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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 16 and valency 2 /divalent and thus forms the ion S2-

A. 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) ,Galena (PbS,Zinc blende(ZnS))and iron pyrites(FeS2) in other parts of the world.

B. 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 170oC 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 115oC.

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.

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It is about 99% pure.

Diagram showing extraction of Sulphur from Fraschs Process

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C. Allotropes of Sulphur.

1. Sulphur exists as two crystalline allotropic forms:

(i)Rhombic sulphur

(ii)Monoclinic sulphur

Rhombic sulphur

Monoclinic sulphur

Bright yellow crystalline solid

Has a melting point of 113oC

Has a density of 2.06gcm-3

Stable below 96oC

Has octahedral structure

Pale yellow crystalline solid

Has a melting point of 119oC

Has a density of 1.96gcm-3

Stable above 96oC

Has a needle-like structure

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

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2. 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.

D. Heating Sulphur.

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

 

 

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On heating the yellow sulphur powder melts at 113oC to clear amber liquid with low viscosity and thus flows easily.

On further heating to 160oC the molten liquid darkens to a brown very viscous liquid that does not flow easily.

This is because the S8 rings break into S8 chain 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 160oC, 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 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2 molecules that solidifies to S8 ring 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 113oC

Amber yellow liquid

 

 

Heat to 160oC

Liquid darkens

 

 

 

Heat to 444oC

Liquid boils to brown vapour

 

 

Cool to room temperature

Yellow sublimate

(Flowers of Sulphur)

Puckered S8 ring

 

Puckered S8 ring in liquid form (low viscosity/flow easily)

 

Puckered S8 ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow)

 

 

Mixture of S8 ,S6 ,S2 vapour

 

 

Puckered S8 ring

 

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/benzene, poor conductor of heat and electricity. It has a melting point of 115oC and a boiling point of 444oC.

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 amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e

(i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid

(ii) 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) (Sulphur(VI)Oxide gas traces)

 SO2(g)  + H2O(l)  ->   H2 SO3 (aq) ( sulphuric(IV)acid) SO3(g)  + H2O(l)  ->   H2 SO4 (aq) ( sulphuric(VI)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 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)

4.The set up below show the reaction of sulphur on heated concentrated sulphuric(VI)acid.

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(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) + 3H2 SO4 (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

6. Revision Practice

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

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com

 

 

 

 

 

(a)Name the substances that passes through:

 M  Superheated water at 170oC and 10 atmosphere pressure

 L  Hot compressed air

 N  Molten sulphur

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

M- Superheated water at 170oC 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 96oC

Below 96oC

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

 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 ,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

B.COMPOUNDS OF SULPHUR

The following are the main compounds of sulphur:

 (i) Sulphur(IV)oxide

 (ii) Sulphur(VI)oxide .

 (iii) Sulphuric(VI)acid

 (iv) Hydrogen Sulphide

 (v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- 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:Using Copper and Sulphuric(VI)acid.

 

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


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(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 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 and choking smell which liquidifies easily. It is about two times denser than air.

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

Image From EcoleBooks.com

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) -> H2 SO3 (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 (H2 SO3 (aq)).

Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions:

  H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)

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

The SO32- (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) SO42- (aq) 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 sulphuric(IV)acid

 SO2(g) + H2O(l) -> H2 SO3 (aq)

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

 Method I. H2 SO3 (aq) + (dye + O) -> H2 SO4 (aq) + dye

(coloured) (colourless)

 Method II. H2 SO3 (aq) + (dye) -> H2 SO4 (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 Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.

 Chemical/ionic equation:

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

SO2(g) + H2O(l) -> H2 SO3 (aq)

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

H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)

(iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq)

3SO32-(aq)
+ Cr2O72-(aq) +8H+(aq) -> 3SO42-(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 Cr2O72- 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 SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.

Chemical/ionic equation:

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

SO2(g) + H2O(l) -> H2 SO3 (aq)

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

H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)

(iii)Oxidation of SO32- (aq)and reduction of MnO4 (aq)

5SO32-(aq)
+ 2MnO4 (aq) +6H+(aq) -> 5SO42-(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.


(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 SO32- ions in sulphuric (IV) acid to SO42- ions in sulphuric(VI) acid.

Chemical/ionic equation:

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

SO2(g) + H2O(l) -> H2 SO3 (aq)

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

H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)

(iii)Oxidation of SO32- (aq)and reduction of MnO4 (aq)

SO32-(aq)
+ Br2 (aq) + H2O(l) -> SO42-(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 SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.

Chemical/ionic equation:

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

SO2(g) + H2O(l) -> H2 SO3 (aq)

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

H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)

(iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq)

SO32-(aq)
+ 2Fe3+ (aq) +3H2O(l) -> SO42-(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 SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.

Chemical/ionic equation:

SO2(g) + 2HNO3 (l)  -> H2 SO4 (l) + NO2 (g)

(brown fumes/gas)

(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 SO32- ions in sulphuric(IV) acid to SO42- 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 SO32- ,SO42- ,CO32- ions. i.e.

Chemical/ionic equation:

SO32-(aq)
+ Ba2+ (aq) -> BaSO3(s)

 white precipitate

SO42-(aq)
+ Ba2+ (aq) -> BaSO4(s)

 white precipitate

CO32-(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 SO42-(aq)ions (BaSO4(s)) is present.

(ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is 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)

I. 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 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.

Or

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)

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 involves 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;

 I. 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)

 II. 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 :

 (i) cheaper/less expensive

 (ii) 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 450oC 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 450oC to ensure that:

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

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

(iv)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)

IV. Uses of Sulphuric(VI)acid

Sulphuric (VI) acid is used:

 (i) in making dyes and paint

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

 (iii) for making soapless detergents

 (iv) for making sulphate agricultural fertilizers

 

VI. Sketch chart diagram showing the Contact process

Image From EcoleBooks.com

(c) Properties of Concentrated sulphuric (VI) acid   (i) Concentrated
sulphuric (VI) acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC.

(ii) 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.

(iii) Concentrated
sulphuric (VI)acid is a covalent compound. It has no free H+ ions.

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.

(iv) 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:

I. As a dehydrating agent

Experiment I;

Put about four spatula end full of brown sugar and glucose in separate 10cm3 beaker.

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.

 10cm3 beaker becomes very hot.

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.

Caution

This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi.

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 170oC.

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

Chemical equation

  C2H5OH(l) –conc.H2SO4/170oC –> 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

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.

 

 

II. 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.

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 non-metals 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.

2. 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 SO42-(aq)ions:

H2SO4 (aq) -> 2H+(aq) + SO42-(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.

(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 with 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.

(b) Preparation

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

 

Image From EcoleBooks.com

(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 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 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) + S2- (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  + 2H+(aq)

 (black)

Fe2+(aq)  + H2S  -> FeS  + 2H+(aq)

 (black)

Zn2+(aq)  + H2S  -> ZnS  + 2H+(aq)

 (black)

Cu2+(aq)  + H2S  -> CuS  + 2H+(aq)

 (black)

2Cu+(aq)  + H2S  -> Cu2S  + 2H+(aq)

 (black)

6. Dry hydrogen sulphide was ignited as below.

Image From EcoleBooks.comImage From EcoleBooks.com    


(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:

(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 Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur.

Chemical/ionic equation:

4H2S(aq)
+ Cr2O72-(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 Cr2O72- 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.


(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:

H2 S(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

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)

(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)

Practice

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

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(a)Name the black solid   Copper(II)sulphide

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

I. Blue solution Cu2+(aq)

 II. Green solution Fe2+(aq)

III. 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)
+ Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l)

3SO32-(aq)
+ Cr2O72-(aq) +8H+(aq) -> 3SO42-(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)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts

1. Sulphate (VI) (SO42-) salts are normal and acid salts derived from Sulphuric (VI)acid H2SO4.

2. Sulphate(IV) (SO32-) salts are normal and acid salts derived from Sulphuric (IV)acid H2SO3.

3. Sulphuric (VI)acid H2SO4 is 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) (SO42-) and hydrogen sulphate (VI) (HSO4) salts.

i.e.

  H2SO4 (aq)
-> 2H+(aq) + SO42-(aq)

  H2SO4 (aq)
-> H+(aq) + HSO4(aq)

All Sulphate (VI) (SO42-) 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).

4. Sulphuric (IV)acid H2SO3 is 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) (SO32-) and hydrogen sulphate (VI) (HSO4) salts.

i.e.

  H2SO3 (aq)
-> 2H+(aq) + SO32-(aq)

  H2SO3 (aq)
-> H+(aq) + HSO3(aq)

 

All Sulphate (IV) (SO32-) 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) (SO42-)and sulphate(IV) (SO32-) 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).

Explanation

(i)All Sulphate (VI) (SO42-) 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) (SO42-) 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 (VI) oxide gas.

Chemical equation

2CuSO4 (s)  ->  CuO(s)
+ SO3(g)

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

6. The following experiments show the test for the presence of sulphate (VI) (SO42-)and sulphate(IV) (SO32-) ions in a sample of a salt/compound:

Experiments/Observations:

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

I. 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- , Clions

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

Observation 1

Observation

Inference

White precipitate/ppt persists

SO42- , Clions

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

SO42-

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

SO32- 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

CO32- ions

Experiments/Observations:

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

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

Observation

Inference

White precipitate/ppt

SO42- , SO32- , CO32- ions

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

 

 

Observation 1

Observation

Inference

White precipitate/ppt persists

SO42- , 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

SO32- 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

CO32- ions

Explanations

Using Lead(II)nitrate(V)

(i)Lead(II)nitrate(V) solution reacts with chlorides(Cl), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) 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)   ->   H2 O (l) + Pb2+(aq)  +   SO2 (g)

 PbCO3 (s) + 2H+(aq)   ->   H2 O (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:

5SO32-(aq)
+ 2MnO4 (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)

(purple) (colourless)

3SO32-(aq)
+ Cr2O72-(aq) +8H+(aq) -> 3SO42-(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.

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

(i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) 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)   ->   H2 O (l) + Ba2+(aq)  +   SO2 (g)

 BaCO3 (s) + 2H+(aq)   ->   H2 O (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:

5SO32-(aq)
+ 2MnO4 (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)

(purple) (colourless)

3SO32-(aq)
+ Cr2O72-(aq) +8H+(aq) -> 3SO42-(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) (SO42-)and Sulphate(IV) (SO32-) salts

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Practice revision question

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

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(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)   +   SO32-(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

3SO32-(aq)
+ Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)

(Orange) (green)

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

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

 I.The roasting furnace (1mk)

 2FeS2 (s) + 5O2 (g)  -> 2FeO(s) + 4SO2 (g)

 II.The absorption tower (1mk)

H2SO4 (l) + SO3 (g)  -> H2S2O7(l)

 III.The diluter (1mk)

H2S2O7(l) + H2 O(l)  -> 2H2SO4 (l)

(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

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 sulphate (IV) and sodium sulphate (VI).

Add acidified Barium nitrate(V)/chloride.

White precipitate formed with sodium sulphate (VI)

No white precipitate formed with sodium sulphate (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 =22.4dm3)

Chemical equation

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

Mole ratio Cu(s: SO2 (g) = 1:1

Method 1

1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3

(120 x 1000) g -> (120 x 1000) g x 22.4.dm3)

63.5 g

= 42330.7087

Method 2  

Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles

63.5

Moles SO2 = Moles of Cu = 1889.7639 moles

Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4)

= 42330.7114

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(a)Identify the:

(i)cation responsible for the green solution T

Cr3+

(ii)possible anions present in white precipitate R


CO32-, SO32-, SO42-

(b)Name gas V

Sulphur (IV)oxide

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

Ba2+ (aq) + CO32- (aq) ->
BaCO3(s)

Ba2+ (aq) + SO32- (aq) ->
BaSO3(s)

Ba2+ (aq) + SO42- (aq) ->
BaSO4(s)


 




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