Chapter 18
Acid-Base Equilibria
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 Relations 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
The Nature of Acids and Bases:
Acids: Acids taste sour.
React with metals and produce H
gas.
turns blue litmus red
pH < 7
 * Bases: Bases taste bitter. They are
slippery.
turns red litmus blue.
pH >7

Arrhenius concept
 Acids
produce H+ ions. Bases
produce OH- ions.
 This concept is limited because it
applies to only aqueous solution and
defines only OH containing bases.
Neutralization:
 acid
+ base _______salt + water.
Acid Dissociation Constant (Ka)
 Write
Ka expression for strong and
weak acids.
Figure 18.1
The extent of dissociation for strong acids.
Strong acid: HA(g or l) + H2O(l)
H2O+(aq) + A-(aq)
Figure 18.2
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l)
H2O+(aq) + A-(aq)
Figure 18.3
1M HCl(aq)
Reaction of zinc with a strong and a weak acid.
1M CH3COOH(aq)
Strong acids dissociate completely into ions in water.
HA(g or l) + H2O(l)
H3O+(aq) + A-(aq)
Kc >> 1
Weak acids dissociate very slightly into ions in water.
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
Kc << 1
The Acid-Dissociation Constant
[H3O+][A-]
Kc =
[H2O][HA]
Kc[H2O] = Ka =
stronger acid
higher [H3O+]
larger Ka
[H3O+][A-]
[HA]
smaller Ka
lower [H3O+]
weaker acid
Classifying the Relative Strengths
of acids and Bases:

Strong acids:
– HCl, HBr, HI
– Oxo acids. HNO3, H2SO4, HClO4

Weak acids:
– HF
– HCN , H2S (H not bonded to O or
halogen)
– Oxo acids. HClO, HNO2, H3PO4

Carboxylic acids. CH3COOH
Classifying the Relative Strengths
of acids and Bases:

Strong bases:
– M2O, MOH
M= (group 1A metal)
– MO , M(OH)2 M=group 2A metal

Weak bases: (N atom and lone pair
of electrons)
– NH3
– Amines.
ACID STRENGTH
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.
Autoionization of water and the
pH scale

Water dissociates into its ions.
Autoionization of Water and the pH Scale
+
H2O(l)
H2O(l)
+
H3O +(a
q)
OH-(aq)
H2O(l) + H2O(l)
Kc =
H3O+(aq) + OH-(aq)
[H3O+][OH-]
[H2O]2
The Ion-Product Constant for Water
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 at 250C
A change in [H3O+] causes an inverse change in [OH-].
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]
Figure 18.4
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
[H3O+]
Divide into Kw
[OH-]
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
ACIDIC
SOLUTION
NEUTRAL
SOLUTION
BASIC
SOLUTION
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 250C 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.
pH scale
 pH
 log[H+]
 pH in water ranges from 0 to 14.
 Kw = 1.00  1014 = [H+] [OH]
 pKw = 14.00 = pH + pOH
 As pH rises, pOH falls (sum =
14.00).
 pH=7-neutral, pH>7-basic, pH<7acidic
pH scale
 Acidic
solns have a higher pOH.
 pK=-log K
 equation reaches equ, mostly
products present, low pK (high K)
 Reverse of the above statement is
also true.
 pH is measure using pH meter, pH
paper or acid-base indicator.
Figure 18.7
Methods for measuring the pH of an aqueous solution
pH (indicator) paper
pH meter
Figure 18.5
The pH values of
some familiar
aqueous solutions
pH = -log [H3O+]
Table 18.3 The Relationship Between Ka and pKa
Acid Name (Formula)
Ka at 250C
pKa
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
1.0x10-10
10.00
Hydrogen sulfate ion (HSO4-)
Phenol (C6H5OH)
Figure 18.6
The relations among [H3O+], pH, [OH-], and pOH.
Problems
 P.3.Calculate
pH and pOH at 25 C
for: 1.0 M H+
 P.4.pH=6.88, Calculate [ H+] and [
OH-] for this sample.
 P.5.Calculate pH of 0.10 M HNO3
 P.6.Calculate pH of 1.0 x 10-10 HCl.
Bronsted-Lowry model:
An acid is a proton (H+ ) donor. A base is
a proton acceptor.
 * A conjugate acid-base pair only differs
by one H.
 HCl +
H2O _________ H3O+ ClAcid
base
conj acid
conj base
 HA(aq) + H2O (l)  H3O+ (aq) +A-(aq)
CA1
CB2
CA2
CB1

Bronsted-Lowry model:
conjugate base: everything that remains
of the acid molecule after a proton is lost.
 Has one H less and one more minus
charge than the acid.
 conjugate acid: formed when the proton
is transferred to the base.
 Has one more H and one less charge than
the base.
 Do Follow up problem-18.4-Page.779

Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species which donates a H+.
A base is a proton acceptor, any species which accepts a H+.
An acid-base reaction can now be viewed from the standpoint
of the reactants AND the products.
An acid reactant will produce a base product and the two will
constitute an acid-base conjugate pair.
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-
+
HSO3-
Proton transfer as the essential feature of a BrønstedLowry acid-base reaction.
Figure 18.8
Lone pair
binds H+
+
+
HCl
H 2O
(acid, H+ donor)
Cl-
H 3 O+
(base, H+ acceptor)
Lone pair
binds H+
+
+
NH3
(base, H+ acceptor)
H2O
(acid, H+ donor)
NH4+
OH-
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
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)
(b) H2O(l) + HS-(aq)
PLAN:
HPO42-(aq) + NH4+(aq)
OH-(aq) + H2S(aq)
Identify the conjugate acid-base pairs and then consult Figure 18.10
(button) to determine the relative strength of each. The stronger the
species, the more preponderant its conjugate.
SOLUTION:
(a) H2PO4-(aq) + NH3(aq)
stronger acid
stronger base
HPO42-(aq) + NH4+(aq)
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 base stronger acid
Net direction is to the left with Kc < 1.
Figure 18.9
Strengths of
conjugate acidbase pairs
Solving Problems Involving
Weak-Acid Equilibria.





