Chemistry Laboratory Manual

DEPARTMENT OF CHEMISTRY 2011-12 Engineering Chemistry Lab Manual A N N A P O O R A N A E N G I N E E R I N G C O L L E G E , S A L E MENGINEERING CHEMISTRY LABORATORY MANUAL (For First Year B.E Students) ANNAPOORANA ENGINEERING COLLEGE NH-47 SANKARI MAIN ROAD, PEERIYASEERAGAPADI SALEM -636 308.ENGINEERING CHEMISTRY LABORATORY MANUAL (For First Year B.E /B.Tech. Students) Compiled by V.S.Saravanamani., M.Sc., M.Phil., B.Ed Assistant Professor DEPARTMENT OF CHEMISTRY Annapoorana Engineering College Name : Branch : Register No :CONTENTS S.NO. NAME OF THE EXPERIMENT PAGE NO. 1. Estimation of Total Hardness of Water sample By EDTA method 02 2. Estimation of Copper In Brass By EDTA Method 08 3. Estimation of Dissolved Oxygen in water (Winkler’s method) 12 4. Estimation Of Chloride Content in Water by Argentometric Method [Mohr’s Method] 16 5. Estimation of Alkalinity of Water Sample 20 6. Determination of Molecular Weight of a Polymer by Viscosity Measurements 26 7. Conductometric titration of strong acid With strong base 30 8. Conductometric Titration Mixture of Acids (HCL & CH3COOH) 34 9. Contuctometric precipitation titration Using BaCl2 – Na2SO4 38 10. Potentiometric titration – Estimation of Ferrous iron 42 11. PH Titration – Determination of Strength of HCl by NaOH 48 12. Determination of water of crystallization of a crystalline salt (CuSO4 · 5H2O) 54 13. Estimation of Iron by Spectrophotometric Method (using Thiocyanate) 58INTRODUCTION Chemistry practical mainly deals with the analysis of a given substance. Analysis is broadly classified into (i) Qualitative analysis and (ii) Quantitative analysis. The qualitative analysis is concerned with the identification of constituents of a compound whereas the quantitative analysis is the estimation of constituents of a compound. Volumetric Analysis It is based on the accurate measurement of the volume of a solution of exactly known concentration which is needed to react quantitatively with the solution of the substance to be determined. Indicator The attainment of the end point in a titration is usually recognized by the use of a chemical reagent called indicator. TitrantIn titrimatric analysis, the reagent of known concentration is called the titrant. Standard solution The solution, whose concentration is accurately known, is termed as Standard solution. Titration The process of adding standard solution to an unknown solution until the reaction is just complete is known as titration. End point The pointed which reaction is just complete is called as end point. Standardization It is a process of finding out the strength of a solution by titrating it with a standard solution. TitrandThe substance being titrate against titrant is known as titrand.GENERAL INSTRUCTIONS  Record your observations once you complete the experiment.  Always keep the table clean. Spillage of chemicals /reagents should be cleaned without delay.  Wash the glass apparatus using dilute soap solution before use.  Stopper the reagent bottles instantly after use. Never interchange the stoppers.  Never pour hot acids directly into the sink. Dilute it with water before trash. Flush the sink with flow of water.  Wash the conical flasks and beakers right away after use.  Close the tap when not in use. Avoid wastage of water.  In case of any accident viz., swallowing /splashing of chemicals in the eyes /body, report immediately to the lab in -charge.  