SULPHUR AND ITS COMPOUND

SULPHUR AND ITS COMPOUND:A.SULPHUR (S)

Sulphur is an element in Group VI(Group 16)of the Periodic table . It has atomic number 16 and electronic configuration 16 and valency 2 /divalent and thus forms the ion S²A. Occurrence.

Sulphur mainly occurs :

(i) as free element in Texas and Louisiana in USA and Sicily in Italy.

(ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near

Naivasha in Kenya

(iii)as copper pyrites(CuFeS₂) and iron pyrites(FeS₂) in other parts of the world.

Extraction of Sulphur from Fraschs process

Suphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in.

Digging of tunnels is thus discouraged in trying to extract the mineral. Sulphur is extracted by drilling three concentric /round pipes of diameter of ratios 2:8: 18 centimeters.

Superheated water at 170^oC and 10atmosphere pressure is forced through the outermost pipe.

The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas.

The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115^oC.

A 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.

SULPHUR AND ITS COMPOUND

Diagram showing extraction of Sulphur from Fraschs Process

Allotropes of Sulphur.
Sulphur exist as two crystalline allotropic forms:

(i)Rhombic Sulphur

(ii)Monoclinic sulphur

Rhombic sulphur

Monoclinic sulphur

Bright yellow crystalline solid

Has a melting point of 113^oC

Has a density of 2.06gcm^-3

Stable below 96^oC

Has octahedral structure

Pale yellow crystalline solid Has a melting point of 119^oC

Has a density of 1.96gcm^-3Stable above 96^oC

Has a needle-like structure

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

Sulphur exists in non-crystalline forms as:

(i)Plastic sulphur-

Plastic sulphur is prepared from heating powdered sulphur to boil

then pouring a thin continuous stream in a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright yellow crystalline rhombic sulphur.

(ii)Colloidal sulphur-

Colloidal sulphur is formed when sodium thiosulphate (Na₂S₂O₃) is added hydrochloric acid to form a yellow precipitate.

Heating Sulphur.

A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S₈.On heating the yellow sulphur powder melts at 113^oC to clear amber liquid with low viscosity and thus flows easily. On further heating to 160^oC the molten liquid darkens to a brown very viscous liquid that does not flow easily.

This is because the S₈rings break into S₈chain that join together to form very long chains made of over 100000 atoms of Sulphur. The long chains entangle each other reducing their mobility /flow and hence increases their viscosity. On continued further heating to above 160^oC, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous. This is because the long chains break to smaller/shorter chains.

At 444^oC, the liquid boils and forms brown vapour of a mixture of S₈,S₆,S₂molecules that solidifies to S₈ring of “flowers of sulphur” on the cooler parts.

Summary of changes on heating sulphur

Observation on heating

Explanation/structure of Sulphur

Solid sulphur

Heat to 113^oC

Amber yellow liquid

Heat to 160^oC

Liquid darkens

Heat to 444^oC

Liquid boils to brown vapour

Puckered S₈ring

Puckered S₈ring in liquid form (low viscosity/flow easily)

Puckered S₈ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow)

Mixture of S₈,S₆,S₂vapour

Puckered S₈ring

Cool to room temperature

Yellow sublimate

(Flowers of Sulphur)

Physical and Chemical properties of Sulphur.
SULPHUR AND ITS COMPOUND

(Questions)

State three physical properties unique to Sulphur Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity. It has a melting point of 115^oC and a boiling point of 444^o
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 amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e

Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid
Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid Chemical equation

S(s) + O₂(g)	-> SO₂(g) (Sulphur(IV)Oxide gas)
2S(s) + 3O₂(g)	-> 2SO₃(g) (Traces of Sulphur(VI)Oxide gas)
SO₂(g) + H₂O(l)	-> H₂SO₃(aq) ( sulphuric(IV)acid)
SO₃(g) + H₂O(l)	-> H₂SO₄(aq) ( sulphuric(VI)acid).

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 to 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)

SULPHUR AND ITS COMPOUND

4.The set up below show 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 to 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 is 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 set up.

Observation

Yellow colour of sulphur fades Colurless 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)

NB:

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:

(i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid.

(ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable.

(iii)Making gun powder and match stick heads

(iv) As ointments to treat fungal infections

Revision Practice

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

(a)Name the substances that passes through:

M Superheated water at 170oC and 10 atmosphere pressure

Hot compressed air N Molten sulphur

(b)What is the purpose of the substance that passes through L and M?

Superheated water at 170^oC 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.

	A	B
Appearance	Bright yellow	Pale yellow
Density(gcm^-3)	1.93	2.08
Melting point(^oC)	119	113
Stability	Above 96^oC	Below 96^oC

I.What are allotropes?

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

Identify allotrope:A. Monoclinic sulphur

B . Rhombic sulphur

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 ,I 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 g x 24 dm3 = 0.3 dm3

32.0 g

Method 2

Moles of sulphur used = Mass of sulphur => 0.4 = 0.0125 moles

Molar mass of sulphur 32

Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one mole of gas =>0.0125 moles x 24 dm3 = 0.3 dm3

SULPHUR AND ITS COMPOUND

B.COMPOUNDS OF SULPHUR

The following are the main compounds of sulphur: (i) Sulphur(IV)oxide

Sulphur(VI)oxide .
Sulphuric(VI)acid
Hydrogen Sulphide
Sulphate(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 geysersand 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:Copper and Sulphuric(VI)acid.

Method 2:Sodium Sulphate(IV) and hydrochloric acid.

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

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 amount/quantity of sulphur (IV)oxide gas.

(ii)Method 2

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

K₂SO₃(aq) + HCl(aq) -> KCl(aq ) + SO₂(g) + 2H₂O(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 smell and choking smell which liquidifies easily. It is about two times denser than air.

The diagram below show 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 sulphuric (IV)acid is formed which turns litmus solution red.

Chemical equation

SO₂(g) + H₂O(l) -> H₂SO₃(aq) ( sulphuric(IV)acid turn 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

(i)Dry Blue litmus paper remains blue.

Dry red litmus paper remains red.

(ii) 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+(aq)ions and thus has no effect on dry blue/red litmus papers.

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

Weak sulphuric(IV)acid(H₂SO₃(aq)) dissociates /ionizes into free H+(aq)ions:

H₂SO₃(aq) -> 2H+(aq) + SO₃^2-(aq)

The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic.

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

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

SULPHUR AND ITS COMPOUND

(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 sulphuric(IV)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. This is because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its original colour. e.g.

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:

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or;

(ii)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 sulphuric(IV) acid to SO₄^2-ions in sulphuric(VI) acid.

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO₂(g) + H₂O(l) -> H₂SO₃(aq)

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H₂SO₃(aq) -> 2H+(aq) + SO₃^2-(aq)

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

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:

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;

(ii)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 to 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 sulphuric(IV) acid to SO₄^2-ions in sulphuric(VI) acid.

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO₂(g) + H₂O(l) -> H₂SO₃(aq)

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H₂SO₃(aq) -> 2H+(aq) + SO₃^2-(aq)

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

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

(purple) (colourless)

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.

SULPHUR AND ITS COMPOUND

(d)Decolorizes bromine water

Experiment:

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or;

(ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.

Observation:

Yellow bromine water turns to 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 sulphuric(IV) acid to SO₄^2-ions in sulphuric(VI) acid.

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO₂(g) + H₂O(l) -> H₂SO₃(aq)

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H₂SO₃(aq) -> 2H+(aq) + SO₃^2-(aq)

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

SO₃^2-(aq) + Br₂(aq) + H₂O(l) -> SO₄^2-(aq) + 2HBr(aq) (yellow) (colourless)

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) salts Fe²⁺Experiment:

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;

(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl.

Observation:

Yellow/brown Iron (III)chloride solution turns to 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 SO₃²ions in sulphuric(IV) acid to SO₄^2-ions in sulphuric(VI) acid.

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO₂(g) + H₂O(l) -> H₂SO₃(aq)

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H₂SO₃(aq) -> 2H+(aq) + SO₃^2-(aq)

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

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

(yellow) (green)

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

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;

(ii)Place about 3cm3 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 sulphuric(IV) acid to SO₄^2-ions in sulphuric(VI) acid.

Chemical/ionic equation:

SO₂(g) + 2HNO₃(l) -> H₂SO₄(l) + NO₂(g)

(brown fumes/gas)

SULPHUR AND ITS COMPOUND

(g)Reduces Hydrogen peroxide to water Experiment:

(i)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 persist /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 sulphuric(IV) acid to SO₄^2-ions in sulphuric(VI) acid.

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

Chemical/ionic equation:

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:

(i) persist/remains if SO₄^2-(aq)ions (BaSO₄(s)) is present.

(ii)dissolves if SO₃^2-(aq)ions (BaSO₃(s)) and CO₃^2-(aq)ions (BaCO₃(s))is present. This is because:

BaSO₃(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO₂gas 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)

BaCO₃(s) reacts with Nitric(V)/hydrochloric acid to produce acidic CO₂gas 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 act as an oxidizing agent as in the following examples.

(a)Reduction by burning Magnesium

Experiment

Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas

Observation

Magnesium ribbon continues to burn with difficulty.

White ash and yellow powder/speck

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 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.

Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas

Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl

Observation

Yellow powder/speck

Explanation

Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar.

A little moisture/water act as 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.

(i)In the contact process for the manufacture of Sulphuric(VI)acid

(ii)As a bleaching agent of pulp and paper.

(_iii)As a fungicide to kill microbes’

(iv)As a preservative of jam, juices to prevent fermentation

(ii) Sulphur(VI)oxide(SO₃)

(a) Occurrence

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

(b) Preparation

In a Chemistry school laboratory Sulphur (VI)oxide may 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 solidify 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 /easily solidify 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)

SULPHUR AND ITS COMPOUND

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(H₂SO₄) is one of the three mineral acids. There are three mineral acids;

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: (i)Sulphur from Fraschs process or from heating metal sulphide ore like

Galena(PbS),Zinc blende(ZnS)

(ii)Oxygen from fractional distillation of air

(iii)Water from rivers/lakes

II. Chemical processes

The contact process involve four main chemical processes:

(i)Production of Sulphur(IV)oxide

As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources;

Burning/roasting sulphur in air.

Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners Chemical equation

S(s) + O₂(g) –> SO₂(g)

Burning/roasting sulphide ores in air.

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 amount /quantity of sulphur(IV)oxide is generated/produced.

Chemical equation

(i)2PbS (s) + 3O₂(g) -> 2PbO(s) + 2SO₂(g)

(ii)2ZnS (s) + 3O₂(g) -> 2ZnO(s) + 2SO₂(g)

(ii)2CuFeS₂(s) + 4O₂(g) -> 2FeO(s) + 3SO₂(g) + Cu₂O(s)

Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely.

(ii)Purification of Sulphur(IV)oxide

Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/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^oC 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 give the process its name.

Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas.

Chemical equation

2SO₂(g) + O₂(g) — V₂O₅–> 2SO₂(g)

This reaction is exothermic (-∆H) and the temperatures need to be maintained at around 450^oC to ensure that:

(i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450^oC

(ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450^oC.

SULPHUR AND ITS COMPOUND

(iii)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid

Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form 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 a 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 Contact process:

(i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers.

(ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water.

This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid.

(iii)scrubbing-This involves 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)sulphate

(IV) /CaSO₃

Chemical equation

Ca(OH)₂(aq) + SO₂(g) –> CaSO₃(aq) + H₂O (g)

III. Uses of Sulphuric(VI)acid

Sulphuric (VI) acid is used:

(i) in making dyes and paint

(ii)as acid in Lead-acid accumulator/battery

for making soapless detergents
for making sulphate agricultural fertilizers

III. 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 1.84gcm^-3.It has a boiling point of 338^oC.
Concentrated sulphuric(VI)acid is very soluble in water.

The solubility /dissolution of the acid very 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⁺

Free H⁺ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change the blue litmus paper red.

Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution.

This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases.

(v)The following are some chemical properties of concentrated sulphuric (VI) acid:

As a dehydrating agent

Experiment I;

Put about two spatula end fulls of brown sugar and glucose in separate boiling tubes .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.

Observation;

Colour in brown sugar change from brown to black.

Colour in glucose change from white to black.

Boiling tube becomes very hot.

SULPHUR AND ITS COMPOUND

Explanation

Concentrated sulphuric (VI) acid is 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 take place.

The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon.

Chemical equation

Glucose: C₆H₁₂O₆(s)	–conc.H₂SO₄–> 6C (s) + 6H₂O(l)
(white)	(black)
Sugar: C₁₂H₂₂O₁₁(s)	–conc.H₂SO₄–> 12C (s) +11H₂O(l)
(brown)	(black)

Experiment II;

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

Observation;

Colour change from blue to white.

Explanation

Concentrated sulphuric (VI) acid is 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) (blue) (white)

Experiment III;

Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 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 to green.

Explanation

Concentrated sulphuric (VI) acid is 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^oC.

Ethene with =C=C= double bond turns orange acidified potassium dichromate (VI) paper turns to green.

Chemical equation

C₂H₅OH(l) –conc.H₂SO₄/170^oC –> C₂H₄(g) + H₂O(l)

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

Experiment IV;

Put about 4cm3 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 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)

NB: This reaction is used for the school laboratory preparation of small amount carbon (II)oxide gas

SULPHUR AND ITS COMPOUND

Experiment V;

Put about 4cm3 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 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)

NB: This reaction is also used for the school laboratory preparation of small amount carbon (II) oxide gas.

Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution.

I As an Oxidizing agent

Experiment I

Put about 2cm3 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 the boiling tube. Put about 0.5g of

Copper turnings, Zinc granule 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 turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.

Explanation

Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes metals to metallic sulphate(VI) salts and itself reduced to sulphur(IV)oxide gas.

Sulphur (IV) oxide gas turn 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 equation

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 2cm3 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 the boiling tube.

Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube separately.

Warm.

SULPHUR AND ITS COMPOUND

Observation;

Black solid charcoal dissolves/decrease

Yellow solid sulphur dissolves/decrease

Colourless gas produced that has a pungent irritating choking smell. Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.

Explanation

Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes nonmetals to non metallic oxides and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn 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 equation

C(s) + 2H₂SO₄(aq) –>	CO₂(aq) + 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

Mineral acid	Relative molecula mass	Boiling point(^oC)
Hydrochloric acid(HCl)	36.5	35.0
Nitric(V)acid(HNO₃)	63.0	83.0
Sulphuric(VI)acid(H₂SO₄)	98.0	333

1.Which is the least volatile acid? Explain

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

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

(i)Chemical equation

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 equation

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 10cm3 of concentrated sulphuric (VI) acid is carefully added to about 90cm3 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⁺(aq) and SO₄^2-(aq)ions:

H₂SO₄(aq) -> 2H⁺(aq) + SO₄^2-(aq)

The presence of free H⁺(aq)ions is responsible for ;

(i)turn litmus red because of the presence of free H⁺(aq)ions

(ii)have pH 1/2/3 because of the presence of many free H⁺(aq)ions hence a strongly acidic solution.

(iii)Reaction with metals

Experiment:

Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test tubes. Add about 0.1g 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) salts 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 take place.

Copper thus do not react with dilute sulphuric(VI)acid.

Chemical/ionic equation

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) + H⁺(aq) –> Fe²⁺(aq) + H₂(g)

NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.

(ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid

(iv)Reaction with metal carbonates and hydrogen carbonates

Experiment:

Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g 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 the 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 carbonates 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) salts is formed.

Chemical/ionic equation

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) + H⁺(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)

NB:

Calcium, Lead and Barium carbonates forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.

SULPHUR AND ITS COMPOUND

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

Experiment I:

Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir.

Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)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 equation

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) + H⁺(aq) –> 2K⁺(aq) + H₂O(l)

NB:

Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides.

Experiment II:

Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of

0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium 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 equation:

2NaOH(aq) + H₂SO₄(aq) –> Na₂SO₄(aq) + H₂O(l) OH^–(s) + H⁺(aq) –> H₂O(l)

2KOH(aq) + H₂SO₄(aq) –> K₂SO₄(aq) + H₂O(l) OH^–(s) + H⁺(aq) –> H₂O(l)

2NH₄OH(aq) + H₂SO₄(aq) –> (NH₄)₂SO₄(aq) + H₂O(l) OH^–(s) + 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.

SULPHUR AND ITS COMPOUND

(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)

Write the equation for the reaction for the school laboratory preparationof Hydrogen sulphide.

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

State three physical properties unique to Hydrogen sulphide. Hydrogen sulphide is a colourless gas with 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.
Hydrogen sulphide exist 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

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.

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 (black)	+ 2H+(aq)
Fe²⁺(aq) + H₂S	-> FeS (black)	+ 2H+(aq)
Zn²⁺(aq) + H₂S	-> ZnS (black)	+ 2H+(aq)
Cu²⁺(aq) + H₂S	-> CuS	+ 2H+(aq)

(black)

2Cu⁺(aq) + H₂S -> Cu₂S + 2H+(aq)

(black)

SULPHUR AND ITS COMPOUND

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)

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:

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution. or;

(ii)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(aq) + 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:

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;

(ii)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 to 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.

SULPHUR AND ITS COMPOUND

(c)Decolorizes bromine water

Experiment:

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water . or;

(ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide gas. Swirl.

Observation:

Yellow bromine water turns to 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) salts Fe²⁺Experiment:

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;

(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Hydrogen sulphide gas. Swirl.

Observation:

Yellow/brown Iron (III)chloride solution turns to green. Yellow solid 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) + Fe²⁺(aq) + 2H⁺(aq) (yellow solution) (yellow residue) (green)

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

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;

(ii)Place about 3cm3 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)

SULPHUR AND ITS COMPOUND

(f)Reduces sulphuric(VI)acid to Sulphur Experiment:

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or;

(ii)Place about 3cm3 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:

(i)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 (l) -> H₂O(l) + NaHS (aq)

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

Sodium sulphide

Chemical/ionic equation:

H₂S(g) + 2NaOH (l) -> 2H₂O(l) + Na₂S (aq

SULPHUR AND ITS COMPOUND

Practice

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

(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 changes the Orange colour of acidified Potassium dichromate(VI) solution to green.

-Yellow solid residue/deposit is formed with Hydrogen sulphide Chemical/ionic equation:

4H₂S(aq) + Cr₂O₇^2-(aq) +6H+(aq) -> 4S(aq) + 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.

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

Sulphate (VI) (SO₄^2-) salts are normal and acid salts derived from Sulphuric

(VI)acid H₂SO₄.

Sulphate(IV) (SO₃^2-) salts are normal and acid salts derived from Sulphuric

(IV)acid H₂SO₃.

Sulphuric (VI)acid H₂SO₄is formed when sulphur(VI)oxide gas is bubbled in water.

The acid exist as a dibasic acid with two ionisable hydrogen. 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) (HSO₃^–) 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) exist also 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).

Sulphuric (IV)acid H₂SO₃is formed when sulphur(IV)oxide gas is bubbled in water.

The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO₃^2-) and hydrogen sulphate (VI) (HSO₄^–) salts. i.e.

H₂SO₃(aq) -> 2H⁺(aq) + SO₃^2-(aq)

H₂SO₃(aq) -> H⁺(aq) + HSO₃^–(aq)

All Sulphate (IV) (SO₃^2-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO₃), Barium (II) sulphate (IV) (BaSO₃) and Lead (II) sulphate (IV) (PbSO₃)

All Hydrogen sulphate (IV) (HSO₃^–) salts exist in solution/dissolved in water.

Sodium (I) hydrogen sulphate (IV) (NaHSO₃), Potassium (I) hydrogen sulphate (IV) (KHSO₃) and Ammonium hydrogen sulphate (IV) (NH₄HSO₃) exist also as solids.

Other Hydrogen sulphate (IV) (HSO₃^–) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO₃)₂) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO₃)₂).

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

Experiment:

In a clean dry test tube place separately about 1.0g of :

Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers.

Observations:

-Colourless droplets of liquid forms 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) sulphate (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 remain and blue and red litmus paper remain red in case of

Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV) -Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI).

SULPHUR AND ITS COMPOUND

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).

(i)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 amount of sulphur(VI)oxide gas.

Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water.

Sulphur (IV) oxide does not.

(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (IV) oxide gas.

Chemical equation

2CuSO₄(s) -> CuO(s) + SO₂(g)

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

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

Experiments/Observations:

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

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

Observation	Inference
White precipitate/ppt	SO42- , SO32- , CO32- , Cl- ions

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

Observation 1

Observation	Inference
White precipitate/ppt persists	SO₄^2-, Cl^–ions
Observation 2
Observation	Inference
White precipitate/ppt dissolves	SO32- , CO32- , ions

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

Observation 1

Observation	Inference
White precipitate/ppt persists on boiling	SO₄^2-ions
Observation 2
Observation	Inference
White precipitate/ppt dissolves on boiling	Cl ^–ions

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

Observation 1

Observation

Inference

(i)acidified potassium manganate(VII)decolorized

(ii)Orange colour of acidified potassium dichromate(VI) turns to green

SO₃^2-ions

Observation 2

Observation

Inference

(i)acidified potassium manganate(VII) not decolorized

(ii)Orange colour of acidified potassium dichromate(VI) does not turns to green

CO₃^2-ions

Experiments/Observations:

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

To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve.
SULPHUR AND ITS COMPOUND

Observation	Inference
White precipitate/ppt	SO42- , SO32- , CO32- ions

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

Observation 1

Observation	Inference
White precipitate/ppt persists	SO₄^2-, ions
Observation 2
Observation	Inference
White precipitate/ppt dissolves	SO32- , CO32- , ions

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

Observation 1

Observation

Inference

(i)acidified potassium manganate(VII)decolorized

(ii)Orange colour of acidified potassium dichromate(VI) turns to green

SO₃^2-ions

Observation 2

Observation

Inference

(i)acidified potassium manganate(VII) not decolorized

(ii)Orange colour of acidified potassium dichromate(VI) does not turns to green

CO₃^2-ions

Explanations

Using Lead(II)nitrate(V)

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

Chemical/ionic equation:
Pb²⁺(aq) + Cl^–(aq)	->	PbCl₂(s)
Pb²⁺(aq) + SO₄²⁺(aq)	->	PbSO₄(s)
Pb²⁺(aq) + SO₃²⁺(aq)	->	PbSO₃(s)
Pb²⁺(aq) + CO₃²⁺(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/ persists.
Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.

. Chemical/ionic equation:
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) do 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 turns to green. Carbon(IV)oxide will not.

Chemical equation:

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

(purple) (colourless)

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

(Orange) (green)

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.

SULPHUR AND ITS COMPOUND

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-), Sulphate (IV)salts (SO₃^2-) and carbonates(CO₃^2-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV).

Chemical/ionic equation:

Ba²⁺(aq) + SO₄²⁺(aq) -> BaSO₄(s)

Ba²⁺(aq) + SO₃²⁺(aq) -> BaSO₃(s)

Ba²⁺(aq) + CO₃²⁺(aq) -> BaCO₃(s)

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

Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists.
Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.

. Chemical/ionic equation:

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 turns to green. Carbon(IV)oxide will not.

Chemical equation:
5SO₃^2-(aq) + 2MnO₄^–(aq) +6H+(aq)	-> 5SO₄^2-(aq) + 2Mn²⁺(aq) + 3H₂O(l)
(purple)	(colourless)
3SO₃^2-(aq) + Cr₂O₇^2-(aq) +8H+(aq)	-> 3SO₄^2-(aq) + 2Cr³⁺(aq) + 4H₂O(l)
(Orange)	(green)

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 Sulphate(IV) (SO₃^2-) salts

Practice

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

(a)Identify the:
I: Sodium salt solution	Sodium sulphate(IV)/Na₂SO₃
II: White precipitate	Barium sulphate(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)

Study the flow chart below and answer the questions that follow. (i)Write equation for the reaction taking place at:

I.The roasting furnace	(1mk)
II.The absorption tower	(1mk)
III.The diluter	(1mk)

(ii) The reaction taking place in chamber K is

SO2 (g) + 1/2O2 (g) SO3 (g)

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

II.Name another substance used in chamber K

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