Chemistry
The Science in Context
Chapter 16
Equilibrium in the Aqueous Phase
HCl(aq) reacting with NH3(aq)
The color of hydrangea flowers depends on the acid content of the soil
Acid rain forms when volatile, nonmetal
oxides react with water vapor.
SO3 + H2O ↔ H2SO4
Acid Rain Tutorial
»PC version
This tutorial explores the effects of fossil fuel burning
on the pH of rainwater, as well as the resulting
environmental and industrial consequences. Includes
practice exercises.
Strong Acid
HNO3
Weak Acid
HNO2
Acids…A molecular view.
Acids (proton donors) react with bases
(proton acceptors) forming a conjugate acid
(H3O+) and conjugate base (Cl-). Note that
the conjugate base of a strong acid like HCl is
a weak base.
Water molecules in acid
solutions cluster around
the Hydronium ion.
These “species” have the
formula:
H(H2O)n+
Autoionization of Water
Water molecules have the ability to ionize each other.
Keq = [H3O+][OH-] = 1.0E-14 (25°C)
This important equilibrium constant is
usually denoted Kw
Ammonia is a weak base in water, Kb = 1.8E-5
conj. acid
conj. base
Trends in acid strength
relative to the strength
of their conjugate
bases.
In water the strongest
base is OH-; stronger
bases will ionize water
to produce hydroxide
ions.
Problem
Benzoic acid is used as a preservative in
foods. Calculate the concentration of H+
ions at equilibrium in a 0.100 M solution of
Benzoic acid. Ka = 6.5E-5
In Solution, to what degree is benzoic acid
ionized?
The pH of a solution is defined
as the negative logarithm of
the hydronium ion
concentration:
pH = -log([H+])
Note that Ka and Kb values are
frequently reported as pKa and
pKb. This avoids writing the
values as exponentials.
16.30. Calculate the pH of a 0.00500 M
solution of HNO3.
Answer: 2.301 (4 S.F.)
Auto-ionization and pH
H2O ↔ H+ + OHKw = 1.0E-14; pKw = 14.00
at 298K
In pure water,
[H+] = [OH-]
So
[H+]2 = 1.0E-14
Thus
[H+] = 1.0E-7
Or
pH = 7.0
16.31. Calculate the pH and pOH of a
0.0450 M solution of NaOH.
Answer: pOH=1.347;
pH=14.000 – 1.347 = 12.65
Problem: A solution of HF has a pH=2.30.
Calculate the equilibrium concentration of all
species present in this solution, and the
original concentration of the HF (i.e. before
dissociation). pKa(HF) = 3.14
For the reaction:
HA ↔ H+ + A-
The concentration of H+ is a function of the
strength of the H-A bond
Acid Strength and Molecular Structure
Sulfuric acid is a stronger acid than sulfurous
acid due to the decrease in electron density
on the O-H bond.
The oxyacids
of chlorine
increase in
strength (Ka)
with
increasing
numbers of
oxygen atoms
bound to the
central
chlorine atom.
Blue color indicates increasing
positive charge on the proton
HClO
HBrO
HIO
rH-O= 0.961Å
rH-O= 0.957Å
rH-O= 0.955Å
Increased H-O bond distance is due to decreased electron
density
Problem
The pH of a 0.10M solution of chloroacetic acid is
found to be 1.95. Calculate Ka for this acid and
compare it to Ka for acetic acid.
Polyprotic Acid Ionization
Table 16.1. Ionization of Diprotic Acids
Acid
Formula
Ka1
Ka2
Carbonic
H2CO3
4.3E-7
4.7E-11
Sulfurous
H2SO3
4.3E-3
6.2E-8
Sulfuric
H2SO4
>>1
1.2E-2
The H+ concentration due to the second
dissociation is generally insignificant, i.e.
compared with the first dissociation.
16.59.
What is the pH of a 0.300 M solution
of H2SO4 (Ka2 = 1.2  10–2)?
Problem
Methylamine is a weak base (Kb=4.4E-4).
Calculate the OH- concentration in a 0.200M
aqueous solution of CH3NH2.
What is the pH of this solution?
Acid and Base Ionization Tutorial
»PC version
This tutorial explores the differences among BrønstedLowry acids, Brønsted-Lowry bases, Lewis acids and
Lewis Bases. Includes practice exercises.
Acid Strength and Molecular Structure Tutorial
»PC version
Learn to determine relative acid strength based on the
molecular and electronic structure of the acid. Includes
practice exercises.
pH Scale Tutorial
»PC version
This tutorial introduces the pH scale and uses interactive
graphs to explain the relationship between pH, pOH
[H3O+], and [OH-]. Includes practice exercises.
