Transcript Chapter 16

Chapter
16
Applications of Aqueous
Equilibria
Chemistry 4th Edition
McMurry/Fay
Buffer Solutions
01
A Buffer Solution: is a solution of a weak
acid and a weak base (usually conjugate pair);
both components must be present.
A buffer solution has the ability to resist
changes in pH upon the addition of small
amounts of either acid or base.
Buffers are very important to biological
systems!
Slide 2
Buffer Solutions
Base is
neutralized
by the
weak acid.
02
Acid is
neutralized
by the
weak base.
Slide 3
Buffer Solutions
03
Buffer solutions must contain relatively high
concentrations of weak acid and weak base
components to provide a high “buffering
capacity”.
The acid and base components must not
neutralize each other.
The simplest buffer is prepared from equal
concentrations of an acid and its conjugate
base.
Slide 4
Buffer Solutions
04
a) Calculate the pH of a buffer system containing 1.0
M CH3COOH and 1.0 M CH3COONa.
b) What is the pH of the system after the addition of
0.10 mole of HCl to 1.0 L of buffer solution?
a) pH of a buffer system containing 1.0 M
CH3COOH and 1.0 M CH3COO– Na+ is: 4.74
(see earlier slide)
Slide 5
Buffer Solutions
04
b) What is the pH of the system after the addition of
0.10 mole of HCl to 1.0 L of buffer solution?
CH3CO2H(aq) + H2O(aq)
I
CH3CO2–(aq) + H3O+(aq)
1.0
1.0
0
– 0.10
-x
0.9 + x
+x
x
add 0.10 mol HCl
C
E
+ 0.10
-x
1.1 – x
(0.9+x)(x)
(0.9)x
Ka = 1.8 x 10-5 =
=
(1.1 – x)
(1.1)
pH = 4.66
x = (1.1/0.9) 1.8 x 10-5
x = 2.2 x 10-5
Slide 6
Water – No Buffer
What is the pH after the addition of 0.10 mole of
HCl to 1.0 L of pure water?
HCl – strong acid, completely ionized.
H+ concentration will be 0.10 molar.
pH will be –log(0.10) = 1.0
The power of buffers!
Adding acid to water:
ΔpH = 7.0 - 1.0
= 6.0 pH units
Adding acid to buffer:
ΔpH = 4.74 - 4.66
= 0.09 pH units!!
Slide 7
Acid–Base Titrations
01
Titration: a procedure for determining the
concentration of a solution using another
solution of known concentration.
Titrations involving strong acids or strong bases
are straightforward, and give clear endpoints.
Titration of a weak acid and a weak base may be
difficult and give endpoints that are less well
defined.
Slide 8
Acid–Base Titrations
02
The equivalence point of a titration is the point at
which equimolar amounts of acid and base have
reacted. (The acid and base have neutralized
each other.)
neutral
H+ (aq) + OH– (aq) →
H2O (l)
For a strong acid/strong base titration, the
equivalence point should be at pH 7.
Slide 9
Acid–Base Titrations
02
The equivalence point of a titration is the point at
which equimolar amounts of acid and base have
reacted. (The acid and base have neutralized
each other.)
basic
HA (aq) + OH– (aq) → H2O (l) + A– (aq)
Titration of a weak acid with a strong base
gives an equivalence point with pH > 7.
Slide 10
Acid–Base Titration Examples
03
Titration curve for strong acid–strong base:
Note the very
sharp endpoint
(vertical line)
seen with strong
acid – strong
base titrations.
add one drop of
base: get a BIG
change in pH
The pH is
changing
very rapidly in
this region.
Slide 11
Titration of 0.10 M HCl
Volume of Added NaOH
03
pH
1. Starting pH (no NaOH added)
1.00
2. 20.0 mL (total) of 0.10 M NaOH.
1.48
3. 30.0 mL (total) of 0.10 M NaOH.
1.85
4. 39.0 mL (total) of 0.10 M NaOH.
