IQ1 SYLLABUS POINTS
1.1 investigate the correct IUPAC nomenclature and properties of common inorganic acids and bases
1.2 predict the products of acid reactions and write balanced equations to represent:
– acids and bases
– acids and carbonates
– acids and metals
1.3 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
1.4 investigate applications of neutralisation reactions in everyday life and industrial processes
1.5 conduct a practical investigation to measure the enthalpy of neutralisation
1.6 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:
a) Arrhenius’ theory
b) Brønsted–Lowry theory
1.1 investigate the correct IUPAC nomenclature and properties of common inorganic acids and bases
Nomenclature of inorganic acids and bases
The system of nomenclature used for the naming of inorganic acids and bases follows the IUPAC system in the Red Book (Connelly, Damhus, Hartshorn, & Hutton, 2005). There are a number of common substances which may be found in a home or school laboratory which have the properties of an acid or a base. These include:
Common Acids
Hydrochloric acid (HCl)
Hydrofluoric acid (HF)
Sulfurous acid (H2SO3)
Sulfuric acid (H2SO4)
Nitrous acid (HNO2)
Nitric acid (HNO3)
Carbonic acid (H2CO3)
Phosphoric acid (H3PO4)
Common bases
Sodium hydroxide (NaOH)
Calcium hydroxide (Ca(OH)2)
Barium hydroxide (Ba(OH)2)
Sodium carbonate (Na2CO3)
Sodium bicarbonate (NaHCO3)
Ammonia (NH3)
These are acids that that contain hydrogen bonded with a non-metal element.
NOTE: You won’t find oxygen in binary acids. This is because that’s a different type of acids (oxyacids) which we explore later.
A common example of a binary acid that you have came across is hydrochloric acid, HCl.
To name binary acid, it is simple. Here are the steps.
Step 1: Name the hydrogen first. You do this simplifying the word to just ‘hydro’.
Step 2: Find the root name of the non-metal.
Step 3: Combine ‘hydro’ and the root name of the non-metal. Start with hydro first, followed by the root name.
Step 4: Combine the hydro + rootname with ‘ic’
Step 5: Following the word you generated at the end of step 4, write ‘acid‘ after leaving a space between the two words.
Nitrogen (N) – Nitr
Phosphorus (P) – Phosph
Oxygen (O) – Ox
Iodine (I) – Iod
Fluorine (F) – Fluor
Chlorine (Cl) – Chlor
Bromine (Br) – Brom
Sulfur (S) – Sulf
Hydrogen (H) – Hydr
Carbon (C) – carb
Transition metal – Chromium (Cr) – Chrom
Transition metal – Manganese (Mn) – Mangan
Let’s do an example. How do you name the HCl compound?
Step 1: ‘Hydro’
Step 2: The non-metal here is chloride. The root name for chloride is ‘chloro’
Step 3: Combine hydro and the root name. This gives ‘hydrochloro’
Step 4: Combine the root name with ‘ic’. This gives ‘hydrochloric’
Step 5: Add the word ‘acid’ following the name of the inorganic acid. This gives ‘hydrochloric acid’.
Hence, the answer to naming the compound, HCl, is hydrochloric acid.
Another type of inorganic acid is oxyacids. Sometimes, you may see this type of acid being referred to as oxoacids.
Oxyacids contain oxygen atom(s).
In terms of common features shared between oxoacids and binary acids, both types have a hydrogen and non-metal component to them.
Step 1: Ignore the hydrogen. This means that we don’t need to name it, ie don’t write hydro.
Step 2: Find the root name of the non-metal group. Sometimes, instead of a non-metal, there may be a transitional metal instead. Such as chromium or manganese.
Step 3: Combine ‘ous’ or ‘ic’ with the rootname. Use ‘ous’ if the non-metal is in lower oxidation state. Use ‘ic’ if the non-metal is in the higher oxidation state. Generally, acids with ‘ic’ are more common than those with ‘ous’.
Step 4: Write acid following the acid’s name.
HNO3.
Step 1: We ignore the hydrogen.
Step 2: The root name of the non-metal, nitrogen, is ‘nitr’
Step 3:
The oxidation state of nitrogen here, in HNO3, is +5.
The alternative is HNO2 where nitrogen has an oxidation state of 3+.
This gives us ‘Nitric’, the one where the central nitrogen atom has the higher oxidation state.
Step 4: We add ‘acid’ to ‘nitric’. This gives us nitric acid.
So, our answer in naming the oxyacid, HNO3, is nitric acid.
For oxyacids that have the phosphorus or sulfur non-metals, we need to modify their root names after Step 4.
For oxyacids containing phosphate central atom, we need to add ‘or‘ after its root name.
For oxyacids that have sulfur as the central atom, we need to add ‘ur‘ after its root name.
This means that INSTEAD OF:
H3PO4 = Phosphic acid, we call it phosphoric acid instead.
H2SO4 = Sulfic acid, we call it sulfuric acid instead.
Sometimes, you may encounter the need to use prefixes when naming oxyacids. However, in most cases, these oxyacids with prefixes are incorporated in a question without requiring you to name them.
So, either way, it is good to understand how the acids looks like if they carry a prefix in front of their name.
You will see two possible prefixes that is incorporated in front the name of oxyacids.
These prefixes are ‘hypo‘ and ‘per‘.
Previously, we touched on the fact that we can have ‘ous‘ or ‘ic‘ after the root name of the non-metal based on its oxidation state. The one with the higher oxidation state have ‘ic’ after its root name whereas the other have ‘ous’.
If we remove an oxygen from chlorous acid, so that it becomes HClO, we need to add the prefix ‘hypo‘ in front of chlorous acid. Oxidation state of chlorine decreased from 3+ to 1+.
If we add an oxygen atom to chloric acid, so that it becomes HClO4, we add the prefix ‘per‘ in front of chloric acid. Oxidation state of chlorine increased from 5+ to 7+.
Typically, you will see bases either written with a metal or ammonium ion to begin with. This is then followed by a hydroxide ion.
