Chemistry Unit 3B - By Maple Leaf International School

Unit 3B Notes Prepared By Adnan Chowdhury

Civil Engineer, BUET, A Level Chemistry Course Instructor Maple Leaf International

PREPARED BY ADNAN CHOWDHURY

Flame tests: Procedure:

*



First the test sample is converted into a chloride salt by adding concentrated HCl acid. This is because chloride salts are more volatile than any other compounds and hence some of the unknown goes into the gas phase readily when heated in the hot flame.



A platinum or nichrome wire or a specially marketed flame test rod is taken. Platinum or nichrome wire is chosen because they have high heat conductivity. A spatula or a wooden splint cannot be taken. The wire is cleaned by dipping it in some concentrated HCl acid on a watch glass and then placing it in the hottest part (blue part) of a Bunsen flame. The flame should not be coloured. If it is, the treatment will be repeated until the flame is not coloured.



Once again, the clean wire is dipped in concentrated HCl acid and then some of the solid under test.



This is then placed in the hottest part (blue part) of the flame and the colour of the flame is observed. Flame colour

Inference

yellow

sodium ion, Na+

lilac

potassium ion, K+

yellow-red / brick red*

calcium ion, Ca2+

crimson red*

lithium ion, Li+

carmine red*

strontium ion, Sr2+

pale / apple green

barium ion, Ba2+

blue

copper (II) ion, Cu2+

no colour

magnesium ion, Mg2+

Further tests would be needed to distinguish these ions. 

The flame test is the only test for a group-1 metal cation in a compound.



The flame test cannot be used on mixtures containing two of these ions because the colour produced by one of the ions will mask the colour produced by the other metal ion.

MAPLE LEAF INTERNATIONAL SCHOOL

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PREPARED BY ADNAN CHOWDHURY 

Origin of Flame Colour: The heat energy of the flame causes the compound to vaporize and promotes an electron in the metal ion into a higher energy level or excited state. The electron falls back to its normal shell or ground state and as it does so, energy in the form of visible light is emitted. The light that is emitted of a characteristic frequency, and hence colour, is dependent on the energy level difference between the two shells.

Action of heating a Solid: Gas or vapour or other carbon dioxide

oxygen

Possible source 

carbonates of metals (including Li2CO3) other than group 1. Li2CO3(s)  Li2O(s) + CO2(g) MCO3(s)  MO(s) + CO2(g) M = Group 2 metal  when a hydrated carbonate such as sodium carbonate is heated. Na2CO3.10H2O(s)  2NaOH(s) + CO2(g) + 9H2O(l) group 1 nitrates (other than LiNO3) 2MNO3(s)  2MNO2(s) + O2(g) M = Group 1 metal

oxygen and nitrogen dioxide

nitrates of group 2 (and LiNO3) 4LiNO3(s)  2Li2O(s) + 4NO2(g) + O2(g) 2M(NO3)2(s)  2MO(s) + 4NO2(g) + O2(g) M = Group 2 metal

water

hydrated salts or hydrogen carbonate Na2CO3.10H2O(s)  2NaOH(s) + CO2(g) + 9H2O(l) 2MHCO3(s)  M2CO3(s) + CO2(g) + H2O(l)M = Group 1 metal

a white solid sublimes on the ammonium salt present cooler part of the tube e.g. NH4Cl (s)  NH3(g) + HCl(g)


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Recognition and identification of common gases: Gas

Observations

oxygen, O2

a colourless, odourless gas which relights a glowing splint. Oxygen can be produced by:  Heating a group-1 nitrate (apart from LiNO3) 2MNO3(s)  2MNO2(s) + O2(g) M = Group 1 metal  Heating other nitrate, but NO2 is also present. 4LiNO3(s)  2Li2O(s) + 4NO2(g) + O2(g) 2M(NO3)2(s)  2MO(s) + 4NO2(g) + O2(g) M = Group 2 metal  The catalytic decomposition of H2O2. H2O2.(aq)  2H2O(l) + O2(g)

carbon dioxide, CO2

a colourless, odourless gas which gives a white precipitate with limewater (calcium hydroxide solution) i.e. it turns limewater milky. If excess CO2 is passed, the precipitates dissolves giving a colourless solution. Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(l) CaCO3(s) + H2O(l) + CO2(g)  Ca(HCO3)2(aq) Carbon dioxide is produced by:  A reaction between an acid and a carbonate or hydrogen carbonate. CO32-(s)/(aq) + 2H+(aq)  CO2(g) + H2O(l) HCO3-(s)/(aq) + H+(aq)  CO2(g) + H2O(l)  Heating a carbonate (apart from sodium, potassium or barium carbonates). Li2CO3(s)  Li2O(s) + CO2(g) MCO3(s)  MO(s) + CO2(g) M = Group 2 metal  Heating a group-1 hydrogen carbonate. 2MHCO3(s)  M2O(s) + CO2(g) + H2O(l) M = Group 1 metal  When a hydrated carbonate such as sodium carbonate is heated. Na2CO3.10H2O(s)  2NaOH(s) + CO2(g) + 9H2O(l)  Heating a hydrogen carbonate 2MHCO3(s)  M2CO3(s) + CO2(g) + H2O(l) M = Group 1 metal


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PREPARED BY ADNAN CHOWDHURY Gas

Observations

sulfur dioxide, SO2

a colourless gas which is acidic and decolourises acidified potassium dichromate(VI) paper or solution from orange [Cr2O72- (aq)] to green [Cr3+(aq)]. Sulfur dioxide is produced by:  Warming a acid with a sulfite. SO32-(aq)/(s) + 2H+(aq)  SO2 (g) + H2O(l)  Burning sulfur. S(s) + O2(g)  SO2 (g)  Reducing concentrated sulfuric acid. Br-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HBr(g) 2HBr(g) + H2SO4(aq)  Br2(g) + SO2 (g) + 2H2O(l) I-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HI(g) 2HI(g) + H2SO4(aq)  I2(s) + SO2 (g) + 2H2O(l) 6HI(g) + SO2 (g)  H2S(g) + 3I2(s) + 2H2O(l)

ammonia, NH3

a colourless, pungent smelling gas which turns moist red litmus paper blue and forms a white smoke with hydrogen chloride gas. NH3(g) + HCl(g)  NH4Cl(s) Ammonia is produced by:  Heating sodium hydroxide with an ammonium salt. NH4+(aq) + OH-(aq)  NH3(g) + H2O(l)  Adding sodium hydroxide and aluminium powder to a nitrate. 3NO3-(aq) + 8Al(s) + 5OH-(aq) + 18H2O(l)  3NH3(g) + 8[Al(OH)4 ]-(aq)

*nitrogen dioxide, NO2

a brown gas. Nitrogen dioxide is produced by:  Heating a group-2 nitrate or LiNO3 4LiNO3(s)  2Li2O(s) + 4NO2(g) + O2(g) 2M(NO3)2(s)  2MO(s) + 4NO2(g) + O2(g) M = Group 2 metal a colourless, odourless gas which ignites a lighted splint with a ‘pop’ sound. Hydrogen is produced by:  The reaction between an acid and a reactive metal. M(s) + nH+(aq)  Mn+(aq) + n/2 H2(g)  The reaction between an alcohol and sodium ROH(l) + Na(s)  RO-Na+(l) + ½H2O(l)  The reaction between water and either a group-1 metal, or calcium, strontium, or barium. M(s) + H2O(l)  MOH(aq) + ½ H2(g) M = Group 1 metal M(s) + 2H2O(l)  M(OH)2(aq) + H2(g) M = Ca, Sr or Ba

hydrogen, H2


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Observations

hydrogen chloride, HCl steamy fumes on exposure to moist air, acidic and forms white smoke with ammonia. NH3(g) + HCl(g)  NH4Cl(s) Hydrogen chloride is produced by:  The reaction between concentrated sulfuric acid a chloride. Cl-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HCl(g)  The reaction between phosphorus pentachloride and either an alcohol or carboxylic acid. ROH(l) + PCl5(s)  RCl(l) + POCl3(l) + HCl(g) RCOOH(l) + PCl5(s)  ROCl(l) + POCl3(l) + HCl(g) water vapour, H2O

it turns blue cobalt chloride paper pink and white anhydrous copper (II) sulfate to blue hydrated copper (II) sulfate. CuSO4(s) + 5H2O(l)  CuSO4. 5H2O(aq)

chlorine, Cl2

a pale green gas which turns moist red litmus paper first blue and then bleaches it rapidly. If chlorine gas is passed into a solution of potassium bromide, the colourless solution becomes brown. 2KBr(aq) + Cl2(g)  2KCl(aq) + Br2(aq) The brown colour is due to Br2(aq). A solution of chlorine can be tested in the same way. Chlorine is produced by:  Electrolysis of a solution of chloride.  Adding dilute HCl acid to a solution containing chlorate(I) ions

*bromine, Br2

a brown gas which turns moist red litmus paper first blue and then bleaches it but it does more slowly. If bromine gas is passed into a solution of excess potassium iodide, the colourless solution becomes deep red brown. This is because iodine is liberated, and then reacts with excess I- ions to form the red brown I3-. Excess bromine would give a grey black precipitate of iodine. 2KI(aq) + Br2(g)  2KBr(aq) + I2(aq)/(s) I2(aq) + I-(aq)  I3-(aq) *Bromine is a brown fuming liquid at rtp. *Bromine dissolves in organic solvents to form a brown solution whereas nitrogen dioxide is insoluble.


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Observations

iodine, I2

a purple vapour which has no effect on moist red litmus paper. It turns starch solution dark blue or blue black. Copper (II) ions, Cu2+ in copper (II) sulfate react with iodide ions, I- in potassium iodide and forms a whitish brown precipitate. 2Cu2+ (aq) + 4I- (aq)  2CuI (s) + I2 (aq) white brown Iodine is grey-black solid at rtp. Solutions in aqueous potassium iodide are red brown and aqueous solutions are pale brown.

Action of dilute acids: •

When dilute sulfuric or hydrochloric acid is added to a substance a gas may be evolved or there may be a colour change in the solution.

Action of acid

Possible source

effervescence of a colourless, odourless gas which gives a white precipitate with limewater (calcium hydroxide solution) i.e. it turns limewater milky. Carbon dioxide evolved a colourless, odourless gas evolves which ignites a lighted splint with a ‘pop’ sound. Hydrogen evolved

carbonate or hydrogencarbonate CO32-(s)/(aq) + 2H+(aq)  CO2(g) + H2O(l) HCO3-(s)/(aq) + H+(aq)  CO2(g) + H2O(l)

yellow solution turns orange

chromate(VI) to dichromate(VI) 2CrO42-(aq) + 2H+(aq)  Cr2O72- (aq) + H2O(l)

sulfur dioxide evolved which is a colourless, acidic gas and decolourises acidified potassium dichromate(VI) paper or solution from orange [Cr2O72- (aq)] to green [Cr3+(aq)]. Also pale yellow sulfur precipitate is formed sulfur dioxide evolved on warming which is a colourless, acidic gas and decolourises acidified potassium dichromate(VI) paper or solution orange [Cr2O72- (aq)] to green [Cr3+(aq)].

thiosulfate S2O32-(aq) + 2H+(aq)  S(s) + SO2 (g) + H2O(l)


a metal M(s) + 2H+(aq)  2M+(aq)/M2+(aq) + H2 (g)

sulfite SO32-(aq)/(s) + 2H+(aq)  SO2 (g) + H2O(l)

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Tests for oxidizing and reducing agents: Reducing agents usually: decolourise aqueous acidified potassium manganate(VII) from purple to colourless and may also turn aqueous, acidified potassium dichromate(VI) from orange to green. Reducing agents include: iron(II) ions iodide ions hydrogen peroxide**

sulphite ions Oxidizing agents usually: liberate iodine as a brown solution or black solid from aqueous potassium iodide. Iodine solution gives a blue-black coloration with starch. Oxidizing agents include: acidified manganate(VII) ions acidified dichromate(VI) ions hydrogen peroxide** copper(II)ions aqueous chlorine aqueous bromine. ** Hydrogen peroxide acts both as an oxidizing agent and reducing agent.

Hydrogen peroxide: Aqueous hydrogen peroxide (H2O2) can act as both an oxidizing and a reducing agent often with the evolution of oxygen, although this may be unreliable. Observation on adding H2O2

Inference

brown precipitate

manganate(VII), brown precipitate is MnO2

purple solution is decolourised

manganate(VII) in acid solution

pale green solution turns yellow

iron(II) to iron(III) in acid solution


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PREPARED BY ADNAN CHOWDHURY Observation on adding H2O2

Inference

green precipitate turns brown

iron(II) hydroxide to iron(III) hydroxide

green alkaline solution goes yellow

chromium(III) to chromate(VI)

brown solution or black precipitate

iodine from iodide in acid solution

brown precipitate in alkaline solution

lead (II); brown precipitate is PbO2

Solubility: Soluble ionic compounds include: 

All group-1 salts.



All ammonium salts.



All nitrates.



All chlorides, apart from silver chloride and lead(II) chloride. (The solubility of bromides and iodides is similar to that of chlorides.)



All sulfates, apart from barium sulfate, strontium sulfate and lead(II) sulfate. Calcium sulfate and silver sulfate are slightly soluble.

Insoluble ionic compounds include: 

All carbonates apart from group-1 carbonates and ammonium carbonates.



All hydroxides, apart from group-1 hydroxides, barium hydroxide and ammonium hydroxide. Calcium and strontium hydroxides are slightly soluble.


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PRECIPITATION REACTIONS: When two aqueous solutions are mixed together and an insoluble compound is formed this is known as a precipitate not a suspension. The observation that a precipitate is formed should always be accompanied by the colour of the precipitate, even if this is white. Some reagents should be added until they are in excess. This may result in a precipitate forming then dissolving in excess reagent.

SODIUM HYDROXIDE AND AMMONIA SOLUTION: 

When small amount of dilute sodium hydroxide (NaOH) solution is added to a solution containing a metal ion a precipitate of the insoluble hydroxide is usually formed. Precipitates which are amphoteric hydroxides will dissolve in excess sodium hydroxide to give a solution containing a complex ion.



Small amount of dilute aqueous ammonia (NH3), when added to a solution containing a cation, will form the same hydroxide precipitate as dilute sodium hydroxide solution. Excess aqueous ammonia may dissolve the precipitate to form a complex ion.



Aqueous sodium hydroxide or aqueous ammonia should be added until it is in excess even if this is not explicitly stated in the instructions.

Likely ion

Observation on adding small Observation on amount of dilute NaOH or adding excess dilute NaOH dilute aqueous NH3

Observation on adding excess dilute aqueous NH3

zinc(II), [Zn(H20)6]2+

a white precipitate of [Zn(OH)2(H2O)4]

precipitate dissolves to a colourless solution of [Zn(OH)4(H2O)2]2-

precipitate dissolves to give a colourless solution of [Zn(H2O)2(NH3)4]2+

magnesium, [Mg(H20)6]2+

a white precipitate of [Mg(OH)2(H2O)4]

precipitate is insoluble


silver, [Ag(H20)6]+

a white precipitate of [Ag(OH)(H2O)5]+ which is dehydrated to form a brown precipitate of Ag2O


precipitate dissolves to give a colourless solution of [Ag(NH3)2]+. The brown precipitate of Ag2O is often not seen because the ammonia complex forms very easily.

lead(II), Pb2+

a white precipitate of Pb(OH)2


precipitate dissolves to a colourless solution of [Pb(OH)4(H2O)2]2-


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PREPARED BY ADNAN CHOWDHURY Likely ion

Observation on adding small Observation on amount of dilute NaOH or adding excess dilute NaOH dilute aqueous NH3

Observation on adding excess dilute aqueous NH3

barium, Ba2+

a white precipitate of Ba(OH)2



strontium, Sr2+

a white precipitate of Sr(OH)2 a white precipitate of Ca(OH)2









sodium, Na+

a white precipitate of Al(OH)3 a white precipitate of LiOH no precipitate

potassium, K+

no precipitate

ammonium, NH4+

no precipitate.

calcium, Ca2+ aluminium, Al3+ lithium, Li+

On heating, a colourless, pungent smelling gas evolves that turns a damp or moist red litmus paper blue. The gas is ammonia, NH3. NH4+(aq) + OH-(aq)  NH3(g) + H2O(l) The test for NH4+ ion is carried out with dilute NaOH only.


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PREPARED BY ADNAN CHOWDHURY Lead (II) Nitrate or Lead (II) Ethanoate Solution: 

Most lead(II) compounds are insoluble. When aqueous lead ions (either lead (II) nitrate or lead (II) ethanoate) are added to a solution containing the appropriate anion a precipitate will form. Anion

Inference

carbonate, CO32-/ hydrogencarbonate, HCO3-

a white precipitate forms which dissolves on addition of an acid with effervescence of a colourless, odourless gas which gives a white precipitate with limewater (calcium hydroxide solution) i.e. it turns limewater milky. If excess CO2 is passed, the precipitates dissolves giving a colourless solution. Carbon dioxide evolved Pb2+(aq) + CO32-(aq)  PbCO3(s) PbCO3(s)+ 2H+(aq)  CO2(g) + H2O(l) + Pb2+(aq) Pb2+(aq) + 2HCO3-(aq)  Pb(HCO3)2(s) Pb(HCO3)2 (s)+ 2H+(aq)  2CO2(g) + 2H2O(l) + Pb2+(aq) Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(l) CaCO3(s) + H2O(l) + CO2(g)  Ca(HCO3)2(aq)

sulfite, SO32-

a white precipitate forms which dissolves on addition of an acid with effervescence of a colourless, acidic gas which decolourises acidified potassium dichromate (VI) solution or paper from orange [Cr2O72- (aq)] to green [Cr3+(aq)]. Pb2+(aq) + SO32-(aq)  PbSO3(s) PbSO3(s) + 2H+(aq)  SO2 (g) + H2O(l) + Pb2+(aq)

sulphate, SO42-

a white precipitate forms which does not dissolve on addition of an acid. Pb2+(aq) + SO42-(aq)  PbSO4(s)

chloride, Cl-

a white precipitate forms which dissolves on heating and reappears on cooling. Pb2+(aq) + 2Cl-(aq)  PbCl2(s)

bromide, Br-

a white precipitate forms which dissolves on heating and reappears on cooling. Pb2+(aq) + 2Br-(aq)  PbBr2(s)


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PREPARED BY ADNAN CHOWDHURY Anion

Inference

iodide, I-

a yellow precipitate forms which has no effect on heating. Pb2+(aq) + 2I-(aq)  PbI2(s)

nitrate, NO3ethanoate, CH3COO-

no precipitate no precipitate

Barium chloride solution: 

Aqueous barium chloride forms precipitates of insoluble barium salts with a number of anions but is usually used as the test for the sulfate, SO42−, ion. Aqueous barium chloride is usually used with dilute hydrochloric acid.

Test-1: To a small portion of the aqueous test sample, a few drops of Barium chloride, BaCl2 solution is added followed by addition of few drops of dilute hydrochloric acid, HCl. Anion

Inference

sulfate, SO42-

a white precipitate forms which does not dissolve on addition of HCl acid. Ba2+(aq) + SO42-(aq)  BaSO4(s)

sulfite, SO32-

a white precipitate forms which dissolves on addition of HCl acid with effervescence of a colourless, acidic gas which decolourises acidified potassium dichromate (VI) solution or paper from orange [Cr2O72- (aq)] to green [Cr3+(aq)]. Ba2+(aq) + SO32-(aq)  BaSO3(s) BaSO3(s)+ 2H+(aq)  SO2(g) + H2O(l) + Ba2+(aq)

carbonate, CO32-/

hydrogencarbonate, H

CO3-

a white precipitate forms which dissolves on addition of HCl acid with effervescence of a colourless, odourless gas which gives a white precipitate with limewater (calcium hydroxide solution) i.e. it turns limewater milky. Carbon dioxide evolved Ba2+(aq) + CO32-(aq)  BaCO3(s) BaCO3(s)+ 2H+(aq)  CO2(g) + H2O(l) + Ba2+(aq) Ba2+(aq) + 2HCO3-(aq)  Ba(HCO3)2(s) Ba(HCO3)2 (s)+ 2H+(aq)  2CO2(g) + 2H2O(l) + Ba2+(aq)


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Inference

chromate(VI), CrO42-

a yellow precipitate forms which dissolves to give a orange solution Ba2+(aq) + CrO42-(aq)  BaCrO4(s) 2BaCrO4(s) + 2H+(aq)  Cr2O72-(aq) + Ba2+(aq)

Test-2: To a small portion of the aqueous test sample, a few drops of dilute hydrochloric acid, HCl is added followed by addition of few drops of Barium chloride, BaCl2 solution Anion

Inference

sulfate, SO42-

a white precipitate forms Ba2+(aq) + SO42-(aq)  BaSO4(s)

sulfite, SO32-

effervescence of a colourless, acidic gas which decolourises acidified potassium manganate(VII) paper or solution from orange to green. SO32-(aq) + 2H+(aq)  SO2 (g) + H2O(l)

carbonate, CO32-/

hydrogencarbonate, H

CO3-

chromate(VI), CrO42-

effervescence of a colourless, odourless gas which gives a white precipitate with limewater (calcium hydroxide solution) i.e. it turns limewater milky. Carbon dioxide evolved CO32-(aq) + 2H+(aq)  CO2(g) + H2O(l) HCO3-(aq) + H+(aq)  CO2(g) + H2O(l) yellow solution turns orange. 2CrO42-(aq) + 2H+(aq)  Cr2O72- (aq) + H2O(l)

Silver nitrate solution: •

Aqueous silver nitrate is commonly used to test for the presence of halide ions in solution. Anions which would interfere with the test (eg carbonate, sulphite) are removed by adding dilute nitric acid before the aqueous silver nitrate.

•

The identity of a halide may be confirmed by the addition of aqueous ammonia, (NH3), both dilute and concentrated.

•

Silver halides which dissolve in ammonia do so to form a colourless solution of the complex ion, [Ag(NH3)2]+.

• •

AgX(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + X-(aq), X = Cl, Br HCl(aq) or H2SO4(aq) acid cannot be used as these acids react with AgNO3. Ionic equation: Ag+(aq) + X-(aq)  AgX(s)


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Precipitate

Addition of aqueous NH3

colour

formula

dilute

chloride

white

AgCl

soluble

bromide

cream

AgBr

soluble in excess

soluble

iodide

yellow

AgI

insoluble

insoluble

fluoride •

concentrated

no precipitate

Silver halides decompose when light shines on them producing silver and the halogen. 2AgX(s)  2Ag(s) + X2(g)

Concentrated sulfuric acid: •

When a few drops of concentrated sulfuric acid (H2SO4) are added to a solid or aqueous halide the observed reaction products may be used to identify the particular halide ion present. This is a potentially hazardous reaction.

•

It must be carried out on a small scale and in a fume cupboard.

•

The products in brackets will not be observed since they are colourless gases. The halide ion may be identified without the need to test for these gases. No attempt should ever be made to detect these gases by smell.

Halide

Observations on adding concentrated H2SO4

Observed reaction products

fluoride

steamy fumes (HF), vigorous reaction

F-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HF(g)

chloride

steamy fumes (HCl), vigorous reaction Cl-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HCl(g)

bromide

steamy fumes (HBr), brown vapour,(Br2) vigorous reaction

Br-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HBr(g) 2HBr(g) + H2SO4(aq)  Br2(g) + SO2 (g) + 2H2O(l)

iodide

steamy fumes (HI), black solid [I2(s)], purple vapour [I2(g)], yellow solid (S), smell of rotten egg (H2S), vigorous reaction

I-(aq)/(s) + H2SO4(aq)  HSO4-(aq) + HI(g) 2HI(g) + H2SO4(aq)  I2(s) + SO2 (g) + 2H2O(l) 6HI(g) + SO2 (g)  H2S(g) + 3I2(s) + 2H2O(l)

•

Reactions with fluoride and chloride are not redox as oxidation number of sulfur does not change.


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PREPARED BY ADNAN CHOWDHURY Test for the nitrate ion: Since all nitrates are soluble no precipitation test is possible for the nitrate ion. A solid which is suspected of being a nitrate then the following tests should be carried out: •

warmed with aqueous sodium hydroxide and aluminium or zinc powder or Devadra’s alloy

•

if the solid is a nitrate then ammonia gas will be evolved. This will turn damp red litmus blue and a white smoke HCl gas 3NO3-(aq) + 8Al(s) + 5OH-(aq) + 18H2O(l)  3NH3(g) + 8[Al(OH)4 ]-(aq) NH3(g) + HCl(g)  NH4Cl(s)

Organic compounds: It will always be told if a compound, or mixture of compounds, to be identified is organic. Often the molecular formula, or the number of carbon atoms in a molecule, of a compound will be given. Chemical tests may be followed by spectroscopic information.

Appearance: Simple organic compounds are usually colourless liquids or white solids. It is unlikely that appearance alone will provide firm evidence for identification.

Solubility: Test and solubility of compound

Possible identity

pH of solution

Possible identity

dissolve in water

simple alcohols, simple carboxylic acids, propanone, simple aldehydes, simple amines and their salts

above 7

amines

below 7

carboxylic acids, phenols

dissolve in dilute acid but may not dissolve in water

Inference • •

It forms hydrogen bonds with water -OH group is present (for alcohol or carboxylic acid)

amines

two layers are formed

• •


It does not form hydrogen bonds with water. No –OH group or an acid or alcohol or carbonyl compound with at least four carbon atoms

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PREPARED BY ADNAN CHOWDHURY Test and solubility of compound

Possible identity

pH of solution

dissolve in aqueous alkali but may not dissolve in water

carboxylic acids, phenols

Possible identity

Inference

It is a neutral substance.

a small amount of universal indicator is added and solution turns green blue litmus paper or solution is added and it turns red

It is an acid.

carboxylic acids

Ignition: Igniting an organic unknown on a crucible lid may help in identifying it.

Observation

Possible inferences

burns with a smoky flame

• high carbon to hydrogen ratio. • aromatic eg benzene, unsaturated eg alkene. • saturated low molar mass compound • low carbon to hydrogen ratio. most lower molar mass compounds

burns with a clear non-smoky flame no residue

Chemical tests: The details of how these tests are to be carried out will be included in the instructions to students in the assessment activities. Test

Observation

bromine water is added and shaken

 

red brown or yellow colour is decolourised i.e it turns colourless two layers are formed


Inferences alkene (C=C)

Type of Reaction & Equation electrophilic addition

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PREPARED BY ADNAN CHOWDHURY Test

Observation

potassium manganate (VII) solution and a dilute acid (eg sulfuric acid) or a dilute alkali (eg sodium hydroxide) are added and warmed neutral potassium manganate (VII) solution is added dilute sodium hydroxide solution or ethanol (it acts as a solvent) is added and warmed. Then excess nitric acid is added to neutralise the sodium hydroxide.Then silver nitrate solution is added. A precipitate is formed whose solubility is tested with diute or concentrated ammonia

Inferences

Type of Reaction & Equation

alkene (C=C)

oxidation

purple solution turns to a brown precipitate

alkene (C=C) or aldehyde (CHO)

oxidation

White precipitate, soluble in dilute ammonia solution

C-Cl present.

 

purple solution turns colourless two layers are formed

RCl(l) + OH-(aq)  ROH(l) + Cl-(aq) Cl-(aq) + Ag+(aq)  AgCl(s) AgCl(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Cl-(aq)

Cream precipitate, insoluble in dilute but soluble in concentrated ammonia solution.

C-Br present.

RBr(l) + OH-(aq)  ROH(l) + Br-(aq) Br-(aq) + Ag+(aq)  AgBr(s) AgBr(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Br-(aq)

Yellow precipitate, insoluble in concentrated ammonia solution.

C-I present.

RI(l) + OH-(aq)  ROH(l) + I-(aq) I-(aq) + Ag+(aq)  AgCl(s) AgI(s) + 2NH3(aq)  no reaction

Note:  All are precipitation reactions.  Nitric acid is added because sodium hydroxide will react with silver nitrate to give a black precipitate of silver oxide. AgNO3(aq) + NaOH(aq)  AgOH(s) + NaNO3(aq) 2AgOH(s)  Ag2O(s)+H2O(l)  To compare the reactivities, three test tubes are set up each containing a different halogenoalkane (RCl, RBr, RI), ethanol (as a solvent), nitric acid and silver nitrate solution in a beaker of hot water.  It will be observed that yellow precipitate appears first, then creamy and finally white. So rate of hydrolysis is I->Br->Cl If the experiment is repeated with three different chloro or bromo or iodoalkanes (1º, 2º, 3º) then rate of hydrolysis is 3º>2º>1º


Page 17


Observation

Inferences


potassium dichromate (VI) solution and dilute sulfuric acid are added and heated under reflux

orange to green solution

1º alcohol 2º alcohol aldehyde

oxidation

to distinguish between 1º and 2º alcohol, the final product will be treated with a small amount of any alcohol, concentrated sulfuric acid and warmed. Then the solution is poured into a beaker containing some sodium carbonate solution and cautiously the product is smell. If a fruity smell is formed with fizzing then the product is a carboxylic acid and the reactant is a 1º alcohol.

no change

3º alcohol ketone

potassium orange to green solution 1º alcohol dichromate (VI) 2º alcohol solution and dilute sulfuric acid are added and the product is immediately distilled to distinguish between 1º and 2º alcohol, the final product will be treated with Tollen’s reagent. If a silver mirror is formed then the final product is an aldehyde and the reactant is a 1º alcohol.

no change a small piece of sodium is added solid phosphorous (V) chloride is added and any gas evolved is tested with a glass rod dipped in concentrated ammonia or damp blue litmus paper

 

bubbles evolved sodium disappears or a white solid forms

 steamy fumes evolved  white smoke formed NH3(g) + HCl(g)  NH4Cl(s)  litmus goes red


3º alcohol ketone -OH group in alcohol or carboxylic acid -OH group in alcohol or carboxylic acid

no reaction oxidation

no reaction ROH(l) + Na(s)  RO-Na+(l) + ½ H2(g) RCOOH(l) + Na(s)  RCOO-Na+(l) + ½H2(g) ROH(l) + PCl5(s)  RCl(l) + POCl3(l) + HCl(g) RCOOH(l) + PCl5(s)  RCOCl(l) + POCl3(l) + HCl(g)

Page 18


Observation

solid sodium carbonate or sodium hydrogen carbonate (or solution) is added and any gas evolved is tested with limewater

blue litmus solution or paper a small amount of any alcohol, concentrated sulfuric acid are added and warmed. Then the solution is poured into a beaker containing some sodium carbonate solution and cautiously the product is smelled a small amount of any carboxylic acid, concentrated sulfuric acid are added and warmed. Then the solution is poured into a beaker containing some sodium carbonate solution and cautiously the product is smelled

Inferences


Note: Alternative reagent: SOCl2(s) ROH(l) + SOCl2(s)  RCl(l) + SO2(g) + HCl(g) RCOOH(l) + SOCl2(s)  RCOCl(l) + SO2(g) + HCl(g) This is better because two inorganic gases are produced which remove themselves so no separation technique is required a colourless, odourless gas Carboxylic acid 2RCOOH(l) + Na2CO3(s)/(aq)  which gives a white (COOH) 2RCOO-Na+(l) + CO2(g) + H2O(l) precipitate with limewater RCOOH(l) + NaHCO3(s)/(aq)  (calcium hydroxide solution) RCOO-Na+(l) + CO2(g) + H2O(l) i.e. it turns limewater milky. If excess CO2 is passed, the precipitates dissolves giving a colourless solution. Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(l) CaCO3(s) + H2O(l) + CO2(g)  Ca(HCO3)2(aq) litmus goes red  

fizzing smell of fruity smell

 

fizzing smell of fruity smell


Carboxylic acid (COOH) Carboxylic acid (COOH)

alcohol

Page 19


Observation

Inferences


a small magnesium ribbon is added

• bubbles evolved • magnesium disappears or a white solid forms blue solution gives red precipitate

Carboxylic acid (COOH)

2RCOOH(l) + Mg(s)  (RCOO-)2Mg2+(l) + H2(g)

aldehyde

oxidation

blue solution remains

ketone

no reaction

a small amount of Fehling’s (or Benedict’s) solution is added and warmed

Note: Fehling solution consists of Fehling A (Cu2+ eg CuSO4 complexed with an alkali) and Fehling B (sodium potassium tartrate) a small amount of Tollen’s reagent is added and warmed

silver mirror formed

aldehyde

oxidation

solution stays colourless

ketone

no reaction

Note: Tollen’s reagent is prepared by mixing small amount of sodium hydroxide with small amount of silver nitrate. A black precipitate forms. AgNO3(aq) + NaOH(aq)  AgOH(s) + NaNO3(aq) 2AgOH(s)  Ag2O(s)+H2O(l) The precipitate is dissolved in minimum volume of dilute ammonia which gives colourless [Ag(NH3)2]+ a small amount of potassium manganate (VII) solution and dilute sulfuric acid is added and warmed

purple solution turns colourless

aldehyde

oxidation

solution stays purple

ketone

no reaction


Page 20


Observation

Inferences


a small amount iodine in an alkaline solution (sodium hydroxide) are added and warmed

pale yellow precipitate

methyl ketone or 21thanol,

Oxidation RCOCH3 + 3I2 + 4NaOH  CHI3(s) + RCOO-Na+ + 3NaI + 3H2O

C CH3 O methyl secondary alcohol or ethanol

CH CH3 OH 

If all the above tests are negative, the unknown is probably an alkane.

Heating under reflux:

  

It is used when the reaction is slow and one of the reactants is volatile. The organic vapours that boil off as the reaction mixture is heated are condensed and flow back into the reaction vessel. As most organic compounds are flammable, it is safer to heat the mixture using an electric heater or a water bath rather than direct heating.


Page 21

PREPARED BY ADNAN CHOWDHURY Distillation:

   

This is used to remove a volatile substance from a mixture containing non-volatile inorganic species, such as acid or alkali. This can also be carried out to separate two volatile organic substances present in a homogeneous mixture only when there is a large enough difference in the boiling temperature of the organic substances. The product must not decompose at the boiling temperature The mixture is carefully heated and the vapour that comes over at ± 2ºC of the boiling temperature (obtained from a data booklet) of the particular substance is condensed and collected.

Steam Distillation:

 

It is used to extract a volatile substance that is insoluble in water from a reaction mixture that contains an immiscible liquid as well as a solution. it is particularly useful for obtaining a substance that would decompose at its boiling point.


Page 22

PREPARED BY ADNAN CHOWDHURY Safety Precautions:     

Distillation (simple or steam) and heating under reflux must be carried out in a fume cupboard if the vapour of one of the reactants or products is harmful, poisonous (toxic) or irritant. The thermometer bulb must be placed adjacent to the mouth of the joint at the neck of round bottom flask. If the mixture is being heated under reflux or distilled, there must be some outlet to the air. If there is not, pressure will build up in the apparatus, which will then fly apart, spraying hot, flammable, and often corrosive, liquid around. Gloves must be worn when corrosive substances are used. Such substances are must always be handled with care. The flask should never be heated with a naked flame. This is because almost all organic substances are flammable and if the liquid being heated were to spill over or the flask to crack, a fire would result.

Solvent Extraction:   

An organic product can often be separated from inorganic substances by solvent extraction. This is useful when the components of a mixture have similar boiling temperatures and fractional distillation is not possible. The organic compound is separated by adding a suitable organic solvent which dissolves the organic compound but not the inorganic one. The organic compound and the solvent should have a large enough difference in boiling temperature so that they can be later separated by distillation.

Purification of a liquid that is insoluble in water: The product is distilled out of the reaction mixture and the following processes are carried out:  The distillate is washed with sodium carbonate or sodium hydrogen carbonate solution in a separating funnel. This removes any acidic impurities. The pressure must be released from time to time to let out the carbon dioxide. This washing is repeated until no more fizzing is seen. CO32-(aq) + 2H+(aq)  CO2(g) + H2O(l) HCO3-(aq) + H+(aq)  CO2(g) + H2O(l)  The aqueous layer is discarded and the organic layer is washed with water. This removes any unreacted sodium salts and any soluble organic substances, such as ethanol.


Page 23


Less dense liquid More dense liquid 

The aqueous layer is discarded and the organic layer is dried, usually with lumps of anhydrous calcium chloride or calcium oxide or silica gel. Solid potassium hydroxide is used to dry amines and alcohols as they forms complex ions with calcium chloride

Purification of a solid by Recrystallisation: The solid is filtered off from the reaction mixture and purified by recrystallisation.  A suitable solvent is chosen in which the solid is soluble when hot and almost insoluble at room temperature.  The solid is dissolved in minimum amount of hot solvent. Minimum volume is used to get a saturated solution.  The solution is filtered through a pre-heated glass filter funnel fitted with a fluted paper. This removes any insoluble impurities.

 

The filtrate is allowed to cool and the crystals of the pure solid appears. Then filtration is carried out again using a Buchner funnel under reduced pressure. This removes any soluble impurities.



The solid is washed with a little cold solvent in order to remove any remaining insoluble impurities.


Page 24

PREPARED BY ADNAN CHOWDHURY 

The solid is then dried either between packs of filter paper or else a vacuum desiccator or an electric oven. Using filter papers will reduce the yield as some of the solids will get stick to the papers.

Hazards and Risks:  

Hazard is the potential of a substance or activity to do harm. Risk is the chance that a substance or activity will cause harm.

Types of Hazards:      

Toxicity Absorption through the skin Irritation if inhaled Corrosive compounds High flammability Carcinogenic compounds

Risks can be reduced by:      

Using less material – the reaction is easier to contain and to control, and the risk of spillage is reduced. Using lower concentrations of solutions – diluted corrosive solutions can become irritants, still a hazard but a much reduced one. Using specific protective clothing – e.g. gloves when handling corrosive liquids. Doing a reaction in a fume cupboard – thus removing harmful vapours from the work area. Reducing the temperature at which the procedure is carried out – thus slowing the reaction and reducing the risk of overheating and too many fumes being produced. Changing the materials used – less hazardous material may not react as quickly or give as much as product, but they will allow the same reaction to be studied.

Yield: Yields are less than 100 percent because of:  Competing / side reactions.  Handling losses during transfer and purification.

Enthalpy of Combustion: Ways to increase the accuracy of the experiment include the following:  The copper calorimeter should be first weighed empty and then when containing water. Alternatively, water could be added to the calorimeter using a pipette, not a measuring cylinder. If the volume of water is measured, the mass is calculated using the density of water, which is 1 gcm-3.  A screen should be placed around the calorimeter to maximize the transfer of heat from the hot combustion gases to the beaker of water. MAPLE LEAF INTERNATIONAL SCHOOL

Page 25

PREPARED BY ADNAN CHOWDHURY    

To ensure an even temperature throughout, the water in the calorimeter must be stirred continually. The temperature of the water should be measured for several minutes before lighting the fuel and for several minutes after putting out the burner flame. The temperature-time measurements are used to plot a graph from which the theoretical temperature rise is estimated by extrapolation. This reduces the error caused by heat loss from the beaker to the surroundings. The burner and its contents should be weighed before and immediately after the experiment, using a balance that reads to an accuracy of 0.01 g or better.

The calculation is carried out in three steps: 

 

Heat produced by the combustion of the fuel in Joules = m × c × ΔT m = mass of water in the beaker (not the mass of fuel burnt). c = specific heat capacity of water, 4.18 J g-1 ºC-1 ΔT = change in temperature. Amount / Number of moles of fuel burnt = (mass before – mass after) ÷ molar mass of fuel Enthalpy of Combustion, ΔHc = - heat produced in kilo Joules ÷ moles of fuel burnt The negative sign is due to the fact the reaction is exothermic.

Displacement Reactions and Enthalpies of Solution:  

In both displacement reactions and experiments to determine the enthalpy of solution, a solid is added to a liquid or solution and temperature change is measured. Temperature – time graphs are necessary because the reactions are not instantaneous.

Errors can be reduced by:  

    

Using powdered solids rather than lumps. This speeds up the reaction, so there is less time for cooling. Making sure that, for displacement reactions, enough metal is taken to ensure that the solution of the salt of the less reactive metal is the limiting reagent (so that it reacts completely). For enthalpy of solution experiments, the water must be in large excess to ensure that all the solid dissolves. Measuring the temperature for several minutes before the start of the reaction and for several minutes after the reaction has finished. The measurements are used to plot a graph, which is extrapolated to find the theoretical temperature rise. Continually stirring the contents of the expanded polystyrene cup. Placing a lid on the cup to prevent heat loss through evaporation. Weighing the cup empty and then, before the reaction starts, weighing it containing the solution. This gives an accurate value of the mass of the solution. The assumption that the density of a solution is 1 gcm-3 is not wholly accurate. Measuring the volume of solution using a pipette rather than a measuring cylinder, so that the amount (moles) can be accurately determined. This is not necessary for enthalpy of solution determinations.


Page 26

PREPARED BY ADNAN CHOWDHURY The calculation is carried out in three steps: 





Heat produced or lost by the reaction in Joules = m × c × ΔT m = mass of solution in the cup (not the mass of solute reacted). c = specific heat capacity of water, 4.18 J g-1 ºC-1 ΔT = change in temperature. Amount / Number of moles of solute reacted = concentration (moldm-3) × volume (dm3) For enthalpy of solution determination, Amount / Number of moles of solute reacted = (mass before – mass after) ÷ molar mass of fuel Enthalpy of Reaction, ΔHr = heat produced or lost in kilo Joules ÷ moles of solute reacted If there is a temperature rise, ΔH is negative; if the temperature falls, ΔH is positive.

Instantaneous Reactions: Neutralisation and precipitation reactions are neutralisation reactions. Errors can be reduced by:      

Using pipettes, rather measuring cylinders, to measure out the volume of the two liquids. Making sure that one of the reactants is in excess. The value of ΔH can then be worked out using the amount in moles of the limiting reagent. For neutralisation reactions only, weighing the expanded polystyrene cup empty and after the reaction. This is a more accurate way of obtaining the mass of solution than using a pipette and assuming that the solution has a density of 1 gcm-3. Measuring the temperature of both liquids before mixing and averaging the two values. Stirring immediately on mixing the two solutions. Reading the maximum temperature reached.

The calculation is carried out in three steps: 

 

Heat produced or lost by the reaction in Joules = m × c × ΔT m = total mass of the two solutions in the cup (not the mass of solute reacted). c = specific heat capacity of water, 4.18 J g-1 ºC-1 ΔT = change in temperature. Amount / Number of moles of solute reacted = concentration (moldm-3) × volume (dm3) Enthalpy of Reaction, ΔHr = heat produced or lost in kilo Joules ÷ moles of solute reacted If there is a temperature rise, ΔH is negative; if the temperature falls, ΔH is positive.


Page 27

PREPARED BY ADNAN CHOWDHURY Acid – Base Titration: Preparation of a Standard Solution: In any titration, the concentration of one of the solutions must be accurately known. The method is as follows:   

  

 

The mass of the solid needed to make a solution of the required concentration is calculated. A weighing bottle is placed on a top–pan balance. The tare button is pressed, so that the scale reads zero. The solid is added to the weighing bottle until the required mass is reached. The best way to do this is to remove the bottle from the pan and then the solid is added, checking the mass until the correct amount has been added. This prevents errors caused by spilling solid onto the pan of the balance. The contents of the weighing bottle is transferred into a beaker. Any remaining solid is washed from the bottle into the beaker. Some distilled water is added to the beaker containing the solid. Using a glass rod, the solution is stirred until all the solid has dissolved. In order to dissolve the solid completely, it may be necessary to heat the beaker. The solution is transferred through a funnel into a volumetric flask (250 cm3 or 100 cm3). The stirring rod and the beaker is washed, making sure that all the washings go through the funnel into the volumetric flask.

More distilled water is added to the solution until the bottom of the meniscus is level with the mark on the standard flask. The stopper is placed on the flask and mixed thoroughly by inverting and shaking several times.


Page 28

PREPARED BY ADNAN CHOWDHURY Performing a Titration: Apparatus required:   

Burette Pipette (25 cm3 or 10 cm3) Conical Flask

Chemicals required:   

Standard solution The solution of unknown concentration Suitable indicator

Procedure:       

 

A small amount of one of solution (normally acid) is drawn into a pipette using a pipette filler and it is rinsed with the solution. The rinsings are then discarded. Using a pipette filler, the pipette is filled so that the bottom of the meniscus is on the mark. The pipette is allowed to discharge into a washed conical flask. When the pipette has emptied, the surface of the liquid is touched in the flask with the tip of the pipette. Making sure that the tap is shut, a burette is rinsed out with a small amount of the other solution (normally alkali) and the rinsings are discarded. Using a funnel, the burette is filled to above the zero mark and the liquid is ran out until the meniscus is on the scale. It is checked that the burette below the tap is filled with liquid and that there are no air bubbles. The funnel is then removed. The initial volume is recorded by looking at where the bottom of the meniscus is on the burette scale. The liquid is ran slowly from the burette into the conical flask, continually mixing the solutions by swirling the liquid in the flask. The liquid is added dropwise as the end point is neared and stopped when the indicator shows the end point colour. The burette reading is recorded to the nearest 0.05 cm3. The titration is repeated until three concordant (it means that the difference between the highest and the lowest titre is not more than 0.2 cm3) are obtained. Any non-concordant titres are ignored and average of the concordant values is calculated, to get the mean titre.

Worked example A student produced the following results from a titration: burette reading

1

2

3

final cm3 start cm3 volume used cm3

23.40 0.00 23.40

23.67 0.05 23.62

23.40 0.05 23.35


Page 29

PREPARED BY ADNAN CHOWDHURY Mean titre = 1/3 × (23.40 + 23.62 + 23.35) = 23.4566 cm3 What are the three errors in the student’s work? Answers (a) 23.67cm3 is incorrect because the burette cannot be read to that level of accuracy. (b) 23.62cm3 should not have been used to calculate the mean titre because it is not in the range of accuracy of the other two values. Two titres are required that are the same or ± 0.20cm3 of each other. (c) Too many significant figures in the mean titre answer. Worked Example: A sample of 2.65g of pure sodium carbonate, Na2CO3, was weighed out, dissolved in water and made up to 250cm3 in a standard flask. Some of this was placed in a burette and used to titrate 25cm3 portions of a solution of hydrochloric acid. The equation for the reaction is: 2HCl(aq) + Na2CO3(aq)  2NaCl(aq) + H2O(l) + CO2(g) The titres obtained are shown in the table.

a) b) c) d) e)

Experiment

Titre / cm3

1

22.35

2

22.40

3

21.85

4

22.50

Calculate Calculate Calculate Calculate Calculate

the the the the the

concentration of the sodium carbonate solution. mean titre. amount (in moles) of sodium carbonate solution in the mean titre. amount (in moles) of hydrochloric acid that reacted. concentration of the hydrochloric acid solution.

Worked Example 25 ibuprofen tablets were reacted with 50cm3 of 1 moldm-3 NaOH solution. C12H17COOH + NaOH  C12H17COO-Na+ + H2O When the reaction was over, the solution containing excess sodium hydroxide, was made up to 250cm3 with distilled water. 25cm3 samples were titrated against 0.110 moldm-3 hydrochloric acid solution: NaOH + HCl  NaCl + H2O Calculate the mass, in mg, of ibuprofen in one tablet. MAPLE LEAF INTERNATIONAL SCHOOL

Page 30

PREPARED BY ADNAN CHOWDHURY Worked Example A fertiliser contains ammonium sulphate and potassium sulphate. When0.50g of the fertiliser was warmed with sodium hydroxide solution, ammonia gas was evolved. The ammonia required 15.0cm3 of 0.2 M hydrochloric acid to neutralise it. What is the percentage by mass of ammonium sulphate in the fertiliser? Worked Example A marble chip of mass 5.0g required 40cm3 of 1.5 moldm-3 hydrochloric acid to react with all the calcium carbonate it contained. What is the percentage of calcium carbonate in the marble chip? Worked Example 13.73g of sodium carbonate crystals were dissolved in 1 dm of water. 25cm3 of the solution were neutralised by 24cm3 of 0.10 mol dm-3 hydrochloric acid. What is the value of n in the formula Na2CO3.nH2O for sodium carbonate crystals?


Page 31

PREPARED BY ADNAN CHOWDHURY Indicators:

1 2 3 4 5

Methyl violet Malachite green Thymol blue (acid) Methyl yellow (in ethanol) Methyl orange—xylene cyanole solution

6 7 8 9 10

Methyl orangeblue Bromophenol Congo red Bromocresol green Methyl red

11 1 12 1 13 1 14 15

Azolitmin (litmus) Azolitmin (litmus) Bromocresol purple Bromothymol blue Phenol red Thymol blue (base)

16 1 17 6 18

Phenolphthalein (in ethanol) Phenolphthalein (in ethanol) Thymolphthalein Alizarin yellow R 

pKin (at 298 K) 0.8 1.0 1.7 3.5 3.7

acid colour yellow yellow red red purple

pH range 0.0—1.6 0.2—1.8 1.2—2.8 2.9—4.0 3.2—4.2

alkaline colour blue blue/green yellow yellow green

3.7 4.0 4.0 4.7 5.1

yellow re violet d yellow red

3.2—4.4 2.8—4.6 3.0—5.0 3.8—5.4 4.2—6.3

yellow blue red blue yellow

6.3 7.0 7.9 8.9

red yellow re yellow d yellow yellow

5.0—8.0 5.0—8.0 5.2—6.8 6.0—7.6 6.8—8.4 8.0—9.6

blue blue purple blue red blue

9.3 9.3 9.7 12.5

colourless colourless colourless yellow

8.2—10.0 8.2—10.0 8.3—10.6 10.1—13.0

red red blue orange/red

neutral colour

orange

green green orange

green

pale pink

Thymol blue also changes from red to yellow around a pH of 2


Page 32

PREPARED BY ADNAN CHOWDHURY Redox Titrations:     

A substance which is found at a high state of purity is known as a primary standard substance. The crystals of a primary standard substance are neither deliquescent (it does not become a liquid by absorbing moisture from air) nor efflorescent (it does not loose water of crystallisation, if any, to atmosphere). The water of crystallisation in a primary standard substance is fixed. The molarity (concentration) of a primary standard substance can be calculated accurately if the mass or number of moles and the volume of the solution is known. Substances which are not primary standard are usually standardised by titrating them against a primary standard substance. These titrations are normally redox titrations.

Iodometric Titration:   

 

Sodium thiosulfate, Na2S2O3.nH2O is not a primary standard substance as the water of crystallisation is variable (maximum value of n = 5). So it is standardised against a solution of iodine, I2 or potassium iodate(V), KIO3 or potassium dichromate(VI), K2Cr2O7. Thiosulfate reduces iodine to iodide ions, I- and forms tetrathionate, S4O62-. 2S2O32-(aq)  S4O62-(aq) + 2eI2(aq) + 2e-  2I-(aq) 2S2O32-(aq) + I2(aq)  S4O62-(aq) + 2I-(aq) The standardised sodium thiosulfate solution can then be used to determine the percentage purity of copper in a substance, the percentage purity or concentration of KIO3, H2O2, K2Cr2O7, etc. The substance to be analysed must be an oxidising agent and will produces iodine in another reaction by oxidation of iodide ions.

Procedure:       

A known volume of the solution of the oxidising agent is transferred into a conical flask using a pipette. Dilute sulfuric acid is then added to the conical flask using a measuring cylinder (as it is in excess). Iodine is liberated and the solution becomes brown. It must be ensured that if any solid particles produced must be completely dissolved. The liberated iodine is then titrated against standardised sodium thiosulfate solution added from a beaker. The brown colour fades to a pale yellow or pale straw colour. At this point, freshly prepared starch solution is added and the solution becomes blue black. If the starch solution is not freshly prepared then blue black colour may not appear. Starch solution should not be added too early because enough iodine may not be liberated nor too late because at that event, the end point will be missing. Without starch solution, the colour would gradually fade away and no sharp end point will be


Page 33


 

obtained. The blue black complex is formed because the remaining unreacted iodine will react with the starch reversibly. The thiosulfate solution is continued adding drop by drop until the blue black colour disappears and the solution becomes colourless. The procedure is repeated until at least two concordant titres are obtained.

Determination of Copper:       

Points 1, 2 and 3 are as before. Iodine is liberated and solution becomes milky brown. 2Cu2+ (aq) + 4I- (aq)  2CuI (s) + I2 (aq) white brown Point 5 as before. The milky brown or white brown fades to a whitish yellow colour. Point 6 as before (except for the fact that the solution now becomes whitish or milky blue black). Point 7 as before (except for the fact that the solution changes colour from milky blue black to colourless.) Point 8 as before.

Worked Example 3.22g of iodine and 7g of potassium iodide are dissolved in distilled water and made up to 250cm3. A 25.0cm3 portion of this solution required 19.0cm3 of sodium thiosulfate solution in a titration. What is the concentration of the sodium thiosulfate solution? Worked Example 5.65g of a copper (II) salt is dissolved in water and made-up to 250 cm3. A 25.0 cm3 sample of solution is added to an excess of potassium iodide, KI. The iodine formed by the reaction required 21.0cm3 of a 0.10 mol dm-3 solution of sodium thiosulfate for its reduction. What is the percentage by mass of copper in the salt? Worked Example A commercial medication contains potassium iodate. 1.20 g of the medication were dissolved in water and made up to 250 cm3. A 25 cm3 sample was added to an excess of potassium iodide, KI. The iodine formed by the reaction required 19.6 cm3 of a 0.05 mol dm-3 solution of sodium thiosulfate for its reduction. What is the percentage by mass of potassium iodate in the medication? (K = 39, I = 127, O = 16) IO3- + 5I- + 6H+ → 3I2 + 3H2O


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PREPARED BY ADNAN CHOWDHURY Worked Example 25 cm3 of liquid bleach, in which the active ingredient is NaClO, are made up to 250 cm3 with distilled water. 25 cm3 of this solution were added to an excess of potassium iodide. The iodine formed by this reaction required 20.30 cm3 of a 0.02 moldm-3 solution of sodium thiosulfate for its reduction. Find the concentration of ClO- ions in the bleach. 2I- + 2H+ + ClO- → I2 + H2O + Cl-


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Chemistry Unit 3B - By Maple Leaf International School

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