2.90
5. 39.9 mL (total) of 0.10 M NaOH.
3.90
6. 40.0 mL (total) of 0.10 M NaOH.
7.00
7. 40.1 mL (total) of 0.10 M NaOH.
10.10
8. 41.0 mL (total) of 0.10 M NaOH.
11.08
9. 50.0 mL (total) of 0.10 M NaOH
12.05
Slide 12
Acid–Base Titrations
04
Titration curve for weak acid–strong base:
The endpoint
(vertical line) is
less sharp with
weak acid –
strong base
titrations.
weak acid equivalence point
strong acid equivalence point
weak
acid
strong acid
pH at the
equivalence
point will always
be >7 w/ weak
acid/strong base
Slide 13
Acid–Base Titration Curves
very
weak
acid
With a very
weak acid, the
endpoint may
be difficult to
detect.
weak
acid
Slide 14
Acid–Base Titrations
09
Strong Acid–Weak Base:
The (conjugate) acid
hydrolyzes to form
weak base and H3O+.
At equivalence point
only the (conjugate)
acid is present.
pH at equivalence
point
will always be <7.
Slide 15
Solubility Equilibria
01
Aqueous Solubility Rules for Ionic Compounds
A compound is probably soluble if it contains the cations:
+, K+, Rb+
a. Li+, NaOld
(Group
1A on periodic
table)
way to
analyze
solubility.
b. NH4+
Answer
is soluble
eitherif “yes”
orthe“no”.
A compound
is probably
it contains
anions:
a. NO3– (nitrate), CH3CO2– (acetate, also written C2H3O2–)
b. Cl–, Br–, I– (halides) except Ag+, Hg22+, Pb2+ halides
c. SO42– (sulfate) except Ca2+, Sr2+, Ba2+, and Pb2+ sulfates
Other ionic compounds are probably insoluble.
Slide 16
Solubility Equilibria
02
New method to measure solubility:
Consider solution formation an equilibrium process:
MCl2(s)  M2+(aq) + 2 Cl–(aq)
Give equilibrium expression Kc for this equation:
KC = [M2+][Cl–]2
This type of equilibrium constant Kc that measures
solubility is called: Ksp
Slide 17
Solubility Equilibria
03
Solubility Product: is the product
of the molar concentrations of the
ions and provides a measure of a
compound’s solubility.
MX2(s)  M2+(aq) + 2 X–(aq)
Ksp = [M2+][X–]2
“Solubility Product Constant”
Slide 18
Solubility Equilibria - Ksp Values 04
Al(OH)3
BaCO3
BaF2
BaSO4
Bi2S3
CdS
CaCO3
CaF2
Ca(OH)2
Ca3(PO4)2
Cr(OH)3
CoS
CuBr
1.8 x 10–33
8.1 x 10–9
1.7 x 10–6
1.1 x 10–10
1.6 x 10–72
8.0 x 10–28
8.7 x 10–9
4.0 x 10–11
8.0 x 10–6
1.2 x 10–26
3.0 x 10–29
4.0 x 10–21
4.2 x 10–8
CuI
Cu(OH)2
CuS
Fe(OH)2
Fe(OH)3
FeS
PbCO3
PbCl2
PbCrO4
PbF2
PbI2
PbS
MgCO3
Mg(OH)2
5.1 x 10–12
2.2 x 10–20
6.0 x 10–37
1.6 x 10–14
1.1 x 10–36
6.0 x 10–19
3.3 x 10–14
2.4 x 10–4
2.0 x 10–14
4.1 x 10–8
1.4 x 10–8
3.4 x 10–28
4.0 x 10–5
1.2 x 10–11
MnS
3.0 x 10–14
Hg2Cl2 3.5 x 10–18
HgS
4.0 x 10–54
NiS
1.4 x 10–24
AgBr
7.7 x 10–13
Ag2CO3 8.1 x 10–12
AgCl
1.6 x 10–10
Ag2SO4 1.4 x 10–5
Ag2S
6.0 x 10–51
SrCO3 1.6 x 10–9
SrSO4 3.8 x 10–7
SnS
1.0 x 10–26
Zn(OH)2 1.8 x 10–14
ZnS
3.0 x 10–23
Slide 19
Solubility Equilibria
05
•
The solubility of calcium sulfate (CaSO4) is found
experimentally to be 0.67 g/L. Calculate the value
of Ksp for calcium sulfate.
•
The solubility of lead chromate (PbCrO4) is
4.5 x 10–5 g/L. Calculate the solubility product of
this compound.
•
Calculate the solubility of copper(II) hydroxide,
Cu(OH)2, in g/L.
Slide 20
Equilibrium Constants - Review
The reaction quotient (Qc) is obtained by
substituting initial concentrations into the
equilibrium constant. Predicts reaction direction.
Qc < Kc
System forms more products (right)
Qc = Kc
System is at equilibrium
Qc > Kc
System forms more reactants (left)
Slide 21
Equilibrium Constants - Qc
Predicting the direction of a reaction.
Qc < Kc
Q c > Kc
Slide 22
Solubility Equilibria
06
We use the reaction quotient (Qc) to determine if
a chemical reaction is at equilibrium:
compare Qc and Kc
Ksp values are also a type of equilibrium constant,
but are valid for saturated solutions only.
We can use “ion product” (IP) to determine whether
a precipitate will form:
compare IP and Ksp
Slide 23
Solubility Equilibria
06
Ion Product (IP): solubility equivalent of reaction
quotient (Qc). It is used to determine whether a
precipitate will form.
IP < Ksp
Unsaturated (more solute can dissolve)
IP = Ksp
Saturated solution
IP > Ksp
Supersaturated; precipitate forms.
Slide 24
Solubility Equilibria
07
A BaCl2 solution (200 mL of 0.0040 M) is added to
600 mL of 0.0080 M K2SO4. Will precipitate form?
(Ksp for BaSO4 is 1.1 x 10-10)
[ Ba2+ ]
0.200 L x 0.0040 mol/L = .00080 moles Ba2+
[Ba2+] = 0.00080 mol  0.800 L = 0.0010 M
[ SO42– ]
0.600 L x 0.0080 mol/L = .00480 moles SO42–
[SO42–] = 0.00480 mol  0.800 L = 0.0060 M
Slide 25
Solubility Equilibria
07
A BaCl2 solution (200 mL of 0.0040 M) is added to
600 mL of 0.0080 M K2SO4. Will precipitate form?
(Ksp for BaSO4 is 1.1 x 10-10)
IP = [ Ba2+ ]1 x [ SO42– ]1
= (0.0010) x (0.0060)
=
6.00 x 10-6
IP > Ksp, so ppt forms
Slide 26
Solubility Equilibria
07
Exactly 200 mL of 0.0040 M BaCl2 are added to
exactly 600 mL of 0.0080 M K2SO4. Will a
precipitate form?
If 2.00 mL of 0.200 M NaOH are added to 1.00 L
of 0.100 M CaCl2, will precipitation occur?
Slide 27
The Common-Ion Effect and Solubility
The solubility product (Ksp) is an equilibrium
constant; precipitation will occur when the ion
product (IP) exceeds the Ksp for a compound.
If AgNO3 is added to saturated AgCl, the
increase in [Ag+] will cause AgCl to precipitate.
IP = [Ag+]0 [Cl–]0 > Ksp
Slide 28
The Common-Ion Effect and Solubility
MgF2
becomes
less soluble
as F- conc.
increses
Slide 29
The Common-Ion Effect and Solubility
CaCO3 is more
soluble at low pH.
Slide 30
The Common-Ion Effect and Solubility
Calculate the solubility of silver chloride (in
g/L) in a 6.5 x 10–3 M silver chloride solution.
Calculate the solubility of AgBr (in g/L) in:
(a) pure water
(b) 0.0010 M NaBr
Slide 31