibria
SAMPLE PROBLEM 18.7:
PROBLEM:
PLAN:
Determining Concentrations from Ka and
Initial [HA]
Propanoic acid (CH3CH2COOH, which we simplify and HPr) is
an organic acid whose salts are used to retard mold growth in
foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)?
Write out the dissociation equation and expression; make whatever
assumptions about concentration which are necessary; substitute.
Assumptions:
For HPr(aq) + H2O(l)
H3O+(aq) + Pr-(aq)
x = [HPr]diss = [H3O+]from HPr= [Pr-]
Ka =
[H3O+][Pr-]
[HPr]
SOLUTION:
Concentration(M)
Initial
Change
Equilibrium
HPr(aq) + H2O(l)
H3O+(aq) + Pr-(aq)
0.10
-
0
0
-x
-
+x
+x
0.10-x
-
x
x
Since Ka is small, we will assume that x << 0.10
21
18.4 – Solving Problems Involving Weak-Acid Equilibria
SAMPLE PROBLEM 18.7:
Determining Concentrations from Ka and
Initial [HA]
continued
1.3x10-5
=
[H3O+][Pr-]
[HPr]
x  (0.10)(1.3x105 )
(x)(x)
=
0.10
= 1.1x10-3 M = [H3O+]
Check: [HPr]diss = 1.1x10-3M/0.10 M x 100 = 1.1%
22
18.4 – Solving Problems Involving Weak-Acid Equilibria
[HA]dissociated
Percent HA dissociation =
x 100
[HA]initial
23
18.4 – Solving Problems Involving Weak-Acid Equilibria
Polyprotic acids
acids with more than more ionizable proton
H3PO4(aq) + H2O(l)
H2PO4
-(aq)
+ H3
O+(aq)
Ka1 =
[H3O+][H2PO4-]
[H3PO4]
= 7.2x10-3
H2PO4-(aq) + H2O(l)
HPO42-(aq) + H3O+(aq)
Ka2 =
[H3O+][HPO42-]
[H2PO4-]
= 6.3x10-8
HPO4
2-(aq)
+ H2O(l)
PO4
3-(aq)
+ H3
O+(aq)
Ka3 =
[H3O+][PO43-]
[HPO42-]
= 4.2x10-13
Ka1 > Ka2 > Ka3
24
ACID STRENGTH
18.4 – Solving Problems Involving Weak-Acid Equilibria
25
18.4 – Solving Problems Involving Weak-Acid Equilibria
SAMPLE PROBLEM 18.8:
PROBLEM:
PLAN:
Calculating Equilibrium Concentrations for a
Polyprotic Acid
Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as
vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12)
found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the
pH of 0.050M H2Asc.
Write out expressions for both dissociations and make assumptions.
Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+.
Ka1 is small so [H2Asc]initial ≈ [H2Asc]diss
After finding the concentrations of various species for the first dissociation,
we can use them as initial concentrations for the second dissociation.
SOLUTION:
H2Asc(aq) + H2O(l)
HAsc-(aq)
+ H2O(l)
HAsc-(aq) + H3O+(aq)
Asc2-(aq)
+ H3
O+(aq)
Ka1 =
Ka2 =
[HAsc-][H3O+]
[H2Asc]
[Asc2-][H3O+]
[HAsc-]
= 1.0x10-5
= 5x10-12
26
18.4 – Solving Problems Involving Weak-Acid Equilibria
continued
Concentration(M)
Initial
H2Asc(aq) + H2O(l)
0.050
-
0
0
-x
-
+x
+x
0.050 - x
-
x
x
Change
Equilibrium
HAsc-(aq) + H3O+(aq)
Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0x10-5 = (x)(x)/0.050 M
