Jeffrey MackCalifornia State University,
Sacramento
Chapter 17
Principles of Chemical Reactivity:
The Chemistry of Acids and Bases
• In Chapter 3, you were introduced to two definitions of acids and bases: the Arrhenius and the Brønsted–Lowry definition.
• Arrhenius acid: Any substance that when dissolved in water increases the concentration of hydrogen ions, H+.
• Arrhenius base: Any substance that increases the concentration of hydroxide ions, OH, when dissolved in water.
• A Brønsted–Lowry acid is a proton (H+) donor.• A Brønsted–Lowry base is a proton acceptor.
Acids & Bases: A Review
• Generally divide acids and bases into STRONG or WEAK ones.
STRONG ACID: HNO3(aq) + H2O(liq) H3O+(aq) + NO3
-(aq)HNO3 is about 100% dissociated in water.
Strong & Weak Acids/Bases
HNO3, HCl, H2SO4 and HClO4 are classified as strong acids.
Strong & Weak Acids/Bases
• Strong Base: 100% dissociated in water.
NaOH(aq) Na+(aq) + OH-(aq)
Other common strong bases include KOH and Ca(OH)2.
CaO (lime) + H2O
Ca(OH)2 (slaked lime)CaO
Strong & Weak Acids/Bases
• Weak base: less than 100% ionized in waterAn example of a weak base is ammoniaNH3(aq) + H2O(liq) NH4
+(aq) + OH-(aq)
Strong & Weak Acids/Bases
Weak acids are much less than 100% ionized in water.Example: acetic acid = CH3CO2H
Strong & Weak Acids/Bases
• Proton donors may be molecular compounds, cations or anions.
3
3 2 3 3
4 2 3 3
22 3 3
HNO (aq) H O(l) NO (aq) H O (aq)
NH (aq) H O(l) NH (aq) H O (aq)
HCO (aq) H O(l) CO (aq) H O (aq)
The Brønsted–Lowry Concept of Acids & Bases Extended
• Proton acceptors may be molecular compounds, cations or anions.
3
3
3 2
2
2 25
3
2 6
23 2
NH (aq) H O(l) NH (aq) OH (aq)
Al H O OH (aq) H O(l)
Al H O (aq) OH (aq)
CO (aq) H O(l) HCO (aq) OH (aq)
The Brønsted–Lowry Concept of Acids & Bases Extended
Using the Brønsted definition, NH3 is a BASE in water and water is itself an ACID
Proton acceptor
Proton donor
The Brønsted–Lowry Concept of Acids & Bases Extended
• Acids such as HF, HCl, HNO3, and CH3CO2H (acetic acid) are all capable of donating one proton and so are called monoprotic acids.
• Other acids, called polyprotic acids are capable of donating two or more protons.
The Brønsted–Lowry Concept of Acids & Bases Extended
• A conjugate acid–base pair consists of two species that differ from each other by the presence of one hydrogen ion.
• Every reaction between a Brønsted acid and a Brønsted base involves two conjugate acid–base pairs
Conjugate Acid–Base Pairs
Conjugate Acid–Base Pairs
Water Autoionization and the Water Ionization Constant, Kw:
The water autoionization equilibrium lies far to the left side. In fact, in pure water at 25 °C, only about two out of a billion (109) water molecules are ionized at any instant.
Even in pure water, there is a small concentration of ions present at all times. [H3O+] = [OH] = 1.00 107
2 2 3H O(l) H O(l) H O (aq) OH (aq)
Water & the pH Scale
H2O can function as both an ACID and a BASE.In pure water there can be AUTOIONIZATION.
Equilibrium constant for autoionization = Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 °C
Water & the pH Scale
• In a neutral solution, [H3O+] = [OH]• Both are equal to 1.00 10 7 M• In an acidic solution, [H3O+] > [OH]• [H3O+] > 1.00 10 7 M and [OH] < 1.00 10
7 M• In a basic solution, [H3O+] < [OH]• [H3O+] < 1.00 10 7 M and [OH] > 1.00 10
7 M
Water & the pH Scale
The pH Scale
• The pH of a solution is defined as the negative of the base (10) logarithm (log) of the hydronium ion concentration.
pH = log[H3O+]
• In a similar way, we can define the pOH of a solution as the negative of the base - 10 logarithm of the hydroxide ion concentration.
pOH = log[OH]pH + pOH = pKw = 14
The pH Scale
• The concentration of acid, [H3O+] is found by taking the antilog of the solutions pH.
