NaCl Find LE f

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NaClFind LEf

Module 3

Practical: Observation and Deduction2002-7

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F Scullion. JustChemy.Com

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Inorganic Observation and Deduction

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The observations and deductions come in pairs of activities. One involves inorganic chemistry and the other organic chemistry.

The inorganic exercise typically involves the analysis of a so-called “Double Salt”. One of the ions in these mixed salts is common to both salts in most cases. Here are some examples: -

Sodium Sulphate and Sodium Chloride

Calcium Carbonate and Calcium Chloride

Na+ SO42- Cl-

Ca2+ CO32- Cl-

Deductions.

A is Ammonium Chloride NH4Cl

Observations

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Identifying halide ions in solution

The equations for the reactions are represented thus:

NaX(aq) + AgNO3(aq) = AgX(s) + NaNO3(aq)

Na

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NaCl(aq) + AgNO3(aq) = AgCl(s) + NaNO3(aq)

white ppt

NaBr(aq) + AgNO3(aq) = AgBr(s) + NaNO3(aq) cream ppt

NaI(aq) + AgNO3(aq) = AgI(s) + NaNO3(aq) yellow ppt

Why

Can’t

You

hite

ream

ellow

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Both AgCl and AgBr are light sensitive.

They have darkened noticeably after 5 minutes

AgBr

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Solubility of Silver Halides in Ammonia solution.

Silver halide

ColourSolubility in NH3(aq)

Dilute Concentrated

AgCl White √ √

AgBr Cream x√

AgI Yellow x x

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Ammonium Compounds

warm

Ammonium + Alkali Salt + NH3 + H2OCompound warm

NH4Cl + NaOH NaCl + NH3 + H2O

warm

(NH4)2SO4 + Ca(OH)2 CaSO4 + 2NH3 + 2H2O

(ii) Describe how you would carry out the test for hydrogen chloride gas and state what you would observe.

White [1] fumes/smoke [1] glass rod [1] dipped in conc [1] ammonia [1] (max [4])

AMMONIUM + ALKALI SALT + AMMONIA + WATERCOMPD


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Observations

Observations

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20It is Potassium Carbonate

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Potassium flame test &Emission spectrum

Lilac

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Sample of dilute HCl (aq)

Limewater

Sign of CO2 gas

Effervescence

Using HCl & Limewater to test for CO32-

23CO3

2-(aq) + Mg2+

(aq) MgCO3(s)

Using Magnesium Nitrate to test for the presence of CO32-

(aq)

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Reactions of the Carbonate Ion.Metal Carbonate + Acid Salt + CO2 + H2OPartial Ionic EquationCO3

2- + H+ H2O + CO2

Balanced Symbol EquationK2CO3 + 2HCl 2KCl + H2O + CO2

--------------------------------------------------------

Precipitation ReactionAB(aq) + CD(aq) AD(s) + CB(aq)

Partial Ionic EquationCO3

2- + Mg2+ MgCO3

Balanced Symbol EquationK2CO3 + MgCl2 MgCO3 + 2KCl

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A is a “double salt”

Observation

A is a “double salt”

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Above is from the Chemguide Website

Potassium chromate above

Nickel chloride opposite

Cobalt chloride above

Colour is a feature of the compounds of TMs

Colour of Aqueous Ions

Copper(II) ion, Cu2+

(aq)

Chromium(III) ion, Cr3+

(aq)

Iron(II) ion, Fe2+

(aq)

Chromate ion, CrO4

2-(aq)

Iron(III) ion, Fe3+

(aq)

Dichromate ion, Cr2O7

2-(aq)

Cobalt(II) ion, Co2+

(aq)

Manganese(II) ion, Mn2+

(aq)

Nickel(II) ion, Ni2+

(aq)

Permanganate ion, MnO4

-(aq)

Colour of Aqueous Ions

Copper(II) ion, Cu2+

(aq)

blueblue Chromium(III) ion, Cr3+

(aq)

deep green

Iron(II) ion, Fe2+

(aq)

greengreen Chromate ion, CrO4

2-(aq)

yellowyellow

Iron(III) ion, Fe3+

(aq)

YellowYellow Dichromate ion, Cr2O7

2-(aq)

orangeorange

Cobalt(II) ion, Co2+

(aq)

pinkpink Manganese(II) ion, Mn2+

(aq)

very pale pink / colourless

Nickel(II) ion, Ni2+

(aq)

deep green

Permanganate ion, MnO4

-(aq)

deep purple

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Ammonium Compounds

warm



warm






33Dilute ammonia solution with UI: about pH 11

NH3(aq)

A weak alkali

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Lithium CalciumCrimson Red Brick Red

StrontiumRed

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Reactions of the Halide Ions with AgNO3(aq).

Silver + Sodium Silver + SodiumNitrate Halide Halide Nitrate

Partial Ionic Equation using X- for Halide IonsAg+ + X- AgCl

Balanced Symbol Equation

AgNO3 + NaX AgX + NaNO3

AgNO3 + NaCl AgCl + NaNO3

AgNO3 + NaBr AgBr + NaNO3

AgNO3 + NaI AgI + NaNO3

36Mixture of Ammonium Chloride and Li/Sr/Ca Chloride

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40Ba2+(aq) + SO4

2-(aq) BaSO4(s)

Acidified Barium Chloride (or Nitrate) is used to test for SO 42-

(aq)

41Has Ammonium Sulphate and Sodium Sulphate

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Ammonium Compounds

warm



warm






44Dilute ammonia solution with UI: about pH 11

NH3(aq)

A weak alkali

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Flame test for Sodium

Orange-yellow

Na+ NH4+ ?

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Ba2+

Cl- Cl-Cu2+

SO42-

Barium nitrate Copper(II) sulphate

BaCl2 CuSO4Complete Formula Equation: BaCl2(aq) + CuSO4(aq) BaSO4(s) + CuCl2(aq)

Complete Ionic Equation:Ba2+

(aq) + 2 Cl-(aq) + Cu2+(aq) + SO4

2-(aq) BaSO4(s) + Cu2+

(aq) + 2 Cl-(aq)

Net Ionic Equation:Ba2+ + 2 Cl- + Cu2+ + SO4

2- BaSO4(s) + Cu2+ + 2 Cl-

Ba2+(aq) + SO4

2-(aq) BaSO4(s) Na+ NH4

+ SO42-

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A is a mixture of 2 salts

Observations

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A is a mixture of 2 salts

Observations

A is a White solid

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Observations

X is a double salt

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Observations

X is a white solidX is a double salt

55Has sodium chloride and ammonium chloride

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A is a double salt

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A is a double salt

A has NaBrandNa2SO4

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Oxidising ability of Concentrated Sulphuric Acid.Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides.

[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]

1)NaCl(s) + H2SO4(l) NaHSO4(aq) + HCl(g)

The concentrated sulphuric acid will not oxidise HCl

Dip a glass rod into concentrated ammonia solution and then into thetest tube where you suspect the presence of HCl. The HCl will form a white smoke with NH3 if it is present.

NH3 + HCl NH4Cl

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Oxidising ability of Concentrated Sulphuric Acid.Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides.[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]

2a) KBr(s) + H2SO4(l) KHSO4(aq) + HBr(g)

However, the concentrated sulphuric acid will oxidise some of the HBr as follows:

-1 0

2b) 2HBr + H2SO4 Br2 + SO2 + 2H2O

Therefore, one will observe a Reddish Vapour due to some bromine being present. The SO2 is and acidic gas. It is also a reducing agent and will, for example, decolourise purple potassium permanganate solution

An increase in O.N.

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Oxidising ability of Concentrated Sulphuric Acid.Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides.[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]

3a) KI(s) + H2SO4(l) KHSO4(aq) + HI(g)

The concentrated sulphuric acid will oxidise some of the HI as follows: 3b) H2SO4(l) + 2HI(g) + SO2 + I2 + 2H20 3c) H2SO4(l) + 6HI(g) + S + 3I2 + 4 H20 3d) H2SO4(l) + 8HI(g) + H2S + 4I2 + 4 H20

During the reaction one will observe: -

· Violet Iodine Vapour being evolved,· The violet vapour cooling and subliming to form dark solid iodine,· A smell of rotten eggs (H2S)· Some free yellow sulphur· Some HI(g) which could be identified in the way one shows the presence of HCl. Use concentrated NH3 solution

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SUMMARYOxidising ability of Concentrated Sulphuric Acid.Its reactions with halide ions in aqueous solution during the formation of the Hydrogen Halides.[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]1) NaCl(s) + H2SO4(l) = NaHSO4(aq) + HCl(g)The concentrated sulphuric acid will not oxidise HCl2a) KBr(s) + H2SO4(l) = KHSO4(aq) + HBr(g)The concentrated sulphuric acid will oxidise some of the HBr as follows: 2b) 2HBr + H2SO4 + Br2 + SO2 + 2H2OTherefore, one will observe a Reddish Vapour due to some bromine being present. 3a) KI(s) + H2SO4(l) à KHSO4 + HI(g)The concentrated sulphuric acid will oxidise some of the HI as follows: 3b) H2SO4(l) + 2HI(g) + SO2 + I2 + 2H20 3c) H2SO4(l) + 6HI(g) + S + 3I2 + 4 H20 3d) H2SO4(l) + 8HI(g) + H2S + 4I2 + 4 H20 During the reaction one will observe: -

· Violet Iodine Vapour being evolved,· The violet vapour cooling and subliming to form dark solid iodine,· A smell of rotten eggs (H2S)· Some free yellow sulphur· Some HI(g) which could be identified in the way one shows the presence of HCl. Use concentrated NH3 solution

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Organic Observation and Deduction

We are informed that B is a mixture of hydrocarbons

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73B has a C=C functional group

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Cyclohexene

Only a very weak permanent dipole.Essentially non-polarMolecules held together by only van der Waals forces

Memo: “Like dissolves Like.”

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Alcohols, carboxylic acids, aldehydes and ketones are miscible with water. In other words, they are soluble in water.

As the organic molecule increases in length, masking of the functional group by the hydrocarbon chain occurs. This reduces solubility.

H-bonding between alcohol and water molecules is shown to the left.

δδ

δδ

δ δ

δδ

δ δ

δ δ

Ethanal H-bonded to water

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The hydrogen/carbon ratio has an influence on how cleanly a fuel burns. In general, the higher this ratio, the cleaner the flame.

Methane CH4 (Ratio = 4/1 = 4.00)

Ethane C2H6 (Ratio = 6/2 = 3.00)

Ethanol C2H5OH (Ratio = 6/2 = 3.00)

Propane C3H8 (Ratio = 8/3 = 2.67)

Large alkane C30H62 (Ratio = 62/30 = 2.07)

Cyclohexane C6H12 (Ration = 12/6 = 2.00)

Ethene C2H4 (Ration = 4/2 = 2.00)

Cyclohexene C6H10 (Ration = 10/6 = 1.67)

Methylbenzene C6H5CH3 (Ration = 8/6 = 1.34)

Benzene C6H6 (Ration = 6/6 = 1.00)

The black smoke and soot is caused by unburnt carbon

Decre

asin

g h

yd

rog

en/

carb

on ra

tio a

nd

more

sm

oke

an

d s

oot

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Ethanol Hexane Cyclohexane Cyclohexene

Increasing carbon/hydrogen ratio = Increasing smoke and soot

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+ Br2 Br

Br

Cyclohexene + Bromine 1,2-dibromocyclohexaneClear Clear Clearcolourless brown colourlessLiquid liquid liquid

Ethene + Bromine 1,2-dibromoethane

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George Bushgets the smell of this chemical

Spirit burner

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A primary or secondary alcohol

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Alcohols, carboxylic acids, aldehydes and ketones are miscible with water. In other words, they are soluble in water.

As the organic molecule increases in length, masking of the functional group by the hydrocarbon chain occurs. This reduces solubility.

H-bonding between alcohol and water molecules is shown to the left.

δδ

δδ

δ δ

δδ

δ δ

δ δ

Ethanal H-bonded to water

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Alcohols undergo combustion in air, burning with a clean blue flame.

The hydrogen/carbon ratio has an influence on how cleanly a fuel burns. In general, the higher this ratio, the cleaner the flame.

Methane CH4 (Ratio = 4/1 = 4.00)

Ethane C2H6 (Ratio = 6/2 = 3.00)

Ethanol C2H5OH (Ratio = 6/2 = 3.00)

Propane C3H8 (Ratio = 8/3 = 2.67)

Large alkane C30H62 (Ratio = 62/30 = 2.07)

Cyclohexane C6H12 (Ration = 12/6 = 2.00)

Ethene C2H4 (Ration = 4/2 = 2.00)

Benzene C6H6 (Ration = 6/6 = 1.00)

The alcohol has its on “inbuilt oxygen” that helps it to burn.

ROH + O2 = CO2 + H2O

C2H5OH + 3O2

2CO2 + 3H2O

Decre

asin

g h

yd

rog

en/

carb

on ra

tio a

nd

more

sm

oke

an

d s

oot

Soot is unburnt carbon

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Potassium dichromate (acidified) is an oxidising agent.

Its formula is K2Cr2O7.

The following equation shows it accepting electrons: -

Cr2O72- + 6 e- = 2 Cr3+

The role of the acid in “mopping up the oxygens” is seen in this next equation:

Cr2O72- + 6 e- + 14H+ = 2 Cr3+ + 7H2O

At this stage (AS) one only need to learn the following: - H H H I I I H – C – C – O-H + [O] H – C – C = O + H2O I I I I H H H H ethanal

Warm

ethanol

Cr(VI)

Cr (III)

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Magnesium reacting with 2M Ethanoic Acid

2CH3COOH + Mg Mg(CH3COO)2 + H2

Deduction

B is a carboxylic acid RCOOH

Vinegar is a diluteaqueous solution ofethanoic acid.

It is approximately 5% CH3COOH

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2CH3COOH + Na2CO3 = 2CH3COO-Na+ + CO2 + H2O

2CH3COOH + Mg = Mg(CH3COO)2 + H2

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THE TRIIODOMETHANE (IODOFORM) REACTION

Gives a positive result for two groupings: -

H IWITH CH3 C - R ALCOHOL

I OH where R = H, CH3, C2H5, etc

-------------------------------------- O You will see that the alcohol above is oxidised

II to this carbonyl structure in step 1 of 3 steps!

WITH CH3 C - R Ethanal and Methyl Ketones

therefore also give +ve tests where R = H, CH3, C2H5, etc

Iodoform molecule HCI3

TriiodomethaneYellow crystals with an antiseptic smell.

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H H H I I IH – C – OH H – C – C - OH I I I H H H

H H H H H H I I I I I I H – C – C – C – OH H – C – C – C - H I I I I I I H H H H HO H

Which of these alcohols would give a +ve iodoform test?

Ethanol is the only primary alcohol to give the iodoform

If "R"is a hydrocarbon group, then you have a secondary alcohol. Lots of secondary alcohols give this reaction, but those that do all have a methyl group attached to the carbon with the -OH group.

No tertiary alcohols can contain this group because no tertiary alcohols can have a hydrogen atom attached to the carbon with the -OH group. No tertiary alcohols give the triiodomethane (iodoform) reaction.

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H I H – C = O H – C – C = O I I I H H H

H H H H I I I I H – C – C – C = O H – C – C – C - H I I I I II I H H H H O H

Which of these carbonyls would give a +ve iodoform test?

Ethanal is the only aldehyde that gives a +ve iodoform test

All the 2-ones of the ketones will give a +ve iodoform test

Carbonyls not

considered

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Version 1 of the Iodoform Test

Pure iodoform

Iodine in KI(aq)

Sample

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Version 1 of the Iodoform Test I2/NaOH

To about 5 cm3 of a saturated solution of iodine in potassium iodide in a test-tube add 5 drops ofalcohol.

Iodine in KI(aq)

Add alcohol

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To about 5 cm3 of a saturated solution of iodine in potassium iodide in a test-tube add 5 drops ofethanol.

Add sodium hydroxide solution carefully until the colour has almost gone.

The cloudiness is a sign of precipitation. Iodoform is a pale yellow solid with an antiseptic smell

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3 REACTIONS or STEPS occur:

1. OXIDATION: Alcohol + [O] Carbonyl

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2. All 3 of the H atoms of the methyl group are substituted by I atoms

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3. The CI3COR formed then goes on to form HCI3 and RCOO-Na+

Stand the test-tube in water at about 70 oC for two or three minutes, then remove and allow to cool.

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To about 5 cm3 of a saturated solution of iodine in potassium iodide in a test-tube add 5 drops ofethanol. Add sodium hydroxide solution carefully until the colour has almost gone. 3 REACTIONS or STEPS occur:



3. The CI3COR formed then goes on to form HCI3 and RCOO-Na+

Stand the test-tube in water at about 70 oC for two or three minutes, then remove and allow to cool.

Yellow crystals of iodoform separate out on standing and the smell is like that of antiseptic

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Reactions taking place.

H I

CH3 C - OH + I2 + 2OH- CH3 C = O + 2I- + 2H2O I I R R

CH3 C = O + 3I2 + 3OH- CI3 C = O + 3 I- + 3H2O I I R R

CI3 C = O + NaOH R – C = O + HCI3

I I

R O-Na+

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Iodoform.

Triiodomethane

CHI3

A fine yellow precipitate

Antiseptic smell

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Version 2. KI/NaClO

Using potassium iodide and sodium chlorate(I) solutionsSodium chlorate(I) is also known as sodium hypochlorite.

Sodium chlorate is the active ingredient in most household bleaches. It is an oxidising agent and will oxidise Iodide ions to Iodine (I2).

Potassium Iodide Iodine liberated

NaClO

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Version 2. KI/NaClOUsing potassium iodide and sodium chlorate(I) solutionsSodium chlorate(I) is also known as sodium hypochlorite.

Potassium iodide solution is added to a small amount of organic sample,

Sample + KI

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Version 2. KI/NaClOUsing potassium iodide and sodium chlorate(I) solutionsSodium chlorate(I) is also known as sodium hypochlorite.


This is followed by sodium chlorate(I) soln.

Sample + KI + NaClO

Iodoform

NaClO is alkaline (source of OH-) and oxidises I- to I2 As a result KI/NaClO is equivalent to using I2/NaOH

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Summary of Version 2. KI/NaClOUsing potassium iodide and sodium chlorate(I) solutionsSodium chlorate(I) is also known as sodium hypochlorite.


This is followed by sodium chlorate(I) soln.

NOTE THAT THE NaClO OXIDISES IODIDE (I-) TO IODINE (I2)So as well as any possible yellow precipitate, you will also see the typical reddish-brown colour of iodine solution being formed during the reaction.Note also, that sodium chlorate(I) solution is alkaline and contains a

sufficietly high [OH-] to carry out the second half of the reaction.

In effect you are making I2 “in situ” so the tests are essentially the same.

If no precipitate is formed in the cold, it may be necessary to warm the mixture very gently.

Look for the formation of a pale yellow precipitate with antiseptic smell

Sample + KI + NaClO

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