After finishing your experiment, immediately switch off your equipment and disconnect the connection. EXPRESSION OF CONCENTRATION Mole: Weight of chemical substance divided by its molecular weight is termed as mole. Mole Fraction: It is the ratio of the moles of the component to the total number of moles in the mixture. Normality: Normality is defined as the number of gram equivalents of solute per litre of the solution Normality (N)= Weight of the solute (g) x 1000 Equivalent mass x Volume of the solution (ml) Molarity (M): It is defined as the number of gram moles of solute per litre in the solution. Molarity (M) = Number of gram moles of solute Volume of solution in litre (OR) = Weight of the solute x 1000 Molecular mass x Volume of solution (ml)Equivalent Weights of Important Compounds S.No Name of the Chemical Formula Equivalent weight 1. Hydrochloric acid HCl 36.5 2. Sulphuric acid H2SO4 49 3. Nitric Acid HNO3 63 4. Sodium Hydroxide NaOH 40 5. Potassium Carbonate K2CO3 69 6. Calcium carbonate CaCO3 50 7. Sodium carbonate Na2CO3 53 8. Potassium Hydroxide KOH 56 9. Oxalic acid H2C2O4.2H2O 63 10. Potassium permanganate KMnO4 31.6 11. Potassium dichromate K2Cr2O7 49 12. Ferrous Sulphate FeSO4.7H2O 278 13. Ferrous Ammonium sulphate FeSO4(NH4)2SO4.6H2O 392 14. Sodium thiosulphate Na2S2O3.5H2O 248 15. Copper sulphate CuSO4.5H2O 249.54 16. Sodium bicarbonate NaHCO3 84 17. Sodium oxalate (COONa)2 67 18. Ferrous/Ferric Fe3+/Fe2+ 55.84 19. Chlorine Cl 35.46 20 Zinc ion Zn2+ 65.37 21. Copper ion Cu2+ 63.5V.S.Saravanamani Structure of EDTA (Figure -I) HOOCH2C CH2COO-Na+ .. .. N-CH2-CH2-N Na+ -OOCH2C CH2COOH Step I: STANDARDISATION OF EDTA TITRATION – I Standard hard water Vs. EDTA S.No. Volume of Standard hard Water (ml) Burette Readings (ml) Concordant value Volume of EDTA (V1) (ml) Indicator Initial Final 1. 20 0 EBT 2. 20 0 3. 20 0 Calculation Volume of standard hard water = 20 ml Volume of EDTA solution consumed, V1 = …………… ml 1 ml of standard hard water contains 1 mg of calcium carbonate (CaCO3) 20 ml of standard water contains 20 mg of CaCO3 20 ml of standard water consumes V1 ml of EDTA ie., V1 ml of EDTA solution = 20 mg of CaCO3 ... 1 ml of EDTA solution = 20 mg of CaCO3 equivalent. V1Engineering Chemistry Laboratory I & II 2 1. ESTIMATION OF TOTAL, PERMANENT, TEMPORARY HARDNESS OF WATER SAMPLE (EDTA METHOD) AIM To estimate the total hardness, permanent hardness, temporary hardness in the given of hard water by EDTA method. Standard hard water and an EDTA solution are provided. PRINCIPLE Disodium salt of Ethylene Diamine Tetra Acetic Acid (EDTA) is a well known complexing agent. Its structure is shown in the figure I. Disodium salt of EDTA is used to estimate the various hardness of the given hard water containing Ca2+ and Mg2+ ions. When EDTA is added to hard water, it reacts with calcium and magnesium ions present in hard water to form stable EDTA metal complexes. From the volume of EDTA consumed the hardness can be calculated. Eriochrome Black – T is used as an indicator. The indicator forms a weak complex with the metal ions present in the hard water and gives wine red colour. Ca2+ pH = 9 – 10 Ca2+ + EBT EBT complex Mg2+ indicator Mg2+ Wine red colour When EDTA is added into the hard water, the metal ions from a stable metal complex with EDTA by leaving the indicator. When all the metal ions are taken by EDTA from the indicator metal ion complex, the wine red colour changes into steel blue, which denotes the end point. The metal EDTA complex is stable at pH 8-10. This pH range can be maintained by adding ammoniacal buffer (NH4Cl + NH4OH). Ca2+ pH = 9 – 10 Ca2+ EBT + EBT EDTA complex + EBT Mg2+ indicator Mg2+ (Steel Blue Colour) Stable (Colourless) V.S.Saravanamani Step II: ESTIMATION OF TOTAL HARDNESS OF HARDWATER SAMPLE TITRATION – II Hardwater sample Vs Std EDTA S.No. Volume of hard Water (ml) Burette Readings (ml) Concordant value Volume of EDTA (V2) (ml) Indicator Initial Final 1. 20 0 EBT 2. 20 0 3. 20 0 Calculation of the total hardness of hard water Volume of hard water = 20ml Volume of EDTA consumed, V2 = …………………………… ml (titre value) 20 ml of the given hard water sample consumes V2 ml of EDTA 20 ie., 20 ml of the given hardwater sample contains -----X V2 mg of CaCO3 V1 ... 1000 ml of the given hard water sample contains 20 1000 = ---X V2 X -----mg of CaCO3 V1 20 V2 = 1000 -----V1 = 1000 X -----------------mg of CaCO3 = …………..ppm ... Total hardness of the given sample of hard water = ……………………. ppm. Engineering Chemistry Laboratory I & II 4 PROCEDURE TITRATION – I Standardisation of EDTA STEP IThe burette is washed well with the distilled water and then rinsed with a little amount of the given EDTA solution. It is then filled with the same EDTA solution upto zero level without air bubbles. Initial reading of the burette is noted.20ml of standard hard water solution is pipetted out into a clean conical flask. 