# Gather and process information from secondary sources to trace developments in understanding and describing acid/base reactions

Although covered in detail in later dotpoints, this question appears often enough to warrant a quick overview before launching into the specifics.

• Lavoisier defined acids by taking into account their Defining acids as substances containing oxygen, this clearly excluded acids such as hydrochloric acid.
• Davy developed our understanding of acids by defining them by their reactions. Stating that acids were simply substances that contained hydrogen that could be replaced by a metal. Distinctly acidic metals such as silver are excluded by this
• Arrhenius defined acids and bases through their ability to ionise, contributing greatly to the concept of weak and strong acids and This also served to explain the relative conduc- tivities of acids and bases. However, this definition did not include carbonates which do not dissolve to produce hydroxide ions, yet are distinctly basic.
• Br¨onsted and Lowry defined acids and bases through their properties relative to their solvents. Explaining much in terms of acid/base equilibria, particularly the pH of salts, this definition still fails to acknowledge the acidic characteristics of metals such as

# Outline the historical development of ideas about acids including those of: Lavoisier, Davy, and Arrhenius

It is necessary to learn exactly how each scientist contributed to the development of the acid-base definitions, including what exactly they believed defined acids, as well as any flaws in their definitions.

In 1779, Antoine Lavoisier asserted that acids could be defined as substances which contained oxygen. However, one limitation of this definition was that it clearly excluded acids such as hydrochloric acid, HCl, yet included bases such as calcium oxide, CaO. However, this definition aimed at defining acids through the composition of substances.

In 1815, Humphrey Davy made the observation that acids contained hydrogen which would be replaced with a metal. This definition aided in the development of acid definitions by relating their nature to their reactions with other substances, yet failed to take into account acidic metals such as silver.

In 1884, Svante Arrhenius proposed that acids were substances which ionised to produce hydrogen ions ($H^+$), and that bases were substances which ionised to produce hydroxide ions ( $OH^-$ ). Although this definition did much by way of explaining the relative strengths of acids and bases, and their respective conductivities, it excluded carbonates which do not dissolve to produce hydroxide ions, yet are still clearly basic.

Remember- Each individual extended the definition of an acid by relating their definitions to the physical and chemical properties of known acids based upon their observations, and the work of previous scientists.

# Outline the Br¨onsted-Lowry theory of acids and bases

The Br¨onsted-Lowry theory of acids and bases is not the most recent definition,, but is certainly the most common definition you will encounter in the HSC course. It hinges on the nature of an acid as a proton donor.

The Br¨onsted-Lowry theory of acids and bases was developed independently by two scientists, defining acids as proton donors and bases as proton acceptors. This definition is important as it identifies the importance of the substance’s properties relative to the solvent, and did much to explain variances in the pH of salts.

Despite this, the Br¨onsted-Lowry acid-base definition does not explain the nature of acidic metals such as silver.

Although not strictly a part of this dotpoint, it is important that you understand what is meant by the statement ’An acid is a proton donor.’ As such, the concept will be explained again in the following paragraph.

The hydrogen atom, H, has one proton and one electron. When the hydrogen atom is ionised, then it loses an electron, becoming positively charged, $H^+$. However, as we already stated that the hydrogen atom has one proton and one electron, then when ionised, only a proton is left. As such, when an acid ionises in solution, it donates a ‘proton’.

See the dotpoint below for an expansion upon the concept of proton donors and proton acceptors.

Remember -The Br¨onsted-Lowry theory of acids and bases states that acids are proton donors and bases are proton acceptors, where a hydrogen ion is a proton.

# Describe the relationship between an acid and its conjugate base and a base and its conjugate acid

If an acid is a proton donor, then after it donates the proton, what is left can essentially accept a proton, thus defining a base as a proton acceptor. As such, the base that is formed is the acid’s conjugate base. A similar relationship exists between a base and its conjugate acid.

If an acid, for example hydrochloric acid, donates a proton, its conjugate base is formed.

Conversely, if a base, for example water, accepts a proton, its conjugate acid is formed

The conjugate acid of a strong base is a weak acid, and the conjugate acid of a weak base is a strong acid. Conversely, the conjugate base of a strong acid is a weak base, and the conjugate base of a weak acid is a strong base.

Remember- The base formed when an acid donates a proton is that acid’s conjugate base, and turns back into the acid when it accepts a proton. In reverse, a base has a corresponding conjugate acid formed when it accepts a proton.

# Identify conjugate acid/base pairs

You can easily identify which substances are paired by the structure of each substance. For example, $HCl$ and $Cl^-$ are one pair, where as $H_2O$ and $H_3O^+$ are another pair. It is then simply down the identifying which is a conjugate acid, and which is a conjugate base. Alternatively, you can identify the conjugate acid as the product with the positive charge, and the conjugate base as the product with a negative charge.

With reference to the previous dotpoint [Describe the relationship between an acid and its conjugate base and a base and its conjugate acid], the conjugate acid/base pairs are identified as $HCl$ / $Cl^-$ and $H_2O$ / $H_3O^+$ respectively.

Another example is as follows:

Note that in a neutralisation reaction, there will be two conjugate pairs. Don’t forget to state both if required.

# Choose equipment and perform a first-hand investigation to identify the pH of a range of salt solutions

This experiment simply seeks to demonstrate the effect on the pH of salts when strong and weak acids and bases are used to neutralise one another. As such, prepare the four combinations below and review the expected results below. A pH probe can be used for quick results.

Expected Results:

 Acid Concentration Base Concentration Nature of Salt Strong Weak Weak Acid Weak Strong Weak Base Strong Strong Relatively Neutral Weak Weak Relatively Neutral

# Identify a range of salts which form acidic, basic or neutral solutions and explain their acidic, neutral or basic nature

Although the common table salt forms a neutral solution, this is not the case for all salts. Rather, the pH of the salt depends upon the solutions which react to form it. These solutions are in turn an acid and a base. Four basic scenarios are possible, and are listed below.

When a strong acid neutralises a weak base, such as HCl reacting with $NH_4OH$ (Ammonium hydroxide), the cation formed is a weak acid.

For example:

The salt dissociates in water, forming ammonium cations and chloride anions.

The chloride anion is the weak conjugate base of the strong hydrochloric acid. Thus, it is relatively inert in water. The ammonium cation is the strong conjugate acid of the weak base ammonia. Thus, the ammonium ions react in water, acting as an acid.

The production of hydronium ions in this process is responsible for the acidity of the salt formed.

Remember, conjugate acid-base pairs have inversion in their relative strengths. As such, the conjugate acid of a strong base is a weak acid, and the conjugate base of a weak base is a strong base.

Similarly, when a strong base neutralises a weak acid, such as NaOH reacting with HCOOH (Formic acid), the anion formed is a weak base. The conjugate acid of the strong base is weak, and is thus inert. The weak acid produces a strong conjugate base, which react with water to produce a basic solution.

When a strong acid and a strong base react, the salt formed is relatively neutral. This is because the conjugate acid and conjugate base of a strong base and strong acid respectively are weak, and do not react significantly with water.

When a weak acid and a weak base react, the salt formed is also relatively neutral. Both the conjugate acid and the conjugate base are strong, and react with water to form hydroxide ions and hydronium ions respectively. The net effect of this is the formation of water. As such, the effects are largely cancelled out between the anion and the cation.

Reviewing the results of the experiment:

 Acid Concentration Base Concentration Nature of Salt Strong Weak Weak Acid Weak Strong Weak Base Strong Strong Relatively Neutral Weak Weak Relatively Neutral

# Identify amphiprotic substances and construct equations to describe their behaviour in acidic and basic solutions

Make sure you don’t confuse the term amphiprotic with the term amphoteric, which refers to a substance’s ability to act as either an acid or a base. The difference is subtle, but does exist.

Acting as an acid:A substance that can behave as both a proton donor and a proton acceptor is known as an amphiprotic substance. An example of an amphiprotic substance is the hydrogen carbonate ion, $HCO^-$ .

Acting as an acid:

Acting as a base:

Other examples of amphiprotic substances include water and the hydrogen sulfate ion. Water acting as an acid:

Water acting as a base:

Remember- An amphiprotic substance is one which can act as either a proton donor or proton acceptor under different conditions.

# Identify neutralisation as a proton transfer reaction which is exothermic

This dotpoint is relatively straightforward as we have established an acid as a proton donor and a base as a proton acceptor.

