Transcript Acids and Bases

Acids and Bases
Chapter 15
Chapter 15 Topics
Acids and Bases
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Acid-Base Definitions
Acid-Base Properties of water
The Ion Product of water (Kw)
pH and pOH
Strong Acids and Bases
Weak Acids and Bases
Percent Ionization
The Relationship of the Ionization constants of Acids and Their
Conjugate Bases
Lewis Acids and Bases
Dr. Ali Bumajdad
Acid-Base Definitions
Acid- Base Equilibria
1) Arrhenius Definition:
+
 Acid: Substance when dissolved in water increase H
 Base: Substance when dissolved in water increase OH
2) BrØnsted-Lowry Definition:
+
 Acid: Substance that donate H
+
 Base: Substance that accept H
 Acid: Electron-pair acceptor
 Base: Electron pair donor
H
F
F B
F
+
••
3) Lewis Definition:
N H
H
Dr. Ali Bumajdad
Sa Ex.
(a) What is the conjugate base of the following acids:
HClO4; H2S; PH4+; HCO3(b) What is the conjugate acid of the following bases:
CN-; SO4-; H2O; HCO3(c) Write the formula of the conjugate acid of each of
the following: HSO3-; F-; PO4-; CO.
Sa. Ex.
The hydrogen sufite (HSO3-) is amphoteric.
(a) Write an equation for the reaction of HSO3- with water in which
the ion act as acid.
(b) Write an equation for the reaction of HSO3- with water in which
the ion act as acid
(c) Identifiy the conjugate acid-pase pair.
Acid-Base Properties of Water
H+ (aq) + OH- (aq)
H2O (l)
autoionization of water
H
O
H
+ H
[H
O
H
]
H
+ H
H
base
H2O + H2O
acid
O
+
conjugate
acid
H3O+ + OHconjugate
base
O
-
The Ion Product of Water (Kw)
H2O (l)
H+ (aq) + OH- (aq)
[H+][OH-]
Kc =
[H2O]
[H2O] = constant
Kc[H2O] = Kw = [H+][OH-]
•Ion-product constant (Kw) is the product of the molar
concentrations of H+ and OH- ions at a particular temperature.
Solution Is
At 250C
Kw = [H+][OH-] = 1.0 x 10-14 (1)
[H+] = Kw / [OH-]
[OH-] = Kw / [H+]
[H+] = [OH-]
neutral
[H+] > [OH-]
acidic
[H+] < [OH-]
basic
Q) What is the concentration of OH- ions in a HCl solution
whose hydrogen ion concentration is 1.3 M?
Kw = [H+][OH-] = 1.0 x 10-14
[H+] = 1.3 M
-14
K
1
x
10
w
-15 M
=
=
7.7
x
10
[OH-] =
[H+]
1.3
pH and pOH
pH = -log [H+]
log 110-7
pOH = -log [OH-] (3)
(2)

=
pH + pOH = 14.00
(4)
At 250C
Solution is
neutral
[H+] = [OH-]
[H+] = 1 x 10-7
pH = 7
acidic
[H+] > [OH-]
[H+] > 1 x 10-7
pH < 7
basic
[H+] < [OH-]
[H+] < 1 x 10-7
pH > 7
pH
[H+] 1M
pH 0
pOH 14
110-1
1
13
[H+]
110-7
7
7
110-14
14
0
[H+][OH-] = Kw = 1.0 x 10-14
-log [H+] – log [OH-] = 14.00
pH + pOH = 14.00
[H+]
= 10
7
-pH
log
shift

shift
[OH-] = 10-pOH
(5)
7

=
log
=
(6)
Q) The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was 4.82.
What is the H+ ion concentration of the rainwater?
pH = -log [H+]