In some cases, an oxide may be present rather than hydroxide ion. We will touch on oxides later.
Step #1: Name the metal or ammonium ion first, depending on the compound whether it has an ammonium ion or not.
Step #2: Add the word ‘hydroxide’ following the name of the metal or ammonium ion.
Mg(OH)2.
Step #1: The name of the metal in the compound is magnesium, Mg.
Step #2: We add the word ‘hydroxide’ after magnesium.
So, the name of the base, Mg(OH)2, is magnesium hydroxide.
Note the following:
So how do you name these acids and bases? This was taught in preliminary HSC chemistry course. But we will quickly go through it again because we can.
So, we first name the metallic or non-metallic by stating the name of the element that is located more towards the left of the periodic table first. Take the metallic oxide, CaO as an example.
For example, N2O4 is called dinitrogen tetroxide. ‘Di‘ prefix is used because there are two nitrogen atoms in a N2O4 compound and the ‘tetra‘* prefix is used because each N2O4 compound has four oxygen atoms.
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TASK 1.1.1
Using the flow chart below, walk through the list of acid names above and verify the names.
Some exceptions:
TASK 1.1.2
a) Name the following:
TASK 1.1.2
b) Write formulas for the following:
TASK 1.1.2
c) Learn a) and b). Really!
REVIEW 1.1
HSC [Qs WITH NO MARKS INDICATED ARE MULTIPLE CHOICE]
1.2 predict the products of acid reactions and write balanced equations to represent:
a) acids and bases
There are reactions involving acids and bases with which you MUST be familiar.
a) Acid + Base (neutralisation) reaction.
Acid and base reactions are called neutralisation reactions if they involve hydrogen (or hydronium ions) reacting with hydroxide ions to produce water. This can be expressed as:
H+(aq) + OH-(aq) ⟶ H2O(l)
This is why water is produced as a product in the equation involving the reaction of hydrochloric acid with sodium hydroxide.
General word equation:
Acid plus Base gives Salt plus Water
Example:
Hydrochloric acid + sodium hydroxide ⟶ sodium chloride + water
Balanced species neutral chemical equation:
HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2O(l)
Net ionic equation:
H+(aq) + OH-(aq) ⟶ H2O(l)
(Apply the NAAG SG CHPS solubility rules to determine whether the salt formed is soluble or insoluble.)
The salt formed from neutralisation can be neutral, acidic or basic depending on the strength of the acid and base used in the reaction.
To determine the acidity or basicity of a salt, dissolve it in water.
Suppose you mix the following acid and base solutions.
Equation 1: HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2O(l)
Sodium chloride will be the salt in the above scenario. It is a neutral salt as it yields a pH of 7 when you dissolve the ionic compound in water.
NOTE: It does not really matter if you use hydrogen ions or hydronium ions. This is because most of the acid and solutions that you will be handling are diluted with water.
For example, HCl(aq) -> H+(aq) + Cl-(aq)
The hydrogen ions can then react with water to form hydronium ions. This can be expressed as:
HCl + H2O(l) -> H3O+(aq) + Cl-(aq)
This is why pH can be expressed in two forms:
In general:
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TASK 1.2.1
1. Write the products of the following acid + base reactions, then balance the equations.
(A) HCl(aq) + Ca(OH)2(aq) ⟶
(B) HNO3(aq) + KOH(aq) ⟶
(C) H2SO4(aq) + NaOH(aq) ⟶
(D) H3PO4(aq) + Ba(OH)2(aq) ⟶
2. Write balanced ionic equations and, through elimination of common ions, the net ionic equations for the following reactions.
(A) HCl(aq) + Ca(OH)2(aq) ⟶
(B) HNO3(aq) + KOH(aq) ⟶
(C) H2SO4(aq) + NaOH(aq) ⟶
(D) H3PO4(aq) + Ba(OH)2(aq) ⟶
1.2 predict the products of acid reactions and write balanced equations to represent:
b) acids and carbonates
General word equation:
Acid plus Carbonate gives Carbon dioxide plus Water plus a Salt
Example:
Hydrochloric acid + sodium carbonate ⟶ sodium chloride + water + carbon dioxide
Sodium chloride is formed as a salt. You will also see carbon dioxide bubbles as it evolves in the solution. Water will also be a product.
Balanced chemical equation:
2HCl(aq) + Na2CO3(aq) ⟶ 2NaCl(aq) + H2O(l) + CO2(g)
Net Ionic Equation:
2H+(aq) + CO32-(aq) ⟶ H2O(l) + CO2(g)
Carbonates are substances that contain the carbonate polyatomic ion CO32-.
Some examples of carbonates include:
When an acid reacts with a carbonate, you will get a salt, water and carbon dioxide gas as products.
Usually in the HSC course, you will see an acid reacting with a metal carbonate or ammonium carbonate.
If you have a metal hydrogen carbonate such as sodium hydrogen carbonate rather than sodium carbonate, the products produced are the same. That is, salt, carbon dioxide and water.
HCl(aq) + NaHCO3(s) -> NaCl(s) + CO2(g) + H2O(l)
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TASK 1.2.2
1. Write the products of the following acid + carbonate reactions, then balance the equations.
(A) HCl(aq) + CaCO3(s) ⟶
(B) HNO3(aq) + Na2CO3(s) ⟶
(C) H2SO4(aq) + CuCO3(s) ⟶
(D) H3PO4(aq) + NaHCO3(aq) ⟶
2. Write balanced ionic equations and, through elimination of spectator ions, the net ionic equations for the following reactions.
(A) HCl(aq) + CaCO3(s) ⟶
(B) HNO3(aq) + Na2CO3(s) ⟶
(C) H2SO4(aq) + CuCO3(s) ⟶
(D) H3PO4(aq) + NaHCO3(aq) ⟶
3. Write balanced ionic equations and, through elimination of spectator ions, the net ionic equations for the following reaction.