5ml of ammonia buffer solution and 2 drop of Eriochrome Black – T indicator are added. The solution turns wine red in colour and it is then titrated against EDTA taken in the burette. The change of wine red colour to steel blue colour is the end point. The final reading in the burette is noted. The difference in the burette reading gives the volume of the EDTA solution. The titration is repeated to get concordant values. Let the volume of EDTA be V1 ml. TITRATION – II Estimation of Total Hardness of Hard water sample STEP II 20ml of the given hard water sample is pipetted out into a clean conical flask. 5ml of ammonia buffer solution and 2 drops of Eriochrome Black – T indicator are added. The solution is titrated against ETDA solution taken in the burette. The change of wine red colour into steel blue colour is the end point. The titration is repeated to get concordant values. Let the volume of EDTA be V2 ml V.S.Saravanamani TITRATION – III Step III: ESTIMATION OF PERMANENT HARDNESS Boiled & filtered hard water sample Vs Std EDTA S.No. Volume of Boiled & filtered hard Water (ml) Burette Readings (ml) Concordant value Volume of EDTA (V3) (ml) Indicator Initial Final 1. 20 0 EBT 2. 20 0 3. 20 0 Calculation of the permanent hardness of the hard water Volume of the boiled hard water = 20 ml Volume of EDTA solution consumed, V3 = …………………… ml (titre value) 20 ml of boiled hard water sample consumes V3 ml of EDTA 20 ie., 20 ml of boiled hard water sample contains ----X V3 mg of CaCO3 V1 20 1000 ... 1000 ml of boiled hard water sample contains ----X V3 X -----mg of CaCO3 V1 20 V3 = -----X 1000 mg of CaCO3 V1 . = ---------------------X 1000 mg of CaCo3 =………………. Mg of CaCO3 equivalent ... Permanent hardness of hard water sample = ………………………….. ppm. Step IV: Estimation of the Temporary hardness of the hard water Temporary hardness of the given sample of water = Total Hardness – Permanent Hardness = ………………. -……………. = ………………………. ppm.Engineering Chemistry Laboratory I & II 6 TITRATION – III Estimation of Permanent Hardness Step III100ml of the given sample of hard water is taken in a clean 250ml beaker and boiled for 10-15 minutes. It is then cooled and filtered. The filtrate is collected in a 100ml standard flask. 5ml of ammonia buffer solution and 2 drops of Eriochrome Black – T indicator are added. The solution turns wine red in colour. This solution is titrated against the EDTA taken in the burette. The change in colour from wine red to steel blue is the end point. The titration is repeated to get concordant values. Let the volume of EDTA be V3 ml Step IV Estimation of Temporary hardness Temporary hardness of the water sample is calculated by subtracting permanent hardness from total hardness. Temporary hardness = Total hardness – Permanent hardness RESULT 1. Amount of Total hardness of the given sample of water = --------------------ppm. 2. Amount of Permanent hardness of the given sample of water = --------------------ppm 3. Amount of Temporary hardness of the given sample of water = --------------------ppmV.S.Saravanamani Step I: STANDARDISATION OF EDTA TITRATION I Std.Zn2+ solution Vs EDTA S.No Volume of Std.Zn2+ solution (ml) Burette Readings (ml) Concordant value Volume of EDTA (ml) Indicator Initial Final 1. 20 0 EBT 2. 20 0 3. 