• In a similar way, [OH] can be found from:
The pH Scale
Once [H3O+] is known, [OH] can be found from:
And vice versa.
The pH Scale
• In Chapter 3, it was stated that acids and bases can be divided roughly into those that are strong electrolytes (such as HCl, HNO3, and NaOH) and those that are weak electrolytes (such as CH3CO2H and NH3)
• In this chapter we will discuss the quantitative aspects of dissociation of weak acids and bases.
• The relative strengths of weak acids and bases can be ranked based on the magnitude of individual equilibrium constants.
Equilibrium Constants for Acids & Bases
• Strong acids and bases almost completely ionize in water (~100%):
Kstrong >> 1(product favored)
• Weak acids and bases almost completely ionize in water (<<100%):
Kweak << 1(Reactant favored)
Equilibrium Constants for Acids & Bases
• The relative strength of an acid or base can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general acid HA, we can write:
Conjugate acid
Conjugate base
Equilibrium Constants for Acids & Bases
• The relative strength of an acid or base can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general base B, we can write:
Conjugate base
Conjugate Acid
Equilibrium Constants for Acids & Bases
Acids ConjugateBases
Increase strength
Increase strength
Ionization Constants for Acids/Bases
• The strongest acids are at the upper left. They have the largest Ka values.
• Ka values become smaller on descending the chart as the acid strength declines.
• The strongest bases are at the lower right. They have the largest Kb values.
• Kb values become larger on descending the chart as base strength increases.
Equilibrium Constants for Acids & Bases
• The weaker the acid, the stronger its conjugate base: The smaller the value of Ka, the larger the value of Kb.
• Aqueous acids that are stronger than H3O+ are completely ionized.
• Their conjugate bases (such as NO3) do not
produce meaningful concentrations of OH ions, their Kb values are “very small.”
• Similar arguments follow for strong bases and their conjugate acids.
Equilibrium Constants for Acids & Bases
Acid HCO3 HClO HF
Ka 4.8 1011 3.5 108 7.2 104
Base CO32 ClO F
Kb 2.1 104 2.9 107 1.4 1011
Equilibrium Constants for Acids & Bases
Equilibrium Constants for Acids & Bases
Ka Values for Polyprotic Acids
In general, each successive dissociation produces a weaker acid.
2 2 3
7a(1)
22 3
19a(2)
H S(aq) H O(l) H O (aq) HS (aq)
K 1 10
HS (aq) H O(l) H O (aq) S (aq)
K 1 10
Equilibrium Constants for Acids & Bases
Logarithmic Scale of Relative Acid Strength, pKa
• Many chemists use a logarithmic scale to report and compare relative acid strengths.
pKa = log(Ka)
The lower the pKa, the stronger the acid.
Acid HCO3 HClO HF
pKa 10.32 7.46 3.14
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an Acid and Its Conjugate Base
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an Acid and Its Conjugate Base
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an Acid and Its Conjugate Base
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an Acid and Its Conjugate Base
3 2
a b 3 w2
H O HS H S OHK K H O OH K
H S HS
Equilibrium Constants for Acids & Bases
Relating the Ionization Constants for an Acid and Its Conjugate Base
When adding equilibria, multiply the K values.
Equilibrium Constants for Acids & Bases
Acid–Base Properties of Salts
Anions that are conjugate bases of strong acids (for examples, Cl or NO3
.
These species are such weak bases that they have no effect on solution pH.
3 2NO (aq) H O(l) No Reaction
Acid–Base Properties of Salts
Anions such as CO3 that are the conjugate
bases of weak acids will raise the pH of a solution.
Hydroxide ions are produced via “Hydrolysis”.
23 2 3CO (aq) H O(l) HCO (aq) OH (aq)
Acid–Base Properties of Salts
Anions such as CO3 that are the conjugate
bases of weak acids will raise the pH of a solution.
Hydroxide ions are produced via “Hydrolysis”.A partially deprotonated anion (such as HCO3
) is amphiprotic. Its behavior will depend on the other species in the reaction.
23 2 3CO (aq) H O(l) HCO (aq) OH (aq)
Acid–Base Properties of Salts
Alkali metal and alkaline earth cations have no measurable effect on solution pH.
Since these cations are conjugate acids of strong bases, hydrolysis does not occur.
