Transcript Acid Base Equilibria

Acid Base Equilibria
Complications: Beyond pH of strong acids or
strong bases
Strong acid
HCl(aq)
Weak acid
HF(aq)
Weak conjugate base
H+(aq) + Cl- (aq)
Ka > 1
stronger conjugate base
H+(aq) + F- (aq)
Mullis
Ka = 7.2 x 10-4
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Conjugate Bases
•
•
•
Bronsted-Lowry Model
Conjugate base remains after acid loses its proton
Strongest conjugate base = best at getting
available protons
• Factors affecting conjugate base strength:
1. Electronegativity
S- > Cl2. Size if acids were binary F- > I3. Number of oxygens if acids were oxyacids
ClO- > ClO2Mullis
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pH of Mixture of Weak Acids
• Find pH of a solution that contains 1.00M
HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2
(Ka = 4.0 x 10-4).
• Because these are weak acids, dissociation
is not complete and [H+] ≠ [HNO2].
• You must use ICE table strategy to solve
pH problems with weak acids or bases.
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How to Determine [H+] in an Acid Mixture
• HCN
• HNO2
• H2O
H+ + CNH+ + NO2H+ + OH-
Ka = 6.2 x 10-10
Ka = 4.0 x 10-4
Ka = 1.0 x 10-14
• Compare species and find the most likely
contributor of protons. Here it is by far
HNO2: Ka is 6 orders of magnitude greater
than the closest possible H+ contributor.
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Equilibrium Method to Find
R
I
C
E
H+
0
+x
x
HNO2
5.00 M
-x
5.00 – x
• Ka = x2
(5.00 –x)
+
[H ]
NO20
+x
x
= 4.0 x 10-4
Small: reactants favored
Change must be very small relative to initial conc.: Ignore x in this term
x2 = 20. x 10-4
x = 4.5 x 10-2
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Accuracy check
• When dealing with weak acids and bases, Ka
values are typically valid to a ±5% accuracy.
Verify that our assumption in ignoring x in the
reduction of [HNO2]0 was valid:
• x
x 100% = 4.5 x 10-2 (100%) = 0.90 %
[HNO2]
5.00
• Since 0.90% is less than 5%, the assumption is
valid---in other words, our answer for x is good.
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What was the Original Question?
•
•
•
•
Find pH of this solution:
[H+] = 4.5 x 10-2
pH = -log[H+] = -log(4.5 x 10-2) = 1.35
Number of decimals in pH = Number of
significant figures in concentration (2 here)
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Find [CN-] in that problem.
• [H+] = 4.5 x 10-2 M
• HCN
H+ + CNKa = 6.2 x 10-10
• 6.2 x 10-10 = [4.5 x 10-2 ][CN-]
[HCN]
Remember our assumption that since Ka is small, we
ignore the small reduction in initial concentration
of product? Be consistent and make that
assumption here, too. So:
• 6.2 x 10-10 = [4.5 x 10-2 ][CN-]
1.00
[CN-] = 1.4 x 10-8 M (That’s a small number!)
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% Dissociation
Ex: HNO2
H+ + NO2[H+]
x 100% = % dissociation
[HNO2]
Or…
[NO2-]
x 100% = % dissociation
[HNO2]
If you know % dissociation and original
concentration of acid [HNO2], can solve for [H+]
or [NO2-] (or x, the change in concentration).
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% Dissociation and Dilution
More Concentrated
Diluted
Less % dissociation
More % dissociation
(more chance of
recombination)
If Q = K for this reaction
Q < Ka for this one, so
equilibrium shifts right
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Ka from % Dissociation
A 0.100 M solution of lactic acid (HC3H5O3) is 3.7%
dissociated. Calculate the value of Ka.
R
I
C
E
H+
0
+x
x
HC3H5O3
0.100 M
-x
0.100 – x
% dissociation = x
x (100%)
0.10
3.7% (0.10) = x
x = 0.0037 M
Mullis
C3H5O3 0
+x
x
Ka = x2
(0.100 –x)
Ka = 0.00372
(0.100 - .0037)
Ka = 1.4 x 10-4
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Organic Acids
(Acid = proton donor)
Compound
Formula
Alcohol
Alkyl Halide
Ether
R-OH
-X
R-O-R’
Aldehyde
O
||
R-C-H
O
||
R-C-R’
O
||
R-C-OH
O
||
R-C-O-R’
Ketone
Carboxylic Acid
Ester
Example
hydroxyl group
X = any halide
one oxygen bonded to 2
hydrocarbon groups
Carbonyl group attached
to end carbon
Carbonyl group attached
to a middle carbon
Carboxyl group
Carboxyl group without
the H
1-propanol
1,2-dibromopropane
diethyl ether
CH3-CH2 –O--CH2--CH3
Ethanal
O
||
CH3—C--H
2-propanone O
||
CH3—C-- CH3
ethanoic acid O
||
CH3—C—OH
methyl ethanoate O
||
CH3—C—O-- CH3
-OH phenol
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Bases without
• B (aq) + H2O
base
acid
• NH3 (aq) + H2O
• Kb = [BH+][OH-]
[B]
OH
BH+ (aq) + OH- (aq)
conjugate acid
conjugate base
NH4+ (aq) + OH- (aq)
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Amines
RxNH(3-x)
• Ammonia with one or more of the H atoms
replaced by another group.
• Many organic molecules involve amines.
OH
..
HO
CHCH2NHCH3
N
N
..
pyridine
C2H5
H
HO
adrenaline
H
ethylamine
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Find the pH of 1.0M methylamine (Kb=4.38x10-4).
R
I
C
E
CH3NH3+
0
+x
x
CH3NH2
1.0 M
-x
1.0 – x
OH0
+x
x
CH3NH2 + H2O
CH3NH3+ + OHKb = x2
= 4.38x10-4
x2 ≈ 4.38x10-4
1.0-x
x≈ 2.1x10-2
[OH-] = x = 2.1x10-2 M
pOH=-log[2.1x10-2 ]
pOH = 1.68 pH= 14 -1.68= 12.32
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PolyProtic Acids
• Lose one proton at a time: H3PO4  H2PO4-  HPO42- PO43• The reactions written to express this loss have
successively smaller Ka values:
• Ka1 > Ka2 > Ka3
• …..makes sense: The loss of a second proton occurs
less readily than the loss of the first one.
In fact, Ka1 is usually so much larger than the others,
the loss of the first proton is the only reaction that
significantly contributes [H+].
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Sequential Loss of Protons
H3PO4
H2PO4HPO42-
H+ + H2PO4H+ + HPO42H+ + PO43-
Mullis
Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.8 x 10-13
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pH of a Polyprotic Acid
• Find pH of 5.0M H3PO4 and the equilibrium concentrations
of each species: H3PO4, H2PO4-, HPO42-, PO43• Ka1 = [H+][H2PO4-] = x2 = 7.5 x 10-3
[H3PO4]
5.0-x
(see previous problems’ RICE tables & pH calculations for details)
x2 = (5.0)7.5 x 10-3 x ≈ 0.19 M= [H2PO4-] = [H+] pH = 0.72
[H3PO4] = 5.0 – x = 4.8 M
Ka2 = [H+][HPO42-] = (0.19) [HPO42-] = 6.2 x 10-8
[H2PO4-]
0.19
[HPO42-]= 6.2 x 10-8 M
Ka3 = [H+][PO43-] = (0.19) [PO43-] = 4.8 x 10-13
[HPO42-]
6.2 x 10-8
[PO43-]= 1.6 x 10-19 M
Mullis
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Sulfuric acid: Relatively High Conc.
•
•
•
•
Find pH of 1.0 M H2SO4.
H2SO4  H+ + HSO4Ka1 > 1
HSO4- H+ + SO42Ka2 = 1.2 x 10-2
In this case, does the 2nd reaction contribute
significantly to [H+]?
• [H+] is at least 1.0 since the first step is total
dissociation.
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Sulfuric acid: Relatively High Conc.
R
I
C
E
HSO41.0 M
-x
1.0 – x
H+
1.0
+x
1.0 + x
SO420
+x
x
1.2 x 10-2= [H+][SO42-] = (1.0 + x)(x)
[HSO4-]
(1.0 – x)
Assume change is small compared to 1.0, so: 1.0(x)
1.0
X = 1.2 x 10-2, or 1.2% of 1.0 M. If this is added to 1.0,
result is 1.0012M, but with 2 sig. figs., [H+] is 1.0 M. (pH = 0.00)
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0.01 M H2SO4 (Relatively Low Conc.)
R
I
C
E
HSO40.01 M
-x
0.01 – x
H+
0.01
+x
0.01 + x
SO420
+x
x
1.2 x 10-2= [H+][SO42-] = (.01 + x)(x)
[HSO4-] (.01 – x)
If assume change is small compared to .01, .01(x)
.01
x = 1.2 x 10-2, or more than 0.01 M! Therefore, cannot ignore the subtraction
or addition of x—use the quadratic to find x = .0045.
Total [H+] = 0.01(from 1st dissociation step) + .0045 = 0.0145.
pH = -log(0.0145) = 1.84
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Salts of Weak Acids or Bases
• Cations of strong bases do not change pH of a solution.
– (Ex: Na+ will not attract or contribute H+)
• Salts in which cations do not change pH and the anion is
the conjugate base of a weak acid will produce basic
solutions.
– Ex. KC2H3O2 dissociates to produce C2H3O2- and OH• Salts in which anions are not a base and the cation is a
conjugate acid of a weak base will produce acidic
solutions.
– Ex. NH4Cl dissociates to produce NH3 and H+.
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Finding Kb for Conjugates
• Look up the Ka for the weak acid that makes the
conjugate base.
• Ex: NaC2H3O2 (aq)
• C2H3O2- + H2O  HC2H3O2 + OH• Ka(acetic acid) = 1.8 x 10-5
• KaKb = Kw
• Kb = 1x10-14
= 5.6 x 10-10
1.8 x 10-5
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Salt as a Weak Base
• Find pH of a 0.30 M NaF solution.
• Na+ does not change pH.
• F- is the conjugate base of HF. Water has to
be the proton donor for this solution.
• F- (aq) + H2O (aq)  HF (aq) + OH- (aq)
• Ka for HF is 7.2 x 10-4.
• Kb = 1.0 x 10-14 = 1.4 x 10-11
7.2 x 10-4
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Salt as a Weak Base
H2O (l)
FHF
OH0.30
-0
≈0
-x
-+x
+x
0.30 - x -+x
x
1.4 x 10-11 = [HF][OH-] = x2
[F-]
.30-x
x2 =.30(1.4 x 10-11)
x = 2.0 x 10-6 (<5% of .30)
[OH-] = 2.0x10-6
pOH = 5.69
pH= 14.00 – 5.69 = 8.31
R
I
C
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Kb =
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Salt as a Weak Acid
• Find pH of a 0.10 M NH4Cl solution.
• Cl- does not change pH.
• NH4+ produces H+ in water. It is the
conjugate acid of NH3.
• NH4+ (aq)  NH3 (aq) + H+ (aq)
• Kb for NH3 is 1.8 x 10-5.
• Ka for NH+ = 1.0 x 10-14 = 5.6 x 10-10
1.8 x 10-5
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Salt as a Weak Acid
NH4+
NH3 +
H+
0.10
0
≈0
-x
+x
+x
0.10 - x
+x
x
5.6 x 10-10 = [NH3][H+] = x2
[NH4+]
.10-x
x2 =.10(5.6 x 10-10 )
x = 7.5 x 10-6
[H+] = 7.5 x 10-6
pH = -log(7.5 x 10-6 ) = 5.13
R
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C
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Kb =
Mullis
(<5% of .10)
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Acid-Base Properties of Salts
Salt
Example
pH
+ from strong base
- from strong acid
NaCl
neutral
+ from strong base
- from weak acid
NaC2H3O2
Anion acts as
base
basic
+ conj. acid fm. weak base
- from strong acid
NH4Cl
Cation acts as
acid
acidic
+ conj. acid fm. weak base
- conj. base fm. weak acid
NH4C2H3O2 Anion acts as Acidic
base, cation as if K >
a
acid
+ from metal ion
- from strong acid
FeCl3
Mullis
Hydrated cation
acts as acid
Kb
acidic
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Practice
1.
2.
3.
4.
5.
A 0.15 M solution of a weak acid is 3.0% dissociated.
Calculate Ka.
What are the major species present in a 0.150 M NH3
solution? Calculate the [OH-] and the pH of this soln.
Calculate the pH of a 5.0 x 10-3 M solution of H2SO4.
Sodium azide is sometimes added to water to kill
bacteria. Calculate the concentration of all species in a
0.010 M solution of NaN3. The Ka value for HN3
(hydrazoic acid) is 1.9 x 10-5.
Calculate the pH of 0.10 M CH3NH3Cl. Kb of CH3NH3+
is 4.38 x 10-4.
Mullis
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