Buffers, Titrations, Solubility, and
other useless information
Chapter 16
From Kotz,
Chemistry &
Chemical
Reactivity, 5th
edition
Illustrates the
transformation
of one
insoluble lead
compound
into an even
less soluble
compound.
16.1 Common Ion Effect
• According to Le Chatelier’s principle: if
the concentration of a substance
involved in an equilibrium is changed,
the system will adjust to accommodate
the change and maintain the value of
the K
• Common ion effect: refers to the
equilibrium disturbance involving a weak
acid or base and its conjugate partner
Common Ion Effect
• CH3CO2H + OH-  H2O + CH3CO2• The addition of more acetate ion
suppresses the ionization of the acetic
acid and causes the system to shift left.
• To calculate the pH of a solution
containing both a weak acid and its
conjugate base, the small changes in
initial concentration can (usually) be
ignored
Common Ion Effect
• Ex. Suppose a solution is 0.10M acetic acid
and 0.050M sodium acetate. The Ka of
acetic acid is 1.8 x 10-5. Calculate pH.
• Ka = [H3O+][CH3CO2-]
[CH3CO2H]
1.8 x 10-5 = [H3O+][0.050] [H3O+] = 3.6x10-5
[0.10]
pH = 4.44
Common Ion – a special case
• What is the pH when 25.0mL of 0.0500M sodium
hydroxide is added to 25.0mL of 0.100M lactic
acid? (Ka of lactic acid is 1.4x10-4)
• 0.00125 mol of NaOH is present
• 0.00250 mol of lactic acid is present
• All of the base combines with half of the acid.
0.00125 mol of lactate ion is produced and
0.00125 mol of lactic acid remain
• This is the half equivalence point! Ka = [H+]
16.2 Buffer Solutions:
Controlling pH
• A buffer solution is resistant to change in
pH when an acid or base is added to the
solution
• Buffer solutions are simply an example of
the common ion effect; they consist of a
weak acid and its conjugate base (the
common ion) or a weak base and its
conjugate acid in solution, in nearly
equimolar quantities
Buffers
• There are two requirements for a buffer:
– Two substances are needed: an acid capable
of reacting with added OH- ions and a base
that can consume added H3O+ ions
– The acid and base must not react with each
other
Buffers
• From a list of weak acids (or bases), one
is chosen that has a pKa close to the pH
value desired
• The relative amounts of the weak acid and
its conjugate base are adjusted to achieve
exactly the pH required
pH of a Buffer
• General buffer equation Ka = [H+][A-]
[HA]
pH = -log [H+]
Check 100Ka < [HAinitial] to be sure you can
use the short cut
Some Commonly Used Buffer Systems
Weak acid
Conjugate
Base
Acid Ka
(pKa)
Phthalic acid Hydrogen
1.3x10-3
phthalate ion (2.89)
Acetic acid
Acetate ion
1.8x10-5
(4.74)
Dihydrogen
Hydrogen
6.2x10-8
phosphate ion phosphate ion (7.21)
Useful pH
range
1.9-3.9
Hydrogen
Phosphate ion 3.6x10-13
phosphate ion
(12.44)
11.3-13.3
3.7-5.7
6.2-8.2
Buffers
• It is the relative quantities of weak acid
and conjugate base that are important; the
actual concentrations do not matter
• Diluting a buffer solution will not change its
pH
• An increase in the concentration of the
buffer components increases the buffer
capacity – more acid or base can be
added without change in pH
Henderson-Hasselbalch Equations
(a.k.a. David Hasselhoff)
• pH = pKa + log [A-]
[HA]
pOH = pKb + log [HB+]
[B]
The Henderson-Hasselbalch equation is
generally valid when the ratio of
[conj base]/[acid] is less than 10 and
greater than 0.1
Using Henderson-Hasselbalch
• Ex. Suppose you dissolve 15.0g of NaHCO3
and 18.0g of Na2CO3 in enough water to
make 1.00L of solution. Calculate the pH
using Henderson-Hasselbalch.
• 15.0g NaHCO3 = 0.179 mol NaHCO3/ 1 L
• 18.0g Na2CO3 = 0.170 mol Na2CO3 / 1 L
• Ka = 4.8x10-11 (from a table)
• pH = -log(4.8x10-11) + log (.170/.179)
• pH = 10.3
Preparing a Buffer Solution
• Ex. Prepare a 1.0L buffer solution with a
pH of 4.30
• Select an acid whose pKa is close to the
desired pH (from a table)
• Use either the general equation for a
buffer or Henderson-Hasselbalch to
calculate the concentration ratio of
acid/base needed
• 2.8/1 acid/base concentration
• ratio
How Does a Buffer Maintain pH?
• If you add 1.0 mL of 1.0M HCl to 1.0L of
water, how does the pH change?
• Water has pH = 7
• 0.001 M H+ has pH = 3
• If you add 1.0mL of 1.0M HCl to 1.0L of
0.7M acetic acid/0.6M sodium acetate
buffer, how does the pH change?