HNO3(aq) + NaHCO3(s) ⟶
1.2 predict the products of acid reactions and write balanced equations to represent:
c) acids and metals
Reactions involving metals and acids are dependent on the activity of the metal and the concentration of the acid.
General word equation:
Dilute Acid plus Active Metal gives Salt plus Hydrogen gas
Example:
Hydrochloric acid + magnesium ⟶ magnesium chloride + hydrogen gas
Balanced chemical equation:
2HCl(aq) + Mg(s) ⟶ MgCl2(aq) + H2(g)
Net Ionic Equation:
2H+(aq) + Mg(s) ⟶ H2(g) + Mg2+(aq)
Acids will react with metals IF the metal is active enough to displace the hydrogen ions from the solution. This would mean that the solid metal will become metal ions by donating electrons, and the hydrogen ions will accept such electrons to become hydrogen gas. If the metal is not active enough to displace the hydrogen ions from solution, no reaction will occur.
The reaction between hydrochloric acid and magnesium metal can be expressed as follows:
2HCl (aq) + Mg(s) -> MgCl2 (aq) + H2 (g)
Magnesium chloride will be the salt in this case. The magnesium metal will displace the hydrogen ions from solution to form hydrogen gas, where the hydrogen ions are derived from the dissociation of hydrochloric acid.
HCl(aq) -> H+(aq) + Cl-(aq)
Each hydrogen ion will accept an electron donated from magnesium metal where two hydrogen atoms can bond to form hydrogen gas.
This means that, after displacement, the overall reaction will be:
Mg(s) + 2H+(aq) -> Mg2+(aq)+ H2 (g)
In preliminary HSC Chemistry, you have learnt about the metal reactivity series.
Reactive metals are those that readily and quickly react with acids. These reactive metals are positioned above hydrogen in the activity series table.
By losing electrons to form metal cations in solution, these metals achieve a more stable state (octet).
As you learnt in the Y11 course, as you move down a group on the periodic table, the number of electrons of elements increases significantly. These electrons occupy energy shells that are further away from the nucleus. As the number of energy shells increases, the effect of ‘shielding’ increases.
This means that the element can achieve a more stable state by losing those electrons that are loosely attracted to the nucleus in the high energy shells.
This means the metal is capable of donating electrons to hydrogen ions. This results in metal ions in solution and the displacing hydrogen ions into hydrogen gas.
Unreactive metals normally do not react with acids. The donation of electrons from these unreactive metals is less common. If the acid is a strong enough reducing agent, the unreactive metal may react with the acid.
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PRACTICAL 1.2.1
Practical Activity 6.3 Reactions of Hydrochloric Acid PP80-82
PRACTICAL 1.2.2
There are numbers of general reactions involving acids and/or bases with which we need to become familiar:
HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2O(l)
2HCl(aq) + Na2CO3(s) ⟶ 2NaCl(aq) + H2O(l) + CO2(g)
2HCl(aq) + Mg(s) ⟶ MgCl2(aq) + H2(g)
In this experiment, you will carry out a series of reactions involving acids and bases, record your observations and, where possible, identify any products.
Gas tests are used to test for the most common gas production such as
Sometimes the products of a reaction depend on the concentration of one or more species present. In some reactions, the concentration of the acid can affect the products of the reaction.
Method
Warning: Wear safety glasses!
1 Transfer 5 drops of 1 mol L–1 acid solution (eg HCl) into each of three clean, dry test tubes.
2 Add 5 drops of a basic solution (eg NaOH) into one of the test tubes containing the acid.
3 Record your observations. Identify any products of this reaction.
4 Add 1 spatula-ful of a carbonate (eg CuCO3) into one of the test tubes containing the acid.
5 Record your observations. Identify any products of this reaction.
6 Add a small piece of a metallic solid (eg Mg) into one of the test tubes containing the acid.
7 Record your observations. Identify any products of this reaction.
TASK 1.2.3
1. Write the products of the following acid + metal reactions, then balance the equations.
(A) HCl(aq) + Ca(s) ⟶
(B) HNO3(aq) + Mg(s) ⟶
(C) H2SO4(aq) + Fe(s) ⟶
(D) H3PO4(aq) + Zn(s) ⟶
2. Write balanced ionic equations and, through elimination of common (spectator) ions, the net ionic equations for the following reactions.
(A) HCl(aq) + Ca(s) ⟶
(B) HNO3(aq) + Mg(s) ⟶
(C) H2SO4(aq) + Fe(s) ⟶
(D) H3PO4(aq) + Zn(s) ⟶
1.3 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
Indicators are plant dyes which change colour in response to the degree of acidity or basicity of a solution, ie through specific pH values. Most acid-base indicators are weak acids whose corresponding (conjugate) base is a different colour. Both the acid and the base are in equilibrium with each other, meaning the equilibrium can be shifted to favour one colour or another based on the addition of hydrogen ions or hydroxide ions.
This will make more sense when you come back to it in revision. For now, indicators change colour across a range of acidity or basicity and one indicator, or several in combination, can be used to identify this. Indicators are organic compounds and some, such as litmus, are found naturally in plant pigments.
Notes
pH is a unitless value that is assigned to solutions in order to compare the relative acidity between solutions.
pH values can range from 0 to 14.
NOTE: pH values can exceed 14 or be less than 0 (negative), if the concentration of the strong acid or base is very high.
Indicators are used because their colour changes are useful in assisting in determining the approximate pH of a substance such as a solution. They are also useful to compare the relative acidity between different solutions.
Indicators are usually weak acids or bases, meaning they form an equilibrium in solution. In most cases, the indicator molecule reacts with water.
H2In(aq) + H2O(l) <-> 2H3O+(aq) + In2- (aq)
‘ln’ is short for indicator. ‘H2ln’ just means that the indicator has is bonded to two hydrogen atoms.
If an acid is present or added into solution, the equilibrium position for the above reaction will shift to the left to minimise the increase in [H+]. The solution will develop the colour that H2In gives in solution.
If a base is present or added in solution, the equilibrium position will shift to the right as the concentration of H3O+ ions decrease as they react with the hydroxide ion from the base to produce water.