20 0 Calculation of strength of Sodium thiosulphate Volume of Std.Zn2+ solution V1 = 20 ml Strength of Std.Zn2+ solution N1 = ……….N Volume of EDTA V2 = …………ml Strength of EDTA N2 = ? According to the law of volumetric analysis N1N1 = V2N2 V1 X N1 N2 = ---------------V2 20 X ……….N = ----------------------………..ml ... Strength of EDTA = …………….NEngineering Chemistry Laboratory I & II 8 2.ESTIMATION OF COPPER IN BRASS BY EDTA METHOD AIM To estimate the amount of copper present in 100ml of the given solution being supplied with standard solution of ZnSO4 of strength 0.01M and an approximately 0.01M EDTA solution. PRINCIPLE Brass is an alloy of 55% Copper and 33% Zinc. The purpose of alloying Copper and Zinc is to improve machinability and strength. Brass may also contain small amounts lead, tin or aluminium. In this method, brass is dissolved in con.HNO3, so that Cu present in the brass is brought into the solution in the form of cupric ions. Then the Cu2+ ions are determined by the usual complexometric method using EDTA. Cu2+ + HD2-CuD-+ H+ (EBT) (Red colour) CuD-+ EDTA Cu-EDTA + HD2-(Colurless) (Blue) PROCEDURE TITRATION – I Standardisation of EDTA STEP IThe burette is washed well with the distilled water and then rinsed with a little amount of the given EDTA solution. It is then filled with the same EDTA solution upto zero level without air bubbles. Initial reading of the burette is noted.20ml of Zn2+ solution is pipetted out into a clean conical flask. 5ml of ammonia buffer solution and 2 drops of Eriochrome Black – T indicator are added. The solution turns wine red in colour and it is then titrated against EDTA taken in the burette. The change of wine red colour to steel blue colour is the end point. The final reading in the burette is noted. The difference in the burette reading gives the volume of the EDTA solution. The titration is repeated to get concordant values. V.S.Saravanamani Step II: ESTIMATION OF COPPER TITRATION II Copper solution Vs EDTA S.No Volume of Copper solution (ml) Burette Readings (ml) Concordant value Volume of EDTA (ml) Indicator Initial Final 1. 20 0 2. 20 0 3. 20 0 Calculation of strength of Copper solution Volume of Copper solution V1 = 20 ml Strength of Copper solution N1 = ? Volume of EDTA V2 = …………ml Strength of EDTA N2 = ………. N According to the law of volumetric analysis V1N1 = V2N2 V2 X N2 N1 = ---------------V1 20 X ……….N = ----------------------………..ml ... Strength of Copper solution = …………….N STEP III: CALCULATION OF AMOUNT OF THE COPPER Amount of copper present in 1 litre of the given brass solution = Strength of brass solution X Eq.wt of copper = 63.5 X …………N ∙∙∙ Amount of copper present in 100ml 100 of the given brass solution = 63.5 X ……..N X ---------1000 ie,gms of brass contains = ……………gms of copper ∙∙∙ Percentage of copper in ………….gms X 100 the given brass sample = -----------------------------………gram of brass = ……………….%Engineering Chemistry Laboratory I & II 10 TITRATION – II Estimation of Copper STEP II The given Copper solution is made upto 100ml in a STD flask. 20ml of the solution is pipetted out into a clean conical flask. 5ml of ammonia buffer solution and 2 drops of Eriochrome Black – T (CuD) indicator are added. The solution is titrated against ETDA solution taken in the burette. The change of wine red colour into steel blue colour is the end point. The titration is repeated to get concordant values. RESULT (i) Amount of copper in 100ml of the given brass solution = ……………gms (ii) Percentage of copper in the given brass sample =…………%V.S.Saravanamani Step I: STANDARDISATION OF SODIUM THIOSULPHATE TITRATION I Potassium dichromateVs Sodium thiosulphate S.No Volume of K2Cr2O7 (ml) Burette Readings (ml) Concordant value Volume of Sodium thiosulphate (ml) Indicator Initial Final 1. 20 0 starch 2. 20 0 3. 20 0 Calculation Volume of potassium dichromate V1 = 20 ml Strength of potassium dichromate N1 = 0.0125N Volume of Sodium thiosulphate V2 = …………ml Strength of Sodium thiosulphate N2 =? According to the law of volumetric analysis N1N1 = V2N2 V1 X N1 N2 = ---------------V2 = 20 X0.0125 V2 ... Strength of Sodium thiosulphate =…………….NEngineering Chemistry Laboratory I & II 12 3. ESTIMATION OF DISSOLVED OXYGEN IN WATER (Winkler’s Method) AIM: To determine the amount of dissolved oxygen in the given water sample by Winkler’s method. A standard solution of K2Cr2O7 of strength 0.0125 N is given. PRINCIPLE: Oxygen dissolves in water to the extend of 7-9 mgs/lit at a temperature range of 250-350C. The