Chapter 12 Acids and Bases
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Transcript of Chapter 12 Acids and Bases
Chapter 12Acids and Bases
12.1 The Nature of Acids and Bases 12.2 Acid Strength 12.3 The pH Scale 12.4 Calculating the pH of Strong Acid Solutions 12.5 Calculating the pH of Weak Acid Solutions 12.6 Bases 12.7 Polyprotic Acids 12.8 Acid-Base Properties of Salts 12.9 Acid Solutions in Which Water Contributes to the H+
Concentration (skip)12.10 Strong Acid Solutions in Which Water Contributes to
the H+ Concentration (skip)12.11 Strategy for Solving Acid-Base Problems: A Summary
Acids and Bases and Their Reactions
Definitions1. Arrhenius Acids and Bases Acids are
• H+ donors• Bases are OH- donors
2. Arrhenius Broadened Definition • Acids increase H+ concentration or [H+]
increases• Bases increase OH- concentration or [OH-]
increases
– Brønsted-Lowry Acids and Bases (1923)• Acids donate H+
• Bases accept H+
Arrhenius 1903
Nobel Prize
Acids-BasesBrønsted-Lowry Acids and Bases
A Brønsted-Lowry acid is a substance that can donate a hydrogen ion (aka H+, proton).
A Brønsted-Lowry base is a substance that can accept a hydrogen ion.
Acids and bases occur as conjugate acid - base pairs.
Conjugate Base - subtract an H+ from the acidConjugate Acid add H+ to the base
Examples
1.OH– is the conjugate base of
2.H2O is the conjugated base of
3.H2O is the conjugated acid of
4.H3O+ (or often shown as H+) is the conjugate acid of
Pairs 1 2 2 1
Acetic Acid
Point of View #1
acid base Conjugate acid of H2O
Conjugate base of CH3CO2H
CH3CO2H + H2O H3O+ + CH3CO2-
Point of View #2
acid baseConjugate base of H3O+
Conjugate acid
of CH3CO2-
CH3CO2H + H2O H3O+ + CH3CO2-
basebaseacid acid
Acetate Ion
CH3CO2H + H2O H3O+ + CH3CO2-
Nomenclature
When H+ is hydrated, it is H3O+ and is called a hydronium ion. A hydronium ion has the same molecular geometry as NH3.
111.7°
+
• There is Competition for the proton between two bases H2O and A–
• If H2O is a much stronger base than A– the equilibrium lies far to the right.
• If A– is a much stronger base than H2O the equilibrium lies far to the left.
HA + H2O H3O+ + A –
HA = generic acid
Acid Strength: graphical representation of the behavior of acids of different strengths in aqueous solution.
A strong acid: equilibrium lies far to the right
HA + H2O H3O+ + A –
A weak acid: equilibrium lies far to the left
HA + H2O H3O+ + A –
A weak acid yields a relatively strong conjugate base
Relationship of acid strength and conjugate base strength
HA + H2O H3O+ + A-
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Ka =[H3O
+][A−]
[HA]
HA + H2O H3O+ + A–
HA H+ + A–
SAME AS
Ka is the Acidity constant
H2O + H2O H3O+ + OH-
Pairs 1 2 2 1basebaseacid acid
Autoionization of H2O
Point of View #2
acid baseConjugate base of H3O+
Conjugate acid
of OH-
H2O + H2O H3O+ + OH-
Point of View #1
acid base Conjugate acid of H2O #2
Conjugate base of H2O #1
H2O + H2O H3O+ + OH-# 1 # 2
Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions
Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions
H2O + NH3 NH4+ + OH-
1.) Water in NH3 serves as an acid
acid acid basebase
2.) Water in acetic acid serves as a base
H2O + CH3CO2H H3O+ + CH3CO2-
base acid acid base
Conjugate acid
of H2O
Conjugate base
of acetic acid
Water as an Acid and a Base
Kw = [H3O+][OH-] = [H+][OH-] [H3O+][OH-]
[H2O(l)]2
Autoionization of water:
2 H2O (l) H3O+ (aq) + OH- (aq)
