Module 2: The Acidic Environment
1) Indicators were identified with the observation that the colour of some
flowers depends on soil composition.
1.1) Classify common substances as acidic, basic or neutral.
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Acids are substances which, in solution, produces H+ ions.
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Bases are substances which either contains oxide O2- or OH- or produces OH ion in solution.
Soluble bases are called alkalis.
Common acids:
o Vinegar (acetic acid), fruit juices (citric acid), carbonated soft drinks (carbonic acid),
car battery acid, vitamin C (absorbic acid) and lactic acid.
o Sour taste, stings and burns skin.
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o
Conducts electricity.
o
Turns blue litmus red.
Common bases:
o Drain cleaners (sodium hydroxide), household cleaners (ammonia solutions), baking
soda solutions, washing soda solutions, oven cleaners and limewater.
o Bitter taste, soapy feel.
o Good conductors of electricity.
o Turns red litmus blue.
Neutral substances
o Water, table salt, glucose solution, alcohol-water solutions, lactose solutions.
Tip: A general rule for most multiple choice questions on this dot-point is: Cleaning products are
generally basic, Foods are usually acidic, Water and NaCl are neutral
1.2) Identify that indicators such as litmus, phenolphthalein, methyl orange and bromothymol
blue can be used to determine the acidic or basic nature of a material over a range, and
that the range is identified by change in indicator colour.
1.2.2) Identify data and choose resources to gather information about the colour changes of
a range of indicators.
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Acid–base indicator: a solution of a pigment or dye that changes colour in the presence of
acids and bases, they are usually weak acids or bases.
This means they can exist in two forms in an equilibrium reaction (different colours), which mix
together to provide the solution colour at different pH values.
Summary of indicators
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Litmus is an aqueous extract from a lichen.
o For neutral substances it is its natural blue-purple colour.
o Blue litmus papers are prepared by soaking white paper in litmus solution and would
stay blue for neutral or basic substances.
o Red litmus paper is prepared by soaking white papers in litmus solution and dilute acid
and would stay red for neutral or acidic substances.
o Remains yellow in basic substances.
Indicator
pH Range
Lower Colour
Upper Colour
Methyl Orange
Litmus
Bromothymol
Blue
Phenolphthalein
3.1-4.4
5.0-7.6
6.0-7.6
Red
Red
Yellow
Yellow
Blue
Blue
8.3-10.0
Colourless
Pink
1.3) Identify and describe some everyday uses of indicators including the testing of soil
acidity/basicity.
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Chemist routinely use indicators in their analytical work.
They use indicators in titrations to signal the point at which an acid neutralizes a fixed amount
of base.
Soil testing: Some soils are naturally acidic or basic.
o Acids may also be produced by the decay of organic material in the soil.
o Because soils are usually dark, the soil must be moistened with a little water before
adding indicator.
o White barium sulphate powder is then added onto the surface to clearly show the
indicator colour.
o Acidity problems can then be solved by adding bases to the soil.
o By testing soil, the types of plants or crops that will grow in it and soil conditioners
required can be determined.
Monitoring pool acidity
o The acidity levels in pools need to be carefully controlled to avoid growth of microbes
and to ensure that skin, membranes and eyes are not irritated.
o Pool water is sampled regularly and tested with an indicator to monitor acidity levels.
o Pool Chlorine (sodium hypochlorite (NaOCl) and hydrochloric acid are used to
maintain correct chlorine and acidity balance).
Tip: This is not a very common exam question, however, ensure you have a solid understanding of
the use and always make sure to try add some chemical explanations (i.e. what types of chemicals
you use and relating it back to pH) as this will distinguish you from students who provide general
and basic statements.
1.2.1) Perform a first-hand investigation to prepare and test a natural indicator.
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Cabbage is cut into fine shreds and placed in a larger beaker (500ml).
Pour water into the beaker and stir the cabbage while boiling.
Decant the extracts into a container and discard of the cabbage shreds.
Divide the cabbage extract into 5 test tubes.
o One of these tubes will be a control.
To the remaining tubes add several drops of each of the following 0.01 molar acids and bases:
hydrochloric acid; acetic acid; ammonia solution; sodium hydroxide solution.
Record the results and repeat for flower petals.
Tip: For questions based on practicals, always ensure that you provide specific numerical values
for how much of a material you are using, as this shows the examiner you understand the practical
in detail. You don’t have to memorise exact values but just ensure they are of a reasonable amount.
2) While we usually think of the air around us as neutral, the atmosphere
naturally contains acidic oxides of carbon, nitrogen and sulfur. The
concentrations of these acidic oxides have been increasing since the
Industrial Revolution.
2.1) Identify oxides of non-metals which act as acids and describe the conditions under
which they act as acids.
2.2) Analyse the position of these non-metals in the Periodic Table and outline the relationship
between position of elements in the Periodic Table and acidity/basicity of oxides.
The left hand side of the periodic table contains the most electropositive elements i.e. metals. Metal
oxides are basic oxides due to them dissolving in water to produce OH- and reacting with acids to
produce salts.
MgO(s)+ H2O(l) → Mg(OH)2(aq)
CaO(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)
The right hand side of the periodic table contains the most electronegative elements, i.e. non metals.
Non-metal oxides are acidic due to them dissolving in water to produce H3O+ and reacting with bases
to form salts.
CO2(g) + H2O(l) ⇋ H2CO3(aq)
CO2(g) + Ca(OH)2(aq) → CaCO3(aq) + H2O(l)
Elements with moderate electronegativity, on the mid-right of the periodic table form amphoteric
oxides (eg. Al, Zn, Sn, Pb, Be). These oxides can react with both acids and bases.
Acting as base: ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(l)
Acting as acid: ZnO(s) + 2NaOH(aq) → Na2ZnO2(aq) + H2O(l)
Noble oxides in recent research highly unstable oxides of xenon and radon have been synthesised. Eg.
XeO3 (acidic) dissolves in water producing xenic acid and reacts with alkaline solutions producing
xenate salts.
XeO3(g) + H2O(l) ⇋ H2XeO4(aq)
Neutral oxides do not react with bases or acids (eg. CO, NO, N2O).
Across a period, acidity generally increases
while down a group basicity increases, this is
consistent with the fact that electronegativity
(tendency of a substance to give up
electrons) is higher for non-metals as shown in
the increase in electronegativity across a
period and decrease in electronegativity
down a group.
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The alkalinity of these oxides increases
down each group.
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Thus barium oxide is a stronger base
than magnesium oxide.
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Generally, the higher the oxidation
state of the metal or semi-metal, the
more amphoteric or acidic the oxide.
Tip: This information provided above would be perfect for a 6 marker. Ensure for questions that
you identify what type of oxides are associated with a certain pH, provide chemical equations
supporting your claims and also importantly you must discuss more than just acidic, basic and
amphoteric oxides, (i.e. neutral and noble gas oxides to receive full marks as this was the case in
the markers comments for HSC).
2.3) Define Le Chatelier’s principle.
Recalling from the preliminary course, not all reactions reach completion. Some substances
can be in a state of dynamic equilibrium, where the forward reaction is occurring at the same
rate as the reverse reaction, therefore resulting in no net change from a macroscopic
perspective.
The conditions required for equilibrium are:
o Closed system.
○ Reversible reactions (most important).
○ Macroscopic (observable) properties are constant.
○ Concentration of reactants and products are constant (NOT EQUAL).
○ Rate of forward reaction equal to rate of reverse reaction.
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Le Chatelier’s principle
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‘If a system is at equilibrium and a change is made that disturbs the equilibrium, then the
system responds in such a way as to counteract the change and eventually a new equilibrium
is established.’
2.4) Identify factors which can affect the equilibrium in a reversible reaction.
Temperature
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The rules regarding temperature are in accordance to whether the reaction is exothermic or
endothermic.
The formal ways of defining the resultant shift is more complicated to apply.
Rather, recall Le Chatelier’s principle while working and understand that the reaction will
oppose the change.
For exothermic reactions heat is released. This can be depicted in the equation on the right
hand side.
If the temperature of the system was increased, the system will wish to reduce the temperature
and will therefore favour the exothermic reaction, that is the reverse reaction. Therefore, it will
shift left as it wishes to remove heat from the right.
If the temperature of the system was decreased, the system will wish to increase the
temperature and will therefore favour the exothermic reaction. Therefore, it will shift right as it
wishes to bring more heat to the right hand side.
For endothermic reactions heat is absorbed. This can be depicted in the equation on the left
hand side.
If the temperature of the system was increased, the system will reduce the temperature by
favouring the endothermic reaction. Therefore, the system will shift right as the endothermic will
“work harder”.
If the temperature of the system decreases, the system will increase the temperature by
