Paper Practical Essay

SEMESTER- I PRACTICAL- I VOLUMETRIC ANALYSIS The method to determine the exact amount of the substance in a given sample is termed as quantitative analysis volumetric analysis is a branch of quantitative analysis involving accurate measurement of volumes of reacting solutions. The volumetric analysis is very much in use due to simplicity rapidity accuracy and wide applicability. The reacting substances are taken in the form of solutions and made to react. The concentration of one solution is determined using another suitable solution whose concentration is accurately known.

A known volume of one solution is measured with a pipette and taken in a conical flask. The other solution is taken in a burette and run into the first solution till the chemical reaction is just complete. The volume of the second solution is read from the burette and the two volumes are compared. Various terms used in volumetric analysis are given below: Titration The process of adding one solution from the burette to another in the conical flask in order to complete the chemical reaction is termed titration.

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Endpoint It is the exact stage at which chemical reaction involved in the titration is just complete Indicator It is a substance which will show the end point of the reaction by change of colour. For example phenolphthalein and methyl orange are indicators used in acid alkali titrations. Potassium permanganate itself acts as an indicator in potassium permanganate titrations. Acidimetry and Alkalimetry Titration: Acidimentry refers to the titration of alkali with a standard acid and alkalimetry refers to the titration of an acid with a standard alkali. 123

Permanganimetry Titration: The titration involving KMnO4 is called permanganimetry titration. In presence of dilute H2SO4 KMnO4 oxidizes ferrous sulphate to ferric sulphate and oxidizes oxalic acid to CO2 and H2O. Normality: The strength of a solution is expressed in terms of normality. Normality is the number gram equivalent mass of solute dissolved in one litre of solution. Standard solution: A solution of known strength (Normality) is called a standard solution. Decinormal Solution: A solution having the strength (Normality) of 0. 1 N is called decinormal solution. Law f volumetric analysis: Whenever two Substances react together, they react in the ratio of their equivalent mass. One litre of a normal solution of a substance will react exactly with same volume of a normal solution of another substance. In other words equal volumes of equal normal solutions will exactly react with each other. This result is stated in the form law of volumetric analysis If V1 ml of a solution of strength N1 is required or complete reaction by V2 ml of the second solution of strength N2 then V1N1=V2N2 If any three factors (V1V2 & N1) are known, the fourth factor N2 can be calculated.

The following are the important formula used in all volumetric estimations Mass of solute per litre of the solution = Equivalent mass x Normality 124 Equivalent mass of some important compounds Name of the compound Hydrochloric acid Sulphuric acid Oxalic acid Sodium carbonate Sodium hydroxide Potassium hydroxide Potassium permanganate Ferrous sulphate Ferrous ammonium sulphate Potassium dichromate Copper sulphate EDTA (disodium salt) Equulvalent Mass 36. 5 49 63 53 40 56. 1 31. 6 278 392 49. 04 249. 54 372 125 1 ESTIMATION OF SULPHURIC ACID EX. NO………………. Date…………..

Aim To estimate the amount of Sulphuric acid present in 400 ml of the given solution. You are provided with a standard solution of oxalic acid of normality ………….. N and an approximately decinormal solution of Sodium hydroxide. (Test solution should be made up to 100 ml) Principle The titration is based on the neutralisation reaction between oxalic acid and Sodium hydroxide in titration I and Sulphuric acid and Sodium hydroxide in titration II. Procedure Titration I: Standardisation of Sodium hydroxide The burette is washed with water, rinsed with distilled water and then with the given oxalic acid.

It is filled with same acid up to zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water,rinsed with distilled water and then with the given Sodium hydroxide solution. 20 ml of Sodium hydroxide is pipetted out in to a clean conical flask. Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against oxalic acid taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value.

From the titre value,the normality of Sodium hydroxide is calculated. Titration II: Standardisation of Sulphuric acid The given Sulphuric acid solution is made upto 100 ml in a 100 ml standard flask. The solution is thoroughly shaken to get a uniformly concentrated solution. The burette is washed with water,rinsed with distilled water and then with the given Sulphuric acid from the standard 126 flask. It is filled with same acid upto zero mark. The initial reading of the burette is noted. Exactly 20 ml of the sodium hydroxide is pipetted out in to a clean conical flask.

To this solution two drops of phenolphthalein indicator is added. The solution becomes pink in colour. The solution is titrated against Sulphuric acid taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value,the normality of Sulphuric acid and the amount of sulphuric acid present in 400 ml of the given solution is calculated. Result Normality of Sodium hydroxide =……………………… N Normality of Sulphuric acid =………………………

N Amount of Sulphuric acid present in 400ml of the given solution =……………………… g Titration-I : Sodium hydroxide Vs Oxalic acid S. No Volume of sodium hydroxide(ml) Burette reading (ml) initial final Volume of oxalic acid(ml) Indicator Concordant value= Calculation: Volume of oxalic acid (V1) Normality of oxalic acid (N1) Volume of sodium hydroxide (V2) Normality of sodium hydroxide (N2) By the principle of volumetric analysis, V1N1 = = = 20 ml = ? = V2N2 N 2 = V1N 1 V2 = X 20 = – – – – -N Normality of sodium hydroxide (N2) = ___________ N 127 Titration-II : Sulphuric acid Vs Sodium hydroxide

S. No Volume of sodium hydroxide(ml) Burette reading (ml) initial final Volume of sulphuric acid(ml) Indicator Concordant value= Calculation: Volume of sulphuric acid (V1) Normality of sulphuric acid (N1) Volume of sodium hydroxide (V2) Normality of sodium hydroxide (N2) By the principle of volumetric analysis, V1N1 VN x 20 N1 = 2 2 = = – – – – -N V1 = = ? = 20ml = = V2N2 Normality of sulphuric acid (N1)= ____________ N Amount of sulphuric acid present in 400 ml of the given solution =Equivalent mass x Normality of sulphuric acid x 400 / 1000 =49x———-x400/1000 =———g SHORT PROCEDURE

Description Burette solution Pipette solution Reagents added Indicator End point Titration I Titration II Oxalic acid Sodium hydroxide ……. Phenolphthalein Disappearance of pink colour Sulphuric acid Sodium hydroxide ……. Phenolphthalein Disppearance of pink colour Equivalent mass of sulphuric acid = 49 128 2. ESTIMATION OF SODIUM HYDROXIDE EX. NO………………. Date………….. Aim To estimate the amount of Sodium hydroxide present in one litre of the given solution. You are provided with a standard solution of Sodium carbonate of normality ………….. N and an approximately decinormal solution of Sulphuric acid. Test solution should be made upto 100 ml ) Principle The titrat ion is based on the neutralisation reaction between Sulphuric acid and Sodium carbonate in titration I and Sulphuric acid and Sodium hydroxide in titration II. Procedure Titration I: Standardisation of Sulphuric acid The burette is washed with water, rinsed with distilled water and then with the given Sulphuric acid. It is filled with same acid upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, rinsed with distilled water and then with the given Sodium carbonate solution. 0 ml of Sodium carbonate is pipetted out in to a clean conical flask. Two drops of methyl orange indicator is added into the flask. The solution becomes pale yellow in colour. The solution is titrated against Sulphuric acid taken in the burette. The end point of the titration is the change in colour from yellow to permanent pale pink. The titration is repeated to get the concordant value. From the titre value, the normality of Sulphuric acid is calculated. Titration II: Standardisation of Sodium hydroxide The given Sodium hydroxide solution is made upto 100 ml in a 100 ml standard flask.

The solution is thoroughly shaken to get a uniformly concentrated solution. A 20 ml pipette is washed with water, rinsed with distilled water and then with the given Sodium hydroxide. Using the rinsed pipette, exactly 20 ml of the made-up solution is transferred into a clean conical flask. To this solution two drops of methyl orange indicator is added. The solution becomes pale yellow in colour. The solution is titrated against Sulphuric acid taken in the burette. The end point of the titration is 129 the change in colour from yellow to permanent pale pink.

The titration is repeated to get the concordant value. From the titre value, the normality of Sodium hydroxide is calculated. Result (i) Normality of Sulphuric acid (ii) Normality of Sodium hydroxide (iii) Amount of Sodium hydroxide present in one litre of given solution Titration-I : Sodium carbonate Vs sulphuric acid =…………………. N =…………………. N =…………………. g S. No Volume of sodium carbonate(ml) Burette reading (ml) initial final Volume of sulphuric acid(ml) Indicator Concordant value= Calculation: Volume of sulphuric acid (V1) = Normality of sulphuric acid (N1) = ?

Volume of sodium carbonate (V2) = 20ml Normality of sodium carbonate (N2) = By the principle of volumetric analysis,V1N1 = V2N2 Normality of sulphuric acid (N1)= ___________ N VN X 20 N1 = 2 2 = = – – – – -N V1 Titration-II : Sulphuric acid Vs Sodium hydroxide S. No Volume of sodium hydroxide(ml) Burette reading (ml) initial final Volume of sulphuric acid(ml) Indicator Concordant value= 130 Calculation: Volume of sulphuric acid Normality of sulphuric acid Volume of sodium hydroxide Normality of sodium hydroxide By the principle of volumetric analysis, N 2 = V1N 1 V2 = X 20 (V1) = (N1) = (V2) = 20ml (N2) = ? V1N1 = V2N2 – – – – -N Normality of Sodium hydroxide(N2)= ____________ N Amount of Sodium hydroxide present in One litre of the given solution =Equivalent mass x Normality of sodium hydroxide = 40 x ———= ———g SHORT PROCEDURE Description Burette solution Pipette solution Reagents added Indicator End point Titration I Titration II Sulphuric acid Sodium carbonate …. Methyl orange Appearance of permanent pink colour Sulphuric acid Sodium hydroxide …. Methyl orange Appearance of permanent pink colour Equivalent mass of sodium hydroxide=40 131 3. COMPARISON OF STRENGTHS OF TWO ACIDS EX. NO………………. Date…………..

Aim To compare the strengths of two hydrochloric acids solutions in bottles A and B and estimate the amount of hydrochloric acid present in 250 ml of the weaker solution. You are provided with a standard solution of sodium hydroxide of normality ………….. N. Principle The experiment is based on the neutralisation reaction between hydrochloric acid A and Sodium hydroxide in titration I and hydrochloric acid B and Sodium hydroxide in titration II. Procedure Titration I: Standardisation of hydrochloric acid A The burette is washed with water, rinsed with distilled water and then with the given hydrochloric acid in bottle A.

It is filled with same acid upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, rinsed with distilled water and then with the given Sodium hydroxide solution. 20 ml of Sodium hydroxide is pipetted out in to a clean conical flask. Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against hydrochloric acid A taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value.

From the titre value, the normality of hydrochloric acid A is calculated. Titration II: Standardisation of hydrochloric acid B The burette is washed with water, rinsed with distilled water and then with the given hydrochloric acid in bottle B. It is filled with same acid upto zero mark. The initial reading of the burette is noted. 20 ml of standardised Sodium hydroxide is pipetted out in to a clean conical flask. Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against hydrochloric acid B taken in the burette.

The end point of the titration is the disappearance of pink 132 colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value, the normality of hydrochloric acid B is calculated Result 1. Normality of hydrochloric acid A =…………………. N 2. Normality of hydrochloric acid B =…………………. N 3. Hydrochloric acid in bottle——-is weaker then hydrochloric acid in bottle——-4. Amount of hydrochloric acid present in 250 ml of the weaker solution =…………………. g Titration-I : Sodium hydroxide Vs Hydrochloric acid A S. No

Volume of sodium hydroxide(ml) Burette reading (ml) initial final Volume of Hydrochlolic Indicator acid(ml) Concordant value= Calculation: Volume of hydrochloric acid A Normality of hydrochloric acid A Volume of sodium hydroxide Normality of sodium hydroxide By the principle of volumetric analysis, N1 = (V1) (N1) (V2) (N2) V1N1 = = ? = 20 ml = = V2N2 V2N2 X 20 = = – – – – -N V1 Normality of Hydrochloric acid A =…….. N Titration-II : Hydrochloric acid B Vs Sodium hydroxide S. No Volume of sodium hydroxide(ml) Burette reading (ml) initial final Volume of Hydrochlolic Indicator acid(ml) Concordant value= 133

Calculation: Volume of hydrochloric acid B (V1) = Normality of hydrochloric acid B (N1) = ? Volume of sodium hydroxide (V2) = 20 ml Normality of sodium hydroxide (N2) = By the principle of volumetric analysis, V1N1 = V2N2 VN X 20 N1 = 2 2 = = – – – – -N V1 Normality of Hydrochloric acid B=…….. N Hydrochloric acid in bottle. ——-is weaker than hydrochloric acid in bottle——-Amount of hydrochloric acid present = Equivalent mass of HCl x In 250 ml of the weaker solution SHORT PROCEDURE Normality of HCl x 250/1000 =……….. g Description Burette solution Pipette solution Reagents added Indicator End point

Titration I Titration II Hydrochloric acid A Sodium hydroxide ……. Phenolphthalein Disappearance of pink colour Hydrochloric acid B Sodium hydroxide …. Phenolphthalein Disappearance of pink colour Equivalent mass of hydrochloric acid = 36. 5 134 4. COMPARISON OF STRENGTHS OF TWO BASES EX. NO………………. Date………….. Aim To compare the normalities of Sodium hydroxide solutions supplied in bottles A and B and to estimate the amount of Sodium hydroxide present in 500 ml of the stronger solution. You are provided with a standard solution of Oxalic acid of normality …………..

N. Principle The titration is based on the neutralisation reaction between Oxalic acid and Sodium hydroxide. Procedure Titration I: Standardisation of Sodium hydroxide (A) The burette is washed with water, rinsed with distilled water and then with the given oxalic acid. It is filled with same acid upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, ,rinsed with distilled water and then with the given Sodium hydroxide solution in bottle A. 20 ml of Sodium hydroxide A is pipetted out in to a clean conical flask.

Two drops of phenolphthalein indicator is added into the flask. The solution becomes pink in colour. The solution is titrated against Oxalic acid taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value and normality of Oxalic acid. The normality of Sodium hydroxide in bottle A is calculated. Titration II: Standardisation of Sodium hydroxide (B) 20 ml of Sodium hydroxide from bottle B is pipette out into a clean conical flask using clean rinsed pipette.

To this solution two drops of phenolphthalein indicator is added. The solution becomes pink in colour. The solution is titrated against Oxalic acid taken in the burette. The end point of the titration is the disappearance of pink colour to give colourless solution. The titration is repeated to get the concordant value. From the titre value, the normality of Sodium hydroxide(B) is calculated. 135 The two normalities are compared and the amount of Sodium hydroxide in 500ml of the stronger solution is calculated. Result Normality of Sodium hydroxide in bottleA =…………………

N Normality of Sodium hydroxide in bottleB =………………… N Sodium hydroxide in bottle————–is stronger then Sodium hydroxide in bottle————–Amount of Sodium hydroxide present in 500ml of the stronger solution =—————–g Titration-I : Sodium hydroxide (A) vs Oxalic acid Volume of sodium S. No hydroxide A (ml) Burette reading (ml) initial final Volume of oxalic acid(ml) Indicator Concordant value= Calculation: Volume of oxalic acid Normality of oxalic acid Volume of sodium hydroxide(A) Normality of sodium hydroxide(A) By the principle of volumetric analysis, V1N 1 X N 2 = = = –V2 20 V1) = (N1) = (V2) = 20 ml (N2) = ? V1N1 = V2N2 – -N Normality of Sodium hydroxide(A) (N2) = ____________N Titration-II : Sodium hydroxideB Vs Oxalic acid Burette Volume of Volume of reading (ml) oxalic Indicator sodium S. No acid(ml) hydroxide B(ml) initial final Concordant value= 136 Calculation: Volume of sulphuric acid Normality of sulphuric acid Volume of sodium hydroxide Normality of sodium hydroxide By the principle of volumetric analysis V1N 1 X N 2 = = V2 20 (V1) = (N1) = (V2) = 20ml (N2) = ? V1N1 = V2N2 = – – – – -N

Normality of Sodium hydroxide (B) (N2) = ____________ N Sodium hydroxide in bottle————–is stronger then Sodium hydroxide in bottle————–Amount of Sodium hydroxide present =Equivalent mass xNormality in 500 ml of the stronger solution of sodium hydroxide ? 1 2 = 40 x ———- ? 1 2 = ———g SHORT PROCEDURE Description Titration I Titration II Burette solution Pipette solution Reagents added Indicator End point Oxalic acid Sodium hydroxideA …. Phenolphthalein Disappearance of pink colour Oxalic acid Sodium hydroxideB …. phenolphthalein Disappearance of pink colour Equivalent mass of sodium hydroxide = 40 137 . ESTIMATION OF MOHR’S SALT EX. NO………………. Date………….. Aim To estimate the amount of crystalline ferrous ammonium sulphate present in 100 ml of the given solution. You are provided with a standard solution of crystalline ferrous sulphate of normality ………….. N and an approximately decinormal solution of potassium permanganate. (Test solution should be made upto 100 ml ) Principle The titration is based on the oxidation and the reduction reaction. The oxidising agent i. e Potassium permanganate oxidises the reducing agent ferrous sulphate and ferrous ammonium sulphate in acidic medium to ferric sulphate.

Procedure Titration I: Standardisation of Potassium permanganate The burette is washed with water, rinsed with distilled water and then with the given Potassium permanganate solution. It is filled with same solution upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, rinsed with distilled water and then with the given ferrous sulphate solution. 20 ml of ferrous sulphate solution is pipetted out in to a clean conical flask. One test tube full of dilute sulphuric acid (20 ml) is added to it. It is titrated against Potassium permanganate taken in the burette.

Potassium permanganate acts as the self indicator. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value, the normality of Potassium permanganate solution is calculated. Titration II: Standardisation of Mohr’s salt (ferrous ammonium sulphate) The given Mohr’s salt solution is made upto 100 ml in a 100 ml standard flask. The solution is thoroughly shaken to get a uniformly concentrated solution. A 20 ml pipette is washed with water, rinsed with distilled water and then with the given Mohr’s salt solution .

Using the rinsed pipette,exactly 20 ml of the made-up solution is transferred into a clean conical flask. To this solution one test tube full of dilute sulphuric acid (20 ml) is added. The solution is titrated against standardised 138 potassium permanganate taken in the burette. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value, the normality of given Mohr’s salt solution and the amount of Mohr’s salt in 100 ml of the given solution is calculated. Result 1.

Normality of Potassium permanganate =…….. N 2. Normality of Mohr’s salt (ferrous ammonium sulphate) =…….. N 3. Amount of Mohr’s salt present in 100 ml of given solution =……. g Titration-I : Potassium permanganate Vs Ferrous sulphate S. No Volume of Ferrous sulphate (ml) Burette reading (ml) initial Volume of Potassium permanganate Indicator final (ml) Concordant value= Calculation: Volume of Ferrous sulphate (V1) = 20 ml Normality of Ferrous sulphate (N1) = Volume of Potassium permanganate (V2) = Normality of Potassium permanganate (N2) = ?

By the principle of volumetric analysis, V1N1 = V2N2 Normality of Potassium permanganate(N2)=——————N VN X 20 N2 = 1 1 = = – – – – -N V2 Titration-II : Potassium permanganate Vs Mohr’s salt S. No Volume of FAS (ml) Burette reading (ml) initial Volume of Potassium permanganate Indicator final (ml) Concordant value= 139 Calculation: Volume of Mohr’s salt Normality of Mohr’s salt Volume of Potassium permanganate (V1) =20 ml (N1) = (V2) = ? Normality of Potassium permanganate (N2) = By the principle of volumetric analysis, N1 = V1N1 = V2N2 V2N2 X = = – – – – -N V1 20 Normality of Mohr’s salt = ___________ N Amount of Mohr’s salt present in 100 ml of the given solution SHORT PROCEDURE = equivalent mass x normality of mohr’s salt x 100/1000 =———g Description Burette solution Pipette solution Reagents added Indicator End point Titration I Titration II Potassium permanganate Ferrous sulphate One test tube of dilute sulphuric acid self Appearance of pale permanent pink colour Potassium permanganate Ferrous ammonium sulphate (mohr’s salt) One test tube of dilute sulphuric acid self Appearance of pale permanent pink colour Equivalent mass of Ferrous ammonium sulphate = 392 40 6. ESTIMATION OF FERROUS SULPHATE EX. NO………………. Date………….. Aim To estimate the amount of crystalline ferrous sulphate present in 500 ml of the given solution. You are provided with a standard solution of crystalline ferrous ammonium sulphate of normality ………….. N and an approximately decinormal solution of potassium permanganate. Principle The titration is based on the oxidation and the reduction reaction. The oxidising agent i. e Potassium permanganate oxidises both ferrous sulphate and ferrous ammonium sulphate in acidic medium to ferric sulphate.

Procedure Titration I: Standardisation of Potassium permanganate The burette is washed with water, rinsed with distilled water and then with the given Potassium permanganate solution. It is filled with same solution upto zero mark. The initial reading of the burette is noted. A 20ml pipette is washed with water, rinsed with distilled water and then with the given ferrous ammonium sulphate solution. 20 ml of ferrous ammonium sulphate solution is pipetted out in to a clean conical flask. One test tube full of dilute sulphuric acid is added to it. It is titrated against Potassium permanganate taken in the burette.

Potassium permanganate acts as the self indicator. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value, the normality of Potassium permanganate solution is calculated. Titration II: Standardisation of ferrous sulphate The given ferrous sulphate solution is made upto 100 ml in a 100 ml standard flask. The solution is thoroughly shaken to get a uniformly concentrated solution. A 20 ml pipette is washed with water, rinsed with distilled water and then with the given ferrous sulphate solution .

Using the rinsed pipette, exactly 20 ml of the made-up solution is transferred into a 141 clean conical flask. To this solution one test tube full of dilute sulphuric acid (20 ml) is added. The solution is titrated against standardised potassium permanganate taken in the burette. The end point of the titration is the appearance of permanent pale pink colour. The titration is repeated to get the concordant value. From the titre value,the normality of given ferrous sulphate solution and the amount of ferrous sulphate in 500 ml of the given solution is calculated. Result 1.

Normality of Potassium permanganate =…………………. N 2. Normality of ferrous sulphate =…………………. N 3. Amount of ferrous sulphate present in 500 ml of given solution=… g Titration-I : Potassium permanganate Vs Mohr’s salt (FAS) S. No Volume of FAS (ml) Burette reading (ml) initial Solume of Potassium permanganate Indicator final (ml) Concordant value= Calculation: Volume of FAS Normality of FAS Volume of Potassium permanganate Normality of Potassium permanganate By the principle of volumetric analysis, N2 = (V1) =20 ml (N1) = (V2) = (N2) = ? ;gt; V1N1 = V2N2

V1N1 X20 = = – – – – -N V2 Normality of Potassium permanganate (N2) = 142 Titration-II : Potassium permanganate Vs ferrous sulphate S. No Volume Ferrous sulphate (ml) Burette reading (ml) initial Solume of Potassium permanganate Indicator final (ml) Concordant value= Calculation: Volume of ferrous sulphate (V1) Normality of ferrous sulphate (N1) Volume of Potassium permanganate (V2) Normality of Potassium permanganate (N2) By the principle of volumetric analysis, Normality of ferrous sulphate N1 = V2N2 X = = – – – – -N V1 20 = 20 ml = ? = = V1N1 = V2N2 = ____________ N

Amount of ferrous sulphate present = Equivalent mass x Normality in 500 ml of the given solution of ferrous sulphate x 500/1000 = ———g SHORT PROCEDURE Description Burettesolution Pipette solution Reagents added Indicator End point Titration I Titration II Potassium permanganate Ferrous ammonium sulphate One test tube of dilute sulphuric acid self Appearance of pale permanent pink colour Potassium permanganate Ferrous sulphate One test tube of dilute sulphuric acid self Appearance of pale permanent pink colour Equivalent mass of Ferrous sulphate = 278 143 7.

ESTIMATION OF TOTAL HARDNESSS OF WATER Ex no………. Date…………. Aim: To estimate the total hardness of the given sample of water by EDTA titration. Principle: The total harness of water can be determined by titrating a known volume of hard water against EDTA solution using Erriochrome Black – T indicator. The estimation is based on the complexometric titration. Procedure: Titration- I Standarisation of EDTA solution 50 ml of the given calcium chloride solution is pipetted into a clean conical flask. Half a test tube of ammonia buffer solution is added into the conical flask.

A pinch of Erriochrome Black – T indicator is added into the conical flask. The solution turns wine red in colour. It is tirated against EDTA solution taken in a clean, rinsed burette. The end point is the change in colour from wine red to steel blue. The tiration is repeated to get concordant values. From the titre values and the molarity of calcium chloride solution, the molarity of EDTA is calculated. Tiration- II Standarisation of hard water 50 ml of the given sample of hard water is pipetted into a clean conical flask. Half a test tube of ammonia buffer solution is added into the conical flask.

A pinch of Erriochrome Black – T indicator is added into the conical flask. The solution turns wine red in colour. It is tirated against EDTA solution taken in a clean, rinsed burette. The end point is the change in colour from wine red to steel blue. The tiration is repeated to get concordant values. From the titre values, the hardness of the given sample of water is calculated in ppm. Result: The total harness of the given sample of water = ……………….. Ppm 144 Titration-I: Standard calcium chloride solution Vs EDTA Volume of Standard S. No calcium chloride sulphate (ml)

Burette reading (ml) initial final Volume of EDTA (ml) Indicator Concordant value= Calculation Volume of standard calcium chloride solution(V1 ) Molarity of standard calcium chloride solution(M1) Volume of EDTA solution (V2 ) Molarity of EDTA solution (M2) By the principle of volumetric analysis, M2 = V1M1 / V2 = 50 ? 0. 01 —- = 50 (ml) =0. 01 M = x (ml) = V1M1 = V2M2 The molarity of EDTA solution = ……………….. M Standardised EDTA Vs Hard water sample S. No Volume of water sample (ml) Burette reading (ml) initial final Volume of EDTA (ml) Indicator Concordant value= 145

Calculation Volume of hard water sample(Vhardwater ) = 50 (ml) Volume of EDTA solution (VEDTA ) = x (ml) Molarity of EDTA solution (M EDTA) = Total hardness SHORT PROCEDURE Description = VEDTA x (M EDTA / Vhardwater )x 10 = ppm 6 Titration I EDTA Calcium chloride solution Buffer solution Eriochrome black-T Change in colour from wine red to steel blue EDTA Titration II Burette solution Pipette solution Reagents added Indicator End point Hard water Buffer solution Eriochrome black-T Change in colour from wine red to steel blue 146 8. DETERMINATION OF pH AND CALCULATION OF HYDROGEN ION CONCENTRATION Ex.

No………. Date………. Aim: To find out 1. The pH of given solutions in bottles A,B,C ,D and E 2. To calculate hydrogen ion concentration of the solutions. Principle: The pH of the solution can be directly measured using a pH meter. Acids give hydrogen ions in solution. The acidic nature of the solution depends on the hydrogen ion concentration which is expressed as gram ions per litre. The pH of the solution varies with concentration of hydrogen ions. + PH = – log 10 [ H ] Procedure: Exactly 50 ml of the given five sample solutions are taken in five 150 ml beakers and labeled as A,B,C,D and E.

The pH meter is standardized using a known buffer solution. The electrodes are then washed with distilled water and then immersed in the solution taken in the beaker. The pH readings are noted. The pH of all the other solutions are to be determined similarly. The electrodes are washed well with distilled water before the electrodes are immersed in next solution. The amount of hydrogen ions present in the solutions are then calculated from the pH. Result: The amount of pH and hydrogen ion concentration of the given five sample solutions are pH and [H+]CONCENTRATION (1) Sample A …………….. ions / lit (2) Sample B …………….. g ions / lit (3) Sample C …………….. g ions / lit (4) Sample D …………….. g ions / lit (5) Sample E …………….. g ions / lit 147 DETERMINATION OF pH AND CALCULATION OF HYDROGEN ION CONCENTRATION S. No 1 2 3 4 5 Sample solution A B C D E pH Hydrogen ion concentration g ions per litre Calculations Sample A Sample B Sample C Sample D Sample E 148 MODEL QUESTION PAPER MODEL: 1 1. Estimate the mass of Sulphuric acid Present in 500 ml of the given solution. You are supplied with a standard solution of oxalic acid of strength 0. 098N and an approximately decinormal solution of Sodium hydroxide.

MODEL: 2 2. Calculate the total hardness of the given sample of water. You are given a standard Hard water Solution of 0. 01M and an approximately 0. 01M EDTA solution. MODEL: 3 3. Calculate pH of given five samples, using pH meter and Calculate the H+ ion Concentration of all the samples. (Any two Students only in a batch). List of Apparatus to be supplied for each student for Board Exam 1. 2. 3. 4. 5. 6. 7. 8. 9. Burette 50ml Pipette 20ml (with safety bulb) Conical Flask 250ml Funnel Porcelain Tile 6×6” Burette stand Standard flask 100 ml Beakers 250 ml Wash Bottle -1 -1 -1 -1 -1 -1 -1 -1 -1 149

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