Unit 8 – Acids and
Bases
Part 1: SL Material Only
Part 2: HL Material Only
Lesson 1
Topic 8.1 Theories of Acids and Bases
Aim: DWBAT identify Bronsted-Lowry Acids and Bases
Acids and Bases
• The word acid derives itself from the Latin (yay!) word acetum
meaning “sour tasting”
• Alkali, another word for basic, derives from the Arabic word
alkalja
• We encounter acids and bases throughout the day. Citrus fruits,
soda, vinegar, and more are common acids. Bases include
baking soda, antacids, and ammonia.
IB Understandings
• A Brønsted–Lowry acid is a proton/H+ donor and a Brønsted–
Lowry base is a proton/H+ acceptor. Guidance
• Students should know the representation of a proton in aqueous
solution as both H+(aq) and H3O+(aq).
• Amphiprotic species can act as both Brønsted–Lowry acids and
bases.
• A pair of species differing by a single proton is called a
conjugate acid–base pair.
IB Applications and Skills
• Deduction of the Brønsted–Lowry acid and base in a chemical
reaction.
• The location of the proton transferred should be clearly indicated.
For example, CH3COOH/CH3COO– rather than C2H4O2/C2H3O2–
.
• Deduction of the conjugate acid or conjugate base in a
chemical reaction.
• Lewis theory is not required here but will be covered later in the
HL material
Arrhenius Theory
• Svante August Arrhenius won the Nobel Prize in Chemistry in
1903 for his work on acids and bases
• He defined an acid as a substance that ionizes in water to
produce H+ ions
• He defined a base as a substance that ionizes in water to
produce OH- ions
• The combination of an Arrhenius acid and base produces a
neutralization reaction the products of which are always a salt
and pure water
Neutralization Reaction
• An example of a neutralization reaction is:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Bronsted-Lowry Theory
• Despite the name, these two scientists came up with this theory
independently!
• Bronsted-Lowry Acids are substances that donate a proton
(H+) ion
• Bronsted-Lowry Bases are substances that accept a proton
(H+) ion
• H+ is equivalent to a proton because hydrogen originally only
had 1 proton and 1 electron; once it loses the electron it is just a
proton!
• Nota Bene: The H+ ion can be denoted by the hydronium ion (H3O+) if
the reaction is occurring in water
Conjugate Acid-Base Pairs
• Let’s look at an example of a Bronsted-Lowry acid and base
reacting:
HCl + NH3  NH4+ + Cl–
• Here, HCl is acting as the acid (as it is donating the H+ ion to
NH3) and NH3 is acting as the base (as it is accepting the H+
ion)
• Since this reaction is reversible, which substance is the acid and
which substance is the base for the reverse reaction?
Conjugate Acid-Base Pairs
• Acids react to form bases and bases react to form acids.
• The acid–base pairs related to each other in this way are called
conjugate acid–base pairs, and you can see that they differ by
just one proton.
• It is important to be able to recognize these pairs in a Brønsted–
Lowry acid– base reaction.
Let’s Practice
• Label the conjugate acid–base pairs in the following reaction:
CH3COOH(aq) + H2O(l)  CH3COO–(aq) + H3O+(aq)
Tips
• The conjugate base of an acid always has one fewer proton
• The conjugate acid of a base always has one more proton
• Example: The conjugate base of H3SO4 is H2SO41- NOT SO43-
SNEAK PEAK: HL Info
• We will go into this more when we move into the HL level
material for this unit but…
• A Lewis acid is an electron pair acceptor
• A Lewis base is an electron pair donator
More Practice
1 Write the conjugate base for each of the following.
(a) H3O+ (b) NH3 (c) H2CO3
2 Write the conjugate acid for each of the following.
(a) NO2– (b) OH– (c) CO32–
Amphiprotic/Amphoteric
• Substances that can act as either acids or bases are called
amphiprotic or amphoteric
• An example of an amphiprotic species is water
• To act as a Brønsted–Lowry acid, they must be able to dissociate and
release H+. To act as a Brønsted–Lowry base, they must be able to
accept H+, which means they must have a lone pair of electrons.
• So substances that are amphiprotic according to Brønsted–Lowry
theory must possess both a lone pair of electrons and hydrogen that
can be released as H+.
Let’s Practice
• Write equations to show HCO3– acting (a) as a Brønsted–
Lowry acid and (b) as a Brønsted–Lowry base.
More Practice
Answers
Lesson 2
Topic 8.2 Properties of Acids and Bases
DWBAT describe the properties of acids and bases
IB Understandings
• Most acids have observable characteristic chemical reactions
with reactive metals, metal oxides, metal hydroxides, hydrogen
carbonates, and carbonates.
• Bases which are not hydroxides, such as ammonia, soluble
carbonates, and hydrogen carbonates should be covered.
• Salt and water are produced in exothermic neutralization
reactions.
Applications and Skills
• Balancing chemical equations for the reaction of acids.
• Identification of the acid and base needed to make different
salts.
• Candidates should have experience of acid–base titrations with
different indicators.
• The color changes of different indicators are given in the data
booklet in section 22.
Property 1: Acid Reactions
• Property 1: Acids react with metals, bases, and carbonates to
form salts
• Salt refers to an ionic compound formed when the hydrogen of
the acid is replaced by a metal or other positive ion
• We can use the term parent acid and parent base to describe
the original compounds that combined to make the salt
Three Main Acid Reactions
1. Acid + Metal  Salt + Hydrogen
•
We’ve done this one many times in lab! (Remember from
Sophomore year, only some metals react. Nonreactive metals like
silver, copper, and gold do not! More on this in redox)
• Special Note: Ions that do not change in reactants and products are
called spectator ions.
Three Main Acid Reactions
2. Acid + Base  Salt + Water
•
Your standard neutralization reaction!
•
Neutralization is an exothermic process. The enthalpy of
neutralization is defined as the enthalpy change that occurs when
an acid and a base react together to form one mole of water
•
When a strong acid and base react, the enthalpy of neutralization is
around ∆H= –57 kJ mol–1 because the net reaction is all the same (the
formation of water from its ions)
Three Main Acid Reactions
3. Acid + Carbonate  Salt + Water + Carbon Dioxide
•
You know this one…vinegar and baking soda!
•
These reactions produce visible bubbles which is known as
effervesence
Making Salts
• Generally, the metal part of the salt comes from a metal oxide
or hydroxide and the non-metal part from the acid
Acid
Formula
Name of Salt
Example
hydrochloric
HCl
chloride
HCl
sulfuric
H2SO4
sulfate
H2SO4
nitric
HNO3
nitrate
NH4NO3
carbonic
H2CO3
carbonate
K2CO3
ethanoic
CH3COOH
ethanoate
Ca(CH3COO)2
Property 2: Indicators
• Property 2: Acids and bases can be distinguished using
indicators
• Indicators are substances that turn color depending on the pH
(measure of the hydrogen ion concentration)
• Litmus is the best known indicator (hence the expression “litmus
test”); litmus turn pink in the pressure of an acid and blue in the
presence of a base
Found in data booklet, Section 22
Indicators (cont.)