favouring the exothermic reaction. Therefore, the system will shift left as the exothermic reverse
reaction will “work harder”.
Concentration
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If the concentration of a reactant or product is increased, then the equilibrium will shift to use
up the added chemical. Similarly, if the concentration of a reactant or product is decreased
then the equilibrium will shift to produce more of that chemical.
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This is best visualized as a tank of water, if the right hand of the tank is increased, it will shift
water to the left to balance it.
o The exception to this is the addition of a pure solid and/or liquid since its mass is equal
to its volume.
Pressure/ Volume
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Changing the pressure or volume will only affect the equilibrium when there is at least 1 gas
present.
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If pressure is increased, according to Le Chatelier’s principal, the equilibrium will shift in the
direction that decreases the pressure again.
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This is the side with fewer total moles. (Sum the mole ratios on both side of the reaction, and the
side with fewer moles is where the equilibrium will shift if pressure
decreases.)
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The opposite happens if pressure increases.
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As there are 4 moles to 2 moles, if the pressure increased then the equilibrium will shift to the
side with less moles to relieve the pressure, which in this case is to the right.
o Inert gasses increases the overall pressure of the system but not the partial pressures of
the reacants and products.
Catalyst
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Catalysts do not alter the position of a system already in equilibrium.
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If a system is not at equilibrium then catalysts reduce the time to reach equilibrium; however,
the final equilibrium position is the same whether the catalyst is present or not.
Tip: For questions on this dot-point, always provide the definition for Le Chatelier’s Principle
stated above and follow a scaffold response similar to the below.
Equilibrium response scaffold:
1. …… disturbs the equilibrium
2. According to Le Chatelier’s Principle (DEFINE), the equilibrium will shift to …..
3. This is the forward/reverse reaction.
4. (effect) Therefore, the concentrations of reactants/products increase/decrease.
5. In some cases link to colours, solubility, etc.
Analysing Equilibrium graphs:
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Flat line = equilibrium
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Increase/decrease of one substance = sharp (vertical drop/up) of this.
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Everything sharp decrease = ↑ Volume = ↓ Pressure
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Everything sharp increase = ↓ Volume = ↑ Pressure
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Gradual change of all = Change of temperature
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When a change in pressure occurs, the change in concentration is proportional to the
stoichiometric ratios of the reaction.
2.5) Describe the solubility of carbon dioxide in water under various conditions as an
equilibrium process and explain in terms of Le Chatelier’s principle.
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Carbon dioxide is an acidic oxide. It dissolves in water to
produce a solution of carbonic acid (H2CO3).
Carbonic acid is in equilibrium with hydrogen ions and hydrogen
carbonate ions as shown in the following equations.
This dissolution reaction is exothermic.
The solubility of carbon dioxide in water depends on the carbon dioxide partial pressure above
the water. As this pressure increases, more carbon dioxide dissolves in the water.
Equilibrium (1) is shifted to the right to reduce
the gas pressure. Thus more carbon dioxide is
dissolved.
This in turn affects equilibrium (2), which is also
pushed to the right to make more carbonic
acid.
The increase in carbonic acid concentration
shifts equilibrium (3) to the right and thus the
acidity increases as more H+ ions are formed.
A decrease in pressure reverses this, this is why
soft drinks lose their acidity after being opened.
As the reaction is exothermic, an increase in temperature will favour the endothermic
reactions, causing a shift to the left for all three equations, therefore reducing the solubility of
CO2.
However, if temperature was decreased, the exothermic reaction would be favoured causing
a shift to the right therefore increasing the solubility of CO2.
Increasing the concentration of any of the reactants on the left hand side will favour a shift to
the right.
Hence CO2 is soluble in water at low temperatures and high pressures.
2.6) Identify natural and industrial sources of sulfur dioxide and oxides of nitrogen.
2.7) Describe, using equations, examples of chemical reactions which release sulfur dioxide
and chemical reactions which release oxides of nitrogen.
Natural
(SO2)
(Acidic)
NO
(Neutral)
NO2
(Acidic)
N2O
(Neutral)
Industrial
Natural
Industrial
Volcanoes, geothermal springs, oxidation of H2S by bacteria (3O2(g)+ 2H2S(g) →
2SO2(g) + 2H2O(l))
Combustion of coal with sulfur impurities (S (s) + O2 (g) → SO2 (g))
Extracting metals from sulfide ores (3O2(g) + 2ZnS(s) → 2SO2 (g) + 2ZnO(s))
Lightning and high temperatures in some combustions (N2 (g) + O2 (g) → 2NO (g))
Internal combustion engines in cars and power stations (same equation)
Natural
Industrial
Natural
Industrial
When NO produced from above sources reacts with oxygen (2NO (g) + O2 (g) →
2NO2 (g))
Decomposition of organic matter
Using nitrogen rich fertilisers provides food for these bacteria.
Tip: Note the dot-point after identifying sources is about chemical equations, thus, in exam
questions if only asked for one source if you can choose the source which has a chemical equation
to distinguish yourself.
2.8) Assess the evidence which indicates increases in atmospheric concentration of oxides of
sulfur and nitrogen.
General statement: evidence shows an increase in atmospheric concentration of SO x and NOx
Sources: list some and provide chemical equations from the above for SOx and NOx
Direct measurements: by statutory bodies, such as NSW EPA have shown a gradual increase in SOx and
NOx. However, prior to the 1970s, instruments able to measure the very low concentration of these oxides
were not available, hence evidence before the 1970s may be unreliable.
Indirect measurements: analysis by CSIRO of Antarctic ice core samples has shown that CO 2 has
increased from 280ppm to 360ppm after the Industrial Revolution, as well as 10% increase of N 2O.
However, this assumes levels at Antartica reflect global levels. Also, gases in trapped pockets may diffuse
in and out of gas bubbles. Therefore, this may not provide accurate data on past atmospheric
concentrations.
 Occurrence of acid rains and the observed erosion of buildings, statutes and damage to
forests and aquatic organisms.
 Presence of photochemical smog over densely populated regions. Ensure to mention the
Great Smog of London.
Knowledge of the sources: of nitrogen and sulfur oxides, such as from the increase in combustion engines,
we can deduce that the concentration of oxides must have increased. However, SOx and NOx eventually
form SO42- NO3- ions respectively. These are water soluble and may be “washed out” from the air,
potentially nullifying increases in atmospheric concentrations.
Difficulties regarding validity and attainability of evidence.
Difficulties gathering evidence:
o SO2 and NOx measured around 0.001 ppm, unlike CO2 (360 ppm).
o Instruments measuring such low concentrations have only existed since the 1970s and
hence clear enough trends cannot be established to certify the observations.
o Nitrogen dioxide form nitrogen ions, just as sulphur dioxide form sulphur ions. Both of
these are soluble in water. This means that they easily travel around the biosphere and
hydrosphere, making them difficult to study.
Finally, measurements of effects: such as increased areas of acid rain (add as much information as
possible from the dot-point listed below) and photochemical smog (same as acid rain) is an indicator of
increased SOx and NOx. However, these are generally geographically localised to specific regions.
Assessment: Evidence overall shows an increase in SOx and NOx
Tip: This is a very good sample response to questions on this dot-point. As you can see the dot-point
says ASSESS and NOT STATE. Thus, you cannot just state the evidence, you must assess it as I
have done above.
2.9) Calculate volumes of gases given masses of some substances in reactions, and calculate
masses of substances given gaseous volumes, in reactions involving gases at 0˚C and 100kPa
or 25˚C and 100kPa.
Avogadro’s Law
 Equal volumes of gases at the same temperature and pressure contain equal numbers of
molecules.
 When dealing with 1 mole of any gas, that volume of gas is called its molar volume Vm.
 This differs from molar weight in that it is constant for all gases.
 However, as gas volumes vary as the temperature and pressure change matters are simplified
by specifying the volumes of 1 mole of any
gaseous substance at set standards.
 At 0℃ and 100 KPa Vm = 22.71 L/mol.
 At 25℃ and 100 KPa Vm = 24.79 L/mol.
2.10) Explain the formation and effects of acid rain.
2.2.2) Analyse information from secondary sources to summarise the industrial origins of sulfur
dioxide and oxides of nitrogen and evaluate reasons for concern about their release into the
environment.
Formation
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Due to pollution of the air, the air contains toxic gases such as sulfur dioxide (SO 2), carbon
dioxide (CO2) and nitrogen dioxide (NO2).
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These toxic gases are removed by the air via rain, where
slight acidic water is formed, known as acid rain. Acid rain is
rain with a Ph lower than 5.
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When the oxides of sulfur and nitrogen dioxide dissolve in
water they produce solutions of various acids. Sulfur dioxide forms weak sulfurous acid (H2SO3).
