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. 45 - 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). 46 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. .


46 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.


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. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1.


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. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1. Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl
48 (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.


Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl
48 (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. 49 (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.


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. 49 (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.


49 (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
50 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 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.


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
50 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 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow. (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.


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
50 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 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.

(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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