Transcript Buffered solutions

Chapter 16. Applications of Aqueous Equilibria
16.1 Buffer Solutions
16.2 Capacity and
Preparation of Buffer
Solutions
16.3 Acid-Base Titrations
16.4 Solubility Equilibria
16.5 Complexation Equilibria
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
16.1 Buffer Solutions
Learning objective:
Calculate the pH of a buffered solution
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
16.1 Buffer Solutions
 Buffered solutions – solutions which are resistant to
change in pH upon addition of strong acid or base.
 Usually prepared from a conjugate acid-base pair -
either a weak acid and its conjugate base or a weak
base and its conjugate acid.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 16 - 1
A solution contains 0.125 mol of solid sodium acetate
dissolved in 1.00 L of 0.250 M acetic acid. Determine
the concentration of hydronium ions, acetate ions, and
acetic acid.
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Example 16 - 2
Determine if a solution prepared by mixing 5.0 mL of
concentrated HCl (12 M) with 75 mL of 1.0 M sodium
acetate is a buffer solution.
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Molecular View of a Buffer Solution
A buffer can protect the pH of a solution by reacting with
both strong acids and strong bases, preventing large
changes in [OH-] and [H3O+ ].
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Example 16-3 Molecular View of a Buffer Solution
The molecular picture at below represents a small portion of a
buffer system. Solvent water molecules are omitted for clarity.
Redraw the original figure to show the equilibrium condition that
is established when (a) three hydroxide ions enter the region,
and (b) seven hydronium ions enter the region. Include any
water molecules that are part of the buffer chemistry.
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The Buffer Equation

[A ]intial
pH  pKa  log
[HA]intial
 Also called the Henderson-Hasselbalch Equation
 The pH of the buffer solution is dependent more on pKa
of the buffer than concentrations of acids and bases.
 As rule, this equation is only useful if HA and A- differ by
less than a factor of 10.
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Example 16 - 4
Buffer solutions with pH values around 10 are prepared
using sodium carbonate (Na2CO3) and sodium hydrogen
carbonate (NaHCO3). What is the pH of a solution
prepared by dissolving 10.0 g of each of these two salts
in enough water to make 0.250 L of solution?
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Example 16 - 5
By how much does the pH of the buffer solution in
Example 16 – 4 change on the addition of 3.50 mL of
6.0 M HCl?
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16.2 Capacity and Preparation of Buffer Solutions
Learning objective:
Explain how to prepare a buffered solution of known pH
and capacity
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16.2 Capacity and Preparation of Buffer Solutions
 Buffer Capacity – the amount of added [H3O+] and [OH-]
that the buffer solution can tolerate without exceeding
a specified pH range.
 Technical definition: the amount of acid (or base) that
when added to a buffer changes the pH by 1 unit.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 16 – 6 Buffer Capacity
Biochemists and molecular biologists use phosphate
buffers to match physiological conditions. A buffer
solution that contains H2PO4- as the weak acid and
HPO42- as the weak base has a pH value very close to
7.0. A biochemist prepares 0.250 L of a buffer solution
that contains 0.225 M HPO42- and 0.330 M H2PO4-.
What is the pH of this buffer solution? Is the buffering
action of this solution destroyed by addition of 0.40 g
NaOH?
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Common Buffer Systems
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Example 16 – 7 Buffer Preparation
What mass of sodium acetate (NaCH3CO2, M = 82.04
g/mol) and what volume of concentrated acetic acid
(17.45 M) should be used to prepare 1.5 L of a buffer
solution at pH = 5.00 that is 0.150 M overall?
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Example 16 – 8 Preparing a Buffer
A technician wants to prepare a buffer solution at pH =
9.00 with an overall concentration of 0.125 M. The
technician has solutions of 1.00 M HCl and NaOH and
bottles of all common salts. What regents should be
used, and in what quantities, to prepare 1.00 L of a
suitable buffer?
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16.3 Acid-Base Titrations
Learning objective:
Calculate an acid or base concentration from titration
data
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16.3 Acid – Base Titrations
1.
Strong Acid with a Strong Base
2.
Weak Acid with a Strong Base
3.
Weak Base with a Strong Acid
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Acid – Base Titrations
The “titrant”
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Example 16-9 Acid – Base Titration
Industrial wastewater often is contaminated with strong
acids. Environmental regulations require that such
wastewater be neutralized before it is returned to the
environment. A 150-mL sample of wastewater was
titrated with 0.1250 M sodium hydroxide, and 38.65 mL
of the base was required to reach the stoichiometric
point. What was the molarity of hydronium ions in this
sample of wastewater?
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At the Stoichiometric Point…

Just enough hydroxide ions have been added to react
with all of the acidic protons present in solution.

This allows us to be quantitative.
c1V1  c2 V2
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Standardization

The concentration of the titrant must be accurately
known for quantitative results.

The titrant is therefore titrated against a stable, pure,
weighable standard, often potassium hydrogen
phthalate, KHC8H4O4:
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Example 16-10 Standardization
A biochemist needed to standardize a solution of KOH. A
sample of potassium hydrogen phthalate weighing
0.6745 g was dissolved in 100.0 mL of water and a drop
of indicator was added. The solution was then titrated
with the KOH solution. The titration required 41.75 mL
of base to reach the stoichiometric point. Find the
molarity of the KOH solution.
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Titration of a Strong Acid by OH- Ions
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Titration of a Weak Acid by OH- Ions
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Titration of a Weak Acid by OH- Ions
At the beginning, HA and H2O are the major species, so
we use Ka to determine the pH of the solution.
2. During the titration we are creating a buffer, so we use
the buffer equation to determine pH. At the midpoint
of the titration (when HA = A-), pH = pKa
3. When nearly all the HA is reacted, the only major
species are A- and water, and we use Kb to determine
pH.
4. After all HA molecules have reacted, the solution
contains excess A- and OH-. Here we determine pH
using the excess OH-.
1.
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A flow chart
summarizes the major
species in solution
and the pH
calculations for the
four key regions of a
weak acid titration.
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Example 16 – 11 pH at the Stoichiometric Point
What is the pH at the stoichiometric point of the titration
of 0.150 L of 0.500 M acetic acid with 2.50 M KOH
solution?
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Titration of a Weak Base by H3O+ Ions
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Titration of a Weak Base by H3O+ Ions
At the beginning, B and H2O are the major species, so
we can use Kb to determine the pH of the solution.
2. During the titration we are creating a buffer, so we
use the buffer equation to determine pH. At the
midpoint of the titration (when B = BH+ ), pH = pKa
3. When nearly all the B is reacted, the only major
species are BH+ and water, and we use Ka to
determine pH.
4. After all B molecules have reacted, the solution
contains excess BH and H3O+. Here we determine pH
using the excess H3O+.
1.
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Example 16 – 12 Titration of a Weak Base
Ephedrine, a weak base, is the active ingredient in many
commercial decongestants. To analyze a sample of
ephedrine dissolved in 0.200 L of water, a chemist
carries out a titration with 0.900 M HCl, monitoring the
pH continuously. The data obtained in this titration are
shown in Figure 16 – 6. Calculate Kb for ephedrine and
determine the pH of the solution at the stoichiometric
point.
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Titration of Polyprotic Acids
HA, A2
HA
A2
H2A
H2A, HA
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A2, OH 
[H3O ]  Ka1Ka2
Example 16 – 14 Drawing a Titration Curve
Sulphurous Acid, H2SO3, has two acidic hydrogen atoms,
with pKa values of 1.85 (Ka = 1.40 x 10-2) and 7.20 (Ka =
6.3 x 10-8) . Construct a titration curve for the titration
of 125 mL of 0.150 M sulphurous acid with 0.800 M
NaOH.
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Indicators
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Common Indicators
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Example 16 – 15 Selecting an Indicator
A student wants to titrate a solution of ammonia whose
approximate concentration is 10-2 M. What indicator
would be appropriate?
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16.4 Solubility Equilibria
Learning objective:
Use the concepts of Ksp and the common-ion effect to
calculate solution concentrations
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16.4 Solubility Equilibria
 Insoluble compounds: solubility is less than 0.01 mol
of dissolved material per liter of solution, Ksp << 1
 Slightly soluble: 10-5 < Ksp < 10-2
 Soluble: Ksp > 10-2
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Example 16 – 16 Solubility Products
Gypsum is a relatively soft rock made of calcium
sulphate. Rainwater percolates through gypsum,
dissolves some of the rock, and eventually becomes
saturated with Ca2+ ions and SO42- ions. A geochemist
takes a sample of groundwater from a cave and finds
that it contains 8.4 x 10-3 M SO42- and 5.8 x 10 -3 M Ca2+.
Use the data to determine the solubility product
constant of calcium sulphate.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 16 – 17 Calculating Ksp
When solid PbI2 is added to pure water at 25 oC, the salt
dissolves until the concentration of Pb2+ reaches 1.35 x
10-3 M. After this concentration is reached, excess solid
remains undissolved. What is the Ksp for this salt?
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Example 16 – 18 Solubility Calculations
Cadmium is an extremely toxic metal that finds its way
into the aqueous environment as a result of some
human activities. A major cause of cadmium pollution
is zinc mining and processing, because natural deposits
of ZnS ores usually also contain CdS. During the
processing of these ores, highly insoluble cadmium
sulphide (Ksp = 7.9 x 10-27) may be converted into
considerably less insoluble cadmium hydroxide (Ksp =
7.2 x 10-15). What mass of Cd(OH)2 will dissolve in 1.00
x 102 L of an aqueous solution?
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Precipitation Equilibria
 Let’s examine the reaction quotient, Q
 If
Q = Ksp, the system is at equilibrium, and the
solution is saturated.
 If Q < Ksp, the system is not at equilibrium and the
solution is not saturated.
 If Q > Ksp, the system is not at equilibrium and the
solution is supersaturated.
If a solution is not at equilibrium, the reaction will shift
either left or right until it is at equilibrium.
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Example 16 – 19 Precipitation Reactions
As illustrated in Example 16 – 18, wastewater resulting
from metal processing often contains significant
amounts of toxic heavy metal ions that must be
removed before the water can be returned to the
environment. One method uses sodium hydroxide
solution to precipitate insoluble metal hydroxides.
Suppose that 1.0 x 102 L of wastewater containing 1.2 x
10-5 M Cd2+ is treated with 1.0 L of 6.0 M NaOH
solution. What is the residual concentration of Cd2+
after treatment, and what mass of Cd(OH)2
precipitates?
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The Common Ion Effect
 Common ion effect – adding a “common ion” to a
saturated solution
e.g. Cd(OH)2 (s) Ý Cd+2 (aq) + 2 OH- (aq)
If we add Cd+2 or OH-, the common ion effect would
cause the reaction to shift to the left according to le
Chatelier’s principle.
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Common-Ion Effect
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Example 16 – 20 The Common Ion Effect
The concentration of chloride ion in seawater is around
0.55 M. To compare the solubility of Pb2+ in freshwater
vs. seawater, calculate the solubility in g/L of PbCl2 in
pure water and in 0.55 M NaCl.
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Solubility and pH Effects
Let’s examine the solubility of calcium carbonate
CaCO3(s) Ý Ca2+ (aq) + CO32- (aq) Ksp = 3.8 x 10-9
The carbonate also hydrolyzes:
CO32- (aq) + H2O (l) Ý HCO3- (aq) + OH- (aq) Kb=2.1 x 10-4
What if we added some strong acid to a solution of
calcium carbonate?
H3O+ (aq) + OH- (aq) Ý 2 H2O (l) K = 1.0 x 10-14
Now, if we add all the reactions together
CaCO3(s) + H3O+ (aq) Ý HCO3- (aq) + Ca2+ (aq) + H2O (l)
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Solubility and pH Effects
CaCO3(s) + H3O+ (aq) Ý HCO3- (aq) + Ca2+ (aq) + H2O (l)
 Kw  1  Ksp
Kacidic  Ksp 