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Sulfurous acid can be catalytically oxidised to produce sulfuric acid.
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Nitrogen dioxide produces weak nitrous acid (HNO2) and strong nitric acid (HNO3) when it
dissolves in water.
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Both weaker acids can be oxidised to the stronger acid.
Impacts
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Fall in soil pH, causing damage to vegetation (difficulty to absorb calcium or potassium).
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Damages the tissue of the plant and ionises solid minerals necessary for the plant’s nutrients.
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Damage to leaves and pine forests as waxes are removed.
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Buildings and statues eroding, as carbonates
(concrete, marble, limestone) readily react with acids.
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Aquatic organisms die as pH goes under 5.
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Rivers and lakes become too acidic and the marine life suffer ass acidic lakes are created.
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Sulfate particles have affected the health of those with respiratory issues and has caused lung
cancer as they are carcinogenic.
Photochemical Smog: the release of oxides of nitrogen into the atmosphere also leads to the
production of photochemical smog, which through decomposition due to sunlight results in the
formation of aldehydes, acrolein, peroxyacyl nitrates (PANs) and ozone:
NO2(g) + UV → NO(g) + O.(g)
O.(g) + O2(g) → O3(g),
Ozone in the troposphere is a harmful pollutant as it is a strong oxidising agent that damages plant and
animal life and also causes respiratory issues.
Tip: Another common exam question. Always ensure effects are specific and you include chemical
equations.
2.2.1) Identify data, plan and perform a first-hand investigation to decarbonate soft drink and
gather data to measure the mass changes involved and calculate the volume of gas
released at 25˚C and 100kPa.
 Soda water consists of CO2 gas and H2O in solution.
 H2O boils at 100°C, and CO2 boils at a much lower temperature.
 So to calculate the amount of CO2 gas in the soda water, first weight the mass of the soda
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water, then heat the solution, but keeping it well below 100°C (to avoid evaporation water),
once the bubbles stop rising (this means all the CO2 is gone) weigh the mass of the remaining
soda water.
The mass decreases as CO2 gas has escaped. Using the mass of the original soda water and
the mass of the CO2 lost, the concentration of CO2 in soda water (v/vm) can be calculated.
3) Acids occur in many foods, drinks and even within our stomachs.
3.1) Define acids as proton donors and describe the ionisation of acids in water.
 An acid is a proton donor (hydrogen ion).
 In water acids undergo ionisation – that is, they donate a hydrogen ion H+, a proton, to water
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to form hydronium ions (H3O+). Here, water can be considered the base.
Certain substances are only acidic in the presence of water, such as HCl. Therefore, water is
referred to as an ionizing solvent.
Monoprotic acid: an acid that can donate one
proton per molecule of acid.
Diprotic acid: an acid that can donate two
protons per molecule of acid.
By this definition, a base can be defined as a
proton acceptor.
Example: HNO3(aq) + H2O(l) → NO-3(aq) + H3O+(aq)
Note that the dissolution of acids in water is an exothermic reaction, similar to the reaction of
metals with water.
3.2) Identify acids including acetic (ethanoic), citric (2-hydroxypropane-1,2,3-tricarboxylic),
hydrochloric and sulfuric acid.
Acid
Systematic
Name
Ethanoic Acid
Chemical
Formula
CH3COOH
Citric Acid
2
hydroxypropane1,2,3 tricarboxylic
C3H5O(COOH)3
Hydrochloric
Acid
Hydrochloric
Acid
HCl
Produced by the glands in the
lining of our stomachs to assist
digestion of food. Strong acid that
fully ionises in water. Cleaning
metals and brickwork, neutralising
bases, acidity of pools.
Sulfuric Acid
Sulfuric Acid
H2SO4
Formation of acid rain.
Diprotic acid, ionises in two stages.
Classified as a strong acid. Used
for fertilisers, synthetic fibres, car
batteries and detergents.
Acetic Acid
(Vinegar)
Structural Formula
Found in Nature
Acetic acid is a weak monoprotic
acid. The proton is donated from
the COOH (carboxylic acid)
functional group. The hydrogen
atom in the carboxylic acid group
is more weakly bonded than the
hydrogen atoms attached to
carbon atoms or the hydrogen in
OH groups. Found in bacteria.
Citric acid is a weak triprotic acid
found in citrus fruits. Citric acid is
also formed during the cellular
respiration of sugars. Citric acid is
added to many food products
such as jams to increase
the sour taste of the food, as well
as assisting in the prevention of
mould growth.
Tip: For the HSC you must know the systematic name of citric acid, as well as the structural
diagram for citric acid as this has shown up in past HSC exams such as 2015.
Additionally, ensure when asked to draw any structural diagrams that you always show the full
structure. This means that you must expand all bonds, so it is INCORRECT to write -OH. It MUST
be drawn as -O-H. This also applies to all other structural diagrams within the HSC such as
ethanol and acetic acid. If you do not do this you run the risk of losing easy marks.
3.6) Compare the relative strengths of equal concentrations of citric, acetic and hydrochloric
acids and explain in terms of the degree of ionisation of their molecules.
3.2.2) plan and perform a first-hand investigation to measure the pH of identical
concentrations of strong and weak acids.
Acid
Concentration
pH
Strength
Degree of Ionisation
Citric
0.1M
2.1
Weak
8%
Acetic
0.1M
2.9
Weak
1.3%
Hydrochloric
0.1M
1
Strong
100%
 In the case of citric acid, the three protons are not equally ionised, and the degree of
ionisation is mainly due to step one of the three ionisation steps.
3.2.1) Solve problems and perform a firsthand investigation to use pH meters/probes and
indicators to distinguish between acidic, basic and neutral chemicals.
 A pH meter/probe is an electronic device that measures pH.
 pH metres/probes are much more accurate than indicators, where a pH metre has about 0.01
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accuracy and indicator at best has 1pH accuracy.
Also the pH meter is not affected by the colour of the original substance, unlike indicators
which are.
However, the pH meter must be calibrated and rinsed in water to neutralise it before use.
The pH electrode must be calibrated with solutions of known pH. These solutions are known as
buffers. Typically, a pH 4 and a pH 7 buffer are used to establish a calibration between voltage
and pH.
3.3) Describe the use of the pH scale in comparing acids and bases.
3.5) Identify pH as -log10 [H+] and explain that a change in pH of 1 means a ten-fold change
in [H+].
 Despite water being a poor conductor of electricity, there are still ions present.
 This occurs due to water’s self-ionizing ability.
 If two water molecules collide with the right amount of energy, the proton is transfers from one
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water molecules to another.
Thus, a hydronium ion and a hydroxide ion are formed. There are so few of these ions formed
that the electrical conductivity of water is essentially negligible.
At a temperature of 25℃, the concentration of these are ions are:
As hydronium ions (acids) are added to the water, the system is no
longer in equilibrium.
The system readjusts to remove some of the added hydronium ions
and thus the concentration of hydroxide ions decreases.
Therefore, when an acid or base is added to water, the shift in
equilibrium will ensure that the product of the concentrations will
remain a constant.
As the self-ionisation of water is an endothermic process, the value of
KW will change as the temperature changes. In hot water, the value
of KW is higher than it is in cold water.
pH is the potential of hydrogen.
It is a base 10 logarithmic scale that indicates the concentration of hydronium ions.
Therefore, a change in 1 unit of pH is equivalent to a 10-fold change in the hydronium ion
concentration.
The pH scale is useful for comparing the acidity of acids of equal
molarity. (If there’s a greater molarity than the value of the
concentrations of hydronium ions cannot be compared.)
If a strong acid has a molarity greater than 1 mol/L, the pH is less than zero.
Similarly, a strong base solution with a molarity greater than 1 mol/L has a pH greater than 14.
pOH + pH = 14 (where pOH is a measure of the amount of OH- ions in a solution).
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In pH calculations it should be noted the way significant figures is applied is different to how
you would apply it in normal calculations, as significant figures are ONLY counted after the
decimal place. So take for example a 2.4 pH is 1 significant figure, while a 2.45 pH is 2
significant figures.
3.4) Describe acids and their solutions with the appropriate use of the terms strong, weak,
concentrated and dilute.
Concentrated and Dilute Acids.
 A concentrated acid or base is one which has a high number of moles of acid or base
molecules per litre (almost purely acid molecules).
 A dilute acid or base is one which has a low number of acid molecules moles per litre (mostly
water, or other solvent).
Strong and Weak Acids
 Strong and weak refers to the degree of ionization of the acid molecules.
 Strong acid is an acid that completely ionizes in water solution.
o Eg; HCl, H2SO4, HNO3, HBr, HI
 Weak acids are acids that do not completely ionize in a water solution.
 As the reaction is not completed, these reactions are denoted by a reversible arrow as an
equilibrium is established.
o Eg; H2CO3, CH3COOH, H2SO3
 The above definitions can be applied to bases.
 All group 1 and 2 metal hydroxides are strong bases.
 Chemists can determine the extent to which a weak acid ionises (at a fixed temperature and
acid concentration) by measuring the concentration of hydronium ions formed.
 According to Le Chatlier’s principle, if a substance is diluted, the reaction will shift to the right to
produce more ions and resist the change of concentration.