2Na (aq) H O(l) No Reaction
Acid–Base Properties of Salts
Basic cations are conjugate bases of acidic cations such as [Al(H2O)6]3+.Acidic cations fall into two categories: (a) metal cations with 2+ and 3+ charges and (b) ammonium ions (and their organic derivatives).All metal cations are hydrated in water, forming ions such as [M(H2O)6]n+.
3
2 26
2
2 35
4
Al H O (aq) H O(l)
Al H O (OH ) (aq) H O (aq)
Ka 7.9 10
Acid–Base Properties of Salts
Salt pH of (aq) solution
CaCl2
NH4Br
NH4F
KNO3
KHCO3
Acid–Base Properties of Salts: Practice
Acid–Base Properties of Salts: Practice
Salt pH of (aq) solution
CaCl2 Neutral
NH4Br Acidic
NH4F Basic
KNO3 Neutral
KHCO3 Basic
• According to the Brønsted–Lowry theory, all acid–base reactions can be written as equilibria involving the acid and base and their conjugates.
• All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.
Acid Base Conjugate base of the acid + Conjugate acid of the base
Predicting the Direction of Acid–Base Reactions
• When a weak acid is in solution, the products are a stronger conjugate acid and base. Therefore equilibrium lies to the left.
• All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.
Predicting the Direction of Acid–Base Reactions
• Will the following acid/base reaction occur spontaneously?
3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)
Predicting the Direction of Acid–Base Reactions
• Will the following acid/base reaction occur spontaneously?
Ka = 7.5 105 Ka = 1.8 105
Kb = 5.6 1010 Kb = 1.3 1012
Predicting the Direction of Acid–Base Reactions
3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)
• Will the following acid/base reaction occur spontaneously?
• Equilibrium lies to the right since all proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.
Ka = 7.5 105 Ka = 1.8 105
Kb = 5.6 1010 Kb = 1.3 1012
Stronger Acid + Stronger Base Weaker Base + Weaker Acid
Predicting the Direction of Acid–Base Reactions
3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)
Strong acid (HCl) + Strong base (NaOH)
Net ionic equation
Mixing equal molar quantities of a strong acid and strong base produces a neutral solution.
3HCl (aq) NaOH (aq) H O (aq) NaCl (aq)
3 2H O (aq) OH (aq) 2H O(l)
Types Acids–Base Reactions
Weak acid (HCN) + Strong base (NaOH)
Mixing equal amounts (moles) of a strong base and a weak acid produces a salt whose anion is the conjugate base of the weak acid. The solution is basic, with the pH depending on Kb for the anion.
2HCN (aq) + OH (aq) CN (aq) + H O (l)
2CN (aq) + H O (l) HCN (aq) + OH (aq)
Types Acids–Base Reactions
Strong acid (HCl) + Weak base (NH3)
Mixing equal amounts (moles) of a weak base and a strong acid produces a conjugate acid of the weak base. The solution is basic, with the pH depending on Ka for the acid.
3 3 2 4H O (aq) + NH (aq) H O (l) + NH (aq)
4 2 3 3NH (aq) + H O (l) H O (aq) + NH (aq)
Types Acids–Base Reactions
Weak acid (CH3CO2H) + Weak base (NH3)
Mixing equal amounts (moles) of a weak acid and a weak base produces a salt whose cation is the conjugate acid of the weak base and whose anion is the conjugate base of the weak acid. The solution pH depends on the relative Ka and Kb values.
3 2 3 3 2 4CH CO H (aq) + NH (aq) CH CO + NH (aq)
Types Acids–Base Reactions
Weak acid + Weak base
• Product cation = conjugate acid of weak base.• Product anion = conjugate base of weak acid.• pH of solution depends on relative strengths of
cation and anion.