• The pH of the buffer is 4.68 (how do you
find this?)
How Does a Buffer Maintain pH?
• 0.001 mol of H+ from the HCl combines with
0.001 mol of acetate ion and produces 0.001
mol of acetic acid
• ICE this to get new concentrations from the
reaction (0.599M acetate ion, 0.701M acetic
acid)
• ICE these concentrations to get
equilibrium concentrations and set up buffer
equation to solve for [H+]
• pH = 4.68
16.3 Acid-Base Titrations
• The pH at the equivalence point of a
strong acid/strong base titration is 7
(neutral)
• If weak acid is titrated with strong base
then pH > 7 at equivalence point due to
conj base of weak acid
• If weak base is titrated with strong acid
then pH < 7 at equivalence point due to
conj acid of weak base
Titration of a Strong Acid with a
Strong Base
pH
15
10
5
0
0
50
100
Volume of NaOH added (mL)
150
Titration of a Strong Acid with a Strong Base
• pH of the initial solution is the pH of the acid
• As NaOH is added to the acid solution, amount of
HCl declines, volume of solution increases, so H+
concentration decreases and pH slowly increases
• The equivalence point is the midpoint of the vertical
portion of the curve; pH is 7 here
• After all HCl has been used, pH rises slowly as
more NaOH is added (and volume increases)
• The pH at any other point is found using
stoichiometry and relationship between pH and
[H+]
Titration of a Weak Acid with a
Strong Base
pH
15
10
5
0
0
50
100
Volume of NaOH added (mL)
150
Titration of a Weak Acid with a Strong Base
• The pH before any titration begins is found from
the Ka of the weak acid and the acid
concentration
• Anywhere between the start and the
equivalence point, the Henderson-Hasselbalch
equation can be used
• At the half-equivalence point the concentration
of the weak acid is equal to the concentration of
the conj base, so pH=pKa
• At the equivalence point only the conj base
remains; the pH is controlled by the conj base
Kb and concentration
• Beyond the equivalence the pH is found from the
volume of the excess base added
Titration of a Weak Polyprotic Acid
pH
15
10
5
0
0
100
200
300
Volume of NaOH added (mL)
400
Titration of a Weak Polyprotic Acid
• The curve can be divided into three parts:
• The portion of the curve up to the first
equivalence point has a pH determined by
the excess of the polyprotic acid
• The portion of the curve between the first
and second equivalence points has a pH
determined by the excess of the amphiprotic
substance
• The portion of the curve after the second
equivalence point is has a pH determined by
the excess of the fully deprotonated conj
base
Titration of a Weak Base with a
Strong Acid
12
10
pH
8
6
4
2
0
0
50
100
Volume of Titrant added (mL)
150
Titration of a Weak Base with a
Strong Acid
• At the half-equivalence point, [OH-] = Kb
of the weak base
• The pH at the equivalence point is weakly
acidic due to the conj acid of the weak
base
16.4 pH Indicators
• Usually a weak acid or base (treat it as
such mathematically)
• Often a large organic molecule that has
different shapes in acid and base solution
• The different structures have different
colors that allows for monitoring changing
pH
• Choose an indicator with a Ka near that of
the acid being titrated so that the color
change occurs at the right stage in the
titration
Acid-Base
Indicators
Figure 18.8
Indicators for Acid-Base
Titrations
Natural Indicators
Red rose extract at different pH’s and with Al3+ ions
Rose extract
In CH3OH
Add Al3+
Add HCl Add NH3 Add NH3/NH4+
See pages 848–849
16.5 Solubility of Salts
• Salts are considered to be insoluble if less
than 0.01 moles can be dissolved per liter of
water
• The equilibrium constant for the solubility of a
salt is called the solubility product (Ksp), and
from this molar solubility can be calculated
• Addition of a common ion depresses the
solubility of a salt (Le Chatelier)
• Direct comparisons of the solubility of two
salts on the basis of their Ksp values can only
be made for salts having the same ion ratio
BaCl2  Ba2+ + 2Clx
2x
Ksp = [Ba2+][2Cl-]2
Ksp = (x)(4x2)
Ksp = 4x3
In solubility problems, s is often substituted for x
when solving for molar solubility.
•
•
•
•
NaCl Ksp = s2
BaCl2 Ksp = 4s3
AlCl3 Ksp = 27s4
Al2(SO4)3 Ksp = 108s5
Barium
Sulfate
Ksp = 1.1 x 10-10
(a) BaSO4 is a
common mineral,
appearing a white
powder or colorless
crystals.
(b) BaSO4 is opaque to
x-rays. Drinking a
BaSO4 cocktail
enables a physician
to exam the
intestines.
Common Ion Effect
PbCl2(s)  Pb2+(aq) + 2 Cl(aq)
How will the -5
addition of
Ksp = lead(II)
1.9 xion10or chloride ion
impact the solubility of
lead(II) chloride?
If a saturated solution of
lead(II) chloride is prepared,
what will happen when
sodium chloride solution is
added to the mixture?
Common Ion Effect and Salt Solubility
• Ex. If solid AgCl is
placed in 1.00L of
0.55M NaCl, what
mass of AgCl will
dissolve?
• AgCl  Ag+ + ClKsp = 1.8x10-10
• s = 1.3x10-5 mol/L
AgCl Ag+ ClI
0
0.55
C
+x
+x
E
x
0.55 + x
Common Ion Effect and Salt Solubility
• Assume x is very small compared to 0.55
(because Ksp is so small)
• Ksp = 1.8x10-10 = (x)(0.55)
• X = 4.4x10-10 mol/L (which is less than
1.3x10-5 mol/L, as predicted by Le
Chatelier’s principle
Effect of Basic Anions on Salt Solubility
• Any salt containing an anion that is the
conjugate base of a weak acid will
dissolve in water to a greater extent than
given by Ksp PbS  Pb2+ + S2• The conj base can hydrolyze water, which
lowers the [conj base] causing more of the
salt to dissolve to reestablish equilibrium
S2- + H2O  HS- + OH• Salts of phosphate, acetate, carbonate,
cyanide, and sulfide can be affected
Effect of Basic Anions on Salt Solubility
• Insoluble salts in which the anion is the conjugate
base of a weak acid will dissolve in strong acids
• Anions such as acetate, carbonate, hydroxide,
phosphate, and sulfide dissolve in strong acids.
Ex. Mg(OH)2 + 2 H3O+  Mg2+ + 4 H2O
• Salts are not soluble in strong acid if the anion is
the conj base of a strong acid.
Ex. AgCl is not soluble in strong acid because Clis a very weak base of a very strong acid
16.6 Precipitation Reactions
• Precipitation is the reverse process of
dissolving
• If you write a dissolving reaction, its
equilibrium constant expression, and its
Ksp, you can write the reverse reaction
and its equilbrium constant expression;
notice that it has 1/Ksp!
Ksp and the Reaction Quotient, Q
• If Q = Ksp the solution is saturated
– The ion concentrations are at equilibrium
values
• If Q<Ksp the solution is not saturated
– If more of the solid is present it will continue to
dissolve until equilibrium is reached; if there is
no solid present, more can be added
• If Q>Ksp the solution is supersaturated
– The ion concentrations are too high and
precipitation will occur until equilibrium is
reached
Solubility and the Reaction
Quotient
• Solid PbI2 (Ksp = 9.8 x 10-9) is placed in a
beaker of water. After a period of time, the
lead(II) concentration is measured and
found to be 1.1 x 10-3 M. Has the system
yet reached equilibrium?
• Q = [Pb2+][2x I-]2
• Q = 5.3 x 10-9 This is less than Ksp, more
PbI2 can dissolve.
• Can you figure out how much more can be
added?
16.7 Solubility and Complex Ions
• Metal ions form complex ions with Lewis
bases, such as ammonia and water
• The formation of complex ions increases
the solubility of metal ions as predicted by
the Ksp
16.8 Solubility, Ion Separations,
and Qualitative Analysis
Separating Salts by Differences
in Ksp
• Add CrO42- to solid PbCl2. The less
soluble salt, PbCrO4, precipitates
• PbCl2(s) + CrO42-  PbCrO4 + 2 Cl• Salt
Ksp
PbCl2
1.7 x 10-5
• PbCrO4
1.8 x 10-14
Separating Salts by
Differences in Ksp
• PbCl2(s) + CrO42-  PbCrO4 + 2 ClSalt
PbCl2
PbCrO4
Ksp
1.7 x 10-5
1.8 x 10-14
PbCl2(s)  Pb2+ + 2 Cl-
K1 = Ksp
Pb2+ + CrO42-  PbCrO4
K2 = 1/Ksp
Knet = K1 • K2 = 9.4 x 108
Net reaction is product-favored
Separating Salts by Differences in
Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add
CrO42- to precipitate red Ag2CrO4 and yellow
PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
The substance whose Ksp is first exceeded
precipitates first.
The ion requiring the lesser amount of CrO42- ppts.
first. You MUST use molar solubility to determine this!
Separating Salts by Differences in
Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate
red Ag2CrO4 and yellow PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
2-] required by each ion.
Calculate
[CrO
4
22+
[CrO4 ] to ppt. PbCrO4 = Ksp / [Pb ]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M
[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2
= 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M
PbCrO4 precipitates first
Dissolving Precipitates
by forming Complex Ions
Examine the solubility of AgCl in ammonia.
AgCl(s)  Ag+ + Cl-
Ksp = 1.8 x 10-10
Ag+ + 2 NH3 --> Ag(NH3)2+
Kform = 1.6 x 107
------------------------------------AgCl(s) + 2 NH3  Ag(NH3)2+ + ClKnet = Ksp • Kform = 2.9 x 10-3
By adding excess NH3, the equilibrium shifts to
the right.