• Many indicators can be mixed together to make a universal
indicator which changes color many times over a specific range
of pH so that you can tell not only if a solution is acidic or
basic but how acidic or basic it is
• There will be a lot more on indicators when we progress to
the HL level material!
Property 3: Titrations
• A neutralization reaction is often used in the lab to calculate
the concentration of an unknown species
• The technique known as acid–base titration involves reacting
together a carefully measured volume of one of the solutions,
and adding the other solution gradually until the so-called
equivalence point is reached where they exactly neutralize each
other.
• We can use an indicator such as phenolphthalein to determine
the equivalence point
Titration Uses
• Titration can be used in the following experiment
• to calculate the concentration of ethanoic acid in vinegar by
titration with a standard solution of aqueous sodium hydroxide,
using phenolphthalein indicator;
• to calculate the concentration of sodium hydroxide by titration
with a standard solution of hydrochloric acid, using methyl orange
indicator.
Other Properties
Acids
Bases
Taste sour
Taste bitter; feel slippery
pH < 7
pH > 7
Litmus is red
Litmus is blue
Phenolphthalein is colorless
Phenolphthalein is pink
Methyl orange is red
Methyl orange is yello
Let’s Practice
Lesson 3
Topic 8.3 The pH Scale
DWBAT solve problems using pH, [H+], [OH-]
Understandings
• pH = –log [H+(aq)] and [H+] = 10–pH.
• A change of one pH unit represents a 10-fold change in the
hydrogen ion concentration [H+].
• Knowing the temperature dependence of Kw is not required (phew!)
• pH values distinguish between acidic, neutral, and alkaline
solutions.
• The ionic product constant, Kw = [H+] [OH–] = 10–14 at 298
K.
Applications and Skills
• Solving problems involving pH, [H+], and [OH–].
• Students should be concerned only with strong acids and bases in this subtopic.
• Students will not be assessed on pOH values.
• Students should be familiar with the use of a pH meter and
universal indicator.
Water Self-Ionization
• Most acid-base reactions happen in aqueous solution so it is
important to consider water and its role in acid-base chemistry
• Water is amphoteric meaning that it can act either as an acid or
a base
• Water to a small degree self-ionizes; meaning that is undergoes
an acid-base reaction with itself as seen in the following
equation:
H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)
Kc and Kw
• Since this is a reversible reaction, we can write an equilibrium
constant for the self-ionization of water as follows:
• Since pure water has a constant concentration, we get:
Kw
• Kw is known as the ionic product constant of water and has a
fixed value at a specified temperature.
• At 298 K, Kw = 1.00 × 10–14.
• Since all the H+ and OH- come from the water they must be equal
to each other, [H+] = [OH-] = √Kw = 10-7
• Keep this in mind as we begin to explore pH!
• The value Kw = 1.00 × 10–14 at 298 K is given
in section 2 of the IB data booklet so does not
have to be learned.
Relationship Between H+ and OH• Since we know that the product of [H+] × [OH–] gives a
constant value, it follows that the concentrations of these ions
must have an inverse relationship in aqueous solution
• Solution behave acidic, basic, or neutral according to the
relative concentrations of these ions
• NOTE: We keep saying H+ ion but remember that this
doesn’t actually exist! H+ is a naked proton and as such is
very unstable and will coordinately covalently bond to make
H3O+(aq) in water!
• We use this relationship to build a scale, the pH scale, to
measure acidity and basicity!
pH – Log Scale
• The pH scale is a measure of the concentration of hydrogen
ions as a measure of acidity; more H+ ions means a more
acidic solution
• pH is the negative logarithm of base 10 of the hydrogen ion
concentration as measured in mol dm-3 in water
pH = –log10 [H+]
Examples
• • A solution that has [H+] = 0.1 mol dm–3 ⇒ [H+] = 10–1 mol
dm–3 ⇒ pH = 1.
• A solution that has [H+] = 0.01 mol dm–3 ⇒ [H+] = 10–2 mol
dm–3 ⇒ pH = 2.
pH Scale
• We define pure water as neutral for convention and base all
other H+ concentrations on that
pH Scale – Important Points
1. pH numbers are usually positive and have no units
•
In theory, the scale is infinite and can be negative, but you will
typically see values between 0-14
2. The pH number is inversely related to the [H+]
3. A change of one pH unit represents a 10-fold change in [H+]
Let’s Practice
• If the pH of a solution is changed from 3 to 5, deduce how the
hydrogen ion concentration changes.
• Decreased by 100
• A sample of lake water was analyzed at 298 K and found to
have [H+] = 3.2 × 10–5 mol dm–3. Calculate the pH of this
water and comment on its value.
• At 298 K this pH < 7, and the lake water is therefore acidic.
• Human blood has a pH of 7.40. Calculate the concentration of
hydrogen ions present.
• [H+] = 4.0 × 10–8 mol dm–3
Let’s Practice - Kw
• A sample of blood at 298 K has [H+] = 4.60 × 10–8 mol dm–3.
Calculate the concentration of OH– and state whether the
blood is acidic, neutral, or basic.
• HINT: Remember that at 298 K, Kw = 1.00 × 10–14 = [H+] [OH–]
Measuring pH
• MANDATORY LAB! Meaning you can be tested on these
techniques
• Indicators: an easy way to test for pH is with universal
indicators which change colors many times over a range of pH
and can be compared to a chart. (Lab note: these do depend on
the users ability to match colors. How accurate would your
results be?)
• pH Meter: is a probe that directly reads the [H+] concentration
through a special electrode. pH meters can record to an
accuracy of several decimal points. They must, however, be
calibrated before each use with a buffer solution, and
standardized for the temperature, as pH is a temperaturedependent measurement.
Lesson 4
Topic 8.4 Strong and Weak Acids
DWBAT compare and contrast strong and weak acids and bases.
IB Understandings
• Strong and weak acids and bases differ in the extent of
ionization.
• The terms ionization and dissociation can be used interchangeably.
• Strong acids and bases of equal concentrations have higher
conductivities than weak acids and bases.
• See section 21 in the data booklet for a list of weak acids and bases.
• A strong acid is a good proton donor and has a weak conjugate
base.
• A strong base is a good proton acceptor and has a weak
conjugate acid.
IB Applications and Skills
• Distinction between strong and weak acids and bases in terms
of the rates of their reactions with metals, metal oxides, metal
hydroxides, metal hydrogen carbonates, and metal carbonates
and their electrical conductivities for solutions of equal
concentrations.
SL Quiz
• Friday, October
th
16
Strength Depends on Ionization
• Since acidity/basicity depends on concentration of the
H+/H3O+ and OH- ions, the overall strength of an acid or a
base depends on how much the species ionizes or dissociates
• For this unit, we will use ionization and dissociation interchangeably
• Let’s look at two acids we have used in the lab – HCl and
CH3COOH
• We know instinctively that HCl is strong and acetic acid is weak
but why?