5
x  (0.050)(1.0x10 )
Concentration(M)
Initial
HAsc-(aq) + H2O(l)
7.1x10-4M
Change
Equilibrium
x
x = 7.1x10-4 M
-x
7.1x10-4 - x
 (7.4x105 )(5x1012 )
pH = -log(7.1x10-4) = 3.15
Asc2-(aq) + H3O+(aq)
-
0
-
+x
+x
-
x
x
= 6x10-8 M
0
27
18.5 – Weak Bases and Their Relation to Weak Acids
[B]
BASE STRENGTH
Kb =
[BH+][OH-]
28
18.5 – Weak Bases and Their Relation to Weak Acids
Lone pair
binds H+
+
CH3NH2
H2O
methylamine
+
CH3NH3+
OH-
methylammonium ion
29
18.5 – Weak Bases and Their Relation to Weak Acids
SAMPLE PROBLEM 18.9:
PROBLEM:
PLAN:
Determining pH from Kb and Initial [B]
Dimethylamine, (CH3)2NH, a key intermediate in detergent
manufacture, has a Kb of 5.9x10-4. What is the pH of 1.5M
(CH3)2NH?
Perform this calculation as you did those for acids. Keep in mind that
you are working with Kb and a base.
(CH3)2NH(aq) + H2O(l)
(CH3)2NH2+(aq) + OH-(aq)
Assumptions:
Kb >> Kw so [OH-]from water is neglible
[(CH3)2NH2+] = [OH-] = x ; [(CH3)2NH2+] - x ≈ [(CH3)2NH]initial
SOLUTION:
Concentration
Initial
Change
Equilibrium
(CH3)2NH(aq) + H2O(l)
(CH3)2NH2+(aq) + OH-(aq)
1.50M
-
0
0
-x
-
+x
+x
1.50 - x
-
x
x
30
18.5 – Weak Bases and Their Relation to Weak Acids
Kb = 5.9x10-4 =
[(CH3)2NH2+][OH-]
(x) (x)
5.9x10-4
=
[(CH3)2NH]
x = 3.0x10-2M = [OH-]
1.5M
Check assumption:
3.0x10-2M/1.5M x 100 = 2%
[H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13M
pH = -log 3.3x10-13 = 12.48
31
18.5 – Weak Bases and Their Relation to Weak Acids
SAMPLE PROBLEM 18.10: Determining the pH of a Solution of APROBLEM:
PLAN:
Sodium acetate (CH3COONa, or NaAc for this problem) has
applications in photographic development and textile dyeing.
What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is
1.8x10-5.
Sodium salts are soluble in water so [Ac-] = 0.25M.
Write the association equation for acetic acid; use the Ka to find the Kb.
SOLUTION:
Concentration
Ac-(aq) + H2O(l)
Initial
0.25M
-
0
0
-x
-
+x
+x
0.25M-x
-
x
x
Change
Equilibrium
Kb =
[HAc][OH-]
[Ac-]
HAc(aq) + OH-(aq)
=
Kw
Ka
Kb =
1.0x10-14
1.8x10-5
= 5.6x10-10M
32
18.5 – Weak Bases and Their Relation to Weak Acids
[Ac-] = 0.25M-x ≈ 0.25M
Kb =
[HAc][OH-]
[Ac-]
5.6x10-10 = x2/0.25M
x = 1.2x10-5M = [OH-]
Check assumption:
1.2x10-5M/0.25M x 100 = 4.8x10-3 %
[H3O+] = Kw/[OH-] = 1.0x10-14/1.2x10-5 = 8.3x10-10M
pH = -log 8.3x10-10M = 9.08
33
18.6 – Molecular Properties and Acid Strength
Bond strength decreases,
acidity increases
Figure 18.12 The effect of atomic and molecular properties on nonmetal hydride acidity.
6A(16)
7A(17)
H2O
HF
H2S
HCl
H2Se
HBr
H2Te
HI
Electronegativity increases,
acidity increases
34
18.6 – Molecular Properties and Acid Strength
Figure 18.13
H
O