Types Acids–Base Reactions
Types Acids–Base Reactions Summary
Determining K from Initial Concentrations and pH
[H2S] [H3O+] [HS]Initial 0.10Change
Equilibrium
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and pH
[H2S] [H3O+] [HS]Initial 0.10 0 0Change 0.10 - x + x + x
Equilibrium 0.10 - x x x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and pH
[H2S] [H3O+] [HS]Initial 0.10 0 0Change 0.10 - x + x + x
Equilibrium 0.10 - x x x
2
3 2a
2
H O NO x x xKHNO 0.10 x 0.10 x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and pH
[H3O+] = [NO2] = 6.76 103
234
a 3
6.76 10K 4.9 10
0.10 6.76 10
Calculations with Equilibrium Constants
2
3 2a
2
H O NO x x xKHNO 0.10 x 0.10 x
Determining K from Initial Concentrations and pH
[H2S] [H3O+] [HS]Initial 1.00Change
Equilibrium
Calculations with Equilibrium Constants
[H2S] [H3O+] [HS]Initial 1.00 0 0
Change - x + x + x
Equilibrium 1.00 - x x x
2
3 7a
2
H O HS x x xK 1.0 10H S 1.00 x 1.00 x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and pH
Determining K from Initial Concentrations and pH
27
Since x << 1.00, the equation reduces to:
x 1.0 101.00
x = 3.2 104 pH = 3.50
2
3 7a
2
H O HS x x xK 1.0 10H S 1.00 x 1.00 x
Calculations with Equilibrium Constants
Determining K from Initial Concentrations and pH
In general, the approximation that
[HA]equilibrium = [HA]initial x [HA]initial
is valid whenever [HA]initial is greater than or
equal to 100 Ka.
If this is not the case, the quadratic equation
must by used.
Calculations with Equilibrium Constants
Determining pH after an acid/base reaction:Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
Calculations with Equilibrium Constants
Calculations with Equilibrium Constants
Determining pH after an acid/base reaction:Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
Solution: From the volume and concentration of each solution, the moles of acid and base can be calculated. Knowing the moles after the reaction and the equilibrium constants, the concentration of H3O+ and pH can be calculated.
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
All of the acetic acid is converted to acetate ion.
3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)
3 23
1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH10 mL 1 L
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
All of the acetic acid is converted to acetate ion.
3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)
3 23
1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH10 mL 1 L
3
3 20.0033 mols 10 mLCH CO 0.075 M
44.0 mL 1 L
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change
Equilibrium
3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change - x + x + x
Equilibrium 0.075 - x x x
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change - x + x + x
Equilibrium 0.075 - x x x
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
Since Kb100 > [CH3CO2-]initial, the quadratic equation
is not needed.
[CH3CO2-] [CH3CO2H] [OH-]
Initial 0.075 0 0
Change - x + x + x
Equilibrium 0.075 - x x x
Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
• Because polyprotic acids are capable of donating more than one proton they present us with additional challenges when predicting the pH of their solutions.
• For many inorganic polyprotic acids, the ionization constant for each successive loss of a proton is about 104 to 106 smaller than the previous step.
• This implies that the pH of many inorganic polyprotic acids depends primarily on the hydronium ion generated in the first ionization step.
• The hydronium ion produced in the second step can be neglected.
Polyprotic Acuids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two H+ ions.(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion, SO3
2- in the 0.45 M solution of H2SO3?
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two H+ ions.(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion, SO3
2- in the 0.45 M solution of H2SO3?2
–2 3 3
2 3
[HSO ][H O ] x1.2 10 [H SO ] 0.45 – x
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two H+ ions.(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion, SO3
2- in the 0.45 M solution of H2SO3?
Since 100 Ka is not << 0.45M, the quadratic equation must be used
2–2 3 3
2 3
[HSO ][H O ] x1.2 10 [H SO ] 0.45 – x
Polyprotic Acids & Bases
Polyprotic Acids & BasesSulfurous acid, H2SO3, is a weak acid capable of providing two H+ ions.(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion, SO3
2- in the 0.45 M solution of H2SO3?
(a)
x = [H3O+] = 0.0677 M
pH = 1.17
Sulfurous acid, H2SO3, is a weak acid capable of providing two H+ ions.(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion, SO3
2- in the 0.45 M solution of H2SO3?
(a) in part a we found that x = [H3O+] = 0.0677 M
(b)
Polyprotic Acids & Bases
Halide Acid Strengths• Experiments show that the acid strength increases in the
order: HF << HCl < HBr < HI.• Stronger acids result when the HX bond is readily broken
(as signaled by a smaller, positive value of H for bond dissociation) and a more negative value for the electron attachment enthalpy of X.
Molecular Structure, Bonding, & Acid–Base Behavior
Comparing Oxoacids: HNO2 and HNO3
• In all the series of related oxoacid compounds, the acid strength increases as the number of oxygen atoms bonded to the central element increases.
