JK-TK/Newdata huggingface dataset:
When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII). (f) Reaction of cation with sulphide / S2- ions All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed. Procedure Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes. Add six drops of Potassium /sodium sulphide solution. Observation Inference Explanation Black ppt S2- ions CuS, FeS,MgS,PbS, ZnS are black insoluble precipitates Cu2+(aq) + S2-(aq) -> CuS(s) Pb2+(aq) + S2-(aq) -> PbS(s) Fe2+(aq) + S2-(aq) -> FeS(s) Zn2+(aq) + S2-(aq) -> ZnS(s) Sample qualitative analysis guide
You are provided with solid Y(aluminium (III)sulphate(VI)hexahydrate).Carry out the following tests and record your observations and inferences in the space provided. 1(a) Appearance Observations inference (1mark) White crystalline solid Coloured ions Cu2+ , Fe2+ ,Fe3+ absent (b)Place about a half spatula full of the solid into a clean dry boiling tube. Heat gently then strongly. Observations inference (1mark) Colourless droplets formed on the cooler Hydrated compound/compound part of the test tube containing water of crystallization Solid remains a white residue (c)Place all the remaining portion of the solid in a test tube .Add about 10cm3 of distilled water. Shake thoroughly. Divide the mixture into five portions. Observation Inference (1mark) Solid dissolves to form Polar soluble compound a colourless solution Cu2+ , Fe2+ ,Fe3+ absent (i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali.
Shake thoroughly. Divide the mixture into five portions. Observation Inference (1mark) Solid dissolves to form Polar soluble compound a colourless solution Cu2+ , Fe2+ ,Fe3+ absent (i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali. Observation Inference (1mark) White ppt, soluble in excess Zn2+ , Pb2+ , Al3+ (ii)To the second portion, add three drops of aqueous ammonia then add excess of the alkali. Observation Inference (1mark) White ppt, insoluble in excess Pb2+ , Al3+ (iii)To the third portion, add three drops of sodium sulphate(VI)solution. Observation Inference (1mark) No white ppt Al3+
(iv)I.To the fourth portion, add three drops of Lead(II)nitrate(IV)solution. Preserve Observation Inference (1mark) White ppt CO32-, SO42-, SO32-, Cl-, II.To the portion in (iv) I above , add five drops of dilute hydrochloric acid. Observation Inference (1mark) White ppt persist/remains SO42-, Cl-, III.To the portion in (iv) II above, heat to boil. Observation Inference (1mark) White ppt persist/remains SO42-, Note that: (i)From test above, it can be deduced that solid Y is hydrated aluminium(III)sulphate(VI) solid (ii)Any ion inferred from an observation below must be derived from previous correct observation and inferences above. e.g. Al3+ in c(iii) must be correctly inferred in either/or in c(ii) or c(i)above SO42- in c(iv)III must be correctly inferred in either/or in c(iv)II or c(iv)I above (iii)Contradiction in observations and inferences should be avoided.e.g. “White ppt soluble in excess” to infer presence of Al3+ ,Ba2+ ,Pb3+ (iv)Symbols of elements/ions should be correctly capitalized. e.g. “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong. Sample solutions of salt were labeled as I,II, III and IV.
e.g. “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong. Sample solutions of salt were labeled as I,II, III and IV. The actual solutions, not in that order are lead nitrate, zinc sulphate potassium chloride and calcium chloride. a)When aqueous sodium carbonate was added to each sample separately, a white precipitate was formed in I, III and IV only. Identify solution II. b)When excess sodium hydroxide was added to each sample separately, a white precipitate was formed in solutions III and I only. Identify solution I 17.When solids/salts /solutes are added to a solvent ,some dissolve to form a solution. Solute + Solvent -> Solvent If a solution has a lot of solute dissolved in a solvent ,it is said to be concentrated. If a solution has little solute dissolved in a solvent ,it is said to be dilute. There is a limit to how much solute can dissolve in a given /specified amount of solvent/water at a given /specified temperature. The maximum mass of salt/solid/solute that dissolve in 100g of solvent/water at a specified temperature is called solubility of a salt. When no more solute can dissolve in a given amount of solvent at a specified temperature, a saturated solution is formed. For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution. The solubility of a salt is thus calculated from the formula Solubility = Mass of solute/salt/solid x 100 Mass/volume of water/solvent Practice examples (a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O Mass/volume of water/solvent 50.0 (b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water.
