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CHEMISTRY MODULE 5 Equilibrium and Acid Reactions

CHEMISTRY MODULE 5: EQUILIBRIUM AND ACID REACTIONS PAGE 1 OF 29

STATIC AND DYNAMIC EQUILIBRIUM

Inquiry question: What happens when chemical reactions do not go through to completion?

Students:

conduct practical investigations to analyse the reversibility of chemical reactions, for example:

– cobalt(II) chloride hydrated and dehydrated

– iron(III) nitrate and potassium thiocyanate

– burning magnesium

– burning steel wool (ACSCH090)

model static and dynamic equilibrium and analyse the differences between open and closed systems (ACSCH079,

ACSCH091)

analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for example:

– combustion reactions

– photosynthesis

investigate the relationship between collision theory and reaction rate in order to analyse chemical equilibrium

reactions (ACSCH070, ACSCH094)

FACTORS THAT AFFECT EQUILIBRIUM

Inquiry question: What factors affect equilibrium and how?

Students:

investigate the effects of temperature, concentration, volume and/or pressure on a system at equilibrium and

explain how Le Chatelier’s principle can be used to predict such effects, for example:

– heating cobalt(II) chloride hydrate

– interaction between nitrogen dioxide and dinitrogen tetroxide

– iron(III) thiocyanate and varying concentration of ions (ACSCH095)

explain the overall observations about equilibrium in terms of the collision theory (ACSCH094)

examine how activation energy and heat of reaction affect the position of equilibrium

CALCULATING THE EQUILIBRIUM CONSTANT ( KE Q)

Inquiry question: How can the position of equilibrium be described and what does the equilibrium constant

represent?

Students:

deduce the equilibrium expression (in terms of Keq) for homogeneous reactions occurring in solution (ACSCH079,

ACSCH096)

perform calculations to find the value of Keq and concentrations of substances within an equilibrium system, and

use these values to make predictions on the direction in which a reaction may proceed (ACSCH096)

qualitatively analyse the effect of temperature on the value of Keq (ACSCH093)

conduct an investigation to determine Keq of a chemical equilibrium system, for example:

– Keq of the iron(III) thiocyanate equilibrium (ACSCH096)

explore the use of Keq for different types of chemical reactions, including but not limited to:

– dissociation of ionic solutions

– dissociation of acids and bases (ACSCH098, ACSCH099)


SOLUTION EQUILIBRIA

Inquiry question: How does solubility relate to chemical equilibrium?

Students:

describe and analyse the processes involved in the dissolution of ionic compounds in water

investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when removing toxicity

from foods, for example:

– toxins in cycad fruit

conduct an investigation to determine solubility rules, and predict and analyse the composition of substances

when two ionic solutions are mixed, for example:

– potassium chloride and silver nitrate

– potassium iodide and lead nitrate

– sodium sulfate and barium nitrate (ACSCH065)

derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic

substance from its Ksp value

predict the formation of a precipitate given the standard reference values for Ksp


REVISION – MODULE 4 – DRIVERS OF REACTIONS

ENTHAPLY CHANGE

Enthalpy (H) is a measure of the heat content of a system. Absolute enthalpy cannot be measured. However, the

change in enthalpy (∆H): the change in the heat content of system during a process, measured at constant pressure.

Δ𝐻 = 𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

- 𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 < 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 , therefore Δ𝐻 is negative and heat energy released by the system (exothermic).

- 𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 > 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 , therefore Δ𝐻 is positive and heat energy absorbed by the system (endothermic).

ENERGY PROFILE DIAGRAMS

HESS’S LAW OF HEAT SUMMATION

‘The total enthalpy change in a chemical reaction is constant, whether the reaction is performed in one step or

several steps.’

Hess’s law is a form of the law of conversation of energy (First law of Thermodynamics).


BOND ENERGIES

Bond energy (or bond enthalpy) is the amount of energy required to break one mole of a bond in a gaseous

molecule.

∆𝐻𝑟𝑥𝑛 = ∑ ∆𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑏𝑜𝑛𝑑𝑠 𝑏𝑟𝑜𝑘𝑒𝑛 + ∑ ∆𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑏𝑜𝑛𝑑𝑠 𝑓𝑜𝑟𝑚𝑒𝑑

∆𝐻𝑟𝑥𝑛 = ∑ ∆𝐻𝑏𝑜𝑛𝑑 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − ∑ ∆𝐻𝑏𝑜𝑛𝑑 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠

ENTROPY

Entropy (S) is a measure of how the available energy is distributed or dispersed amount particles in a system. It is

also a measure of energy dispersal (function of temperature). Generally, low entropy → high entropy. (Chaos)

When energy can be distributed in more ways, there is a greater entropy

- Entropy is sometimes referred to as the measure of disorder or randomness.

- A system with greater possible arrangements (microstates Ω), or greater diversity of movement has higher

entropy.

FACTORS THAT CHANGE ENTROPY

- Increasing the number of particles → Increases microstates → Increase entropy

- Mixing different types of particles → Increases microstates → Increase entropy

- Increasing the volume of a container of gas → Increases microstates → Increases entropy

o The larger the volume, the more ways there are to distribute the energy

- Increasing the number of particles in states with more freedom of movement (gas > liquid > solid)

- Molecules becoming more complex → Increases microstates → Increases entropy

- Increase temperature → Increases microstates → Increases entropy

SECOND LAW OF THERMODYNAMICS

The second law of thermodynamics states that the entropy of the universe is always increasing.

Δ𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 > 0

CALCULATING ENTROPY IN CHEMICAL REACTIONS

Δ𝑆𝑟𝑥𝑛 = ∑ Δ𝑆𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − ∑ Δ𝑆𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

GIBBS FREE ENERGY

In any process, the main form of interaction between the system and the surroundings is the exchange of heat.

- In an exothermic reaction, heat from the system enters the surrounding and increases temperature, which

will increase its entropy. The reverse will be true for an endothermic reaction.

- At a lower temperature, the same amount of heat will cause a greater proportional change in entropy.


Δ𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 > 0

Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 =−Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚

𝑇

Δ𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 −−Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚

𝑇> 0

−𝑇Δ𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚 − 𝑇Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 < 0

Josiah Willard Gibbs redefined the quantity −𝑇Δ𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 as free energy or Gibbs Free Energy, 𝚫𝑮.

Δ𝐺𝑠𝑦𝑠𝑡𝑒𝑚 = Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚 − 𝑇Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚

The equation allows the comparison between the relative contributions of the two driving forces for a reaction,

entropy and enthalpy.

- If 𝚫𝑮 < 𝟎 (Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚 < 𝑇Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚), a reaction is spontaneous

- If 𝚫𝑮 > 𝟎 (Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚 > 𝑇Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚), a reaction is non-spontaneous

- If 𝚫𝑮 = 𝟎, a reaction will occur both in the forward and reverse directions, equilibrium.

EQUATION SUMMARY

𝑞 = 𝑚𝑐∆𝑇

Δ𝐻 = 𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

∆𝐻𝐶𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛 = −𝑞

𝑛

∆𝐻𝑆𝑜𝑙𝑛 = −𝑞

𝑛

∆𝐻𝑟𝑥𝑛 = ∑ ∆𝐻𝑏𝑜𝑛𝑑 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − ∑ ∆𝐻𝑏𝑜𝑛𝑑 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠

Δ𝑆𝑟𝑥𝑛 = ∑ Δ𝑆𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − ∑ Δ𝑆𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

Δ𝐺𝑠𝑦𝑠𝑡𝑒𝑚 = Δ𝐻𝑠𝑦𝑠𝑡𝑒𝑚 − 𝑇Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚


REVERSIBLE REACTIONS

OUTCOMES COVERED

model static and dynamic equilibrium and analyse the differences between open and closed systems

(ACSCH079, ACSCH091)

investigate the relationship between collision theory and reaction rate in order to analyse chemical

equilibrium reactions (ACSCH070, ACSCH094)

explain the overall observations about equilibrium in terms of the collision theory (ACSCH094)

EQUILIBRIUM

For reversible reactions, a reversible arrow is used to indicate that both reactions are capable of proceeding.

Reactants Products

Physical changes are generally reversible

STATIC AND DYNAMIC EQUILIBRIUM

Reactions will proceed until either a static or dynamic equilibrium is reached. Equilibrium refers to the state of a

closed chemical system which:

1) The concentrations of both reactant and products do not change with time

2) The rate of the forward reaction is equal to the rate of the reverse reaction

Irreversible reactions (shown with a forward arrow →) that go to completion reach a static equilibrium.

Reversible reactions (shown with a reversible arrow ) do not go to completion. In a closed system, reversible

reactions will instead reach a state known as dynamic equilibrium.

- At equilibrium, the rates of the forward and reverse reactions are the same, but non-zero.

- The equilibrium is dynamic because there are changes occurring at the microscopic level, even though the

system undergoes no change at the macroscopic level.

There are no macroscopic changes when a closed system is at equilibrium.

TYPES OF SYSTEMS

- An open system is a system where matter and energy can enter and leave.

- A closed system is a system where matter cannot enter and leave, but energy exchange can take place with

the surrounding (in the form of pressure or heat).

- An isolated system is a system where neither matter nor energy can leave.

RATES OF REACTION

1) Concentration or Volume/Pressure

2) Surface Area

3) Temperature

4) Presence of catalyst


5) Reactivity of reactants

COLLISION THEORY

In order for any reaction to proceed, reactants must collide. Particles need to collide with sufficient energy and in

the correct orientation for it to be a successful reaction. A collision with sufficient energy and the correct orientation

is called an effective collision.

The rate of reaction is how rapidly a reaction proceeds. The rate is defined as the change in the concentration of

reactants or products over time. It is dependant on the frequency of effective collisions.

ADDITION/REMOVAL OF A REACTION COMPONENT

If the concentration of one reaction component increases, its rate of reaction will increase as there are more

particles to collide with, thus increasing the frequency of effective collisions. The rate of that reaction will be

relatively greater than that of the rate of the reverse reaction. This means that more products or reactants will being

produced until equilibrium is reached.

Similarly, the reduction in one of the reaction components will reduce its rate of reaction. This occurs as there are

fewer particles to collide with which reduces the frequency of effective collisions. The rate of that reaction will be

relatively less than the rate of the reverse reaction. This means that more products or reactants will being produced

until equilibrium is reached.

CHANGE IN VOLUME OR PRESSURE

A decrease in the volume of a chemical system involving gasses will result in gasses colliding more often. The gas

particles will also be colliding with more energy as pressure is inversely proportional to volume. As particles are

colliding more frequently and with more energy to overcome the activation energy barrier, the frequency of

effective collision increases. Both the rate of the forward and reverse reaction will increase, however, the rate of

reaction that uses the greatest number of moles will be relatively greater than the reverse reaction as there are

more particles that can collide effectively with each other.

When the volume is increased and pressure is decreased, the partial pressures of all gasses will decrease. Both the

rate of the forward and reverse reaction will decrease. The rate of reaction that produces more moles will be

relatively greater than the reverse as the reaction is more likely to occur because it requires fewer particles to

effectively collide.

Changing the overall pressure of a chemical system does not always cause a disturbance in equilibrium. For example,

the addition of inert gasses.

CHANGE IN TEMPERATURE

By increasing the temperature of a chemical system at equilibrium, particles will possess more kinetic energy, which

means that more particles (both reactants and products) have enough energy to collide and overcome the activation

energy of the forward and reverse reactions.

For an exothermic reaction, the activation energy of the reverse reaction is higher than that of the forward reaction.

An increase in temperature means that proportionally more products will be able to collide with enough energy in

reverse reaction than the reactants. This causes the rate of the reverse reaction to occur at a faster rate than the

forward reaction. Therefore, the concentrations of the reactants will increase whereas the concentration of the


products will decrease until a new state of equilibrium is reached.

When the temperature is decreased, all the particles in the system lose energy which decreases both the rate of the

forward and the reverse reaction. However, the rate of the reverse endothermic reaction will be relatively higher

than the rate of the forward reaction.

OUTCOMES COVERED

analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for

example:

– combustion reactions

– photosynthesis

PHOTOSYNTHESIS

Photosynthesis appears to be the reverse reaction of the combustion of glucose and may seem to be a reversible

reaction. However, in nature, the process involves many individual irreversible steps which combine to give the

overall reaction, hence photosynthesis is irreversible.

6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)

- ∆𝐻 > 0 (endothermic)

- ∆𝑆 > 0 (less moles, moving to more order)

- Non-spontaneous at all temperatures

- Chlorophyll is the catalyst for photosynthesis.

- UV rays drives the photosynthesis reaction.

SPONTANEITY AND EQUILIBRIUM

Haber Process

N2(g) + 3H2(g) 2NH3(g) ∆𝐺 = −33.3 𝑘𝐽𝑚𝑜𝑙−1

- The Gibbs free energy change for this reaction is negative, therefore we would predict that the forward

reaction is spontaneous.

- The Gibbs free energy change for the reverse reaction will be the negative of this value, +33.3 𝑘𝐽𝑚𝑜𝑙−1, so

we would predict that the reverse reaction is nonspontaneous.

- Entropy of mixing allows the reaction to be reversible.