Kw = 1.0 x 10-14 (at 25oC)
In pure water [H+] = [OH-]
Kw = [H3O+][OH-]
NH4+ H+ + NH3
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Ka =[H+][NH3]
[NH4+]
NH3 + H2O OH- + NH4+
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Kb =[OH−][NH4
+]
[NH3]
Ka/Kb/Kw
NH4+ H+ + NH3
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Ka =[H+][NH3]
[NH4+]
NH3 + H2O OH- + NH4+
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Kb =[OH−][NH4
+]
[NH3](1)
Ka/Kb/Kw
H2O OH- + H+
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Kw = KaKb =[OH−][H+]
(1)
The pH ScalepH = -log10[H3O+]
pH < 7 acidic solution
[H3O+] > [OH-]
pH = 7 neutral solution
[H3O+] = [OH-]pH > 7 basic solution
[H3O+] < [OH-]
Sig figs: for logs: the number of decimal places in the log is equal to the number of sig figs in the original number
pH = -log10[H+]SAME AS
Calculate the pH of Strong Acid-Base Solutions
pH = -log10[H3O+]EXAMPLE
Calculate the pH (at 25oC) of an aqueous solution that has an
OH-(aq) concentration of 1.2 x 10-6 M (i.e., mol/liter)
Solution
The concentration of H+ (aq) is
[H+][OH-] = KW
[H+] = KW/[OH-] = 10-14/1.2x10-6 = 8.3 x 10-9
pH = -log[8.3 x 10-9] = 8.1
Strong Acids
A strong acid is one that dissociates completely in water to produce H+(aq).
E.g., Hydrochloric acid (HCl) is a strong acid:
HCl (aq) → H+ (aq) + Cl- (aq) (reaction essentially complete)
Dissolving 0.10 mol of HCl in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for H+(aq).
[H+][OH-] = Kw
[OH-] = Kw / [H+] = 10-14/10-1 = 10-13
H2O (l) H+ (aq) + OH- (aq)
Weak Acids
Write the major species in solution and Ka values. Which is dominant?
#1: HA H+ + A –
#2: H2O H+ + OH –
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Kw =[H+ ][OH −]
Compare the value for Ka and Kw. Which is
larger? This is the dominate source of H+
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Ka =[H+ ][A −][ ]HA
Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. The concentration of CH3CO2H is 1 M (initial, or total). The Ka for acetic acid is 1.8 x 10-5
Estimate major species in solution CH3CO2H (a weak acid) and H2O.
CH3CO2H H+ + CH3CO2─ Ka = 1.8 x 10-5
H2O H+ + OH – Kw = 1.0 x 10-14
Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.
CH3CO2H H+ + CH3CO2─
Initial 1.0M ~ 0 0
Change - y + y + y
Equilibrium 1.0 – y y y
Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.
CH3CO2H H+ + CH3CO2─
Initial 1.0M ~ 0 0
Change - y + y + y
Equilibrium 1.0 – y y y
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Ka =( )( )y y
(1.0 - )yassume y << 1.0
Ka =(y)(y)
(1.0)
Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium.
CH3CO2H H+ + CH3CO2─
Initial 1.0M ~ 0 0
Change - y + y + y
Equilibrium 1.0 – y y y
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Ka =( )( )y y
(1.0 - )yassume y << 1.0
Ka =(y)(y)
(1.0)
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Ka =1.8x10−5 =(y)(y)
(1.0)
y = 1.8x10−5
y = 4.3x10−3
check assumption : is y << 1.0?
Effect of dilution on the % dissociation and [H+] and pH
HA H+ + A –
The more dilute the weak acid solution, the greater the percent dissociation
Le Chatelier’s Priniciple:
If a chemical system at equilibrium experiences a change in concentration (or temperature, volume, or partial pressure) then the system shifts to counteract the imposed change.
Strong Acids and Bases
Strong BasesA strong base reacts completely with water to produce OH-(aq) ions.
Sodium hydroxide (NaOH) is a strong base:
Others are KOH–, NH2– (amide ion) and H– (Hydride ion)
NaOH (s) → Na+ (aq) + OH- (aq) (reaction essentially complete)
Dissolving 0.10 mol of NaOH in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for OH- (aq). From this you can calculate pH and pOH.
[H+][OH-] = Kw
[OH-] = 10-1 [H+] = 10-13 pOH = 1 pH = 13
[H+] = Kw/[OH-] =10-14/10-1 = 10-13
Weak Bases
B (aq) + H2O (l) BH+ (aq) + OH– (aq)B = Base
Kb is the basicity constant
Kb = [BH+][OH–]/[B]
Calculations for solutions of weak bases are similar to those for weak acids. Bases (B) compete with OH–, a very strong base, for H+ ions.
B (aq) + H2O (l) BH+ (aq) + OH– (aq)
Acid-Base Equilibria
Base Strength– strong acids have weak
conjugate bases– weak acids have strong
conjugate bases
pKa + pKb = pKwThis equation applies to an acid and its conjugate base.
The strength of a base is inversely related to the strength of its conjugate acid; the weaker the acid, the stronger its conjugate base, and vice versa
Weak Bases
NH3 is a base Kb = 1.8 x 10-5
NH3 + H20 NH4+ + OH-
acid1 base2 acid2 base1
NH4+
is an acid Ka = 5.6 x 10-10
NH4+ NH3 + H+
acid1 base1
Weak Bases
NH3 is a base Kb = 1.8 x 10-5
NH3 + H20 NH4+ + OH-
NH4+
is an acid Ka = 5.6 x 10-10
NH4+ NH3 + H+
Ka Kb = Kw = (5.6 x 10-10) (1.8 x 10-5) = 10-14
pKa + pKb = pKw = (-9.25) + (-4.75) = -14
Weak Bases with Weak Acids
NH4+
is an acid Ka(am) = 5.6 x 10-10
change the direction of the aa reaction and add them together
NH4+ + CH3COO- NH3 + CH3COOH
K = Ka(am)/Ka(aa) = 5.6 x 10-10/1.8 x 10-5
=3.1 x 10-5
CH3COOH is an acid Ka(aa) = 1.8 x 10-5
CH3COOH CH3COO- + H+
NH4+ NH3 + H+
Assuming 0.1M NH3 initial, calculate the pH of the resulting solution
H20 + NH3 NH4+ + OH-
Init. conc. 0.1M 0 ~0
Change - y + y + y
Equil. conc. 0.1– y y y
Assuming 0.1M NH3 initial, calculate the pH of the resulting solution
H20 + NH3 NH4+ + OH-
Init. conc. 0.1M 0 ~0
Change - y + y + y
Equil. conc. 0.1– y y y
Assuming 0.1M NH3 initial, calculate the pH of the resulting solution
H20 + NH3 NH4+ + OH-
- 24
b 3
5b 3
25
b
- -34
[NH ][OH ] yK
[NH ][1] 0.1 y
K for NH 1.8x10
assume y is small
yK 1.8x10
0.1
y [OH ] [NH ] 1.3x10
+
−
−
+
= =−
=
= =≈
= = =
[H+] = Kw/[OH-] =10-14/y
pH=-log[H+]
Polyprotic Acids
Note: Sulfuric acid is a strong acid in its first dissociation step and a weak acid in its second step.
H2SO4 (aq) → H+ (aq) + HSO4– (aq) Ka1 ≈ 1.0x102
HSO4– (aq) H+ (aq) + SO4
2– (aq) Ka2 ≈ 1.2x10-2
It is always easier to remove the first proton in a polyprotic acid than the second. That is, Ka1 > Ka2 > Ka3
Polyprotic Acids1. Polyprotic acids have more than one ionizable proton.2. The protons are removed in steps, not all at once.3. The extent of the steps can be calculated sequentially.4. From the successive acidity constants, the equilibrium concentrations of all
species present can be calculated at any value of the pH.
H2SO3 (aq) + H2O (l) → H3O+ (aq) + HSO3- (aq)
Ka1 = 103
HSO3- (aq) + H2O (l) H3O+ (aq) + SO3
2- (aq)
Ka2 = 1.2 x 10-2
Polyprotic Acids
• pH calculations for solutions of Polyprotic acids appear complicated, most common cases (with weak acids) are surprising straightforward.
• Setup equations, K values, set up ICE type table, assume dissociation is small (check assumption) and solve.
• For typical weak Polyprotic acids.
Ka1 >> Ka2 >> Ka3
• The first dissociation step dominates the H+ concentration and thus the pH
A plot of the fractions of H2CO3, HCO-3 and CO3
2-
At pH = 9.00 H2CO3 ≈ 0%, HCO-3 = 95% and CO3
2- = 5%
At pH = 10.00 H2CO3 ≈ 0%, HCO-3 = 68% and CO3
2- = 32%
Acid-Base Properties of Salts
“Hydrolysis”
Hydrolysis is a Brønsted-Lowry Acid and Base Reaction
Anion: A─ + H2O ↔ HA + OH─
Cation: M+2 + H2O ↔ H3O+ + M(OH)+
E.g., Cations Ni+2 and Fe+2
• Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7
• When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7)
• However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic
• Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.• These interactions between salts and water are called hydrolysis
Example problem:
Suppose a 0.1 mole solution sodium acetate is dissolved in 1 liter of water. What is the pH of the solution?
Init. conc. 0.1M 0 ~0
∆ conc. - y + y + y
Equil. conc. 0.1– y y y
1. Find Kb
2. Find [OH-]
3. Find [H+]
4. Find pH
CH3CO2– + H2O ↔ CH3CO2H + OH–
base baseacid acid
Example problem:
What is the pH of the solution?
CH3CO2– + H2O ↔ CH3CO2H + OH–
Init. conc. 0.1M 0 ~0
∆ conc. - y + y + y
Equil. conc. 0.1– y y y
1. Find Kb
2. Find [OH-]
3. Find [H+]
4. Find pH
Ka x Kb = Kw
• Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7
• When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7)
• However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic (Problem: found at 0.1M NaAc, pH =8.89)
• a non-hydrolyzed cation (Na+)• a hydrolyzed anion (acetate ion)
• Other salts behave similarly, NH4Cl and AlCl3 give acid solutions.
• a non-hydrolyzed anion (Cl-)
• a hydrolyzed cation (NH4+ or Al+3)
• These interactions between salts and water are called hydrolysis
Hydrolysis of Result
Anions
Cations
Raise pH
Lower pH
Non-Hyrolyzed Ions (a few)
7 Anions, not hydrolyzed
Cl –, Br –, I –, HSO4–, NO3
–, ClO3–, ClO4
–
10 Cations, not hydrolyzed
Li+, Na+, K+, Rb+, Sc+, Mg++, Ca++, Sr++, Ba++, Ag+
Hydrolysis of Result
Anions
Cations
Raise pH
Lower pH
Predict pH of salts in water (relative pH)
Na3PO4 is basic (raised pH)
(a non hydrolyzed cation and a hydrolyzed anion)
FeCl3 is acidic (lowers pH)
(a hydrolyzed cation and a non hydrolyzed anion)
Na3PO4?
FeCl3?
Summary
Acid-Base Properties of Salts
“Hydrolysis”
Summary
• Acid-Base Equilibrium Problems– Which major species are present– Does a reaction occur that can be assumed to go to
completion?– Which equilibrium dominates the solution?– Set up ICE type table– Solve for equilibrium concentrations using know K
values– Check any simplifying assumptions– Typically solve for pH or % species in solution
Chapter 7Acids and Bases
7.1 The Nature of Acids and Bases 7.2 Acid Strength 7.3 The pH Scale 7.4 Calculating the pH of Strong Acid Solutions 7.5 Calculating the pH of Weak Acid Solutions 7.6 Bases 7.7 Polyprotic Acids 7.8 Acid-Base Properties of Salts 7.9 Acid Solutions in Which Water Contributes to the H+
Concentration (skip)7.10 Strong Acid Solutions in Which Water Contributes to the
H+ Concentration (skip)7.11 Strategy for Solving Acid-Base Problems: A Summary