As we have already identified the transferring of a hydrogen ion as the transferring of a proton, it is a given that neutralisation- the reaction between an acid and a base- is as proton transfer reaction. This is because a neutralisation reaction involves a proton donor and a proton acceptor, where a proton is effectively transferred from the acid to the base.This reaction usually produces a heat of neutralisation between strong acids and strong bases of approximately 57 kJ per mole of water formed. Thus this reaction is exothermic (Heat is given off during the reaction).

# Describe the correct technique for conducting titrations and preparation of standard solutions

Various methods will exist for conducting a titration and preparing a standard solution. The procedure provided below is but one example, and can be altered in many ways but remain valid. The key for any question is to identify how the method you provide allows for a high degree of experimental reliability and validity.

Titration is an experimental procedure used to determine the concentration of an unknown acid or base. It works by taking the unknown sample, and then neutralising it using a measured quantity of a substance of known concentration. The neutralisation is detected by using an indicator that changes colour at the midpoint of the titration, known as the equivalence point.

When an acid and a base react, they can form a salt that has a pH related to the strengths of the acid and base used. This means that the equivalence point of a titration is not necessarily neutral, with a pH of 7, since the neutralisation reaction may form an acidic or basic salt. Therefore, an indicator must be chosen that changes colour at the expected pH for the equivalence point of the reaction. For example, if a weak base is neutralised by a strong acid, then the solution will be weakly acidic, so the indicator chosen must change colour at a pH of around 35, rather than 7.

The premise of titration is the equation c1v1 = c2v2, where c1 and c2 are the concentrations of the unknown and known solutions, and v1 and v2 are the volumes of the unknown and known solutions. In a titration, everything is known or measured apart from c1, for which the equation is solved. This means two things: Firstly, v1 and v2 need to be precisely measured, and secondly, c2 is known with a high degree of accuracy. A standard solution the term given to a highly pure solution that can be used to provide accurate measurements in volumetric analysis techniques such as those used to perform a titration.

Once the reaction reaches its equivalence point and both reactants are completely consumed, the concentration of the unknown solution c1 can be determined using basic stoichiometric calculations. For example, if 500mL of a 2 mol/L solution of hydrochloric acid is required to completely neutralise a 250mL of sodium hydroxide, then the concentration of sodium hydroxide must be 4 mol/L.

When preparing a standard solution:

1. Weigh the primary standard- The solid dissolved to form the standard This solid must be as pure and as free from moisture as possible. If necessary, use an oven or desiccator to dry or cool the solid respectively.
2. Dissolve the primary standard in a small amount of distilled water in a
1. Pour the contents of the beaker into a volumetric flask, rinsing the beaker slightly to ensure all the contents are
2. Add water to the flask up until the graduation Use a small squirt bottle for the final few drops.

Remember that each step in preparing a standard solution is designed to minimise errors in determining the concentration of the resulting standard solution. If the concentration of the standard solution is incorrect, then the calculated concentration for the unknown solution will also be incorrect.

When conducting titrations:

1. Wash a burette using distilled water, using a small amount of the prepared solution with known concentration to flush out the burette
2. Fill the burette with the rest of the solution of known concentration, taking care to ensure the solution is correctly
3. Place a solution which is being tested in a flask under the
1. Add two drops of the appropriate indicator to the conical
2. Using one hand to twist the burette to allow its contents into the flask below, use the other hand to constantly swirl the
3. When the solution in the flask changes colour- at its equivalence point- ensure that the burette is no longer dripping, and read the volume of solution discharged by the burette. The first reading should be taken as an approximation, and three further readings should be averaged to determine the
4. Use basic stoichiometric calculations to calculate the unknown

Distilled water is used in this experiment because tap water may contain dissolved salts that can affect the titration.

Remember $c_1v_1 = c_2v_2$, where c is equal to concentration and v is equal to volume. Watch what units you use!

# Perform a first-hand investigation and solve problems using titrations and including the preparation of standard solutions, and use available evidence to quantitatively and qualitatively describe the reaction between selected acids and bases

Make use of the methods outlined in the above dotpoint in preparing the standard solution and conducting the titration. Although the process is important, of equal importance are considerations of reliability and validity. Be prepared to answer questions which ask you to improve the experiment, as well as basic risk assessment questions. Although these answers will be common sense, make sure to explain how your recommendation would improve reliability/validity/safety.

You will be required to provide stoichiometric calculations using the formula $c_1v_1 = c_2v_2$, where c is equal to concentration and v is equal to volume. Again, make sure that your two v values are in the same units (mL or L).

