Transcript Unit 10 Powerpointx

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
What is the pH of the same solution
without the common ion added?
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
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
• Stoich 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.