Write balanced equation and Ka
expression.
Make an ICE table.
Make required assumptions.
Substitute values and solve for x.
Verify assumptions by calculating %
error.
Problems
 P.7.
Calculate the pH of 1.00 M
aqueous solution of HF. Ka=7.2 x 104
 P.8 Calculate the pH of 0.100 M aq
solution of HOCl. Ka=3.5 x 10-8
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 0.12M
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+][PAc-]
[HPAc]
H3O+(aq) + PAc-(aq)
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
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
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%
Percent dissociation:
For a weak acid % dissociation
increases as the acid becomes more
dilute.
 For solution of any weak acid, [ H+]
decreases as [HA]0 decreases, but
% dissociation increases as [HA]0
increases.

Problems
P.11.Calculate the % dissociation of
acetic acid (Ka= 1.8 x 10-5 ) 1.00 M
and 0.100 M solutions.
P.12.In a 0.100 M HC3H5O3 aqueous
solution, lactic acid is 3.7 %
dissociated. Calculate ka for the acid.
[HA]dissociated
Percent HA dissociation =
x 100
[HA]initial
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)
HPO4
2-(aq)
+ H3
O+(aq)
Ka2 =
[H3O+][HPO42-]
[H2PO4-]
= 6.3x10-8
HPO4
2-(aq)
+ H2O(l)
PO4
3-(aq)
+ H3
O+(aq)
Ka1 > Ka2 > Ka3
Ka3 =
[H3O+][PO43-]
[HPO42-]
= 4.2x10-13
ACID STRENGTH
Polyprotic Acids
They can furnish more than one proton
(H+) to the solution.
 Characteristics: 1. Ka values are much
smaller than the first value , so only the
first dissociation step makes a significant
contribution to equilibrium concentration
of H+ .
 2. For sulfuric acid, it behaves as a strong
acid in the first dissociation step and a
weak acid in the second step, where its
contribution of H+ ions can be neglected.
Use quadratic equation to solve such kind
of a problem.

Problems
P.13.Calculate the pH of a 5.0 M
H3PO4 solution and the equilibrium
concentration of the species H3PO4 ,
H2PO4-, HPO42-, PO43P.14.Calculate the pH of 1.0 M sulfuric
acid solution.
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
SAMPLE PROBLEM 18.8:
Calculating Equilibrium Concentrations for a
Polyprotic Acid
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
Bases
“Strong” and “weak” are used in the same
sense for bases as for acids.
 strong = complete dissociation (hydroxide
ion supplied to solution)
 NaOH(s)  Na+(aq) + OH(aq)
 weak = very little dissociation (or reaction
with water)
 H3CNH2(aq) + H2O(l)  H3CNH3+(aq)
+ OH(aq)

[B]
BASE STRENGTH
Kb =
[BH+][OH-]
Problems
P.15.Calculate pH of 5.0 x 10-2 M
NaOH solution.
P.16.Calculate the pH for a 1.50 M
solution of ammonia
(Kb=1.8 x 10-5 )
Figure 18.11 Abstraction of a proton from water by methylamine.
Lone pair
binds H+
+
CH3NH2
H 2O
Methylamine
Base in water
+
CH3NH3+
methylammonium ion
OH-
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
SAMPLE PROBLEM 18.9:
Determining pH from Kb and Initial [B]
continued
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
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
SAMPLE PROBLEM 18.10: Determining the pH of a Solution of Acontinued
[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
Molecular Properties and Acid
strength:
 Trends
in acid strength of Nonmetal
Hydrides:
 Across a period nonmetal hydride
acid strength increases.
 Down a group nonmetal hydride acid
strength increases.
Bond strength decreases,
acidity increases
Figure 18.12
The effect of atomic and molecular properties on
nonmetal hydride acidity.
6A(16)
7A(17)
H 2O
HF
H 2S
HCl
H2Se
HBr
H2Te
HI
Electronegativity increases,
acidity increases
Trends in Acid strength of
Oxoacids
1. For oxoacids with same number of
O around E , acid strength increases
with electronegativity of E.
2. For oxoacids with different number
of O around E, acid strength
increases with number of O atoms.
Figure 18.13
H
O


The relative strengths of oxoacids.
I
>
H
O


Br
>
H
O


O
H
O


Cl
<<
H

O

Cl
O
O
Cl
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
Figure 18.13
The acidic behavior of the hydrated Al3+ ion.
Electron density
drawn toward Al3+
Nearby H2O acts
as base
H 3 O+
H 2O
Al(H2O)63+
Al(H2O)5OH2+
Acid-Base properties of Salt:
 Salt
is an ionic compound.
 Ka xKb=Kw
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.
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.
Molecules as Lewis Acids
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
Figure 18.15
The Mg2+ ion as a Lewis acid in the
chlorophyll molecule.
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
BCl4-
acceptor
acceptor
(c) K+ + 6H2O
K(H2O)6+
donor