• Thus, nitric acid (HNO3) is a stronger acid than nitrous acid (HNO2).
Molecular Structure, Bonding, & Acid–Base Behavior
Why Are Carboxylic Acids Brønsted Acids?• There is a large class of organic acids, all like acetic
acid (CH3CO2H) have the carboxylic acid group, CO2H
• They are collectively called carboxylic acids.
Molecular Structure, Bonding, & Acid–Base Behavior
Why Are Carboxylic Acids Brønsted Acids?• The carboxylate anion is stabilized by
resonance.
Molecular Structure, Bonding, & Acid–Base Behavior
Why Are Carboxylic Acids Brønsted Acids?• The acidity of carboxylic acids is enhanced if
electronegative substituents replace the hydrogen atoms in the alkyl (–CH3 or –C2H5) groups.
• Compare, for example, the pKa values of a series of acetic acids in which hydrogen is replaced sequentially by the more electronegative element chlorine.
Molecular Structure, Bonding, & Acid–Base Behavior
• Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl.
• Cl withdraws electrons from the rest of the molecule.
• This makes the O—H bond highly polar. The H of O—H is very positive.
Acetic acid Trichloroacetic acid
Ka = 1.8 x 10-5 Ka = 0.3
• The concept of acid–base behavior advanced by Brønsted and Lowry in the 1920’s works well for reactions involving proton transfer.
• However, a more general acid– base concept, was developed by Gilbert N. Lewis in the 1930’s.
• A Lewis acid is a substance that can accept a pair of electrons from another atom to form a new bond.
• A Lewis base is a substance that can donate a pair of electrons to another atom to form a new bond.
The Lewis Concept of Acids & Bases
A + B: BAAcid Base Adduct
• The product is often called an acid–base adduct. In Section 8.3, this type of chemical bond was called a coordinate covalent bond.
• Lewis acid-base reactions are very common. In general, they involve Lewis acids that are cations or neutral molecules with an available, empty valence orbital and bases that are anions or neutral molecules with a lone electron pair.
The Lewis Concept of Acids & Bases
Lewis acid a substance that accepts an electron pair
Lewis base a substance that donates an electron pair
The Lewis Concept of Acids & Bases
• New bond formed using electron pair from the Lewis base.
• Coordinate covalent bond
• Notice geometry change on reaction.
Reaction of a Lewis Acid & Lewis Base
The formation of a hydronium ion is an example of a Lewis acid / base reaction
H H
H
BASE
••••••
O—HO—HH+
ACID
The H+ is an electron pair acceptor.Water with it’s lone pairs is a Lewis acid donor.
The Lewis Concept of Acids & Bases
Lewis AcidBase Reactions
Metal cations often act as Lewis acids because of open d-orbitals.
Lewis Acids & Bases
The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to Coordinate Complex ions.
Lewis Acids & Bases
Aqueous solutions of Fe3+, Al3+, Cu2+, Pb2+, etc. are acidic through hydrolysis.
This interaction weakens this bond
Another H2O pulls this H away as H+
[Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O+(aq)
Lewis Acids & Bases
• Because oxygen is more electronegative than C, the CO bonding electrons in CO2 are polarized away from carbon and toward oxygen.
• This causes the carbon atom to be slightly positive, and it is this atom that the negatively charged Lewis base OH can attack to give, ultimately, the bicarbonate ion.
Molecular Lewis Acids
• Ammonia is the parent compound of an enormous number of compounds that behave as Lewis and Brønsted bases. These molecules all have an electronegative N atom with a partial negative charge surrounded by three bonds and a lone pair of electrons.
• This partially negative N atom can extract a proton from water.
Molecular Lewis Acids
Many complex ions containing water undergo HYDROLYSIS to give acidic solutions.
2+ ++
Lewis Acids & Bases
Reaction of NH3 with Cu2+(aq)
• The heme group in hemoglobin can interact with O2 and CO.
• The Fe ion in hemoglobin is a Lewis acid
• O2 and CO can act as Lewis bases
Heme group
Lewis Acid–Base Interactions in Biology
Slides from Chang Book
99
pH – A Measure of Acidity
pH = -log [H+]
[H+] = [OH-][H+] > [OH-][H+] < [OH-]
Solution Isneutralacidicbasic
[H+] = 1 x 10-7
[H+] > 1 x 10-7
[H+] < 1 x 10-7
pH = 7pH < 7pH > 7
At 250C
pH [H+]
100
percent ionization = Ionized acid concentration at equilibrium
Initial concentration of acidx 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0x 100% [HA]0 = initial concentration
% Ionization =
101
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Start
End
0.002 M
0.002 M 0.002 M0.0 M
0.0 M 0.0 M
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.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6
102
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF] = 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Initial (M)Change (M)Equilibrium (M)
0.50 0.00-x +x
0.50 - x
0.00+x
x x
Ka =x2
0.50 - x= 7.1 x 10-4
Ka x2
0.50 = 7.1 x 10-4
0.50 – x 0.50Ka << 1
x2 = 3.55 x 10-4 x = 0.019 M
[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72[HF] = 0.50 – x = 0.48 M
103
When can I use the approximation?
0.50 – x 0.50Ka << 1
When x is less than 5% of the value from which it is subtracted.
x = 0.019 0.019 M0.50 Mx 100% = 3.8%
Less than 5%Approximation
ok.What is the pH of a 0.05 M HF solution (at 250C)?
Ka x2
0.05 = 7.1 x 10-4 x = 0.006 M
0.006 M0.05 Mx 100% = 12%
More than 5%Approximation not
ok.Must solve for x exactly using quadratic equation or method of successive approximations.
Ka *100 << [HA]0
104
What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Initial (M)Change (M)Equilibrium (M)
0.122 0.00-x +x
0.122 - x
0.00+x
x x
Ka =x2
0.122 - x= 5.7 x 10-4
Ka x2
0.122= 5.7 x 10-4
0.122 – x 0.122Ka << 1
x2 = 6.95 x 10-5 x = 0.0083 M
0.0083 M0.122 Mx 100% = 6.8%
More than 5%Approximation not
ok.
105
Ka =x2
0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0
ax2 + bx + c =0 -b ± b2 – 4ac 2ax =
x = 0.0081 x = - 0.0081
HA (aq) H+ (aq) + A- (aq)
Initial (M)Change (M)Equilibrium (M)
0.122 0.00-x +x
0.122 - x
0.00+x
x x
[H+] = x = 0.0081 M pH = -log[H+] = 2.09
106
Acid-Base Properties of SaltsNeutral Solutions:
Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3
-).
NaCl (s) Na+ (aq) + Cl- (aq)H2O
Basic Solutions:Salts derived from a strong base and a weak acid.
NaCH3COOH (s) Na+ (aq) + CH3COO- (aq)H2O
CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)
107
Acid-Base Properties of Salts
Acid Solutions:Salts derived from a strong acid and a weak base.
NH4Cl (s) NH4+ (aq) + Cl- (aq)
H2O
NH4+ (aq) NH3 (aq) + H+ (aq)
Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid.
Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+
108
Acid-Base Properties of Salts
Solutions in which both the cation and the anion hydrolyze:• Kb for the anion > Ka for the cation, solution will be basic
• Kb for the anion < Ka for the cation, solution will be acidic
• Kb for the anion Ka for the cation, solution will be neutral
109
What is the molarity of an NH4NO3(aq) solution that has a pH = 4.80?
StrategyAmmonium nitrate is the salt of a strong acid (HNO3) and a weak base (NH3). In NH4NO3(aq), NH4
+ hydrolyzes and NO3– does not. The ICE format
must be based on the hydrolysis equilibrium for NH4+(aq). In that format
[H3O+], derived from the pH, will be a known quantity, and the initial concentration of NH4
+ will be the unknown.Solution
We begin by writing the equation for the hydrolysis equilibrium and theequation for Ka in terms of Kw and Kb.
As usual, we can calculate [H3O+] from the pH of the solution.
log[H3O+] = –pH = –4.80
[H3O+] = 10–4.80 = 1.6 x 10–5 M
110
Example 15.15 continued
Solution continuedIf we assume that all the hydronium ion comes from the hydrolysis reaction, we can set up an ICE format in which x represents the unknown initialconcentration of NH4
+.
We now substitute equilibrium concentrations into the ionization constantexpression for the hydrolysis reaction.
We can assume that the ammonium ion is mostly nonhydrolyzed and that the change in [NH4
+] is much smaller than the initial [NH4+], so that 1.6 x 10–
5 << x and we can replace (x – 1.6 x 10–5) by x. Then we can solve for x.
The solution is 0.46 M NH4NO3.