Science 30
Unit B: Chemistry and the
environment
Chapter 1: Acid Deposition
1.1- Products of Combustion
Reactions
 Combustion
reactions (eg. Cellular
respiration, burning fossil fuels) are useful
but produce emissions.
 Collisions between the methane and
oxygen molecules form new molecules.
 If a hydrocarbon combusts, H20 and CO2
are formed; these are the waste products.
Chemical Equations
a) Oxides of Carbon
 Burning
carbon compounds, or anything
with biomass, or hydrocarbons results in
carbon dioxide.

Examples: C/R, volcanic eruptions
 Carbon
dioxide is important in the carbon
cycle and as a greenhouse gas;
contributes to climate change.
carbon cycle animation
Greenhouse effect
Greenhouse Effect - animated diagram
7
Hole in the ozone layer
Carbon monoxide
 Carbon
monoxide (CO) is produced when
the quantity of oxygen is limited in
combustion.
 Caused by vehicle exhaust, furnaces (in
poor condition); natural concentration of
carbon monoxide in air is around 0.2 parts
per million (ppm).
 Can bind to hemoglobin = decreased
amount of oxygen to tissues.
CO
b) Oxides of sulfur
 Found
when coal and crude oil or tar are
burnt; found in natural gas as sour gas
(hydrogen sulfide).
 Sour gas needs to have the hydrogen
sulfide removed.
 The amount of SO2 released depends on
the sulfur content of coal, normally 0.7% to
2% by weight. High sulfur coal sometimes
contains as much as 6% sulfur by weight.
Air pollution reduction
 Catalytic
Converters
 Catalytic Converters again
 Energy efficient homes
 Reduce recycle reuse
Air Pollution
flaring
 When
low quality natural gas is produced,
it is burned off to produce SO2 and SO3 =
flaring.
 Adds oxygen to hydrogen sulfide to form
products.
c) Oxides of Nitrogen
 Whenever
any fuel is combusted, nitrogen
is present (includes breathing).
 When temperature reaches 650°C,
Nitrogen activates and forms NO and NO2.
 These are referred to as NOx compounds.
 Common sources:


Combustion of fuels in cars and furnaces.
Fossil fuel power plants
Fossil Fuels
d) Monitoring emissions

The government creates standards
to protect environment, organisms
and support sustainability of
resources.
 Monitored using specialized
equipment and by an outside
group.
 Cars are monitored using MAML
labs.
1.2) Acids and Bases
 Acids,
Bases and Neutral solutions have
specific properties that are used to classify
them.
 Acids: conducts a current (electrolyte), pH
= 6 or less, corrosive, reacts with metals,
tastes sour.
 Bases: Conducts current, corrosive, pH = 8
or more, feels slippery, bitter.
 Neutrals: pH = 7, can be electrolytic or not.
There are 4 different types of
solutions
Neutral
molecular
Neutral
ionic
Acid
base
Solutions of
compounds that
are composed
of non-metals
only.
Solutions of
compounds that
are composed
of metals
combined with
non-metals
Solutions of
compounds that
produce
hydrogen ions.
Solutions of
compounds that
produce
hydroxide ions
C12H22O11(aq) NaCl(aq)
HCl(aq)
KOH(aq)
CH3OH(aq)
CaBr2(aq)
H2SO4(aq) Mg(OH)2(aq
No effect
on litmus
No effect
on litmus
Turns
litmus red
Turns
litmus blue
a) Types of deposition
2


types of deposition:
Wet: emissions that
contact precipitation and
return as rain/snow.
Dry: gases and particles
absorbed by the earth;
deposited on any
surface.
• Most deposition in Alberta
is from dry deposition.
b) Acids
 Can
be classified by properties (empirical)
or by chemical composition.
 Classified as a molecular compound but
behave like ionic compounds when
dissolved in water (electrolytic solutions).
 The water molecules break the bonds in
ionic compounds; these charges can now
move in a direction = conduct electricity.
Electrostatic Attraction
 Force
that pulls oppositely charged objects
towards each other.
 Water pulls positive ions towards oxygen;
creates a positive and negative charge.
 Dissociation occurs when 2 ions separate
into different charges.
Arrhenius
Svante Arrhenius formulated a theory in
1887 that all acids had a H+ ion and
bases had an OH- ion.
Problems:


1.
2.
Not all acids and bases have an H+ or OH-.
H+ can not exist in water because it is so
positively charged; actually forms H30
(hydronium ion) with a water particle.
1. Hydronium ion is responsible for acidic properties.
c) Bronsted-Lowry Acid-base
reactions
 Describes
the actions of acids and bases
during a chemical reaction.
 2 roles in the reaction:


Donor (acid) = gives H+ (proton) ion.
Acceptor (base) = accepts H+ (proton) ion.
 Product


is a conjugate acid/base.
Conjugate acid = formed when base accepts
H+.
Conjugate base = formed when acid accepts
H+.
Writing reactions
 Loss

Recognized by no H+ ion in formula.
 Gain

of a H+ ion by acid = conjugate base.
of H+ ion by base = conjugate acid.
Recognized by extra H+ ion in formula.
 Use
the table of acids and bases (page 12
in data book) and follow 5 steps.
Steps in Bronsted-Lowry reactions
Follow these 5 steps to write reaction:

1.
2.
Find the 2 solutions that are reacting.
Identify the acid and base.
1. Stronger acid is higher on the table- always
choose the highest one if both are listed!
2. The base is the non-acid (or weak acid).
3.
4.
5.
Write the reactants side of the equation.
Find conjugate form of acid and base.
Write conjugate forms on products side of the
equation.
HCl
acid-base
EXAMPLE - Conjugate Acids: Write the formula for the conjugate acid of
(a) F-, (b)NH3, (c) HSO4-, and (d) CrO42-.
Solution:
In each case, the formula for the conjugate acid is derived by adding one H+ ion to
the formulas above.
a. HF b. NH4+ c. H2SO4 d. HCrO4EXAMPLE - Conjugate Bases: Write the formula for the conjugate base of
(a) HClO3, (b)H2SO3, (c) H2O, and (d) HCO3-.
Solution:
In each case, the formula for the conjugate base is derived by removing one H+
ion from the formulas above.
a. ClO3b. HSO3c. OHd. CO32-
See page 174 Example problem 1.2/1.3
Proton hopping
 Confirmation
of Bronsted-Lowry theory.
 Used lasers to do this; captured images of
motion in chemical reactions.
 Helped predict outcomes of acid-base
reactions.
d) Acid Deposition
 Emissions
that are from human sources
are anthropogenic; from combustion of
energy sources.
 These emissions combine with water to
form acid rain= acidic precipitation.
 Rain is acidic due to natural and human
sources; the degree of acidity can be
measured using pH.
e) pH and pH scale
 pH
is measuring the amount of hydronium
ions (H30+) in a solution.
 The number of Hydronium ions influences:

Reactivity, amount of base needed to
neutralize/to react.
 pH
scale was developed in 1909 by
Sorenson; designed to measure dilute
acids.
 pH scale measures from 1-14 (1-6 = acid,
7= neutral, 8-14 = base).
pH
pH calculations
- SØren Sørensen came up with “power
of hydrogen” or pH
 pH corresponds to the hydronium ion
concentration in mol/L
 [ H3O+(aq) ] = 10-pH
 pH = 5 then [ H3O+(aq) ] = 10-5
 [ H3O+(aq)] = 0.00001 mol/L or 1.0 x 10-5 mol/L
 1909
36
pH to mol/L and back again …
pH
[
= - log [
+
H3O (aq)
+
H3O (aq)
]=
]
-pH
10
37
Converting pH to H3O+ concentration








What is the concentration of hydrogen ions for an
acid with a pH = 4.56?
You must take the inverse of a log
That is probably the 10x button on your calculator
Type it in your calculator as
10-4.56 = 0.000027542 mol/L
2.8 x 10-5 mol/L (sig figs)
that is the concentration of hydrogen ions
be sure to put the negative sign in before the 4.56
38
Significant figures and pH/pOH