Extension: For difference in strength of citric acid and acetic acid is due to the polarity of the
OH bond which is affected by a variety of factors; notably the electronegativity of the group
attached to the carboxyl group, the greater the electronegativity, the stronger the acid. This is
the reason why CCl3COOH (trichloroacetic acid) which is monoprotic is stronger than citric
acid (triprotic).
3.7) Describe the difference between a strong and a weak acid in terms of an equilibrium
between the intact molecule and its ions.
 We can describe the differences between strong and weak acids in terms of the equilibrium
between its molecules and ions.

The equilibrium of the above equation lies very much to the right, as it ionises completely,
making it a strong acid and resulting in only ions and no intact molecules.

However, for the reaction shown above, the equilibrium lies much to the left because it is a
weak acid and most of the molecules do not ionise and remain intact molecules.
3.2.5) Gather and process information from secondary sources to explain the use of acids as
food additives.
Use
Food
Preservation
Flavouring
Nutrients
Examples
Acids inhibit the growth of micro-organisms such as bacteria and moulds which
decompose food. They do this by lowering the pH as micro-organisms cannot usually
survive below a pH of 5.
Acids used include,
 Acetic acid in vinegar used in pickling.
 Sorbic acid used in cheeses to control yeasts and moulds.
 Acids act as an antioxidant to prevent the spoilage of food through
oxidation.
Acids have a sour taste – acetic acid in vinegar is used in sweet and sour sauces.
Vitamin C (ascorbic acid) is an essential nutrient for humans, promoting the growth
of connective tissue. Vitamin C can be found as an additive in juices.
Tip: Not a common exam question but to distinguish yourself ensure you provide some examples of
that use.
3.2.6) Identify data, gather and process information from secondary sources to identify
examples of naturally occurring acids and bases and their chemical composition.
 For acids refer to syllabus point 3.2.
 Ammonia (NH ) – from bird droppings (called guano)
3
4) Because of the prevalence and importance of acids, they have been used
and studied for hundreds of years. Over time, the definitions of acid and base
have been refined.
4.1) Outline the historical development of ideas about acids including those of:
4.2.1) Gather and process information from secondary sources to trace developments in
understanding and describing acid/base reactions.
– Lavoisier
 Showed that nonmetal compounds with oxygens produced acids when dissolved in water.
 Hypothesized: The presence of oxygen in compounds gave these non-metals acidic
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properties.
Theory: All acids contain oxygen and that causes acidity when dissolved in water.
Weaknesses: Lavoisier’s theory of acids did not explain why oxides of metals were not acidic
(e.g. BaO). Also muriatic acid (HCL) did not contain oxygen.
However, it did stimulate a lot of research into the composition of acids and created
awareness of the need to define an acid.
– Davy
 Observed that metals could displace hydrogen from acids.
 Hypothesis: Davy proposed that all acids contain a removable hydrogen (as opposed to
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oxygen) again attempting to define acids in terms of their composition.
Failed to explain why many compounds of hydrogen were not acidic.
The Davy definition defined an acid in terms of its properties and reactions; it helped classify
substances without trying to interpret properties.
– Arrhenius
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Arrhenius acid: a substance that ionized in solution to produce hydrogen ions.
Arrhenius base: a substance that produces hydroxide ions in aqueous solution.
Hence, an acidic substance ionises in solution to produce H+, and a basic substance ionises in
solution to produce OH–.
 This explained the conductivities of acid and bases.
 Acids were strong if they ionised completely and weak if they ionised only slightly.
Limitations
 Only accounts for substances which already have H+ or OH– in their structure (e.g. NH3 is basic)
 Hence it does not explain the behaviour of some salts (ZnCl2: acidic, NaS: basic)
 Cannot explain how some substances can act as both an acid and a base (amphoteric
substances like H2O, HCO3–, HSO4–, H2PO4–)
 The theory failed to explain the important role of water as an ionizing solvent and it did not
consider neutralization in non-aqueous solvents.
4.2) Outline the Brönsted-Lowry theory of acids and bases.
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Working independently, Bronsted and Lowry defined acids and bases as follows:
o Acids are proton donors.
o Bases are proton acceptors.
Hence, a substance cannot act as an acid without another acting as a base.
The theory recognized the importance of water as an ionizing solvent.
If a substance has greater tendency to give up protons than the solvent, then it is an acid.
o Conversely, if a substance has a greater tendency to accept protons than the solvent,
then it is a base.
The B-L theory stated that ionization is a result of interaction between the acid/base and the
solvent, not just by the acid/ base.
o For example, when hydrochloric acid reacts with ammonia:
o HCl(l) + NH3(l)  NH4Cl(l)
According to Arrhenius theory, this would not be an acid/base reaction because there is no
water, and no H+ ions. However, what happens in reality is that ammonium chloride dissolves:
o NH4Cl(l)  NH4+ + ClHCl has donated a proton to ammonia to form NH4+, while itself is reduced to Cl-. Thus, this HCL
is a proton donor and HH3 is a proton acceptor.
This makes HCl an acid and ammonia a base and this equation an acid/base reaction
according to the B-L theory.
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Molecular acids such as HCl gas dissolve in water to produce ions since a proton is donated
from the molecular acid to the water molecule to produce hydronium ions.
Therefore, HCl is as B-L acid as it is donating a proton
and water is a B-L base as it is a proton acceptor.
The presence of hydronium ions gives the solution its
acidic properties, but what is significant is water is not
just behaving as a solvent, but as a B-L base.
In a similar way, molecular bases such as NH3 can be
explained as water becomes the B-L acid.
Hence, the self-ionization of water can be understood
as one molecule of water acting as a B-L base and
the other as a B-L acid.
The Brönsted–Lowry definition increased our
understanding further by showing that acidity
depends not just upon the structure of the substance
itself but rather on its properties relative to those of the
solvent or other reactant present in the solution.
Tip: This is a very very common exam question. I cannot stress enough how important these 4
theories are. You must know them and also be able to apply them to situations. I recommend for
questions such as outline the 4 theories, that you define the theory, provide a chemical equation
and list weaknesses.
4.3) Describe the relationship between an acid and its conjugate base and a base and its
conjugate acid.
4.5) Identify conjugate acid/base pairs.
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When a B-L acid donates its proton to the base, the anion of the acid is proton deficient. This
species can act as a base as it could accept a proton and reform the original acid.
Likewise, when a B-L base accepts a proton from an acid, it has a surplus
of protons, hence it can give off a proton to form the original base.
Therefore, every acid has a conjugate base and every base has a
conjugate acid.
Strong acids such as hydrochloric acid and
nitric acid have very weak conjugate bases.
The conjugate acids of strong bases such as the
hydroxide ion are also very weak.
This is because if a substance is a strong proton
donator, it will tend not to want to reform itself.
Tip: This is usually a multiple choice question, so all you need to know essentially is underlined.
4.6) Identify amphiprotic substances and construct equations to describe their behaviour in
acidic and basic solutions.
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Amphiprotism is a classification only within the Bronsted–Lowry theory.
Species that can behave as either proton donators or proton acceptors are called
amphiprotic species.
o (Amphoteric refers to only oxides, do not
confuse.)
HCO3- (hydrogen carbonate)
In the presence of a strong acid, HCO3- acts as a proton
acceptor.
In the presence of a base, HCO3- acts as a proton
donator.
HSO3- (hydrogen sulfite)
Water is also amphoteric – H2O acts as acid and
becomes OH- and H2O acts as base and becomes
H3O+.
Notice they react with the ions of water not the water molecules itself. This differs to acidic salts
below.
4.4) Identify a range of salts which form acidic, basic or neutral solutions and explain their
acidic, neutral or basic nature.
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As outlined above, the Arrhenius theory failed to explain why certain salt solutions were acidic
or basic.
This is due to the fact that salts are not necessarily neutral substances upon dissolution.
The reaction of a salt with water to produce a change in pH is called hydrolysis.
Salts that form basic solutions are called basic salts and salts that form acidic solutions are
called acidic salts.
A neutralization reaction is one in which an acid + base  salt + water.
However, the acidity or basicity of a salt in solution is determined by the strength and nature of
the acids and bases that neutralize to form it.
Strong Acid + Strong Base  Neutral Salt + Water
 E.g. NaCl, KNO3, Na2SO4
 E.g., CH3COONH4 (ammonium acetate) (Weak acid and weak base)
HCl + NaOH  NaCl + H2O
 Hydrochloric acid (strong acid) + Sodium hydroxide (strong base)  Neutral salt + Water
 The NaCl ionizes in the water forming Na+ and Cl-. As there are no hydrogen ions or hydroxide
ions present in the solution to react with water the solution is neutral.
Strong Acid + Weak Base  Acidic Salt + Water
 E.g. NH4Cl, NH4NO3
HCl + NH4OH  NH4Cl + H2O
 Hydrochloric acid (strong acid) + Ammonium hydroxide (weak base)  Acidic salt + Water.
 The salt NH4Cl ionizes forming NH4+ and Cl-. The NH4+ then goes on to react with water.
 NH4+ (aq) + H2O(l)  NH3 (aq)+ H3O+(aq).
Weak Acid + Strong Base  Basic Salt + Water
 E.g. CH3COONa, KNO2, NaCO3
CH3COOH + NaOH  Ch3COONa +H2O
 Acetic acid (weak acid) + Sodium hydroxide (Strong base)  Basic salt + Water
 The basic nature of the salt CH3COONa can be shown by the hydroxide ions in the equation.
CH3COONa + H2O  CH3COOH + Na+ + OH-
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Salts are acidic or basic, as ions react with water to form H3O+ or OH- ions.
The basicity or acidity of a salt is determined by the ability of the constituent cations and
anions to act as BL acids/bases respectively.
Equilibrium
 When acids and bases are combined in aqueous solution the position of the equilibrium is
determined by the relative strengths of the conjugate acid-base pairs.
The equilibrium will lie on the side of the weaker acid and base.
Tip: This is a dot-point that will be confusing at first, however, after some practice will become 2nd
nature and free marks in an exam. To prove a salt produced from a neutralisation reaction is
acidic or basic, follow these this scaffold:
1. Write an equation for the neutralisation reaction
2. Dissociate the salt that is produced into its ions
3. Identify which ion is able to react with water
4. Identify what type of ion is it? Is it a conjugate acid or base?
5. Write an equation for the reaction of water with that ion, showing the formation of OH- or
H3O+
6. Relate the basic or acidic ion produces back to what type of salt it is.
4.7) Identify neutralisation as a proton transfer reaction which is exothermic.
4.2.5) Analyse information from secondary sources to assess the use of neutralisation
reactions as a safety measure or to minimise damage in accidents or chemical spills.
Acid + Base  Salt + Water
 Neutralization reactions are exothermic.
 The amount of heat liberated per mole when a strong base is neutralised by a strong acid is
almost the same no matter what acid or base is used.
 Also 1 mole acid + 1 mole base → 1 mole water.
 The acid transfers a hydrogen ion (proton) to the base.
 As bonds are formed heat is released from the kinetic energy of the
particles.
Use of Neutralization in the Laboratory
 Neutralisation reactions can be used in laboratories to clean up after acids or bases that have
been accidentally spilled on the workbench or floor.
 Because of the exothermic nature of the neutralisation process, we should never use
concentrated acids or bases in cleaning up spills. Not only is there the potential to cause
boiling and the evolution of noxious fumes, but the strong base or acid will only cause further
damage.
 Goggles, gloves and rubber aprons should be worn and the base powder added slowly to
allow for heat dissipation.
 The best method of neutralising a small acid spill, following containment, is to perform some
limited dilution and then slowly add a powdered solid base (sodium carbonate or sodium
hydrogen carbonate) in excess.
 For large spillages, once an acid or base spillage is absorbed with sand or vermiculite it can be
cleaned up and neutralized in a safe location.
 Dilution of the acid or base with water can dissipate the heat produced.
 Sodium carbonate or sodium hydrogen carbonate are commonly used.
 As the final solution is neutral it may have disposed of down the drain.
4.8) Describe the correct technique for conducting titrations and preparation of standard
solutions.

