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Properties of Concentrated sulphuric (VI) acid  

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

(ii) Concentratedsulphuric (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) Concentratedsulphuric (VI)acid is a covalent compound. It has no free H+ ions.

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

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

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

(v)The following are some chemical properties of concentratedsulphuric (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 concentratedsulphuric (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

Concentratedsulphuric (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 concentratedsulphuric (VI) acid .Warm .

Observation;

          Colour change from blue to white.

Explanation

Concentratedsulphuric (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 concentratedsulphuric (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

Concentratedsulphuric (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 concentratedsulphuric (VI) acid. Heat gently

Caution:

This should be done in a fume chamber/open

Observation;

 Colourless gas produced.

Explanation

Concentratedsulphuric (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 concentratedsulphuric (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

Concentratedsulphuric (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 Concentratedsulphuric (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

Concentratedsulphuric (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 Concentratedsulphuric (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

Concentratedsulphuric (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 acidRelative molecula massBoiling point(oC)
Hydrochloric acid(HCl)36.535.0
Nitric(V)acid(HNO3)63.083.0
Sulphuric(VI)acid(H2SO4)98.0333

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)