- The position with lowest free energy is somewhere in between pure reactants and pure products. This is

the position of equilibrium.

The sign of ∆𝐺 indicates whether reactants or products will dominate the mixture with lowest free energy.

- Since entropy of mixing always exist, technically no reaction is strictly irreversible.

o However, if the position of equilibrium lies very close to the products, the reaction is called

“irreversible” as the reverse reaction will not occur to any observable extent.

This is when Gibbs free energy is very negative.

- Reversibility of a reaction can also be considered in terms of activation energy.

o Reactions are unlikely to be reversible as molecules will not have sufficient energy.

o In order for a reaction to be reversible, the forward and reverse reaction must have a small

activation energy.


LE CHATELIER’S PRINCIPLE

OUTCOMES COVERED

investigate the effects of temperature, concentration, volume and/or pressure on a system at

equilibrium and explain how Le Chatelier’s principle can be used to predict such effects, for example:




examine how activation energy and heat of reaction affect the position of equilibrium

RATE OF REACTION (FROM MODULE 3)

- The rate of reaction is the speed with which reactants are converted to products, or how rapidly a reaction

proceeds.

- Rate is defined as the change in concentration of reactants or products over time.

- The rate of reaction depends on the frequency of effective collisions.

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 ∝ 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝐶𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠

FACTORS AFFECTING THE RATE OF REACTION

- Nature of reactants

- Concentration

- Surface area

- Temperature

- Catalysts

- Pressure/volume

NATURE OF REACTANTS

Every reaction has its own rate and its own activation energy, depending on the reactivity of the reactants.

Aqueous solutions already have dissociated ions. They do not need to collide in any correct orientation and usually

have very low Ea.

CONCENTRATION

The rate of reaction increases when the concentration of reactants is increased.

- The rate of reaction is directly proportional to the reactant concentration.

- Increasing the concentration increases the number of effective collisions.

- Increases the number of particles in a given space.


SURFACE AREA (PARTICLE SIZE)

The rate of reaction increases when the surface area of reactants is increased.

- Exposes more particles to the reactant. This increases the chance of a successful collision which therefore

increases the rate of reaction.

TEMPERATURE

The rate of reaction increases when the temperature of the reactants is increased.

- The total number of collisions increase.

o Increased KE, particles move faster. There is a greater chance of successful collision.

o When the temperature increases, the average kinetic energy of the molecules increase, thus

molecules move faster which means they collide more frequently.

- The average energy of the collisions increases. Therefore, a higher fraction of collisions exceed activation

energy.

- In general, reaction rate doubles every 10 degrees Celsius.

PRESENCE OF A CATALYST

A catalyst is a substance that increases the rate of reaction without being consumed.

Catalysts work by allowing the reaction to take an alternative reaction pathway with a lower activation energy.


WHAT IS LE CHATELIER’S PRINCIPLE

In 1888, a French chemist called Henri Le Chatelier (1850-1936) put forth the statement known as Le Chatelier’s

Principle:

“If a system at dynamic equilibrium is disturbed by changing the conditions, the system undergoes a reaction which

minimises the effect of the disturbance to attain a new equilibrium”

A chemical system at equilibrium can be disturbed in the following ways:

- Change in concentration

- Change in pressure → Change in volume →Change in concentration

- Change in temperature

Le Chatelier’s principle is a convenient method for predicting equilibrium shifts, but does not explain why it shifts.

Collision theory explains the shift in equilibrium.

THE HABER PROCESS

CHANGE IN CONCENTRATION

N2(g) + 3H2(g) 2NH3(g) ∆𝐻 = −92𝑘𝐽 𝑚𝑜𝑙−1

ADDING A REACTION COMPONENT

Addition of H2(g) increases [H2(g)]. This will result in the rate of the forward reaction to increase, meaning the forward

reaction has been favoured.

Since the rate of the forward reaction is different to the rate of the reverse reaction, the equilibrium has been

disturbed.

Generally, the reaction that counteracts the disturbance will be favoured; in other words, its rate will increase

relative to the other reaction.

REMOVING A REACTION COMPONENT

Removing N2(g) decreases [N2(g)]. This will result in the rate of the reverse reaction to increase, meaning the reverse

reaction has been favoured.

The reaction that counteracts the disturbance will be favoured; in other words, its rate will increase relative to the

other reaction.

CHANGE IN VOLUME (OR PRESSURE)

MOLES, VOLUME AND PRESSURE

The pressure exerted by a gas arises from the force of the gas particles colliding with the walls of the container.

Therefore, the pressure is proportional to the number of gas particles present.

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∝ 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑔𝑎𝑠


Boyle’s Law states that pressure and volume are inversely proportional to each other.

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∝ 1

𝑉𝑜𝑙𝑢𝑚𝑒

CHANGE IN VOLUME

Doubling the volume of the container decreases pressure. To counteract this effect, the reaction will shift in the

direction that produces the most amount of moles. Therefore, the reverse reaction will be favoured.

CHANGE IN PRESSURE

Doubling the pressure of the container decreases volume. To counteract this effect, the reaction will shift in the

direction that produces the least amount of moles. Therefore, the forward reaction will be favoured.

CHANGE IN TEMPERATURE

EXOTHERMIC AND ENDOTHERMIC REACTIONS

The effect of a change in temperature on a reaction at equilibrium depends on whether the forward reaction is

exothermic or endothermic.

N2(g) + 3H2(g) 2NH3(g) ∆𝐻 = −92𝑘𝐽 𝑚𝑜𝑙−1

If temperature of the system is increased, the reverse reaction will be favoured to counteract this effect. The reverse

reaction is endothermic and will be favoured.

ADDITION OF A CATALYST

A catalyst increases the rate of a chemical reaction without being consumed, by providing an alternate pathway of

lower activation energy. The addition of a catalyst reduces the activation energy of both the forward and reverse

reaction by the same amount.

- Therefore the addition of a catalyst will not disturb the equilibrium

- The concentrations of the components are not affected, by the system will reach equilibrium faster

ADDITION OF INERT GAS

The addition of an inert gas will increase pressure, but equilibrium will not be disturbed. This is because the

concentrations of reactants and products remain the same, if the volume of the container does not change.

DIMERISATION OF NITROGEN DIOXIDE

When colourless dinitrogen tetroxide gas (N2O4) is enclosed in a vessel, a brown colour will appear indicating the

formation of nitrogen dioxide (NO2). The intensity of the brown colour indicates the amount of nitrogen dioxide

present in the vessel. The dimerization of nitrogen dioxide is an exothermic process

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NOTES

- A shift in the forward direction is called a shift towards the right side.

- A shift in the reverse direction is called a shift towards the left side.

- Equilibrium will shift to remove an added component (away from component)

- Equilibrium will shift to replace a removed component (towards the component)

- An increase in volume will cause equilibrium to shift towards the side with more gas moles.

- A decrease in volume will cause equilibrium to shift towards the side with fewer gas moles.

ANSWERING LE CHATELIER’S PRINCIPLE QUESTIONS

Clearly explain the effect of changes in conditions on the yield of equilibrium reactions.

1) State that the change in reaction conditions disturbs the equilibrium.

2) “According to Le Chatelier’s principle, the position of equilibrium shifts left/right”

3) Justify the shift:

a. “To replace/remove”

b. “To the side with more/less gas moles, to increase/reduce pressure”

c. “In the exothermic/endothermic direction, to replace/remove heat”

4) … and minimise the disturbance

5) State the effect of the shift, “Therefore, the concentrations of the reactants/products increase/decrease”


QUALITATIVE ANALYSIS OF EQUILIBRIUM

OUTCOMES

investigate the effects of temperature, concentration, volume and/or pressure on a system at

equilibrium and explain how Le Chatelier’s principle can be used to predict such effects, for example:




CONCENTRATION PROFILE DIAGRAMS

The changes occurring in a system can be identified by examining the shape of the line in a concentration profile

diagram during a particular time period.

- The x-axis is time or reaction progress

- The y-axis is concentration or partial pressure

Graph Feature Disturbance

Concentration of one component spikes up, then all concentrations change

Increase [concentration] of a component

Concentration of one component spikes down, then all concentrations change

Decrease [concentration] of a component

Concentrations of all components spike up, then all concentrations change

Decrease volume → Increase pressure → Increase [concentrations] of components

Concentrations of all components spike down, then all concentrations change

Increase volume → Decrease pressure → Decrease [concentrations] of components

There are no spikes in the graph, then all concentrations change

Change in temperature

Concentrations are flat Nothing, at equilibrium


EQUILIBRIUM CONSTANT

OUTCOMES

deduce the equilibrium expression (in terms of Keq) for homogeneous reactions occurring in solution

(ACSCH079, ACSCH096)


– dissociation of ionic solutions

EQUILIBRIUM CONSTANT EXPRESSION

A quantitative way of describing the position of equilibrium is the equilibrium constant (𝐾𝑒𝑞).

𝑎𝐴 + 𝑏𝐵 𝑐𝐶 + 𝑑𝐷

- Capital letters represent chemical substances

- Lower case letters represent the stoichiometric coefficients of the balanced equation

𝐾𝑒𝑞 = 𝐾𝑐 = 𝐾 =[𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠]

[𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]=

[𝐶]𝑐[𝐷]𝑑

[𝐴]𝑎[𝐵]𝑏

In an ideal system, the value of K is constant at constant temperature.

PURE, LIQUIDS AND SOLIDS

In heterogeneous systems, some of the components are in different phases:

𝐶𝑎𝐶𝑂3(𝑠) 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2(𝑔)

The concentrations of pure solids and pure liquids cannot change at a constant temperature.

- Because the equilibrium constant is only concerned with concentrations that change as they approach

equilibrium, eliminate the terms for pure solids or liquids.

- The concentrations disappear from the equilibrium constant expression.

- Only gases and aqueous species appear in the equilibrium, except if all the reactant and products are all

liquids.

OUTCOMES

perform calculations to find the value of Keq and concentrations of substances within an equilibrium

system, and use these values to make predictions on the direction in which a reaction may proceed (ACSCH096)

INTERPRETING THE EQUILIBRIUM CONSTANT

The larger the value of K, the further the equilibrium lies towards the RHS. A reaction with a very large K, proceeds

almost to completion.

The smaller the value of K, the further the equilibrium lies towards the LHS. A reaction with a very small K, proceeds

barely at all.


K AND THE DIRECTION OF REACTION

If concentrations are substituted into the expression at any point, the value is called Q, the reaction quotient.

𝑄 =[𝐶]𝑐[𝐷]𝑑

[𝐴]𝑎[𝐵]𝑏

The relative values of Q and K determines which way the reaction will proceed to reach equilibrium.

- If 𝑄 = 𝐾, then the system is at equilibrium

- If 𝑄 > 𝐾, then the backwards reaction will be favoured

- If 𝑄 < 𝐾, then the forwards reaction will be favoured

RICE CALCULATIONS

Ratio The stoichiometric molar ratio of reactants and products.

[Initial] The initial concentrations of all components. The initial concentrations of the products are zero if not explicitly stated.

Change The change in concentration. Work one entry out from being given one equilibrium concentration and fill in the rest of the row according to the molar ratio.

[End] The concentrations of the components at equilibrium.

CALCULATIONS WITH QUADRATIC EQUATIONS

In some calculations, it will be required to solve a quadratic equation to determine 𝑥.

In these calculations, simplification can be used to make the calculation less complicated to solve:

- If K is very small, then 𝑥 will be very small, and the calculation can be simplified by using the following

approximation: [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑥 ≈ [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡]𝑖𝑛𝑖𝑡𝑖𝑎𝑙

- However, the assumption must be checked to be valid. The calculated value of 𝑥 must be less than 5% of

the reactant’s initial concentration.

If the assumption is invalid, the quadratic formula has to be used: 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

OUTCOMES

qualitatively analyse the effect of temperature on the value of Keq (ACSCH093)

EFFECT OF TEMPERATURE ON 𝑲𝒆𝒒

When temperature is increased, equilibrium shifts so that the endothermic reaction is favoured, whether it is the

forward or the reverse reaction. Therefore, unlike changes in concentration and pressure, a change in temperature is

the only factor that will change the value of K.

For exothermic reactions, K increases with lower temperatures and decreases with higher temperatures.

For endothermic reactions, K decreases with lower temperatures and increases with higher temperatures.


IRON(III) THIOCYANATE

OUTCOMES

conduct an investigation to determine Keq of a chemical equilibrium system, for example:

– Keq of the iron(III) thiocyanate equilibrium (ACSCH096)

IRON (III) THIOCYANATE EQUILIBRIUM REACTION

When mixed together, aqueous solutions of iron(III) nitrate (Fe(NO3)3) and potassium thiocyanate (KSCN) combine,

in a reversible exothermic process, to form the aqueous iron(III) thiocyanate complex ([Fe(SCN)]2+(aq)).

𝐹𝑒3+(𝑎𝑞) + 𝑆𝐶𝑁−

(𝑎𝑞) [𝐹𝑒(𝑆𝐶𝑁)]2+(𝑎𝑞)

Iron(III) ions are very pale yellow colour, while iron(III) thiocyanate complex is an intense deep red colour. When

diluted and at equilibrium, the colour of the mixture containing all three species is amber.

EQUILIBRIUM SHIFTS

A shift to the RHS, increase concentration of [Fe(SCN)]2+ → More intense

- Cool down system: ice water bath

- Increase [Fe3+] and [SCN-]: Add more FeCl3 and KSCN

A shift to the LHS, decreases concentration of [Fe(SCN)]2+ → Less intense

- Heat system: Hot water bath (Approx. 70 degrees Celsius)

- Decrease [Fe3+] and [SCN-]: Add NaOH and AgNO3

NOTES

Since thiocyanate ions bind to iron via the nitrogen atom, the formula of the iron(III) thiocyanate complex is

sometimes written as [Fe(NCS)]2+.

WHAT IS SPECTROPHOTOMETRY

Spectrophotometry is an analytical technique that is used to determine the concentration of a substance in a

solution.

- Iron(III) thiocyanate strongly absorbs light at

a wavelength of 447 nm and reflects light

that is mostly orange-red, hence we see it as

orange-red in colour.

- Iron(III) thiocyanate absorbs light blue light.

- When it absorbs the light, electrons are

transitioning from lower to higher quantised

energy levels. The energy gap between the

levels is equal to the energy of the 447 nm

wavelength light.


A UV/Vis spectrophotometer like the one shown measure the intensity of light passing through a sample solution in

a cuvette.

- A lamp provides white light (continuous source, which is narrowed and aligned into a beam using a slit.

- A prism splits the light into different wavelengths and is rotated so the desired wavelength of light passes

through the monochromator (exit slit).

- The light beam passes through the sample, which absorbs a fraction of the light. The greater the

concentration of iron(III) thiocyanate, the greater the absorption of 447 nm light.

- The remaining light is transmitted through the sample and reaches a detector, which converts the amount

of light to an electrical signal.

A spectrophotometer expresses the intensity of light in absorbance (A).

𝐴 = log10

𝐼0

𝐼

- A: Absorbance, a number typically between 0.3 – 2.5. It is dimensionless but often expressed in absorbance

units (AU)

- I0: The intensity of light passing through the blank (reference) sample.

- I: The intensity of light passing through the analyte sample.

Absorbance is proportional to both concentration and the length of the sample, according to the Beer-Lambert Law.

𝐴 = 𝜀𝑙𝑐

- 𝜺: Extinction coefficient (also known as molar absorptivity), a constant that relates absorbance with

concentration and path length, with the units 𝐿 𝑐𝑚−1 𝑚𝑜𝑙−1

- 𝒍: Path length of sample in cm

- 𝒄: Concentration of the substance in the sample in 𝑚𝑜𝑙 𝐿−1


SOLUBILITY

OUTCOMES

describe and analyse the processes involved in the dissolution of ionic compounds in water

PROPERTIES OF IONIC COMPOUNDS

A process where equilibrium is often established is in the dissolution of ionic compounds in water.

- Ionic compounds contain both cations and anions.

- They have an electrostatic attraction between oppositely charged ions.

- They are:

o Brittle

o High MP, BP

o Solids at room temperature

o Molten and aqueous state conducts electricity

SOLUTIONS OF IONIC COMPOUINDS

Soluble ionic compounds will dissolve in water to form aqueous solutions. When a soluble ionic compound is added

to water, the ions at the surface of the crystal become surrounded by water molecules.

- Some molecules of water separate from one another and are able to pack closer to the cations and anions of

the ionic salt.

o They form ion-dipole forces between the molecules.

- Solute: Ionic salt

- Solvent: Water

When the ion-dipole forces between the ions and the permanent dipoles of the water molecules (adhesive forces)

become stronger than the ionic bonds between the ions and the hydrogen bonding within the water (cohesive

forces), the ions are dislodged from their position in the crystal.

- The ionic compound dissociates into its component ions.

- The ions become solvated.

The solvated or hydrated ions are surrounded by a shell of water molecules known as a solvation layer.


- The solvation layer acts as a cushion and prevents a solvated anion from colliding directly with a solvated

cation, and therefore keeps the ions in the solution.

Many ionic compounds are soluble in water and will dissociate to form aqueous solutions.

- However not all ionic compounds are soluble in water. For example, AgCl and Ca3(PO4)2 are insoluble.

- The insolubility of these ionic compounds is due to their strong ionic bonds which cannot be disrupted by

the adhesive ion-dipole forces between solute ions and water molecules. Hence the ions do not become

dislodged from their positions in the crystal.

Entropy can also contribute to solubility. Most dissolutions are entropically favourable (∆𝑆 > 0), but some

dissolutions are unfavourable.

- The solution consists of the solute and the solvent, so the change in entropy of each part can be considered

separately to determine the overall change in entropy for the dissolution process:

∆𝑆𝑠𝑜𝑙𝑢𝑡𝑒 + ∆𝑆𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 ∆𝑆

- Depending on the relative magnitude of the effects, total entropy can increase or decrease for the overall

system when dissolution occurs, although in majority of cases overall entropy will increase.

- An overall decrease in entropy sometimes occurs for ions with high charge density as they interact strongly

with water molecules, so the water molecules are held tightly around the solvated ions.

INCREASE IN ENTROPY

Solute in a solid state has a fixed ordered arrangement. Dissolved solute has free mobile ions. Therefore, there is an

increase in entropy as the number of possible arrangement increases.

DECREASE IN ENTROPY

Pure water molecules are in random arrangements and are also mobile. When water becomes a solvent, molecules

solvate the solute and has less possible arrangements decreasing entropy.

NOTES

“Clear” means that light can pass through the substance without being scattered.

“Colourless” means the substance is not coloured. Solutions are clear, but not necessarily colourless.”


OUTCOMES

conduct an investigation to determine solubility rules, and predict and analyse the composition of

substances when two ionic solutions are mixed, for example:

– potassium chloride and silver nitrate

– potassium iodide and lead nitrate

– sodium sulfate and barium nitrate (ACSCH065)

PRECIPITATION

The production of an insoluble compound, usually by reacting two soluble compounds.

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

When two clear solutions are mixed together, an insoluble compound is formed. This is called a precipitate reaction.

Precipitation reactions are used to remove minerals from drinking water, to remove heavy metals from wastewater

and in purification plants of reservoirs.

SOLUBILITY RULES

Rule Exceptions

All Group 1 (Li+ Na+ K+ Rb+ Cs+) and ammonium (NH+) salts are soluble

NONE

All nitrate (NO3-), acetate (CH3COO-) and

bicarbonate (HCO3-) salts are soluble

NONE

All chloride (Cl-), bromide (Br-) and iodide (I-) salts are soluble

Ag+, Pb2+ and Hg2+ halides are insoluble

All sulfate (SO42-) salts are soluble Ag+, Pb2+, Hg2+, Ba2+, Ca2+, Sr2+

sulfates are insoluble

All hydroxides (OH-) salts are insoluble Group 1 and NH4+ hydroxides

are soluble,

Ba2+ and Ca2+ are slightly soluble

All carbonate (CO32-) and phosphate (PO4

3-) salts are insoluble

Group 1 and NH4+ carbonates

are soluble


OUTCOMES

derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of

an ionic substance from its Ksp value

SOLUBILITY

The solubility of a compound is the maximum amount of solute that can dissolve in a specific volume of solvent at a

certain temperature.

- Solubility is a physical constant, like melting point or boiling point

Term Solubility Range in Water (𝒈 𝑳−𝟏)

Very soluble ≥ 1000

Freely soluble 100 – 1000

Soluble 33 – 100

Sparingly soluble 10 – 33

Slightly soluble 1 – 10

Very slightly soluble 0.1 – 1

Insoluble < 0.1

EQUILIBRIA IN SATURATED SOLUTIONS

When a solvent has dissolved all the solute it can at a given temperature, the resulting solution is saturated.

- Any solution containing less solute than this is unsaturated.

- If more solute is added to a saturated solution, no more will dissolve.

In a saturated solution, the system is in dynamic equilibrium:

- All of the species – the ionic solid and dissolved ions – are present in the final mixture.

- The forward reaction is occurring at the same rate as the reverse reaction. In other words, when an ion

dissolves, another ion is precipitating at the same time.

The equilibrium constant for these solution equilibria is called the solubility product constant (𝐾𝑠𝑝).

- The equilibrium constant is for the equation written in the direction of the dissolution.


UNITS OF CONCENTRATION

MOLARITY

Molarity is the main unit of concentration used in chemistry:

𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑐) = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑛)

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑉)

- The volume of solution is expressed in litres (L).

o The volume of the total solution (solute + solvent) is used calculate concentration.

PERCENTAGE BY MASS OR WEIGHT (% M/M OR % W/W)

- Percentages by mass and weight are both extensively used in industry.

- % m/m is an abbreviation for the percentage mass of a substance relative to the total mass. This is the

same as percentage weight (% w/w).

o Percentage is given by the formula: % 𝑏𝑦 𝑚𝑎𝑠𝑠 =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛× 100

o % w/w or m/m can be interpreted as grams of solute per 100g.

o Note: the formula requires the total mass of the solution, not solvent.

PERCENTAGE BY VOLUME (% V/V)

- Percentages by mass and weight are both extensively used in industry.

- The Percentage by volume is given by the formula: % 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒 =𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛× 100

MASS PER VOLUME (M/V)

- Mass per volume is used in pharmacy and medicine.

- Mass per volume is used to measure the blood alcohol level of drivers.

o A blood alcohol level of 0.01 refers to 0.010/g / 100mL of blood

PARTS PER MILLION & PARTS PER BILLION

- Parts per million and parts per billion (ppm and ppb) are useful when describing very dilute solutions.

o 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑝𝑝𝑚) =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑔)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑘𝑔)

o 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑝𝑝𝑏) =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝜇𝑔)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑘𝑔)


SUMMARY

Name Unit Solute Unit Solution Unit

Molarity 𝑚𝑜𝑙 per 1 𝐿 𝑚𝑜𝑙 𝐿−1 𝑜𝑟 𝑀

% by mass 𝑔 per 100 𝑔 % 𝑚/𝑚 𝑜𝑟 % 𝑣/𝑣

% by volume 𝑚𝐿 per 100 𝑚𝐿 % 𝑣/𝑣

Parts per million 𝑚𝑔 per 𝑘𝑔 𝑝𝑝𝑚

Parts per billion 𝜇𝑔 per 𝑘𝑔 𝑝𝑝𝑏

OUTCOMES

investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when removing toxicity

from foods, for example:

– toxins in cycad fruit

Many native foods eaten by Aboriginal and Torres Strait Islander people are poisonous and need to be detoxified

before consumption

There are several physical and chemical processes user for detoxification:

COOKING OR ROASTING (CHEMICAL REACTION)

• The food items are heated in an oven or fire

• The heat causes the toxins to decompose

LEACHING (PHYSICAL CHANGE)

• The food items are cut up into smaller pieces and soaked in running water.

• Water-soluble toxins are washed away.

FERMENTATION OR PROLONGED STORAGE (CHEMICAL REACTION)

• The food items are stored for long periods (several months to several years)

• During this time, various biological processes occur that break down the toxins.

o They are digested by fungi, or broken down by the plant's natural enzymes. (Biological catalysts)

CYCADS (MACROZAMIA)

Cycads are palm-like plants that produce seeds in cones. Cycad seeds are a rich source of carbohydrates and have

been eaten in regions of Northern Australia for thousands of years. However, the seeds contains highly toxic

chemicals.

The two main types are cycasin and b-methylamino-l-alanine (BMAA).

To prepare them, Aboriginal and Torres Strait Islander people who ate these seeds would commonly prepare them

by:

- Cooking the seeds in an oven or fire


- Cutting the seeds open and leach them in running water for around a week

- Fermentation or storage for several months, or several years (some groups stored them for more than three

years)

THE POISON, CYCASIN

• Cycasin (C8H16N2O7)

• Methylazoxmethanol glucoside

• Molar Mass = 252.22g/mol

• pKa = 12.21 (measure of acidity)

SOLUBILITY

• 56.6 g/L @ 25 in water/diluted ethanol

• Sparingly soluble in absolute ethanol

• Insoluble in benzene, chloroform + acetone

DECOMPOSITION

• Melting point @144

• Decomposes at 154

• Produces N2

MORETON BAY CHESTNUT (BLACK BEAN)

The Moreton Bay Chestnut is found on the east coast of Australia. It produces pods containing large seeds that are

toxic. Eating unprocessed seeds causes vomiting and diarrhoea. When cooked, processed seeds taste like sweet

chestnuts.


SOLUBILITY CALCULATIONS

OUTCOMES

derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of

an ionic substance from its Ksp value

predict the formation of a precipitate given the standard reference values for Ksp

CALCULATING 𝑲𝒔𝒑 FROM SOLUBILITY

An equilibrium is only present in a saturated solution where the maximum amount of ionic compound has dissolved.

Therefore, the solubility of a substance, whether given in moles per litre or mass per volume, can be used to

calculate the solubility constant 𝐾𝑠𝑝.