The shift of the equilibrium position to the right will give the unique colour In2- which is observed as a colour change.
This is the reason for a colour change when doing titration experiments.
Some indicators change colour over a small pH range, other indicators change colour over a large pH range.
In the diagram below, Indicator A changes colour over a small pH range whereas Indicator B changes colour over a small pH range.
Which indicator is more accurate?
Indicator A changes colour (from red to blue and vice versa) over a smaller pH range than indicator B. It is a more accurate and more effective* indicator as, from the indicator’s colour change, the substance testing the pH for will have a pH of 5.5 – 6.5.
Since indicator B changes colour over a larger pH range (from 5.5 to 10) than indicator A, it is a less effective* and less accurate indicator than indicator A. It is harder to pinpoint the exact pH by observing the indicator colour.
* NOTE: The term ‘effective‘ for indicators depends on the purpose of using the indicator. To see if the substance is acidic or basic and nothing more specific, use an indicator that changes colour over the pH range between 7 to less than 7 to see if the substance is acidic. Alternatively, use an indicator that changes colour over the pH range from 7 to greater than 7 to see if the substance is basic. Another method of determining pH is using a litmus paper to easily distinguish if the substance is acidic or basic.
To use the indicator (A and B) to pinpoint the exact pH of the substance, then indicator A is more effective in doing the job than indicator B. However, indicator A still fails to pinpoint the exact pH as the substance can have a pH of anywhere from 5.5 to 6.5.
To pinpoint the exact pH, a pH probe can be used. It is more accurate than indicators.
Investigate the Gizmo below.
PRACTICAL 1.3.1a
Practical Activity 6.5 Beetroot a natural indicator PP85-7
Improvements
A pH probe can be used to determine the precise pH of the solutions where a calibrating solution is used to calibrate the pH probe to improve accuracy and reliability of results. The calibrating solution is a system that is known as a buffer. We will talk about buffers in next week’s notes.
PRACTICAL 1.3.1b
1. Boil 100 g of shredded raw beetroot in 250 mL of water until the water turns dark red-purple.
2. Allow mixture to cool and decant indicator solution into a clean beaker.
3. Add one drop of each 1M test solution (HCl, CH3COOH, NaCl, NH3 and NaOH ) in separate spots on a spotting plate using a droplet.
4. Add one drop of the indicator to each spot and record colour changes.
5. Repeat Steps 3-4 with universal indicator and compare.
PRACTICAL 1.3.1c
Method [Adapted from (Pearson Education Australia, 2019)]
1 Transfer 200 mL of water into a 600 mL beaker. Add a leaf of shredded red cabbage. Boil the mixture gently until a richly coloured extract is formed. Allow to cool, then pour off the reddish-purple indicator solution.
2 Test the indicator by adding it to 1 mL samples of the following solutions:
· cloudy ammonia
· white vinegar
· 0.1 mol L–1 NaOH
· soda water
· 0.1 mol L–1 HCl
Add the indicator dropwise until a definite colour can be seen. Record your observations in a data table.
3 Repeat step 5 using universal indicator, methyl orange and phenolphthalein solutions. Use a colour chart to determine the pH of each test solution with the universal indicator. Record the colour and/or the pH. Add these results to your data table.
4 Correlate the results and determine a pH/colour scale for red cabbage indicator.
The colour of the red cabbage extract should be purple or dark red in a neutral environment -when no substance is added to the indicator.
When an acidic is added to the indicator, the colour should change from purple to pink (red at very low pH - 1 or less than 1).
When a base is added to the indicator, the indicator’s colour should change from purple to green (or yellow at very high pH, 13-14)
When a neutral solution such as distilled water is added to the indicator, the red cabbage indicator will not change colour and remain purple.
Within the pH range, the indicator’s colour should be purple.
Risk assessment
Improvements
A pH probe can be used to determine the precise pH of the solutions where a calibrating solution is used to calibrate the pH probe to improve accuracy and reliability of results. The calibrating solution is a system that is known as a buffer. We will talk about buffers in next week’s notes.
PRACTICAL 1.3.1d
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Suppose you have the following indicator dissolved in a beaker of water. This indicator can exhibit two colours which are red and yellow.
Equation 1: H2In (aq) + H2O (l) <-> 2H3O+ (aq) + In2- (aq)
Suppose that equation one proceeds to equilibrium when you dissolve the indicator in the solution containing the test solution or you wish to add the test solution into the indicator.
Either way, IF the test solution contains hydronium ions (e.g. adding acid to the beaker of water or the test solution already contains hydrogen ions when indicator was added), it will shift the equilibrium position of equation one to the left according to Le Chatelier’s Principle to minimise the increase in hydronium ion concentration. This would cause the solution change colour from yellow to red.
If the solution contains hydroxide ions, the hydronium ions formed from the dissociation of the indicator molecule will react with the hydroxide ion to form water.
This decreases the concentration of hydronium ions in solution and thus will cause the equilibrium position of equation one to shift to the right as per Le Chatelier’s Principle to minimise the decrease in [H3O+]. This would cause the solution to change colour from red to yellow.
The indicator would undergo a colour change according to the equilibrium equation above.
Sometimes you may see this equation instead:
HIn (aq) + H2O (l) <-> H3O+(aq) + ln- (aq)
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TASK 1.3.1
1. What is an indicator?
2. Name three places or situations where pH might be an important factor.
3. (HSC 2004, Q23, 3 marks)
A natural indicator was made by heating yellow flowers with water. The resulting yellow solution was tested with a range of substances. The results are recorded below.
Substance Colour
0.1 M hydrochloric acid Red
0.1 M ammonia Yellow
Lemon juice Orange
Oven cleaner Yellow
Pure water Yellow
Assess the usefulness of the yellow solution as an acid-base indicator
4. (2002, Q20b, 3 marks)
A student used indicators to determine whether three colourless solutions were acidic or basic. The indicators used are shown in the table.
Indicator Colour Change pH range
Methyl orange Red to yellow 3.2 – 4.4
Methyl red Red to yellow 4.8 – 6.0
Thymol blue Yellow to blue 8.0 – 9.6
Alizarin Red to purple 11.0 – 12.4
Samples of each solution were tested with the indicators. The colours of the resulting solutions are shown in the table.
Indicator Colour of solution A Colour of solution B Colour of solution C
Methyl orange yellow yellow yellow
Methyl red yellow yellow yellow
Thymol blue blue blue yellow
Alizarin purple red red
The student concluded that each of the three solutions tested was basic. Assess the validity of this conclusion.
HSC [Qs WITH NO MARKS INDICATED ARE MULTIPLE CHOICE]
1.4 investigate applications of neutralisation reactions in everyday life and industrial processes
Neutralisation reactions occur when an acidic substance reacts with a basic substance to form a salt and water. The hydrogen ion (proton) from the acid reacts with hydroxide ions from the base to form water.
This reaction has a number of important applications.
A. Applications in Everyday life
1. Antacids:
Neutralising stomach acids with an antacid is both an effective way to relieve heartburn as well as an example of an everyday application of a neutralisation reaction.
Our stomachs secrete an acid for the purpose of protein digestion. This acid is HCl. The protease enzymes in the stomach work most effectively at low pH, however if the stomach secretes HCl at the wrong time or in high quantities, stomach acid levels can rise, causing indigestion or heartburn. A quick way to ease the discomfort is to take an antacid. These may contain magnesium hydroxide Mg(OH)2 or aluminium hydroxide Al(OH)3 but may also contain a carbonate, eg sodium carbonate Na2CO3 in the alka-seltzer. Of course if a carbonate is used, then we know CO2 will also be produced, which is why the drinking of an antacid solution may also bring up a burp.
Mg(OH)2(s) + H2O(l) -> MgCl2(aq) + 2H2O(l)
It is not only our stomachs that are sensitive to particular pH values. Many of the enzymes (biological catalysts) in our bodies have an optimum pH at which they function. If the pH is either too high or too low, the body must seek to neutralise the excess acid or base and restore the system to the correct pH range. This is also true for blood and urine.
2. Baking soda
Baking soda is used in the process of baking cakes and contains sodium hydrogen carbonate. When it makes contact with the buttermilk in the cake, a neutralisation reaction takes place. This is because NaHCO3 in baking soda can act as base and buttermilk is an acid.
3. Ant stings
Similarly, sodium hydrogen carbonate can be used to neutralise ant stings which contain formic acid. Formic acid is corrosive and can burn the skin. By neutralising the formic acid, the pain on the skin can be reduced.
NaHCO3(aq) + CH2O2(aq) -> NaCHO2(aq) + H2O(l) + CO2(g)
3. Gardening
Gardeners will often use their knowledge of neutralisation reactions to change the acidity of the soil. Some plants grow better in acidic or neutral soils whilst others prefer alkaline soils. This is particularly true of hydrangeas which act as living indicators of the soil pH. In alkaline soils, hydrangeas are pink, which the flowers are blue in acidic soils.
Slaked lime (calcium hydroxide) can be added to acidic soils, whilst gypsum or iron salts are added to alkaline soils.
The most common types of fertilisers are also produced in a neutralisation reaction; sulfate of ammonia (NH4)2SO4 and ammonium nitrate.
B. Applications in Industry
Factories may dispose liquid waste that is often acidic. Calcium oxide (a base) is added to neutralise these acidic factory effluents before they are allowed to be disposed of outside the factory, as per government environmental regulations.
Some factories produces sulfur dioxide, which is treated by neutralising it with calcium oxide (lime) before it is allowed to be released from the factory and into the atmosphere. Sulfur dioxide is acidic.
In this case, the acidic oxide (SO2) reacts with an base (CaO) to form salt and water. This is an acid-base reaction.
SO2(g) + CaO -> CaSO3(s)
SIDE NOTE: This reaction does not produce water. According to Arrhenius’s theory, water must be present as a product for neutralisation reactions. However, according to Bronsted Lowry’s definition of acid-base reactions between acid and bases, water is not a compulsory product. The Bronsted-Lowry’s definition of neutralisation requires a conjugate acid and base to be produced.
2. Production of fertilisers that contains ammonium sulfate, (NH4)2SO4.
Ammonium sulfate is produced by reacting ammonia gas with sulfuric acid. This is a multi-step process as you need to combine hydrogen and nitrogen gas to form ammonia. You also need to process the air to extract hydrogen.
2NH3(g) + H2SO4(aq) -> (NH4)2SO4 (aq)
Other industrially important applications of neutralisation reactions include:
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TASK 1.4.1
Choose one application of a neutralisation reaction for everyday life and one for industry and write a page of notes for each to answer the (probable HSC) question:
TASK 1.4.2
1. Explain each of the following:
a) Bile, secreted by the liver and stored in the gall bladder, is alkaline.
b) Too much soft drink can increase the rate of tooth decay
c) Cakes baked with self-raising flour tend to rise in the oven
d) Some rain causes damage to buildings and statues made from limestone or marble
e) Gardeners often add lime to ‘sweeten’ the soil
f) The production of ammonium sulfate used in many fertilisers is the product of a neutralisation reaction
2. The antacid Mylanta is pictured below. Two of the key ingredients in the chewable tablets are aluminium hydroxide and magnesium hydroxide.
a) Write an equation to show how aluminium hydroxide can be used to neutralise the hydrochloric acid in the stomach.
b) Use the information given on the packet to determine the number of moles of stomach acid which could be neutralised by two of these chewable tablets.
c) If the stomach held a volume of 2L, what was the initial concentration of the hydrochloric acid in the stomach?
REVIEW
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HSC [Qs WITH NO MARKS INDICATED ARE MULTIPLE CHOICE]
1.5 conduct a practical investigation to measure the enthalpy of neutralisation
All neutralisations are exothermic.
When the heat of neutralisation is measured for a range of strong acids and strong bases, the amount of heat released is always approximately 57 kJ per mole of water formed.
When a basic solution is added to an acidic solution, a reaction takes place that usually forms an ionic salt and water. The solutions are said to have been neutralised when the concentrations of the hydronium and hydroxide ions within the mixture become equal.
The net ionic equation for the reaction between a strong acid and a strong base, including the enthalpy change is:
H+ + OH– → H2O + 57kJ
A study of the heat energy released for various (strong) acid-base reactions reveal values which are very close to one another, suggesting that the same type of reaction is occurring each time, ie. the production of a salt and water, or more specifically, the transfer of a proton species from a proton donor to a proton acceptor.
A change in the enthalpy of a solution can be determined using the following equation: ΔH = -mcΔT. We will assume the final solution will have the same c value as pure water (a potential source of error).
We will be determining the standard enthalpy of neutralisation, that is, the experiment should be performed under standard conditions: surrounding conditions of 25 degrees Celsius and 1 atmospheric pressure or 100 kiloPascals (kPa).
Under these conditions, we measure change in enthalpy of neutralisation where exactly 1 mole of water is formed (the salt formed is just composed of spectator ions).
For example, suppose there is a neutralisation between sodium hydroxide and hydrochloric acid:
NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
Ionic equation:
Removing the spectator ions:
H+(aq) + OH– (aq) -> H2O (l)
The formula to determine the change in enthalpy in a neutralisation reaction is:
ΔHneutralisation = - m x c x ΔT
ΔH = Δq / n
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TASK 1.5.1
Note the error of Ba written as Ca in the equation above.
PRACTICAL 1.5.1
Text Practical Activity 6.4 Measuring the enthalpy of neutralisation PP83-84
Introduction
We are already familiar with a number of neutralisation reactions, such as the following:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + H2O(l) + CO2(g)
In this experiment, you will carry out a series of reactions involving acids and bases, to determine the empirical value for the enthalpy of neutralisation.
Neutralisation reactions are exothermic reactions.
The enthalpy of neutralisation can be determined by using the following equation:
ΔH = - mcΔT
Where:
ΔH is the change in enthalpy
m is the mass of the solution (specifically water)
c is the specific heat of the solution (we will use the value for water (4.12 J.g-1.K-1))
ΔT is the change in temperature in K.
PRACTICAL 1.5.2
Method
1 Warning: Wear safety glasses!
2 Transfer 5 mL of 1 mol.L–1 acid solution (eg HCl) into each of three clean, dry test tubes.
3 Record the temperatures of each starting solution.
4 Add 5 mL of 1 mol.L–1 basic solution (eg NaOH) into one of the test tubes containing the acid.
5 Record the highest temperature of the solution after all the base has been added.
6 Add 1g of a carbonate (eg CuCO3) into one of the test tubes containing the acid.
7 Record the highest temperature of the solution after all the base has been added.
8 Transfer 5 mL of 1 mol.L–1 acid solution (a different acid to step 1) into each of three clean, dry test tubes.
9 Record the temperatures of each starting solution.
10 Add 5 mL of 1 mol.L–1 basic solution (eg NaOH) into one of the test tubes containing the acid.
11 Record the highest temperature of the solution after all the base has been added.
12 Add 1g of a carbonate (eg CuCO3) into one of the test tubes containing the acid.
13 Record the highest temperature of the solution after all the base has been added.
Adapted from (Pearson Education Australia, 2019)
PRACTICAL 1.5.3
The neutralisation reaction between the acid and base will take place inside a calorimeter (copper or aluminium can, or polystyrene cup in a school).
Procedure:
Step 1: Measure the mass of the calorimeter (or weight of calorimeter for the physics students :P).
Step 2: Set up calorimeter with temperature probe.
Step 3: Measure 50ml of diluted 1M HCl and 50ml of diluted 1M NaOH solutions using a measuring cylinder.
Step 4: Add the HCl into the calorimeter, ensuring that temperature probe is submerged in the acid solution.
Step 5: Record temperature for three minutes (there should be minimal change in temperature).
Step 6: Pour, all at once, the NaOH solution into the calorimeter, containing the HCl solution.
Step 7: Record temperature for three minutes (you should notice a spike in temperature the moment you add the NaOH as exothermic neutralisation reaction occurs).
Step 8: Calculate the molar heat of neutralisation (ΔHmol). Since this is molar change in enthalpy, the units will be in kJ/mol.
The question may ask you to calculate the molar heat of neutralisation. In that case, you need to divide ΔH by the moles of water. This is because standard enthalpy of neutralisation is defined by acid and base reacting to produce one mole of water. Hence, the change in standard enthalpy for neutralisation value will based on one mole of water.
To calculate the moles of water released, you need to identify the limiting reagent (acid or base) and apply the mole ratio.
All neutralisation reactions are exothermic and so will give a negative change in enthalpy, ΔH.
NOTE: If the question asks you to determine the change in enthalpy of HCl (acid), you need to divide ΔH by the number of moles of HCl. Vice versa, if the question asks you to determine the change in enthalpy for the base, you need to divide ΔH by the number moles of the base.
NOTE: Since the acid was added into the calorimeter initially and then the base was added later, a question based off the procedure above will only ask you about the change in enthalpy for the acid. This is because you know the initial temperature of the acid as you submerged the temperature probe as per step 4. You do not know the initial temperature of the base (it was not part of the procedure). Due to this, you cannot calculate the molar change in enthalpy for the base. IF, however, you want to calculate the molar enthalpy for the base, you will need to perform an extra step. That is, you need measure and record initial temperature of the base like you did with the acid.
NOTE: The value of the change in enthalpy of reaction is equal to the value of the change in enthalpy of the calorimeter. The only difference is that they are opposite in signs, i.e. ΔHreaction = – ΔHcalorimeter.
Errors to the deviation in your experimental value of the change in enthalpy of neutralisation from actual or published value
There are many reasons the experimental value of the change in enthalpy of neutralisation deviates from the published value.
