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Ch16 and 17 - Acids and Bases
Brady & Senese, 5th Ed
2
Index
15.1. Brønsted-Lowry acids and bases exchange protons15.2. Strengths of Brønsted acids and bases follow periodic trends15.3. Lewis acids and bases involve coordinate covalent bonds15.4. Elements and their oxides demonstrate acid-base properties15.5. pH is a measure of the acidity of a solution15.6. Strong acids and bases are fully dissociated in solution
15.1. Brønsted-Lowry acids and bases exchange protons3
Brønsted Acid/Base Reactions Transfer H+
• Products differ by one H+ from the reactants to form conjugate
• Conjugate acid-base pairs differ by one H+.
• HCN(aq) + OH-(aq)
↔H2O(l) + CN-(aq)
• Note that in the conjugate pairs, the acid has one more H+ than its conjugate base
Brønsted
acid
Brønsted
base
conjugate
acid
conjugate
base
15.1. Brønsted-Lowry acids and bases exchange protons 4
Learning Check
conjugate base conjugate acid
HCl
NH3
HC2H3O2
CN-
HF
Cl-
NH4+
C2H3O2-
HCN
F-
• Identify the Conjugate Partner for Each
15.1. Brønsted-Lowry acids and bases exchange protons 5
Your Turn!
How many of the following pairs are conjugate pairs:
i. HCN/CN- ii. HCl/Cl- iii. H2S/S2-
A. 1
B. 2
C. 3
D. None of them are conjugate
15.1. Brønsted-Lowry acids and bases exchange protons 6
Amphoteric Substances
• Amphoteric (aka amphiprotic) substances are able to act either as Brønsted acid or Brønsted base
• The following materials are amphoteric: Water Amino acids Anions of polyprotic acids
15.1. Brønsted-Lowry acids and bases exchange protons 7
Acid/Base Strengths In Aqueous Solution
• Hydronium ion (H3O+) is the strongest acid in solution: stronger acids react completely with water to give H3O+
• Hydroxide ion (OH-) is the strongest possible base in solution: stronger bases react completely with water to give OH-
• This leveling effect causes very strong acids/bases to have nearly equal strength in aqueous solution
15.1. Brønsted-Lowry acids and bases exchange protons 8
Conjugate Pairs Have Reciprocal Strengths
• The stronger the acid, the weaker its conjugate base
• The stronger the base, the weaker its conjugate acid
• Strong acids are ionized 100%: their anions are extraordinarily poor bases- most are essentially neutral
15.2. Strengths of Brønsted acids and bases follow periodic trends 9
Acid Strength
Binary Acids:
• Acidity increases from left to right in a period, increasing electronegativity makes the H-X bond more polar (weaker)
• Acidity increases from top to bottom in a group, the increasing length of the H-X bond leads to poor orbital overlap and a weaker bond
Oxoacids:
• Acidity increases as number of oxygens increases (presence of high EN oxygen draws electrons away from other bonds)
• Oxoacids with at least 2 more oxygens than hydrogens tend to be strong. Others tend to be weak.
15.2. Strengths of Brønsted acids and bases follow periodic trends 10
Learning Check
• Which is stronger?