PREDICTING THE FORMATION OF A PRECIPITATE

Solubility constants can be used to predict if a precipitate will form when two solutions are mixed.

𝐵𝑎(𝑂𝐻)2(𝑠) 𝐵𝑎2+(𝑎𝑞) + 2𝑂𝐻−

(𝑎𝑞)

𝐾𝑠𝑝 = [𝐵𝑎2+][𝑂𝐻−]2

- The solubility constant for this reaction at 25 is 2.55 × 10−4

- This means [𝐵𝑎2+][𝑂𝐻−]2 = 2.55 × 10−4 in a saturated solution.

If the concentration of Ba2+ and OH- are higher than the amounts in a saturated solution, 𝑄 > 𝐾𝑠𝑝 and precipitation

will occur.

If the concentration of Ba2+ and OH- are lower than the amounts in a saturated solution, 𝑄 < 𝐾𝑠𝑝 and precipitation

will not occur.

THE COMMON ION EFFECT

In an aqueous solution of an ionic compound, the ions are dissociated.

- This means that the ions separate into individual solvated ions.

- Ions of the same species are indistinguishable, regardless of where they originated.

This means that in a saturated solution, if another substance is added that has an ion in common with the first

substance, it will affect the position of equilibrium, leading to lower solubility. This is known as the common ion

effect.


SOLUBILITY CURVES (SOLUBILITY AND TEMPERATURE)

Since dissolution is a reversible process, the position of equilibrium (extent of dissolution) will depend on

temperature.

The relationship between solubility and temperature can be seen on a solubility curve, which shows the maximum

amount of solute that can dissolve at a range of temperatures.


DISSOCIATION OF ACIDS

OUTCOMES


– dissociation of acids and bases (ACSCH098, ACSCH099)

COMMON ACIDS

Name Formula

Hydrochloric acid HCl

Sulfuric acid H2SO4

Nitric acid HNO3

Phosphoric acid H3PO4

Carbonic acid H2CO3

Acetic acid (Vinegar/Ethanoic acid) CH3COOH

ACID STRENGTH

Acids differ in their strength: the extent of ionisation or dissociation in water.

- A stronger acid will ionise further

Strong acids completely dissociate in water

- Straight arrows are used to indicate that these dissociations are irreversible and proceed to completion.

Weak acids partially dissociate in water.

- Reversible arrows are used indicate that these dissociations proceed to equilibrium.

The acid dissociation constant is the equilibrium constant for the dissociation (ionisation) of an acid into hydrogen

ions (H+) and an anion (𝐾𝑎).

Acids that can produce more than one H+ ion are known as polyprotic acids.

- The dissociation of each H+ ion occurs stepwise, and an acid dissociation constant is assigned to each step.

𝐻3𝑃𝑂4(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐻2𝑃𝑂4

−(𝑎𝑞)

𝐾𝑎1 =[𝐻+][𝐻2𝑃𝑂4

−]

[𝐻3𝑃𝑂4]= 7.52 × 10−3

𝐻2𝑃𝑂4−

(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐻𝑃𝑂4

2−(𝑎𝑞)

𝐾𝑎2 =[𝐻+][𝐻𝑃𝑂4

2−]

[𝐻2𝑃𝑂4−]

= 6.23 × 10−8

𝐻𝑃𝑂42−

(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝑃𝑂4

3−(𝑎𝑞)

𝐾𝑎1 =[𝐻+][𝑃𝑂4

3−]

[𝐻𝑃𝑂42−]

= 2.20 × 10−13

Acid strength depends on the identity of the acid and the extent of its ionisation in water.


STRONG ACIDS

- Perchloric: HClO4

- Hydriodic: HI

- Hydrobromic: HBr

- Hydrochloric: HCl

- Nitric: HNO3

- Sulfuric: H2SO4

PH SCALE

The acidity of a solution is determined by both the strength and the concentration of the acids present.

The concentration of the hydrogen ions in a solution are generally small. The pH scale, a logarithmic scale, is a

convenient way of expressing [H+] as a number generally between 0 (extremely acidic) and 14 (extremely basic).

A pH of 7 represents a neutral solution. The further away from 7, the more acidic or alkaline the solution.

pH can be calculated from the [H+] using the equation:

𝑝𝐻 = −[𝐻+]

- 𝑝 stands for −𝑙𝑜𝑔10, and the concentration of the hydrogen ions are in 𝑚𝑜𝑙 𝐿−1

- The notation pH derives from the French pouvoir hydrogene, meaning the “power of hydrogen”

DEGREE OF IONISATION

To calculate the percentage of any component in a sample, the formula is:

𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 (%) =𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

𝑡𝑜𝑡𝑎𝑙× 100

FOR A WEAK ACID HA:

𝐻𝐴(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐴−

(𝑎𝑞)

The percentage ionised (degree of ionisation):

% 𝑖𝑜𝑛𝑖𝑠𝑒𝑑 =[𝐻+]𝑒𝑞𝑚

[𝐻𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙× 100 =

[𝐻−]𝑒𝑞𝑚

[𝐻𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙× 100

The percentage unionised (percentage of intact molecules):

% 𝑢𝑛𝑖𝑜𝑛𝑖𝑠𝑒𝑑 =[𝐻𝐴]𝑒𝑞𝑚

[𝐻𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙× 100 = 100 − % 𝑖𝑜𝑛𝑖𝑠𝑒𝑑

CHEMISTRY MODULE 6 Acid/Base Reactions

CHEMISTRY MODULE 6: ACID/BASE REACTIONS PAGE 1 OF 40

PROPERTIES OF ACIDS AND BASES

Inquiry question: What is an acid and what is a base?

Students:

investigate the correct IUPAC nomenclature and properties of common inorganic acids and bases (ACSCH067)

conduct an investigation to demonstrate the preparation and use of indicators as illustrators of the

characteristics and properties of acids and bases and their reversible reactions (ACSCH101)

predict the products of acid reactions and write balanced equations to represent:

– acids and bases

– acids and carbonates

– acids and metals (ACSCH067)

investigate applications of neutralisation reactions in everyday life and industrial processes

conduct a practical investigation to measure the enthalpy of neutralisation (ACSCH093)

explore the changes in definitions and models of an acid and a base over time to explain the limitations of each

model, including but not limited to:

– Arrhenius’ theory

– Brønsted–Lowry theory (ACSCH064, ACSCH067)

USING BRØNSTED–LOWRY THEORY

Inquiry question: What is the role of water in solutions of acids and bases?

Students:

conduct a practical investigation to measure the pH of a range of acids and bases

calculate pH, pOH, hydrogen ion concentration ([H+]) and hydroxide ion concentration ([OH–]) for a range of

solutions (ACSCH102)

conduct an investigation to demonstrate the use of pH to indicate the differences between the strength of acids

and bases (ACSCH102)

write ionic equations to represent the dissociation of acids and bases in water, conjugate acid/base pairs in

solution and amphiprotic nature of some salts, for example:

– sodium hydrogen carbonate

– potassium dihydrogen phosphate

construct models and/or animations to communicate the differences between strong, weak, concentrated and

dilute acids and bases (ACSCH099)

calculate the pH of the resultant solution when solutions of acids and/or bases are diluted or mixed

QUANTITATIVE ANALYSIS

Inquiry question: How are solutions of acids and bases analysed?

Students:

conduct practical investigations to analyse the concentration of an unknown acid or base by titration

investigate titration curves and conductivity graphs to analyse data to indicate characteristic reaction profiles,

for example:

– strong acid/strong base

– strong acid/weak base

– weak acid/strong base (ACSCH080, ACSCH102)


model neutralisation of strong and weak acids and bases using a variety of media

calculate and apply the dissociation constant (Ka) and pKa (pKa = -log10 (Ka)) to determine the difference

between strong and weak acids (ACSCH098)

explore acid/base analysis techniques that are applied:

– in industries

– by Aboriginal and Torres Strait Islander Peoples

– using digital probes and instruments

conduct a chemical analysis of a common household substance for its acidity or basicity (ACSCH080) , for

example:

– soft drink

– wine

– juice

– medicine

conduct a practical investigation to prepare a buffer and demonstrate its properties (ACSCH080)

describe the importance of buffers in natural systems (ACSCH098, ACSCH102)


ACIDS/BASES & INDICATORS

OUTCOMES

investigate the correct IUPAC nomenclature and properties of common inorganic acids and bases (ACSCH067)

ACIDS AND BASES

WHAT IS AN ACID?

Acids are a group of commonly used substances that share the following properties:

• They taste sour

• They produce H+ when dissolved in water (Arrhenius definition)

• Acids ionise in solution (𝐻𝑋 + 𝐻2𝑂 → 𝐻3𝑂+ + 𝑋−)

• pH below 7

• They are corrosive and will wear down materials like metal and skin.

• Turns blue litmus indicator red

• Acids conduct electricity as it has mobile ions that are able to conduct charge.

o Conductivity ∝ [ions]

o Strong Acids = Strong electrolyte

o Weak Acids = Weak electrolyte

WHAT IS A BASE?

Bases are another group of commonly used substances. The properties of bases include:

• They taste bitter and have a slippery or soap-like feel

• They produce hydroxide ions in water (Arrhenius definition)

• pH above 7

• They are corrosive (caustic)

• They turn red litmus indicator blue

OUTCOMES

conduct an investigation to demonstrate the preparation and use of indicators as illustrators of the

characteristics and properties of acids and bases and their reversible reactions (ACSCH101)

INDICATORS

• Indicators give a qualitative indication of the [H+]

• Indicators changes colour in response to the surrounding pH

• An acidic solution has a high [H+], whereas a basic solution has low [H+].

The equation for calculating pH is: 𝑝𝐻 = − log10[𝐻+]


HOW DO INDICATORS WORK?

Most indicators are organic weak acids or weak bases. They are a special case of a weak acid.

𝐻𝐼𝑛𝑑(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐼𝑛𝑑−

(𝑎𝑞)

𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑒 (𝐶𝑜𝑙𝑜𝑢𝑟 𝐼) 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 𝐼𝑜𝑛 + 𝐴𝑛𝑖𝑜𝑛 (𝐶𝑜𝑙𝑜𝑢𝑟 𝐼𝐼)

• Ind: An abbreviation for “indicator” and represents the rest of the organic molecule.

• The two forms are different colours (I and II). Changes in [H+] will shift equilibrium.

LOW PH & HIGH PH

• Low pH == High [H+]

o Equilibrium will lie towards the LHS

o Hind will dominate and colour observed will be Colour I

• High pH == Low [H+]

o Equilibrium will lie towards the RHS

o Ind- will dominate and colour observed will be colour II

Indicator Colour at lower pH Colour at higher pH pH of colour change range

Methyl orange Red Yellow 3.1 – 4.4

Bromothymol blue Yellow Blue 6.0 – 7.6

Phenolphthalein Colourless Pink 8.3 – 10.0

Methyl red Red Yellow 4.8 – 6.0

Litmus Red Blue 4.5 – 8.3

Phenol red Yellow Red 6.8 – 8.4

Some limitations of indicators include:

• Approximate pH (not accurate)

• Cannot distinguish between strong/weak acid/base

• Destroys/contaminates solutions

Some advantages include:

• Cheap and relatively easy to use


EVERYDAY USES OF INDICATORS

SWIMMING POOLS

Swimming pool pH needs to be maintained close to 7.4 to avoid irritation of the eyes and mucous membranes

(which are at approximately pH 7.4).

• Bromothymol blue or phenol red can be used to test the pH of the pool.

• If the pH needs to be lowered, acids such as HCl or solid (NaHSO4) can be added.

• If the pH needs to be raised, sodium carbonate (Na2CO3) can be added.


OUTCOMES

predict the products of acid reactions and write balanced equations to represent:

– acids and bases

– acids and carbonates

– acids and metals (ACSCH067)

REACTIONS OF ACIDS

ACID + METAL HYDROXIDE/OXIDE (NEUTRALISATION)

𝐻𝑋(𝑎𝑞) + 𝑀𝑂𝐻(𝑠)/(𝑎𝑞) → 𝑀𝑋(𝑎𝑞) + 𝐻2𝑂(𝑙)

Metal oxides are also bases and react with acids to form a salt and water.

• 2𝐻𝐶𝑙(𝑎𝑞) + 𝑀𝑔𝑂(𝑠) → 𝑀𝑔𝐶𝑙2(𝑎𝑞) + 𝐻2𝑂(𝑙)

Ammonia is a special case of a base that will undergo neutralisation but will not produce water, as it does not

contain oxygen in its formula.

• Ammonia (NH3) is covalent molecular. It is a gas at room temperature and highly soluble at RTP

• Instead it will produce an ammonium salt as the product

• 𝐻𝐵𝑟(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) → 𝑁𝐻4𝐵𝑟(𝑎𝑞)

ACID + METAL CARBONATE / HYDROGEN CARBONATE

𝐻𝑋(𝑎𝑞) + 𝑀𝐶𝑂3(𝑠) → 𝑀𝑋(𝑎𝑞) + 𝐻2𝑂(𝑙) + 𝐶𝑂2(𝑔)

ACID + REACTIVE METAL (REDOX REACTION)

𝐻𝑋(𝑎𝑞) + 𝑀(𝑠) → 𝑀𝑋(𝑎𝑞) + 𝐻2(𝑔)

EVERYDAY USES OF ACIDS/BASES

Bee stings and ant bites are acidic in nature. They can be neutralised using alkaline medicine such as baking powder.

Wasp stings are alkaline in nature. Vinegar can be used to cure wasp stings because vinegar can neutralise the sting.

When stung by a stingray, concentrated vinegar can be used to stop the nematocysts from firing off such that you

won’t get injected with more venom.


THEORIES OF ACIDS AND BASES

OUTCOMES

explore the changes in definitions and models of an acid and a base over time to explain the limitations of each

model, including but not limited to:

– Arrhenius’ theory

– Brønsted–Lowry theory (ACSCH064, ACSCH067)

LAVOISIER – 1776 “FATHER OF CHEMISTRY” (OXIDES)

Antoine Lavoisier was a French chemist who established the quantitative science of chemistry. He investigated

oxides of different elements, and found that non-metal oxides reacted with water, producing acidic solutions. He

concluded that an acid must contain oxygen. Eg:

𝑆𝑂2(𝑔) + 𝐻2𝑂(𝑙) → 𝐻2𝑆𝑂3(𝑎𝑞)

𝐶𝑂2(𝑔) + 𝐻2𝑂(𝑙) → 𝐻2𝐶𝑂3(𝑎𝑞)

DAVY – 1810 (HYDROGEN)

English chemist Humphry Davy (famous for his redox and electrolytes works) demonstrated that hydrochloric acid

did not contain oxygen, thus disproving Lavoisier’s theory. Davy suggested that hydrogen must be the unifying

component of acids rather than oxygen.

LIEBIG – 1838 (ACID + METAL)

The German chemist Justus von Liebig expanded on Davy’s idea. In 1838, he proposed that acids contain hydrogen

which could be displaced by a reaction with a metal. Eg:

𝑀𝑔(𝑠) + 2𝐻𝐶𝑙(𝑎𝑞) → 𝑀𝑔𝐶𝑙2(𝑎𝑞) + 𝐻2(𝑔)

𝑍𝑛(𝑠) + 𝐻2𝑆𝑂4(𝑎𝑞) → 𝑍𝑛𝑆𝑂4(𝑎𝑞) + 𝐻2(𝑔)

ARRHENIUS – 1884 (H+ AND OH-)

Svante Arrhenius proposed the first concept of acids and bases we still use.

- Arrhenius’ work centred on the conductivity of electrolytes. He postulated that electrolytes dissociated in

water into ions.

- He defined acids and bases according to their effect in water.

An Arrhenius acid is a substance that produces a H+(aq) in water.

𝐻𝐶𝑙(𝑎𝑞) → 𝐻+(𝑎𝑞) + 𝐶𝑙−

(𝑎𝑞)

Arrhenius also notated that the most reactive acids also had the highest electrical conductivities. This led to the

concept that the strongest acids were the most dissociated in aqueous solution.

An Arrhenius base is a substance that produces OH-(aq) in water.


LIMITATIONS OF THE ARRHENIUS DEFINITION

Arrhenius’ definition does not explain the basic behaviour of substances like ammonia, which do not contain

hydroxide ions in their formulae and hence should not be able to produce OH-.

It does not explain why neutralisation reactions between some acids and bases produced solutions that were not

neutral.

- The reaction between ammonia and hydrochloric acid produces an acidic solution. (NH4Cl – an acidic salt)

- The reaction between acetic acid and sodium hydroxide produces a basic solution (NaCH3OO – a basic salt)

The Arrhenius definition only covers acids and bases in aqueous solutions.

BRONSTED-LOWRY ACIDS AND BASES (CONCEPTUAL DEFINITION)

In 1923, Danish chemist Johannes Nicolaus Bronsted and English chemist Thomas Martin Lowry independently

proposed a new definition of acids and bases.

Acids and bases are defined by their role in a reaction:

- The proton donor is the Bronsted-Lowry acid

- The proton acceptor is the Bronsted-Lowry base

The Bronsted-Lowry definition allows many more species to be defined as acids or bases. It can explain the basic

behaviour of ammonia. The NH3(aq) is accepting a proton from HCl in the aqueous solution. NH3(aq) is a Bronsted

Lowry base and HCl is a Bronsted Lowry acid.

𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) → 𝑁𝐻4+

(𝑎𝑞) + 𝐶𝑙−(𝑎𝑞)

Bronsted Lowry theory also explains the basic behaviour of ionic compounds in solution.

Soluble carbonates and hydrogen carbonates contain Bronsted Lowry bases. They produce basic solutions.

1) First the compound dissolves in water to produce aqueous ions. This step proceeds completely, since all

Group 1 ionic compounds are soluble:

𝑁𝑎2𝐶𝑂3(𝑠) → 2𝑁𝑎+(𝑎𝑞) + 𝐶𝑂3

2−(𝑎𝑞)

2) The dissolved carbonate or hydrogen carbonate ion is a Bronsted Lowry base which reacts with water to

produce hydroxide ions:

𝐶𝑂32−

(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐻𝐶𝑂3−

(𝑎𝑞) + 𝑂𝐻−(𝑎𝑞)

The Bronsted Lowry definition is broad enough that some species like water can be classified both an acid and a

base.

- For example, when ammonia dissolves in water, water donates H+ to ammonia and is acting as a Bronsted

Lowry acid.

𝑁𝐻3(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝑁𝐻4+



- When HCL dissolves in water, water accepts H+ from HCl and is acting as a Bronsted Lowry base.

𝐻𝐶𝑙(𝑎𝑞) + 𝐻2𝑂(𝑙) → 𝐻3𝑂+(𝑎𝑞) + 𝐶𝑙−

(𝑎𝑞)

A substance that is capable of both donating and accepting protons (depending on the other reagent) is known

as an amphiprotic substance.

HYDRONIUM IONS

- A H+ ion is a bare proton with a +1 charge. This means that any H+ ion in water immediately combines with a

water molecule to form a more stable hydronium ion, H3O+.

- H+ does not technically exist independently in solution.

𝐻+(𝑎𝑞) + 𝐻2𝑂(𝑙) → 𝐻3𝑂+

(𝑎𝑞)

STRENGTHS OF ACIDS AND BASES

STRONG/WEAK ACIDS

A strong acid is a strong electrolyte that ionises completely in water to produce hydronium ions in aqueous

solutions. Eg:

𝐻𝐶𝑙(𝑎𝑞) → 𝐻+(𝑎𝑞) + 𝐶𝑙−

(𝑎𝑞)

List of strong acids:

• 𝐻𝐶𝑙

• 𝐻𝑁𝑂3

• 𝐻2𝑆𝑂4

• 𝐻𝐵𝑟 (Hydrobromic acid)

• 𝐻𝐼 (Hydroiodic acid)

• 𝐻𝐶𝑙𝑂4 (Perchloric acid)

• 𝐻𝐶𝑙𝑂3 (Chloric acid)

All other acids are weak acids, which ionise partially in water to form hydronium ions in solution. They are weak

electrolytes. Eg:

𝐻𝐹(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐹−

(𝑎𝑞)


ACID DISSOCIATION CONSTANT (K A)

The higher the Ka the stronger the acid strength as it favours the RHS.

Most acids are monoprotic: they can only produce one proton. However, some acids have more than one Ka value.

These acids can produce more than one H+ ion and are known as polyprotic acid.

For example, phosphoric acid is a triprotic acid.

𝐻3𝑃𝑂4(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐻2𝑃𝑂4

−(𝑎𝑞)

𝐾𝑎1 =[𝐻+][𝐻2𝑃𝑂4

−]

[𝐻3𝑃𝑂4]= 7.52 × 10−3

𝐻2𝑃𝑂4−

(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐻𝑃𝑂4

2−(𝑎𝑞)

𝐾𝑎2 =[𝐻+][𝐻𝑃𝑂4

2−]

[𝐻2𝑃𝑂4−]

= 6.23 × 10−8

𝐻𝑃𝑂42−

(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝑃𝑂4

3−(𝑎𝑞)

𝐾𝑎3 =[𝐻+][𝑃𝑂4

3−]

[𝐻𝑃𝑂42−]

= 2.20 × 10−13

The number of protons an acid produces is unrelated to its strength.

STRONG/WEAK BASES

Strong bases include Group 1 and 2 metal hydroxides. When they dissolve, they produce free hydroxide ions.

Carbonates are weak bases.

- NaOH (Caustic soda, drain cleaner, oven cleaner)

- KOH

- Ca(OH)2 (Lime water)

- Ba(OH)2

Although Group 2 hydroxides are strong bases, they are poorly soluble in water.

A weak base is one that reacts partially with water to indirectly form hydroxide ions in solution.

𝑁𝐻3(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝑁𝐻4+


STRENGTH VS CONCENTRATION

- Acid strength depends on the identity of the acid and the extent of its ionisation in water.

- Concentration depends on the amount of acid in a given volume of solution

Solution Strong or Weak? Concentrated or Dilute?

10 M HCl Strong Concentrated

10 M HF Weak Concentrated

0.01 M HCl Strong Dilute

0.01 M HF Weak Dilute


CONJUGATE ACIDS AND BASES

The ionisation of an acid can be represented by the following equation:


(𝑎𝑞)

When the equation above is read in reverse, it shows A- accepting a proton and acts as a Bronsted Lowry base.

HA and A- are known as a conjugate acid/base pair.

- The two species in a conjugate pair differ by a proton (H+)

- When a Bronsted Lowry acid loses a proton, it forms the conjugate base of that acid. (𝐻𝑁𝑂3/𝑁𝑂3−)

- When a Bronsted Lowry base gains a proton, it forms the conjugate acid of that base. (𝐹−/𝐻𝐹)

- An acid’s conjugate base has 1 less proton, while a base’s conjugate acid has 1 more proton.

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝐻2𝑂(𝑙) → 𝐶𝐻3𝐶𝑂𝑂−(𝑎𝑞) + 𝐻3𝑂+

(𝑎𝑞)

- Conjugate acid-base pair

- Conjugate base-acid pair

Acid → Donates H+ → Conjugate Base

Base → Accepts H+ → Conjugate Acid

RELATIVE STRENGTH OF CONJUGATE PAIRS

The two species in conjugate pair have inverse strength.

- A strong acid will have an extremely weak (in practice, neutral) conjugate base. (e.g. 𝐻𝐶𝑙/𝐶𝑙−)

- A weak acid will have a relatively strong conjugate base. (e.g. 𝐶𝐻3𝐶𝑂𝑂𝐻/𝐶𝐻3𝐶𝑂𝑂−)

- The same situation applies for bases and their conjugate acids.

A strong acid has a strong tendency to give up a H+, and completely ionise in water.

𝐻𝐶𝑙(𝑎𝑞) + 𝐻2𝑂(𝑙) → 𝐻3𝑂+(𝑎𝑞) + 𝐶𝑙−

(𝑎𝑞)

- The equilibrium lies far to the right

- If the reaction is viewed in reverse, it shows that the chloride ion has a very weak tendency to accept a

proton, as the reverse reaction barely proceeds.

- Therefore, HCl has an extremely weak conjugate base. It practically has negligible proton-accepting ability,

and produces a neutral solution.

A weak acid has a relatively strong conjugate base.

𝐻𝐹(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐻3𝑂+(𝑎𝑞) + 𝐹−

(𝑎𝑞)

- The equilibrium lies far to the left.

- High tendency for the fluoride ion to accept a proton. Therefore, it will be a strong conjugate base.

Rule: The weaker the acid, the stronger the conjugate base. (Vice versa for weak bases)


EXAMPLE QUESTION 1

Explain why the presence of nitrate ions in an aqueous solution will not make it basic. (i.e. will not produce extra OH-

ions).

𝐻𝑁𝑂3(𝑎𝑞) + 𝐻2𝑂(𝑙) → 𝐻3𝑂+(𝑎𝑞) + 𝑁𝑂3

−(𝑎𝑞)

This reaction goes to virtual completion. The reverse reaction does not occur to any significant extent.

∴ 𝑁𝑂3−

(𝑎𝑞) has no tendency to accept a H+ and is a neutral ion. It will not accept any H+ from water to produce OH-

ions. 𝑁𝑂3− is also the conjugate base of a strong acid.

EXAMPLE QUESTION 2

Explain why an aqueous solution containing fluoride ions will be basic.

𝐻𝐹(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐻3𝑂+(𝑎𝑞) + 𝐹−

(𝑎𝑞)

F- has a relatively high tendency to accept H+ as it is a conjugate base of a weak acid HF. It will accept H+ from water

to produce OH- ions resulting in a basic solution of pH > 7.


APPLYING BRONSTED LOWRY THEORY

OUTCOMES

write ionic equations to represent the dissociation of acids and bases in water, conjugate acid/base pairs in

solution and amphiprotic nature of some salts, for example:

– sodium hydrogen carbonate

– potassium dihydrogen phosphate

NON-NEUTRAL SALT SOLUTIONS

Many ions are not neutral when dissolved in water. They will have acidic or basic properties.