A Tale of Two Acids
• HCl when dissolved in water fully dissociates, meaning all of it
becomes H+ and Cl-. Therefore it is a strong acid.
• The reaction is written without the equilibrium sign.
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
• If, on the other hand, the acid dissociates only partially, it produces
an equilibrium mixture in which the undissociated form dominates.
It is said to be a weak acid. For example, ethanoic acid, CH3COOH,
is a weak acid.
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO–(aq)
• If I put equal concentration of acid in the same amount of water, the
HCl would have a much higher concentration of H3O+!!!!!
Strong/Weak Acids
• Strong acids are strong proton donors; their conjugate bases
are poor proton acceptors and the reaction does not readily go
backwards
• Weak acids are weak proton donors; their conjugate bases are
stronger proton acceptors and the backwards reaction rate is
significant enough to reach equilibrium
IB is Annoying! Hints
• When representing a strong acid/base you MUST use a single
arrow forward to denote complete dissociation
• When representing a weak acid/base reaction you MUST use a
double arrow to show that this reaction reaches equilibrium
and take place in the forward and reverse direction
• The two weak acids you should remember are carboxylic acids
and carbonic acid (H2CO3)
SL Quiz
• Thursday, October 15th OR Friday, October 16th
• 4 multiple choice
• 17 points of short answers
Strong/Weak Bases
• In a similar way, bases can be described as strong or weak on
the basis of the extent of their ionization. For example, NaOH
is a strong base because it ionizes fully.
• NaOH(aq) → Na+(aq) + OH–(aq)
• Its dissociation is written without equilibrium signs.
(Note that it is the OH– ions that show Brønsted–Lowry base behavior by
accepting protons.)
• On the other hand, NH3 is a weak base as it ionizes only
partially, so its equilibrium lies to the left and the concentration
of ions is low.
• NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
Base Conjugates
• Strong bases are good proton acceptors and form conjugate
acids that are weak acids or do not show acidic properties at all
• Weak bases are poor proton acceptors and form conjugate
acids that are strong acids and better proton donors
• As we move onto HL level material, we will use the
equilibrium constant (remember Unit 7!) as a way to quantify
the strength of acids and bases
Nota Bene!
• DO NOT CONFUSE strength with concentration! A strong
acid can be made “weak” if it is present in a dilute enough
solution
• Be careful not to confuse two different pairs of opposites:
• strong and weak acids or bases refer to their extent of dissociation;
• concentrated and dilute refer to the ratio of solute and water in the
solution.
Common Strong and Weak Acids
and Bases
Orgo Sneak Peak
• There are hundreds of acids and bases in organic chemistry,
most of them relatively weak
• Organic acids typically have the –COOH functional group or
phosphoric acid
• Organic bases are usually derivatives of NH3 and have amine
functional groups
• Many of these are listed in Section 21 of the IB data booklet
Properties Strong/Weak Acid/Bases
1. Electrical conductivity – since stronger acids have a higher
concentration of ions, they show higher electrical conductivity
(though these comparisons are only valid at the same
concentrations since obviously electrical conductivity is related
to overall concentration of ions)
2. Rate of Reaction – as we learned in Unit 7, rate of reactions
are proportional to concentration so reactions that rely on the
H+ ions will happen faster with strong acids
3. pH – pH can be used to measure strength of acids/bases.
When equal amounts are added together, strong acids will
have lower pH than weak acids and vice versa for bases
Woo hoo! Practice Problems
Answers
16. B
17. A
18
(a) H2CO3
(b) HCOOH
Lesson 5
18.1 Lewis Acids and Bases
DWBAT identify Lewis acids and bases.
HL Material Already?
Understandings:
• A Lewis acid is a lone pair acceptor and a Lewis base is a lone
pair donor.
• Relations between Brønsted–Lowry and Lewis acids and bases should be
discussed.
• Both organic and inorganic examples should be studied.
• When a Lewis base reacts with a Lewis acid a coordinate bond
is formed.
• A nucleophile is a Lewis base and an electrophile is a Lewis
acid.
Applications and Skills
• Application of Lewis acid–base theory to inorganic and
organic chemistry to identify the role of the reacting species.
Lewis Acids
• Lewis (yes, the same Lewis as in the dots…) realized that all
bases needed to have a lone pair to accept a proton expanded
the definition of acids and bases to look at the electron
transferring happening
• He therefore defined a:
• Lewis acid as an electron pair acceptor and a
• Lewis base as an electron pair donators
Examples
• Look at the reaction of ammonia with a proton.
• The curly arrows will become your best friend (or worst enemy)
in Orgo and show us where electrons are being transferred
Expanded Definition
• Bronsted-Lowry bases and Lewis bases are essentially the same
group because both need to have a lone pair
• HOWEVER…now we could have acids that did not have
hydrogen; they only need an empty orbital to accept the lone
pair of electrons
• Lewis acid–base reactions result in the formation of a covalent
bond, which will always be a coordinate bond because both the
electrons come from the base.
• NOTA BENE: When referring to Lewis acids/bases, always
use the word “electron pair” acceptor/donor or else IB will
say you are talking about redox and not give you points!
Another Example
• BF3 has an incomplete octet so it can act as a Lewis acid
• When we studied complex ions, the ligands were acting as
Lewis bases and the transition metals as Lewis acids
• For example, Cu2+ in aqueous solution reacts as follows:
Cu2+(aq) + 6H2O(l) → [Cu(H2O)6]2+(aq)
• Cu2+ is a Lewis acid and H2O is a Lewis base.
Nucleophiles and Electrophiles
• A nucleophile (‘likes nucleus’) is an electron-rich species that
donates a lone pair to form a new covalent bond in a reaction.
(Lewis base)
• An electrophile (‘likes electrons’) is an electron-deficient
species that accepts a lone pair from another reactant to form a
new covalent bond. (Lewis acid)
• These terms are often used to describe reactions in terms of
electron-rich nucleophiles attacking electron-deficient
electrophiles, and are depicted using curly arrows to
show electron movements.
• These terms become especially important in Orgo!
Examples
Orgo Sneak Peek
• The reaction below shows the hydroxide ion, OH–, acting as a
nucleophile on an organic molecule known as a
halogenoalkane.
Comparing Theories
• Although all Brønsted–Lowry acids are Lewis acids, not all
Lewis acids are Brønsted– Lowry acids. The term Lewis acid is
usually reserved for those species which can only be described
by Lewis theory, that is those that do not release H+.
• Many reactions cannot be described as Brønsted–Lowry acid–
base reactions, but do qualify as Lewis acid–base reactions.
These are reactions where no transfer of H+ occurs.
Let’s Practice
Answers
Lesson 6
18.2 Calculations of Acids and Bases
DWBAT solve problems involving [H+ (aq)], [OH-(aq)], pH,
pOH, Ka, pKa, Kb and pKb.
Hold on to your HL seats…
IB Understandings
• The expression for the dissociation constant of a weak acid
(Ka) and a weak base (Kb).
• For a conjugate acid–base pair, Ka × Kb = Kw.
• The value Kw depends on the temperature.
• Only examples involving the transfer of one proton will be assessed.
• Calculations of pH at temperatures other than 298 K can be assessed.
• The relationship between Ka and pKa is pKa = –log Ka and
between Kb and pKb is pKb = –log Kb.
Applications and Skills
• Solution of problems involving [H+(aq)], [OH–(aq)], pH, pOH,
Ka, pKa, Kb, and pKb.
• Students should state when approximations are used in equilibrium
calculations.
• The use of quadratic equations will not be assessed.
• Discussion of the relative strengths of acids and bases using
values of Ka, pKa, Kb, and pKb.
• The calculation of pH in buffer solutions will only be assessed in options
B.7 and D.4.
Kw
• Remember that Kw = [H+] [OH–] = 1.00 × 10–14 at 298 K
• Since Kw is an equilibrium constant, it depends on
temperature. Water dissociating in endothermic so it happens
more readily at higher temperatures
• Increasing temperature will shift the equilibrium right,
increasing the concentration of H+ ions and decreasing pH
• Reducing temperature will shift the equilibrium left, decreasing
the concentration of H+ ions and increasing pH
Implications
• This means that the pH of water is only 7 at 298K!
• It doesn’t mean that pure water is not still neutral at higher or
lower temperatures (the concentration of H+ ions is still equal
to the concentration of OH- ions) but the pH does change
• What it means is that since Kw is temperature dependent,
temperature should always be stated with pH values to get a
sense of the true acidity of the solution
pOH
• Another way to measure the acidity/basicity of a solution is to
measure the concentration of OH- ions in aqueous solution; the
formula for measuring pOH is the same as measuring pH but
we change what concentration we are looking at!
• In aqueous solutions, pH + pOH = 14 (but only at 298K)
pH and pOH
pKw
• In the same way as the negative logarithms to base 10 of H+
and OH– are known as pH and pOH respectively, the same
terminology can be applied to Kw to derive pKw.
pKw = –log10 (Kw)
Kw =10–pKw
• So we can rewrite the expression above in a form that will
apply to all temperatures:
pH + pOH = pKw
Summary of Relationships
Now onto practicing
• Lemon juice has a pH of 2.90 at 25 °C. Calculate its [H+],
[OH–], and pOH.
More Practice
• Calculate the pH of the following at 298 K:
• (a) 0.10 mol dm–3 NaOH(aq) (Hint: get pOH first)
• (b) 0.15 mol dm–3 H2SO4(aq) (Hint: take mole ratio into account)
here!)
More Practice!
Answers
How does this get more difficult?
• Because you know it does…
• Think back to Unit 7 and the equilibrium constant problems
look at initial concentrations, change in concentrations, and
final concentrations. We will need to use these!
• For strong acids and bases, we can safely assume full
dissociation (1 mol of HCl = 1 mol of H+ ions, 1 mol of
H2SO4 = 2 mols of H+ ions, etc).
• For weak acids and bases, we have to calculate final
concentrations of H+ ions or OH- ions (and therefore pH,
pOH) using Keq
Lesson 7
18.2 Calculations of Acids and Bases
DWBAT solve problems involving [H+ (aq)], [OH-(aq)], pH,
pOH, Ka, pKa, Kb and pKb.
Dissociation Constant Expression
• Recall that the equilibrium expression is the product of the
concentration of products raised to the power of their
coefficient in the balanced chemical equation divided by the
product of the concentration of reactants raised to the power of
their coefficient in the balanced chemical equation
• Or, for
aA + bB  cC + dD
Keq = [C]c[D]d
[A]a[B]b
Acid Dissociation Constant
Expression
• Since weak acids and bases do not fully dissociate, we simply
cannot deduce pH from the initial concentrations of acid and
base; we must use the equilibrium expression to first figure out
the final concentrations and go from there
• For a weak acid, a generic equation can be written:
HA(aq) + H2O(l)  H3O+(aq) + A–(aq)
But since
water is a
constant 
Ka
• Ka is known as the acid dissociation constant.
• It will have a fixed value for a particular acid at a specified
temperature.
• The higher the value of Ka at a particular temperature, the
greater the dissociation, and so the stronger the acid.
• Note that because Ka is an equilibrium constant, its value does not
depend on the concentration of the acid or on the presence of other
ions. This will be important when discussing buffers.
Base Dissociation Constant
Expression
• We can also write an equilibrium expression for weak bases in
similar fashion
Kb
• Kb is known as the base dissociation constant.
• It will have a fixed value for a particular base at a specified
temperature.
• As with Ka, the value of Kb relates to the position of the
equilibrium and so in this case to the strength of the base.
• The higher the value of Kb at a particular temperature, the
greater the ionization and so the stronger the base.
Summary
Let’s Do This!
• Write the expressions for Ka and Kb for the following acid and
base. (a) CH3COOH(aq) (b) NH3(aq)
More Practice
Answers
Calculation of Ka and Kb from pH
and initial concentration
• Calculate Ka at 298 K for a 0.01mol dm–3 solution of ethanoic
acid (CH3COOH). It has a pH of 3.4 at this temperature.
Calculate the Kb for a 0.100 mol dm–3
solution of methylamine, CH3NH2 at 25 °C
. Its pH is 11.80 at this temperature.
Calculation of [H+] and pH, [OH–]
and pOH from Ka and Kb
• A 0.75 mol dm–3 solution of ethanoic acid has a value for Ka =
1.8 × 10–5 at a specified temperature. What is its pH at this
temperature?
A 0.20 mol dm–3 aqueous solution of
ammonia has Kb of 1.8 × 10–5 at 298 K .
What is its pH?
Lesson 8
18.2 Calculations of Acids and Bases
DWBAT solve problems involving [H+ (aq)], [OH-(aq)], pH,
pOH, Ka, pKa, Kb and pKb.
REVIEW FROM YESTERDAY
29
30
31
pKa and pKb
• While Ka and Kb are a good measure of how acidic or basic a
substance is, these values tend to be very small and span a large
range
• What do we do with very small numbers that span large ranges?
You guessed it…Log ‘em. Introducing pKa and pKb.
Important Points
1. pKa and pKb numbers are usually positive and have no units.
2. The relationship between Ka and pKa and between Kb and
pKb is inverse.
3. A change of one unit in pKa or pKb represents a 10 fold
change in the value of Ka or Kb.
4. pKa and pKb must be quoted at a specified temperature.
Relationship between Ka and Kb,
pKa and pKb for a conjugate pair
• Consider the Ka and Kb expressions for a conjugate acid–base
pair HA and A–.
pKa/pKb For Conjugate Pair
• Therefore
• What does this show us?
1. The higher the pKa for an acid, the lower the pKb for the
conjugate base!
2. The higher the pKb for a base, the lower the pKa for the
conjugate acid!
Summary
Let’s Practice
Answers
Lesson 9
18.3 Buffers
DWBAT define how buffers work
IB Understandings
• The characteristics of the pH curves produced by the different
combinations of strong and weak acid and bases.
• An acid–base indicator is a weak acid or a weak base where the
components of the conjugate acid–base pair have different colors.
• Examples of indicators are listed in the data booklet in section 22.
• The relationship between the pH range of an acid–base indicator,
which is a weak acid, and its pKa value.
• The colour change can be considered to take place over a range of pKa ± 1.
• The buffer region on the pH curve represents the region where small
additions of acid or base result in little or no change in pH.
• The composition and action of a buffer solution.
IB Skills and Applications
• The general shapes of graphs of pH against volume for titrations
involving strong and weak acids and bases with an explanation of
their important features.
• Selection of an appropriate indicator for a titration, given the
equivalence point of the titration and the end-point of the indicator.
• While the nature of the acid–base buffer always remains the same,
buffer solutions can be prepared by either mixing a weak acid/base
with a solution of a salt containing its conjugate, or by partial
neutralization of a weak acid/base with a strong acid/base.
• Prediction of the relative pH of aqueous salt solutions formed by the
different combinations of strong and weak acid and base.
• Salts formed from the four possible combinations of strong and weak acids and
bases should be considered. Calculations are not required.
• The acidity of hydrated transition metal ions is covered in topic 13. The treatment
of other hydrated metal ions is not required.
Buffer Solutions
• A bufer refers to something that acts to reduce the impact of
one thing on another – a little bit like a shock absorber.
• In acid–base chemistry, a buffer acts to reduce the pH impact of
added acid or base on a chemical system.
• A buffer solution is resistant to changes in pH on the
addition of small amounts of acid or alkali.
pH Changes
• Let’s look at how a small quantity of a strong acid or base can
affect the pH of a solution
• Here, a few drops of acid or base can change the pH by 3!
• Scary considering our enzymes only work over a narrow
range of pH! So how to biological systems maintain order?
Buffers!
How Buffers Work
• Buffers can maintain a pH of 7 or any other pH that is
desirable
• Buffers are always made of a mixture of two solutions that
create an equilibrium that is difficult to disturb
• The mixture of solutions will contain two species of conjugate
acid-base pairs
Types of Buffers: Type 1 Acidic
Buffers
Composition
• Acidic buffers are made by combining a weak acid with its salt
of a strong alkali
Types of Buffers: Type 1 Acidic
Buffers
Composition (cont.)
• Since this mixture contains large quantities of both CH3COOH
and CH3COO–, that is an acid and its conjugate base, these can
be thought of as reserves with which to react with either an
acid or base being added
Types of Buffers: Type 1 Acidic
Buffers
Response
Type of Buffers: Type 2 Basic
Buffers
Composition
• Made by mixing an aqueous solution of a weak base with its
salt of a strong acid.
Type of Buffers: Type 2 Basic
Buffers
Response
Summary
• Buffer solutions are a mixture containing both an acid and a
base of a weak conjugate pair.
• The buffer’s acid neutralizes added alkali, and the buffer’s base
neutralizes added acid, and so pH change is resisted.
Making Buffers
• The pH of a buffer is determined by the interactions of its
components. Specifically it depends on:
1. the pKa or pKb of its acid or base;
2. the ratio of the initial concentrations of acid and salt, or base and
salt, used in its preparation.
• While we will not get into the equations for making buffers
in this unit, we will cover this whether we choose
Biochemistry or Medicinal Chemistry as our option!
Making Buffers (cont.)
• Buffer solutions can be prepared by starting with an acid or
base that has a pKa or pKb value as close as possible to the
required pH or pOH of the buffer. This is then either:
• mixed with a solution of a salt containing its conjugate or
• partially neutralized by a strong base or acid to make sure ~50% of
the acid or base is in its salt form
• After either of these reactions what we have is approximately
50/50 of the starting acid/base and the salt by moles
• Example below:
More To Come
• If you look at older texts, you may see questions on buffer
calculations in the general HL level material; this has now been
moved exclusively to the options (covered in both Biochemistry
and Medicinal Chemistry so we can’t escape it!)
• The equation below is the Henderson-Hasselbach equation and
will come in handy when performing buffer calculations.
Practice
• State whether each of the following mixtures will form a buffer
solution when dissolved in 1.00 dm3 of water.
(a) 0.20 mol NaHCO3 and 0.20 mol Na2CO3
(b) 0.20 mol CH3COOH and 0.10 mol HCl
(c) 0.20 mol NH3 and 0.10 mol HCl
(d) 0.10 mol H3PO4 and 0.20 mol NaOH
Answers
• (a) 0.20 mol NaHCO3 and 0.20 mol Na2CO3
• Solution contains HCO3– and CO32–, a conjugate pair, so it is a
buffer.
• (b) 0.20 mol CH3COOH and 0.10 mol HCl
• Solution contains two acids – it is not a buffer.
• (c) 0.20 mol NH3 and 0.10 mol HCl
• NH3 and HCl react together forming 0.10 mol NH4Cl and 0.10 mol
NH3 unreacted; it is a buffer.
• (d) 0.10 mol H3PO4 and 0.20 mol NaOH
• H3PO4 and NaOH react together forming 0.20 mol Na2HPO4 it is
not a buffer.
Factors That Influence Buffers
Factor 1: Dilution
• Since Ka and Kb are equilibrium constants, they are not
influenced by dilution. The ratio of acid or base to its salt
remains the same. Therefore pH is the same.
• HOWEVER diluting a buffer does weaken a buffer as it lessens
the amount of acid or base it can absorb without significantly
changing the pH. This is called the buffering capacity which
depends on the molar concentrations of the acid or base and
salt
Factors That Influence Buffers
Factor 2: Temperature
• As we learned in our last unit, the equilibrium constant (in this
case Ka and Kb) is influenced by temperature. Therefore, the
pH of a buffer will change at different temperatures
• Look ahead: this is why temperature must be carefully
maintained in many medical procedures because it can
influence the buffers in our body
Let’s Practice
Answers
37. B A buffer solution must contain approximately equal amounts of a weak
acid or weak
base with its conjugate base or acid. In the examples given here, you must
consider how the given components will react together and the proportions of
products and unused reactants that will result. We will consider each in turn.
A: equimolar quantities of weak acid and strong base which react in 1 : 1 ratio,
with the resulting mixture containing salt and water only: not a buffer.
B: weak acid and strong base in 2:1 ratio by moles. These react in 1:1 ratio, so
the resulting mixture contains (unreacted) weak acid and salt in equimolar
amounts: this is a buffer.
C: weak acid and strong base in 1:2 ratio by moles. These react in 1 : 1 ratio, so
the resulting mixture contains salt and (unreacted) strong base: not a buffer.
D: weak acid and strong base in 2:1 ratio by moles, but here the reacting ratio
will be 2:1 as Ba[OH]2 will neutralize 2 moles of CH3COOH. The resulting
mixture contains salt and water only: not a buffer.
Answers
38. B The concept here is similar to Q37. You must work out the
products of each reaction.
I: The components do not react together. The mixture will contain
equal moles of the weak acid CH3COOH and its salt
CH3COONa. This is a buffer.
II: There is a 2 : 1 ratio of weak acid to strong base, so after
reaction the mixture will contain equal quantities of weak acid
and its salt. This is a buffer.
III: There is a 1:1 ratio of weak acid and strong base, so after
reaction the mixture contains salt and water only. This is not a
buffer.
Answers
39 (i), because it has a higher concentration of the acid and its
conjugate base, and so has the capability of buffering to a greater
extent than the mixture with the lower concentrations of solution.
Lesson 10
18.3 Salt Hydrolysis
Neutralization Leads To Salts
• A neutralization between an acid and a base always produces
water and a salt as the products
• A salt is an ionic compound which gets the cation from the
base and the anion from the acid
• However, even though salts are part of a neutralization reaction
not all solution of salts are completely neutral; it depends on
the strength of the parent acid and base!
Strength of Salts
• Remember that the stronger the acid, the weaker the conjugate
base will be and the strong the base, the weaker the conjugate
acid will be (and vice versa!)
• The relative strengths of the conjugate acid and base will
determine the pH of a solution of the resulting salt
Strong Acid-Strong Base
• When a strong acid and a strong base react, both of their
conjugates are weak; therefore there will be no hydrolysis and
the resulting salt will produce a neutral solution
Weak Acid-Strong Base
• A weak acid will produce a relatively stronger conjugate base
(the anion in the salt) than the parent base will produce a
conjugate acid (the cation)
• When the acid is weak this conjugate base is strong enough to
cause hydrolysis:
• This causes an increase in the concentration of OH- and an
increase in the pH of the solution
Strong Acid-Weak Base
• In this case, the weak base has the relatively stronger conjugate
acid (which will be the cation in the salt) which will cause
water to hydrolyze
• This reaction will increase the concentration of H+ in the
solution and decrease the pH
• NOTE: Metal ions with higher charge densities are better at
carrying out hydrolysis than those with lower (Al3+ > Na1+)
Weak Acid-Weak Base
• Since both conjugates of a weak acid and a weak base are
relatively strong, they both facilitate hydrolysis
• The pH of the solution therefore depends on the relative Ka
and Kb values of the acids and bases involved.
Summary
Let’s Practice
Answers
Answers
Lesson 11
18.3 pH Curves
DWBAT perform calculations involving titrations in relation to pH
curves.
Titrations
• We should be familiar with titrations, which are used to
determine an unknown concentration of an acid or base by
slowly neutralizing it with a known concentration of the
opposite
• Make sure you
know the
names of the
equipment and
all that fun
stuff for IB!
Procedure (Review)
• Controlled volumes of one reactant are added from a burette to
a fixed volume of the other reactant that has been carefully
measured using a pipette and placed in a conical flask.
• The reaction between acid and base takes place in the flask
until the equivalence point or stoichiometric point is reached,
where they exactly neutralize each other.
• An indicator or a pH meter are used to detect the exact volume
needed to reach equivalence.
pH Curves
• As a base is added to an acid (or vice versa) the pH of the
solution starts to change as the neutralization reaction happens
• However this chance is not linear (which you would expect as
pH is a log function)
• If we plot the change in pH as varying concentrations of
acid/base are added to a base/acid, the resulting curve is
called a pH curve!
pH Curve Shape
• The easiest way to follow the reaction is to record pH using a
pH meter or data-logging device as a function of volume of
base added, and plot these values as pH curves.
• It is found that in most titrations a big jump in pH occurs at
equivalence, and this is known as the point of inflection. The
equivalence point is determined as being half- way up this
jump.
Strong Acid and Strong Base
• **Note that we are going to look at examples where the acid and base
have the same molarity and react in a 1:1 ratio so that at the
equivalence point, the pH is determined only by water and the salt
• Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
• pH at equivalence = 7 as neither ion hydrolyses appreciably
Strong Acid and Strong Base
Key Points
1. The initial pH is going to be close to 1. Why? We start with a
strong acid
2. pH changes only gradually until equivalence
3. There is a very sharp jump at the equivalence point; in this
case from pH 3 to pH 11
4. After equivalence the curve flattens out until it reaches the pH
of a strong base (high value)
5. pH at equivalence is 7
Weak Acid and Strong Base
• Example: CH3COOH(aq) + NaOH(aq) s NaCH3COO(aq) +
H2O(l)
• pH at equivalence > 7 as the anion hydrolysis releases OH–
Weak Acid and Strong Base
Key Points
• initial pH fairly high (pH of weak acid)
• pH stays relatively constant until equivalence – labeled as buffer
region;
• jump in pH at equivalence from about pH 7.0–11.0, which is not
as much of a
• jump as for a strong acid–strong base titration;
• after equivalence the curve flattens out at a high value (pH of
strong base);
• 5 pH at equivalence is > 7
Half-Equivalence
• If we start with 50cm3 acid, the half-equivalence point occurs
when 25cm3 of the base has been added. This point is where
exactly half of the acid has been neutralized and turned into
the salt.
• Since now this solution has ½ weak acid and ½ the conjugate
salt, it is a BUFFER! That is why this area of the curve is
called the buffer region
Buffer Region Calculations
• This may be an area where you will see calculation questions.
• The pH at the half-equivalence point gives us an easy way to
calculate pKa. Because at this point [acid] = [salt], we can
substitute these values into the equilibrium expression of the
acid:
• Since the [acid] or [HA] at this point equal [salt] or [A-], these
terms cancel out and Ka=[H+] and therefore pKa=pH
• We can read pH from the curve and calculate pKa!!!
Strong Acid and Weak Base
• Example: HCl(aq) + NH3(aq) s NH4Cl(aq)
• pH at equivalence < 7 as cation hydrolysis releases H+
Strong Acid and Weak Base
Key Points
1. Initial pH =1 (strong acid);
2. pH stays relatively constant through the buffer region to
equivalence;
3. Jump in pH at equivalence from about pH 3.0–7.0;
4. After equivalence the curve flattens out at a fairly low pH (pH
of weak base);
5. pH at equivalence is < 7.
Weak Base and Weak Acid
• Example: CH3COOH(aq) + NH3(aq)  NH4CH3COO(aq)
• pH at equivalence is difficult to define.
Weak Acid and Weak Base
Key Points
1. initial pH is fairly high (pH of weak acid);
2. addition of base causes the pH to rise steadily;
3. change in pH at the equivalence point is much less sharp than
in the other titrations;
4. after equivalence the curve flattens out at a fairly low pH (pH
of weak base).