• H2S or H2O
• CH4 or NH3
• HF or HI
δ-
δ+δ+
δ-δ-
δ+δ+
δ-
δ+
δ-
δ+
15.3. Lewis acids and bases involve coordinate covalent bonds 11
Lewis Acids and Bases
• Lewis acids (electron pair acceptors) molecules & ions with incomplete valence shells molecules or ions with central atoms that can
accommodate additional electrons
• Lewis bases (electron pair donors) molecules & ions that have complete valence shells with
unshared electrons generally donate lone pairs or pi bond electrons
15.3. Lewis acids and bases involve coordinate covalent bonds 12
Lewis Acid/Base Reactions
• Lewis acids accept an electron pair to form coordinate covalent bonds
• Lewis bases donate lone pairs of electron to form coordinate covalent bonds
• Neutralization is the formation of a coordinate covalent bond between the donor and acceptor
15.3. Lewis acids and bases involve coordinate covalent bonds 13
Learning Check
Identify the Lewis acid and base in the following
• NH3 + H+ ↔NH4+
• F- + BF3 ↔ BF4
-
+
:::
: --
15.4. Elements and their oxides demonstrate acid-base properties 14
Hydrated Metal Ions Can Act as Weak Acids
• Electron deficiency of metal cations causes them to inductively attract electrons from the hydrating water molecules
• M(H2O)m n+ + H2O↔M(H2O)m-1OH(n-1)+ + H3O+
• The higher the charge density (greater charge concentrated in smaller radius) the more acidic the metal.
• Acidity increases left to right in a period.• Acidity decreases top to bottom in a group.
volumeioniccharge ionic
density charge
Al(H2O)5OH2+Al(H2O)63+
The acidic behavior of the hydrated Al3+ ion.
H2O H3O+
Electron density drawn toward Al3+
Nearby H2O acts as base
15.4. Elements and their oxides demonstrate acid-base properties 16
Oxides Can Be Acidic or Basic
• Most metal oxides are basic• Non-metal oxides are acidic anhydrides
SO2+H2O →H2SO3
15.5. pH is a measure of the acidity of a solution 17
• Water ionizes to a very small extent (Kw=1.0x10-
14 at room temperature) according to the following reaction: H2O(l) + H2O(l) ↔ H3O+
(aq) + OH-(aq)
• Since water is present in all aqueous solutions, the Kw equilibrium exists in all aqueous solutions. Kw=[H3O+ ][OH-]
Kw = 1.0 x 10-14 at 25°C
• When [H3O+]=[OH-], the solution is neutral.
Auto-ionization of Water (Kw)
15.5. pH is a measure of the acidity of a solution 18
pH and Kw
• p -log• pH is defined for aqueous solutions only, and is
temperature dependent• pH=-log[H3O+]• 10-pH = [H3O+]• It derives from the auto ionization of water.
Kw=[H3O][OH-] pKw= pH + pOH
• pH>7 is basic; pH=7 is neutral; pH<7 is acidic
Figure 18.5
The pH values of some familiar
aqueous solutions
pH = -log [H3O+]
15.5. pH is a measure of the acidity of a solution 20
Indicators Help Us Estimate pH
15.5. pH is a measure of the acidity of a solution 21
Learning Check
[OH-] [H3O+] pH
3.2 × 10-3 M
2.3 × 10-5 M
1.5 × 10-2 M
2.55 × 10-6 M
Complete the following with the missing data at 25 deg C
4.64
12.17
5.593
11.513.1×10-12
4.3×10-10
6.7×10-13
3.9×10-9
15.6. Strong acids and bases are fully dissociated in solution 22
Strong Acids Ionize 100% in Water
• As the substances are placed into water, they form H3O+ .
• Water’s contribution to [H3O+] is negligible• The pH can be calculated from the concentration
of H3O+ produced by the strong acid• The reaction of strong acids occurs irreversibly, so
we show the reaction with a → instead of using a double arrow
15.6. Strong acids and bases are fully dissociated in solution 23
Learning Check
What is the pH of 0.1M HCl
•HCl(aq) + H2O(l) →H3O+(aq) + Cl-
(aq)
•0.1 - 0 0 I•-0.1 - 0.1 0.1 C•0 - 0.1 0.1 end•pH = -log(0.1) = 1.0
15.6. Strong acids and bases are fully dissociated in solution 24
Strong Bases Dissociate 100% In Water
• They are strong electrolytes that form OH- when dissolved
• pOH can be calculated from the [OH-] from the solution
• Water’s contribution is negligible if the base is sufficiently concentrated [OH-]>10-7M
15.6. Strong acids and bases are fully dissociated in solution 25
Learning Check
What is the pH of 0.5M Ca(OH)2?