- This means that in neutralisation reactions, the salt product is not necessarily neutral (pH = 7).

- pH depends on the nature of the salt.

To show that an ionic compound will form an acidic, basic or neutral solution:

1) Write an equation showing the dissociation of the compound into its two ions.

2) Determine if either ion is acidic or basic by examining their conjugates.

3) Write an equation showing the acidic or basic ion reacting with water to form H3O+ or OH-.

- Strong acid + Weak base → Acidic

- Strong acid + Strong base → Neutral

- Weak acid + Strong base → Basic

EXAMPLE QUESTION 1

Will a solution of potassium fluoride be acidic or basic?

𝐾𝐹(𝑎𝑞) → 𝐾+(𝑎𝑞) + 𝐹−

(𝑎𝑞)

F- is a conjugate base of a weak acid HF. Therefore, it will have a relatively high tendency to react with water to form

OH- resulting in a basic solution of pH > 7. (𝐹−(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝑂𝐻−

(𝑎𝑞) + 𝐻𝐹(𝑎𝑞))

AMPHIPROTIC SUBSTANCE

Water can be classified as both a Bronsted-Lowry acid and base. A molecule or ion that can accept and donate

protons is called amphiprotic. Amphiprotic species will lose or gain a proton depending on the other reactant.

- If the other species is a stronger base, it will act as a Bronsted-Lowry acid

- If the other species is a stronger acid, it will act as a Bronsted-Lowry base.

Examples include: HCO3-, HSO4

-, H2PO4-, HPO4

2- and NH3


Amphiprotic substances are a subset of amphoteric substances, which are substances that can react with both acids

and bases.

- In laboratories, NaHCO3 is commonly used for neutralising acid spills as it is a weak, amphoteric base that

produces CO2 during neutralisation.

o It is a weak base therefore neutralisation is not as exothermic.

o Amphoteric/amphiprotic therefore will absorb OH- if too much is added.

o Produces CO2 therefore it can be seen when neutralisation is complete

o Solid, therefore will not contribute to the size of the spill

ACID/BASE BEHAVIOUR OF OXIDES

An acidic oxide is one which either reacts with water to form an acidic solution or reacts with bases to form acidic

salts. Common acidic oxides are CO2 and P4O10 (diphosphorus pentoxide) and SO2.

𝐶𝑂2(𝑔) + 𝐻2𝑂(𝑙) 𝐻2𝐶𝑂3(𝑎𝑞) 𝐻+(𝑎𝑞) + 𝐻𝐶𝑂3

−(𝑎𝑞)

𝐶𝑂2(𝑔) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎2𝐶𝑂3(𝑎𝑞) + 𝐻2𝑂(𝑙)

Non-metal from the RHS of the periodic table tend to form acidic oxides. These elements have high

electronegativity and share electrons when bonding with oxygen, so non-metal oxides are covalent.

A basic oxide is one that reacts with water to form an alkaline solution or reacts with acids to form basic salts.

Metals from the LHS of the periodic table tend to form basic oxides. These elements have low electronegativity, so

metal oxides are ionic.


PH SCALE

OUTCOMES



calculate and apply the dissociation constant (Ka) and pKa (pKa = -log10 (Ka)) to determine the difference

between strong and weak acids (ACSCH098)

PH AND H+ CONNECTION

- pH 7 is neutral at 25 ([H3O+] = [OH-])

The pH (potential of hydrogen) scale is logarithmic (base 10), not linear. Significant figures for logs are those after

the decimal point.

- 𝑝𝐻 = − log10[𝐻+]

- [𝐻+] = 10−𝑝𝐻

PH OF STRONG AND WEAK ACIDS

pH can deduce the relative strength and concentration of different acidic solutions.

- ↑ Acid strength → ↑ Degree of ionisation → ↑ [H+] → ↓ pH

- ↑ [Acid] → ↑ [H+] → ↓ pH

- Acid strength and concentration both affect the pH of a solution.

For a monoprotic strong acid, the concentration of H+ is the same as the concentration of the acid (𝐻𝑋 → 𝐻+ + 𝑋−)

For weak acids, the concentration of H+ will depend on its strength and concentration of the acid solution.

𝐻𝐴 𝐻+ + 𝐴− Where 𝐾𝑎 =[𝐻+][𝐴−]

[𝐻𝐴]


POH OF STRONG AND WEAK BASES

pOH is a measure of OH- concentration that is similar to pH.

- 𝑝𝑂𝐻 = − log10[𝑂𝐻−]

- [𝑂𝐻−] = 10−𝑝𝑂𝐻

Strong bases are Group 1 and 2 hydroxides. The concentration of hydroxide ions depends on the number of

hydroxide ions in the formula.

For weak bases, the concentration of OH- will depend on its strength and the concentration of the base solution.

AUTOIONISATION OF WATER

The concentration of H+ and OH- in any aqueous solution are directly related. This is because water is a very weak

acid which forms the following exothermic equilibrium reaction.

- The equilibrium constant for this equilibrium is called the ionisation constant for water.

𝐾𝑤 = [𝐻3𝑂+][𝑂𝐻−] = 1.0 × 10−14 at 25

𝑝𝐾𝑤 = 𝑝𝐻 + 𝑝𝑂𝐻

𝑝𝐻 + 𝑝𝑂𝐻 = 14

𝑝𝐻 = 14 + log10[𝑂𝐻−]

PKA AND PKB

Since Ka values are usually very small, pKa values are often cited instead:

𝑝𝐾𝑎 = − log10 𝐾𝑎

𝐾𝑎 = 10−𝑝𝐾𝑎

↑ Strong acid → ↑ Ka → ↓pKa


PH CALCULATIONS

OUTCOMES



EXAMPLE QUESTION 1

Benzoic acid, a food preservative, is a weak monoprotic acid. The pH of 0.50M benzoic acid solution is 2.25. What is

the Ka of benzoic acid?


(𝑎𝑞)

[𝐻+] = 10−𝑝𝐻 = 10−2.25 = 5.62341325 × 10−3 𝑚𝑜𝑙/𝐿

𝑯𝑨 𝐻+ 𝑨−

Mole Ratio 1 1 1

[Initial] 0.50 0 0

[Change] −10−2.25 +10−2.25 +10−2.25 [Final] 0.50 − 10−2.25 10−2.25 10−2.25

𝐾𝑎 =[𝐻+][𝐴−]

[𝐻𝐴]=

(10−2.25)(10−2.25)

0.50 − 10−2.25= 6.4 × 10−5 (2 𝑠. 𝑓)

EXAMPLE QUESTION 2

Acetic acid has a Ka of 1.8 x 10-5. Calculate the pH of a 0.20M solution of acetic acid.

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) 𝐶𝐻3𝐶𝑂𝑂−(𝑎𝑞) + 𝐻+

(𝑎𝑞)

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) 𝐶𝐻3𝐶𝑂𝑂−(𝑎𝑞) 𝐻+

(𝑎𝑞)

Mole Ratio 1 1 1

[Initial] 0.20 0 0

[Change] −𝑥 +𝑥 +𝑥

[Final] 0.20 − 𝑥 𝑥 𝑥

𝐾𝑎 =[𝐶𝐻3𝐶𝑂𝑂−][𝐻+]

[𝐶𝐻3𝐶𝑂𝑂𝐻]=

(𝑥)(𝑥)

(0.20 − 𝑥)= 1.8 × 10−5

Assume 0.20 − 𝑥 ≈ 0.20

(𝑥)(𝑥)

(0.20)= 1.8 × 10−5

𝑥 = 1.879736660 × 10−3

𝑥

0.20−𝑥= 0.96% Therefore, assumption is valid

𝑝𝐻 = −log10[𝐻+] = −log10[𝑥] = 2.72 (2 𝑠. 𝑓)


DEGREE OF IONISATION

All acids ionise in water, the degree of which is different for individual acids.

To calculate the percentage of any component in a sample, the formula is:

𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 (%) =𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

𝑡𝑜𝑡𝑎𝑙× 100

FOR A WEAK ACID HA


(𝑎𝑞)

The percentage ionised (degree of ionisation):

% 𝑖𝑜𝑛𝑖𝑠𝑒𝑑 =[𝐻+]𝑒𝑞𝑚

[𝐻𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙× 100 =

[𝐴−]𝑒𝑞𝑚

[𝐻𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙× 100

The percentage unionised (percentage of intact molecules):

% 𝑢𝑛𝑖𝑜𝑛𝑖𝑠𝑒𝑑 =[𝐻𝐴]𝑒𝑞𝑚

[𝐻𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙× 100 = 100 − % 𝑖𝑜𝑛𝑖𝑠𝑒𝑑

PH MEASUREMENTS

PH PROBE

A pH probe or pH meter can be used to measure pH.

- Advantages: Precision, non-destructive

- Disadvantages: Initial costs, requires calibration before use


PH OF MIXED SOLUTIONS

OUTCOMES

calculate the pH of the resultant solution when solutions of acids and/or bases are diluted or mixed

DILUTION CALCULATIONS

The total moles of solute in a concentrated solution and the diluted solution are the same.

- 𝑐1𝑉1 = 𝑐2𝑉2

EXAMPLE QUESTION 1

Calculate pH when 1.0 mL of 1.0M HCl is diluted to 1.25L.

𝑐1𝑉1 = 𝑐2𝑉2

𝑐2 =(0.001)(1.0)

1.25= 0.0008 𝑚𝑜𝑙/𝐿

[𝐻𝐶𝑙] = [𝐻+] = 0.0008 𝑀

𝑝𝐻 = − log[𝐻+] = − log(0.0008) = 3.10 (2 𝑠. 𝑓)

PH OF SOLUTIONS AFTER MIXING

EXAMPLE QUESTION 2

Calculate the pH of the resultant solutions when 50.0 mL of 0.020 M NaOH solution and 100 mL of 0.012 M Ba(OH)2

𝑛(𝑁𝑎𝑂𝐻) = (0.050)(0.020) = 0.001 𝑚𝑜𝑙𝑒𝑠

𝑛(𝑂𝐻−) = 𝑛(𝑁𝑎𝑂𝐻) = 0.001 𝑚𝑜𝑙𝑒𝑠

𝑛(𝐵𝑎(𝑂𝐻)2) = (0.100)(0.012) = 0.0012 𝑚𝑜𝑙𝑒𝑠

𝑛(𝑂𝐻−) = 2 × 𝑛(𝐵𝑎(𝑂𝐻)2) = 2 × 0.0012 = 0.0024 𝑚𝑜𝑙𝑒𝑠

𝑛(𝑂𝐻𝑇𝑜𝑡𝑎𝑙− ) = 0.0024 + 0.001 = 0.0034 𝑚𝑜𝑙𝑒𝑠

𝑉 = 50.0 + 100 = 150 𝑚𝐿 = 0.150 𝐿

[𝑂𝐻−] =0.0034

0.150= 0.0226667 𝑚𝑜𝑙/𝐿

𝑝𝐻 = 14 + log[𝑂𝐻−] = 12.36 (2 𝑠. 𝑓)


PH AFTER NEUTRALISATION

When a strong acid and strong base react and undergo a neutralisation reaction, the pH of the final solution will

depend on the concentration of the reactant in excess.

EXAMPLE QUESTION 1

Calculate the pH of the resultant solution when 35.0mL of 0.250M HCl and 45.0mL of 0.350M NaOH are mixed.

𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐻2𝑂(𝑙)

𝑛(𝐻𝐶𝑙) = 0.250 × 0.035 = 0.00875 𝑚𝑜𝑙

𝑛(𝑁𝑎𝑂𝐻) = 0.350 × 0.045 = 0.01575 𝑚𝑜𝑙

Therefore, HCl is the limiting reagent and NaOH will be in excess. pH > 7 & NaCl is a neutral salt.

𝑛(𝑁𝑎𝑂𝐻 𝑙𝑒𝑓𝑡 𝑜𝑣𝑒𝑟) = 0.01575 − 0.00875 = 0.007 𝑚𝑜𝑙

[𝑁𝑎𝑂𝐻] =0.007

0.08= 0.0875 𝑀 = [𝑂𝐻−]

𝑝𝐻 = 14 + log10[0.0875] = 12.942


BUFFERS

OUTCOMES

describe the importance of buffers in natural systems (ACSCH098, ACSCH102)

HOW DO BUFFERS WORK?

A buffer solution is a solution that can resist pH change when small amounts of an acid or base are added. In order

for it to work, it must be able to compensate for the addition of either of an acid or base, otherwise the pH will

change significantly.

Buffers are commonly made by mixing a weak acid and its conjugate base (or vice versa) typically in equimolar

amounts of each.

- For example, a buffer solution containing 𝐶𝐻3𝐶𝑂𝑂𝐻 and 𝐶𝐻3𝐶𝑂𝑂− (in the form of 𝑁𝑎𝐶𝐻3𝐶𝑂𝑂)

- The mixture exists in equilibrium, so all of the species in the equation are present.

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐶𝐻3𝐶𝑂𝑂(𝑎𝑞)− + 𝐻3𝑂(𝑎𝑞)

+

When a small amount of acid is added, the concentration of H3O+ increases.