A primary standard is a solution made by dissolving an accurately measured mass of solute in a
known volume of solvent.
 Must be of high purity, soluble, stable in air and reacts instantaneously and completely.
o HCl is not a viable option as it is in gaseous form.
o Sodium hydroxide and concentrated sulfuric acid reacts with water in the atmosphere.
o Hydrated sodium carbonate is also unsuitable as a primary standard as it effloresces
(loses water) as it is being weighed. Its composition is therefore uncertain (we use
anhydrous sodium carbonate).
 A secondary standard is a solution whose concentration has
been determined through titration against a primary standard.
Preparation of a primary standard
 A beaker is placed on a scale and the scale is zeroed.
 Sodium carbonate is poured into the beaker and its mass to 3
d.p. is determined.
 Distilled water is added to the beaker and a glass stirring rod is
used to dissolve the sodium carbonate.
 The sodium carbonate solution is then transferred into a
volumetric flask through a funnel.
o The volumetric flask should be cleaned with distilled
water. This isn’t a problem as more distilled water will be
added to dilute the solution to 250ml.
o When transferring ensure to rinse the edges of the
beaker so the whole salt solution is transferred.
 Distilled water is then added to the mark of the volumetric flask.
Titration
The primary standard can now be used to analyse unknown acids.
Preparing the burette and filling it with the unknown acid
 The burette is an accurate piece of volumetric glassware. It has markings from 0 to 50 mL with
0.1 mL divisions.
 In reality, it does not matter whether the acid or base goes within the burette. However, when
doing a double titration in which the determined secondary standard of acid is then used to
find a secondary standard of base, it is ideal to fill the burette with acid as it saves the need to
clean piece of equipment.
 The burette should be rinsed with distilled water multiple times before use.
o This includes allowing some of the water to leave through the tap and then swirling the
burette as water is let out the top.
 The burette should then be rinsed with the unknown concentration of acid to remove water
that would dilute the solution.
o The acid of this procedure should be disposed of as it is now diluted.
 The burette is then clamped to a retort stand.
 The burette is then filled with the acid and excess volume of the acid is released through the
tap into the beaker.
o Ensure that no bubbles exist in the region of the tap.
o The solution added to the burette is the titrant.
Preparing the pipette and transferring the standard base into the conical flask
 A pipette is an accurate piece of volumetric glassware with only one engraved line for a given
volume.
 The pipette is rinsed several times with distilled water.
 A pipette filler is used to draw up small volumes of the base to rinse the pipette from residue
water molecules.
 The pipette can now be filled with the standardized base solution known as the aliquot.
 Ensure that the bottom of the meniscus is on the engraved line.
 The aliquot can now be transferred into a conical flask.
o The flask should be rinsed with distilled water but it can be left wet as the pipette will
allow us to determine the number of moles present regardless of dilution.
 Do not shake out or blow out the remaining drop in the pipette as it has been accounted for in
manufacturing.
 A second conical flask may be used as reference.
Performing the titration
 Three drops of a suitable indicator are added to the conical flask.
 The flask is placed on a white tile or white paper as to easily distinguish colour changes.
 During the first run a “rough” volume is established by allowing significant amounts of volume to
pass at once.
 During the proper trials, the tap is open as to be perpendicular to the bench and small volumes
of water are released into the flask.
 While this is being done constant swirling of the flask should be done with the free hand.
 Rinse the edges of the flask to ensure that all the acid is reacting with the base.
 Continue this process until a slight but permanent colour change occurs.
 The titre is the volume of solution delivered from the burette that achieves an end-point
 The trials should be done all within 0.1 ml of each other.
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Now a known concentration and volume of one substance is reacted with a known volume of
another.
Using stoichiometric ratios the concentration of the unknown reactant can be determined.
Choosing a suitable Indicator
 The range of the indicator used within a titration must encompass the equivalence point of the
neutralization reaction.
 End point — the point in the titration when the indicator just changes colour.
 Equivalence point — the point where the acid has stoichiometrically reacted with the base
(sometimes called the stoichiometric end-point).
 Because indicators change colour over a narrow pH range rather than at an exact pH, it is
important to achieve a close match of the equivalence point with the indicator’s pH range.
Indicators are weak acids
 Acid–base indicators are weak acids and therefore weakly ionize in water solution.
 If we represent the un-ionised form of an indicator as HIn and its conjugate base as In–, then
the equilibrium is:
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This indicator equilibrium is influenced by the pH of the solution.
When the pH is low then the equilibrium shifts to the left.
When the pH is high the equilibrium shifts to the right.
The colour changes that we observe occur because the un-ionised molecule (HIn) has a
different colour from that of the conjugate base (In–).
End Points
Strong acid and strong base
 As a neutral salt is formed between the two no hydrolysis will occur and therefore the
equivalence point will be around 7 pH.
Strong acid- weak base
 As an acidic salt is formed the solution will be acidic and therefore the equivalence point will
be below 7 pH.
Weak acid- strong base
 As a basic salt is formed the solution will be basic and the equivalence point will be above 7
pH.
Tip: This is a crucial dot-point to HSC and is tested in some form (calculation, method, technique,
titration curves) almost every single HSC, so ensure you do a lot of practice and understand this
area. For calculations this is difficult to teach because there is a variety, through a document, so I
would recommend you use a textbook such as Conquering Chemistry and view a few examples and
the steps they use.
4.9) Qualitatively describe the effect of buffers with reference to a specific example in a
natural system.
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A buffer is a solution that resists changes in pH when an acid or a base is added.
They are made by mixing together a weak acid and the salt of that weak acid, or a weak base
and the salt of that weak base.
In other word, the solution will be that of a B-L acid and its conjugate base or a weak B-L base
and its conjugate acid.
o (Note: it only works when these conditions of weak acid/base + conjugate and
equimolar concentrations are present. Otherwise no buffer forms).
By choosing the correct amounts of the weak acid and weak base in the solution, the pH of
the solution can be fixed to within narrow limits.
(Similar to the inflection of equivalence points).
When acid is added the equilibrium of the weak
acid will shift left.
When base is added to increase the hydroxide
ion concentration the equilibrium of the
conjugate base shifts left.
Consider the weak acid HA and its conjugate
base, A-. If HA is added to water, the following
equilibrium is established.
As the hydronium ion is a stronger acid than HA,
and A- is a stronger base than water, the
equilibrium lies well to the left.
If A- is added to water(e.g. as the sodium salt),
then the following equilibrium is established.
As the hydroxide ion is a stronger base than A-, and HA is a stronger acid than water, then the
equilibrium also lies well to the left.
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A buffer solution contains a weak acid and its conjugate base. The addition of small amounts
of strong acids or bases causes little change in pH as the excess acid or base is removed by
either the conjugate base or the B–L acid.
Bicarbonate Buffer System