Adding Acid to Base
• In all of these examples, we have added a base to an acid but
pH curves can be examined adding an acid to a base as well
• This is the way a titration must be performed to calculate Kb of
a weak base.
Let’s Do This!
Answers
Lesson 12
18.3 pH Curves
DWBAT perform more calculations using pH curves and choose
an indicator based on the end point of titration
Indicators
• Indicators are substances that change color at a certain pH
• Indicators are weak acids and bases themselves; the external
concentration of H+ ions causes a shift in their own equilibrium and
the acid/base and conjugate base/acid have different colors
• Another example of Le Chatelier’s principle in action!
• Increasing [H ]: the equilibrium will shift to the left in favor of HIn.
+
• Decreasing [H ]: the equilibrium will shift to the right in favor of In .
+
–
End-Point
• The change in color of an indicator is know as its end-point or
change-point
• What determines this point? The pKa of the weak acid or base. Why?
Note!
• Do not confuse the term end-point with equivalence point
• The equivalence point is where stoichiometrically equal amounts
of acid and base have neutralized each other.
• The end-point is the pH at which the indicator changes color.
Indicators and Titrations
• Indicators can be used during titration
• An indicator will be effective in signaling the equivalence point
of a titration when its end-point coincides with the pH at the
equivalence point.
• This means that different indicators must be used depending on
the acid-base reaction going on
• Up until this point, we have mostly done strong-strong which
neutralized at 7 so phenolphthalein is appropriate because the pH
rapidly goes from very low to very high
Steps For Determining Indicator
1. Determine what combination of weak and strong acid and
base are reacting together.
2. Deduce the pH of the salt solution at equivalence from the
nature of the parent acid and base.
3. Choose an indicator with an end-point in the range of the
equivalence point by consulting data tables.
Example
• If we titrate a weak acid and a strong base, the equivalence might between
pH 7 and 11
• Therefore phenolphthalein (end-point 8.2-10) might be appropriate
Indicator and Range
Why a Range?
• In Regents Chemistry, everyone used to ask why there was a
range over which color changes and “what happens in the
middle?”
• Now that we know how weak acids and bases work, we know
that over this range the equilibrium is slowly shifting from
reactants to products or vice versa; HOWEVER our eyes are
not perfect and since there is a mixture of both colors in the
solution it is not until the ratio is approximately 10:1 that our
eyes perceive one color verse another
• Since pH is a log base 10 scale that is about 1 pH above and
below the point where pKa = pH or end-point
Let’s Practice
46 Which statement about indicators is always correct?
A The mid-point of an indicator’s colour change is at pH = 7.
B The pH range is greater for indicators with higher pK values.
C The colour red indicates an acidic solution.
D The pK value of an indicator is within its pH range.
a
a
47 Bromocresol green has a pH range of 3.8–5.4 and changes
colour from yellow to blue as the pH increases.
(a) Of the four types of titration shown in the table above, state in
which two of these this indicator could be used.
(b) Suggest a value for the pK of this indicator. (c) What colour
will the indicator be at pH 3.6?
a
Answers
• 46. D Different indicators have different pH values at which
they change colour (so A is wrong). The size of the pH range
over which the colour change occurs and the pKa of an
indicator are not related (so B is wrong).
• The colour observed in acidic solution will depend on the
specific indicator (so C is wrong). The pH range of the colour
change always includes the pKa of the indicator (so D is
correct).
Answers - 47
Lesson 13
Topic 8.5 Acid Deposition
DWBAT examine the environmental effects of industrialization
and acid rain
Understandings
• Rain is naturally acidic because of dissolved CO2 and has a pH
of 5.6. Acid deposition has a pH below 5.6.
• Acid deposition is formed when nitrogen or sulfur oxides
dissolve in water to form HNO3, HNO2, H2SO4, and H2SO3.
• Sources of the oxides of sulfur and nitrogen and the effects of
acid deposition should be covered.
Applications
• Balancing the equations that describe the combustion of sulfur
and nitrogen to their oxides and the subsequent formation of
H2SO3, H2SO4, HNO2, and HNO3.
• Distinction between the pre-combustion and post-combustion
methods of reducing sulfur oxide emissions.
• Deduction of acid deposition equations for acid deposition
with reactive metals and carbonates.
Acid Rain
• All rain is naturally acidic because as it falls through the
atmosphere, some of the CO2 is dissolved which causes the
concentration of H+ ions to increase according to the following
reaction that occurs:
• We only call rain “acid rain” when the pH is below 5.6, and
which therefore contain additional acids
Acid Rain
• https://www.youtube.com/watch?v=MqHw1hMEkAQ
Acid Deposition
• Acid deposition is a broader term than acid rain and includes
all processes by which acidic components as precipitates or
gases leave the atmosphere.
• There are two main types of acid deposition:
• wet acid deposition: rain, snow, sleet, hail, fog, mist, dew fall to
ground as aqueous precipitates;
• dry acid deposition: acidifying particles, gases fall to ground as dust
and smoke, later dissolve in water to form acids.
Do I Really Have To…
• …memorize these equations?
• YES! File it along with your 10,000 declension and conjugation
endings at stuff you just need to memorize. Life is tough.
Pollution: Sulfur Dioxide
• Sulfur dioxide, SO2, is produced from the burning of fossil fuels,
particularly coal and heavy oil in power plants used to generate
electricity.
• It is also released in industrial processes of smelting where metals are
extracted from their ores.
• Sulfur dioxide is a colourless gas with a sharp smell. It dissolves in
water to form sulfurous acid, H2SO3(aq).
Mechanism
• There are several mechanisms that might occur in these
reactions
• During sunlight hours photo-oxidation may occur, and oxidation
may also be catalyzed by tiny particles of metal present in the
clouds, such as iron or manganese.
• Ozone (O3) or hydrogen peroxide (H2O2) present as pollutants in
the atmosphere, can be involved.
• Hydroxyl free radicals, •HO, which form by reactions between
water and atomic oxygen or ozone may also be involved
• •HO + SO2 → •HOSO2
• •HOSO2 + O2 → •HO2 + SO3
Nitrogen Oxides
• NO is produced mainly from car engines (we’ve touched on
this before with ozone depletion)
• N2(g) + O2(g) → 2NO(g) ∆H = +181 kJ mol–1 OR
• N2(g) + 2O2(g) → 2NO2(g)
• NO2 can also be formed from NO: 2NO(g) + O2(g) → 2NO2(g)
• Nitrogen dioxide can react with water to form a mixture of
nitrous acid and nitric acid:
• H2O(l) + 2NO2(g) → HNO2(aq) + HNO3(aq)
• It can also be oxided to form just nitric acid:
• 2H2O(l) + 4NO2(g) + O2(g) → 4HNO3(aq)
Mechanism
• Photo-oxidation, the presence of ozone, and hydroxyl free
radicals (•HO) all contribute to the production of nitrous acid
and nitric acid.
• •HO + NO → HNO2
• •HO + NO2 → HNO3
Effects of Acid Rain/Deposition
1. Destruction of buildings and statues
Marble and limestone are both forms of calcium carbonate
CaCO3 which reacts with sulfuric acid to form the more soluble
calcium sulfate CaSO4 which can then wash away
•
• 2CaCO3(s) + 2SO2(g) + O2(g) → 2CaSO4(aq) + 2CO2(g)
• CaCO3(s) + H2SO4(aq) → CaSO4(aq) + H2O(l) + CO2(g)
Calcium sulfate also has a higher molar volume which can cause
cracks and and expansion in stonework
A similar reaction happens with nitric acid
•
•
•
•
CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2O(l) + CO2(g)
This can happen during wet or dry deposition
Effects of Acid Rain/Deposition
1. Destruction of buildings and statues
Metals can also be destroyed by acid deposition (both dry and
wet)
Acids react with metals to form salts which quickens corrosion
and rust
•
•
•
•
Acid is also able to remove the protective layer on the surface of
metals
•
•
•
Fe(s) + SO2(g) + O2(g) → FeSO4(s)
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
Al2O3(s) + 6HNO3(aq) → 2Al(NO3)3(aq) + 3H2O(l)
Since metals make up many of our buildings and bridges, this is
not a good thing!
Effects of Acid Rain/Deposition
2. Destruction of Plant Life
•
•
•
•
Acid rain has been shown to be a direct cause of slower growth,
injury, or death of plants.
One of its effects is to cause important minerals such as Mg2+,
Ca2+, and K+ held in the soil to become soluble and so wash
away in a process called leaching, before they can be absorbed by
plants. Without sufficient Mg2+ ions, for example, a plant cannot
synthesize chlorophyll and so cannot make its food through
photosynthesis.
At the same time, acid rain causes the release of substances that
are toxic to plants, such as Al3+, which damage plant roots.
In addition, dry deposition can directly affect plants by blocking
the pores for gas exchange, known as stomata.
Effects of Acid Rain/Deposition
3. Destruction of Bodies of Water
•
•
•
Many fish cannot live if the pH dips below 5
Below 4, a body of water is considered dead as toxic Al3+ ions
normally trapped in the rock as insoluble aluminum hydroxide
leach out under acid conditions: Al(OH)3(s) + 3H+(aq) →
Al3+(aq) + 3H2O(l)
Acid rain also contributes to an additional problem known as
eutrophication. This is over-fertilization of bodies of water, and
can be caused by nitrates present in acid rain. It results in algal
blooms leading to oxygen depletion, and sometimes the death of
the lake or stream.
Algal Bloom – Lake Eerie 2011
What About Us?
• Other than destroying our buildings, food supply, and more,
sulfate and phosphate particles can irritate our respiratory
tracks and eyes
• Metal ions can also pose health risks
So What Do We Do?!
1. Limit SO2 Emissions
a) Pre-Combustion Methods
• Since much of the sulfates come from burning coal or oil, we can try to
eliminate the sulfur in the fossil fuel before we burn it (the original focus of
“clean coal” technologies before we realized we also need to minimize the
amount of CO2 we churn into the environment.
• We can try to eliminate sulfur when it is present as a metal sulfide by
crushing coal and washing; higher density compounds sink to the bottom.
• Hydrodesulfurization (HDS) is a catalytic process that removes sulfur from
refined petroleum products by reacting it with hydrogen to form hydrogen
sulfide, H2S. This is a highly toxic gas, so it is captured and later converted
into elemental sulfur for use in the manufacture of sulfuric acid, H2SO4.
So What Do We Do?!
1. Limit SO2 Emissions
b) Post-Combustion Methods
• Flue-gas desulfurization can remove up to 90% of SO2 from flue gas in the
smoke stacks of coal-fired power stations before it is released into the
atmosphere. The process uses a wet slurry of CaO and CaCO3 which reacts
with SO2 to form the neutral product calcium sulfate, CaSO4.
• CaO(s) + SO2(g) → CaSO3(s)
• CaCO3(s) + SO2(g) → CaSO3(s) + CO2(g)
• 2CaSO3(s) + O2(g) → 2CaSO4(s)
So What Do We Do?!
2. Limit NOx Emissions
a) Catalytic Converters
• Exhaust gases can be controlled by the use of catalytic converters in which
the hot gases are mixed with air and passed over a platinum- or palladiumbased catalyst.
• The reaction converts toxic emissions into relatively harmless products.
• 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)
So What Do We Do?!
2. Limit NOx Emissions
b) Lower Temperature Combustion
• Think about N2 and O2 just for a hot second…why would lowering the
temperatures prevent Nox emissions?
• Recirculating the exhaust gases back into the engine lowers the temperature
to reduce the nitrogen oxide in the emissions.
So What Do We Do?!
3) Lower our demand for burning fossil fuels
Less…
More..
4) Restore ecosystems damaged by using calcium oxide CaO to
neutralize acid
CaO(s) + H2SO4(aq) → CaSO4(s) + H2O(l)
Ca(OH)2(s) + H2SO4(aq) → CaSO4(s) + 2H2O(l)
Let’s Practice
Let’s Practice
Lesson 14
Review Problems!
Review
Describe two different properties that
could be used to distinguish between
a 1.00 mol dm-3 solution of a strong
monoprotic acid and a 1.00 mol dm-3
solution of a weak monoprotic acid.
Taken from May 2011 Exam
1.
2.
What does the command term
describe mean you have to do?
If this is worth two points, there
are at least three different
answers you could put. Let’s
brainstorm these!
• You could measure pH using a
universal indicator or pH probe.
The stronger acid will dissociate
more and therefore have a lower
pH due to the higher concentration
of H+ ions in solution.
• Conductivity (measurement): the
strong acid will be a better
conductor
• The strong acid will react more
vigorously with metals/carbonates
• The heat change when it is
neutralized with a base will be
different; heat of neutralization;
Review
Ethanoic acid, CH3COOH, is a weak
acid.
(a) Define the term weak acid and state
the equation for the reaction of
ethanoic acid with water.
(b) Vinegar, which contains ethanoic
acid, can be used to clean deposits of
calcium carbonate from the elements
of electric kettles. State the equations
for the reaction of enthanoic acid
with calcium carbonate.
Take from IB May 2009
You do not have to memorize every
reaction as long as you understand
what is happening!
Notice that you do need to remember
charges on polyatomic ions. They
won’t tell you the formula for calcium
carbonate. CO32-
Review
• 0.100 mol of ammonia, NH3, was dissolved in water to make 1.00 dm-3 of
solution. This solution has a hydroxide ion concentration of 1.28 * 10-3 mol
dm-3
• (a) Determine the pH of the solution [2]
• (b) Calculate the base dissociation constant, Kb, for ammonia. [3]
IB November 2009