JHARKHAND RAI UNIVERSITY

LAB MANUAL of B.Tech –I semester

1. Study of indicator (methyl orange)

Theory:indicators are generally weak organic compound which have

different colour in different media. It is used in volumetric analysis where

change of colour indicate the end point of the reaction .the change of colour

also indicates whether the solution is acidic , base , or neutral. An indicator

does not take part in a chemical reaction but it change in colour raptly on the

neutral point or equivalence.

Example: Methyl orange, Phenolphthalein, Methyl red, Methyl blue

Colour change of indicator in different media

Indicator Colour in

alkaline

media(Base)

Colour in

neutral

media

Colour in

Acidic media

Methyl orange Yellow Orange pink

Phenolphthalein pink colourless colourless

Selection of indicator

Titrating of Indicator used

Acid Base

Strong(HCL,H2SO4,HNO3) Strong(NaOH,KOH) Use any indicator

Strong Weak (Fe(OH)3,Cu(OH)2 Methyl orange

Weak (oxalic acid ,acetic acid) Strong Phenolphthalein

Weak Weak No indicator is

suitable, do not

titrate.

Chemical required:

(i) NaOH (N/10) solution

(ii) HCl (N/10) solution

(iii) Methyl orange

Apparatus required

(i) Burette stand

(ii) Burette

(iii) Pipette

(iv) Beaker

(v) Conical flask

(vi) Conical funnel

(vii) Dropper

Procedure:

(a) Take a burette and wash it with distilled water.

(b) Rinse and fill the solutionHCl N/10 with the help of conical funnel

and set the initial burette reading as zero.

© Clamp it vertically to the burette stand.

(c) Rinse the pipette with water and then with the given NaOH solution.

(d) Pipette out 10ml of given NaOH (N/10) solution into a conical flask

and add one or two drops of methyl orange.

(e) Titrate it against the HCl(N/10) solution taken in the burette till the

colour of the solution in the conical flask changes from colour

yellowish to pink colour

(f) Note down the final burette reading.

(g) Repeat the titration until concordant values are obtained.

Observation:

No of

observation

Volume of

NaOH

solution in

mL

Burette

reading

Different Constant Indicator

used

Initial Final

Calculation:

2. Determination of strength of KMnO4 using standard solution of

Mohr's salt

Theory:

In this titration, potassium permanganate is the oxidizing agent and Mohr’s

salt is the reducing agent. Mohr’s salt is a double salt of ferrous sulphate and

ammonium sulphate and its composition is FeSO4.(NH4)2SO4.6H2O. It is a

primary standard. Therefore, its standard solution can be prepared directly.

Ferrous ions of Mohr’s salt undergo hydrolysis in aqueous solution. To

prevent the hydrolysis, Conc. H2SO4 needs to be added to the Mohr’s salt

crystals during the preparation of its standard solution.

Chemical required:

(i) Potassium permanganate (KMnO4)

(ii) Mohr’s salt (FeSO4.(NH4)2SO4.6H2O)

(iii) Half test tube of H2SO4

Apparatus required

(viii) Burette stand

(ix) Burette

(x) Pipette

(xi) Beaker

(xii) Conical flask

(xiii) Conical funnel

(xiv) Dropper

(xv) Measuring flask 10mL.

Procedure:

(a) Take a burette and wash it with distilled water.

(b) Rinse and fill the solution KMnO4 with the help of conical funnel

and set the initial burette reading as zero.

(c) Clamp it vertically to the burette stands.

(d) Rinse the pipette or measuring flask with water and then with the

given solution.

(e) Pipette out or measure 10ml of given Mohr’s salt solution into a

conical flask and add half test tube of H2SO4 solution

(f) Titrate it against the KMnO4 solution taken in the burette till the

colour of the solution in the conical flask changes from colourless to

light pinkcolour.

(g) Note down the final burette reading.

(h) Repeat the titration until concordant values are obtained.

Observation:

No of

observation

Volume of

NaOH

solution in

mL

Burette

reading

Different Constant Indicator

used

Initial Final

Calculation:

3. Estimation of Hydrochloric Acid solution Supplied titrating against

standard Sodium Hydroxide Solution

Objective: — the main theme of this experiment is to determine the strength

of ‘Commercial Hydrochloric acid’ with a standard Sodium Hydroxide

solution. As Sodium Hydroxide is itself a secondary standard substance, so it

is standardized by the standard Oxalic-acid solution.

Theory: — in this experiment we shall determine the Strength of commercial

Hydrochloric Acid solution by a secondary standard solution of NaOH. This

is done by means of ‘Titration’. The important matters that are related with

the experiment are stated below:

Titration: — in presence of a suitable indicator, the volumetric analysis in

which a standard solution is added in another solution (whose strength is not

known) to reach its end point to determine the strength of that solution is

called ‘titration’.

Standard Solution:A solution of known concentration is called a ‘standard

solution’.

Secondary Standard Solution: — A ‘Secondary Standard’ is a substance

which may be used for standardizations, and whose content of active

substance has been found by comparison against a primary standard. On the

other hand ‘primary standard’ is a compound of sufficient purity from which

a standard solution can be prepared by direct weighing of a quantity if it,

followed by dilution to give a defined volume of solution.

Indicator: — in our acid-base titration there is an important use of indicator.

An ‘indicator’ is a chemical substance that detects the Equivalent point (i.e.

the end point) of reaction by changing its colour. Indicators have different

structures in acidic and in basic solution.

Equivalent Point:—

The ‘equivalent point’ is the point in a titration when a stoichiometric amount

of reactant has been added.

Normality: The number of gram equivalent weight of a solute per liter of

solution is called normality.

Normality (N) = gm. equivalent of solute /liters per solution.

It is known to us that both alkalimetry and acidimetry are based on

neutralization reaction.

Acid produces H+ and alkali produces OH-ions (Hydrogen and Hydroxyl ion

respectively) and when they come in contact with each other, they create

H2O (water) by combining with one another.

H+ + OH

- =H2O

It is the basic concept of neutralization and by using this idea a certain

volume of standard solution is allowed to react with a known volume of

another solution until neutral point is achieved. Some chemical reagents

called indicators are used to determine the neutral point.

If an acid-base reaction is such like that,

aACID + b BASE = PRODUCT

Then we know that

V base X S base= V acid X S acid

So, S acid = (V base X S base) / V acid

As NaOH is a secondary substance it will be first standardized using primary

standard substance ‘Oxalic-acid’

Chemical required:

(iv) NaOH (N/10) solution

(v) HCl (N/10) solution

(vi) Methyl orange

Apparatus required

(xvi) Burette stand

(xvii) Burette

(xviii) Pipette

(xix) Beaker

(xx) Conical flask

(xxi) Conical funnel

(xxii) Dropper

Procedure:

(a) Take a burette and wash it with distilled water.

(i) Rinse and fill the solution HCl N/10 with the help of conical funnel

and set the initial burette reading as zero.

© Clamp it vertically to the burette stand.

(j) Rinse the pipette with water and then with the given NaOH solution.

(k) Pipette out 10ml of given NaOH (N/10) solution into a conical flask

and add one or two drops of methyl orange.

(l) Titrate it against the HCl(N/10) solution taken in the burette till the

colour of the solution in the conical flask changes from colour

yellowish to pink colour

(m) Note down the final burette reading.

(n) Repeat the titration until concordant values are obtained.

Observation:

No of

observation

Volume of

NaOH

solution in

mL

Burette

reading

Different Constant Indicator

used

Initial Final

Calculation:

4. Estimation of NaOH solution Titrating against(N/10) Oxalic acid

solution.

Theory:

Estimation means to determine the amount of the substance in gram present in 1L

of its solution

Gram per liter = equivalent wt. X Normality of solution

Where normality is determine from the relation.

N1V1=N2V2

Chemical required:

(vii) NaOH (N/10) solution

(viii) Oxalic (N/10) solution

(ix) Phenolphthalein

Apparatus required

(xxiii) Burette stand

(xxiv) Burette

(xxv) Pipette

(xxvi) Beaker

(xxvii) Conical flask

(xxviii) Conical funnel

(xxix) Dropper

Procedure:

(a) Take a burette and wash it with distilled water.

(o) Rinse and fill the solution Oxalic acid N/10 with the help of conical

funnel and set the initial burette reading as zero.

© Clamp it vertically to the burette stand.

(p) Rinse the pipette with water and then with the given NaOH solution.

(q) Pipette out 10ml of given NaOH (N/10) solution into a conical flask

and add one or two drops of methyl orange.

(r) Titrate it against the oxalic acid (N/10) solution taken in the burette

till the colour of the solution in the conical flask changes from pink

colour tocolourless.

(s) Note down the final burette reading.

(t) Repeat the titration until concordant values are obtained.

Observation:

No of

observation

Volume of

NaOH

solution in

mL

Burette

reading

Different Constant Indicator

used

Initial Final

Calculation:

5. Preparation of potass alum

Theory:

Crystallization is always carried out with saturated solution .potass

alum in the commercial name its chemical name is potassium sulphate

aluminum sulphate 24.H2O

[K2SO4.Al2(SO4)3.24H2O]

Chemical required:

(i) K2SO4O

(ii) Al2(SO4)3

Apparatus required

(i) Copper water bath

(ii) China Basin

(iii) Glass rod

(iv) Beaker

(v) Spatula

(vi) Watch glass

(vii) Tripod

(viii) Filter paper

Procedure:

1. I g of K2SO4was dissolve in minimum volume of

Water accompanied with stirring and warming in 100ml beaker.

2. 2g of Al2(SO4)3 was dissolve with similar fashion in another 100mL

beaker.

3. The two solutions mixed together and taken in basin and evaporate on

the cupper water bath.

4. When crystal appeared on the solution the basin was made float on

water taken in a beaker up to it top. It was cooled at the room

temperature.

5. The mother liquor was decanted out and crystal were press between

layer of filter paper to soak dry

6. Salt analysis :

Group I (dilute H2SO4 group) –CO32-

(carbonate), SO32-

(sulphite), S2-

(sulphide), NO2- (nitrite): Salt + dilH2SO4

1. No reaction: Group I anion notpresent. Continue to group II.

2. Carbonate: Colourless andodorless gas (CO2) 1. WE

1 + MgSO4 = white ppt

3. Sulphite: Colourless gas withpungent smell

1. WE + BaCl2 (aq) = white ppt soluble in dilHCl

Group II (conc H2SO4 group) –Cl-(chloride), Br

- (bromide), I

- (iodide),

NO3- (nitrate), CH3COO

- (acetate), C2O4

2-(oxalate): Salt + conc H2SO4

1. No reaction: Group II anion notpresent. Continue to group III.

2. Chloride: Colourless whitepungent fumes (HCl); intensify when glass

rod dipped in NH4OH is brought near mouth of test tube

1. Silver nitrate test: WE +AgNO3 = white ppt soluble in NH4OH

2. Chromyl chloride test: Salt +solid K2Cr2O7 + 2-3 drops conc H2SO4 = orange / red fumes of chromyl chloride

6

1. Vapours + NaOH (aq) = yellow solution 2. Yellow solution + acetic acid + lead acetate sol = yellow ppt

3. Bromide: Reddish brown vapour 1. Silver nitrate test: WE +AgNO3 (aq) = yellow ppt partially soluble

in NH4OH 2. Organic layer test: WE + CCl4+1 drop conc HNO3 = Upper layer

aqueous; lower layer organic of orange / brown colour (bromine is

soluble in non-polar solvent) 4. Iodide: Violet vapours

1. Silver nitrate test: WE +AgNO3 (aq) = yellow ppt insoluble in

NH4OH 2. Organic layer test: WE + CCl4 + 1 drop conc HNO3 = Upper layer

aqueous; lower layer organic of violet colour (iodine is soluble in non-

polar solvent) 5. Nitrate: Brown fumes withpungent smell, which intensify on adding

paper pellets (may need heating)

1. Brown ring test: WE + freshlyprepared FeSO4 sol + 1 drop conc

HNO3 added alongside of test tube = brown ring formed at junction of

sol and acid

Group III anions (special group) – SO4

2- (sulphate), PO4

3- (phosphate): Nogroup reagent

1. Sulphate:

1. Barium chloride test: WE +BaCl2 (aq) = white ppt insoluble in

concHCl 2. Lead acetate test: WE + leadacetate (aq) + acetic acid = white ppt

soluble in CH3COONH (ammonium acetate)

Cations

Group 0 –NH4

+(ammonium):Nogroup reagent

1. Sodium hydroxide test: Salt +NaOH = pungent smelling gas; gives

white fumes when a glass rod dipped in concHCl is brought near

mouth of test tube 2. Nessler's reagent

8 test: OS

9+NaOH + Nessler's reagent = Brown /

yellow ppt Group I –Pb

2+(lead):

OS + dilHCl = white ppt; add water, boil, and divide into three parts

1. Leave OS undisturbed = white crystals formed on cooling 2. Potassium iodide test: OS + KI =yellow ppt 3. Potassium chromate test: OS +K2CrO4 = yellow ppt

Group II –Cu

2+(copper): OS + dilHCl + H2S = black ppt

1. Throw off extra sol, retain ppt, and dissolve in a few drops of conc

HNO3 = bluish green sol, ppt dissolves; divide into two parts

1. Part 1 + excess NH4OH = blue coloured sol 2. Potassium ferrocyanide test:

Part 2 + acetic acid + K4[Fe(CN)6] = reddish brown / chocolate

colouredppt (Note – this test is difficult to get) Group III –Fe

2+, Fe

3+(iron), Al

3+(aluminium):

OS + solid NH4Cl + excess NH4OH

1. Fe2+

/ Fe3+

: Note – ferrous saltsare green in colour, ferric salts are brown in colour.

1. If ferrous salt has been given, convert to ferric: OS + concHNO3 +

heat = brown ppt; then do reaction with group reagent

2. Brown ppt + HCl; then divide into two parts 1. Potassium ferrocyanide test: Part 1 + K4[Fe(CN)6] =blue ppt /

colour 2. Potassium thiocyanate test: Part 2 + KCNS = bloodred colour

2. Al

3+: Gelatinous white ppt

1. Blue lake test: Retain ppt + dilHCl = clear sol 1. Sol + blue litmus and NH4OH (drop-by-drop) = blue colour

layer ('lake') floats over colourless sol Group IV –Co

2+(cobalt), Ni

2+(nickel),Mn

2+ (manganese), Zn

2+ (zinc):

OS + solid NH4Cl + excess NH4OH + pass H2S gas

1. Co2+

/ Ni2+

: Black ppt; dissolveppt in aqua regia10

and evaporate sol

to dryness to get residue

1. Co

2+: Blue residue; turnspink / purple when dissolved in water;

divide into two parts 1. Part 1 + dil acetic acid + KNO2 + warm = yellow ppt 2. Part 2 + ether (1 mL) + solid NH4CNS

11 = blue colour in

ether

2. Ni2+

: Yellow residue; turnsgreen when dissolved in water; divide into two parts