For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution. The solubility of a salt is thus calculated from the formula Solubility = Mass of solute/salt/solid x 100 Mass/volume of water/solvent Practice examples (a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O Mass/volume of water/solvent 50.0 (b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 50.0 x 100 ) = 20.0 g /100g H2O
Mass/volume of water/solvent 250.0 (c)If the solubility of potassium chlorate(V) is 5g/100g H2O at 80oC,how much can dissolve in 5cm3 of water at 80oC . Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 5 x 5 = 0.25g of KClO3 dissolve 100 (d)If the solubility of potassium chlorate(V) is 72g/100g H2O at 20oC,how much can saturate 25g of water at 20oC . Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 72 x 25 = 18.0g of KClO3 dissolve/saturate 100 (e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC. Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O Mass/volume of water/solvent 40.0.
Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 72 x 25 = 18.0g of KClO3 dissolve/saturate 100 (e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC. Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O Mass/volume of water/solvent 40.0. (f)What volume of water should be added to 22.0g of water at 10oC if the solubility of KNO3 at 10oC is 5.0g/100g H2O? Solubility is mass/100g H2O => 22.0g + x = 100cm3/100g H2O X= 100 – 22 = 78 cm3 of H2O 18. A graph of solubility against temperature is called solubility curve. It shows the influence of temperature on solubility of different substances/solids/salts. Some substances dissolve more with increase in temperature while for others dissolve less with increase in temperature KNO3 KClO3 Saturated solution of KClO3 Solubility
Note: (i)solubility of KNO3 and KClO3 increase with increase in temperature. (ii)solubility of KNO3 is always higher than that of KClO3 at any specified temperature. (iii)solubility of NaCl decrease with increase in temperature. (iv)NaCl has the highest solubility at low temperature while KClO3 has the lowest solubility at low temperature. (v)At point A both NaCl and KNO3 are equally soluble. (vi)At point B both NaCl and KClO3 are equally soluble. (vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution. (viii) An area below the solubility curve of the salt shows an unsaturated solution. 19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature.
(vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution. (viii) An area below the solubility curve of the salt shows an unsaturated solution. 19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature. (b) For salts whose solubility decreases with increase in temperature, crystals form when the salt solution at lower temperatures is heated to a higher temperature. The examples below shows determination of the mass of crystals deposited with changes in temperature. 1.The solubility of KClO3 at 100oC is 60g/100g water .What mass of KClO3 will be deposited at: (i)75 oC if the solubility is now 39g/100g water. At 100oC = 60.0g Less at 75oC = - 39.0g Mass of crystallized out 21.0g (i)35 oC if the solubility is now 28 g/100g water. At 100oC = 60.0g Less at 35oC = - 28.0.0g Mass of crystallized out 32.0g 2. KNO3 has a solubility of 42 g/100g water at 20oC.The salt was heated and added 38g more of the solute which dissolved at100oC. Calculate the solubility of KNO3 at 100oC. Solubility of KNO3 at 100oC = solubility at 20oC + mass of KNO3 added => 42g + 38g = 80g KNO3 /100g H2O 3. A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved.