The relative strengths of oxoacids.
I
>
H
O


Br
>
H
O
Cl


O
H
O


Cl
<<
H

O

Cl
O
O
35
18.6 – Molecular Properties and Acid Strength
Table 18.7 Ka Values of Some Hydrated Metal Ions at 250C
Hydrated Ion
Ka
Fe3+
Fe(H2O)63+(aq)
6 x 10-3
Sn2+
Sn(H2O)62+(aq)
4 x 10-4
Cr3+
Cr(H2O)63+(aq)
1 x 10-4
Al3+
Al(H2O)63+(aq)
1 x 10-5
Cu2+
Cu(H2O)62+(aq)
3 x 10-8
Pb2+
Pb(H2O)62+(aq)
3 x 10-8
Zn2+
Zn(H2O)62+(aq)
1 x 10-9
Co2+
Co(H2O)62+(aq)
2 x 10-10
Ni2+
Ni(H2O)62+(aq)
1 x 10-10
ACID STRENGTH
Free Ion
36
18.6 – Molecular Properties and Acid Strength
Figure 18.13
The acidic behavior of the hydrated Al3+ ion.
Electron density drawn
toward Al3+
Nearby H2O acts
as base
H3O+
H2O
Al(H2O)63+
Al(H2O)5OH2+
37
18.7 – Acid-Base Properties of Salt Solutions
38
18.7 – Acid-Base Properties of Salt Solutions
SAMPLE PROBLEM 18.11: Predicting Relative Acidity of Salt Solutions
PROBLEM:
Predict whether aqueous solutions of the following are acidic,
basic, or neutral, and write an equation for the reaction of any
ion with water:
(a) Potassium perchlorate, KClO4
(b) Sodium benzoate, C6H5COONa
(c) Chromium trichloride, CrCl3
PLAN:
(d) Sodium hydrogen sulfate, NaHSO4
Consider the acid-base nature of the anions and cations. Strong
acid-strong base combinations produce a neutral solution; strong
acid-weak base, acidic; weak acid-strong base, basic.
SOLUTION: (a) The ions are K+ and ClO4- , both of which come from a strong
base(KOH) and a strong acid(HClO4). Therefore the solution will be neutral.
(b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a
weak organic acid. The salt solution will be basic.
(c) Cr3+ is a small cation with a large + charge, so it’s hydrated form will react
with water to produce H3O+. Cl- comes from the strong acid HCl. Acidic solution.
(d) Na+ comes from a strong base. HSO4- can react with water to form H3O+.
So the salt solution will be acidic.
39
18.7 – Acid-Base Properties of Salt Solutions
SAMPLE PROBLEM 18.12: Predicting the Relative Acidity of Salt
Solutions from Ka and Kb of the Ions
PROBLEM:
PLAN:
Determine whether an aqueous solution of zinc formate,
Zn(HCOO)2, is acidic, basic, or neutral.
Both Zn2+ and HCOO- come from weak conjugates. In order to find
the relatively acidity, write out the dissociation reactions and use the
information in Tables 18.2 and 18.7.
SOLUTION:
Zn(H2O)62+(aq) + H2O(l)
HCOO-(aq) + H2O(l)
Zn(H2O)5OH+(aq) + H3O+(aq)
HCOOH(aq) + OH-(aq)
Ka Zn(H2O)62+ = 1x10-9
Ka HCOO- = 1.8x10-4 ; Kb = Kw/Ka = 1.0x10-14/1.8x10-4 = 5.6x10-11
Ka for Zn(H2O)62+ >>> Kb HCOO-, therefore the solution is acidic.
40
18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
F
B
F
F
acid
F
H
+
H
N
HH
base
B
F
F
N
HH
adduct
M(H2O)42+(aq)
M2+
H2O(l)
adduct
41
18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition
Figure 18.15
The Mg2+ ion as a Lewis acid in the
chlorophyll molecule.
42
18.9 – Electron-Pair Donation and the Lewis Acid-Base Definition
SAMPLE PROBLEM 18.13: Identifying Lewis Acids and Bases
PROBLEM:
PLAN:
Identify the Lewis acids and Lewis bases in the following reactions:
(a) H+ + OH-
H2O
(b) Cl- + BCl3
BCl4-
(c) K+ + 6H2O
K(H2O)6+
Look for electron pair acceptors (acids) and donors (bases).
SOLUTION:
acceptor
(a) H+ + OHdonor
H2O
donor
(b) Cl- + BCl3
acceptor
BCl4-
acceptor
(c) K+ + 6H2O
donor
K(H2O)6+
43