Our body fluids and secretions must be maintained in a narrow pH range of about 7.4 in order
for our biochemical processes to occur at an optimal rate.

The system consists of a carbonic acid – hydrogen carbonate buffer linked to the haemoglobin
oxyhaemoglobin equilibrium that is used to maintain blood pH.

An increase in oxygen upon inhaling causes a shift to the right that forms oxyhaemglobin and
hydronium ions.
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To buffer the system the carbonic acid- hydrogen carbonate equilibrium shifts to the left in
order to reduce the excess hydronium acid molecules.
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The formed carbonic acid breaks down to release carbon dioxide.

As tissue begins to respire oxygen is used up causing a shift to the left, therefore increasing
alkalinity.
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This causes a shift to the right in the first equation.
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When the pH of blood is decreased when acid is added, H3O+ concentration increases. This causes the
equilibrium to shift to the left, favouring the reaction that removes H3O+, according to LCP. Thus,
minimising the change in pH due to acid addition.
When pH is increased by the addition of OH-, this reacts with H3O+ . This decrease in H3O+ disturbs the
equilibrium and as a result of LCP, the system shifts to the right hand side, thus increasing H 3O+
concentration. Thus, minimising the change in pH due to base addition.
Tip: It should be noted that the change in pH is minimised and NOT COMPLETELY buffered.
However, the overall change is very small. For questions on this dot-point you must provide a
definition of buffers, an example of a buffer, the role of the buffer in the natural environment (very
important a lot of people lose marks on this for not stating this), a chemical equation for the buffer
and a discussion of the addition of acid and base using Le Chatelier’s Principle.
Also, the reason why buffers don’t form with a strong acid such as HCl and its conjugate Clrelates to the dot-point of the conjugate acids and bases, as Cl- is an extremely weak conjugate base
it is unable to react with water and thus there is no possibility of an equilibrium (buffering)
reaction forming.
4.2.2) Choose equipment and perform a first-hand investigation to identify the pH of a range
of salt solutions.
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Calibrate the pH meter probe with the buffers supplied.
Use the pH meter and probe to measure the pH of each of the supplied solutions. Recalibration
of the pH meter with buffers will probably be required after measuring several salt solutions.
4.2.4) Perform a first-hand investigation to determine the concentration of a domestic acidic
substance using computer-based technologies.
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A titration was carried out with vinegar (acetic or ethanoic acid) and Sodium Hydroxide.
It was done with a pH meter and date logger on top of a magnetic stirrer.
The pH proves are calibrated using a pH 7 buffer.
The Phenolphthalein was used as the indicator.
The data logger will be set to record the pH over time.
5) Esterification is a naturally occurring process which can be performed in
the laboratory.
5.1) Describe the differences between the alkanol and alkanoic acid functional groups in
carbon compounds.
5.3) Explain the difference in melting point and boiling point caused by straight-chained
alkanoic acid and straight-chained primary alkanol structures.
Alkanoic compounds are weak acids that contain the carboxylic acid functional group.
(-COOH)
Revision regarding the nature of alkanols

Alkanols are polar molecules due to the presence of the alcohol functional group.

The melting and boiling points of the alkanols are higher than their corresponding alkanes
because of dipole–dipole attractions

Dispersion forces are the only forces that bind hydrocarbon molecules together.

This is explained by the hydrogen bonding between the electropositive hydrogen atoms and
the electronegative oxygen atoms of alcohol groups on neighbouring molecules.

Alkanols are essentially neutral molecules. The alcohol functional group is strongly bonded to
the carbon chain and there is no tendency for the loss of hydroxide ions when alkanols dissolve
in water.
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Alkanoic acids
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General formula:
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Naming System: “Identify the number of carbons present in the straight chain. Select
the correct stem to name the parent alkane. Remove the ‘e’ and replace it with
the suffix ‘oic acid’.”

The preferred name for the first two members of the alkanoic acid homologous series is their
common name rather than their systematic name. Thus methanoic acid is called formic acid
and ethanoic acid is called acetic acid.
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Alkanoic acids have greater molar weights than their equivalent alkanols or alkanes and thus
the dispersion forces between their molecules are also greater.

Alkanoic acids are usually slightly more polar than alkanols due to
the dipole-dipole interactions of the C=O bonds and thus the
dipole–dipole forces are greater. When an alkanol and an
alkanoic acid of the same molar weight are compared, the
alkanoic acid has the higher melting point or boiling point.

This is mainly due to the more extensive hydrogen bonding
between alkanoic acid molecules.

As the carbon chain lengthens, the increasing dispersion forces
dominate and there is a general trend of increasing boiling point
with increasing molar weight.
Esters

Comparatively to esters, both alkanols and alkanoic acids of the same molecular weight have
a significantly higher melting and boiling point.

Despite all three being polar, alkanols and alkanoic acids exhibit hydrogen bonding between
their molecules which therefore requires significantly more energy to break the bonds.

Alkanoic acids have the highest boiling and melting as they are more polar as they contain a
COOH group and a OH group and thus are able to form twice as many hydrogen bonds
between molecules.

The ester cannot form hydrogen-bonds as there are no OH functional groups in the molecule.
Consequently its boiling point is much lower
Tip: Since this dot-point specifies STRUCTURE, in your answer you should draw the diagrams of
the molecules and the type of intermolecular forces they experience shown in the image above.
5.2) Identify the IUPAC nomenclature for describing the esters produced by reactions of
straight-chained alkanoic acids from C1 to C8 and straight-chained primary alkanols from
C1 to C8.
Naming conventions

Count the number of carbon atoms in the alkyl group that is
derived from the original alkanol. Name this alkyl group by
deleting the ‘anol’ suffix from the alkanol and replacing it with
the suffix ‘yl’ (e.g. butanol becomes butyl).