The Self-Ionization of Water Tutorial
»PC version
This tutorial illustrates the process by which water
molecules act as both a proton acceptor (base) and a proton
donor (acid), and explores the equilibrium constant (Kw) for
the self-ionization of water. Includes practice exercises.
Salts of weak acids and bases.
Many naturally
occurring
compounds used
as drugs act as
weak bases
(due to amine
groups).
For this reason
they are often
referred to as
alkaloids…they
produce alkaline
solution.
Problem 63. Which of the following salts
produce an acidic solution in water?
Ammonium acetate
NH4Cl
Sodium formate
Problem 63. Which of the following salts
produce an basic solution in water?
NaF
KCl
Sodium bicarbonate
Problem 66
Codeine is a widely-prescribed pain killer
because it is much less addictive than morphine
(which is much less addictive than heroin).
Codeine contains a basic nitrogen atom that
can be protonated to form the conjugate acid .
Calculate the pH of a 3.97E-4 M solution of
codeine if the pKa of the conjugate acid is 8.21.
Problem. For a 6.75E-3 M solution of
sodium benzoate, determine the following:
Identify the equilibrium reaction that
determines the pH.
Calculate the pH. pKa(benzoic acid) = 4.20
Lewis Acids and Bases
• A Lewis Acid is a substance that accepts a
pair of electrons.
• A Lewis Base is a substance that donates a
pair of electrons.
A Lewis Acid/base adduct
Buffer Solutions are solutions that contain
significant amounts of both an acid and
it’s conjugate base.
For example the following solutions prepared
by adding equivalent amounts of the
acid/conj.base pairs:
CH3COOH/CH3COOH2PO4-/HPO42HCO3-/CO32The presence of both the acid and base means
that the pH will resist change went additional
acid or base are added to the solution.
Common Buffer Systems
Acid
Conj.base
pKa
H2PO4-
2.16
1-3
CH3CO2H CH3CO2 -
4.75
4-6
H2PO4-
HPO4 2-
7.21
6-8
HCO3 -
CO3 2-
10.33
9-11
H3PO4
pH range
Acididosis can be caused by extreme changes in diet as
well as chronic respiratory diseases
Problem 80 (pH buffer problem)
Determine the pH and pOH of 0.250 L of a
buffer containing 0.0200M boric acid and
0.0250M sodium borate.
The pKa for B(OH)3 = 9.00 at 25°C.
Buffers Tutorial
»PC version
Use the Henderson-Haselbach equation to predict the
pH of a buffer. The tutorial concludes with practice
exercises and an interactive titration experiment.
Alkalinity Titrations to determine total CO32-
Acid/Base Titrations
Strong acid with a strong base
Weak acid with a strong base
Problem 93. A 25.0 mL sample of 0.100 M
acetic acid is titrated with 0.125 M NaOH.
Calculate the pH of the of the reaction
solution after 10.0, 20.0, and 30.0 mL of
base have been added.
Acid/Base Titrations
Strong base with a strong acid
Weak base with a strong acid
Problem 100
In an alkalinity titration of 100.0 mL sample of
water from a hot spring, 2.56mL of 0.0355 M
HCl is needed to reach the first equivalence
point (pH=8.3) and another 10.42mL is needed
to reach the second equivalence point (pH=4).
If the alkalinity of the spring is due only to the
presence of carbonate and bicarbonate, what
are the concentrations of each of them?
Problem. Which of the following solutions
show buffer properties. Compute the pH of
each solution that is buffered
(a)0.100 L of 0.25 M NaCH3CO2 + 0.150 L of
0.25 M HCl
(b)0.100 L of 0.25 M NaCH3CO2 + 0.050 L of
0.25 M HCl
(c)0.100 L of 0.25 M NaCH3CO2 + 0.050 L of
0.25 M NaOH
Acid/base Indicators are weak organic acids that
change color when ionized.
The pKas of the Indicators determine the pH
range that they can be used in titrations
Strong Acid and Strong Base Titrations Tutorial
»PC version
This interactive virtual titration lab introduces the
titration apparatus and challenges you to determine the
concentration of an unknown acid from the volume of
basic solution added. Includes practice exercises.
Titrations of Weak Acids Tutorial
»PC version
Learn to read and understand the different stages of a
titration curve for a weak acid or polyprotic acid, and
understand what is happening at a molecular level.
Includes practice exercises.