[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M
Q) The OH- ion concentration of a blood sample is 2.5 x 10-7
M. What is the pH of the blood?
pH + pOH = 14.00
pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60
pH = 14.00 – pOH = 14.00 – 6.60 = 7.40
Strong Acids and Bases
Strong Electrolyte – 100% dissociation
NaCl (s)
H 2O
Na+ (aq) + Cl- (aq)
Weak Electrolyte – not completely dissociated
CH3COOH
CH3COO- (aq) + H+ (aq)
Strong Acids are strong electrolytes
HCl (aq) + H2O (l)
H3O+ (aq) + Cl- (aq)
HNO3 (aq) + H2O (l)
H3O+ (aq) + NO3- (aq)
HClO4 (aq) + H2O (l)
H3O+ (aq) + ClO4- (aq)
H2SO4 (aq) + H2O (l)
H3O+ (aq) + HSO4- (aq)
Weak Acids are weak electrolytes
H3O+ (aq) + F- (aq)
HF (aq) + H2O (l)
HNO2 (aq) + H2O (l)
H3O+ (aq) + NO2- (aq)
HSO4- (aq) + H2O (l)
H3O+ (aq) + SO42- (aq)
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)
Strong Bases are strong electrolytes
NaOH (s)
KOH (s)
H 2O
H 2O
Ba(OH)2 (s)
Na+ (aq) + OH- (aq)
K+ (aq) + OH- (aq)
H 2O
Ba2+ (aq) + 2OH- (aq)
Weak Bases are weak electrolytes
F- (aq) + H2O (l)
NO2- (aq) + H2O (l)
OH- (aq) + HF (aq)
OH- (aq) + HNO2 (aq)
Conjugate acid-base pairs:
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The conjugate base of a strong acid has no measurable
strength (extremely weak base).
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H3O+ is the strongest acid that can exist in aqueous
solution.
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The OH- ion is the strongest base that can exist in aqeous
solution.
Strong Acid
Weak Acid
Sa. Ex.
Indicate whether solutions with each of the following ion concentrations is
neutral, acidic or basic:
(a) [H+] = 4  10-9 M
(b) [OH-] = 1  10-7 M
(c) [OH-] = 7  10-13 M
(d)
Sa Ex.
Calculate [H+] in
(a) a solution in which [OH-] = 0.010 M
(b) a solution in which [OH-] = 1.8  10-9 M
Sa. Ex.
Calculate [OH-] in
(c) a solution in which [H+] = 2  10-6 M
(d) a solution in which [H+] = [OH-]
(e) a solution in which [H+] = 100  [OH-]
Sa Ex.
Calculate the pH for the two solutions in Sa Ex. 16.5
Sa. Ex.
In a sample of lemon juice [H+] = 3.8  10-4 M. What is the pH
Sa Ex.
A sample of freshly pressed apple juice has a pH of 3.76. Calculate [H+]
Sa. Ex.
A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate
[H+].
Sa Ex.
What is the pH of a 0.040 M solution of HClO4?
Sa.EX.
Aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the
acid?
Sa Ex.
What is the pH of
(a) 0.028 M solution of NaOH
(b) 0.0011 M solution of Ca(OH)2
Sa.Ex.
What is the concentration of a solution of
(a) KOH for which pH is 11.89
(b) Ca(OH)2 for which the pH is 11.68
Q) What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
Start 0.002 M
HNO3 (aq) + H2O (l)
End 0.0 M
0.0 M
0.0 M
H3O+ (aq) + NO3- (aq)
0.002 M 0.002 M
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Q) What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Start 0.018 M
Ba(OH)2 (s)
End 0.0 M
0.0 M
0.0 M
Ba2+ (aq) + 2OH- (aq)
0.018 M 0.036 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6
Weak of Acids and Bases
Weak Acids (HA) and Acid Ionization Constants
HA (aq) + H2O (l)
HA (aq)
H3O+ (aq) + A- (aq)
H+ (aq) + A- (aq)
[H+][A-]
Ka =
[HA]
(7)
Ka is the acid ionization constant
Ka
weak acid
strength
Weak Bases and Base Ionization Constants
NH3 (aq) + H2O (l)
NH4+ (aq) + OH- (aq)
[NH4+][OH-]
Kb =
[NH3]
(8)
Kb is the base ionization constant
Kb
weak base
strength
Solve weak base problems like weak acids
except solve for [OH-] instead of [H+].
Q) What is the pH of a 0.5 M HF solution (at 250C)?
+][F-]
[H
= 7.1 x 10-4
Ka =
HF (aq)
H+ (aq) + F- (aq)
[HF]
HF (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50
0.00
0.00
-x
+x
+x
0.50 - x
x
x
x2
= 7.1 x 10-4
Ka =
0.50 - x
Ka 
H+ (aq) + F- (aq)
x2
= 7.1 x 10-4
0.50
[H+] = [F-] = 0.019 M
[HF] = 0.50 – x = 0.48 M
Ka << 1
0.50 – x  0.50
x2 = 3.55 x 10-4
x = 0.019 M
pH = -log [H+] = 1.72
Q) When can I use the approximation?
