Chemistry Form 3 Notes : CHEMISTRY OF SULPHUR

Chemistry of Sulphur

A. Sulphur (S)

Sulphur is an element in Group VI (Group 16) of the Periodic Table. It has atomic number 16 and electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁴. It has valency 2 (divalent) and thus forms the ion S^2-.

A. Occurrence

Sulphur mainly occurs:

• As a free element in Texas and Louisiana in the USA and Sicily in Italy.
• As hydrogen sulphide gas in active volcanic areas, e.g., Olkaria near Naivasha in Kenya.
• As copper pyrites (CuFeS₂), galena (PbS), zinc blende (ZnS), and iron pyrites (FeS₂) in other parts of the world.

B. Extraction of Sulphur by Frasch Process

Sulphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in. Digging tunnels is thus discouraged in trying to extract the mineral.

Sulphur is extracted by drilling three concentric round pipes of diameters in the ratio 2:8:18 centimeters.

Superheated water at 170°C and 10 atmospheres pressure is forced through the outermost pipe. The high pressure ensures the water remains liquid at high temperatures instead of vapor/gas.

The superheated water melts the sulphur because the melting point of sulphur is about 115°C.

Compressed air at 15 atmospheres is forced/pumped through the innermost pipe. The hot air forces the molten sulphur up the middle pipe where it is collected and solidifies in a large tank. It is about 99% pure.

Diagram showing extraction of Sulphur from Frasch Process:

C. Allotropes of Sulphur

1. Sulphur exists as two crystalline allotropic forms:

• Rhombic sulphur
• Monoclinic sulphur

Rhombic sulphur and Monoclinic sulphur have a transition temperature of 96°C. This is the temperature at which one allotrope changes to the other.

2. Sulphur exists in non-crystalline forms as:

• Plastic sulphur – prepared by heating powdered sulphur to boiling, then pouring a thin continuous stream into a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed. If left for long, it turns to bright yellow crystalline rhombic sulphur.
• Colloidal sulphur – formed when sodium thiosulphate (Na₂S₂O₃) is added to hydrochloric acid to form a yellow precipitate.

D. Heating Sulphur

A molecule of sulphur exists as a puckered ring of eight atoms joined by covalent bonds as S₈.

On heating, yellow sulphur powder melts at 113°C to a clear amber liquid with low viscosity and flows easily.

On further heating to 160°C, the molten liquid darkens to a brown very viscous liquid that does not flow easily. This is because the S₈ rings break into S₈ chains that join together to form very long chains made of over 100,000 atoms of sulphur.

The long chains entangle each other, reducing their mobility/flow and hence increasing viscosity.

On continued heating above 160°C, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous. This is because the long chains break into smaller/shorter chains.

At 444°C, the liquid boils and forms brown vapour of a mixture of S₈, S₆, S₂ molecules that solidify to S₈ rings of “flowers of sulphur” on cooler parts.

Summary of changes on heating sulphur

E. Physical and Chemical Properties of Sulphur (Questions)

1. State three physical properties unique to Sulphur

Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide, tetrachloromethane, and benzene. It is a poor conductor of heat and electricity. It has a melting point of 115°C and a boiling point of 444°C.

2. Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made.

Observations:

• Sulphur melts then burns with a blue flame.
• Colourless gas produced that has a pungent smell.
• Red litmus paper remains red. Blue litmus paper turns red.

Explanation:

Sulphur burns in air and faster in oxygen to form Sulphur(IV) Oxide gas and traces/small amounts of Sulphur(VI) Oxide gas. Both oxides react with water to form the corresponding acidic solutions:

• Sulphur(IV) Oxide gas reacts with water to form sulphurous acid (H₂SO₃).
• Sulphur(VI) Oxide gas reacts with water to form sulphuric acid (H₂SO₄).

Chemical equations:

S(s) + O₂(g) → SO₂(g) (Sulphur(IV) Oxide gas)

2S(s) + 3O₂(g) → 2SO₃(g) (Sulphur(VI) Oxide gas traces)

SO₂(g) + H₂O(l) → H₂SO₃(aq) (sulphurous acid)

SO₃(g) + H₂O(l) → H₂SO₄(aq) (sulphuric acid)

3. Iron filings were put in a test tube containing powdered sulphur then heated on a Bunsen flame. Stop heating when reaction starts. State and explain the observations made. Test the effects of a magnet on the mixture before and after heating. Explain.

Observations:

• Before heating, the magnet attracts iron filings leaving sulphur.
• After heating, the magnet does not attract the mixture.
• After heating, a red glow is observed that continues even when heating is stopped.
• Black solid is formed.

Explanation:

Iron is attracted to a magnet because it is ferromagnetic. When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture continue glowing even after stopping heating. Black Iron(II) sulphide is formed which is a compound and thus not ferromagnetic.

Chemical equation: Fe(s) + S(s) → FeS(s) (Exothermic reaction, -∆H)

Heated powdered heavy metals combine with sulphur to form black sulphides:

• Cu(s) + S(s) → CuS(s)
• Zn(s) + S(s) → ZnS(s)
• Pb(s) + S(s) → PbS(s)

4. The setup below shows the reaction of sulphur on heated concentrated sulphuric(VI) acid.

(i) State and explain the observation made.

Observation:

• Yellow colour of sulphur fades.
• Orange colour of potassium dichromate(VI) paper turns green.

Explanation:

Hot concentrated sulphuric(VI) acid oxidizes sulphur to sulphur (IV) oxide gas. The oxide is also reduced to water. Traces of sulphur (VI) oxide are formed.

Chemical equation:

S(s) + 3H₂SO₄(l) → 3SO₂(g) + 3H₂O(l) + SO₃(g)

Sulphur (IV) oxide gas turns orange potassium dichromate(VI) paper to green.

(ii) State and explain the observation made if concentrated sulphuric (VI) acid is replaced with concentrated nitric (V) acid in the above setup.

Observation:

• Yellow colour of sulphur fades.
• Colourless solution formed.
• Brown fumes/gas produced.

Explanation:

Hot concentrated nitric(V) acid oxidizes sulphur to sulphuric (VI) acid. The nitric (V) acid is reduced to brown nitrogen(IV) oxide gas.

Chemical equation:

S(s) + 6HNO₃(l) → 6NO₂(g) + 2H₂O(l) + H₂SO₄(l)

Note: Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur like the other mineral acids.

5. State three main uses of sulphur.

Sulphur is mainly used in:

• Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI) acid.
• Vulcanization of rubber to make it harder, tougher, stronger, and more durable.
• Making gunpowder and matchstick heads.
• As ointments to treat fungal infections.

6. Revision Practice

The diagram below represents the extraction of sulphur by Frasch process. Use it to answer the questions that follow.

(a) Name the substances that pass through:

• M – Superheated water at 170°C and 10 atmosphere pressure
• L – Hot compressed air
• N – Molten sulphur

(b) What is the purpose of the substances that pass through L and M?

• M – Superheated water at 170°C and 10 atmosphere pressure is used to melt the sulphur.
• L – Hot compressed air is used to force up the molten sulphur.

(c) The properties of the two main allotropes of sulphur represented by letters A and B are given in the table below. Use it to answer the questions that follow.

I. What are allotropes?

Different forms of the same element existing at the same temperature and pressure without change of state.

II. Identify allotrope:

1. Monoclinic sulphur – A

B – Rhombic sulphur

III. State two main uses of sulphur.

• Manufacture of sulphuric(VI) acid.
• As fungicide.
• In vulcanization of rubber to make it harder, tougher, stronger.
• Manufacture of dyes/fibres.

(d) Calculate the volume of sulphur (IV) oxide produced when 0.4 g of sulphur is completely burnt in excess air. (S = 32.0, 1 mole of a gas occupies 24 dm³ at room temperature)

Chemical equation:

S(s) + O₂(g) → SO₂(g)

Mole ratio S : SO₂ = 1 : 1

Method 1

32.0 g of sulphur → 24 dm³ of SO₂(g)

0.4 g of sulphur → 0.4 × 24 / 32.0 = 0.3 dm³

Method 2

Moles of sulphur used = Mass of sulphur / Molar mass = 0.4 / 32 = 0.0125 moles

Moles of sulphur used = Moles of sulphur(IV) oxide produced = 0.0125 moles

Volume of sulphur(IV) oxide produced = Moles × volume of one mole of gas = 0.0125 × 24 = 0.3 dm³

B. Compounds of Sulphur

The following are the main compounds of sulphur:

• Sulphur(IV) oxide
• Sulphur(VI) oxide
• Sulphuric(VI) acid
• Hydrogen sulphide
• Sulphite (IV) / SO₃^2- and Sulphate (VI) / SO₄^2- salts

(i) Sulphur(IV) oxide (SO₂)

(a) Occurrence

Sulphur(IV) oxide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world, e.g., Olkaria and Hells Gate near Naivasha in Kenya.