estimation of dissolved oxygen in water is useful in studying corrosion effects of boiler feed water and in studying water pollution. The amount of dissolved oxygen in water is estimated using Winkler’s reagent (Potassium bromide + Potassium bromate). Water sample is collected carefully avoiding aeration /deaeration in grount stoppered flssk. Initially manganese sulphate and alkali-iodide are added and the reactions occur as follows: Mn2+ + 2 OH ¯ Mn(OH)2 (White) Mn(OH)2 + ½ O2 MnO(OH)2 (Yellow brown) The precipitate dissolves in concentrated sulphuric acid liberating iodine and the liberated iodine is titrated against Na2S2O3 . MnO(OH)2 + 2H2SO4 Mn(SO4)2 +3H2O Mn(SO4)2 + 2KI MnSO4 + K2SO4+ I2 2Na2S2O3+ I2 Na2S4O6 + 2NaIV.S.Saravanamani Step II: ESTIMATION OF DISSOLVED OXYGEN TITRATION II Water sample Vs Sodium thiosulphate S.No Volume of Water sample (ml) Burette Readings (ml) Concordant value Volume of Sodium thiosulphate (ml) Indicator Initial Final 1. 20 0 starch 2. 20 0 3. 20 0 Calculation Volume of Sodium thiosulphate V1 = ………. ml Strength of Sodium thiosulphate N1 = ……….N Volume of water sample V2 = 100 ml Strength of water sample (or) N2 =? Strength of dissolved oxygen in water According to the law of volumetric analysis V1N1 = V2N2 V1 X N1 N2 = ---------------V2 X = ----------------------100 Amount of dissolved oxygen in 1 litre of tap water = Normality x Eq.Wt of O2 = ………. x 8 x 1000 = ………mg/lit PROCEDURE:Engineering Chemistry Laboratory I & II 14 TITRATION – I Standardisation of Sodium Thiosulphate The burette is washed and rinsed with Sodium Thiosulphate solution. Then the burette is filled with the given Sodium Thiosulphate solution. 20ml of 0.0125 N Potassium dichromate solutions is pipetted out into a clean conical flask. To this, 5ml of Sulphuric acid and 50ml of 5 % Potassium Iodide are added. This is titrated against Sodium thiosulphate solution. When the solution becomes straw yellow colour, starch indicator is added and then titration is continued. The end point is disappearance of blue colour and the appearance of light green colour. The titration is repeated to get concordant value. TITRATION-II Estimation of dissolved oxygen 100-150ml of the water is taken in the iodine flask, 2ml of Manganese Sulphate and 2ml of Alkali-Iodide are added. The stopper is replaced and the flask is inverted and shakes several times for the rough mixing of reagents. The flask is left aside for some time. When half of the precipitate settles down, the stopper is removed and 2 ml of concentrated Sulphuric acid is added. The stopper is replaced and the flask is inverted several times for complete dissolution of the precipitate. 100ml of this solution is pipetted out and titrated against standardised Sodium Thiosulphate solution. Starch indicator is added when the solution becomes light yellow. The titration is continued until the blue colour disappears. From the titre value the strength of dissolved oxygen is calculated and hence the amount of dissolved oxygen in the water sample is calculated RESULT: Amount of dissolved oxygen in tap water = ……..mg/litV.S.Saravanamani Step I:STANDARDISATION OF SILVER NITRATE TITRATION – I Standard NaCl Vs Silver nitrateAgNO3 Sl.No Volume of Std NaCl (ml) Burette Readings (ml) Concordant value Volume of AgNO3 (ml) Indicator Initial Final 1. 20 0 K2CrO4 2. 20 0 3. 20 0 Calculation of the normality of Silver Nitrate Volume of Sodium chloride V1 = 20 ml Strength of Sodium chloride N1 = …………….. N Volume of Silver Nitrate V2 = ……………. ml Strength of Silver Nitrate N2 =? According to volumetric law: V1 N1 = V2 N2 V1 N1 N2 = --------V2 20 X = -------------------Strength of Silver Nitrate = ……………………. . N Engineering Chemistry Laboratory I & II 16 4. ESTIMATION OF CHLORIDE CONTENT IN WATER BY ARGENTOMETRIC METHOD [Mohr’s Method] AIM To estimate the amount of chloride present in the given water sample, being supplied with standard solution chloride of strength …………. N and an approximately N/20 solution of silver nitrate. PRINCIPLE Generally water contains chloride ions (Cl¯) in the form of NaCl, KCl, CaCl2 and MgCl2. the concentration of chloride ions more than 250 ppm is not desirable for drinking purpose. The total chloride ions can be determined by Argentometric method (Mohr’s Method). In this method Cl¯ ion solution is directly titrated against AgNO3 using potassium chromate (K2CrO4) as the indicator. AgNO3 + Cl ¯ AgCl + NO3¯ (In water) (White precipitate) At the end point, when all the Cl ¯ ions are removed. The yellow colour of chromate changes into reddish brown due to the following reaction. 