2
Friday Midterm Feb 25th
Rooms to be announced.
Chemical equilibrium
Acids and Bases
Additional Aspects of Acids and Bases
Sample midterm posted outside of Room 370, Chemistry.
Bronsted-Lowry Theory:
An acid is a proton H+ donor
A base is a proton H+ acceptor
Lewis Theory:
An acid accepts electron pairs
A base donates an electron pair.
Last Lecture
1. Polyprotic acids and bases
2. Lewis Acids/Bases
3. Ionization percentages (these
relate simply to pKa and pH)
Ka
percent ionized = 100i
+
K a + [ H 3O ]
[H O+ ]
3
percent not ionized = 100i
+
K a + [ H 3O ]
Lewis Acids and Bases
Lewis Acid - A molecule that is an electron pair acceptor.
Lewis Base - A molecule that is an electron pair donor.
B: + A
e.g.
B:A
OH- and NH3 are Lewis bases (lone pairs on O or N)
H+ is a Lewis acid (can accept lone pairs). H3O+ provides H+.
base
acid
adduct
Degree of Ionization Varies with pH
HA + H2O
Percent Ionized: =
Ka
Percent ionization =
+
K a + ( H 3O )
Percent Not Ionized: =
Degree of ionization =
H3 O+ + Ai100
( H 3O + )
K a + ( H 3O )
[H3O+] from HA
[HA] originally
[H3O+] from HA
[HA] originally
+
100%
i100
Protonated form:
Unprotonated form:
Percent of HA and A- for a Closed System with Monoprotic Acid
100
80
HL
Percentage i
Li
60
50% ionized,
or [HA]=[A-]
40
20
pH=pKa
0
0
5
10
xindi
pH
15
Polyprotic Acids
Monoprotic 1 H+ donation : HA
Example : HCl, HOAc, HNO3
Diprotic 2 H+ donation : H2A
Example: H2CO3, H2SO4
Two equilibria : Ka1, Ka2
Triprotic 3 H+ donation : H3A
Example : H3PO4
H3PO4 + H2O H3
O+
H2PO4 + H2O H3
-
+ H2PO4
O+
+ HPO4
HPO42- + H2O H3O+ + PO43 Ka1 >> Ka2 >> Ka3
One equilibrium: Ka
-
2-
Three equilibria : Ka1, Ka2, Ka3
[H3O+] [H2PO4-]
Ka1=
[H3PO4]
= 7.110-3
[H3O+] [HPO42-]
Ka2=
[H2PO4-]
= 6.310-8
[H3O+] [PO43-]
Ka3=
[HPO42-]
= 4.210-13
Almost all H3O+ formed in first ionization step
Example From Text: Phosphoric Acid
For a 3.0 M H3PO4 solution, calculate:
(a) [H3O+];
(b) [H2PO4-];
(c) [HPO42-]
H3PO4 +
Initial conc.
Changes
Eqlbrm conc.
Ka1=
[H3O+] [H2PO4-]
[H3PO4]
H2O
=
(3.0 x)
0
+x M
xM
= 7.110-3
But H2PO4 essentially does not ionize further
a,b) answers:
H2PO4- +
3.0 M
-x M
(3.0-x) M
x2
(d) [PO43-]
H3 O +
0
+x M
xM
x = 0.14 M
[H2PO4-] [H3O+]
[H2PO4-] [H3O+] 0.14 M
c) [HPO42-]:
H2PO4- + H2O HPO42- + H3O+
Because [H2PO4-] [H3O+]
Ka2=
[H3O+] [HPO42-]
[HPO42] = 6.310-8
[H2PO4-]
Do the ICE table for Ka2 and see:
H2PO4- + H2O HPO42- + H3O+
Initial conc.
Changes
Eqlbrm conc.
Ka2=
0.14 M
-y M
(0.14 - y) M
[H3O+] [HPO42-]
[H2PO4-]
Solve:
=
0
+y M
yM
(0.14 + y) y
(0.14 - y)
y = [HPO42-] = 6.310-8
0.14 M
+y M
(0.14 +y) M
= 6.310-8
d) [PO43-]:
HPO4- +
H2O
PO43-
+
H3O +
Because [H3O+] 0.14 M and [HPO42-] Ka2= 6.310-8 :
Ka3 =
4.210-13 =
[H3O+] [PO43-]
[HPO4
0.14 [PO43-]
2-]
6.310-8
[PO43-] = 4.210-13 6.310-8 / 0.14 = 1.910-19 M
Do the ICE table for Ka2 and see:
Initial conc.
Changes
Eqlbrm conc.
Ka3=
HPO4- + H2O
6.310-8 M
-z M
(6.310-8 - z) M
[H3O+] [PO43-]
Solve:
[HPO42-]
=
PO430
+z M
zM
(0.14 + z) z
(6.310-8 - z)
z = [PO43-] = 1.910-19 M
+
H 3 O+
0.14 M
+z M
(0.14 +z) M
= 4.210-13
Chapter 17: Additional Aspects of Acid-Base Equilibria
The Common-Ion Effect
The effect on an equilibrium by a second substance that produces
ions that can participate in that equilibrium. The added ions are
said to be common to the equilibrium.
CH3CO2H + H2O H3O+ + CH3CO2Common ion
HCl + H2O
H3 O +
+
Cl-
Le Chateliers Principle
CH3CO2H + H2O H3O+ + CH3CO2Initial
0.100 mol/L
0
0
Change
-x
+x
+x
Eqlbrm
(0.100 - x)
x
x
[H3O+] [CH3CO2-]
x2
= 1.810-5
Ka =
=
[CH3CO2H]
(0.100 - x)
x = [H3O+ ] = [CH3CO2-]= 1.310-3
Add 0.1 M HCl:
Initial concs.
Change
Eqlbrm
Ka =
CH3CO2H + H2O
0.100
-x
(0.100 - x)
[H3O+] [CH3CO2-]
[CH3CO2H]
=
x (0.100 + x)
[H3O+ ]= 0.100 + x 0.100 M
[CH3CO2-]=x= 1.810-5
(0.100 - x)
pH=2.9
H3O+ +
0.100
+x
(0.100 + x)
CH3CO2
0
+x
x
x = 1.810-5
pH = 1
Ionization of Acetic acid suppressed!
Similarly, Suppression of Ionization
of a Weak Base:
Weak Acid its and salt:
Strong electrolyte
Also, Weak Base and its Salt: e.g. NH3 and NH4Cl
Protonated form:
Unprotonated form:
Percent of HA and A- for a Closed System with Monoprotic Acid
100
80
HL
Percentage i
Li
60
50% ionized,
or [HA]=[A-]
40
20
pH=pKa
0
0
5
10
xindi
pH
15
Buffer Solutions
Solutions that resist pH changes when acids or bases are added
(pure water has no buffering capability. e.g. add 0.001 M HCl. ?
Weak acid+conjugate base or weak base+conjugate acid
Base accepts protons to resist added acid
Acid donates protons to resist added base
Consider [CH3CO2H] = [CH3CO2-] in a solution (e.g. 0.1M)
Ka=
[H3O+] [CH3CO2-]
[CH3CO2H]
= 1.810-5
[HCH3CO2]
[H3O+] = Ka
= 1.810-5
[CH3CO2-]
pH = -log[H3O+] = -logKa = -log(1.810-5) = 4.74
Now add a small amount of strong acid. (e.g. 0.001M)
CH3CO2 + H3O+
CH3CO2H + H2O
Acetate will decrease slightly, acetic acid increase slightly and H3O+
will have hardly changed
buffer in action
[HCH3CO2]
= 1.810-5
[H3O+] = Ka
[CH3CO2-]
Buffer Preparation with desired pH
Consider a weak acid, HA, and its salt NaA:
HA + H2O
A- + H3O+
-logKa= -log[H3O+] - log
pKa = pH - log
pH = pKa + log
Ka=
[H3O+] [A-]
[HA]
[A-]
[HA]
[A-]
[HA]
[conjugate base]
[acid]
If equal, pH=pKa
Henderson-Hasselbalch Equation
pH = pKa + log
[conjugate base]
[acid]
For the buffer to work, the acid and conjugate base must have
similar concentrations:
[A-]
0.1 <
< 10
To keep pH within 1:
[HA]
Need the concentrations of acid and conjugate base to be both
much larger than Ka.
[A-] > 100Ka and [HA] > 100Ka
Example Henderson-Hasselbalch Equation
What concentration of sodium acetate must be added to a 0.2 M
acetic acid solution (Ka=1.810-5) to produce pH=5?
HC2H3O2 + H2O
C2H3O2- + H3O+
[C2H3O2-][H3O+]
Ka=
= 1.810-5
[HC2H3O2]
Solution Henderson-Hasselbalch Equation
pH = pKa + log
5=
Target pH
pKa value
[conjugate base]
[acid]
-log(1.810-5)
+ log
x = 0.36 M.
x
0.2
Q:
Why must a buffer contain weak acid/conj base?
A1:
Need a lot of available acid to react with added base
(or base to react with added acid).
10 M HCl pH=-1
1 M HCl
pH=0
0.01 M HCl
pH=2
0.00001 M HCl
pH=5
10 M HOAc / OAcpH~5
1 M HOAc / OAcpH~5
0.01 M HOAc / OAcpH~5
0.00001 M HOAc / OAcpH~5
Percent of HA and A- for a Closed System with Monoprotic Acid
100
pKa = pH - log
80
HLi
60
Li
40
20
0
0
5
10
xindi
pH
15
[A-]
[HA]
Q:
Why must a buffer contain weak acid/conj base?
Weak acid ionization equilibrium allows pH to be almost
independent of concentrations in a useful range, near the pKa.
However, the pKa for strong acids and bases is Kw=10-14 and is
useless for controlling the solution pH:
[A-]
pH = pKa + log
[HA]
If I add 0.1M NaOH to a 0.00001M HCl
because of:
OH- + H3O+
2H2O
pH change from 5 to 13
Buffer Capacity and Range
Buffer capacity is the amount of acid or base that a
buffer can neutralize before its pH changes
appreciably.
Maximum buffer capacity exists when [HA] and [A-] are
large and approximately equal to each other.
Buffer range is the pH range over which a buffer
effectively neutralizes added acids and bases.
Practically, range is 2 pH units (i.e. 1 around pKa)
e.g.
Acetic acid / sodium acetate:
Buffer Range:
3.7 < pH < 5.7
pKa = 4.7
Example Buffer
Can a buffer solution be created from Ammonia and nitric acid?
NH3 + H3O+
NH4+ + H2O
Will produce both an acid and conj base in solution
and can thus act as a buffer.
Add the nitric acid until [NH3] and [NH4+] give desired ratio.
pH = pKa + log
[NH3]
[NH4+]
Six Methods of Preparing Buffer Solutions
Extra Slides: Use only if there is extra time.
Neutralization of strong and weak
Q: How much strong acid is required to neutralize 0.1M NaOH?
A: 0.1M of course!
Q: How much strong acid is required to neutralize 0.1M NH3?
NH3 + H3O+
NH4+ + H2O
NH3 + H2O
Kb=1.8x10-5
NH4+ + H2O
K= 1/Ka
NH3 + H3O+
Ka = Kw/Kb
NH4+ + OH-
Kb
Ka=Kw/Kb=5.6x10-10
K = 1/Ka = 1.8x10+9 Completion!
amount required also = 0.1M!
Volume of acid need to neutralize the base is independent
of the strength of the base (and vice versa)