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Homework Problems
Chapter 15 Homework Problems: 4, 6, 8, 24, 34, 36, 38, 39, 42, 52,
56a, 62, 68, 70, 74, 80a, 89, 93, 100, 103a, 116, 117, 124
CHAPTER 15
Acids and Bases
Acids and Bases (Arrhenius)
There are several definitions we can use for acids and bases:
Arrhenius definition
acid - A substance which, when added to water, forms H+ ion
base - A substance which, when added to water, forms OH- ion
Examples:
HCl(aq)  H+(aq) + Cl-(aq)
HF(aq)  H+(aq) + F-(aq)
NaOH(s)  Na+(aq) + OH-(aq)
NH3(aq) + H2O()  NH4+(aq) + OH-(aq)
The advantages of the Arrhenius definition are that it is simple
and easy to implement. The disadvantages are that it is tied in to a
particular solvent (water) and is not a general definition.
Acids and Bases (Bronsted-Lowry)
Bronsted-Lowry definition
acid - a proton (H+) donor; forms a conjugate base
base - a proton (H+) acceptor; forms a conjugate acid
H+
HF(aq) + H2O()  H3O+(aq) + F-(aq)
acid
base
conjugate acid
conjugate base
In the Bronsted theory, in an acid-base reaction an acid donates a
proton to form a conjugate base, while a base accepts a proton to form a
conjugate acid.
In addition, in Bronsted theory acids form hydronium ion (H3O+
ion) instead of hydrogen ion (H+ ion) when added to water.
Examples of Acids
HCl(aq) + H2O()  H3O+(aq) + Cl-(aq)
acid:
HCl
base: H2O
conjugate base: Cl-
conjugate acid: H3O+
CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq)
acid:
CH3COOH
base: H2O
conjugate base: CH3COOconjugate acid: H3O+
The second reaction goes in both directions, so has an equilibrium constant, the acid ionization constant
KC = Ka = [H3O+] [CH3COO-]
[CH3COOH]
Examples of Bases
NaOH(s)  Na+(aq) + OH-(aq)
Considered an ionization reaction in Bronsted theory, not an acidbase reaction.
NH3(aq) + H2O()  NH4+(aq) + OH-(aq)
base: NH3
conjugate acid: NH4+
acid:
conjugate base: OH-
H2O
The reaction goes in both directions, so has an equilibrium constant, the base ionization constant
KC = Kb = [NH4+] [OH-]
[NH3]
Amphoteric Substances
Some substances can act as either Bronsted acids or Bronsted
bases. Substances that can act as either acids or bases depending on
what they are reacting with are called amphoteric. For example:
H2O (water)
(acid) NH3(aq) + H2O()  NH4+(aq) + OH-(aq)
(base) CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq)
HCO3- (hydrogen carbonate ion)
(acid) NH3(aq) + HCO3-(aq)  NH4+(aq) + CO32-(aq)
(base) CH3COOH(aq) + HCO3-(aq)  H2CO3(aq) + CH3COO-(aq)
We can picture the reaction of a Bronsted acid with a Bronsted
base in several ways.
Given any substance, the conjugate base for the substance is
formed by removing a proton (H+), and the conjugate acid is formed by
adding a proton (H+).
Example: H2O
The conjugate base of H2O is OHThe conjugate acid of H2O is H3O+
Example: HSO4The conjugate base of HSO4- is SO42The conjugate acid of HSO4- is H2SO4
Example
Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a
weak base. Indicate the behavior of these two substances when added to
water, according to Bronsted theory.
Example
Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a
weak base. Indicate the behavior of these two substances when added to
water, according to Bronsted theory.
HClO2(aq) + H2O()  H3O+(aq) + ClO2-(aq)
C5H5N(aq) + H2O()  C5H5NH+(aq) + OH-(aq)
Strong and Weak Acids
There are seven common strong acids:
Binary strong acids
Ternary strong acids
HCl
HClO3
HNO3
HBr
HClO4
H2SO4 (1st proton)
HI
Sulfuric acid is special in that it is a strong acid with respect to
the first proton and a weak acid with respect to the second proton.
H2SO4(aq) + H2O()  HSO4-(aq) + H3O+(aq)
HSO4-(aq) + H2O()  SO42-(aq) + H3O+(aq)
Ka2 = [SO42-] [H3O+] = 1.2 x 10-2
[HSO4-]
All other acids are weak acids.
HNO2(aq) + H2O()  H3O+(aq) + NO2-(aq)
Ka = [H3O+] [NO2-] = 4.5 x 10-4
[HNO2]
CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq)
Ka = [H3O+] [CH3COO-] = 1.8 x 10-5
[CH3COOH]
The larger the value of Ka, the stronger the acid. Therefore,
HNO2 is a stronger acid than CH3COOH.
Strong and Weak Bases
There are seven common strong soluble bases:
Group 1A strong bases
Group 2A strong bases
LiOH
Sr(OH)2
NaOH
Ba(OH)2
KOH
RbOH
CsOH
All other metal hydroxides are insoluble bases. They do not
dissolve in water to an appreciable extent, but react as bases in acid-base
reactions. Examples: Cu(OH)2, Al(OH)3, Pb(OH)2.
Cu(OH)2(s) + 2 HCl(aq)  Cu2+(aq) + 2 Cl-(aq) + 2 H2O()
Weak bases establish an equilibrium in water.
NH3(aq) + H2O()  NH4+(aq) + OH-(aq)
Kb = [NH4+] [OH-] = 1.8 x 10-5
[NH3]
N2H4(aq) + H2O()  N2H5+(aq) + OH-(aq)
Kb = [N2H5+] [OH-] = 8.9 x 10-7
[N2H4]
Since Kb is larger for NH3 than for N2H4, NH3 is a stronger base
than N2H4.
Autoionization of Water
In pure water there will be a small number of H3O+ and OH- ions.
This is due to the autoionization reaction:
H2O() + H2O()  H3O+(aq) + OH-(aq)
Kw = [H3O+] [OH-] = 1.0 x 10-14 (at T = 25 C)
Because the above reaction is endothermic (Hrxn = + 55.8
kJ/mol) the value for Kw increases as T increases.
Equilibrium in Pure Water
Since H2O() + H2O()  H3O+(aq) + OH-(aq)
Kw = [H3O+] [OH-] = 1.0 x 10-14 (at T = 25 C)
Substance
Initial
Change
Equilibrium
H3O+
0.0
x
x
OH-
0.0
x
x
So (x) (x) = x2 = 1.0 x 10-14
x = 1.0 x 10-7
So in pure water at equilibrium at T = 25 C, [H3O+] = [OH-] =
1.0 x 10-7 M.
Acidic, Basic, and Neutral Solutions
We can use the above results to define what we mean by an
acidic, basic, and neutral solution, using [H3O+] [OH-] = 1.0 x 10-14.
acidic solution
[H3O+] > [OH-]
[H3O+] > 1.0 x 10-7 M
[OH-] < 1.0 x 10-7 M
neutral solution
[H3O+] = [OH-]
[H3O+] = 1.0 x 10-7 M
[OH-] = 1.0 x 10-7 M
basic solution
[H3O+] < [OH-]
[H3O+] < 1.0 x 10-7 M
[OH-] > 1.0 x 10-7 M
pH
pH represents a convenient way of representing the concentration
of hydronium ion in an aqueous solution. pH is defined as follows:
pH = - log10[H+] = - log10[H3O+]
For a neutral solution at T = 25 C
pH = - log10[H3O+] = - log10(1.0 x 10-7) = 7.00
Note that the number of digits to the right of the decimal point is
equal to the number of significant figures in the H3O+ concentration.
For acidic and basic solutions
acidic solution
[H3O+] > 1.0 x 10-7 M means pH < 7.00
basic solution
[H3O+] < 1.0 x 10-7 M means pH > 7.00
The further away the pH is from 7.00 the more acidic (if less than
7.00) or basic (if greater than 7.00) the solution.
pOH and pK
We can use the “p” notation as a general symbol to indicate that
we are taking - log10 of something. In particular
pH = - log10[H3O+]
pOH = - log10[OH-]
pK = - log10 K
By reversing the above definitions we get the following
relationships
[H3O+] = 10-pH
[OH-] = 10-pOH
K = 10-pK
Relationship Between pH and pOH
There is a simple relationship between pH and pOH.
[H3O+] [OH-] = 1.0 x 10-14
-log10[H3O+] + ( - log10[OH-]) = - log10(1.0 x 10-14)
pH + pOH = 14.00 (at 25 C)
The more general relationship, true at all temperatures, is
pH + pOH = pKw
Relationship Between pH and Concentration
If we know the concentration of hydronium ions in solution we
can find the pH of the solution (and vice versa). We can also find pOH
and the concentration of hydroxide ions in solution.
Example: A particular solution has pH = 4.62 at T = 25. C.
What are the concentrations of H3O+ and OH- in the solution?
Relationship Between pH and Concentration
Example: A particular solution has pH = 4.62 at T = 25. C.
What are the concentrations of H3O+ and OH- in the solution?
[H3O+] = 10-pH = 10-4.62 = 2.4 x 10-5 M
[H3O+] [OH-] = 1.0 x 10-14, so
[OH-] = 1.0 x 10-14 = 1.0 x 10-14 = 4.2 x 10-10 M
[H3O+]
2.4 x 10-5
We could also find the OH- concentration as follows:
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 4.62 = 9.38
[OH-] = 10-pOH = 10-9.38 = 4.2 x 10-10 M
pH for Solutions of Strong Acids or Strong Bases
Because strong acids and strong soluble bases are strong
electrolytes, and so completely dissociate in solution, finding the value
for pH is relatively simple. We may use the following procedure:
1) Use the information in the problem to find the concentration of
H3O+ (strong acid) or OH- (strong base).
2) Find the pH
a) For strong acids, find the pH directly
b) For strong bases, find the pOH, then use (at T = 25 C)
pH + pOH = 14.00 ; pH = 14.00 - pOH
to find the pH
Example: Find the pH for the following solutions, at T = 25 C
a) A 3.5 x 10-3 M solution of HBr, a strong acid
b) A solution formed by dissolving 0.200 moles of Sr(OH)2, a
strong soluble base, in water, to form a solution with a final volume of
500.0 mL.
a) A 3.5 x 10-3 M solution of HBr, a strong acid
Reaction is
HBr(aq) + H2O()  H3O+(aq) + Br-(aq)
[H3O+] = 3.5 x 10-3 mol HBr
L soln
pH = - log10(3.5 x 10-3) = 2.46
1 mol H3O+ = 3.5 x 10-3 M
1 mol HBr
b) A solution formed by dissolving 0.200 moles of Sr(OH)2, a
strong soluble base, in water, to form a solution with a final volume of
500.0 mL.
Reaction is
Sr(OH)2(s)  Sr2+(aq) + 2 OH-(aq)
[OH-] = 0.200 mol Sr(OH)2
2 mol OH- = 0.800 M
0.5000 L soln
1 mol Sr(OH)2
pOH = - log10(0.800) = 0.097
pH = 14.00 - pOH = 14.00 - 0.097 = 13.90
Weak Acids or Weak Bases
For problems involving solutions containing a single weak acid
or weak base we proceed as we do other equilibrium problems.
1) Start with the following information
Balanced chemical reaction
Initial conditions
Value for Ka (weak acid) or Kb (weak base)
2) Set up the problem using the “ICE” method
3) Find [H3O+] (weak acid) or [OH-] (weak base)
4) Find pH (for weak base, first find pOH, then use pH + pOH =
14.00 to find pH)
Example: Find the pH of a 0.100 M solution of HNO2, a weak
acid (Ka = 4.5 x 10-4), at T = 25 C
Find the pH of a 0.100 M solution of HNO2, a weak acid (Ka =
4.5 x 10-4), at T = 25 C
HNO2(aq) + H2O()  H3O+(aq) + NO2-(aq)
Ka = [H3O+] [NO2-] = 4.5 x 10-4
[HNO2]
Substance
Initial
Change
Equilibrium
H3O+
0.00
x
x
NO2-
0.00
x
x
HNO2
0.100
-x
0.100 - x
(x) (x)
(0.100 - x)
= 4.5 x 10-4
There are two ways to proceed...
1) Assume x is small (in this case, x << 0.100)
(x) (x)

x2
= 4.5 x 10-4
(0.100 - x)
0.100
so x2 = (0.100)(4.5 x 10-4) = 4.5 x 10-5
x = (4.5 x 10-5)1/2 = 6.7 x 10-3
[H3O+] = 6.7 x 10-3 M ; pH = - log10(6.7 x 10-3) = 2.17
Is 6.7 x 10-3 << 0.100? Yes (at least 10 times smaller).
2) Solve the quadratic
x2 = (0.100 - x)(4.5 x 10-4) = (4.5 x 10-5) - (4.5 x 10-4) x
x2 + (4.5 x 10-4) x - (4.5 x 10-5) = 0
x = 6.5 x 10-3 ; - 6.9 x 10-3
[H3O+] = 6.5 x 10-3 M ; pH = - log10(6.5 x 10-3) = 2.19
Percent Dissociation
The percent dissociation of a weak acid is defined as
% dissociation = amount dissociated x 100%
initial amount
For strong acids the percent dissociation is ~ 100%.
For weak acids we can use the equilibrium concentrations to find
the percent dissociation.
For the above example,
% dissociation = 6.7 x 10-3
0.100
initial HNO2 = 0.100 M
amount dissociated = 6.7 x 10-3 M
x 100% = 6.7 %
pH Calculations involving Weak Bases
Just as we can find concentrations and pH values for solutions of
weak acids, we can do the same for solutions of weak bases.
Example: Find the pH of a 0.100 M solution of NH3, a weak base
(Kb = 1.8 x 10-5), at T = 25 C
Find the pH of a 0.100 M solution of NH3, a weak base (Kb = 1.8
x 10-5), at T = 25 C
NH3(aq) + H2O()  NH4+(aq) + OH-(aq)
Kb = [NH4+] [OH-] = 1.8 x 10-5
[NH3]
Substance
Initial
Change
Equilibrium
NH4+
0.00
x
x
OH-
0.00
x
x
NH3
0.100
-x
0.100 - x
(x) (x)
(0.100 - x)
= 1.8 x 10-5
Assume x is small (in this case, x << 0.100)
(x) (x)  x2 = 1.8 x 10-5
(0.100 - x)
0.100
so x2 = (0.100)(1.8 x 10-5) = 1.8 x 10-6
x = (1.8 x 10-6)1/2 = 1.3 x 10-3
[OH-] = 1.3 x 10-3 M ; pOH = - log10(1.3 x 10-3) = 2.87
pH = 14.00 - 2.87 = 11.13
Is 1.3 x 10-3 << 0.100? Yes (at least ten times smaller)
If we solve using the quadratic formula, we get pH = 11.12
Polyprotic Acid
A polyprotic acid has two or more ionizable protons that can be
donated in an acid-base reaction.
Monoprotic (one ionizable proton)
HCl, HNO2, CH3COOH
Diprotic (two ionizable protons)
H2CO3, H2SO4
Triprotic (three ionizable protons)
H3PO3
For polyprotic acids one can talk about the acid dissociation
constant for each proton.
Example: H2CO3
1st proton
H2CO3(aq) + H2O()  H3O+(aq) + HCO3-(aq)
Ka1 = [H3O+][HCO3-] = 4.3 x 10-7
[H2CO3]
2nd proton
HCO3-(aq) + H2O()  H3O+(aq) + CO32-(aq)
Ka2 = [H3O+][CO32-] = 5.6 x 10-11
[HCO3-]
For a polyprotic acid Ka1 > Ka2 > Ka3...
Calculations Involving Polyprotic Acids
It would seem like calculations with polyprotic acids should be
complicated, since there are several sources of H3O+ ions.
However, there is usually a big enough difference in the values of
the Kas (acid dissociation constants) that only the first dissociation needs
to be considered for polyprotic acid solutions.
Procedure (diprotic acid):
1) Calculate equilibrium concentrations using the first ionization
constant.
2) Calculate equilibrium concentrations using the second
ionization constant, using the results from the first calculation for the
initial conditions.
3) If there are any significant changes in concentrations involving
the first ionization, go back and recalculate concentrations using the
previous results for the initial conditions.
Relationship Between Ka and Kb
We may find a general relationship between the values for Ka and
Kb for a Bronsted acid/conjugate base pair of substances. We proceed as
follows:
Let HA be a weak monoprotic acid. A- is the conjugate base.
HA(aq) + H2O()  H3O+(aq) + A-(aq) Ka = [H3O+] [A-]
[HA]
A-(aq) + H2O()  HA(aq) + OH-(aq)
Kb = [HA] [OH-]
[A-]
Ka Kb = [H3O+] [A-] [HA] [OH-] = [H3O+] [OH-] = Kw
[HA]
[A-]
Ka Kb = Kw
pKa + pKb = pKw
Ka Kb = 1.0 x 10-14
pKa + pKb = 14.00, at T = 25. C
Example: The value for the acid ionization constant for acetic
acid (CH3COOH) is Ka = 1.8 x 10-5 at T = 25. C. What is the value for
Kb for the acetate ion (CH3COO-)?
Example: The value for the acid ionization constant for acetic
acid (CH3COOH) is Ka = 1.8 x 10-5 at T = 25. C. What is the value for
Kb for the acetate ion (CH3COO-)?
Ka Kb = Kw = 1.0 x 10-14 for an acid/conjugate base pair
Kb = Kw = (1.0 x 10-14) = 5.6 x 10-10
Ka
(1.8 x 10-5)
Note that we could prepare a solution that initially only contained
the conjugate base if we added a salt such as NaCH3COO, KCH3COO,
etc.
NaCH3COO(aq)  Na+(aq) + CH3COO-(aq)
General Statements About Acid and Base Strength
We may use the above results to make the following general
statement concerning the relative strengths of acids and their conjugate
bases. This is based on the relationship Ka Kb = Kw = 1.0 x 10-14.
The stronger the acid the weaker the conjugate base.
Example: Which is a stronger base, F- or CN-?
Ka(HF) = 3.5 x 10-4
Ka(HCN) = 4.9 x 10-10
Since HF is a stronger acid than HCN, F- is a weaker base than
CN-.
Kb(F-) = (1.0 x 10-14)/(3.5 x 10-4) = 2.9 x 10-11
Kb(CN-) = (1.0 x 10-14)/(4.9 x 10-10) = 2.0 x 10-5
Acid-Base Properties of Salts
Recall that a salt is an ionic compound formed from the reaction
of an acid with a base. We have the following four possibilities:
1) Salt of a strong acid and a strong base.
Example: HCl + NaOH  NaCl + H2O
(Na+ and Cl-)
No acid-base properties for the salt. Solutions will be neutral.
2) Salt of a strong acid and a weak base.
Example: HCl + NH3  NH4Cl
(NH4+ and Cl-)
NH4+ will act as a weak acid. Solutions will be acidic.
3) Salt of a weak acid and a strong base.
Example: HF + KOH  KF + H2O
(K+ and F-)
F- will act as a weak base. Solutions will be basic.
4) Salt of a weak acid and a weak base.
Example: HF + NH3  NH4F
(NH4+ and F-)
NH4+ will act as a weak acid. F- will act as a weak base.
Solutions will be approximately neutral.
Ka(HF) = 3.5 x 10-4
so Kb(F-) = 2.9 x 10-11
Kb(NH3) = 1.8 x 10-5
so Ka(NH4+) = 5.6 x 10-10
Since NH4+ is a stronger acid than F- is a base, the solution will
be slightly acidic.
Acid-Base Calculations for Salts
We do acid-base problems for salts the same way as we do other
weak acid or weak base problems.
Example: What is the pH of a 0.100 M solution of ammonium
chloride (NH4Cl), the salt of a strong acid and a weak base. Kb(NH3) =
1.8 x 10-5. Assume T = 25 C
Example: What is the pH of a 0.100 M solution of ammonium
chloride (NH4Cl), the salt of a strong acid and a weak base. Kb(NH3) =
1.8 x 10-5. Assume T = 25 C
NH4Cl(s)  NH4+(aq) + Cl-(aq)
NH4+(aq) + H2O()  H3O+(aq) + NH3(aq)
Ka Kb = 1.0 x 10-14, so Ka = 1.0 x 10-14 = (1.0 x 10-14) = 5.6 x 10-10
Kb
(1.8 x 10-5)
Ka = [H3O+] [NH3] = 5.6 x 10-10
[NH4+]
Substance
Initial
Change
Equilibrium
H3O+
0.00
x
x
NH3
0.00
x
x
NH4+
0.100
-x
0.100 - x
Ka = [H3O+] [NH3] =
(x) (x)
= 5.6 x 10-10
[NH4+]
(0.100 - x)
Assume x << 0.100. Then
x2
= 5.6 x 10-10
(0.100)
x2 = (0.100) (5.6 x 10-10) = 5.6 x 10-11
x = (5.6 x 10-11)1/2 = 7.5 x 10-6
pH = - log10(7.5 x 10-6) = 5.13
Note that 7.5 x 10-6 << 0.100, so x is in fact small.
Acid-Base Properties of Cations
Metal cations can act as weak acids in solution. There are several
ways in which these reactions can be written
Example: Al3+ ion
Al3+(aq) + H2O()  H+(aq) + Al(OH)2+(aq)
Al(H2O)63+(aq) + H2O()  H3O+(aq) + Al(H2O)5(OH)2+
Generally speaking, metal
cations act as weak acids when they
are small and have multiple positive
charges. Common examples of metal
cations acting as weak acids are Co2+,
Ni2+, Zn2+, Fe2+, Fe3+, and Al3+ ions.
Factors Affecting Acid Strength
We can often predict the relative strengths of weak acids by
focussing on the factors that control acid strength.
Binary acids
1) In the same column (group). Acid strength increases from top
to bottom.
Reason: The H - X bond strength decreases from top to bottom,
so it is easier to break the bond and form H+ (or H3O+).
Example: Group 7 binary acids.
HF
567 kJ/mol
Ka = 3.5 x 10-4
HCl
431 kJ/mol
“strong acid” (Ka ~ 107)
HBr
366 kJ/mol
“strong acid” (Ka ~ 109)
HI
299 kJ/mol
“strong acid” (Ka ~ 1011)
2) In the same row. Acid strength increases from left to right.
Reason: The electronegativity of the nonmetal increases from left
to right, making the conjugate base more stable.
Example: Second row.
acid
conj. base
nonmetal EN
CH4
CH3-
EN(C) = 2.5
NH3
NH2-
EN(N) = 3.0 “weak base” (Ka ~ 10-33)
H2O
OH-
EN(O) = 3.5 “amphoteric” (Ka = 1.0 x 10-14)
HF
F-
EN(F) = 4.0
insoluble in water (Ka ~ 10-49)
“weak acid” (Ka = 3.5 x 10-4)
Ternary acids
1) Same nonmetal, different number of oxygens. The more oxygens, the stronger the acid.
Reason: The more oxygens in the oxyacid the more stable the
conjugate base, and so the easier it will form.
acid
conj. base
HClO
ClO-
Ka = 3.5 x 10-8
HClO2
ClO2-
Ka = 1.1 x 10-2
HClO3
ClO3-
“strong acid” (Ka ~ 5 x 102)
HClO4
ClO4-
“strong acid” (Ka = 1 x 103)
2) Same number of oxygens, different nonmetal in the same
column (group). Acid strength increases from bottom to top.
Reason: The electronegativity of the nonmetal increases from
bottom to top, making the conjugate base more stable, and so easier to
form.
acid
conj. base
nonmetal EN
HClO
ClO-
EN(Cl) = 3.0
Ka = 3.5 x 10-8
HBrO
BrO-
EN(Br) = 2.8
Ka = 2.0 x 10-9
HIO
IO-
EN(I) = 2.5
Ka = 2.3 x 10-11
Lewis Acids and Bases
There is a third way in which acids and bases are defined, developed by G. N. Lewis.
Lewis definition
acid - an electron pair acceptor
base - an electron pair donor
Example:
BF3(g) + NH3(g)  BF3NH3(s)
In the above reaction BF3 is an electron pair acceptor (Lewis
acid) and NH3 is an electron pair donor (Lewis base). Note that the
above reaction would not be an acid-base reaction under the Bronsted
definition.
Also note that in the above reaction a coordinate covalent bond is
formed (a bond where both electrons in the covalent bond come from the
same atom).
Identifying Lewis Acids and Bases
To identify the Lewis acid and Lewis base in a reaction one needs
only to identify the species that is accepting an electron pair (Lewis acid)
and donating an electron pair (Lewis base).
Example:
CO2 + OH-  HCO3-
CO2 accepts an electron pair and so is a Lewis acid; OH- donates
an electron pair and so is a Lewis base.
End of Chapter 15
“Things should be made as simple as possible, but not any
simpler.” – Albert Einstein
“Never express yourself more clearly than you think.” – Neils
Bohr