16.9. Solubility of Minerals and other
Compounds
Minerals in contact with ground water will
dissolve to some extent.
CaCO3(s) ↔ Ca2+(aq) + CO32-(aq)
Write the equilibrium expression for this
dissolution.
Solubility Equilibria
CaCO3(s) ↔ Ca2+(aq) + CO32-(aq)
K = [Ca2+(aq)][CO32-(aq)] = Ksp
The equilibrium expression is called the
solubility product (sp), because it only
involves products of the concentrations of the
dissolved species and not the solid.
If Ksp is known, the solubility (at equilibrium)
of the solid can be calculated
Ksps of some common salts
HgS(s)
Fe(OH)3(s)
AgI(s)
Ksp = [Hg2+][S2-] = 4.0 x 10-53
Ksp = [Fe3+][OH-]3 = 2.8 x 10-39
Ksp = [Ag1+][I1-] = 8.5 x 10-17
CaCO3(s)
CaSO4(s)
Ag2SO3(s)
Ksp = [Ca2+][CO32-] = 9.8 x 10-9
Ksp = [Ca2+][SO42-] = 4.9 x 10-5
Ksp = [Ag1+]2[SO32-] = 1.2 x 10-5
NaCl(s)
Ksp = [Na1+][Cl1-] = 6.2
Solubility Problem
2.75 grams of BaF2 is placed in enough water
to make 1.00 L at 25°C. After equilibrium
has been established…the F- concentration
equal 0.0150 M, what is the Ksp for BaF2.
Solubility Problem
50 mg of PbSO4 is placed in 250 mL of pure
water; What percentage of the solid
dissolves?
Solubility Problem 116
Calculate the pH of a saturated solution of
zinc hydroxide, Ksp = 4.0E-17
Solubility Problem 120
Calculate the solubility of silver chloride in
seawater with a chloride concentration of
0.547 M. Ksp(AgCl) = 1.8E-10
16.10. Complex Ions
Dissolved metal ions are Lewis Acids, and form
complexes with Lewis Bases
Metal ions as Lewis Acid also promote hydrolysis
of water, and the formation of H3O+
Metal cations (Fe3+, Cr3+ and Al3+)with large
positive charges are more likely to cause
hydrolysis.
Formation Reactions
The equilibrium constants associated with
complexation are called formation constants
Kf = [Cu(NH3)42+]/[Cu2+][NH3]4
= 5.0E+13
Chlorophylls such as Chl a,
and Chl b absorb visible
light in a process that
creates an electrical
potential that drives
phosphorylation
Heme Group of
Hemoglobin
If a single drop of water containing 0.02
mmol of HCl is added to 1.0 L of a 10-5 M
solution of AgNO3, will a precipitate form?
AgCl(s)
Ag+(aq) + Cl-(aq)
A) Yes
B) No
Ksp = 1.8×10-10
C) Can’t tell
CQ16-10.3a-Adding a Drop of HCl to a AgNO Solution
Consider the following arguments for each answer and
vote again:
A. After only 1 drop of HCl is added, the Cl-(aq)
concentration will be 2×10-5 M, which is high enough
to induce the precipitation of AgCl(s).
B. Far more than 1 drop of HCl is required to raise the Clconcentration to the point where Cl-(aq) and Ag+(aq)
are in equilibrium with AgCl(s).
C. It is not clear whether the ΔG° of the drop increases or
decreases when it enters the solution.
CQ16-10.3b-Adding a Drop of HCl to a AgNO Solution
Given that Ksp(AgCl) > Ksp(AgBr),
which of the following salts, when
added in excess to an aqueous 0.1 M
AgNO3 solution, will result in the
lowest concentration of Ag+(aq)?
A) AgNO3
B) NaCl
CQ16-10.4a-Concentration of Ag+ In Ionic Solutions
C) AgBr
Consider the following arguments for each answer
and vote again:
A. Because of the common-ion effect, the addition of
AgNO3(s) will cause a net decrease in the
concentration of Ag+(aq).
B. Adding NaCl will induce the precipitation of AgCl(s)
from the solution, thus lowering the Ag+(aq)
concentration.
C. AgBr(s) is less soluble than AgCl(s), and so its
addition will cause the greatest decrease in the
Ag+(aq) concentration.
CQ16-10.4b-Concentration of Ag+ In Ionic Solutions
Suppose water is slowly added to a
vessel containing a speck of the
sparingly soluble salt BaSO4(s).
Which of the following plots shows
the equilibrium concentration of
Ba2+(aq) in the resulting solution
versus the amount of water added?