(b) School laboratory preparation

In a Chemistry school laboratory, Sulphur(IV) oxide is prepared from the reaction of:

Method 1: Using Copper and Sulphuric(VI) acid

Method 2: Using Sodium Sulphite and Hydrochloric acid

(c) Properties of Sulphur(IV) oxide (Questions)

1. Write the equations for the reaction for the formation of sulphur (IV) oxide using:

(i) Method 1

Cu(s) + 2H₂SO₄(l) → CuSO₄(aq) + SO₂(g) + 2H₂O(l)

Zn(s) + 2H₂SO₄(l) → ZnSO₄(aq) + SO₂(g) + 2H₂O(l)

Mg(s) + 2H₂SO₄(l) → MgSO₄(aq) + SO₂(g) + 2H₂O(l)

Fe(s) + 2H₂SO₄(l) → FeSO₄(aq) + SO₂(g) + 2H₂O(l)

Calcium, Lead, and Barium will form insoluble sulphate(VI) salts that will cover unreacted metals stopping further reaction, thus producing very small amounts of sulphur (IV) oxide gas.

(ii) Method 2

Na₂SO₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + H₂O(l)

K₂SO₃(aq) + 2HCl(aq) → 2KCl(aq) + SO₂(g) + H₂O(l)

BaSO₃(s) + 2HCl(aq) → BaCl₂(aq) + SO₂(g) + H₂O(l)

CaSO₃(s) + 2HCl(aq) → CaCl₂(aq) + SO₂(g) + H₂O(l)

PbSO₃(s) + 2HCl(aq) → PbCl₂(s) + SO₂(g) + H₂O(l)

Lead(II) chloride is soluble on heating; thus, reactants should be heated to prevent it coating/covering unreacted PbSO₃(s).

2. State the physical properties unique to sulphur (IV) oxide gas.

Sulphur (IV) oxide gas is a colourless gas with a pungent, irritating, and choking smell which liquefies easily. It is about two times denser than air.

3. The diagram below shows the solubility of sulphur (IV) oxide gas. Explain.

Sulphur(IV) oxide is very soluble in water. One drop of water dissolves all the sulphur (IV) oxide in the flask leaving a vacuum. If the clip is removed, atmospheric pressure forces the water up through the narrow tube to form a fountain to occupy the vacuum.

An acidic solution of sulphurous acid (H₂SO₃) is formed which turns litmus solution red.

Chemical equation:

SO₂(g) + H₂O(l) → H₂SO₃(aq) (sulphurous acid turns litmus red)

4. Dry litmus papers and wet/damp/moist litmus papers were put in a gas jar containing sulphur(IV) oxide gas. State and explain the observations made.

Observations:

• Dry blue litmus paper remains blue.
• Dry red litmus paper remains red.
• Wet/damp/moist blue litmus paper turns red.
• Moist/damp/wet red litmus paper remains red.
• Both litmus papers are then bleached/decolorized.

Explanation:

Dry sulphur(IV) oxide gas is a molecular compound that does not dissociate/ionize to release H⁺ ions and thus has no effect on dry blue/red litmus papers.

Wet/damp/moist litmus papers contain water that dissolves/reacts with dry sulphur(IV) oxide gas to form a solution of weak sulphurous acid (H₂SO₃).

Weak sulphurous acid dissociates/ionizes into free H⁺ ions:

H₂SO₃(aq) → 2H⁺(aq) + SO₃^2-(aq)

The free H⁺ ions are responsible for turning blue litmus paper red, showing the gas is acidic.

The SO₃^2- ions in wet/damp/moist sulphur(IV) oxide gas are responsible for many reactions of the gas.

It is easily oxidized to sulphate(VI) SO₄^2- ions, making sulphur(IV) oxide gas act as a reducing agent as in the following examples:

(a) Bleaching agent

Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas. This is because sulphur(IV) oxide removes atomic oxygen from the coloured dye/material to form sulphuric(VI) acid.

Chemical equations:

(i) Formation of sulphurous acid:

SO₂(g) + H₂O(l) → H₂SO₃(aq)

(ii) Decolorization/bleaching of the dye/removal of atomic oxygen:

Method I: H₂SO₃(aq) + (dye + O) → H₂SO₄(aq) + dye (coloured → colourless)

Method II: H₂SO₃(aq) + (dye) → H₂SO₄(aq) + (dye – O) (coloured → colourless)

Sulphur(IV) oxide gas therefore bleaches by reduction/removing oxygen from a dye, unlike chlorine that bleaches by oxidation/adding oxygen.

The bleaching by removing oxygen from sulphur(IV) oxide gas is temporary because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst, thus regaining/restoring its original colour. For example, old newspapers turn brown on exposure to air on regaining the atomic oxygen.

The bleaching through adding oxygen by chlorine gas is permanent.

(b) Turns orange acidified potassium dichromate(VI) to green

Experiment:

• Pass a stream of sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution; or
• Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing sulphur(IV) oxide gas.

Observation: Orange acidified potassium dichromate(VI) turns to green.

Explanation: Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr₂O₇^2- ions to green Cr³⁺ ions without leaving a residue, itself oxidized from SO₃^2- ions in sulphurous acid to SO₄^2- ions in sulphuric acid.

Chemical/ionic equations:

(i) Reaction of sulphur(IV) oxide gas with water:

SO₂(g) + H₂O(l) → H₂SO₃(aq)

(ii) Dissociation/ionization of sulphurous acid:

H₂SO₃(aq) → 2H⁺(aq) + SO₃^2-(aq)

(iii) Oxidation of SO₃^2- and reduction of Cr₂O₇^2-:

3SO₃^2-(aq) + Cr₂O₇^2-(aq) + 8H⁺(aq) → 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

This is a confirmatory test for the presence of sulphur(IV) oxide gas.

Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr₂O₇^2- ions to green Cr³⁺ ions leaving a yellow residue.

(c) Decolorizes acidified potassium manganate(VII)

Experiment:

• Pass a stream of sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution; or
• Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing sulphur(IV) oxide gas.

Observation: Purple acidified potassium manganate(VII) turns colourless/acidified potassium manganate(VII) is decolorized.

Explanation: Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO₄^– ions to green Mn²⁺ ions without leaving a residue, itself oxidized from SO₃^2- ions in sulphurous acid to SO₄^2- ions in sulphuric acid.

Chemical/ionic equations:

(i) Reaction of sulphur(IV) oxide gas with water:

SO₂(g) + H₂O(l) → H₂SO₃(aq)

(ii) Dissociation/ionization of sulphurous acid:

H₂SO₃(aq) → 2H⁺(aq) + SO₃^2-(aq)

(iii) Oxidation of SO₃^2- and reduction of MnO₄^–:

5SO₃^2-(aq) + 2MnO₄^–(aq) + 6H⁺(aq) → 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3H₂O(l)

This is another test for the presence of sulphur(IV) oxide gas.

Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO₄^– ions to colourless Mn²⁺ ions leaving a yellow residue.

(d) Decolorizes bromine water

Experiment:

• Pass a stream of sulphur(IV) oxide gas in a test tube containing bromine water; or
• Put three drops of bromine water into a gas jar containing sulphur(IV) oxide gas. Swirl.

Observation: Yellow bromine water turns colourless/bromine water is decolorized.

Explanation: Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue, itself oxidized from SO₃^2- ions in sulphurous acid to SO₄^2- ions in sulphuric acid.

Chemical/ionic equation:

(i) Reaction of sulphur(IV) oxide gas with water:

SO₂(g) + H₂O(l) → H₂SO₃(aq)

(ii) Dissociation/ionization of sulphurous acid:

H₂SO₃(aq) → 2H⁺(aq) + SO₃^2-(aq)

(iii) Oxidation of SO₃^2- and reduction of Br₂:

SO₃^2-(aq) + Br₂(aq) + H₂O(l) → SO₄^2-(aq) + 2HBr(aq)

This can also be used as another test for the presence of sulphur(IV) oxide gas.

Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue.

(e) Reduces Iron(III) Fe³⁺ salts to Iron(II) Fe²⁺

Experiment:

• Pass a stream of sulphur(IV) oxide gas in a test tube containing about 3 cm³ of Iron (III) chloride solution; or
• Place about 3 cm³ of Iron (III) chloride solution into a gas jar containing sulphur(IV) oxide gas. Swirl.

Observation: Yellow/brown Iron (III) chloride solution turns green.

Explanation: Sulphur(IV) oxide gas reduces Iron (III) chloride solution from yellow/brown Fe³⁺ ions to green Fe²⁺ ions without leaving a residue, itself oxidized from SO₃^2- ions in sulphurous acid to SO₄^2- ions in sulphuric acid.

Chemical/ionic equations:

(i) Reaction of sulphur(IV) oxide gas with water:

SO₂(g) + H₂O(l) → H₂SO₃(aq)

(ii) Dissociation/ionization of sulphurous acid:

H₂SO₃(aq) → 2H⁺(aq) + SO₃^2-(aq)

(iii) Oxidation of SO₃^2- and reduction of Fe³⁺:

SO₃^2-(aq) + 2Fe³⁺(aq) + 3H₂O(l) → SO₄^2-(aq) + 2Fe²⁺(aq) + 2H⁺(aq)

(f) Reduces Nitric(V) acid to Nitrogen(IV) oxide gas

Experiment:

• Pass a stream of sulphur(IV) oxide gas in a test tube containing about 3 cm³ of concentrated nitric(V) acid; or
• Place about 3 cm³ of concentrated nitric(V) acid into a gas jar containing sulphur(IV) oxide gas. Swirl.

Observation: Brown fumes of a gas evolved/produced.

Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V) acid to brown nitrogen(IV) oxide gas, itself oxidized from SO₃^2- ions in sulphurous acid to SO₄^2- ions in sulphuric acid.

Chemical/ionic equation:

SO₂(g) + 2HNO₃(l) → H₂SO₄(l) + NO₂(g)

(g) Reduces Hydrogen peroxide to water

Experiment:

• Pass a stream of sulphur(IV) oxide gas in a test tube containing about 3 cm³ of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V) or Barium chloride followed by five drops of 2M hydrochloric acid/2M nitric(V) acid.

Observation: A white precipitate is formed that persists/remains on adding 2M hydrochloric acid/2M nitric(V) acid.

Explanation: Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself oxidized from SO₃^2- ions in sulphurous acid to SO₄^2- ions in sulphuric acid.

When Ba²⁺ ions in Barium nitrate(V) or Barium chloride solution are added, a white precipitate of insoluble Barium salts is formed showing the presence of either SO₃^2-, SO₄^2-, or CO₃^2- ions.

Chemical/ionic equations:

SO₃^2-(aq) + Ba²⁺(aq) → BaSO₃(s) (white precipitate)

SO₄^2-(aq) + Ba²⁺(aq) → BaSO₄(s) (white precipitate)

CO₃^2-(aq) + Ba²⁺(aq) → BaCO₃(s) (white precipitate)

If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate:

• Persists/remains if SO₄^2- ions (BaSO₄(s)) are present.
• Dissolves if SO₃^2- ions (BaSO₃(s)) and CO₃^2- ions (BaCO₃(s)) are present. This is because:

I. BaSO₃(s) reacts with nitric(V)/hydrochloric acid to produce acidic SO₂ gas that turns orange moist filter paper dipped in acidified potassium dichromate to green.

Chemical equation:

BaSO₃(s) + 2H⁺(aq) → Ba²⁺(aq) + SO₂(g) + H₂O(l)

II. BaCO₃(s) reacts with nitric(V)/hydrochloric acid to produce acidic CO₂ gas that forms a white precipitate when bubbled in lime water.

Chemical equation:

BaCO₃(s) + 2H⁺(aq) → Ba²⁺(aq) + CO₂(g) + H₂O(l)

5. Sulphur(IV) oxide also acts as an oxidizing agent as in the following examples.

(a) Reduction by burning Magnesium

Experiment: Lower a burning magnesium ribbon into a gas jar containing sulphur(IV) oxide gas.

Observation: Magnesium ribbon continues to burn with difficulty. White ash and yellow powder/specks form.

Explanation: Sulphur(IV) oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose sulphur(IV) oxide to sulphur and oxygen. The metal continues to burn on oxygen forming white magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of the reaction flask/gas jar. During the reaction, sulphur(IV) oxide is reduced (oxidizing agent) while the metal is oxidized (reducing agent).

Chemical equation:

SO₂(g) + 2Mg(s) → 2MgO(s) + S(s)

(white ash/solid) (yellow speck/powder)

(b) Reduction by Hydrogen sulphide gas

Experiment: Put two drops of water into a gas jar containing dry sulphur(IV) oxide gas. Bubble hydrogen sulphide gas into the gas jar containing sulphur(IV) oxide gas. Or invert a gas jar full of hydrogen sulphide gas over the gas jar containing sulphur(IV) oxide gas. Swirl.

Observation: Yellow powder/specks form.

Explanation: Sulphur(IV) oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and is itself reduced to sulphur that forms on the sides of the reaction flask/gas jar. A little moisture/water acts as a catalyst/speeds up the reaction.

Chemical equation:

SO₂(g) + 2H₂S(g) → 2H₂O(l) + 3S(s)

(yellow speck/powder)

6. Sulphur(IV) oxide has many industrial uses. State three.

• In the contact process for the manufacture of sulphuric(VI) acid.
• As a bleaching agent of pulp and paper.
• As a fungicide to kill microbes.
• As a preservative of jam, juices to prevent fermentation.

(ii) Sulphur(VI) oxide (SO₃)

(a) Occurrence

Sulphur(VI) oxide does not occur free in nature/atmosphere.

(b) Preparation

In a Chemistry school laboratory, sulphur(VI) oxide may be prepared from:

Method 1: Catalytic oxidation of sulphur(IV) oxide gas

Sulphur(IV) oxide gas and oxygen mixture are first dried by being passed through concentrated sulphuric(VI) acid. The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form sulphur(VI) oxide gas.

Sulphur(VI) oxide gas readily solidifies as silky white needles if passed through a freezing mixture/ice cold water. The solid fumes out on heating to a highly acidic poisonous gas.

Chemical equation:

2SO₂(g) + O₂(g) –platinised asbestos–> 2SO₃(g)

Method 2: Heating Iron(II) sulphate(VI) heptahydrate

When green hydrated Iron(II) sulphate(VI) heptahydrate crystals are heated in a boiling tube, it loses the water of crystallization and colour changes from green to white.

Chemical equation:

FeSO₄.7H₂O(s) → FeSO₄(s) + 7H₂O(l)

(green solid) (white solid)

On further heating, the white anhydrous Iron(II) sulphate(VI) solid decomposes to a mixture of sulphur(VI) oxide and sulphur(IV) oxide gas. Sulphur(VI) oxide readily solidifies as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III) oxide is left as a brown residue/solid.

Chemical equation:

2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃(g)

(green solid) (brown solid)

Caution: On exposure to air, sulphur(VI) oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI) acid and thus its preparation in a school laboratory is very risky.

(c) Uses of sulphur(VI) oxide

One of the main uses of sulphur(VI) oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale production of sulphuric(VI) acid.

(iii) Sulphuric(VI) acid (H₂SO₄)

(a) Occurrence

Sulphuric(VI) acid is one of the three mineral acids. The three mineral acids are:

• Nitric(V) acid
• Sulphuric(VI) acid
• Hydrochloric acid

Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.

(b) The Contact process for industrial manufacture of H₂SO₄

I. Raw materials

The main raw materials for industrial preparation of sulphuric(VI) acid include:

• Sulphur from Frasch process or from heating metal sulphide ores like Galena (PbS), Zinc blende (ZnS).
• Oxygen from fractional distillation of air.
• Water from rivers/lakes.

II. Chemical processes

The contact process involves four main chemical processes:

(i) Production of Sulphur(IV) oxide

As one of the raw materials, sulphur(IV) oxide gas is obtained from the following sources:

• Burning/roasting sulphur in air.
• Burning/roasting sulphide ores in air.

Sulphur from Frasch process is roasted/burnt in air to form sulphur(IV) oxide gas in the burners.

Chemical equation:

S(s) + O₂(g) → SO₂(g)

Sulphur(IV) oxide gas is produced as a by-product in extraction of some metals like:

• Lead from Lead(II) sulphide/Galena (PbS)
• Zinc from Zinc(II) sulphide/Zinc blende (ZnS)
• Copper from Copper iron sulphide/Copper pyrites (CuFeS₂)

On roasting/burning, large quantities of sulphur(IV) oxide are generated/produced.

Chemical equations:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)

2CuFeS₂(s) + 4O₂(g) → 2FeO(s) + 3SO₂(g) + Cu₂O(s)

Sulphur(IV) oxide easily liquefies and thus can be transported safely over long distances.

(ii) Purification of Sulphur(IV) oxide

Sulphur(IV) oxide gas contains dust particles and Arsenic(IV) oxide as impurities. These impurities “poison”/impair the catalyst by adhering to/covering its surface.

The impurities are removed by electrostatic precipitation method.

In the contact process, Platinum or Vanadium(V) oxide may be used. Vanadium(V) oxide is preferred because it is:

• Cheaper/less expensive.
• Less easily poisoned by impurities.

(iii) Catalytic conversion of Sulphur(IV) oxide to Sulphur(VI) oxide

Pure and dry mixture of sulphur(IV) oxide gas and oxygen is heated to 450°C in a heat exchanger. The heated mixture is passed through long pipes coated with pellets of Vanadium(V) oxide catalyst.

The close “contact” between the reacting gases and catalyst gives the process its name.

Vanadium(V) oxide catalyses the conversion/oxidation of sulphur(IV) oxide to sulphur(VI) oxide gas.

Chemical equation:

2SO₂(g) + O₂(g) –V₂O₅–> 2SO₃(g)

This reaction is exothermic (-∆H) and the temperature needs to be maintained at around 450°C to ensure that:

• The reaction rate/time taken for the formation of sulphur(VI) oxide is not too slow/long at lower temperatures below 450°C.
• Sulphur(VI) oxide gas does not decompose back to sulphur(IV) oxide gas and oxygen gas at higher temperatures than 450°C.

(iv) Conversion of Sulphur(VI) oxide to Sulphuric(VI) acid

Sulphur(VI) oxide is the acid anhydride of concentrated sulphuric(VI) acid. Sulphur(VI) oxide reacts with water to form a thick mist of fine droplets of very/highly corrosive concentrated sulphuric(VI) acid because the reaction is highly exothermic.

To prevent this, sulphur(VI) oxide is passed up to meet downward flow of 98% sulphuric(VI) acid in the absorption chamber/tower.

The reaction forms a very viscous oily liquid called Oleum/fuming sulphuric(VI) acid/pyrosulphuric(VI) acid.

Chemical equation:

H₂SO₄(aq) + SO₃(g) → H₂S₂O₇(l)

Oleum/fuming sulphuric(VI) acid/pyrosulphuric(VI) acid is diluted carefully with distilled water to give concentrated sulphuric(VI) acid.

Chemical equation:

H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(l)

The acid is stored ready for market/sale.

III. Environmental effects of contact process

Sulphur(VI) oxide and sulphur(IV) oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere.

In the contact process, about 2% of these gases do not form sulphuric(VI) acid.

The following precautions prevent/minimize pollution from the contact process:

• Recycling back any unreacted sulphur(IV) oxide gas back to the heat exchangers.
• Dissolving sulphur(VI) oxide gas in concentrated sulphuric(VI) acid instead of water. This prevents the formation of fine droplets of corrosive/toxic/poisonous fumes of concentrated sulphuric(VI) acid.
• Scrubbing – passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with sulphur(VI) oxide gas forming harmless calcium(II) sulphite (CaSO₃).

Chemical equation:

Ca(OH)₂(aq) + SO₂(g) → CaSO₃(aq) + H₂O(g)

IV. Uses of Sulphuric(VI) acid

Sulphuric(VI) acid is used:

• In making dyes and paint.
• As acid in lead-acid accumulator/battery.
• For making soapless detergents.
• For making sulphate agricultural fertilizers.

VI. Sketch chart diagram showing the Contact process

(c) Properties of Concentrated sulphuric (VI) acid

• Concentrated sulphuric (VI) acid is a colourless oily liquid with a density of 1.84 g/cm³. It has a boiling point of 338°C.
• It is very soluble in water. The solubility/dissolution of the acid is highly exothermic. The concentrated acid should thus be diluted slowly in excess water. Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container.
• Concentrated sulphuric (VI) acid is a covalent compound. It has no free H⁺ ions. Free H⁺ ions are responsible for turning blue litmus paper red. Concentrated sulphuric (VI) acid thus does not change blue litmus paper red.
• It is hygroscopic. It absorbs water from the atmosphere and does not form a solution. This makes concentrated sulphuric (VI) acid very suitable as a drying agent during preparation of gases.
• Chemical properties of concentrated sulphuric (VI) acid include:

I. As a dehydrating agent

Experiment I:

Put about four spatula end full of brown sugar and glucose in separate 10 cm³ beakers. Carefully add about 10 cm³ of concentrated sulphuric (VI) acid. Allow to stand for about 10 minutes.

Observation:

• Colour in brown sugar changes from brown to black.
• Colour in glucose changes from white to black.
• 10 cm³ beaker becomes very hot.

Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent. It removes chemically and physically combined elements of water (hydrogen and oxygen in ratio 2:1) from compounds. When added to sugar/glucose, a vigorous reaction that is highly exothermic takes place. The sugar/glucose is charred to a black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon.

Caution: This reaction is highly exothermic that starts slowly but produces fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi.

Chemical equations:

Glucose: C₆H₁₂O₆(s) –conc. H₂SO₄–> 6C(s) + 6H₂O(l)

Sugar: C₁₂H₂₂O₁₁(s) –conc. H₂SO₄–> 12C(s) + 11H₂O(l)

Experiment II:

Put about two spatula end full of hydrated copper(II) sulphate(VI) crystals in a boiling tube. Carefully add about 10 cm³ of concentrated sulphuric (VI) acid. Warm.

Observation: Colour changes from blue to white.

Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent. It removes physically combined elements of water (hydrogen and oxygen in ratio 2:1) from hydrated compounds. The acid dehydrates blue copper(II) sulphate to white anhydrous copper(II) sulphate.

Chemical equation:

CuSO₄.5H₂O(s) –conc. H₂SO₄–> CuSO₄(s) + 5H₂O(l)

Experiment III:

Put about 4 cm³ of absolute ethanol in a boiling tube. Carefully add about 10 cm³ of concentrated sulphuric (VI) acid. Place moist/damp/wet filter paper dipped in acidified potassium dichromate(VI) solution on the mouth of the boiling tube. Heat strongly.

Caution: Absolute ethanol is highly flammable.

Observation: Colourless gas produced. Orange acidified potassium dichromate (VI) paper turns green.

Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent. It removes chemically combined elements of water (hydrogen and oxygen in ratio 2:1) from compounds. The acid dehydrates ethanol to ethene gas at about 170°C. Ethene with =C=C= double bond turns orange acidified potassium dichromate (VI) paper green.

Chemical equation:

C₂H₅OH(l) –conc. H₂SO₄/170°C–> C₂H₄(g) + H₂O(l)

Note: This reaction is used for the school laboratory preparation of ethene gas.

Experiment IV:

Put about 4 cm³ of methanoic acid in a boiling tube. Carefully add about 6 cm³ of concentrated sulphuric (VI) acid. Heat gently.

Caution: This should be done in a fume chamber/open.

Observation: Colourless gas produced.

Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent. It removes chemically combined elements of water (hydrogen and oxygen in ratio 2:1) from compounds. The acid dehydrates methanoic acid to poisonous/toxic carbon(II) oxide gas.

Chemical equation:

HCOOH(l) –conc. H₂SO₄–> CO(g) + H₂O(l)

Note: This reaction is used for the school laboratory preparation of small amounts of carbon (II) oxide gas.

Experiment V:

Put about 4 cm³ of ethan-1,2-dioic/oxalic acid in a boiling tube. Carefully add about 6 cm³ of concentrated sulphuric (VI) acid. Pass any gaseous product through lime water. Heat gently.

Caution: This should be done in a fume chamber/open.

Observation: Colourless gas produced. Gas produced forms a white precipitate with lime water.

Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent. It removes chemically combined elements of water (hydrogen and oxygen in ratio 2:1) from compounds. The acid dehydrates ethan-1,2-dioic/oxalic acid to a mixture of poisonous/toxic carbon(II) oxide and carbon(IV) oxide gases.

Chemical equation:

HOOCCOOH(l) –conc. H₂SO₄–> CO(g) + CO₂(g) + H₂O(l)

Note: This reaction is also used for the school laboratory preparation of small amounts of carbon (II) oxide gas. Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution.

II. As an Oxidizing agent

Experiment I

Put about 2 cm³ of concentrated sulphuric (VI) acid into three separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI) solution on the mouth of each boiling tube. Put about 0.5 g of copper turnings, zinc granules, and iron filings to each boiling tube separately.

Observation:

• Effervescence/fizzing/bubbles.
• Blue solution formed with copper.
• Green solution formed with iron.
• Colourless solution formed with zinc.
• Colourless gas produced that has a pungent irritating choking smell.
• Gas produced turns orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI) solution to green.

Explanation: Concentrated sulphuric (VI) acid is a strong oxidizing agent. It oxidizes metals to metallic sulphate(VI) salts and itself is reduced to sulphur(IV) oxide gas. Sulphur(IV) oxide gas turns orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI) solution to green. CuSO₄(aq) is a blue solution. ZnSO₄(aq) is a colourless solution. FeSO₄(aq) is a green solution.

Chemical equations:

Cu(s) + 2H₂SO₄(aq) → CuSO₄(aq) + SO₂(g) + 2H₂O(l)

Zn(s) + 2H₂SO₄(aq) → ZnSO₄(aq) + SO₂(g) + 2H₂O(l)

Fe(s) + 2H₂SO₄(aq) → FeSO₄(aq) + SO₂(g) + 2H₂O(l)

Experiment II

Put about 2 cm³ of concentrated sulphuric (VI) acid into two separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI) solution on the mouth of each boiling tube. Put about 0.5 g of powdered charcoal and sulphur powder to each boiling tube separately. Warm.

Observation:

• Black solid charcoal dissolves/decreases.
• Yellow solid sulphur dissolves/decreases.
• Colourless gas produced that has a pungent irritating choking smell.
• Gas produced turns orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI) solution to green.

Explanation: Concentrated sulphuric (VI) acid is a strong oxidizing agent. It oxidizes non-metals to non-metallic oxides and itself is reduced to sulphur(IV) oxide gas. Sulphur(IV) oxide gas turns orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI) solution to green. Charcoal is oxidized to carbon(IV) oxide. Sulphur is oxidized to sulphur(IV) oxide.

Chemical equations:

C(s) + 2H₂SO₄(aq) → CO₂(g) + 2SO₂(g) + 2H₂O(l)

S(s) + 2H₂SO₄(aq) → 3SO₂(g) + 2H₂O(l)

III. As the least volatile acid

Study the table below showing a comparison in boiling points of the three mineral acids:

1. Which is the least volatile acid? Explain.

Sulphuric(VI) acid (H₂SO₄) because it has the largest molecule and is joined by hydrogen bonds, making it have the highest boiling point/least volatile.

2. Using chemical equations, explain how sulphuric(VI) acid displaces the less volatile mineral acids.

(i) Chemical equations:

KNO₃(s) + H₂SO₄(aq) → KHSO₄(l) + HNO₃(g)

NaNO₃(s) + H₂SO₄(aq) → NaHSO₄(l) + HNO₃(g)

This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO₃).

(ii) Chemical equations:

KCl(s) + H₂SO₄(aq) → KHSO₄(s) + HCl(g)

NaCl(s) + H₂SO₄(aq) → NaHSO₄(s) + HCl(g)

This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl).

(d) Properties of dilute sulphuric(VI) acid

Dilute sulphuric(VI) acid is made when about 10 cm³ of concentrated sulphuric(VI) acid is carefully added to about 90 cm³ of distilled water.

Diluting concentrated sulphuric(VI) acid should be done carefully because the reaction is highly exothermic. Diluting concentrated sulphuric(VI) acid decreases the number of moles present in a given volume of solution which makes the acid less corrosive.

On diluting concentrated sulphuric(VI) acid, water ionizes/dissociates the acid fully/wholly into two (dibasic) free H⁺ and SO₄^2- ions:

H₂SO₄(aq) → 2H⁺(aq) + SO₄^2-(aq)

The presence of free H⁺ ions is responsible for:

• Turning litmus red because of the presence of free H⁺ ions.
• Having pH 1/2/3 because of the presence of many free H⁺ ions, hence a strongly acidic solution.
• Reaction with metals.

Experiment:

Place 5 cm³ of 0.2M dilute sulphuric(VI) acid into four separate clean test tubes. Add about 0.1 g of magnesium ribbon to one test tube. Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger “stopper”. Repeat by adding zinc, copper, and iron instead of the magnesium ribbon.

Observation:

• No effervescence/bubbles/fizzing with copper.
• Effervescence/bubbles/fizzing with iron, zinc, and magnesium.
• Colourless gas produced that extinguishes burning splint with a “pop” sound.
• Colourless solution formed with zinc and magnesium.
• Green solution formed with iron.

Explanation:

When a metal higher than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI) acid, effervescence/bubbling/fizzing takes place with evolution of hydrogen gas. Impure hydrogen gas extinguishes burning splint with a “pop” sound. A sulphate(VI) salt is formed. Iron, zinc, and magnesium are higher than hydrogen in the reactivity/electrochemical series. They form iron(II) sulphate(VI), magnesium sulphate(VI), and zinc sulphate(VI).

When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI) acid, there is no effervescence/bubbling/fizzing that takes place. Copper thus does not react with dilute sulphuric(VI) acid.

Chemical/ionic equations:

Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)

Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g)

Note:

• Calcium, lead, and barium form insoluble sulphate(VI) salts that cover/coat the unreacted metals.
• Sodium and potassium react explosively with dilute sulphuric(VI) acid.

(iv) Reaction with metal carbonates and hydrogen carbonates

Experiment:

Place 5 cm³ of 0.2M dilute sulphuric(VI) acid into four separate clean boiling tubes. Add about 0.1 g of sodium carbonate to one boiling tube. Introduce a burning splint on top of the boiling tube. Repeat by adding zinc carbonate, copper(II) carbonate, and iron(II) carbonate in place of sodium hydrogen carbonate.

Observation:

• Effervescence/bubbles/fizzing.
• Colourless gas produced that extinguishes burning splint.
• Colourless solution formed with zinc carbonate, sodium hydrogen carbonate, and sodium carbonate.
• Green solution formed with iron(II) carbonate.
• Blue solution formed with copper(II) carbonate.

Explanation:

When a metal carbonate or a hydrogen carbonate is put in a test tube containing dilute sulphuric(VI) acid, effervescence/bubbling/fizzing takes place with evolution of carbon(IV) oxide gas. Carbon(IV) oxide gas extinguishes a burning splint and forms a white precipitate when bubbled in lime water. A sulphate(VI) salt is formed.

Chemical/ionic equations:

ZnCO₃(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂O(l) + CO₂(g)

ZnCO₃(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂O(l) + CO₂(g)

CuCO₃(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) + CO₂(g)

CuCO₃(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂O(l) + CO₂(g)

FeCO₃(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂O(l) + CO₂(g)

FeCO₃(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂O(l) + CO₂(g)

2NaHCO₃(s) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l) + 2CO₂(g)

NaHCO₃(s) + H⁺(aq) → Na⁺(aq) + H₂O(l) + CO₂(g)

Na₂CO₃(s) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)

NaHCO₃(s) + H⁺(aq) → Na⁺(aq) + H₂O(l) + CO₂(g)

(NH₄)₂CO₃(s) + H₂SO₄(aq) → (NH₄)₂SO₄(aq) + H₂O(l) + CO₂(g)

(NH₄)₂CO₃(s) + 2H⁺(aq) → NH₄⁺(aq) + H₂O(l) + CO₂(g)

2NH₄HCO₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq) + H₂O(l) + CO₂(g)

NH₄HCO₃(aq) + H⁺(aq) → NH₄⁺(aq) + H₂O(l) + CO₂(g)

Note: Calcium, lead, and barium carbonates form insoluble sulphate(VI) salts that cover/coat the unreacted metals.

(v) Neutralization – reaction of metal oxides and alkalis/bases

Experiment I:

Place 5 cm³ of 0.2M dilute sulphuric(VI) acid into four separate clean boiling tubes. Add about 0.1 g of copper(II) oxide to one boiling tube. Stir. Repeat by adding zinc oxide, calcium carbonate, and sodium oxide in place of the copper(II) oxide.

Observation:

• Blue solution formed with copper(II) oxide.
• Colourless solution formed with other oxides.

Explanation: When a metal oxide is put in a test tube containing dilute sulphuric(VI) acid, the oxide dissolves forming a sulphate(VI) salt.

Chemical/ionic equations:

ZnO(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂O(l)

ZnO(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂O(l)

CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)

CuO(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂O(l)

MgO(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂O(l)

MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l)

Na₂O(s) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)

Na₂O(s) + 2H⁺(aq) → 2Na⁺(aq) + H₂O(l)

K₂CO₃(s) + H₂SO₄(aq) → K₂SO₄(aq) + H₂O(l)

K₂O(s) + 2H⁺(aq) → 2K⁺(aq) + H₂O(l)

Note: Calcium, lead, and barium oxides form insoluble sulphate(VI) salts that cover/coat the unreacted metal oxides.

Experiment II:

Fill a burette with 0.1M dilute sulphuric(VI) acid. Pipette 20.0 cm³ of 0.1M sodium hydroxide solution into a 250 cm³ conical flask. Add three drops of phenolphthalein indicator. Titrate the acid to get a permanent colour change. Repeat with 0.1M potassium hydroxide solution in place of 0.1M sodium hydroxide solution.

Observation: Colour of phenolphthalein changes from pink to colourless at the end point.

Explanation: Like other mineral acids, dilute sulphuric(VI) acid neutralizes bases/alkalis to a sulphate salt and water only. Colour of the indicator used changes when a slight excess of acid is added to the base at the end point.

Chemical equations:

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

OH^–(aq) + H⁺(aq) → H₂O(l)

2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)

OH^–(aq) + H⁺(aq) → H₂O(l)

2NH₄OH(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq) + 2H₂O(l)

OH^–(aq) + H⁺(aq) → H₂O(l)

(iv) Hydrogen sulphide (H₂S)

(a) Occurrence

Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world, e.g., Olkaria and Hells Gate near Naivasha in Kenya. It is present in rotten eggs and human excreta.

(b) Preparation

Hydrogen sulphide is prepared in a school laboratory by heating iron(II) sulphide with dilute hydrochloric acid.

(c) Properties of Hydrogen sulphide (Questions)

1. Write the equation for the reaction for the school laboratory preparation of Hydrogen sulphide.

Chemical equation: FeS(s) + 2HCl(aq) → H₂S(g) + FeCl₂(aq)

2. State three physical properties unique to Hydrogen sulphide.

Hydrogen sulphide is a colourless gas with a characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is denser than water and turns blue litmus paper red.

3. Hydrogen sulphide exists as a dibasic acid when dissolved in water. Using a chemical equation, show how it ionizes in aqueous state.

H₂S(aq) → H⁺(aq) + HS^–(aq)

H₂S(aq) → 2H⁺(aq) + S^2-(aq)

Hydrogen sulphide therefore can form both normal and acid salts, e.g., sodium hydrogen sulphide and sodium sulphide both exist.

4. State and explain one gaseous impurity likely to be present in the gas jar containing hydrogen sulphide above.

Hydrogen (H₂)

Iron(II) sulphide contains iron as impurity. The iron will react with dilute hydrochloric acid to form iron(II) chloride and produce hydrogen gas that mixes with hydrogen sulphide gas.

5. State and explain the observations made when a filter paper dipped in Lead(II) ethanoate / Lead(II) nitrate(V) solution is put in a gas jar containing hydrogen sulphide gas.

Observations:

• Moist Lead(II) ethanoate / Lead(II) nitrate(V) paper turns black.

Explanation:

When hydrogen sulphide is bubbled in a metallic salt solution, a metallic sulphide is formed. All sulphides are insoluble black salts except sodium sulphide, potassium sulphide, and ammonium sulphides. Hydrogen sulphide gas blackens moist Lead(II) ethanoate / Lead(II) nitrate(V) paper. The gas reacts with Pb²⁺ in the paper to form black Lead(II) sulphide. This is the chemical test for the presence of H₂S other than the physical smell of rotten eggs.

Chemical equations:

Pb²⁺(aq) + H₂S → PbS(s) + 2H⁺(aq)

Fe²⁺(aq) + H₂S → FeS(s) + 2H⁺(aq)

Zn²⁺(aq) + H₂S → ZnS(s) + 2H⁺(aq)

Cu²⁺(aq) + H₂S → CuS(s) + 2H⁺(aq)

2Cu⁺(aq) + H₂S → Cu₂S(s) + 2H⁺(aq)

6. Dry hydrogen sulphide was ignited as below.

(i) State the observations made in flame A

Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV) oxide gas and water.

Chemical equation: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g)

Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water.

Chemical equation: 2H₂S(g) + O₂(g) → 2H₂O(l) + 2S(s)

7. Hydrogen sulphide is a strong reducing agent that is oxidized to yellow solid sulphur as precipitate. The following experiments illustrate the reducing properties of Hydrogen sulphide.

(a) Turns orange acidified potassium dichromate(VI) to green

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution; or
• Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas jar containing hydrogen sulphide gas.

Observation: Orange acidified potassium dichromate (VI) turns to green. Yellow solid residue.

Explanation: Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr₂O₇^2- ions to green Cr³⁺ ions leaving a yellow solid residue as itself is oxidized to sulphur.

Chemical/ionic equation:

4H₂S(aq) + Cr₂O₇^2-(aq) + 6H⁺(aq) → 4S(s) + 2Cr³⁺(aq) + 7H₂O(l)

This test is used for differentiating hydrogen sulphide and sulphur(IV) oxide gas.

Sulphur(IV) oxide also reduces acidified potassium dichromate(VI) from orange Cr₂O₇^2- ions to green Cr³⁺ ions without leaving a yellow residue.

(b) Decolorizes acidified potassium manganate(VII)

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing acidified potassium manganate(VII) solution; or
• Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing hydrogen sulphide gas.

Observation: Purple acidified potassium manganate(VII) turns colourless/acidified potassium manganate(VII) is decolorized. Yellow solid residue.

Explanation: Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple MnO₄^– ions to green Mn²⁺ ions leaving a residue as the gas itself is oxidized to sulphur.

Chemical/ionic equation:

5H₂S(g) + 2MnO₄^–(aq) + 6H⁺(aq) → 5S(s) + 2Mn²⁺(aq) + 8H₂O(l)

(purple) (colourless)

This is another test for differentiating hydrogen sulphide and sulphur(IV) oxide gas.

Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO₄^– ions to colourless Mn²⁺ ions leaving no yellow residue.

(c) Decolorizes bromine water

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing bromine water; or
• Put three drops of bromine water into a gas jar containing hydrogen sulphide gas. Swirl.

Observation: Yellow bromine water turns colourless/bromine water is decolorized. Yellow solid residue.

Explanation: Hydrogen sulphide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) leaving a yellow residue as the gas itself is oxidized to sulphur.

Chemical/ionic equation:

H₂S(g) + Br₂(aq) → S(s) + 2HBr(aq)

(yellow solution) (yellow solid) (colourless)

This is another test for differentiating hydrogen sulphide and sulphur(IV) oxide gas.

Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO₄^– ions to colourless Mn²⁺ ions leaving no yellow residue.

(d) Reduces Iron(III) Fe³⁺ salts to Iron(II) Fe²⁺

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing about 3 cm³ of Iron (III) chloride solution; or
• Place about 3 cm³ of Iron (III) chloride solution into a gas jar containing hydrogen sulphide gas. Swirl.

Observation: Yellow/brown Iron (III) chloride solution turns green. Yellow solid.

Explanation: Hydrogen sulphide gas reduces Iron (III) chloride solution from yellow/brown Fe³⁺ ions to green Fe²⁺ ions leaving a yellow residue. The gas is itself oxidized to sulphur.

Chemical/ionic equation:

H₂S(aq) + 2Fe³⁺(aq) → S(s) + 2Fe²⁺(aq) + 2H⁺(aq)

(yellow solution) (yellow residue) (green)

(e) Reduces Nitric(V) acid to Nitrogen(IV) oxide gas

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing about 3 cm³ of concentrated nitric(V) acid; or
• Place about 3 cm³ of concentrated nitric(V) acid into a gas jar containing hydrogen sulphide gas. Swirl.

Observation: Brown fumes of a gas evolved/produced. Yellow solid residue.

Explanation: Hydrogen sulphide gas reduces concentrated nitric(V) acid to brown nitrogen(IV) oxide gas, itself oxidized to yellow sulphur.

Chemical/ionic equation:

H₂S(g) + 2HNO₃(l) → 2H₂O(l) + S(s) + 2NO₂(g)

(yellow residue) (brown fumes)

(f) Reduces sulphuric(VI) acid to sulphur

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing about 3 cm³ of concentrated sulphuric(VI) acid; or
• Place about 3 cm³ of concentrated sulphuric(VI) acid into a gas jar containing hydrogen sulphide gas. Swirl.

Observation: Yellow solid residue.

Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI) acid to yellow sulphur.

Chemical/ionic equation:

3H₂S(g) + H₂SO₄(l) → 4H₂O(l) + 4S(s)

(yellow residue)

(g) Reduces Hydrogen peroxide to water

Experiment:

• Pass a stream of hydrogen sulphide gas in a test tube containing about 3 cm³ of 20 volume hydrogen peroxide.

Observation: Yellow solid residue.

Explanation: Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and itself oxidized to yellow sulphur.

Chemical/ionic equation:

H₂S(g) + H₂O₂(l) → 2H₂O(l) + S(s)

(yellow residue)

8. Name the salt formed when:

(i) Equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution:

Sodium hydrogen sulphide.

Chemical/ionic equation:

H₂S(g) + NaOH(aq) → H₂O(l) + NaHS(aq)

(ii) Hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution:

Sodium sulphide.

Chemical/ionic equation:

H₂S(g) + 2NaOH(aq) → 2H₂O(l) + Na₂S(aq)

Practice

Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V) salts as in the flow chart below:

(a) Name the black solid

Copper(II) sulphide.

(b) Identify the cation responsible for the formation of:

• Blue solution: Cu²⁺(aq)
• Green solution: Fe²⁺(aq)
• Brown solution: Fe³⁺(aq)

(c) Using acidified potassium dichromate(VI), describe how you would differentiate between sulphur(IV) oxide and hydrogen sulphide.

• Bubble the gases in separate test tubes containing acidified potassium dichromate(VI) solution.
• Both change the orange colour of acidified potassium dichromate(VI) solution to green.
• Yellow solid residue/deposit is formed with hydrogen sulphide.

Chemical/ionic equations:

4H₂S(aq) + Cr₂O₇^2-(aq) + 6H⁺(aq) → 4S(s) + 2Cr³⁺(aq) + 7H₂O(l)

3SO₃^2-(aq) + Cr₂O₇^2-(aq) + 8H⁺(aq) → 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

(d) State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator.

Observation: Gas continues burning with a blue flame.

Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV) oxide gas and water.

Chemical equation: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g)

(v) Sulphite (IV) (SO₃^2-) and Sulphate (VI) (SO₄^2-) salts

1. Sulphate (VI) (SO₄^2-) salts are normal and acid salts derived from sulphuric (VI) acid H₂SO₄.

2. Sulphite (IV) (SO₃^2-) salts are normal and acid salts derived from sulphurous acid H₂SO₃.

3. Sulphuric (VI) acid H₂SO₄ is formed when sulphur(VI) oxide gas is bubbled in water. The acid exists as a dibasic acid with two ionisable hydrogens. It forms therefore the sulphate (VI) (SO₄^2-) and hydrogen sulphate (VI) (HSO₄^–) salts.

i.e.,

H₂SO₄(aq) → 2H⁺(aq) + SO₄^2-(aq)

H₂SO₄(aq) → H⁺(aq) + HSO₄^–(aq)

All sulphate (VI) (SO₄^2-) salts dissolve in water/are soluble except calcium (II) sulphate (VI) (CaSO₄), barium (II) sulphate (VI) (BaSO₄), and lead (II) sulphate (VI) (PbSO₄).

All hydrogen sulphate (VI) (HSO₄^–) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO₄), potassium (I) hydrogen sulphate (VI) (KHSO₄), and ammonium hydrogen sulphate (VI) (NH₄HSO₄) also exist as solids.

Other hydrogen sulphate (VI) (HSO₄^–) salts do not exist except those of calcium (II) hydrogen sulphate (VI) (Ca(HSO₄)₂) and magnesium (II) hydrogen sulphate (VI) (Mg(HSO₄)₂).

4. Sulphurous acid H₂SO₃ is formed when sulphur(IV) oxide gas is bubbled in water. The acid exists as a dibasic acid with two ionisable hydrogens. It forms therefore the sulphite (IV) (SO₃^2-) and hydrogen sulphite (IV) (HSO₃^–) salts.

i.e.,

H₂SO₃(aq) → 2H⁺(aq) + SO₃^2-(aq)

H₂SO₃(aq) → H⁺(aq) + HSO₃^–(aq)

All sulphite (IV) (SO₃^2-) salts dissolve in water/are soluble except calcium (II) sulphite (IV) (CaSO₃), barium (II) sulphite (IV) (BaSO₃), and lead (II) sulphite (IV) (PbSO₃).

All hydrogen sulphite (IV) (HSO₃^–) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphite (IV) (NaHSO₃), potassium (I) hydrogen sulphite (IV) (KHSO₃), and ammonium hydrogen sulphite (IV) (NH₄HSO₃) also exist as solids.

Other hydrogen sulphite (IV) (HSO₃^–) salts do not exist except those of calcium (II) hydrogen sulphite (IV) (Ca(HSO₃)₂) and magnesium (II) hydrogen sulphite (IV) (Mg(HSO₃)₂).

5. The following experiments show the effect of heat on sulphate(VI) (SO₄^2-) and sulphite (IV) (SO₃^2-) salts:

Experiment:

In a clean dry test tube, place separately about 1.0 g of zinc(II) sulphate (VI), iron(II) sulphate (VI), copper(II) sulphate (VI), sodium (I) sulphate (VI), sodium (I) sulphite (IV). Heat gently then strongly. Test any gases produced using litmus papers.

Observations:

• Colourless droplets of liquid form on the cooler parts of the test tube in all cases.
• White solid residue is left in case of zinc(II) sulphate (VI), sodium (I) sulphate (VI), and sodium (I) sulphite (IV).
• Colour changes from green to brown/yellow in case of iron(II) sulphate (VI).
• Colour changes from blue to white then black in case of copper(II) sulphate (VI).
• Blue litmus paper remains blue and red litmus paper remains red in case of zinc(II) sulphate (VI), sodium (I) sulphate (VI), and sodium (I) sulphite (IV).
• Blue litmus paper turns red and red litmus paper remains red in case of iron(II) sulphate (VI) and copper(II) sulphate (VI).

Explanation:

(i) All sulphate (VI) (SO₄^2-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating, e.g.,

K₂SO₄.10H₂O(s) → K₂SO₄(s) + 10H₂O(l)

Na₂SO₄.10H₂O(s) → Na₂SO₄(s) + 10H₂O(l)

MgSO₄.7H₂O(s) → MgSO₄(s) + 7H₂O(l)

CaSO₄.7H₂O(s) → CaSO₄(s) + 7H₂O(l)

ZnSO₄.7H₂O(s) → ZnSO₄(s) + 7H₂O(l)

FeSO₄.7H₂O(s) → FeSO₄(s) + 7H₂O(l)

Al₂(SO₄)₃.6H₂O(s) → Al₂(SO₄)₃(s) + 6H₂O(l)

CuSO₄.5H₂O(s) → CuSO₄(s) + 5H₂O(l)

All sulphate (VI) (SO₄^2-) salts do not decompose on heating except iron(II) sulphate (VI) and copper(II) sulphate (VI).

(ii) Iron(II) sulphate (VI) decomposes on strong heating to produce acidic sulphur(IV) oxide and sulphur(VI) oxide gases. Iron(III) oxide is formed as a brown/yellow residue.

Chemical equation:

2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃(g)

This reaction is used for the school laboratory preparation of small amounts of sulphur(VI) oxide gas.

Sulphur(VI) oxide readily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water. Sulphur(IV) oxide does not.

(iii) Copper(II) sulphate (VI) decomposes on strong heating to black copper(II) oxide and sulphur(VI) oxide gas.

Chemical equation:

2CuSO₄(s) → CuO(s) + SO₃(g)

This reaction is used for the school laboratory preparation of small amounts of sulphur(VI) oxide gas.

6. The following experiments show the test for the presence of sulphate (VI) (SO₄^2-) and sulphite (IV) (SO₃^2-) ions in a sample of a salt/compound:

Experiments/Observations:

(a) Using Lead(II) nitrate(V)

I. To about 5 cm³ of a salt solution in a test tube add four drops of Lead(II) nitrate(V) solution. Preserve.

II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid. Preserve.

Observation 1

Observation 2

III. (a) To the preserved sample observation 1 in (II) above, heat to boil.

Observation 1

Observation 2

(b) To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII)/dichromate(VI).

Observation 1

Observation 2

Explanations

Using Lead(II) nitrate(V)

(i) Lead(II) nitrate(V) solution reacts with chlorides (Cl^–), sulphate (VI) salts (SO₄^2-), sulphite (IV) salts (SO₃^2-), and carbonates (CO₃^2-) to form the insoluble white precipitate of Lead(II) chloride, Lead(II) sulphate(VI), Lead(II) sulphite(IV), and Lead(II) carbonate(IV).

Chemical/ionic equations:

Pb²⁺(aq) + Cl^–(aq) → PbCl₂(s)

Pb²⁺(aq) + SO₄^2-(aq) → PbSO₄(s)

Pb²⁺(aq) + SO₃^2-(aq) → PbSO₃(s)

Pb²⁺(aq) + CO₃^2-(aq) → PbCO₃(s)

(ii) When the insoluble precipitates are acidified with nitric(V) acid:

• Lead(II) chloride and Lead(II) sulphate(VI) do not react with the acid and thus their white precipitates remain/persist.
• Lead(II) sulphite(IV) and Lead(II) carbonate(IV) react with the acid to form soluble Lead(II) nitrate(V) and produce/effervesce/fizz/bubble out sulphur(IV) oxide and carbon(IV) oxide gases respectively.

Chemical/ionic equations:

PbSO₃(s) + 2H⁺(aq) → H₂O(l) + Pb²⁺(aq) + SO₂(g)

PbCO₃(s) + 2H⁺(aq) → H₂O(l) + Pb²⁺(aq) + CO₂(g)

(iii) When Lead(II) chloride and Lead(II) sulphate(VI) are heated/warmed:

• Lead(II) chloride dissolves in hot water/on boiling (recrystallizes on cooling).
• Lead(II) sulphate(VI) does not dissolve in hot water; thus, its white precipitate persists/remains on heating/boiling.

(iv) When sulphur(IV) oxide and carbon(IV) oxide gases are produced:

• Sulphur(IV) oxide will decolorize acidified potassium manganate(VII) and/or orange colour of acidified potassium dichromate(VI) will turn to green. Carbon(IV) oxide will not.

Chemical equations:

5SO₃^2-(aq) + 2MnO₄^–(aq) + 6H⁺(aq) → 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3H₂O(l)

3SO₃^2-(aq) + Cr₂O₇^2-(aq) + 8H⁺(aq) → 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

Carbon(IV) oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV) oxide will not.

Chemical equation:

Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)

These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.

Using Barium(II) nitrate(V)/ Barium(II) chloride

(i) Barium(II) nitrate(V) and/or Barium(II) chloride solution reacts with sulphate (VI) salts (SO₄^2-), sulphite (IV) salts (SO₃^2-), and carbonates (CO₃^2-) to form the insoluble white precipitate of Barium(II) sulphate(VI), Barium(II) sulphite(IV), and Barium(II) carbonate(IV).

Chemical/ionic equations:

Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)

Ba²⁺(aq) + SO₃^2-(aq) → BaSO₃(s)

Ba²⁺(aq) + CO₃^2-(aq) → BaCO₃(s)

(ii) When the insoluble precipitates are acidified with nitric(V) acid:

• Barium(II) sulphate(VI) does not react with the acid and thus its white precipitate remains/persists.
• Barium(II) sulphite(IV) and Barium(II) carbonate(IV) react with the acid to form soluble Barium(II) nitrate(V) and produce/effervesce/fizz/bubble out sulphur(IV) oxide and carbon(IV) oxide gases respectively.

Chemical/ionic equations:

BaSO₃(s) + 2H⁺(aq) → H₂O(l) + Ba²⁺(aq) + SO₂(g)

BaCO₃(s) + 2H⁺(aq) → H₂O(l) + Ba²⁺(aq) + CO₂(g)

(iii) When sulphur(IV) oxide and carbon(IV) oxide gases are produced:

• Sulphur(IV) oxide will decolorize acidified potassium manganate(VII) and/or orange colour of acidified potassium dichromate(VI) will turn to green. Carbon(IV) oxide will not.

Chemical equations:

5SO₃^2-(aq) + 2MnO₄^–(aq) + 6H⁺(aq) → 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3H₂O(l)

3SO₃^2-(aq) + Cr₂O₇^2-(aq) + 8H⁺(aq) → 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

Carbon(IV) oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV) oxide will not.

Chemical equation:

Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)

These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.

Summary test for Sulphate (VI) (SO₄^2-) and Sulphite (IV) (SO₃^2-) salts

Practice revision question

1. Study the flow chart below and use it to answer the questions that follow

(a) Identify the:

• I: Sodium salt solution – Sodium sulphite (IV) / Na₂SO₃
• II: White precipitate – Barium sulphite (IV) / BaSO₃
• III: Gas G – Sulphur (IV) Oxide / SO₂
• IV: Colourless solution H – Barium chloride / BaCl₂

(b) Write an ionic equation for the formation of:

• I. White precipitate

Ionic equation: Ba²⁺(aq) + SO₃^2-(aq) → BaSO₃(s)

• II. Gas G

Ionic equation: BaSO₃(s) + 2H⁺(aq) → SO₂(g) + H₂O(l) + Ba²⁺(aq)

• III. Green solution from the orange solution

3SO₃^2-(aq) + Cr₂O₇^2-(aq) + 8H⁺(aq) → 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)

(Orange) (green)

2. Study the flow chart below and answer the questions that follow.

(i) Write equation for the reaction taking place at:

I. The roasting furnace (1 mark)

2FeS₂(s) + 5O₂(g) → 2FeO(s) + 4SO₂(g)

II. The absorption tower (1 mark)

H₂SO₄(l) + SO₃(g) → H₂S₂O₇(l)

III. The diluter (1 mark)

H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(l)

(ii) The reaction taking place in chamber K is:

SO₂(g) + 1/2 O₂(g) → SO₃(g)

I. Explain why it is necessary to use excess air in chamber K

To ensure all the SO₂ reacts.

II. Name another substance used in chamber K

Vanadium(V) oxide.

3. (a) Describe a chemical test that can be used to differentiate between sodium sulphite (IV) and sodium sulphate (VI).

Add acidified Barium nitrate(V)/chloride.

White precipitate formed with sodium sulphate (VI).

No white precipitate formed with sodium sulphite (IV).

(b) Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI) acid. (Cu = 63.5, 1 mole of a gas at s.t.p = 22.4 dm³)

Chemical equation:

Cu(s) + 2H₂SO₄(l) → CuSO₄(aq) + H₂O(l) + SO₂(g)

Mole ratio Cu : SO₂ = 1 : 1

Method 1

1 mole Cu = 63.5 g → 1 mole SO₂ = 22.4 dm³

(120 × 1000) g → (120 × 1000) × 22.4 / 63.5 dm³

= 42330.71 dm³

Method 2

Moles of Cu = (120 × 1000) / 63.5 = 1889.76 moles

Moles SO₂ = Moles of Cu = 1889.76 moles

Volume of SO₂ = Moles × molar gas volume = 1889.76 × 22.4 = 42330.71 dm³

(a) Identify the:

(i) Cation responsible for the green solution T

Cr³⁺

(ii) Possible anions present in white precipitate R

CO₃^2-, SO₃^2-, SO₄^2-

(b) Name gas V

Sulphur (IV) oxide

(c) Write a possible ionic equation for the formation of white precipitate R.

Ba²⁺(aq) + CO₃^2-(aq) → BaCO₃(s)

Ba²⁺(aq) + SO₃^2-(aq) → BaSO₃(s)

Ba²⁺(aq) + SO₄