lecture 2
TRANSCRIPT
Lecture #2: Ionic EquilibriumLecture #2: Ionic Equilibrium
Overview: AcidOverview: Acid--Base ConceptBase Concept
There exist different definitions for an acid and a base
Arrhenius Definition
An acid is a substance that produces hydroxonium ion
(H3O+) as the only positively charged ion
A base is a substance the produces (OH-) as the only
negatively charged ion
There exist different definitions for an acid and a base
Arrhenius Definition
An acid is a substance that produces hydroxonium ion
(H3O+) as the only positively charged ion
A base is a substance the produces (OH-) as the only
negatively charged ion
AcidAcid--Base ConceptBase Concept
The Arrhenius theory accounts for the many acids
and bases that exist
One limitation with the theory is that it is only
applicable to aqueous solutions. As a result a
molecule like ammonia (alkaline gas) would not be
classified as a base
A more general definition for acid and base were
later proposed by 2 Scientists, Johannes BrØnsted
and Thomas Lowry
The Arrhenius theory accounts for the many acids
and bases that exist
One limitation with the theory is that it is only
applicable to aqueous solutions. As a result a
molecule like ammonia (alkaline gas) would not be
classified as a base
A more general definition for acid and base were
later proposed by 2 Scientists, Johannes BrØnsted
and Thomas Lowry
AcidAcid--Base ConceptBase Concept BrØnsted-Lowry Theory
An acid is any substance (molecule or ion) that
transfers a proton to another substance
A base –is any substance (molecule or ion) that accepts
a proton
HA + B BH+ + A-
H+ donor H+ acceptor H+ donor H+ acceptor
BrØnsted-Lowry Theory
An acid is any substance (molecule or ion) that
transfers a proton to another substance
A base –is any substance (molecule or ion) that accepts
a proton
HA + B BH+ + A-
H+ donor H+ acceptor H+ donor H+ acceptor
AcidAcid--Base ConceptBase Concept
• Chemical species whose formulae differ by one proton
are said to be conjugate acid-base pairs
AcidAcid--Base ConceptBase Concept
Exercise
Write balanced equation for the dissociation
of each of the BrØnsted-Lowry acids below
(a) H2SO4 sulphuric acid
What is the conjugate base of the acid in each
case
Exercise
Write balanced equation for the dissociation
of each of the BrØnsted-Lowry acids below
(a) H2SO4 sulphuric acid
What is the conjugate base of the acid in each
case
AcidAcid--Base ConceptBase Concept
Write balanced equation for the dissociation of each of
the BrØnsted-Lowry bases below
(a) HCO32-
What is the conjugate acid for the base in each case
Write balanced equation for the dissociation of each of
the BrØnsted-Lowry bases below
(a) HCO32-
What is the conjugate acid for the base in each case
Acid and Base StrengthAcid and Base Strength A strong acid (strong base) is completely dissociated in
aqueous solution
Therefore the acid (base) dissociation equilibrium of a
strong acid (strong base) lies entirely to the right and as
such only ions are present in the solution
HCl (aq) → H+(aq) + Cl- (aq)
NaOH (aq)→ Na+(aq) + OH-
(aq)
A strong acid (strong base) is completely dissociated in
aqueous solution
Therefore the acid (base) dissociation equilibrium of a
strong acid (strong base) lies entirely to the right and as
such only ions are present in the solution
HCl (aq) → H+(aq) + Cl- (aq)
NaOH (aq)→ Na+(aq) + OH-
(aq)
Source: http://wikis.lawrence.edu/display/CHEM/Acids+and+Bases-Davis
Acid and Base StrengthAcid and Base Strength
A weak acid (weak base) is partially dissociated in
aqueous solution
Only a small fraction of the of the molecule is
transferred to ions
CH3COOH (aq) CH3COO-(aq) + H+
(aq)
A weak acid (weak base) is partially dissociated in
aqueous solution
Only a small fraction of the of the molecule is
transferred to ions
CH3COOH (aq) CH3COO-(aq) + H+
(aq)
Acid and Base StrengthAcid and Base Strength
Source: http://wwwchem.csustan.edu/chem3070/images/acetic.gif
Relative Strength of Conjugate AcidRelative Strength of Conjugate Acid--Base PairsBase Pairs
Dissociation of WaterDissociation of Water Water is amphoteric, i.e. it can act both as an
acid and a base In the presence of an acid water act as a base
and in the present of a base water act as a acid
HCl (aq) + H2O(l) → H3O+(aq) + Cl- (aq)
proton donor proton acceptor
NH3(aq) + H2O(l) NH4+
(aq) + OH-(aq)
proton acceptor proton donor
Water is amphoteric, i.e. it can act both as anacid and a base In the presence of an acid water act as a base
and in the present of a base water act as a acid
HCl (aq) + H2O(l) → H3O+(aq) + Cl- (aq)
proton donor proton acceptor
NH3(aq) + H2O(l) NH4+
(aq) + OH-(aq)
proton acceptor proton donor
Pure water can also act as an acid and a base with itself
H2O(l) + H2O (l) H3O+(aq) + OH-
(aq)
2H2O (l) H3O+(aq) + OH-
(aq)
Changing the equilibrium constant Kc to Kw
Kw = [H3O+][OH-]
Experiment done at 25ºC have shown that [H3O+] =
1.0 x 10-7
Since the dissociation of water produces equal volumes of
[H3O+] and [OH-] then
Kw = (1.0 x 10-7)(1.0 x 10-7)
= 1.0 x 10-14
Pure water can also act as an acid and a base with itself
H2O(l) + H2O (l) H3O+(aq) + OH-
(aq)
2H2O (l) H3O+(aq) + OH-
(aq)
Changing the equilibrium constant Kc to Kw
Kw = [H3O+][OH-]
Experiment done at 25ºC have shown that [H3O+] =
1.0 x 10-7
Since the dissociation of water produces equal volumes of
[H3O+] and [OH-] then
Kw = (1.0 x 10-7)(1.0 x 10-7)
= 1.0 x 10-14
A neutral solution - [H3O+] = [OH-] = 1.0 x 10-7
Acidic solutions - [H3O+] > [OH-][H3O+] > 1.0 x 10-7
[OH-] < 1.0 x 10-7
Basic solutions - [H3O+] < [OH-][H3O+] < 1.0 x 10-7
[OH-] > 1.0 x 10-7
In all solutions at 25ºC - [H3O+] [OH-] = Kw = 1.0 x 10-14
A neutral solution - [H3O+] = [OH-] = 1.0 x 10-7
Acidic solutions - [H3O+] > [OH-][H3O+] > 1.0 x 10-7
[OH-] < 1.0 x 10-7
Basic solutions - [H3O+] < [OH-][H3O+] < 1.0 x 10-7
[OH-] > 1.0 x 10-7
In all solutions at 25ºC - [H3O+] [OH-] = Kw = 1.0 x 10-14
Exercise
1. The concentration of H3O+ ions in a sample of lemon
juice is 2.5 x 10-3 M. Calculate the concentration of OH-
ions, and classify the solution as neutral, acidic, or basic
2. The concentration of OH- in a sample of sea water is
5 x 10-6 M. Calculate the concentration of H3O+ ions,
and classify the solution
3. At 50ºC the Kw is 5.5 x 10-14. What are the
concentrations of H3O+ and OH- ions in a neutral
solution at 50ºC.
Exercise
1. The concentration of H3O+ ions in a sample of lemon
juice is 2.5 x 10-3 M. Calculate the concentration of OH-
ions, and classify the solution as neutral, acidic, or basic
2. The concentration of OH- in a sample of sea water is
5 x 10-6 M. Calculate the concentration of H3O+ ions,
and classify the solution
3. At 50ºC the Kw is 5.5 x 10-14. What are the
concentrations of H3O+ and OH- ions in a neutral
solution at 50ºC.
pH ScalepH Scale
pH (power of hydrogen)
pH = -log [H3O+]
It is a logarithmic scale that is used to specify
the acidity of a solution
pH < 7 – acidic
pH > 7 – alkaline
pH = 7 - neutral
pH (power of hydrogen)
pH = -log [H3O+]
It is a logarithmic scale that is used to specify
the acidity of a solution
pH < 7 – acidic
pH > 7 – alkaline
pH = 7 - neutral
pH ScalepH Scale
pOH – power of hydroxide ion pH + pOH = 14
Source: http://academic.cuesta.edu/gbaxley/chem1A/notes/ph_scale_2.jpg
Calculate the pH of the following solution and
indicate whether the solution is acidic or basic
[H3O+] = 1.8 x 10-4 M
pH = -log [H3O+]
pH = -log (1.8 x 10-4 )
pH = -(- 3.74)
pH = 3.74
N.B when you take the log of a quantity the result should
have the same number of decimal places as significant
figures of the original quantity
Calculate the pH of the following solution and
indicate whether the solution is acidic or basic
[H3O+] = 1.8 x 10-4 M
pH = -log [H3O+]
pH = -log (1.8 x 10-4 )
pH = -(- 3.74)
pH = 3.74
N.B when you take the log of a quantity the result should
have the same number of decimal places as significant
figures of the original quantity
Exercise
1. Calculate the pH of a solution with a [H3O+] of
9.5 x 10-9 M
2. Calculate the [H3O+] for a solution of a pH of
4.80
3. Calculate the [OH-] for a solution with a pH of
3.66
Exercise
1. Calculate the pH of a solution with a [H3O+] of
9.5 x 10-9 M
2. Calculate the [H3O+] for a solution of a pH of
4.80
3. Calculate the [OH-] for a solution with a pH of
3.66
[H3O+] [OH-] pH Acidic or basic
1 x 10-14 4.00
5.5 x 10-3
Complete the table below
5.5 x 10-3
3.2 x 10-6
4.8 x 10-9
7.55
EquilibriaEquilibria Solutions of Weak AcidsSolutions of Weak Acids
Recap: strong acids are completely dissociated in aqueous
solutions and weak acids are partially dissociated in
aqueous solutions
HA (aq) + H2O(l) H3O+(aq) + A-
(aq)
Kc = [H3O+][A-]
[HA][H2O]
Ka = acid-dissociation constant
Ka = Kc [H2O] = [H3O+][A-]
[HA]The larger the Ka value the stronger the acid
Recap: strong acids are completely dissociated in aqueous
solutions and weak acids are partially dissociated in
aqueous solutions
HA (aq) + H2O(l) H3O+(aq) + A-
(aq)
Kc = [H3O+][A-]
[HA][H2O]
Ka = acid-dissociation constant
Ka = Kc [H2O] = [H3O+][A-]
[HA]The larger the Ka value the stronger the acid
The pH of a 0.250 M HF is 2.036. What is the value of Ka
for the hydrofluoric acid solution?
HF (aq) + H2O(l) H3O+(aq) + F-
(aq)
Ka = [H3O+][F-]
[HF]
pH = - log [H3O+]
[H3O+] = antilog (-pH)
= 10 –pH
= 9.2 x 10-3 M
Since mole ration of H3O+ : F- is 1:1; [F-] = 9.2 x 10-3 M
The [HF] at equilibrium = 0.25 - 0.00920 = 0.241 M
The pH of a 0.250 M HF is 2.036. What is the value of Ka
for the hydrofluoric acid solution?
HF (aq) + H2O(l) H3O+(aq) + F-
(aq)
Ka = [H3O+][F-]
[HF]
pH = - log [H3O+]
[H3O+] = antilog (-pH)
= 10 –pH
= 9.2 x 10-3 M
Since mole ration of H3O+ : F- is 1:1; [F-] = 9.2 x 10-3 M
The [HF] at equilibrium = 0.25 - 0.00920 = 0.241 M
Ka = (9.20 x 10-3)(9.20 x 10-3)0.241
Ka = 3.52 x 10-4
Exercise1. The pH of 0.1 M HClO is 4.23. Calculate the Ka for the
hyperchlorous acid.
2. The pH of 0.040 M hypobromous acid (HOBr) is 5.05.
Set up the equilibrium equation for the dissociation of
HOBr, and calculate the acid dissociation constant.
Exercise1. The pH of 0.1 M HClO is 4.23. Calculate the Ka for the
hyperchlorous acid.
2. The pH of 0.040 M hypobromous acid (HOBr) is 5.05.
Set up the equilibrium equation for the dissociation of
HOBr, and calculate the acid dissociation constant.
EquilibriaEquilibria Solutions of Weak BasesSolutions of Weak Bases
Kb – dissociation constant of a weak base
NH3 (aq) + H2O (l) NH4+
(aq) + OH-(aq)
Kb = [NH4+ ][OH- ]
[NH3]
Kb – dissociation constant of a weak base
NH3 (aq) + H2O (l) NH4+
(aq) + OH-(aq)
Kb = [NH4+ ][OH- ]
[NH3]
Exercise
1. Codeine (C8H21NO3), a drug used in painkillers and
cough medicines, is a naturally occurring amine that has
a Kb = 1.6 x 10-6. Calculate the pH and the
concentration of all species present in a 0.0012 M
solution of codeine.
Use cod as the abbreviation for codeine and codH+ for
the conjugated acid.
2. Calculate the pH and the concentration of all the
species present in 0.40 M NH3 (Kb = 1.8 x 10-5)
Exercise
1. Codeine (C8H21NO3), a drug used in painkillers and
cough medicines, is a naturally occurring amine that has
a Kb = 1.6 x 10-6. Calculate the pH and the
concentration of all species present in a 0.0012 M
solution of codeine.
Use cod as the abbreviation for codeine and codH+ for
the conjugated acid.
2. Calculate the pH and the concentration of all the
species present in 0.40 M NH3 (Kb = 1.8 x 10-5)
AcidAcid –– Base TitrationsBase Titrations
When an acid reacts with a base, salt and water is
formed
This type of reaction is called neutralization
There exist four types of neutralization reactions
(a) Strong acid – strong base
(b) Weak acid – strong base
(c) Strong acid – weak base
(d) Weak acid – weak base
When an acid reacts with a base, salt and water is
formed
This type of reaction is called neutralization
There exist four types of neutralization reactions
(a) Strong acid – strong base
(b) Weak acid – strong base
(c) Strong acid – weak base
(d) Weak acid – weak base
AcidAcid –– Base TitrationsBase Titrations
An acid-base titration is a neutralization reaction that is
performed in the lab to determine an unknown
concentration of acid or base
The solution containing a known concentration is added
from a burette to a second solution containing an
unknown concentration of acid or base
The progress of the reaction is monitored using a pH
meter or by observing the colour of a suitable indicator
An acid-base titration is a neutralization reaction that is
performed in the lab to determine an unknown
concentration of acid or base
The solution containing a known concentration is added
from a burette to a second solution containing an
unknown concentration of acid or base
The progress of the reaction is monitored using a pH
meter or by observing the colour of a suitable indicator
AcidAcid –– Base TitrationsBase Titrations
A typical pH graph has a
sigmoid shape
Mid way the vertical region of
the graph is known as the
equivalence point
A typical pH graph has a
sigmoid shape
Mid way the vertical region of
the graph is known as the
equivalence point
Source: http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
AcidAcid –– Base TitrationsBase Titrations
At the equivalence point there are equal numbers of
H+ and OH- ion concentration
The equivalence point varies depending on the acid-
base combination used
As a result an appropriate indicator has to be used
At the equivalence point there are equal numbers of
H+ and OH- ion concentration
The equivalence point varies depending on the acid-
base combination used
As a result an appropriate indicator has to be used
AcidAcid –– Base TitrationsBase Titrations
Indicators are organic compounds that change
colour in response to a pH change
The strength of the acid and base used in
neutralization will determine the type of indicator
used
Indicators are organic compounds that change
colour in response to a pH change
The strength of the acid and base used in
neutralization will determine the type of indicator
used
Indicator pH Neutral colour Acid – basecombination
Litmus solution 7 Purple Strong acid –strong base
Phenolphthalein 9.7 Colourless Strong base – weakacidStrong base – weakacid
Methyl orange 3.7 Yellow Strong acid – weakbase
Solubility ProductSolubility Product The equilibrium equation for the dissolution reaction is called the
equilibrium constant Ksp
Ksp – solubility product constant
CaF2(s) Ca2+(aq) + 2F-
(aq) (Heterogeneous mixture)
The Ksp = [Ca2+][F-]2
The Ksp is always equal to the product of the equilibrium
concentrations of all the ions on the right side of the chemical
equation, with the concentration of each ion raised to the power of
its coefficient in a balanced equation
The concept of solubility product constant applies only to sparingly
soluble compounds. It would not apply to vey soluble compounds.
The equilibrium equation for the dissolution reaction is called the
equilibrium constant Ksp
Ksp – solubility product constant
CaF2(s) Ca2+(aq) + 2F-
(aq) (Heterogeneous mixture)
The Ksp = [Ca2+][F-]2
The Ksp is always equal to the product of the equilibrium
concentrations of all the ions on the right side of the chemical
equation, with the concentration of each ion raised to the power of
its coefficient in a balanced equation
The concept of solubility product constant applies only to sparingly
soluble compounds. It would not apply to vey soluble compounds.
Exercise1. A particular saturated solution of silver chromate,
Ag2CrO4, has [Ag+] = 5.0 x 10-5M and
[CrO42-] = 4.4 x 10-4 M. What is the value of Ksp for
Ag2CrO4?
2. A saturated solution of Ca3(PO4)2 has
[Ca2+] = 2.01 x 10-8 M and PO43- = 1.6 x 10-5M.
Calculate the Ksp for Ca3(PO4)2
3. The solubility of lead(II)chloride at 298 K is 0.834g
per 100 cm3 of water. Find the solubility product of
lead(II)chloride at this temperature
Exercise1. A particular saturated solution of silver chromate,
Ag2CrO4, has [Ag+] = 5.0 x 10-5M and
[CrO42-] = 4.4 x 10-4 M. What is the value of Ksp for
Ag2CrO4?
2. A saturated solution of Ca3(PO4)2 has
[Ca2+] = 2.01 x 10-8 M and PO43- = 1.6 x 10-5M.
Calculate the Ksp for Ca3(PO4)2
3. The solubility of lead(II)chloride at 298 K is 0.834g
per 100 cm3 of water. Find the solubility product of
lead(II)chloride at this temperature
Common Ion EffectCommon Ion Effect
The common ion effect can be defined as that
effect which causes the repression of the
dissociation of a compound by the addition of a
common cation or anion
The presence of the common ion affect the solubility of
a salt
The common ion effect can be defined as that
effect which causes the repression of the
dissociation of a compound by the addition of a
common cation or anion
The presence of the common ion affect the solubility of
a salt
Common Ion EffectCommon Ion Effect
e.g. CaF2(s) Ca2+(aq) + 2F-
(aq)
If F- is added from another source (e.g NaF) then this
will result in an increase in the concentration of F- and
according to Le Chatelier’s principle the equilibrium will
shift to the left producing more CaF2(s)
As a result the solubility of the salt CaF2(s) decreases
This is known as the common ion effect
e.g. CaF2(s) Ca2+(aq) + 2F-
(aq)
If F- is added from another source (e.g NaF) then this
will result in an increase in the concentration of F- and
according to Le Chatelier’s principle the equilibrium will
shift to the left producing more CaF2(s)
As a result the solubility of the salt CaF2(s) decreases
This is known as the common ion effect
BuffersBuffers
Buffers are solutions that resist changes in pH
Buffers do this by neutralizing added acid or base
Blood is a buffer that resists changes in pH. If the pH of
blood drops below 7.0 or increase beyond 7.8 we would die
Like all buffers blood contains a significant amounts of both
weak acid and conjugate base
When additional base is added to blood the acid reacts with
the base neutralizing it. In addition if acid is added to blood
the conjugate base reacts with the acid neutralizing it
Buffers are solutions that resist changes in pH
Buffers do this by neutralizing added acid or base
Blood is a buffer that resists changes in pH. If the pH of
blood drops below 7.0 or increase beyond 7.8 we would die
Like all buffers blood contains a significant amounts of both
weak acid and conjugate base
When additional base is added to blood the acid reacts with
the base neutralizing it. In addition if acid is added to blood
the conjugate base reacts with the acid neutralizing it
BuffersBuffers
Blood contains an carbonic acid- bicarbonate buffer
H+(aq) + HCO3
-(aq) H20(l) + CO2(g)
The Henderson-Hasselbalch equation is used to
calculate the pH of a buffer solution
pH = pKa + log ([base]/[acid]
pKa = -log Ka
Blood contains an carbonic acid- bicarbonate buffer
H+(aq) + HCO3
-(aq) H20(l) + CO2(g)
The Henderson-Hasselbalch equation is used to
calculate the pH of a buffer solution
pH = pKa + log ([base]/[acid]
pKa = -log Ka
Exercise
1. Calculate the pH of a buffer solution made from 0.20 M
HC2H3O2 and 0.50 M C2H3O2- that has an acid
dissociation constant for HC2H3O2 of 1.8 x 10-5.
pH = pKa + log ([A-]/[HA]) pH = pKa + log ([C2H3O2
-] / [HC2H3O2]) pH = -log (1.8 x 10-5) + log (0.50 / 0.20) pH = -log (1.8 x 10-5) + log (2.5) pH = 4.7 + 0.40 pH = 5.1
pH = pKa + log ([A-]/[HA]) pH = pKa + log ([C2H3O2
-] / [HC2H3O2]) pH = -log (1.8 x 10-5) + log (0.50 / 0.20) pH = -log (1.8 x 10-5) + log (2.5) pH = 4.7 + 0.40 pH = 5.1
2. Use the Henderson-Hasselbalch equation to calculate
the pH of a buffer solution that is 0.25 M in formic acid
(HCOOH) and 0.50 M in sodium formate (HCO2Na).
Ka of formic acid = 1.80 x 10-4.
2. Use the Henderson-Hasselbalch equation to calculate
the pH of a buffer solution that is 0.25 M in formic acid
(HCOOH) and 0.50 M in sodium formate (HCO2Na).
Ka of formic acid = 1.80 x 10-4.