• Ca(OH)2(aq) → Ca2+(aq) + 2OH-
(aq)
• 0.5 0 0 I•-0.5 +0.5 + 0.5×2 C•0 0.5 1 end•pOH = -log(1) =0.0•pH = 14.0
26
Index
16.1. Ionization constants can be defined for weak acids and bases 16.2. Calculations can involve finding or using Ka and Kb 16.3. Salt solutions are not neutral if the ions are weak acids
or bases 16.4. Simplifications fail for some equilibrium calculations16.5. Buffers enable the control of pH 16.6. Polyprotic acids ionize in two or more steps 16.7. Acid-base titrations have sharp changes in pH at the
equivalence point
16.1. Ionization constants can be defined for weak acids and bases 27
Weak Acids & (Ka)
• Weak acids only partially ionize
• The equilibrium constant Ka is used to indicate the degree of ionization
• Ka are tabulated in table 16.1, as are pKa
• Ka=10-pKa pKa=-log(Ka)
• Acid strength increases as Ka increases and pKa decreases
• For Conjugate Pairs Kw=Ka x Kb
(aq)(aq)3 (l)2(aq) AOHOHHA
16.1. Ionization constants can be defined for weak acids and bases 28
Bases & Kb
• Weak bases only partially ionize
• The equilibrium constant Kb is used to indicate the degree of ionization
• Kb and pKb are tabulated in table 16.2
• Kb=10-pKb pKb=-log(Kb)
• Base strength increases as Kb increases and pKb decreases
• For Conjugate Pairs Kw=Ka x Kb
)()()(2)( aqaqlaq OHBHOHB
16.2. Calculations can involve finding or using Ka and Kb 29
Determining the pH Of Aqueous Weak Acid Solutions
• Dominant equilibrium is Ka reaction
write the net ionic equation look up the Ka value for the acid set up ICE table solve for x
• Calculate pH from the hydronium concentration at equilibrium
16.2. Calculations can involve finding or using Ka and Kb 30
Simplifications: Dropping x Term
• Solving ICE tables sometimes requires the quadratic equation
• Other times simplifications may be made• If the K for the reaction is small, the degree of
ionization will also be small• The concentration of a reactant may sometimes be
simplified to the initial concentration• You must perform a proof to show that the
dropped x was justified (less than 5% error)
05.0binomial ofconstant
x term dropped ?
16.2. Calculations can involve finding or using Ka and Kb 31
Learning Check
Determine the pH of 0.1M solutions of:
• HC2H3O2
• Ka=1.8×10-5
• HCN
• Ka=6.2×10-10
HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2
-(aq)
I
C
E
HCN + H2O↔ H3O+(aq) + CN-
(aq)
I
C
E
0.1M N/A 0 0
-x +x +x
(0.1-x) ≈ 0.1 N/A x
-x
x
0.1M N/A 0 0
-x +x +x
(0.1-x)≈0.1 N/A x
-x
x
52
108.10.1
x
pH=2.87
102
102.60.1
x
pH=5.10
X=1.34(10-3)M
X=7.87(10-6)M
16.2. Calculations can involve finding or using Ka and Kb 32
Determining The pH Of Base Solutions:
• Dominant equilibrium is Kb reaction write the net ionic equation set up ICE table for starting quantities solve for x
• Calculate pOH from the [OH-] at equilibrium, and convert to pH
16.2. Calculations can involve finding or using Ka and Kb 33
Learning Check:
Determine the pH of 0.1M solutions of:• N2H4
• Kb=1.7×10-6
• NH3
• Kb=1.8×10-5
N2H4 + H2O ↔ OH-(aq) + N2H5
+(aq)
I
C
E
NH3 + H2O↔ OH-(aq) + NH4
+(aq)
I
C
E
0.1M N/A 0 0
-x +x +x
(0.1-x) ≈ 0.1 N/A x
-x
x
0.1M N/A 0 0
-x +x +x
(0.1-x) ≈ 0.1 N/A x
-x
x
62
107.10.1x
pOH=3.38
52
108.10.1
x
pH=11.13
pH=10.62
pOH=2.87
X=4.12(10-4)M
X=1.34(10-3)M
16.2. Calculations can involve finding or using Ka and Kb 34
Percent Ionization
• Weak acids ionize some amount less than 100%.