Calculate the pH of a solution where the [H+] is 0.00100 M.
That's pretty easy, the answer is 3. After all 0.00100 is 10¯3 and
the negative log of 10¯3 is 3.
But the pH is not written to reflect the number of significant
figures in the concentration.
Notice that there are three sig figs in 0.00100 M.
So, our pH value should also reflect three significant figures.
Let's phrase that another way: in a pH (and a pOH), the only
place where significant figures are contained is in the decimal
portion.
So, the correct answer to the above problem is 3.000. Three sig
figs and they are all in the decimal portion, NOT (I repeat NOT)
in the whole number portion.
Example ...
 What
is the pH of a solution with a
[H3O+] of:
1.89 x 10-4 mol/L
Answer
= - log [1.89 x 10-4]
 pH = 3.723 this is an acidic solution
because the pH is less than 7
 pH
Practice
1 a)
pH = 7
= 10-7 = 1 x 10-7 mol/L
1 b)
pH = 11
= 10-11 = 1 x 10-11 mol/L
1 c)
pH = 2
= 10-2 = 1 x 10-2 mol/L
1 d)
pH = 4
= 10-4 = 1 x 10-4 mol/L
1 e)
pH = 14
= 10-14 = 1 x 10-14 mol/L
41
Practice
2 a) [H3O+(aq)] = 10-3 mol/L
pH = - log [ 10-3 ] = 3
2 b) [H3O+(aq)] = 10-5 mol/L
pH = - log [ 10-5 ] = 5
2 c)
[H3O+(aq)] = 10-7 mol/L
pH = - log [ 10-7 ] = 7
2 d)
[H3O+(aq)]
pH = - log [ 10-10 ] =
10
=
10-10
mol/L
42
f) Indicators





First nations used natural acids to adjust the
color of the dyes made from leaves, berries and
bark.
An indicator is anything that changes color in
response to a change in pH.
Common indicators are shown in the table on
page 12 of your booklet.
Used to measure the pH of a substance.
pH meter is more accurate and gives exact
measure of pH.
The Rainbow
Connection
Demonstration
Examples:
According to the acid-base indicator table, what is the color of
each of the following indicators in the solutions of given pH?
(a) Phenolphthalein in a solution with a pH = 12.7. RED
(b) Bromothymol blue in a solution of pH = 2.8
YELLOW
(c) Methyl orange in a solution of pH = 3.
RED
(d) Thymol blue in a pH = 5.0 solution
(e) Litmus in a solution with a pH of 8.2
YELLOW
BLUE
Example Problem:
 Separate
samples of an unknown
solution turned both methyl orange
and bromothymol blue to yellow,
and turned bromocresol green to
blue. The pH of the unknown solution
is likely __________
Example Problem:
 methyl
orange = yellow = 4.4+
 bromothymol
 bromocresol
 The
blue = yellow = 6.0-
green = blue = 5.4+
pH of the unknown solution is
likely between 5.4 and 6.0.
1.3) Impact of Acid Deposition
 Higher
levels of sulphates and nitrates in
rainwater = higher concentration of
hydronium ions and lower pH in water.
 Wind patterns affect the deposition;
provides way to trace pollution.
 Alberta soil is slightly basic (alkaline) from
the carbonate caused by erosion of
limestone; neutralizes the acid deposition.
a) Buffering

A buffer is used to resist pH change in soil or lake
water.
 The buffering capacity is the ability of a
substance to resistance pH change when an
acid/base is added.
 Specific plants need to be used in acidic soils;
most are not able to grow due to lack of nutrients.
 Nutrients are deposited in soil from
biogeochemical cycles and through neutralization
reactions.
b) pH and plants
 Plants
absorb nutrients into the roots from
the soil, up to the leaves.
 Acid deposition changes nutrients so they
are insoluble; plants can not use them.
 Nutrient deficiencies causes plants to die
or become diseased.

A
Chlorosis = calcium deficiency causes
decreased chlorophyll = yellow leaves.
neutral pH is best for plant growth; too
acidic or basic results in death.
c) Leaching

acid in soil makes metal ions (aluminum and
mercury) available to plants.
 Plants take those nutrients into their roots and
they dissolve in the soil or groundwater =
leaching.
 Affects ecosystem by:





Decreased root growth.
Prevents absorption of calcium.
Reduce decomposing soil bacteria.
Mercury causes damage to gills.
Methyl mercury traps in tissues = bioaccumulation
Biomagnification
A
pollutant increases in concentration up a
food chain.
 Causes disease and death.
Measured in
ppm (106) or
ppb (109) or
ppt (1012)
d) Effect on Biotic factors
 The
abiotic (non-living) factors in an
ecosystem affect the biotic (living).
 Acid deposition decreases the biodiversity
of the system.
 Acid deposition affects the ecosystem in:



Decreasing soil bacteria.
Destroying waxy coating on plants.
Damaging aquatic ecosystems.
1.4) Monitoring effects of acid
deposition
 Classification


of acids is done in 2 ways:
Quantitatively = involves measurement (pH,
titration)
Qualitative = involves properties,
characteristics, attributes (color, observations)
a) Titrations
 Used
to determine the amount of
acid/base present.
 Uses an acid, base and an indicator.
 When the “end point” is reached, the
indicator will change color and the amount
of acid/base needed to neutralize the
solution is known.
Calculating the Concentration of Acids and
Bases

Calculating concentrations: Calculate the concentration, in
moles per litre, of 250 mL of a solution containing 0.243 mol of
potassium hydroxide, KOH (s), used to analyze the concentration
of an acid solution.

Given:
n = .243 mol
V= 250 mL (0.250 L)

Formula: C = n/V
= 0.243 mol/0.250 L
= 0.972 mol/L

Answer: The KOH has a concentration of 0.972
mol/L

The CONCENTRATION of an acid can easily be changed by
adding more solvent (water).

What would the concentration of KOH be if an additional 150 mL
of water was added to the solution?
Technique
Solution #1: “standard solution”
Will know the concentration and the volume
used.
 Solution #2: “unknown”
Will know volume used but NOT the
concentration
 Acid / Base titrations will involve a
NEUTRALIZATION Reaction
 Neutralization Example:
acid + base → salt + water
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Titration Set-up
TITRANT


solution with known
concentration
goes in the buret
SAMPLE


solution with
unknown
concentration
goes in Erlenmeyer
Flask
Titration Process

The titrant will be added to sample drop by drop until
they have reacted fully (there is an equal amount of acid
and base) and the ENDPOINT is reached.

How do you know you have reached the endpoint?
The Erlenmeyer flask will also contain an INDICATOR
that will change color when the reaction has reached the
endpoint.

The indicator changes colour at the endpoint because
the [H3O+] and therefore pH will have changed
sufficiently.
How do you determine the volume of
titrant used?

Read the volume of titrant in the buret
before the titration begins AND once the
endpoint has been reached
 Subtract the volumes
 REMEMBER - read the bottom of the
meniscus!!
b) Acid Base Stoichiometry

KOH(aq) + HCl(aq)  KCl(aq) + HOH(aq)

Calculate ‘n’ for the given substance
Use a molar ratio to calculate ‘n’ for the required
substance
Make the appropriate calculation to answer the
question you are asked.


Calculate:
Find the concentration of a solution of potassium
hydroxide, KOH(aq), if it requires 8.32 mL of a 0.100
mol/L standard solution of hydrochloric acid to
neutralize 10.0 mL of the potassium hydroxide
solution.
Strong and Weak Acids and
Bases






Strength ≠ Concentration
Strength refers to the % of the acid or base that
dissociates in water
You can change the concentration but you can
NOT change strength
Strong acids and strong bases dissociate
completely in aqueous solutions
Weak acids - only a portion of the acid
molecules release protons
Weak base – only a portion of the base
molecules accept protons
Examples of Strong and Weak Acids /
Bases

Example:
 NaOH is a strong base
 NaOH dissociates completely in H2O
+
 NaOH  Na and OH

Example:
 Acetic Acid is a weak acid
 Most acetic acid molecules DO NOT release
protons but instead remain in undissociated form
CH3COOH + H2O  CH3COO- + H3O+


For every weak acid there is a conjugate base
(See page 12 of Data Booklet)
What is the conjugate base of acetic acid?
c) Buffers

A Buffer is a weak acid / weak base
conjugate pair
 Most effective when there are equal amount
of acid and base
 Example: H2CO3 in Blood


CO2 and H2O react to make H2CO3 (carbonic
acid)
Carbonic acid resists changes in blood pH
H2CO3 + H2O → H3O+ + HCO3 -
Sources of buffering
 Natural
sources include hydrogen
carbonate in your blood, levels of
carbonate in the soil and limestone
deposits.
 Buffers do not work forever however;
Eastern Canada has reached the capacity
to buffer soil and has acidic deposition
currently.