A)
B)
CQ16-10.5a-Dissolution of a Speck of BaSO in H O
C)
Consider the following arguments for each answer
and vote again:
A. As water is added and more BaSO4(s) is dissolved,
the concentration of Ba2+(aq) will increase until the
solution becomes saturated.
B. The concentration of Ba2+(aq) will increase until all
the BaSO4(s) has dissolved, after which additional
water will decrease the Ba2+(aq) concentration.
C. Until the BaSO4(s) has completely dissolved, the
concentration of Ba2+(aq) will remain constant.
CQ16-10.5b-Dissolution of a Speck of BaSO in H O
The conductivity of an aqueous solution
is directly proportional to the
concentration of the ions present.
Given this fact, which of the following
plots shows the conductivity of a NaCl
solution as a function of the amount of
AgNO3(s) added?
A)
B)
CQ16-10.6a-Conductivity of a NaCl + AgNO Solution
C)
Consider the following arguments for each answer
and vote again:
A. Adding AgNO3(s) will increase the total ion
concentration, so the conductivity will increase until
the solution is saturated.
B. As AgNO3(s) is added, the conductivity of the
solution will decrease because of the precipitation of
AgCl(s) until all of the Cl-(aq) is consumed.
C. Although AgCl(s) will precipitate as AgNO3(s) is
added, the total concentration of ions will remain
constant until the Cl-(aq) is depleted.
CQ16-10.6b-Conductivity of a NaCl + AgNO Solution
To the left is a plot of the
autoionization constant, Kw,
versus temperature. What is
the pH of hot water?
A) < 7
B) 7
CQ16-11.5a-Dependence of pH on Temperature
C) > 7
Consider the following arguments for each answer
and vote again:
A. At higher temperatures, the concentrations of H3O+
and OH- increase. Therefore, the pH of hot water is
less than 7.
B. Regardless of temperature, the concentrations of H3O+
and OH- remain equal, so the pH remains 7, which is
neutral.
C. At higher temperatures, H+ ions acquire enough
kinetic energy to escape the solution, leaving a
predominance of OH- ions.
CQ16-11.5b-Dependence of pH on Temperature
Which of the following, when added to
an NH3(aq) solution, will form a basic
buffer?
A) NaOH
CQ16-11.6a-NH Buffer Solution
B) HCl
C) NaCl
Consider the following arguments for each answer
and vote again:
A. NH3, a weak base, is normally an acidic buffer, so to
create a basic buffer, one must add NaOH.
B. By adding HCl to the NH3 solution to form some
NH4+, the solution will become a basic buffer.
C. NH3 is already a weak base, so to create a basic buffer
solution, one need only add a neutral buffering salt
like NaCl.
CQ16-11.6b-NH Buffer Solution
To the left is a plot that shows
the pH of a weak acid as it is
titrated with 0.01 M NaOH.
Which of the following plots
would correspond to the same
titration if the same weak acid
were diluted with water and
then titrated with 0.01 M
NaOH?
A)
B)
CQ16-11.7a-Titration of a Diluted Weak Base
C)
Consider the following arguments for each answer
and vote again:
A. Diluting a weak acid with water will increase the
initial pH of the solution and decrease the final pH of
the solution.
B. The dilution would have little effect on the initial pH
of the weak acid, especially in the buffer region.
However, the pH after the equivalence point will be
lower.
C. If the weak acid is diluted, the titration will reach the
equivalence point sooner, since the concentration of
the acid will be lower.
CQ16-11.7b-Titration of a Diluted Weak Base
Given that the conductivity of an aqueous
solution depends on the concentration of
the ions present, which of the following
graphs shows conductivity (y-axis) plotted
against the acid added (x-axis) for the
titration of the strong base Ba(OH)2 with
the strong acid H2SO4?
A)
B)
CQ16-11.8a-Conductivity of a H SO /Ba(OH) Solution
C)
Consider the following arguments for each answer
and vote again:
A. This is a titration of a strong base with a strong acid,
so the conductivity will track the pH of the solution.
B. Although BaSO4(s) will precipitate as H2SO4 is
added, the total concentration of ions will remain
constant until the Ba2+(aq) is depleted.
C. The conductivity will decrease as BaSO4(s) and
H2O(λ) are formed, after which excess H2SO4 will
increase the conductivity.
CQ16-11.8b-Conductivity of a H SO /Ba(OH) Solution
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This concludes the Norton Media Library
slide set for chapter 16
Chemistry
The Science in Context
by
Thomas Gilbert,
Rein V. Kirss, &
Geoffrey Davies