A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved. Mass of solvent at 20oC = mass of solution – mass of solute => 65 - 5 = 60g Solubility before adding salt = mass of solute x 100 Volume of solvent => 5 x 100 = 8.3333g/100g water 60 Mass of solute to equalize with solubility = 25 – 8.3333g = 16.6667g Mass of solute undissolved = 60.0 - 16.6667g = 43.3333 g 4. Study the table below Salt Solubility in gram at
50oC 20oC KNO3 90 30 KClO3 20 6 (i)What happens when the two salts are dissolved in water then cooled from 50oC to 20oC. (90 – 30) = 60.0 g of KNO3 crystals precipitate (20 – 6) = 14.0 g of KClO3 crystals precipitate (ii)State the assumption made in (i) above. Solubility of one salt has no effect on the solubility of the other. 5. 10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate. (a)Calculate the mass of anhydrous salt obtained. Hydrated on heating leave anhydrous = 7.93 g (b)Calculate the mass of water of crystallization in the hydrated salt Mass of water of crystallization = hydrated – anhydrous => 10.0 - 7.93 = 2.07 g (c)How many moles of anhydrous salt are there in 10of hydrate?
10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate. (a)Calculate the mass of anhydrous salt obtained. Hydrated on heating leave anhydrous = 7.93 g (b)Calculate the mass of water of crystallization in the hydrated salt Mass of water of crystallization = hydrated – anhydrous => 10.0 - 7.93 = 2.07 g (c)How many moles of anhydrous salt are there in 10of hydrate? (K= 39.0,C=12.0.O= 16.0) Molar mass K2CO3= 138 Moles K2CO3 = mass of K2CO3 => 7.93 = 0.0515 moles Molar mass K2CO3 138 (d)How many moles of water are present in the hydrate for every one mole of K2CO3 ? (H=1.0.O= 16.0) Molar mass H2O = 18 Moles H2O = mass of H2O => 2.07 = 0.115 moles Molar mass H2O 18 Mole ratio H2O : K2CO3 = 0.115 moles 2 = 2 0.0515 moles 1 (e)What is the formula of the hydrated salt? K2CO3 .2 H2O 6. The table below shows the solubility of Potassium nitrate(V) at different temperatures. Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0 mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0 (a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC
(b)From the graph show and determine (i)the mass of KNO3 dissolved at: I. 20oC From a correctly plotted graph = 32g II. 35oC From a correctly plotted graph = 57g III.
Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0 mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0 (a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC
(b)From the graph show and determine (i)the mass of KNO3 dissolved at: I. 20oC From a correctly plotted graph = 32g II. 35oC From a correctly plotted graph = 57g III. 55oC From a correctly plotted graph = 104g (ii)the temperature at which the following mass of KNO3 dissolved: I. 22g From a correctly plotted graph =13.0oC II. 30g From a correctly plotted graph =17.5oC III.100g From a correctly plotted graph =54.5oC (c)Explain the shape of your graph. Solubility of KNO3 increase with increase in temperature/More KNO3 dissolve as temperature rises. (d)Show on the graph the supersaturated and unsaturated solutions. Above the solubility curve write; “supersaturated” Below the solubility curve write; “unsaturated” (e)From your graph, calculate the amount of crystals obtained when a saturated solution of KNO3 containing 180g of the salt is cooled from 80oC to: I. 20oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 32 g Mass of KNO3crystals = 148 g II. 35oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 58 g Mass of KNO3crystals = 122 g III. 55oC
Solubility before heating = 180 g Less Solubility after heating(from the graph) = 102 g Mass of KNO3crystals = 78 g 7. The table below shows the solubility of salts A and B at various temperatures.
35oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 58 g Mass of KNO3crystals = 122 g III. 55oC
Solubility before heating = 180 g Less Solubility after heating(from the graph) = 102 g Mass of KNO3crystals = 78 g 7. The table below shows the solubility of salts A and B at various temperatures. Temperature(oC) 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 Solubility of A 28.0 31.0 34.0 37.0 40.0 43.0 45.0 48.0 51.0 Solubility of B 13.0 21.0 32.0 46.0 64.0 85.0 110.0 138.0 169.0 (a)On the same axis plot a graph of solubility (y-axis) against temperature for each salt. (b)At what temperature are the two salts equally soluble. The point of intersection of the two curves = 24oC (c)What happens when a mixture of 100g of salt B with 100g if water is heated to 80oC From the graph, the solubility of B at 80oC is 169g /100g water. All the 100g crystals of B dissolve. (d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC. Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II.
(d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC. Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens.
Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8.
Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves.
Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted.