Topic 18.5: Indicators

18.5.1 Describe qualitatively the action of an acid-base indicator

18.5.2 State and explain how the pH range of an acid-base indicator relates to its pKa value.

18.5.3 Identify an appropriate indicator for a titration, given the equivalence point of the titration and the pH range of the indicator

Topic 18.4: Acid-base titrations

18.4.1 Sketch the general shapes of graphs of pH against volume for titrations involving strong and weak acids and bases, and explain their important features.

Topic 18.3: Salt hydrolysis

18.3.1 Deduce whether salts form acidic, alkaline or neutral aqueous solutions

Topic 18.2: Buffer solutions

18.2.1 Describe the composition of a buffer solution and explain its action

A buffer refers to something that acts to reduce the impact of one thing on another - little bit like a shock absorber. For example, buffer in the computer world are areas shared by hardware devices that operate at different speeds. In acid-base chemistry, a buffer acts to reduce the impact on pH of adding acid or base to a chemical system.

18.2.2 Solve problems involving the composition and pH of a specified buffer system

Factors that can influence buffers are Dilution and Temperature.

Dilution, Ka and Kb as equilibrium constants are not changed by dilution. Nor is the ratio of acid to salt concentration, as both components will be decreased by the same amount. Therefore, diluting a buffer does not change its pH.

Nonetheless, diluting a buffer does alter the amount of acid or base it can absorb without significant changes in pH - the so-called buffering capacity. This depends on the molar concentration of its components, so decrease as they are lowered by dilution.

Temperature affects the values of Ka and Kb. it accordingly affects the pH of the buffer. This is why a constant temperature should be maintained in all work involving buffers such as calibration of pH meters. Temperature fluctuations must also be minimized in many medical procedures such as blood transfusions, due to the effect on the buffers in the blood.

Topic 18.1: Calculations involving acids and bases

18.1.1 State the expression for the ionic product constant of water (Kw).

18.1.2 Deduce [H+(aq)] and [OH-(aq)] for water at different temperatures given Kw values

Since we know that

As the temperature increases, the equilibrium has shifted to the right. When it is shifted to the right this means the concentration of Hydrogen ions has increased thus lowering the pH. Thus the solution has a lower pH at higher temperature.


18.1.3 Solve problems involving [H+(aq)], [OH-(aq)], pH and pOH

18.1.4 State the equation for the reaction of any weak acid or weak base with water, and hence deduce the expressions of Ka and Kb

18.1.5 Solve problems involving solutions of weak acids and bases using the expression:

Ka x Kb = Kw
pKa + pKb = pKw
pH + pOH = pKw

18.1.6 Identify the relative strengths of acids and bases using values of Ka, Kb, pKa and pKb

Topic 18: Acids and bases

Topic 18 of the IB HL Chemistry syllabus is the Acids and bases. IBO recommends to spend 10 hours on this topic.

This topic has 5 sub-chapters: "Calculations involving acids and bases", "Buffer solutions", "Salt hydrolysis", "Acid-base titrations" and "Indicators". Each are separated with numerical values in order of mentioned.

These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.

Topic 8.4: The pH scale

8.4.1 Distinguish between aqueous solutions that are acidic, neutral or alkaline using the pH scale

The pH scale is created through this equation

On the pH scale, 7 is neutral, Anything below 7 is acidic and anything above 7 is alkaline.


8.4.2 Identify which of two or more aqueous solutions is more acidic or alkaline using pH values

To identify which of two solution is more acidic or alkaline, we have to understand that the lower the pH value, the more acidic it is. The high the pH value, the more alkaline it is.

8.4.3 State that each change of one pH unit represents a 10-fold change in the hydrogen ion concentration [H+(aq)]

A change of one pH unit represents a 10-fold change in the concentration of hydrogen ions.

This means that increasing the pH by one unit represents a decrease in the concentration of hydrogen by 10 times; decreasing by one pH unit represents an increase in the concentration of hydrogen by 10 times.

8.4.4 Deduce changes in [H+(aq)] when the pH of a solution changes by more than one pH unit

If the pH of a solution is changed from 3 to 5, deduce how the hydrogen ion concentration changes.

pH = 3, so the concentration of hydrogen is 10^-3 mol dm^-3

pH = 5, so the concentration of hydrogen is 10^-5 mol dm^-5

Therefore the concentration of hydrogen has decreased by 100.

Topic 8.3: Strong and weak acids and bases

8.3.1 Distinguish between strong and weak acids and bases in terms of the extent of dissociation, reaction with water and electrical conductivity

Acids produce H+ ions and bases produce OH- ions in aqueous solution.

Strong acid and bases have complete dissociation. Weak acid and bases have partial dissociation, often forms an equilibrium between the OH- ions and the base itself.

The reaction with water causes the final result to change the pH value. This is due to release of Hydrogen ion and hydroxide ion from the acid and base respectively.

Electricity conductivity of a solution depends on the concentration of mobile ions. Strong acids and strong bases will therefore show higher conductivity than weak acid and bases. This can be measured using a conductivity probe.

8.3.2 State whether a given acid or base is strong or weak

Strong acids include Hydrochloric acid, Nitric acid and Sulphuric acid

Strong bases include Lithium hydroxide, Sodium hydroxide, Potassium hydroxide and Barium hydroxide

Weak acids include Ethanoic acid, Carbonic acid and Phosphoric acid

Weak bases include Ammonia and Ethylamine

8.3.3 Distinguish between strong and weak acids and bases, and determine the relative strengths of acids and bases, using experimental data

Electrical conductivity of a solution as mentioned above depends on the availability of hydrogen ions. This can be measured using a conductivity meter.

The reactions of acids described depends on the concentration of hydrogen ions. They will therefore happen at a faster rate with stronger acids. This may be an important consideration, for example, regarding safety in the labouratory, but usually does not provide an easy means of quantifying data to distinguish between weak and strong acids.

As we will learn in the next section, the pH scale is a measure of the hydrogen ions concentration and so can be used directly to compare the strengths of acids. It is a scale in which the higher the hydrogen ion concentration, the lower the pH value.

Topic 8.2: Properties of acids and bases

8.2.1 Outline the characteristic properties of acids and bases in aqueous solution

We will look here at some typical reactions of acids and bases in aqueous solutions where H+ is the ion common to all acids. The bases considered here are those that neutralize acids to produce water and these include metal oxides and hydroxides, ammonia, soluble carbonates and hydrogencarbonates.

Acids react with metal, bases and carbonates to form salts.

Acid + Metal -> Salt + Hydrogen

Acid + Base -> Salt + Water

Acid + Carbonate -> Salt + Water + Carbon Dioxide

Topic 8.1: Theories of acids and bases

8.1.1 Define acids and bases according to the Bronsted-Lowry and Lewis theories

A Bronsted-Lowry acid is a proton donor

A Bronsted-Lowry base is a proton acceptor

A Lewis acid is an electron pair acceptor

A Lewis base is an electron pair donor.

8.1.2 Deduce whether or not a species could act as a Bronsted-Lowry and/or a Lewis acid or base

All Bronsted-Lowry acids are Lewis acids

Not all Lewis acids are Bronsted-Lowry acids

The term Lewis acid is usually reserved for those species which can only be described by Lewis theory, that is those that do not release H+.

8.1.3 Deduce the formula of the conjugate acid (or base) of any Bronsted-Lowry base (or acid)

Topic 8: Acids and bases

Topic 8 of the IB HL Chemistry syllabus is the Acids and bases. IBO recommends to spend 6 hours on this topic.

This topic has 4 sub-chapters: "Theories of acids and bases", "Properties of acids and bases", "Strong and weak acids and bases" and "The pH scale". Each are separated with numerical values in order of mentioned.

These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.

Topic 17.2: The equilibrium law

17.2.1 Solve homogeneous equilibrium problems using the expression for Kc.

There are three different types of equilibrium questions the IBO can ask.


Calculating the equilibrium constant from initial and equilibrium concentrations.

If we know the equilibrium concentations of all reactants and products in a reaction, we can simply substitute these into the equilibrium expression to calculate Kc. The first step in such a calculation is always to write the equilibrium expression from the chemical equation.

Once you have your equation, just plug in the numbers of the equilibrium concentrations and you will receive Kc. If you were given only the initial concentration, we just need to convert it back to the equilibrium concentration for it to work.


Calculating equilibrium concentration from the equilibrium constant

If we know the value of Kc and the equilibrium concentrations of all but one of the components, we can calculate the remaining equilibrium concentration simply by substituting the values into the equilibrium expression.

Finding out what x is equal to would show the equilibrium concentrations. To find x, simply equate the Kc value with the equation containing the x. Simple maths and some rearrangement will provide the answer for x.

Topic 17.1: Liquid-vapour equilibrium

17.1.1 Describe the equilibrium established between a liquid and its own vapour and how it is affected by temperature changes

If we place some liquid in a container in which none of its molecules is in the vapour phase, evaporation will occur as vapour particles escape from the surface. In an open system, the vapour particles will leave so the liquid will continue to evaporate and equilibrium will not be reached. In this situation, the liquid will eventually all be converted into vapour - this is what happens, for example, when a puddle of water dries up or clothes are hung up to dry outside.

Temperature

The kinetic theory of matter gives us a model to describe the behaviour of liquids and gases. As temperature increases, there will be more vapour particles due to the increase rate of evaporation. Thus vapour pressure increases,



17.1.2 Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms of kinetic theory

17.1.3 State and explain the relationship between enthalpy of vapourization, boiling point and intermolecular forces

Stronger intermolecular forces
Higher enthalpy of vapourization
Lower vapour pressure
Higher boiling point

Weaker intermolecular forcese
Lower enthalpy of vapourization
Higher vapour pressure
Lower boiling point

The vapour pressure of a substance depends on its temperature and on the strength of the intermolecular forces present. The higher the temperature and the weaker the intermolecular forces, the higher the vapour pressure.

Topic 17: Equilibrium

Topic 17 of the IB HL Chemistry syllabus is the Equilibrium. IBO recommends to spend 4 hours on this topic.

This topic has 2 sub-chapters: "Liquid-vapour equilibrium" and "The equilibrium law". Each are separated with numerical values in order of mentioned.

These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.

Topic 7.2: The position of the equilibrium

7.2.1 Deduce the equilibrium constant expression (Kc) from the equation for a homogeneous reaction

7.2.2 Deduce the extent of a reaction from the magnitude of the equilibrium constant

Different reactions have different values of Kc.

As the equilibrium constant expression puts products on the numerator and reactants on the denominator, a high value of Kc will mean that at equilibrium there are proportionately more products than reactants. In other words, such an equilibrium mixture lies to the right and the reaction goes almost to completion. By contrast a low value of Kc must mean that there are proportionately less products with respect to reactants, so the equilibrium mixture lies to the left and the reaction has barely taken place.

7.2.3 Apply Le Chatelier's principle to predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and on the value of equilibrium constant

The Le Chatelier's principle states that a system at equilibrium when subjected to a change will respond in such a way as to minimize the effect of the change. Simply put, this means that whatever we do to a system at equilibrium, the system will respond in the opposite way. Add something and the system will adjust accordingly.

Changes in temperature

Kc is temperature dependent, so changing the temperature will change Kc. However in order to predict how it will change the equilibrium, we must examine the enthalpy changes of the forward and backward reactions.

If this reaction at equilibrium is subjected to a decrease in temperature, the system will respond by producing heat and favouring the exothermic reaction. This means when temperature decrease, the exothermic reaction is favoured and vice versa.

Changes in pressure

Equilibira involving gases will be affected by a change in pressure if the reaction involves a change in the number of molecules. This is because there is a direct relationship between the number of molecules and the pressure exerted by a gas in a fixed volume.

If a reaction at equilibrium is subject to an increase in pressure, the system responds to decrease this pressure by favouring the side with the smaller number of molecules. Conversely, a decrease in pressure will cause a shift in the equilibrium position to the side with the larger number of molecules. Kc will remain constant as long as temperature has not changed.

Changes in concentration

Suppose an equilibrium is disrupted by an increase in the concentration of one of the reactants. This will cause the rate of the forward reaction to increase, while the backward reaction will not be affected, so the reaction rates will no longer be equal. The value of Kc will be unchanged.

7.2.4 State and explain the effect of a catalyst on an equilibrium reaction

A catalyst speeds up the rate of a reaction by lowering its activational energy (Ea) and so increase the number particles that have sufficient energy to react without raising the temperature.

In the equilibrium, both the rate of forward reaction and back reaction increases equally from the catalyst. Thus it has overall no net effect on the reaction.


7.2.5 Apply the concepts of kinetics and equilibrium to industrial processes

A common industrial process used in the IB exams include the production of ammonia.

Haber Process:

Concentration -  the reactants nitrogen and hydrogen are supplied in the molar ratio 1:3 in accordance with their stoichiometry in the equation. The product ammonia is removed as it forms, thus helping to pull the equilibrium to the right and increasing the yield.

Pressure - As the forward reaction involves a decrease in the number of molecules, it will be favoured by high pressure. However, anything above 200 atmospheric pressure will not be economically efficient.

Temperature - As the forward reaction is exothermic, it will be favoured by a lower temperature. However, too low a temperature would cause the reaction to by economically slow and so a moderate temperature of 450 degrees Celcius is used.

Catalyst -  Thought the catalyst will not increase the yield of ammonia, it will speed up the rate of production and so help to compensate for the moderate temperature used. A catalyst of finely divided iron is used, with small amounts of aluminium and magnesium oxides added to improve its activities.

Topic 7.1: Dynamic equilibrium

7.1.1 Outline the characteristics of chemical and physical systems in a state of equilibrium

Physical systems

As bromine is a volatile liquid, with a boiling point close to room temperature, a significant number of particles will have enough energy to escape from the liquid state and form vapour in the process known as evaporation. At the same time, some of these vapour molecules will collide witht he surface of the liquid, lose energy and become liquid in the process known as condensation.

There will however come a time when the rate of evaporation is equal to the rate of condensation and at this point there will be no net change in the amounts of liquid and gas present. We say that the system has reached equilibrium. This will only occur in a closed system, where the gas can't escape.

Chemical systems

Consider the reaction of dissociation between hydrogen iodide HI and its elements hydrogen and iodine.

If were to analyse the contents of the flask at this point, we could find HI, hydrogen and iodine would all be present and that if there were no change in conditions, their concentration would remain constant over time.


Characteristics of equilibrium state

Equilibrium is dynamic - The reaction has not stopped but both forward and backward reactions are occuring at the the same rate

Equilibrium is achieved in a closed system - A closed system prevents exchange of matter with the surroundings, so equilibrium is achieved where both reactants and products can react and recombine with each other.

The concentration of reactants and products remain constant at equilibrium - They are being produced and destroyed at an equal rate.

At equilibrium there is no change in marcoscopic properties - This refers to observable properties such as colour and density. These do not change as they depend on the concentrations of the components of the mixture

Equilibrium can be reached from either direction - The same equilibrium mixture will result under the same conditions, no matter whether the reaction is started with all reactants, all products or a mixture of both.

Topic 7: Equilibrium

Topic 7 of the IB HL Chemistry syllabus is the Equilibrium. IBO recommends to spend 5 hours on this topic.

This topic has 2 sub-chapters: "Dynamic Equilibrium" and "The position of the equilibrium". Each are separated with numerical values in order of mentioned.

These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.

Option D10: Mind-altering drugs

D.10.1 Describe the effects of lysergic acid diethylamide (LSD), mescaline, psilocybin and tetrahydrocannabinol

LSD

• Potent hallucinogen that creates distortions of the body and crawling geometric patterns
• Causes impaired judgement, hypertension, dilated pupils and changes to body temperature and heart rate
• Can cause unpredictable mood swings from euphoria to depression and panic

Mescaline

• About 1000-3000 times less potent than LSD; causes subjective hallucinations dependent on the individual
• Effects include anxiety, static tremors, and psychic disturbances with vivid hallucinations
• Abdominal pain, nausea and diarrhoea are also common.

Psilocybin

• Produces subjective hallucinations similar to mescaline but milder
• Some people experience a pleasant mood, others become apprehensive
• Compulsive movement and inappropriate laughter may occur; also vertigo and dizziness
• Numbness, muscle weakness and drowsiness are also common

Tetrahydrocannabinol (THC)

• It acts to depress the central nervous system and cause mental relaxation and euphoria
• Can lead to loss of inhibitions and an alteration of the perception of time and space
• Gives a loss of concentration, light-headedness, weakness and a sense of floating are commonly experienced
• Can cause depression of respiration and can lead to collapse 

D.10.2 Discuss the structural similarities and differences between LSD, mescaline and psilocybin

LSD

• Benzene ring
• Alkene
• Secondary amine
• Tertiary amine
• Tertiary amide

Mescaline

• Benzene ring
• Primary amine
• Ether

Psilocybin

• Benzene ring
• Secondary amine
• Tertiary amine ion
• Phosphate

D.10.3 Discuss the arguments for and against the legalization of cannabis