Rajesh Chem

MOHR's SALT

AIM – (a) To prepare 250ml of M/20 solution of Mohr’s salt.

(b) Using this calculate the molarity and strength of the given KMnO4 solution.

APPARATUS AND CHEMICALS REQUIRED- Mohr’s salt, weighing bottle, weight box, volumetric flask, funnel, distilled water, chemical balance, dilute H2SO4, beakers, conical flask, funnel, burette, pipette, clamp stand, tile, KMnO4 solution.

THEORY- (a) Mohr’s salt having the formula FeSO4.(NH4)2SO4.6H2O has molar mass 392gmol-1. It is a primary standard.

Its equivalent mass is 392/1 = 392 as its n factor is 1 as per the following reaction:

Fe2+ → Fe3+ + e-

PROCEDURE:

1. Weigh a clean dry bottle using a chemical balance.2. Add 4.9g more weights to the pan containing the weights for the weighing bottle.3. Add Mohr’s salt in small amounts to the weighing bottle, so that the pans are balanced.4. Remove the weighing bottle from the pan.5. Using a funnel, transfer the Mohr’s salt to the volumetric flask. 6. Add about 5ml. of dilute H2SO4 to the flask followed by distilled water and dissolve the

Mohr’s salt.7. Make up the volume to the required level using distilled water.8. The standard solution is prepared.

(b) THEORY-

1. The reaction between KMnO4 and Mohr’s salt is a redox reaction and the titration is therefore called a redox titration.

2. Mohr’s salt is the reducing agent and KMnO4 is the oxidizing agent.3. KMnO4 acts as an oxidizing agent in all the mediums; i.e. acidic, basic and neutral

medium.4. KMnO4 acts as the strongest oxidizing agent in the acidic medium and therefore dil.

H2SO4 is added to the conical flask before starting the titration.

IONIC EQUATIONS INVOLVED:

Reduction Half: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Oxidation Half: 5Fe2+ → 5Fe3+ + 5e-

Overall Equation: MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

INDICATOR- KMnO4 acts as a self indicator.

END POINT- Colourless to light pink (KMnO4 in the burette)

PROCEDURE-

1. Fill the burette with KMnO4 solution.2. Pipette out 10ml. of Mohr’s salt solution into the conical flask.3. Add half a test tube of dil. H2SO4.4. Keep a glazed tile under the burette and place the conical flask on it.5. Note down the initial reading of the burette.6. Run down the KMnO4 solution into the conical flask drop wise with shaking.7. Stop the titration when a permanent pink colour is obtained in the solution.8. This is the end point. Note down the final burette reading.9. Repeat the experiment until three concordant values are obtained.10.

OBSERVATION TABLE: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL)

Volume of Mohr’s salt solution taken =

S.No BURETTE READINGS VOLUME OF

KMnO4

INITIAL FINAL USED (ml)1 10 18.8 8.8

2 18.8 27.7 8.93 27.7 36.5 8.8

Concordant Value = 8.8mL

CALCULATIONS: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL)

Calculation of amount of Mohr’s Salt to be weighed to prepare 100ml M/20 solution:

Molecular Mass of Mohr’s Salt = 392g/mole

1000 cm3 of 1M KMnO4 require 392g Mohr’s Salt.

250 cm3 of M/40 KMnO4 require =392/40g = 4.9g

Using formula:

N1M1V1 = N2M2V2

Where N1=5 (for KMnO4), V1=8.8mL , M1 =?

N2 =1 (for Mohr’s salt), V2 = 10ml, M2 = 1/20M

M1 = [1*(1/20)*10]/[5*8.8] = 1/88M = 0.01M

Strength = M X Molar Mass = 158 *( 1/88) = 1.79g/L

RESULT- (ON RULED SIDE ) - The Molarity of KMnO4 = 0.01M

And the strength of KMnO4 = 1.79g/L

Volumetric analysis 2

Volumetric Analysis Acid-Base TitrationBack to Top

The organic compounds containing the acidic group are analyzed by this titration. This is based on the neutralization reaction between acid and base. An acid-base indicator is used to detect the end point which indicates the completion of reaction between acid and base by changing the color of the mixed solution.

The selection of indicator is very important in acid-base titration because the equivalence point is determined by the stoichiometric of the reaction whereas the endpoint is measured by changing the color from indicator. The reaction is neutralized but it is not necessary to come to the end point at pH = 7. The completion of the reaction depends on the strength of acid and base.

For example: The acid-base titration between HCl (strong acid) and NaOH (strong base) in which phenolphthalein indicator is used.

Chemical reaction - HCl + NaOH → NaCl + H2O

Similar titration between NH4OH (weak base) and HCl (strong acid) in which methyl orange is used as indicator.