Ka << 1
0.50 – x  0.50
When x is less than 5% of the value from which it is subtracted.
x = 0.019
0.019 M
x 100% = 3.8%
0.50 M
Less than 5%
Approximation ok.
Q) What is the pH of a 0.05 M HF solution (at 250C)?
x2
Ka 
= 7.1 x 10-4 x = 0.006 M
0.05
More than 5%
0.006 M
x 100% = 12%
0.05 M
Approximation not ok.
Must solve for x exactly using quadratic equation (not required
by this course) or method of successive approximation.
Solving weak acid ionization problems:
1. Identify the major species that can affect the pH.
•
In most cases, you can ignore the autoionization of
water.
•
Ignore [OH-] because it is determined by [H+].
2. Use ICE to express the equilibrium concentrations in terms
of single unknown x.
3. Write Ka in terms of equilibrium concentrations. Solve for x
by the approximation method. If approximation is not valid,
solve for x exactly.
4. Calculate concentrations of all species and/or pH of the
solution.
Q) What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4?
HA (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122
0.00
0.00
-x
+x
+x
0.122 - x
x
x
x2
= 5.7 x 10-4
Ka =
0.122 - x
Ka 
H+ (aq) + A- (aq)
x2
= 5.7 x 10-4
0.122
0.0083 M
x 100% = 6.8%
0.122 M
Ka << 1
0.122 – x  0.122
x2 = 6.95 x 10-5
x = 0.0083 M
More than 5%
Approximation not ok.
x2
= 5.7 x 10-4
Ka =
0.122 - x
ax2 + bx + c =0
x = 0.0081
HA (aq)
Initial (M)
Change (M)
Equilibrium (M)
x2 + 0.00057x – 6.95 x 10-5 = 0
-b ± b2 – 4ac
x=
2a
x = - 0.0081
H+ (aq) + A- (aq)
0.122
0.00
0.00
-x
+x
+x
0.122 - x
x
x
[H+] = x = 0.0081 M
pH = -log[H+] = 2.09
Percent Ionization
(9)
Ionized acid concentration at equilibrium
percent ionization =
x 100%
Initial concentration of acid
For a monoprotic acid HA
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration
Sa Ex.
0.10 M solution of formic acid (HCHO2) of pH = 2.38.
(a) Calculate Ka for formic acid
(b) What percentage of the acid is ionized in this 0.1 M solution.
Sa. Ex.
A 0.020 M solution of niacin (a week acid) has pH of 3.26.
(a) What percentage of the acid is ionized in this solution?
(b) What is the acid-dissociation constant Ka for niacin?
Sa Ex.
Calculate the pH of 0.20 M solution of HCN. (Ka = 4.9  10-10)
Sa. Ex.
The Ka for niacin is 1.6  10-5. What is the pH of 0.010 M solution of niacin?
The Relationship of the Ionization constants of Acids and Their
Conjugate Bases
HA (aq)
A- (aq) + H2O (l)
H2O (l)
H+ (aq) + A- (aq)
OH- (aq) + HA (aq)
H+ (aq) + OH- (aq)
KaKb = Kw
Ka
Kb
Kw
(10)
Weak Acid and Its Conjugate Base
Kw
Ka =
Kb
Kw
Kb =
Ka
Sa Ex.
Calculate :
(a) The base-dissociation constant Kb for F- if Ka for HF = 6.8  10-4
(b) The acid disociation constant Ka for the ammonium ion NH4+ if Kb for
NH3 = 1.8  10-5.
Acid- Base Equilibria
Dr. Ali Bumajdad
Kw = [H+] [OH-] = 1.0  10-14 (at 25°C)
pH = - log [H+]
[H+] = 10-pH
pOH = - log [OH-]
[OH-] = 10-pOH
Ka  Kb = Kw = 1.0  10-14 (at 25°C)
 Where a is the acid and b is its conjugate base
 As Ka , Kb of its conjugate base
pKa + pKb = pKw = 14 (at 25°C)
Lewis acids and Bases
A Lewis acid is a substance that can accept a pair of electrons
A Lewis base is a substance that can donate a pair of electrons
••
•• + OH
••
acid base
H+
H+ +
acid
••
H
N H
H
base
••
H O H
••
H
+
H N H
H
Lewis Acids and Bases
+
F B
••
H
F
N H
F
H
acid
base
F
F B
F
H
N H
H
No protons donated or accepted!