For example, imagine you have conducted a titration using- for the sake of simplicity- hydrochloric acid and sodium hydroxide.

1. Write down the equation

1. Identify what variables you know in the equation $c_1v_1 = c_2v_2$. You should be given, or be able to work out, three of the four necessary variables. Let us assume that the concentration of HCl was 1M, and 50mL of the solution was used to react 45mL of NaOH. The concentration of the NaOH is
1. Work out the number of moles of the substance you have the concentration and volume of using the formula n = cV .

The number of moles of HCl was therefore 1 × 0.05 which equals 0.05 moles.

Keep in mind that V is measured in litres.

1. Work out the ratio of moles between the two In this case, one mole of HCl reacts one mole of NaOH.

If a ratio other than a one-to-one ratio is obtained, then you must take into account the fact that X moles of the acid reacts Y moles of the base, which would clearly influence your later calculations of the unknown variable.

1. Work out the missing variable. At this point you have the number of moles of the substance with an unknown variable, as well as one known Using the c = nV formula, you can easily work out the last variable. In our example, if c = nV , then n = c . The number of moles has been determined to be 0.05, and V = 0.045. Therefore the concentration must be 0.05 × 0.045 = 0.00225M.

# Perform a first-hand investigation to determine the concentration of a do- mestic acidic substance using computer-based technologies

This experiment simply involves the use of pH probes in solutions. This experiment is unlikely to appear in any questions, but simply remember to calibrate the probe before using it for ideal results.

Procedure:

1. Prepare beakers of a variety of domestic acidic substances such as vinegar, lemon juice and milk.
2. Calibrate a pH
3. Place the probe gently into each beaker in turn, taking care to dip the probe in the buffer in between each test so as to avoid distorted

# Analyse information from secondary sources to assess the use of neutralisation reactions as a safety measure or to minimise damage in accidents or chemical spills

It is highly recommended that not only should you note at least one substance used to neutralise acids/bases, but be able to provide reasons as to why it would be an effective means of minimising danger.

With many acids and bases being highly corrosive, neutralisation reactions form an integral part of safety protocols within scientific and industrial settings. Finding uses ranging from the management of photographic and sewerage wastes, neutralisation reactions offer a quick means of minimising danger if correctly done.

One such example is sodium hydrogen carbonate. Advantages of using $NaHCO_3$ include:

• It can be stored and transported easily given its stable
• It is amphiprotic, meaning that it can be used to neutralise both acid and alkali
• It is relatively cheap, providing an incentive for industrial
• It is not so strong that excess usage would be extremely detrimental. If excess $NaHCO_3$ is used, the result will only be mildly reactive. In contrast, excess $HCl$ or $NaOH$ could serve to only worsen the situation as the result would be strongly acidic or

Remember- The ideal neutralising agent should be safe to store and use, effective against a variety of situations (such as both acid and base spillages), and should be economically viable.

# Qualitatively describe the effect of buffers with reference to a specific example in a natural system

Keep in mind Le Chatelier’s principle when trying to grasp what a buffer does. This guide will use the bicarbonate ion as an example of a buffer- in particular its use in maintaining the pH of blood.

A buffer is a substance that will work to maintain the pH of a system even if small amounts of acids or bases are added. A buffer is normally a combination of chemicals, commonly consisting of a weak acid and its conjugate base (although it can also be formed from a weak base and its conjugate acid).

An example of a buffer that can be found in a natural system is the bicarbonate ion in human blood. Although this natural system must maintain a pH of 7.4, carbon dioxide frequently makes its way into the bloodstream, potentially reducing pH below 7.0.

In response to this, the bicarbonate ion found in blood plasma forms a state of equilibrium as shown below.

In the above equation, if an acid is added, the increase in hydrogen ions (pH falls) will result in a favouring of the forwards reaction according to Le Chatelier’s principle, consuming the additional hydrogen ions (pH rises). Conversely, if a base is added, the hydroxide ions (pH rises) will react with the hydrogen ions, such that the system will work to correct itself by favouring the backwards reaction as stated by Le Chatelier’s principle (pH falls).

In both cases, it can be seen that a buffer works to lessen changes in the pH of a system.

Remember- As a buffer is essentially an application of Le Chatelier’s principle, it can only minimise the effect on a system rather than completely remove the effects.