NH4OH +HCl → NH4Cl + H2O

Steps for Acid base Titration

To determine the concentration of HCl, we take 20 ml of HCl and neutralize it with 0.150M of NaOH.

The steps for titration process are given below.

1. The titrant, NaOH solution of known concentration, is added from the burette.2. 20ml HCl is taken in volumetric flask with the help of pipette and 2-3 drop of phenolphthalein

indicator is added.3. The solution is color less due to the acidic medium.4. The solution of NaOH is added to the HCl drop by drop. As it comes in the contact with HCl acid,

the color of solution becomes pink which quickly disappears. This is due to the formation of OH- ions from NaOH which reacts with indicator and changes the color. The solution becomes color less again due to the neutralization of OH- ions with H+ ions (from HCl).

5. After adding more NaOH, the equivalence point of titration is reached when the equal number of OH- ions and H+ ions react. At this point the pH of the solution in the flask is equal to 7 and the indicator is colorless.

6. After this, adding more NaOH drop by drop in the solution makes the solution pink from color less as the solution becomes basic now. So, indicator changes its color in basic medium. This permanent change shows the end point of titration. Note this reading which is the volume of base required to neutralize 20 ml of HCl acid. If 50ml is the volume of NaOH (end point reading from burette).

http://chemistry.tutorvista.com/analytical-chemistry/volumetric-analysis.html#top

Now to calculate the concentration of HCl ;HCl + NaOH → NaCl + H2O

the molar ratio is 1:1 for this reaction

So mole of HCl = moles of NaOH(molarity = moles/ volume)

[Molarity x Volume] of acid = [Molarity x Volume] of baseM x 20 = 0.150 x 50 ml

M = 0.150 x 50 /20 = 0.375M is the concentration of acid HCl.Indicators Used For Various TitrationBack to Top

Strong Acid Against a Strong Base

For example, titration of HCl and NaOH. At the initial stage of titration, the pH changes very slowly and increases up to pH= 4. But with addition of small amount of alkali like 0.01 mL approx., pH value reaches to about 7. This shows the neutralization of acid. Further addition of small amount of alkali increases the concentration of hydrogen ions and the pH value rises to about 9. There is a rapid increase of pH from about 4 to 9 when the end point approaches. Indicators like methyl orange, methyl red and phenolphthalein could be used for these kind of neutralization reactions which show the color change within the pH range of 4 to 10.

Weak Acid against Strong Base

For example, titration of acetic acid against NaOH. The end point of this titration lies between pH 8 and 10 due to the formation of sodium acetate and water. So, phenolphthalein is a good indicator for this.

Strong Acid against Weak Base

For example, the titration of ammonium hydroxide with HCl. The end point is in the range of 6 to 4 which shows the acidic pH of solution. Methyl orange is a good indicator for this titration.

IndicatorColor change on acidic

side Range of color

change Color change on basic

side

Methyl Violet Yellow 0.0 - 1.6 Violet

Bromophenol Blue Yellow 3.0 - 4.6 Blue

Methyl Orange Red 3.1 - 4.4 Yellow

Methyl Red Red 4.4 - 6.3 Yellow

Litmus Red 5.0 - 8.0 Blue

Bromo thymol Yellow 6.0 - 7.6 Blue


Blue

Phenolphthalein Colorless 8.3 - 10.0 Pink

Alizarin Yellow Yellow 10.1 - 12.0 Red

Methods to Determine the EndpointBack to Top

1. pH indicator - This is a substance which shows the chemical change by changing the color.2. A potentiometer - This is used to measure the electrode potential of the solution. These are

used for redox titration. They show the end point with changing potential of the working electrode.3. pH meter - It is a ion-selective electrode. In pH meter the potential of electrode depends on the

amount of H+ ion present in the solution. The pH of the solution can be measured in the whole titration. This gives more correct result than indicators.

4. Conductance - The conductivity of solutions is also changed in titration and it depends on the ions present in the solution, mobility of ions and ions concentration.

5. Color change - In the redox reactions, the color of solution changes without use of indicator. This is due to different oxidation states of the product.

The volumetric analysis can be categorized into four types which are based on the type of reaction involved in the process.

Types Chemicals to be analyzed Reagents to be used Indicators to be used

Acid / Base acid or base alkali or acid pH indicator

Precipitation ion that form insoluble salt

compound containingthe other ion needed toform the insoluble salt

conductivity

Redox oxidizing or reducing agent reducing oroxidizing agent

natural color change orredox indicator

Complexometric metal ion that form complexes

complexing agent metal ion indicator

Precautions of Volumetric analysisBack to Top

Some important points should be remembered in doing volumetric analysis to get the accurate results.



1. All the equipments like burette, beaker, pipette and volumetric flask should be washed properly with distilled water before taking them in use as the presence of any other chemical can be the reason for wrong measurement.

2. The process of filling the pipette should be accurate to avoid excess addition of solutions.3. The flask should be shaking well after adding the indicator and also the titration flask with addition

of each drop of solution from burette.4. The addition of acid should be drop wise.5. Contamination should be avoided.6. The indicator should not be used in excess.7. The flask should be removed as the indicator changes color.8. Sometimes, an air bubble in the nozzle of the burette can be the reason for altering the readings,

so, it must be removed before taking the initial reading.9. The burette should not be leaked during titration.10. Keep eyes in the level of the liquid surface during the time of taking the burette reading or

measuring flask and pipette etc.11. Lower meniscus and upper meniscus are always read in case of color less and colored solutions

respectively.12. Do not blow through the pipette to expel the last drop of solution from it; simply touch the inner

surface of the titration flask with the nozzle of the pipette for this purpose.13. Index finger should be used for pi-petting the solution.

Volumetric Analysis ProblemsBack to Top

Below you could see problem

Solved ExamplesQuestion 1: Calculate the concentration of HCl acid if 50ml of HCl is required to neutralize 25ml of 1.00M NaOH in acid base titration.Solution:

HCl + NaOH -> NaCl + H2O in this chemical reaction the molar ratio is 1:1 between

HCl and NaOH. So mole of HCl = moles of NaOH

[molarity = moles/volume]

MHCl x volume of HCl = MNaOH x volume of NaOH

MHCl = $\frac{M_{NAOH} \times volume\ of\ NaOH} {volume\ of\ HCl}$

MHCl = 25.00ml×1.00M50.00ml

MHCl = 0.50 M HClSo the concentration of HCl is 0.50 M.


Question 2: The titration curve of ammonia and HCl are given below. Find out all the unknown species A, B, C and D.

Solution:At point A, the pH is highly basic so its NH3. At point B, NH3 and NH4 + ammonium ions are in the buffering region.C = NH4

+. This is equivalence point at which all the NH3 has been protonated and water molecules starts to take acidic protons.D =H3O+ more acidic solution.

A = NH3

B = NH3 and NH4+ buffering solution

C = NH4+

D = H3O+

Question 3: Find out the molecular weight of the unknown mono protic acid whose 5 gm is neutralized with 26.23 ml of a 1.008 M NaOH solution in 150.2 ml of solution.Solution: Moles of NaOH = Molarity of NaOH x Volume of NaOHSo moles of NaOH = 1.008 x .02623 = 0.026439molesAs the acid is mono protic so the molar ratio of acid and base is 1:1.So mole of mono protic acid = moles of NaOH Moles of acid = 0.026439 moles.Mole=massmolecular mass

So molecular mass = massmole

Molecular mass of unknown mono protic acid = 50.026439 = 189.1 gm/mole.

Question 4: If the 20 ml of an unknown acid is required 30ml of 0.125 N NaOH to get the equivalence point. Find out the normality of the unknown acid.Solution: equivalents of acid = equivalents of base

Vacid Nacid = Vbase Nbase

So Nacid = (30.0ml)(0.125N)(20.0ml)= 0.1875 N

experiment 3

ALCOHOLPreliminary Test:-

ExperimentObservation Inference

Added bromine water to solution

No Effect

Or

Pink colour of Bromine water changed to colourless.

Compound is saturated.

Or

Compound is unsaturated.

Confirmatory Test:-


Warmed with acetic acid and a few drops of concentrated sulphuric acid.

Fruity smell is formed. Alcohol confirmed.

Chemical Equations:-

R3C-CR3 + Br2 No Effect

OR

R2C=CR2 + Br2 BrR2C-CR2Br

CH3COOH + ROH CH3COOR + H2O

Exp. 4

ALDEHYDEPreliminary Test:-



No Effect

Or



Or


Added drop of blue Litmus to solution.

No Effect. Carboxylic Acid and Phenol absent.

Added a few drops 2,4-dinitrophenylhydrazine to the solution.

Yellowish -orange precipitate formed.

Aldehyde or ketone may be present.

Confirmatory Test:-


Added a few drops ofFehling's solution A & B.

Formation Of red colour. Aldehyde confirmed.



OR


R-CHO + (NO2)2C6H3NHNH2 (NO2)2C6H3NHN=CRH(Yellowish Orange) + H2O

RCHO + 2Cu2+ + 2H2O RCOOH + Cu2O(Red) + 4H+

Exp 5

AMINEAIM: To test the presence of amino group in the given organic compound.

PROCEDURE:

S.NoEXPERIMENT OBSERVATION INFERENCE

1 LITMUS TEST

Organic compound + few drops of red litmus solution.

Red litmus turns blue

Amino group present.

2 SOLUBILITY TEST

Organic compound + 1-2 ml of dil.HCl. Shake well.

Organic compound dissolves.

Amino group present

3 CARBYLAMINE TEST

Organic compound + CHCl3 + Alc.KOH. Heat

An obnoxious smell is obtained.

Primary amine present.

4 AZO DYE TEST

Dissolve organic compound in dil.HCl and cool in ice. Add ice cold NaNO2 solution to it. Mix well. Add ice cold solution of β- naphthol + NaOH.

A red or orange dye is obtained.

Primary aromatic amino group present.

EQUATIONS: (ON BLANK SIDE USING A PENCIL)

1. R-NH2 + HCl → R- NH3+Cl-

amine amine salt

1. R-NH2 + CHCl3 + 3KOH → R- N≡ C + 3KCl + 3H2O

Isocyanide

or

carbylamine

1. NaNO2 + HCl → HNO2 + NaCl

ArNH2 + HNO2 + HCl → Ar- N+≡ N-Cl + 2H2O

Aromatic Aryldiazonium chloride

10amine (stable between 0- 50C)

β- naphthol (draw structure) + Ar- N+≡ N-Cl → (draw structure of the azo dye obtained) + NaCl + H2O

RESULT: : (ON RULED SIDE ) Amino present in the given organic compound.

CARBOXYLIC ACIDPreliminary Test:-



No Effect

Or



Or


Added drop of blue Litmus to solution.

Solution turned Red. Carboxylic Acid may be present.

Confirmatory Test:-


Added sodium bicarbonate to solution.

Gas with brisk effervescence evolved.

Carboxylic acid confirmed.



OR


RCOOH + NaHCO3 RCOONa + H2O + CO2

exp 5

OXALIC ACIDAIM: – (a) To prepare 100ml of M/40 solution of oxalic acid.

(b)Using this calculate the molarity and strength of the given KMnO4 solution.

APPARATUS AND CHEMICALS REQUIRED- Oxalic acid, weighing bottle, weight box, volumetric flask, funnel, distilled water, chemical balance, beakers, conical flask, funnel, burette, pipette, clamp stand, tile, dilute H2SO4, KMnO4 solution.

THEORY- (a) Oxalic acid is a dicarboxylic acid having molar mass 126gmol-1. It is a primary standard and has the molecular formula COOH-COOH.2H2O. Its equivalent mass is 126/2 = 63 as its n factor is 2 as per the following reaction:

COOH-COOH → 2CO2 + 2H+ + 2e-.

PROCEDURE:

1. Weigh a clean dry bottle using a chemical balance.2. Add 3.15g more weights to the pan containing the weights for the weighing bottle.3. Add oxalic acid in small amounts to the weighing bottle, so that the pans are balanced.4. Remove the weighing bottle from the pan.5. Using a funnel, transfer the oxalic acid to the volumetric flask.6. Add a few drops of distilled water to dissolve the oxalic acid.7. Make up the volume to the required level using distilled water.8. The standard solution is prepared.

(b) THEORY-

1. The reaction between KMnO4 and oxalic acid is a redox reaction and the titration is therefore called a redox titration.

2. Oxalic acid is the reducing agent and KMnO4 is the oxidizing agent.3. KMnO4 acts as an oxidizing agent in all the mediums; i.e. acidic, basic and neutral medium.4. KMnO4 acts as the strongest oxidizing agent in the acidic medium and therefore dil. H2SO

added to the conical flask before starting the titration.5. The titration between oxalic acid and KMnO4 is a slow reaction, therefore heat the oxalic acid

solution to about 600C to increase the rate of the reaction.

IONIC EQUATIONS INVOLVED:

Reduction Half: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O] X 2

Oxidation Half: C2O42- → 2CO2

+ 2e- ] X 5

Overall Equation: 2MnO4- + 16H+ + 5C2O4

2- → 2Mn2+ + 10CO2 + 8H2O

INDICATOR- KMnO4 acts as a self indicator.

END POINT- Colourless to light pink (KMnO4 in the burette)

PROCEDURE-

1. Fill the burette with KMnO4 solution.

2. Pipette out 10ml. of oxalic acid solution into the conical flask.

3. Add half a test tube of dil. H2SO4 and heat the solution to about 600C to increase the rate of the reaction.

4. Keep a glazed tile under the burette and place the conical flask on it.

5. Note down the initial reading of the burette.

6. Run down the KMnO4 solution into the conical flask drop wise with shaking.

7. Stop the titration when a permanent pink colour is obtained in the solution.

8. This is the end point. Note down the final burette reading.

9. Repeat the experiment until three concordant values are obtained.

OBSERVATION TABLE: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL)

Volume of Oxalic Acid solution taken = 10mL

S.No

BURETTE

READINGS

VOLUME OF KMnO4

INITIAL FINAL USED (ml)

1 16 26.5 10.52 26.5 36.9 10.43 36.9 47.4 10.5

Concordant Value = 10.5mL

CALCULATIONS: (TO BE PUT UP ON THE BLANK SIDE USING A PENCIL)

Calculation of amount of oxalic acid to be weighed to prepare 100ml M/20 solution:

Molecular Mass of Oxalic Acid = 126g/mole

1000 cm3 of 1M oxalic acid require 126g oxalic acid.

1000 cm3 of M/40 oxalic acid require =126/40g = 3.15g

Using formula:

N1M1V1 = N2M2V2

Where N1=5 (for KMnO4), V1= 10.5 , M1 = ?

N2=2 (for oxalic acid), V2 = 10ml, M2 = 1/40

M1 = [2*(1/40)*10]/[5*10.5] = 1/105M = 0.0095M

Strength = M X Molar Mass = 158 *( 1/105) = 1.504g/L

RESULT- (ON RULED SIDE ) - The Molarity of KMnO4 = 0.0095M

And the strength of KMnO4 = 1.504g/L

CARBOHYDRATE TESTAIM: To test the presence of carbohydrate in the given food sample.

PROCEDURE:


1 CONC H2SO4 TEST

Food sample + conc. H2SO4. Heat

Charring occurs with smell of burnt sugar

Carbohydrate present.

2 MOLISCH’S TEST

Food sample + Molisch’sreagent (1% alcoholic solution of α naphthol) + conc. H2SO4along the sides of the test tube.

A purple ring is obtained at the junction of the two layers.

Carbohydrate present.

3 BENEDICT’S / FEHLING’S TEST

Food sample + Benedict’s reagent/ Fehling’s reagent (A mixture of equal amounts of Fehling’s A and Fehling’s B). Heat.

A red ppt. is obtained.

Carbohydrate

present.

4 TOLLEN’S TEST

Food sample + Tollen’s reagent (amm. silver nitrate solution). Heat on water bath.

A silver mirror is obtained the walls of the test tube.

Carbohydrate

present.


1. CHO(CHOH)4CH2OH + 2Cu2+ + 5OH- → COOH(CHOH)4CH2OH + Cu2O + 3H2O

Glucose Gluconic acid

2. CHO(CHOH)4CH2OH + 2[Ag(NH3)2]+ + 3OH- → COOH(CHOH)4CH2OH + 4NH3

Glucose (Gluconic acid) + 2Ag ↓ + 2H2O

RESULT: : (ON RULED SIDE ) The food sample has been tested for carbohydrate.

OIL FAT TESTAIM: To test the presence of oil or fat in the given food sample.

PROCEDURE:


1 SOLUBILITY TEST

Food sample + water

Food sample + chloroform(CHCl3)

Does not dissolve

Miscible

Oil / fat present.

2 SPOT TEST

Smear the food sample on paper.

A translucent spot is observed.

Oil / fat present.

3 ACROLEIN TEST

Food sample + KHSO4. Heat

An irritating odour is obtained.

Oil / fat present.


Oil/ fat --> glycerol + fatty acid

CH2 (OH)CH(OH) CH2 (OH) + KHSO4, --> CH2═CHCHO (acrolein) + 2H2O

RESULT: (ON RULED SIDE ) The food sample has been tested for oil/fat

MOHR's SALT CRYSTALSAIM: To prepare crystals of Mohr’s salt.

THEORY: Mohr’s salt i.e. ferrous ammonium sulphate [FeSO4.(NH4)2SO4.6H2O] is a double salt. It can be prepared by making equimolar solution of hydrated ferrous sulphate and ammonium sulphate in minimum amount of water. A few ml of dil. H2SO4 is added to prevent the hydrolysis of FeSO4.7H2O. Cooling of the hot saturated solution yields light green crystals of Mohr’s salt.

FeSO4.7H2O + (NH4)2 SO4 → FeSO4.(NH4)2SO4.6H2O + H2O

RESULT- Colour of the crystals: Light green

Shape of the crystals: Monoclinic.

POTASH ALUM CRYSTALAIM: To prepare crystals of Potash alum.

THEORY: Potash alum, a double salt, commonly known as fitkari has the formula K2SO4.Al2(SO4)3.24H2O. It can be prepared by making equimolar solution of potassium sulphate and aluminium sulphate in minimum amount of water. A few ml of dil. H2SO4 is added to prevent the hydrolysis of Al2(SO4)3.18H2O. Cooling of the hot saturated solution yields colourless crystals of Potash alum.

K2SO4 + Al2(SO4)3.18H2O + 6H2O → K2SO4.Al2(SO4)3.24H2O

RESULT- Colour of the crystals: Colourless

Shape of the crystals: Octahedral.

CARBONATEPreliminary Test:-


Added dilute sulphuric acid to given salt.

Gas with brisk effervescence evolved.

Carbonate may be present.

Confirmatory Test:-


Pass the above evolved gas through lime water.

Lime water turns milky. Carbonate confirmed.


Ionic equation:-

CO32-

(s) + 2H+(aq) -->H2O(l) + CO2(g)

CO2(g) + Ca(OH)2(aq) CaCO3(s)(milky) + H2O(l)

SULPHIDESULPHITE

Preliminary Test:-



Gas with smell of burning sulphur evolved.

Sulphite may be present.

Confirmatory Test:-


Add barium chloride to the salt.

A white ppt. of barium sulphite formed which dissolved in excess hydrochloric acid to give a clear colourless solution.

Sulphite confirmed.


Ionic equation:-

Ba2+(aq) + SO3

2-(aq) --> BaSO3(s)

NITRITEPreliminary Test:-



Reddish brown gas with pungent smell evolved.

Nitrite may be present.

Confirmatory Test:-


Added potassium permanganate solution.

Pink colour of potassium permanganate discharged.

Nitrite confirmed.


Ionic equation:-

NO2- + H2SO4(aq) NO(g) + SO4

2- + H2O(l)

2KMnO4(Pink) + 3H2SO4 + 5KNO2 --> K2SO4(Colourless) + 2MnSO4 + 5KNO3 + 3H2O

CHROMATEPreliminary Test:-



Solution turned yellow. Chromate may be present.

Confirmatory Test:-


Added barium chloride solution.

Yellow precipitate formed. Chromate confirmed.


Ionic equation:-

CrO42-

(aq) + 2H+(aq) Cr2O7

2-(aq)(yellow)

Ba2+(aq) + CrO4

2-(aq) BaCrO4(s)(yellow)

CHLORIDEPreliminary Test:-



No effect. Carbonate, sulphide , sulphite, nitrate absent.

Added concentrated sulphuric acid to given salt.

Colourless gas with pungent smell evolved.

Chloride may be present.

Confirmatory Test:-


Added few potassium dichromate crystals and concentrated sulphuric acid and heated. Passed the vapors through the test tube which contains sodium hydroxide solution. To this yellow solution, added dilute CH3COOH and lead acetate solution.

Yellow coloured precipitate is formed.

Chloride confirmed.


K2Cr2O7 + 4NaCl + 6H2SO4 ————→ 2KHSO4 + 4NaHSO3 + 2CrO2Cl2

chromyl chloride

(orange-red vapours)

BROMIDEPreliminary Test:-





Brown colour gas evolved. Nitrate or Bromide may be present.

Added copper turnings. No effect. Bromide may be present.

Confirmatory Test:-


Added dilute nitric acid A pale yellow precipitate Bromide confirmed.

and silver nitrate solution

formed.


Ionic equation:-

Ag+(aq) + Br-

(aq) AgBr(s)(yellow)

IODIDEPreliminary Test:-





Violet colour gas evolved. Iodide may be present.

Confirmatory Test:-


Added dilute nitric acid and silver nitrate solution

A pale yellow precipitate formed.

Iodide confirmed.


Ionic equation:-

Ag+(aq) + I-

(aq) AgI(s)(yellow)

NITRATEPreliminary Test:-





Brown colour gas evolved. Nitrate or Bromide may be present.

Added copper turnings. Brown colour intensifies. Nitrate may be present.

Confirmatory Test:-


Added fresh ferrous sulphate, nitric acid to salt. Added concentrated sulphuric acid alsong the walls of test tube.

Brown coloured ring is formed.

Nitrate confirmed.


Ionic equation:-

NO3- + 3Fe2+ + 4H+ → 3Fe3+ + NO + 2H2O

[Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+

OXALATEPreliminary Test:-





Colourless gas with brisk effervescence evolved.

Oxalate may be present.

Confirmatory Test:-


Passed the above evolved gas through lime water .

Lime water turned milky. Oxalate confirmed.


C2O42- + H2SO4(aq) CO(g) + CO2(g) + H2O(l) + SO4

2-

CO2(g) + Ca(OH)2(aq) CaCO3(s)(milky) + H2O(l)

SULPHATEPreliminary Test:-





No effect. Nitrate,Bromide,

Chloride,Acetate, absent. Sulphate or phosphate may be present.

Confirmatory Test:-


Added barium chloride solution.

White coloured precipitate is formed.

Sulphate confirmed.


Ionic equation:-

Ba2+(aq) + SO4

2-(aq) ==> BaSO4(s)(white)

PHOSPHATEPreliminary Test:-





No effect. Nitrate,Bromide,

Chloride,Acetate, absent. Sulphate or phosphate may be present.

Confirmatory Test:-


Added soda extract1 and

dilute nitric acid and then added Ammonium Molybdate solution.

A canary yellow precipitateformed.

Phosphate confirmed.


(NH4)2(NO3)+2Na2(PO4) --> 2(NH4)(PO4)+2Na2(NO3)

Basic radicals

AMMONIUMPreliminary Test:-


Smelled the salt. Ammonical smell. Ammonium may be present.

Confirmatory Test:-


Added sodium hydroxide solution to the salt.

Ammonical smelling gas evolved.

Ammonium confirmed.


Ionic equation:-

NH4+

(aq) + OH-(aq) NH3(g) + H2O(l)

LEAD1Preliminary Test:-


Smelled the salt. No Ammonical smell. Ammonium absent.Added dilute hydrochloric acid to original solution.


Lead(II) may be present.

Confirmatory Test:-


Added potassium iodide. Yellow coloured precipitate formed.

Lead(II) confirmed.


Ionic equation:-

Pb2+(aq) + HCl(aq) PbCl2(White)

Pb2+(aq) +2I-

(aq) PbI2(s)(Yellow)

SILVERPreliminary Test:-




Lead(II), Ag(I) may be present.

Added precipitate in hot water.

No effect. Ag(I) may be present.

Confirmatory Test:-


Added potassium chromate to original solution.

Brick red coloured precipitate formed.

Ag(I) confirmed.


Ionic equation:-

Ag+ + HCl(aq) --> AgCl(s)(white) + H+

2AgCl(s)(white) + K2CrO4 Ag2CrO4(s)(yellow) + 2KCl(aq)

LEAD2Preliminary Test:-



No Effect. Lead(II)may be present.

Added hydrogen disulphide, hydrochloric

Black coloured precipitate formed.

Lead(II)may be present.

acid to original solution.

Confirmatory Test:-


Added potassium iodide to original solution.

Yellow coloured precipitate formed.

Lead(II) confirmed.


Ionic equation:-

Pb2+ + HCl(aq) --> PbCl2(l) + 2H+

Pb2+ + KI(aq) PbI2(s)(yellow) + 2K+

COPPERPreliminary Test:-




Added hydrogen disulphide, hydrochloric acid to original solution.


Cu(II) ,Pb(II),As(III) may be present.

Added sodium hydroxide to original solution.

Blue coloured precipitate formed.

Cu(II) may be present.

Confirmatory Test:-


Added ammonium hydroxide to original

Blue coloured precipitate formed.

Cu(II) confirmed.

solution.


Ionic equation:-

Cu2+(aq) + 2OH-

(aq) --> Cu(OH)2(s)(blue)

Cu(OH)2(s) + 4NH3(aq) --> [Cu(NH3)4]2+(aq)(blue) + 2OH-

(aq)

ARSENICPreliminary Test:-






As(III) may be present.

Confirmatory Test:-


Added dilute hydrochloric acid to above solution.


As(III) confirmed.


Ionic equation:-

2As3+ + 3H2S(g) As2S3(s)(yellow) + 6H+

2As2S3(s)(yellow) + 12HCl(l) 4AsCl3(s)(yellow) + 6H2S

IRON

Preliminary Test:-





No Effect. Pb(II), Cu(II) , As(III) absent.

Added dilute hydrochloric acid, concentrated nitric acid and boiled solution. Then cooled it and added solid ammonium hydroxide.

Reddish Brown precipitate formed.

Fe(III) may be present.

Confirmatory Test:-


Dissolved the precipitate formed with hydrochloric . Added potassium sulphocyanide.

Blood red colouration. Fe (III) confirmed.


Ionic equation:-

Fe3+(aq) + 3OH-

(aq) Fe(OH)3(s)( Reddish Brown)

Fe3+(aq) + 3HCl(l) FeCl3(aq) + 3H2O(aq)

FeCl3(aq) + 6KCNS(aq) K3[Fe(CNS)6](Blood Red) + 3KCl(aq)

MAGNESIUMPreliminary Test:-







No Effect. Al(III) ,Fe(III) absent.

Passed Hydrogen disulphide gas through above formed solution.

No Effect. Zn (II) ,Mn(II), Ni(II), Co(II) absent.

Added ammonium carbonate to above solution.

No Effect. Mg (II) may be present.

Confirmatory Test:-


Added ammonium chloride, ammonium hydroxide and disodium hydrogen phosphate to original solution.

White precipitate is formed. Mg(II) Confirmed.


Ionic equation:-

Mg2+ + HPO42-

(aq) + NH4+(aq) Mg(NH4)PO4(white) + H+

ALUMINIUMPreliminary Test:-







Gelatinous white precipitate formed.

Al(III) may be present.

Confirmatory Test:-


Dissolved the precipitate formed with hydrochloric acid and added drops of blue litmus. Then added ammonium hydroxide till alkaline.

Blue coloured precipitate floated on surface .

Al(III) confirmed.

MANGANESEPreliminary Test:-








Passed Hydrogen disulphide gas through

Flesh(buff) coloured precipitate formed.

Mn (II) may be present.

above formed solution.

Confirmatory Test:-


Added lead dioxide to above solution, then added concentrated nitric acid. Boiled it.

Pink Colouration formed. Mn (II) Confirmed.


Ionic equation:-

Mn 2+ + H2S(aq) MnS(s)(Flesh) + 2H+

MnS(aq) + 2HCl(aq) MnCl2 + 2H+ + S2-

2Mn 2+ + 4H+ + 5PbO2(s) 2MnO4-(aq) + 5Pb2+

(aq) + 2H2O(aq)

ZINCPreliminary Test:-









White coloured precipitate formed.

Zn (II) may be present.

Confirmatory Test:-


Dissolved the precipitate in concentrated hydrochloric acid and boiled it.Added potassium ferrocyanide to it

Bluish white coloured precipitate formed.

Zn (II) Confirmed.


Ionic equation:-

Zn 2+ + H2S(aq) ZnS(s)(White) + 2H+

ZnS(aq) + 2HCl(aq) ZnCl2 + 2H+ + S2-

2Zn 2+ + K4[Fe(CN)6] Zn2[Fe(CN)6](Bluish White) + 4K+

NICKELPreliminary Test:-






Added dilute hydrochloric acid, concentrated nitric acid and boiled solution. Then cooled it and added solid ammonium


hydroxide.Passed Hydrogen disulphide gas through above formed solution.


Co(II) or Ni(II) may be present.

Dissolved the precipitate in aqua regia.Evaporated residue.

Yellow coloured residue remained.

Ni(II) may be present.

Confirmatory Test:-


Dissolved the yellow residue in water. Added ammonium hydroxide till alkaline then added dimethylglyxoglime solution.

Bright red precipitate formed.

Ni(II) confirmed.


Ionic equation:-

Ni 2+

+ H2S(aq) NiS(s)(black) + 2H+

NiS(s)(black) + 2HCl(aq) + [O] NiCl2(aq) + H2O(l) + S

COBALTPreliminary Test:-










Co(II) or Ni(II) may be present.

Dissolved the precipitate in aqua regia.Evaporated residue.

Blue coloured residue remained.

Co(II) may be present.

Confirmatory Test:-


Dissolved the blue residue in water.Added ammonium sulphocyanide to it.

Blue coloured layer floated on surface .

Co(II)confirmed.


Ionic equation:-

Co2+ + H2S(aq) CoS(s)(black) + 2H+

CoS(s)(black) + 2HCl(aq) + [O] CoCl2(aq) + H2O(l) + S

Co2+ + 4KCNS K2[Co(CNS)4](blue colouration) + 2K+

CALCIUMPreliminary Test:-




Added hydrogen No Effect. Pb(II), Cu(II) , As(III)

disulphide, hydrochloric acid to original solution.

absent.






White precipitate is formed.

Ba(II), Sr(II), Ca(II) may be present.

Confirmatory Test:-


Dissolved precipitate in acetic acid and boiled it. Added few drops of ammonium oxalate to above solution and then added ammonium hydroxide.


Ca(II) Confirmed.


Ionic equation:-

Ca2+ + (NH4)2CO3 (aq) CaCO3(s)(white) + 2NH4+

(aq)

CaCO3(s) + 2CH3COOH (aq) (CH3COO)2Ca + CO2(g) + H2O(l)

Ca2+(aq) + (NH4)2C2O4(aq) CaC2O4(s)(white) + 2NH4

+(aq)

BARIUMPreliminary Test:-











White precipitate is formed. Ba(II), Sr(II), Ca(II) may be present.

Confirmatory Test:-


Dissolved precipitate in acetic acid and boiled it.Added few drops of potassium chromate to above solution.

Yellow precipitate is formed. Ba (II) Confirmed.


Ionic equation:-

Ba2+ + (NH4)2CO3 (aq) BaCO3(s)(white) + 2NH4+(aq)

BaCO3(s) + 2CH3COOH (aq) (CH3COO)2Ba + CO2(g) + H2O(l)

Ba2+(aq) + K2CrO4(aq) BaCrO4(s)(yellow) + 2K+

STRONTIUMPreliminary Test:-












Ba(II), Sr(II), Ca(II) may be present.

Confirmatory Test:-


Dissolved precipitate in acetic acid and boiled it. Added few drops of ammonium sulphate to above solution and then added ammonium hydroxide.


Sr(II) Confirmed.


Ionic equation:-

Sr2+ + (NH4)2CO3 (aq) SrCO3(s)(white) + 2NH4+

(aq)

SrCO3(s) + 2CH3COOH (aq) (CH3COO)2Sr + CO2(g) + H2O(l)

Sr2+(aq) + (NH4)2SO4(aq) SrSO4(s)(white) + 2NH4

+(aq)

Rajesh Chem

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