• Generally calculated from Ka and ICE table
• Percentage ionized varies with starting concentration, even for the same acid
100[HA]
[A-]ionization %
16.2. Calculations can involve finding or using Ka and Kb 35
Learning Check
Determine the % ionization of 0.2M solution of HC2H3O2
• Ka=1.8×10-5
HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2
-(aq)
I
C
E
0.2M N/A 0 0
-x +x +x
(0.2-x) ≈ 0.2 N/A x
-x
x
52
108.10.2
x
x=1.90×10-3M
%1000.2
x 100
literper moles availableliterper ionized moles
ionization %
0.95 % ionized
16.2. Calculations can involve finding or using Ka and Kb 36
Learning Check
Determine the % ionization of 0.1M solution of HC2H3O2
• Ka=1.8×10-5HC2H3O2 + H2O↔ H3O+
(aq) + C2H3O2-(aq)
I
C
E
0.1M N/A 0 0
-x +x +x
(0.1-x) ≈ 0.1 N/A x
-x
x
52
108.10.1x
x=1.34 x 10-3M
%1000.1x 100
literper moles availableliterper ionized moles
ionization %
1.3 % ionized
16.5. Buffers enable the control of pH 37
Buffers enable the control of pH
• Are a combination of a weak acid and its conjugate weak base or a weak base and its conjugate acid
• Are a special case of the common ion effect (the presence of the conjugate base inhibits the ionization of the Brønsted acid)
• Are resistant to changes in the pH Any acid added to the mixture is reacted by the base of the
buffer pair added bases are reacted by the conjugate acid
16.5. Buffers enable the control of pH 38
Simplification Of Buffer pH Calculations – Henderson-Hasslebach Equation
[HA]]][AO[H
Ka
AOHOHHA
3
(aq)(aq)3(l)2(aq)
)[HA]
][Alog()Olog(Hlog(Ka) 3
)log(log(Ka)])Olog([H [HA]][A
3
][
][log
HA
ApKapH
16.5. Buffers enable the control of pH 39
Learning Check
What is the pH of a buffer made from 0.5M HF (Ka=6.8×10-4) and 0.2M NaF?
][][
logHAA
pKapH
]5.0[]2.0[
log17.3pH
pKa=-log(6.8×10-4)
pH=2.8
16.5. Buffers enable the control of pH 40
Buffer Characteristics
• Buffer capacity is the ability to compensate for added acid or base, defined for each component acid buffer capacity=moles of conjugate base present base buffer capacity= moles of acid present.
• If acid is added, remove from base and add to the acid of the pair
• If base is added, remove from acid of the pair and add to the acid of the pair.
acid mol HAmolacid molA mol
logpKapH
base mol - HAmolbase molA mol
logpKapH
16.5. Buffers enable the control of pH 41
Your Turn!
Which of the following could be used as a buffering pair?
A. NH4Cl/NH3
B. HCl/NaCl
C. NaOH/H2O
D. All of these
16.5. Buffers enable the control of pH 42
Buffer CharacteristicsConsider 500. mL of a sodium acetate/acetic acid buffer in which
the concentration of sodium acetate is 0.500M and that of acetic acid is 0.100M. What is the pH after 10.0 mL of 0.100M NaOH is added? (Ka acetic acid= 1.8×10-5)
16.5. Buffers enable the control of pH 43
Selecting A Buffer
• To choose buffer system, try to find a system whose pKa is close to desired system pH (pH = pKa is ideal)
• Then, calculate necessary ratio of components to obtain desired pH.
• Consider toxicity of the materials!
16.5. Buffers enable the control of pH 44
Learning Check
Choose a buffer system to maintain a pH of 5.5. Describe its composition completely. select acid with pKa≈5.5.
thus, look for Ka ≈3.2×10-
6
H2C8H4O4, phthalic acid has Ka2=3.9×10-6
. We could use NaHC8H4O4 and Na2C8H4O4.
448
2448
OH HCmolOHC mol
log41.55.5
HAmolA mol
logpK5.5 a
23.1OH HCmol
OHC mol448
2448
16.6. Polyprotic acids ionize in two or more steps 45
Polyprotic Acids
• Have more than one ionizable hydrogen• Each successive ionization has a specific
ionization constant (Ka)
• Each successive proton is significantly less acidic than the last
• For H3PO4: H3PO4(aq) + H2O(l) ↔H3O+
(aq) + H2PO4-(aq) Ka1
H2PO4-(aq) + H2O(l) ↔H3O+
(aq) + HPO42-
(aq) Ka2
HPO42-
(aq) + H2O(l) ↔H3O+(aq) + PO4
3-(aq) Ka3
Ka1 >> Ka2 >> Ka3
16.6. Polyprotic acids ionize in two or more steps 46
Learning Check:
What is the pH of a 0.1M solution of H3PO4? Ka1=7.1×10-3; Ka2=6.3×10-8; Ka3=4.5×10-13
16.7. Acid-base titrations have sharp changes in pH at the equivalence point 47
Titration
• The controlled addition of one substance (titrant) to a known quantity of another substance (analyte) until the stoichiometric requirements are met
• Equivalence point - the volume of titrant needed to achieve an equimolar concentration of titrant and of titrate
• Endpoint - the volume of titrant necessary to achieve the stoichiometric ratio of reactants
• The endpoint may be the last of several equivalence points in the case of polyprotic acids
Curve for a strong acid-strong base titration
Curve for a weak acid-
strong base titration
Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH
[HPr] = [Pr-]
pH = 8.80 at equivalence point
pKa of HPr = 4.89
methyl red
Calculating the pH During a Weak Acid-Strong Base Titration
PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000M NaOH:
(a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mLPLAN: The amounts of HPr and Pr- will be changing during the titration.
Remember to adjust the total volume of solution after each addition.
SOLUTION: (a) Find the starting pH using the methods of Chapter 18.
Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] [Pr-] = x = [H3O+]
x (1.3x105)(0.10) x = 1.1x10-3 ; pH = 2.96
(b)
Before addition
Addition
After addition
0.004000
0.003000
0.0030000.001000
0 -
-
-0
-
- -
HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)Amount (mol)
Calculating the pH During a Weak Acid-Strong Base Titration
continued
[H3O+] = 1.3x10-50.001000 mol
0.003000 mol= 4.3x10-6M pH = 5.37
(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be (0.004000 mol)
0.04000L + 0.04000L= 0.05000M
Ka x Kb = Kw Kb = Kw / Ka = 1.0x10-14 / 1.3x10-5 = 7.7x10-10
[H3O+] = Kw / = 1.6x10-9M
Kbx[Pr ] pH = 8.80
(d) 50.00mL of NaOH will produce an excess of OH-.
mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100molM = (0.001000mol)
(0.09000L)
M = 0.01111[H3O+] = 1.0x10-14/0.01111 = 9.0x10-11M
pH = 12.05
Curve for a weak base-strong acid
titration
Titration of 40.00mL of 0.1000M NH3 with 0.1000M HCl
pH = 5.27 at equivalence point
pKa of NH4+ =
9.25
pKa = 7.19
pKa = 1.85
Curve for the titration of a weak polyprotic acid.
Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH