Acids and Bases Part II - ricksobers.comricksobers.com/Chemistry_files/Acids and Bases Part...
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Acids and Bases Part IIDr. Sobers’ Lecture Notes
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Acid and Base Ionization Constants Ka and Kb
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Acid Ionization Constant
HAn(aq) + H2O(l) ⇄ H3O+ (aq) + An-1
An acid in water undergoes an ionization reaction. The equilibrium constant is the acid ionization constant, Ka.
Ka=[An-1][H3O
+]HAn⎡⎣ ⎤⎦
Acid + Water Hydronium ion
+ ConjugateBase
⇄
Know how to write the chemical equation and the expression
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Base Ionization ConstantAn base in water undergoes an ionization reaction. The equilibrium constant is the base ionization constant, Kb.
Base + Water Hydroxide ion
+ ConjugateAcid
⇄
An-1(aq) + H2O(l) ⇄ OH- (aq) + HAn
Kb=[HAn ][OH− ]
An-1⎡⎣ ⎤⎦
Know how to write the chemical equation and the expression
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ExamplesWrite the equilibrium reactions and equilibrium expressions for the following acids:
HF and NH4+
HF(aq) + H2O(l) ⇄ H3O+(aq) + F-(aq)Ka=
[F-1][H3O+]
HF[ ]
NH4+(aq) + H2O(l) ⇄ H3O+(aq) + NH3(aq)
Ka=[NH3][H3O
+]NH4
+⎡⎣ ⎤⎦
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ExamplesWrite the equilibrium reactions and equilibrium expressions for the following bases:
F- and CH3NH2
F-(aq) + H2O(l) ⇄ OH-(aq) + HF(aq)
Kb=[CH3NH3
+][OH-]CH3NH2[ ]
CH3NH2(aq) + H2O(l) ⇄ OH-(aq) + CH3NH3+(aq)
Kb=[OH-][HF]
F-⎡⎣ ⎤⎦
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Acid and Base Strength
HAn(aq) + H2O(l) ⇄ H3O+ (aq) + An-1
The equilibrium expressions for each equation may be written:
An-1(aq) + H2O(l) ⇄ OH- (aq) + HAn
Ka=[An-1][H3O
+]HAn⎡⎣ ⎤⎦
Kb=[HAn ][OH− ]
An-1⎡⎣ ⎤⎦
The larger the Ka, the stronger the acid.
The larger the Kb, the stronger the base.
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Sample Ka and Kb Problems
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The pH of a 0.10M formic acid (HCOOH) solution is 2.39. Calculate the Ka of formic acid
Ka=[HCOO− ][H3O
+]HCOOH[ ]
[H3O+]=10− pH
What do we know about final equilibrium state?The pH
[H3O+] = 0.0041
HCOOH(aq) + H2O(l) ⇄ H3O+(aq) + HCOO- (aq)
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The pH of a 0.10M formic acid (HCOOH) solution is 2.39. Calculate the Ka of formic acid
HCOOH(aq) + H2O(l) ⇄ H3O+(aq) + HCOO- (aq)
[ ]initial
Δ[ ]final
0.10M 0M 0M
Ka=[HCOO− ][H3O
+]HCOOH[ ]
[H3O+]=10− pH
What do we know about final equilibrium state? The pH
[H3O+] = 0.0041
0.0041
+0.0041+0.0041
0.0041-0.0041
0.096
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The pH of a 0.10M formic acid (HCOOH) solution is 2.39. Calculate the Ka of formic acid
HCOOH(aq) + H2O(l) ⇄ H3O+(aq) + HCOO-
[ ]initial
Δ[ ]final
0.10M 0M 0M
Ka=[HCOO− ][H3O
+]HCOOH[ ]
0.0041
+0.0041+0.0041
0.0041-0.0041
0.096
= (0.0041)(0.0041)0.096
=1.8x10-4
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The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH and percent ionization of a 0.250M acetic acid solution. Do the same for a 0.100M acetic acid solution.
CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO- (aq)
Ka=[CH3COO
− ][H3O+]
CH3COOH[ ]
[ ]initial
Δ[ ]final
0.250M 0M 0M
x
+x+x
x-x
0.250-x
1.8x10−5= (x)(x)(0.250-x)
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To avoid the quadratic equation, make the assumption that because the constant is so small that x will be small relative to 0.250.
1.8x10−5= (x)(x)(0.250-x)
1.8x10−5= (x)(x)(0.250-x)
≈ x2
(0.250)
This works for (# + x) or (# - x) when the # is > 100 Keq
x = 2.12x10-3
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The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH and percent ionization of a 0.250M acetic acid solution. Do the same for a 0.100M acetic acid solution.
CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO- (aq)
[ ]initial
Δ[ ]final
0.250M 0M 0M
x
+x+x
x-x
0.250-x
x = 2.12x10-3 pH = -log [H3O+]
pH = -log x = 2.67
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Percent Ionization
% Ionization = Amount IonizedTotal amount present
*100
The total amount present is the concentration* regardless of whether it is ionized or not. This is the initial concentration of the weak acid or base in the ICE table.The amount ionized is the concentration* ionized from (products). This is usually defined as x in the equilibrium table. For weak acids it would be the hydronium ion concentration at equilibrium.
Concentrations may be used for amounts because everything is in the same solution volume (molarity instead of moles).
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The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH and percent ionization of a 0.250M acetic acid solution. Do the same for a 0.100M acetic acid solution.
CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO- (aq)
[ ]initial
Δ[ ]final
0.250M 0M 0M
x
+x+x
x-x
0.250-xx = 2.12x10-3
% Ionization = 2.12x10-3 M0.250 M
*100 = 0.848%
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The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH and percent ionization of a 0.250M and 0.100M acetic acid solutions. Similar work produces answers for 0.100M
Acetic AcidMolarity
Equilibrium[H3O+] pH % Ionization
0.100M
0.250M
1.34x10-3M
2.12x10-3M
2.87
2.67
1.34%
0.848%
Increasing the overall weak acid concentration increases the acidity of the solution. The percent ionization is less but there is more acid overall.
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Relative Acid and Base Strengths
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Ka and Kb are like any equilibrium constants. The larger the constant, the more product favored the equilibrium.
The stronger the acid, the larger the Ka.
The stronger the base, the larger the Kb.
The Kb of ammonia is 1.8x10-5 and the Kb of ethyl amine is 5.6x10-4. Which is the stronger base?
Ethyl amine is the stronger base. It has a larger Kb.
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Ka, Kb and Kw
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Acid and Base Strength
The acid in water forms hydronium ion and conjugate base.
HAn(aq) + H2O(l) ⇄ H3O+ (aq) + An-1
Consider a weak acid HAn and its conjugate base of An-1. Each will ionize in water.
The base in water forms hydroxide ion and conjugate acid.
An-1(aq) + H2O(l) ⇄ OH- (aq) + HAn
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Acid and Base StrengthThe sum of the chemical equations is the auto-ionization of water:
Upon addition of equations, the net equilibrium constant is the product of the original constants.
HAn(aq) + H2O(l) ⇄ H3O+ (aq) + An-1
An-1(aq) + H2O(l) ⇄ OH- (aq) + HAn
Ka
Kb
Kw2H2O ⇄ H3O+ + OH-
Kw = Ka • Kb
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A strong acid has a weak conjugate base.
A weak acid has a strong conjugate base.
A weak base has a strong conjugate acid.
A strong base has a weak conjugate acid.
Conjugate Acid-Base Pairs
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Acid and Base StrengthKw = Ka • Kb
Acid Ka KbConjugate Base
H2O OH-1.8x10-16 55.5
CH3CO2H CH3CO2-1.8x10-5 5.6x10-10
NH4+ NH35.6x10-10 1.8x10-5
HF F-7.4x10-4 1.4x10-11
HCl Cl-large small
acid
stre
ngthbase strength
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pKa and pKb
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Acid and Base Strength
The sum of the pKa and pKb values is 14:
-log(Ka • Kb) = -log(1x10-14)pKa + pKb = 14
Kw = Ka • Kb
Kw = 1x10-14 = Ka • Kb
The pKa and pKb functions are “-log” of each respectively:
pKa = - log Ka pKb = - log Kb
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Kw = 1x10-14 = Ka • KbpK
a = -
log
Ka
pKb =
- lo
g K
b
Kb = 10-pKbKa = 10-pKa
Ka Kb
pKa pKbpKa + pKb = 14
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Acid and Base StrengthpKa = - log Ka
Acid Ka
CH3CO2H 1.8x10-5
NH4+ 5.6x10-10
HF 7.4x10-4
HCl 1.3x106
acid
stre
ngth
pKa
4.749.25
3.13-6.1
Stronger acids have smaller pKa values. The same for bases.
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Acid and Base Strength
Acid Conjugate Base
CH3CO2H CH3CO2-
NH4+ NH3
HF F-
HCl Cl-
acid
stre
ngth
base strength
pKa
4.749.25
3.13-6.1
pKb
9.264.75
10.8720.1
pKa = - log Ka pKb = - log Kb
pKa + pKb = 14
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The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH of a 0.100M sodium acetate solution.
The pH of a Salt Solution
The sodium acetate dissolves and dissociates into sodium and acetate ions. The acetate ion is a weak base (the conjugate base of a weak acid).
NaCH3COO (aq) → Na+(aq) + CH3COO-
[CH3COO] = 0.100MThe Kb is found from the Ka:
Kw = 1x10-14 = Ka • Kb
Kb = 1x10-14 / (1.8x10-5) = 5.6x10-10
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CH3COO-(aq) + H2O(l) ⇄ OH-(aq) + CH3COOH(aq)
[ ]initial
Δ[ ]final
0.100M 0M 0M
x
+x+x
x-x
0.100-x
Kb = 5.6x10-10 =[HCOOH][OH− ]HCOO−⎡⎣ ⎤⎦
= (x)(x)0.100 − x
5.6x10-10 ≈ x2
0.100
x = [OH-] = 7.45x10-6
pOH = -log [OH-]pOH = 5.13
pH = 14 - pOH = 8.87
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Common Ion Effect
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The Ka of acetic acid, CH3COOH, is 1.8x10-5. Calculate the pH of a solution that is both 0.250M in acetic acid and 0.350M sodium acetate.
Previously it was determined that the hydronium ion concentration in a 0.250M acetic acid solution is 2.12x10-3M and the pH is 2.67. In this problem there is some conjugate base present too.
CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)
0.250M 0.350M
Consider Le Chȃtelier’s principle. If the acetic acid were in equilibrium and the acetate ion were added, it would shift to the left. The hydronium ion concentration should be less.
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CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO-(aq)
0.250M 0.350M
Instead of just the acid form, the base form is present too. The pH should be higher.
[ ]initial
Δ[ ]final
0M+x+x-x
0.250 -x 0.300 + xx
Ka=[CH3COO
− ][H3O+]
CH3COOH[ ]1.8x10-5= (0.300+x)(x)
(0.250-x)≈ (0.350)x(0.250)
x = 1.29x10-5
pH = 4.89(higher pH with acetate present)