- When HCl is added to the buffer solution, ↑ [H3O+] which is a disturbance

- It will shift to minimise the disturbance by shifting to the LHS and forming the products.

- Role of conjugate base is to react with excess H3O+. It ‘absorbs’ the additional H3O+

When a small amount of base is added, the situation is slightly more complicated.

- There are two acids that could react with the base 𝐶𝐻3𝐶𝑂𝑂𝐻 and 𝐻3𝑂+

- Because the concentration of 𝐻3𝑂+ is far lower than the concentration of 𝐶𝐻3𝐶𝑂𝑂𝐻, the majority of 𝑂𝐻−

will react with 𝑪𝑯𝟑𝑪𝑶𝑶𝑯, producing water and 𝑪𝑯𝟑𝑪𝑶𝑶−

- Thus the concentration of 𝐻3𝑂+does not change significantly, and the pH stays relatively constant.

It can also be explained when 𝐻3𝑂+ and 𝑂𝐻− react, causing equilibrium to shift to right to minimise the disturbance

𝐻3𝑂(𝑎𝑞)+ + 𝑂𝐻(𝑎𝑞)

− 2𝐻2𝑂(𝑙)

BUFFER CAPACITY

The effectiveness of a buffer is known as buffer capacity.

- It is defined as the moles of 𝐻3𝑂+ or 𝑂𝐻− necessary to change the pH of the buffer solution by one unit

- Buffer capacity depends on both the pH of the buffer and the total concentration of the weak acid and

conjugate base (or vice versa)

A buffer is most effective when the amounts of weak acid and conjugate base present are similar (equimolar).

- When pH = pKa, the concentrations of weak acid and conjugate base are equal.

- Therefore, a buffer solution is most effective when the pH is within 1 unit of its pKa.

- When the pH is too high, there is not enough acid to react with the added 𝑂𝐻−. When the pH is too low,

there is not enough conjugate base to react with any 𝐻3𝑂+.


Name Formulae pKa Effective pH Range

Acetic acid / Acetate 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐶𝐻3𝐶𝑂𝑂(𝑎𝑞)− + 𝐻3𝑂(𝑎𝑞)

+

4.76 3.6 − 5.6

Ammonium / Ammonia 𝑁𝐻3(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝑁𝐻4+

(𝑎𝑞)+ 𝑂𝐻(𝑎𝑞)

− 9.25 8.8 − 9.9

Carbonic acid / Hydrogen carbonate

𝐻2𝐶𝑂3(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐻𝐶𝑂3−

(𝑎𝑞) + 𝐻3𝑂(𝑎𝑞)+ 6.35 6.0 − 8.0

Hydrogen carbonate / Carbonate

𝐶𝑂3−2

(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐻𝐶𝑂3−

(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)− 10.33 9.5 − 11.1

Dihydrogen phosphate / Hydrogen phosphate

𝐻𝑃𝑂4−2

(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝐻2𝑃𝑂4−2

(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)− 7.20 5.8 − 8.0


TITRATION CALCULATIONS

OUTCOMES


DIRECT TITRATION CALCULATIONS

The steps for a titration calculation are:

1) Write a balanced chemical equation for the reaction

2) Calculate the number of moles of Reactant A (of known concentration) in the volume used

3) Using the number of moles of A and the mole ratio in the equation, calculate the number of moles of

Reactant B (unknown concentration) used.

4) Calculate the concentration of reactant B.

For neutralisation reactions, the strength of the acid is irrelevant as the base is stronger than water. All of the

protons in a polyprotic acid will be irreversibly removed by the base.

EXAMPLE QUESTION 1

Alex wanted to analyse a solution of NaOH. She first prepared 250 mL of a standard solution using 4.72 g of solid

sodium carbonate. She then reacted 25.0 mL of HCl of unknown concentration with the standard solution. The

following volumes of standard solution were required:

Trial 1 2 3 4

Volume (mL) 31.50 29.85 29.75 29.80

25.0 mL of the HCl solution was then titrated against the NaOH solution. The average volume of NaOH required to

reach the equivalence point was 30.85 mL.

𝑁𝑎2𝐶𝑂3(𝑎𝑞) + 2𝐻𝐶𝑙(𝑎𝑞) → 2𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐶𝑂2(𝑔) + 𝐻2𝑂(𝑙)

𝑛(𝑁𝑎2𝐶𝑂3) =4.72

105.99= 0.0453250307 𝑚𝑜𝑙

[𝑁𝑎2𝐶𝑂3] =𝑛(𝑁𝑎2𝐶𝑂3)

0.250=

0.0453250307

0.250= 0.1781300123 𝑀

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑁𝑎2𝐶𝑂3 𝑢𝑠𝑒𝑑 = 29.8 𝑚𝐿


𝑛(𝑁𝑎2𝐶𝑂3) = 0.1781300123 × 0.0298 = 0.00530827436 𝑚𝑜𝑙

𝑛(𝐻𝐶𝑙) = 2 × 𝑛(𝑁𝑎2𝐶𝑂3) = 0.01061654873 𝑚𝑜𝑙

𝑛(𝑁𝑎𝑂𝐻) = 𝑛(𝐻𝐶𝑙) = 0.01061654873 𝑚𝑜𝑙

[𝑁𝑎𝑂𝐻] =𝑛(𝑁𝑎𝑂𝐻)

0.03085=

0.01061654873

0.03085= 0.344 𝑀


DILUTION – TITRATION CALCULATIONS

When analysing a substance, the concentration may be too high for a direct titration experiment tom be efficiently

carried out. Instead, the substance would be diluted by a known amount, and then the diluted solution would be

titrated.

WORKED EXAMPLE 1

21.55 mL of a hair dye product containing ammonia was diluted to 250.0 mL. 20.0 mL aliquots of the diluted dye

were titrated against 0.048 M sulfuric acid. The average titre was 18.35 mL. The density of the hair product was

measured to be 1.1 g/mL.

Calculate the percentage by weight of ammonia in the hair dye.

2𝑁𝐻3(𝑎𝑞) + 𝐻2𝑆𝑂4(𝑎𝑞) → (𝑁𝐻4)2𝑆𝑂4(𝑎𝑞)

𝑛(𝐻2𝑆𝑂4) = 0.048 × 0.01835 = 0.0008808 𝑚𝑜𝑙

𝑛(𝑁𝐻3) = 2 × 𝑛(𝐻2𝑆𝑂4) = 0.0017616 𝑚𝑜𝑙

[𝑁𝐻3] =𝑛(𝑁𝐻3)

0.020=

0.0017616

0.020= 0.08808 𝑀

𝑛(𝑁𝐻3) = 0.08808 × 0.250 = 0.02202 𝑚𝑜𝑙

𝑚(𝑁𝐻3) = 0.02202 × 17.034 = 0.37508868 𝑔

𝑀𝑎𝑠𝑠 𝑜𝑓 ℎ𝑎𝑖𝑟 𝑑𝑦𝑒 = 21.55 × 1.1 = 23.705𝑔

%𝑤

𝑤=

𝑚(𝑁𝐻3)

23.705=

0.37508868

23.705= 1.6%

BACK TITRATION CALCULATIONS

A back titration, or indirect titration, is a two-stage analysis:

- Reactant A (of unknown concentration) is reacted with an excess of Reactant B (of known concentration and

volume).

- A titration is the performed on the excess Reactant B by determining the moles of Reactant C required to

neutralise the excess.

Summary:

1) Sample is reacted with known excess of reagent. (e.g. known amount of a particular acid)

2) Leftover excess is added

3) Excess is titrated to find moles of reagent reacted with solution.

Back titrations are generally used when:

- One of the reactants is volatile (e.g. ammonia)

- An acid or base is an insoluble salt (e.g. calcium carbonate)

- Direct titration would involve weak acid/weak base titration (making it difficult to determine the

equivalence point).


EXAMPLE QUESTION 1

A student wanted to determine the calcium carbonate present in a 0.250 g sample of chalk. The student placed the

chalk sample in a 250 mL conical flask and pipetted 50.00 mL of 0.100 M sulfuric acid into the flask. The excess acid

was titrated with 0.340 M sodium hydroxide. The results for the volume of sodium hydroxide required are tabulated

below:

Trial 1 2 3 4

Volume NaOH (mL) 23.40 21.95 22.10 21.90

Calculate the mass percentage of calcium carbonate in the chalk sample.

𝐻2𝑆𝑂4(𝑎𝑞) + 𝐶𝑎𝐶𝑂3(𝑎𝑞) → 𝐶𝑎𝑆𝑂4(𝑎𝑞) + 𝐶𝑂2(𝑔) + 𝐻2𝑂(𝑙)

𝐻2𝑆𝑂4(𝑎𝑞) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎2𝑆𝑂4(𝑎𝑞) + 2𝐻2𝑂(𝑙)

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣(𝑁𝑎𝑂𝐻) = 21.9833 𝑚𝐿

𝑛(𝑁𝑎𝑂𝐻) = 0.0219833 × 0.340 = 0.00747433 𝑚𝑜𝑙

𝑛(𝐻2𝑆𝑂4 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑤𝑖𝑡ℎ 𝑁𝑎𝑂𝐻) = 0.5 × 𝑛(𝑁𝑎𝑂𝐻) = 0.0037371666 𝑚𝑜𝑙

𝑛(𝐻2𝑆𝑂4 𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 0.050 × 0.100 = 0.005 𝑚𝑜𝑙

𝑛(𝐻2𝑆𝑂4 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑤𝑖𝑡ℎ 𝐶𝑎𝐶𝑂3) = 0.005 − 0.003737166 = 0.0012628333 𝑚𝑜𝑙

𝑛(𝐶𝑎𝐶𝑂3) = 𝑛(𝐻2𝑆𝑂4 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑤𝑖𝑡ℎ 𝑁𝑎𝑂𝐻) = 0.0037371666 𝑚𝑜𝑙

𝑚(𝐶𝑎𝐶𝑂3) = 100.94 × 0.0037371666 = 0.1263969883 𝑔

%𝑤

𝑤=

0.1263969883

0.250= 50.558795% = 50.6% (3 𝑠. 𝑓)


EXAMPLE QUESTION 2

A student was asked to determine the concentration of ammonia in a commercial sample of “cloudy ammonia

cleaning solution”. The student first pipetted 25.00 mL into a 250 mL volumetric flask and filled it to the mark with

distilled water. She then transferred a 25.00 mL aliquot to a clean conical flask, and added exactly 40.00 mL of 0.100

M HCl. She then titrated the excess HCl with 0.050 M sodium carbonate. 21.30 mL of sodium carbonate solution was

required.

2𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝑎2𝐶𝑂3(𝑎𝑞) → 2𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐶𝑂2(𝑎𝑞) + 𝐻2𝑂(𝑙)

𝑛(𝑁𝑎2𝐶𝑂3) = 0.01230 × 0.050 = 0.0010615 𝑚𝑜𝑙

𝑛(𝐻𝐶𝑙 𝑟𝑒𝑎𝑐𝑡𝑒𝑑) = 2 × 𝑛(𝑁𝑎2𝐶𝑂3) = 0.00213 𝑚𝑜𝑙

𝑛(𝐻𝐶𝑙 𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 0.040 × 0.100 = 0.004 𝑚𝑜𝑙

𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) → 𝑁𝐻4𝐶𝑙(𝑎𝑞)

𝑛(𝐻𝐶𝑙 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑤𝑖𝑡ℎ 𝑁𝐻3) = 0.004 − 0.00213 = 0.00187 𝑚𝑜𝑙

𝑛(𝑁𝐻3) = 𝑛(𝐻𝐶𝑙) = 0.00187 𝑚𝑜𝑙

[𝑁𝐻3] =0.00187

0.025= 0.0748 𝑀

0.025 × 𝐶1 = 0.250 × 0.0748

𝐶1 = 0.748 𝑀 = 0.75 𝑀 (2 𝑠. 𝑓)


TITRATION

OUTCOMES


TITRATION TERMINOLOGY

Term Definition

Titration Volumetric analytical technique used for determining the concentration of a solution, when the concentration of the other reacting solution is known and volumes of the 2 solutions are accurately measured.

Aliquot An accurately known volume of liquid. In titration this usually refers to the liquid transferred via pipette into the conical flask.

Analyte A substance being analysed. This is usually a solution of unknown concentration.

Back Titration A two-stage analysis in which an excess of reactant is added to the analyte, then the excess is determined to calculate the concentration of the analyte.

Burette A graduated piece of glassware which dispenses measured amounts of the solution (the titrant).

Endpoint The point in titration when the indicator permanently changes colour.

Equimolar Same concentration.

Equivalence Point The point in a neutralisation reaction when the amounts of reactants are just sufficient to consume both reactants, without an excess of either. (i.e. in stoichiometric ratio)

Pipette A piece of glassware used to transfer a very accurately measured volume of solution.

Primary standard A substance of sufficiently high purity and stability that a solution of accurately known concentration can be prepared by weighing out the desired mass, dissolving in water and making the volume up to a known value.