 Ka   Kw  Ka
3.36  109

 71
11
4.7  10
Kacidic
71
71
10



2.1

10
Kneutral Ksp 3.36  109
i.e. CaCO3 is 10 orders of magnitude more soluble in
acidic solution!
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Example 16 – 21
The most acid rain on record has pH = 1.87, recorded at
Inverpolley Forest, Scotland in 1983. Calculate the
concentration of Ca2+ cations in a solution formed
when this rain becomes saturated with calcium
carbonate.
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16.5 Complexation Equilibria
Learning objective:
Calculate the concentrations of species involved in
complex-ion formation
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16.5 Complexation Equilibria
 Stoichiometry of Complexes
A species that bonds to a metal cation to form a complex is
known as a ligand.
 The number of ligands (also called coordination number)
provides a shorthand notation for the stoichiometry of the
metal complex.

Ag+ (aq) + 2 NH3 (aq) Ý [Ag(NH3)2]+ (aq)
NH3 is the ligand
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Complexation Calculations
Ag+ (aq) + NH3 (aq) Ý [Ag(NH3)]+ (aq)
K1 
[Ag(NH3 ) ]eq
[Ag ]eq [NH3 ]eq
[Ag(NH3)]+ (aq) + NH3 (aq) Ý [Ag(NH3)2]+ (aq)
K2 
[Ag(NH3 )2 ]eq
[Ag(NH3 ) ]eq [NH3 ]eq
Kf ([Ag(NH3 )2 ]eq )  K1K2 
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[Ag(NH3 )2 ]eq
2
[Ag ]eq [NH3 ]eq
Example 16 – 22 Formation of a Gold Complex
The small amounts of gold contained in low-grade ores can be
extracted using a combination of oxidation and complexation.
Gold is oxidized to Au+, which forms a very strong complex with
cyanide anions.
Au+ (aq) + 2 CN- (aq) Ý [Au(CN)2]- (aq)
Kf = 2 x 1038
Suppose that a sample of ore containing 2.5 x 10-3 mol of gold is
extracted with 1.0 L of 4.0 x 10-2 M aqueous KCN solution.
Calculate the concentrations of the three species involved in the
complexation equilibrium.
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The Chelate Effect
 Ligands that have two or more
donor atoms are chelating
ligands. Ethylenediamine
(H2NCH2CH2NH2) is a common
one and is abbreviated as “en”
 Each nitrogen has a lone pair of
electrons which can be donor
atoms, thus en is said to be
bidentate
 Chelating ligands bond more
tightly to the metal cations.
Ca(EDTA)2- (with the H atoms removed)
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The Chelate Effect
 The stabilization of a metal complex by a ligand with
more than one donor atom is known as the chelate
effect.
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Complex Formation and Solubility
 Complexation can enhance solubility by removing
metal cations from solution causing the equilibrium to
shift to the right, and dissolve more solid.
NH3 (aq)
AgCl (s)
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Ag(NH3)2+ (aq)
Example 16 – 23 Capacity of Thiosulphate Fixer
What mass of AgBr will dissolve
in 1.5L of fixer solution that
contains 0.50 M of
thiosulphate ions?
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Chapter 16 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 16 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 16 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 16 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.
Chapter 16 Visual Summary
Chemistry, 2nd Canadian Edition ©2013 John Wiley & Sons Canada, Ltd.