Here could be some reasons why:
The method assumes that the solution formed in the acid-base reaction consists of mainly water and so it would have the same specific heat capacity as water. In reality, the solution formed in reaction is not pure water.
The method assumes that any heat that evolved from the reaction retained in the system and did not escape from the calorimeter. In reality, convection currents can carry away some heat that was formed as a result of neutralisation. You can use a draught shield to radiate any heat escaped from the calorimeter back towards the can.
The method assumes that all the thermal energy produced as a result of the neutralisation reaction was transferred to the solution, that the calorimeter and temperature probe did not absorb any of the heat that was formed as a result of the neutralisation reaction.
TASK 1.5.2
1. (2007, Q7)
Which graph represents the enthalpy change for an acid-base neutralisation reaction? [D]
TASK 1.5.2 (Answers to Q1 and 2 follow)
Question 1: Suppose you have 100ml of 0.5M sulfuric acid and 200ml of 0.5M KOH. You have both of them reacting in a bomb calorimeter. You measured the temperature rise as 3.5 degrees Celsius. Calculate the enthalpy change of neutralisation.
Question 2:
a) Define the standard enthalpy change due to neutralisation
b) Explain why the value of enthalpy changes due to neutralisation involving any given strong acid reacting with a strong base is always about 57kJ/mol.
Question 3: What does the value of standard enthalpy of the following reactions mean?
Question 4: You have a reaction between 200mL of 1.5M HCl and 200mL of 2M NaOH occurring inside a calorimeter. You recorded that the change in temperature is 14.2 degrees celsius. You also know that the specific heat capacity of the calorimeter to be 30 J/K. Calculate the molar enthalpy change due to the neutralisation reaction.
Question 5: Provide possible reasons why an experimental value of the molar enthalpy change of strong acid and strong base neutralisation reaction can be different from the published value (hint: experimental design issues, limits of accuracy of equipment used, etc).
Question 6: Explain the advantages of two named neutralisation reactions used in everyday life.
Question 7: Explain the advantages of two named neutralisation reactions used in industrial processes.
Question 8: Describe the experimental procedure which you use to determine the change in enthalpy due to neutralisation between HCl and NaOH.
Question 9: What is an indicator? Describe how would you prepare a natural indicator.
Question 10: Explain the factors you would use to assess the accuracy AND effectiveness of an indicator? HINT: Typically an indicator’s accuracy refers to pH range over which it changes colour.
Question 11: Construct models to represent the theories of acids and bases proposed by Arrhenius and Bronsted-Lowry.
HINT: Use a chemical reaction that is relevant to the scientist’s theory of acids and bases and write out the appropriate chemical equation for that reaction you have selected. Use arrows to illustrate your point such as using arrows to show how the an acid molecule in your chemical equation is donating a proton and a base molecule accepting a proton.
True/False: Neutralisation reactions have positive molar change in enthalpy which causes the temperature of the salt solution in the calorimeter to increase.
ANSWERS
Q1
H2SO4 + 2KOH gives 2H2O + K2SO4
Mass of solution = 50.0 mL soln ×1.00g/1mL soln = 50.0 g soln
q = m c delta T
= (100 + 200 g) x 4.18 /1000 x 3.5
= 4.389 kJ
delta H = - q / n
n = 0.5 x 100/1000 divided by 2 (since 1 mol H2SO4 produces 2 moles water, and we need 1 mol water for delta H: kJ per mol) = 0.025
delta H = 4.389 / 0.025 = - 43.9 kJ
1.6 a) explore the changes in definitions and models of an acid and a base over time to explain the limitations of each model
NOTE: YOU DO NOT NEED TO LEARN THESE TWO
Because of the prevalence and importance of acids, they have been used and studied for hundreds of years. Over time, the definitions of acid and base have been refined.
Antoine Lavoisier | Oxygen Theory of Acids
Lavoisier was the first scientist to provide a definition for an acid.
His definition of an acid, based on his experimental observations, proposed that non-metal compounds that contain one or more oxygen atoms act as acids when dissolved in water.
This is true, as most non-metal oxides will produce acidic solutions when dissolved in water, eg sulfur oxide, carbon dioxide, nitrogen dioxide.
Lavoisier proposed the oxygen atom as the reason non-metal oxides produced acidic solutions when dissolved in water. He in fact was the scientist who came up with the name ‘oxygen’.
Limitations of Lavoisier’s theory:
Humphry Davy | Hydrogen Theory of Acids
Humphry Davy performed electrolysis on a solution containing HCl and found that hydrogen and chlorine gas were produced. (Davy did not know the acidic solution contained hydrochloric acid. He just knew that the solution was acidic.) Because of Lavoisier’s definition of acidic solutions, Davy expected oxygen gas to appear. However, only hydrogen and chlorine gas were produced.
There were other experiments performed around the world at this time. One of these included electrolysing acidic solutions such as HCN, which also did not produce oxygen gas.
Davy theorised that acids are substances that contain hydrogen atoms, which gave the acid its acidic properties.
Limitations of Davy’s theory:
As the field of chemistry developed, definitions of acids began to expand into conceptual definitions - definitions that attempt to explain observations. One of these was Svante Arrhenius.
Svante Arrhenius’ (1884) definition of acids was not based on direct observation, as were the ideas of Lavoisier and Davy. It was theoretical, based on the particulate behaviour of acids and bases in water.
1.6 a) explore the changes in definitions and models of an acid and a base over time to explain the limitations of each model
YOU DO NEED TO LEARN THIS ONE
Svante Arrhenius | Hydrogen Ion Theory of Acids
Arrhenius suggested that acids were substances that ionised in solution to produce hydrogen ions.
Arrhenius proposed that all acids are substances that dissociate to produce hydrogen ions in aqueous solution. The hydrogen ion dissociating from the acid molecule gave the substance its acidic properties.
Unlike Lavoisier’s and Davy’s theories, Arrhenius’s theory of acids accounts for the relative strength between different acids.
A substance which provides H+ ions in aqueous solutions (or which ionises water producing hydronium ions H3O+) is termed an Arrhenius acid. The excess [H+] gives the solution its ‘acidic’ properties.
eg. H2SO4(aq) → 2H+(aq) + SO42-(aq)
A substance which produces hydroxide ions in aqueous solutions is called an Arrhenius base. The excess [OH-] gives the solution its ‘basic’ properties.
eg. Ca(OH)2(aq) → Ca2+(aq) + 2OH-(aq)
Arrhenius also suggested that when the acid and base react to form salt and water, it is the H+ ions in the acid which neutralise the OH- ions in the base to form H2O.
Neutralisation: H+(aq) + OH-(aq) → H2O(l)
The salts produced in neutralisation reactions are ionic compounds formed from the union of the cation from the base and the anion from the acid. Salts are classified according to the nature of the parent acid (eg sulfates come from sulfuric acid).
Arrhenius proposed that all bases are substances that dissociate to produce hydroxide ions in aqueous solution. The hydroxide ions in solution gave the substance its basic properties.
He stated that the strength of a base is dependent on the degree of dissociation, where strong bases completely dissociate into their constituent ions when dissolved in water.
Arrhenius’s definitions were able to account for why neutralisation occurred when acids and base solutions mix together. He attributed the neutralisation reaction to the reaction between hydroxide ions with hydrogen ions to produce water as a product in solution. Since water is neutral (pH = 7), neutralisation reactions move the overall pH of the solution towards 7.
Limitations of Arrhenius’s theory:
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1.6 b) 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:
YOU DO NEED TO LEARN THIS ONE
Bronsted - Lowry | Proton Donor Theory of Acids
Whilst the Arrhenius definition worked for many acid base reactions, especially those involving common acids and hydroxides, it did not explain them all.
Problem 1: ZnCO3(s) + 2H+ → Zn2+ + CO2(g) + H2O(l)
The zinc carbonate neutralised the acid (producing water) and so could be classified as a base, yet the zinc carbonate does not produce hydroxide ions in aqueous solution, so does not meet the Arrhenius definition.
Problem 2: HCl (aq) + NH3 (aq) → NH4+(aq) +Cl-(aq)
The ammonia NH3 is a base, but does not contain hydroxide ions.
Lowry (English) and Brönsted (Danish) independently proposed a better explanation, based on proton transfer, for acids and bases:
The Brønsted-Lowry theory of acids and bases took the Arrhenius definition one step further, as a substance no longer needed to be composed of hydrogen (H+) or hydroxide (OH-) ions in order to be classified as an acid or base.
HCl (aq) + NH3 (aq) → NH4+(aq) +Cl-(aq)
Hydrochloric acid (HCl) "donates" a proton (H+) to ammonia (NH3) which "accepts" it , forming a positively charged ammonium ion (NH4+) and a negatively charged chloride ion (Cl-). Therefore, HCl is a Brønsted-Lowry acid (donates a proton) while ammonia is a Brønsted-Lowry base (accepts a proton).
HA(aq) ⇌A-(aq) + H+(aq)
Ka = [A-] [H+] / [HA]
B-(aq) + H2O(l) ⇌ HB+(aq) + OH-(aq)
F-(aq) + H2O(l) ⇌ HF(aq) + OH-
Kb = [HB+] [OH-] / [B]
Limitations of Bronsted-Lowry Theory:
Another theory, developed at around the same time as the Bronsted-Lowry theory was proposed by Lewis. This accommodates the drawbacks of Bronsted-Lowry theories, but is not used as often. Bronsted-Lowry is the more usable, Lewis is used for special cases.
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TASK 1.6.1
1. Write a chemical formula and/or equation to demonstrate an example of the behaviour of a base according to each theory below:
a) Arrhenius definition of a base
b) Brønsted-Lowry definition of a base
2. (2008, Q8) According to the Arrhenius theory of acids and bases, an acid is a substance that
(A) tastes sour.
(B) is capable of donating a hydrogen ion.
(C) can accept a pair of electrons to form a co-ordinate covalent bond.
(D) increases the concentration of hydrogen ions in an aqueous solution.
3. (2006, Q11) In 1884, Svante Arrhenius proposed a definition for acids. His definition was soon accepted as superior to that put forward by earlier chemists.
Why was Arrhenius’ definition seen as a major improvement?
(A) It explained why some acids do not contain oxygen.
(B) It showed how the solvent can affect the strength of an acid.
(C) It showed the relationship between pH and the concentration of H+ ions.
(D) It could be used to explain why some acids are strong and others are weak.
4. (2004, Q5) Which statement best represents Davy’s definition of an acid?
(A) Acids contain oxygen.
(B) Acids are proton donors.
(C) Acids contain replaceable hydrogen.
(D) Acids ionise in solution to form hydrogen ions.
5. (2010, Q8)
In a research report a student wrote, ‘Acids are compounds that contain hydrogen and can dissolve in water to release hydrogen ions into solution.’
Who originally stated this theory of acids?
(A) Arrhenius
(B) Brönsted–Lowry
(C) Davy
(D) Lavoisier
6. (2005, Q21, 5 marks)
Analyse how knowledge of the composition and properties of acids has led to changes in the definition of acids.
ANS:
1.
a) HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)
b) HCl (aq) + NH3 (aq) → NH4+(aq) +Cl-(aq)
2. B 3. D 4. C 5. A
6.
• Describes the correct chronological development of theory of acids • Draws out the implications of each development • Provides at least three definitions of acids • Indicates limitations of earlier definitions 4–5
Better responses showed a clear understanding of the chronological development of knowledge of acid-base theory and drew out and related the implications of each development. They used precise terms with respect to ions, atoms and molecules
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REVIEW 1.6
PAST HSC QS (QS WITH NO MARKS INDICATED ARE MULTIPLE CHOICE. HSC EXAMS IN EARLIER YEARS HAD 15 MC QS, CURRENTLY THERE ARE 20.)
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Complete Text WS 6.1 Knowledge Review PP 61-2