Titrand The solution to which another reagent (titrant) is added during titration (usually in the conical flask)

Titrant The solution that is added during a titration (usually from a burette)

Titre The volume of titrant used in a titration

Volumetric flask A piece of glassware which can hold a set volume of solution very accurately

TYPES OF ANALYSIS

- Qualitative analysis: Involves observations only

- Quantitative analysis: Involves measurements (mass, volume etc.)

o Volumetric analysis: involves measurements of volume

o Gravimetric analysis: involves measurements of mass/weight


GENERAL TITRATION PROCEDURE

In a titration experiment, the number of moles of a target material (the analyte) is determined. This can then be

used to calculate the concentration.

A measured volume of the solution of unknown concentration, the analyte, is usually placed in a conical flask

(titrand) with a burette containing the titrant above it.

A burette is a piece of volumetric glassware. It is a long tube with a tap at one end so measured volumes of titrant

can be accurately added to the titrand.

Before the experiment begins, an indicator will normally also be added to the conical flask to determine the

approximate equivalence point.

To perform the experiment the titrant is slowly added and stepwise to the conical flask, with swirling, until the

indicator undergoes a permanent colour change (the end point).

Acid-base titrations are the most common titrations:

- An acid and a base are reacted in a neutralisation reaction during the titration.

- A suitable acid-base indicator is added to show when the reaction is just complete. However, indicators

change colour over a range of pH, making it difficult to accurately determine the equivalence point.

- For greater accuracy, a pH meter can be used.

The pH changes rapidly towards the end point of the titration and it is easy to add too much titrant. The titrant must

be added very carefully, in small volumes close to the end to successfully determine the exact amount required for

complete reaction.


DETAILED TITRATION PROCEDURE

For titrating HCl against NaOH. HCl is placed into the conical flask and NaOH in the burette.

PREPARING THE TITRAND (IN THE CONICAL FLASK)

- Rinse the inside of the pipette with a small with a small amount of HCl solution 3 times.

- Use a pipette filler to fill the pipette with HCl until the bottom of the meniscus rests on the calibration line.

- Hold the pipette so that its tip is resting against the inside of a clean conical flask. Let the solution run out.

- Once the liquid level has stabilised, leave the pipette tip touching the flask for a few seconds before

removing. (The pipette is calibrated to deliver the correct volume when a small amount of liquid remains in

the tip – do not shake it into the flask).

- Use a wash bottle containing distilled water to wash any solution that might be on the inside wall to the

bottom of the conical flask.

PREPARING THE TITRANT (IN THE BURETTE)

- Rinse the inside of the burette with a small amount of NaOH solution 3 times, including through the tap.

- Clamp the burette vertically

- Pour NaOH solution into the burette

TITRATION

- Add a few drops of indicator into the conical flask and swirl gently.

- Record the initial burette reading

- Place the conical flask under the burette

- Add NaOH to the conical flask until a permanent colour change occurs

- Record the final burette reading

- Repeat the experiment 3 times with fresh aliquots of HCl.

WASHING

Burette and pipette: Used to deliver the solutions used in the titration. Final rinsing with the solution to be

delivered.

Conical flask: Used to hold the aliquot or titrand. Final rinsing with distilled water.

The washing procedure affects the accuracy of the calculated concentration.


TITRATION TECHNIQUE

RINSING

It is important to rinse each piece of glassware with the appropriate solution after cleaning with distilled water and

immediately prior to use.

The solution that is to be transferred using a pipette is of accurately known concentration, or its concentration is to

be accurately determined. If droplets of distilled water are present in the pipette, it will dilute the reagent being

delivered.

- Rinse burette and pipette with solutions

- Rinse conical flask and volumetric flask with distilled water

VOLUMETRIC ERRORS

All glassware in titration is calibrated to be accurate when measurements are taken at the bottom of the meniscus.

Any air bubbles in the liquid must be removed for volumes to be accurate.

PIPETTE CALBIRATION

The pipettes used in titration are calibrated to deliver the specified volume of solution with no additional force.

- They are marked TD (to deliver) or EX (to expel)

- This means that there should be a small volume of liquid left in the tip of the pipette after the aliquot has

been accurately transferred. This should not be shaken out into the conical flask.

DEVIATION

Accessing accuracy – How close you are to the accepted value

% 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = |𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒−𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒|

𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒


STANDARD SOLUTION

A standard solution is a solution containing a precisely known concentration of a substance. They can be categorised

as primary or secondary.

A primary standard is produced when a substance of high purity dissolved in a known volume of solvent.

1) Accurately weigh out a mass of solid close to the required mass in a beaker. Record the actual mass

weighed.

2) Add enough distilled water to dissolve the solid.

3) Carefully transfer all the weighed mass to a clean volumetric flask of the approximate size, using a wash

bottle and funnel. All the equipment that came into contact with weighed mass should be rinsed into the

flask.

4) Add distilled water until the bottom of the meniscus is resting on the line on the neck of the flask. Add the

last few drops with a dropper.

5) Stopper the flask. Firmly holding the stopper in place, invert several times to ensure the solution is

homogeneous.

6) Label the flask with the exact concentration, solution, date and name.

A substance suitable for preparing a primary standard solution should have the following features:

- High purity

- Unaffected by exposure to air

- Non-hygroscopic (does not absorb water from air)

- Have a large molecular mass to reduce percentage errors

- Be a solid for easier weighing

- Cheap and readily available

- Have a high water solubility

A secondary standard is produced when its concentration is determined via stoichiometry.

- The process of producing a secondary standard is called standardisation

EQUIVALENCE POINT

The equivalence point of a titration is the point at which the amount moles of acid and bases added match the

stoichiometric ratio.

- It is the point at which reaction is complete, with no excess reactant

- The pH of the solution at the equivalence point determines the appropriate indicator to be used.

PH OF THE EQUIVALENCE POINT

As the pH of water is neutral, the pH of the equivalence point will depend entirely on the salt produced: whether it is

acidic, basic or neutral.

If an acidic or basic salt is produced by the neutralisation reaction in a titration experiment, the equivalence point

will not be neutral.


- Neutral salts are formed when strong acids react with strong bases.

- Acidic salts are formed when strong acids react with weak bases.

- Basic salts are formed when weak acids react with strong bases.

Note: When strong acids react with metal carbonates (weak bases), the neutral salt is formed, but the resulting

solution is still acidic. This is because the carbon dioxide dissolves in water to produce an acidic solution.

INDICATOR SELECTION

Indicators can be used to find the approximate equivalence point of a titration.

- The equivalence point is when exactly enough moles of titrant have been added to react with all the

titrand.

- The end point is when the indicator first undergoes a permanent colour change.

- An indicator should be selected so that the end point is as close as possible to the equivalence point.

(Systematic error. Will impact validity and accuracy)

Indicator Colour at lower pH Colour at higher pH pH of colour change range

Methyl orange Red Yellow 3.1 – 4.4

Bromothymol blue Yellow Blue 6.0 – 7.6

Phenolphthalein Colourless Pink 8.3 – 10.0

Methyl red Red Yellow 4.8 – 6.0

Litmus Red Blue 4.5 – 8.3

Phenol red Yellow Red 6.8 – 8.4

BROMOTHYMOL BLUE


PHENOLPHTHALEIN

METHYL ORANGE

EXAMPLE QUESTION 1

Explain why 0.20 M acetic acid and 0.20 M hydrochloric acid require the same volume of sodium hydroxide solution

to reach equivalence point, but the pH values at their equivalence points are different.

𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎(𝑎𝑞) + 𝐻2𝑂(𝑙)


𝐶𝐻3𝐶𝑂𝑂(𝑎𝑞)− + 𝐻2𝑂(𝑙) 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑂𝐻(𝑎𝑞)

−

Both acetic acid and hydrochloric acid are monoprotic acids that reacts to completion when reacted with a strong

base. As the concentration are the same, the same amount is needed to neutralise the strong base, NaOH.

𝐶𝐻3𝐶𝑂𝑂− is the conjugate base of a weak acid and will react with water to produce a basic solution. Therefore,

resulting in a pH > 7. 𝐶𝑙− is the conjugate base of a strong acid and will not react hence the solution remains pH = 7.


APPLICATION OF TITRATION

STRONG ACID-STRONG BASE TITRATION

The equivalence point is located at the most vertical point (point of inflection). All three common indicators for

titration are suitable for determining the equivalence point for a strong acid-strong base. This is because there is a

large rapid change in pH near the equivalence point so all of the indicator would change colour when the same

volume of based is added, therefore it is not critical which indicator is used.

WEAK ACID-STRONG BASE TITRATION


STRONG ACID-WEAK BASE TITRATION

WEAK ACID-WEAK BASE TITRATION

Never use indicators for weak acid and weak base.


OUTCOME

investigate titration curves and conductivity graphs to analyse data to indicate characteristic reaction profiles,

for example:

– strong acid/strong base

– strong acid/weak base

– weak acid/strong base (ACSCH080, ACSCH102)

CONDUCTIVITY GRAPHS

During a titration, the conductivity of the solution changes. The equivalence point may be located by plotting the

conductance as a function of the volume of titrant added.

The electrical conductivity of a solution depends on:

- The concentration ions present

- The mobility of the ions present

o More mobile ions, the more conductive it is

o H+ and OH- are highly mobile

- H+ ions are more conductive than OH- ions

Conductometric titrations are useful for titrations of coloured solutions, analysis of dilute solutions, and when

reversible reactions are used (e.g. weak acid-weak base titration).

General rule:

- Strong → Linear

- Weak → Curved

STRONG ACID + STRONG BASE


WEAK ACID + STRONG BASE

1) Conductivity is initially low as acetic acid is a weak acid and only partially ionises in water.

2) [Not shown on the graph] As the base is initially added, conductivity decreases

o H+ is replaced by Na+ which is less conductive

o The presence of the newly formed acetate ions also decreases the ionisation of acetic acid due to

the common ion effect.

o 𝐶𝐻3𝐶𝑂𝑂𝐻(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎(𝑎𝑞) 𝐶𝐻3𝐶𝑂𝑂(𝑎𝑞)− + 𝑁𝑎(𝑎𝑞)

+

The initial production of the CH3COO- ions results in the suppression of the CH3COOH

ionisation due to the common ion effect.

3) Conductivity then increases are more Na+ and CH3COO- are produced.

o There is a minimal change in pH due to the buffer region. The mixture of CH3COO-/CH3COOH

4) Reaches the equivalence point

5) Conductivity increases more rapidly as Na+ and highly conducting OH- are added. (Excess strong base)

Initially, the conductance is low due to the low ionisation of the weak acid. On the addition of the strong base, there

is a decrease in conductance due to the replacement of the H+ by Na+ but also supresses the dissociation of the

acetic acid due to the common ion acetate.

The conductance increases on adding NaOH as it neutralises the undissociated CH3COOH to NaCH3COO which is a

strong electrolyte. Conductivity increases due to the highly conductive OH- ions.


STRONG ACID + WEAK BASE

𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝐻3(𝑎𝑞) → 𝑁𝐻4𝐶𝑙(𝑎𝑞)

𝑁𝐻4+

(𝑎𝑞) + 𝐻2𝑂(𝑙) 𝑁𝐻3(𝑎𝑞) + 𝐻3𝑂(𝑎𝑞)+

Before the equivalence point, conductivity decreases like in the strong acid-strong base graph.

After the equivalence point, the graph is almost horizontal as the excess weak base is not significantly ionised due to

the presence of its conjugate acid.

Initially, the conductance is high due to the strong acid. The conductance decreases due to the replacement of H+.

After the equivalence point has been reached in the graph becomes almost horizontal, since the excess weak base

(aqueous ammonia) is not easily ionised in the presence of the salt.


WEAK ACID + WEAK BASE

[Not shown on the graph]

1) CH3COOH is a weak acid and therefore only partially ionise.

2) [H+] gets used up. CH3COO- gets produced which supresses the ionisation of CH3COOH.

3) Production of more ions

4) Equivalence point

5) Excess NH3 is suppressed due to the common ion effect


OUTCOMES

conduct a chemical analysis of a common household substance for its acidity or basicity (ACSCH080) , for

example:

– soft drink

– wine

– juice

– medicine

ANALYSIS OF COMMON HOUSEHOLD SUBSTANCES

Substance Major Acidic or Basic Species

Soft Drink Carbonic acid, phosphoric acid

Wine Tartaric acid

Orange juice Citric acid

Aspirin Acetylsalicylic acid

Hair dye Ammonia

Vinegar Acetic acid

Antacids Metal hydroxides and carbonates

Chalk Calcium carbonate

The exact procedures will vary depending on the nature of the substance to be analysed:

- Indicators may not be suitable for intensely coloured substances

- Insoluble solids may need to be dissolved before titration

- Many substances will require dilution before titration.

OUTCOMES

explore acid/base analysis techniques that are applied:

– in industries

– by Aboriginal and Torres Strait Islander Peoples

USE IN THE WIDER WORLD

Industry Aboriginal and Torres Strait Islander

Analysis of contents of wine. (Ethanoic acid) Big face plants (carpobrotus)

Determining the strength of antacids

Analysis of juice/fruits (Vitamin C)

chemistry module 5€¦ · chemistry module 5: equilibrium and acid reactions page 3 of 29 revision...

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