JK-TK/Newdata huggingface dataset:
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. 26 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.
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. 27 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.
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
28 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 nonmetals to non metallic oxides and itself reduced to sulphur(IV)oxide gas.
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. 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)
29 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.
(i)Chemical equation KNO3(s) + H2SO4(aq) --> KHSO4(l) + HNO3(g) NaNO3(s) + H2SO4(aq) --> NaHSO4(l) + HNO3(g)
29 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.
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,
30 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.
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
31 extinguishes a burning splint and forms a white precipitate when bubbled in lime water. A sulphate (VI) salts is formed.
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
31 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)
32 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.
Chemical/ionic equation ZnO(s) + H2SO4(aq) --> ZnSO4(aq) + H2O(l) ZnO(s) + 2H+(aq) --> Zn2+ (aq) + H2O(l)
32 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.
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)
33 (iv) Hydrogen sulphide(H2S) (a) Occurrence Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya. It is present in rotten eggs and human excreta. (b) Preparation Hydrogen sulphide is prepared in a school laboratory by heating Iron (II) sulphide with dilute hydrochloric acid. (c) Properties of Hydrogen sulphide(Questions) 1. Write the equation for the reaction for the school laboratory preparation of Hydrogen sulphide. 34 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.
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.
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)
35 2Cu+(aq) + H2S -> Cu2S + 2H+(aq) (black) 6. Dry hydrogen sulphide was ignited as below. (i) State the observations made in flame A Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water. Chemical equation: 2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g) Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water. Chemical equation: 2H2S(g) + O2(g) -> 2H2O(l) + 2S(s) 7. Hydrogen sulphide is a strong reducing agent that is oxidized to yellow solid sulphur as precipitate. The following experiments illustrate the reducing properties of Hydrogen sulphide. (a)Turns Orange acidified potassium dichromate(VI) to green Experiment: (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.
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. Flame A Dry Hydrogen sulphide gas
36 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 .
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. 37 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 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.