Identify the number of carbon atoms in the alkanoate chain that
is derived from the alkanoic acid. Name this alkanoate chain by
deleting the ‘oic acid’ suffix and replacing it with the suffix
‘oate’ (e.g. propanoic acid becomes propanoate).
 The name of the ester is two separate words. The first name
comes from the alkanol and the second from the alkanoic acid.
5.4) Identify esterification as the reaction between an acid and
an alkanol and describe, using equations, examples of esterification.
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The fragrances of flowers and the smell and taste of their
fruits are produced by a group of organic compounds
called esters
Chemists have developed a range of synthetic esters
that can be used as food flavorings and perfumes.
Esters are produced by the acid-catalysed reaction
between an alcohol and a carboxylic acid in a process
referred to as esterification.
They never proceed to completion and as a result
forward and reverse reactions must always be included.
Esterification reactions are also quite slow and therefore
a suitable catalyst and sufficient heat must be applied
during the reaction.
As the reactants and products of this process are volatile
and therefore vaporize on heating, the mixture is
enclosed in a reflux apparatus.
Water is a by-product of the reaction.
This reaction is not an acid–base reaction. It is classified
as a condensation reaction.
The apparatus is open to the atmosphere to reduce the
pressure while the reaction vessel occurs in a hot-water
bath to avoid the direct heating of organic liquids that
could explode.
Boiling chips are normally pieces of crushed ceramic
that prevent a process called ‘bumping’ as they provide
a large surface area on which vaporisation can occur
without the risks of sudden superheating and the
explosive ejection of vapours.
Normally the reaction is not done in equal stoichiometric
ratios as to have greater concentrations of a reactant to
shift the reaction.
Tip: For these types of structural diagram and naming questions practice is your best friend do as
many questions as you can until it is 2nd nature. Also, any of the following must be noted or it will
result in mark losses:
 For the structural equation you must draw all bonds expanded.
 You must include the equilibrium arrow (a forward arrow will result in a one mark
reduction)
 Concentrated H2SO4 (the catalyst) must be written on the arrow
 Water must also be drawn and not simply written in formula as H2O
5.5) Describe the purpose of using acid in esterification for catalysis.
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Concentrated sulfuric acid is used to increase the rate of reaction by reducing the activation
energy.
Sulfuric acid also acts a dehydrating agent that removes water from the equilibrium resulting in
a shift to the right, therefore increasing the yield of the ester.
Donates a proton to the unshared electron pairs of the oxygen of either the acid or alkanol.
This makes them more reactive, speeding up the reaction. Eventually the proton is returned to
the catalyst.
5.6) Explain the need for refluxing during esterification.
Refluxing is the process of where a condenser is attached to the esterification apparatus. It is
essential in esterification because:
o The boiling point of alkanols and alkanoic acid are relatively low, thus they will vaporize
and escape. The cool water in the condenser condenses the vapour back into liquid.
This prevents the loss of reactants.
o Thus, refluxing allows esterification at high temperatures, which increases collisions and
thus the rate of reaction, reaching equilibrium faster.
o Esterification is a slow process that can take several hours. Because it is an
endothermic reaction, at high temperatures the equilibrium shifts RHS, increasing the
yield of esters.
The other option to create high temperatures without loss of reactant is using a closed vessel, physically
containing the vapours. However, this leads to a buildup of pressure and volatile organic compounds.
This could lead to an explosion.
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Other safety precaution in esterification:
Using a water bath instead of a Bunsen burner to promote even heating.
Using boiling chips to promote even heating and to prevent buildup of large bubbles that
could splatter and damage equipment.
5.7) Outline some examples of the occurrence, production and uses of esters.
5.2.2) Process information from secondary sources to identify and describe the uses of esters
as flavours and perfumes in processed foods and cosmetics.
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Esters are produced naturally in plants and animals eg fats, oils, natural waxes.
Manufactured as jet engine lubricants due to their low viscosity at low temperatures as well as
their clean high-temperature operation.
The use of esters as lubricants stems from their molecular polarity. Polar ester molecules have
stronger intermolecular attractions than other nonpolar oils, so they have lower rates of
evaporation and higher flash points. Polar ester molecules are attracted to metal surfaces,
forming a film that requires more energy to penetrate.
Solvents and coatings due to strong attraction to metal objects, have high volatility due to low
mol. Weight.
o Solvents (dissolve some organic compounds) Ethyl ethanoate (nail polish remover,
plastics)
They produce the fragrance of many fruits and flowers. For example:
o Pentyl ethanoate (bananas)
o Pentyl butanoate (apricot)
o Ethyl methanoate (rum)
Tip: Not a common exam question but to distinguish yourself ensure you provide some specific
examples of esters that are used for certain applications.
5.2.1) Identify data, plan, select equipment and perform a first-hand investigation to prepare
an ester using reflux.
Risk Assessment:
➢ Volatile, flammable and toxic reactants, so use fume-cupboard and keep clear of any open
flames.
Method:
1) Set up the equipment as shown.
2) Add 10 mL of butan-1-ol and 10 mL of butanoic acid to a pear
shaped flask.
3) Add a 5 drops of concentrated sulfuric acid and boiling chips.
4) Attach a condenser vertically and run water through it.
5) Heat the reaction flask at reflux for 2 hours in a water bath with a
hot plate.
6) Transfer the final mixture to a separating funnel and add water.
Discard the lower layer (top layer is ester) into a toxic waste bottle.
Safety
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Wear safety glasses throughout this experiment.
Sulfuric acid is corrosive. Clean up spills immediately. If the acid is spilled on the skin then wash
the area with large quantities of water- Seek assistance.
Organic chemicals are flammable. Do not allow liquids or vapours.
Tip: For questions on esterification that ask for a method, always include a diagram.
If immiscible then use a separating funnel
EMF
electromotive force or voltage of a galvanic cell is the difference in the reduction
potentials for the 2 couples making up the cell
Capital e to the power of theta
Oxidation potentials is found by flipping the sign
The reference cell
A measure of the reduction potential for a half cell can only be obtained by going it
with a reference couple.
The chosen reference couple is hydrogen. because
This half-cell consists of an inert platinum electrode covered in ‘platinum black’ which
is a highly porous form of platinum metal powered.
This electrode is placed in a 1.0mol/l solution of hydrogen ions and pure hydrogen
gas is bubbled over the surface of the electrode under conditions of standard
standard temperature and pressure.
The standard cell potential is defined as the sum of the standard half-cell reduction
potential and the standard half-cell oxidation potential.
https://dc.edu.au/hsc-chemistry-acidic-environment/#Strong_and_Weak_Bases
Basic sides
The metals of Group 1 and 2 all form basic oxides
The basicity of these oxides increases down each group
Thus barium oxide is a stronger base than magnesium oxide.
Acidic oxides
Most non-metals (other than the noble gases) form acidic oxides
The acidity of these oxides decrease down each group as the elements become
more metallic in character
In addition the non-metallic oxides with the highest oxidation states in the non-metal
tend to be more acidic
Thus sulfur trioxide (SO3) is more acidic than sulfur dioxide (SO2)
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Volumes of gases at standard temperature and pressures
Avogadro’s Law
Equal volumes of gases at the same temperature and pressure contain equal numbers of
molecules.
Gas volumes vary as the temperature and pressure change.
To simply matters, the following standard conditions are used.
Pressure: 100KPa
Temperature: O or 25 degrees Celsius
When dealing with 1 mole of any gas, that volume of gas is called its molar volume Vm
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This differs from molar weight in that it is constant for all gases.
The Vm of any gas has the following values at the stated temperatures and pressures.
At 0 degrees it is 22.71 L/mol
At 25 degrees it is 24.79 L/mol
The relationship between the number of moles (n( of a gas and its volume (V) is n=V/Vm.