2 AgNO3 + K2CrO4 Ag2CrO4 + 2 KNO3 (Yellow) (Reddish brown) PROCEDURE Step I TITRATION – I Standardisation of AgNO3 The burette is washed well with distilled water and rinsed with the small amount of AgNO3 solution. It is then filled with the same solution upto the zero mark without any air bubbles.The pipette is washed with distilled water and rinsed with the small amount of standard NaCl solution. 20ml of this solution is pipetted out into a clean conical flask. 1 ml of 2% K2CrO4 indicator solution is added and titrated against AgNo3 solution taken in the burette. The end point is the change of colour from yellow to reddish brown. The titration is repeated for concordant values.V.S.Saravanamani Step II: ESTIMATION OF CHLORIDE TITRATION – II Water sample vs. Std. AgNO3 Sl.No Volume of Water sample (ml) Burette Readings (ml) Concordant value Volume of AgNO3 (ml) Indicator Initial Final 1. 20 0 K2CrO4 2. 20 0 3. 20 0 Calculation of the normality of the water sample (chloride ion) Volume of water sample V1 = 20 ml Strength of water sample (chloride ion) N1 = ? Volume of Silver Nitrate V2 = ……………. ml Strength of Silver Nitrate, N2 = …………….N According to the law of volumetric analysis: V1 N1 = V2 N2 V2 N2 ……… X ………. N1 = ____________ = _____________________ V1 20 ... Strength of the water sample (chloride ion) = ……………………..N Calculation of amount of the chloride Amount of chloride ion present in Equivalent weight of chloride ion 1 lit (1000) of the water sample = X strength of the chloride ion. .... Amount of the chloride ion present in 1000 ml of the given water sample = 35.46 X Normality of the chloride ion = 35.46 X ………… X 1000 = …………………mg/lit. Engineering Chemistry Laboratory I & II 18 Step II TITRATION II Estimation of chloride ion 20ml of the given water sample is pipetted out into a clean conical flask and 1ml of 2% K2CrO4 indicator solution is added. It is then titrated against standardised AgNo3 solution taken in the burette. The end point is the change of colour from yellow to reddish brown. RESULT (i) Amount of chloride ion present in the 1000 ml of the given water sample = -----------------------mg/litV.S.Saravanamani TABLE – I Titre values and different alkalinities S.No. Result of Titration of [P] and [M] Hardness causing ions OH-CO32-HCO3-1. [P] = 0 0 0 [M] 2. [P] = [M] [P] or [M] 0 0 3. [P] = ½ [M] 0 2 [P] or [M] 0 4. [P] > ½ [M] 2 [P] -[M] 2 [M] -2 [P] 0 5. [P] < ½ [M] 0 2 [P] [M -2 [P] ] TABLE – II Water sample Vs Std. HCl S.No. Volume of Water sample (ml) Volume of 0.1 N HCl (ml) Phenolphthalein end point [P] Methyl Orange end point [M] 1. 20 2. 20Engineering Chemistry Laboratory I & II 20 5. ESTIMATION OF ALKALINITY OF WATER SAMPLE AIM To estimate the type and amount of alkalinity present in the given water sample. A standard solution of hydrochloric acid of strength 0.1 N is given. PRINCIPLE Alkalinity is caused by the presence of hydroxide, carbonate and bicarbonate. There are five alkalinity conditions possible in a given sample of water, hydroxide only, carbonate only, bicarbonate only, combination of carbonate and hydroxide or carbonate and bicarbonate. The various alkalinities can be estimated by titrating with a standard acid using phenolphthalein and methyl orange indicators successively. (i) Phenolphthalein end point Hydroxide alkalinity is completely neutralised and carbonate alkalinity is partially neutralised during phenolphthalein end point. OH¯ + H+ H2O CO32¯ + H+ HCO3¯ (ii) Methyl Orange end point Bicarbonate neutralization occurs during methyl orange end point HCO3¯ + H+ CO2 + H2O From the two titre values the different alkalinities are calculated. When, P = M, hydroxide alkalinity 2P = M, carbonate alkalinity P = 0, bicarbonate alkalinity P< 1/2 M carbonate and bicarbonate alkalinity P> 1/2 M hydroxide and carbonateV.S.Saravanamani Calculation If the data satisfies the condition P > ½ M i. Volume of HCl required for [OH ¯] alkalinity = 2[P] – [M] = 2 X ………… _ ……………. = ………………ml ii. Volume of HCl required for [CO3¯] alkalinity = 2[M] – [P] = 2 X …………. – 2 X ………… = ………………. Ml iii. HCO3¯ is not present. Note: 1000 cc of 1 N HCl = 1 gm equivalent of CaCO3 1000 cc of 1 N HCl = 50 gm of CaCO3. 1. calculation of OH -alkalinity Volume of HCl (V1) = ………………. Strength of HCl (N1) = 0.1 N Volume of water sample (V2) = 20 ml Strength of water sample (OH ¯ alkalinity) (N2) =? According to volumetric law: V1 N1 = V2 N2 V1 N1 = N2 V2 X 0.1 N2 = = …………N 20 ... Amount of OH – content present in 1 litre of Strength of water sample x water sample, interms of CaCO3 equivalent = equ. wt of CaCO3. i.e., OH ¯ alkalinity interms of CaCO3 equivalent = ……………… N X 50 X 103 ppm or mg/litEngineering Chemistry Laboratory I & II 22 Alkalinity due to OH¯ ion = ………………… ppm CO32-= 2P HCO3-= M – 2P CO32-= 2(M-P) OH -= 2P -M Alkalinity values are expressed in terms of milligram per litre as calcium carbonate. PROCEDURE TITRATION – I (With phenolphthalein indicator) The burette is washed and rinsed with the given hydrochloric acid. Then the burette is filled with 0.1 N hydrochloric acid. 20ml of water sample is pipetted out in a clean conical flask. A drop of phenolphthalein indicator is added. Pink colour is observed. This solution is titrated against the standard acid until pink colour is disappeared. The end point is noted. This titre value corresponds to phenolphthalein end point (P).V.S.Saravanamani 2. Calculation of CO32-alkalinity Volume of HCl (V1) = ………………. Strength of HCl (N1) = 0.1 N Volume of water sample (V2) = 20 ml Strength of water sample (CO32¯ alkalinity) (N2) =? According to volumetric law: V1 N1 = V2 N2 V1 N1 _____________ = N2 V2 X 0.1 N2 = ________________________ = …………N 20 Amount of CO32¯ content present in Strength of water sample X 1 litre interms of CaCO3 equivalent = equivalent wt of CaCO3. i.e., OH – alkalinity interms of CaCO3 Equivalent = …………… N X 50 X 103 ppm or mg/lit Alkalinity due to CO32¯ ion = ………………… ppm.Engineering Chemistry Laboratory I & II 24 TITRATION – II (With methyl orange indicator) Few drop of methyl orange indicator is added to the same solution after the phenolphthalein end point. The titration is continued until the solution becomes red orange. The total titre value is noted. This titre value corresponds to methyl orange end point (M).The titration is repeated for concordant values. From the titre values the amount of each alkalinity present in given water sample is calculated. RESULT Water sample contains the following alkalinity (i) Hydroxide alkalinity = _______________ ppm (ii) Carbonate alkalinity = _______________ ppm (iii) Bicarbonate alkalinity = _______________ ppm V.S.Saravanamani TABLE – I Preparation of various polymer solutions Stock solution N1 = 5 %; Total volume V2 = 50 ml S.No Volume of 5% polymer solution (Stock Solutions) (V1 ml) Volume of water Concentration C (%) (N2) I 5 ml 45 ml V1N1 = V2N2 (5X5) ÷ 50 = 0.5 II 10 ml 40 ml (5X10) ÷ 50 = 1.0 III 15 ml 35 ml (5X15) ÷ 50 = 1.5 IV 20 ml 30 ml (5X20) ÷ 50 = 2.0 V 25 ml 25 ml (5X25) ÷ 50 = 2.5Engineering Chemistry Laboratory I & II 26 6. DETERMINATION OF MOLECULAR WEIGHT OF A POLYMER BY VISCOSITY MEASURMENTS AIM To determine the molecular weight of a given polymer using Ostwald’s viscometer, a 5% polymer solution is provided. PRINCIPLE Molecular weight of a polymer is nothing but the average molecular weight. This can be determined by measuring the intrinsic viscosity (ηi) of a dilute polymer solution. This intrinsic viscosity is related to the molecular weight by the following relationship. ηi = KMa (Mark Hownik equation) Where, ηi = Intrinsic viscosity. K & a = Constants for a given polymer-solvent combination at a given temperature. M = Average molecular weight. Molecular weight constants (K and a) for polyvinyl alcohol /solvent systems are given below: Polymer Solvent K X 10 -5 gm/dl a Polyvinyl alcohol water 45.3 0.64 MATERIALS REQUIRED 1. Polymer material 2. Solvent 3. Viscometer 4. 50ml std. flask 5. 20ml graduated pipette 6. Stop watch.V.S.Saravanamani TABLE – II Viscosity data for a polymer /solvent Flow time of the pure solvent (t0) = ……..sec S.No. Concentration (C) gm /100 ml Flow time t (sec) η/ηo= t/to = ηr t/to – 1 = ηsp ηsp /C = ηred I 0.5 II 1.0 III 1.5 IV 2.0 V 2.5 Graph ● ● ● ● ● ηsp /C ● Concentration Calculation Mark – Hownik equation is given by ηi = KMa log ηi = log K + a log M log ηi – log K log M = a log ηi – log K M = A ∙ log a Engineering Chemistry Laboratory I & II 28 Important Viscosity definitions Where, M = Molecular weight of the polymer Where, ηi = Absolute viscosity of a polymer solution. ηo = Absolute viscosity of a pure solvent. t = Flow time for the polymer solution. to = Flow time for the solvent. PROCEDURE Step I: preparation of polymer solutions of different concentrations. Polymer solutions of different concentrations, say 0.5%, 1.0%, 2.0% and 2.5% are prepared from the given polymer stock solution as shown in the Table-I. Step II: Flow time of solvent 20ml of the solvent is taken into the viscometer and is sucked through the capillary tube upto the uppermark, without any air bubbles. Now note the flow time of the solvent to flow from the uppermark (M1) to lowermark (M2). Step III: Flow time of polymer solutions Now fill the viscometer with 20ml of one of the polymer solution (Say I) into the viscometer and determine the flow time as before, similarly the flow time of the other polymer solutions (IInd , IIInd etc.,) are determined. From the flow time, reduced viscosity (ηsp/C) can be calculated. Graph is plotted between ηsp /C Vs concentration straight line obtained with an intercept called intrinsic viscosity (ηi). RESULT The molecular weight of the given polymer = ………………… 1. Relative viscosity(ηr) ηr = η /ηo = t /to 2. Specific viscosity (ηsp) ηsp = η /ηo -1 = t /to -1 3. Reduced viscosity (ηred) ηred = ηsp /C 4. Intrinsic viscosity (ηi) ηi = lim ηsp /C C → O Note: For each polymer solution, wash and rinse the viscometer with the solvent.V.S.Saravanamani Step I TABLE – I Titration of Standard HCl Vs NaOH Volume of HCl taken = 20 ml S.No Volume of NaOH added (ml) Conductance (mho) 1. 0 Decreases ● End point Increases 2. 1 3. 2 4. 3 5. 4 6. 5 7. 6 8. 7 9. 8 10. 9 11. 10 12. 11 13. 12 14. 13 15. 14 16. 15 17. 16 18. 17 19. 18 20. 19 21. 20Engineering Chemistry Laboratory I & II 30 7. CONDUCTOMETRIC TITRATION OF STRONG ACID WITH STRONG BASE AIM: To determine the amount of Hydrochloric acid present in 1000ml of the given solution by conductometric titration using standard Sodium hydroxide of --------------N PRINCIPLE: Solution of electrolytes conducts electricity due to the presence of ions. The specific conductance of a solution is proportional to the concentrations of ions in it. The reaction between HCl and NaOH may be represented as, HCl + NaOH NaCl + H2O When a solution of hydrochloric acid is titrated with NaOH, the fast moving hydrogen ions are progressively replaced by slow moving Sodium ions. As a result conductance of the solution decreases. This decrease in conductance will take place until the end point is reached. Further addition of alkali raises the conductance sharply as there is an excess of hydroxide ions. A graph is drawn between volume of NaOH added and the conductance of solution. The exact end point is intersection of two curves. MATERIAL REQUIRED: 1. Conductivity bridge 2. Conductivity cell 3. Beaker 4. Standard N/10 HCl 5. Given NaOH solution 6. Burette,pipette,glass rod etc.,V.S.Saravanamani Graph: Conductance Vs Volume of NaOH . Equivalent point Conductance (1/R) Volume of NaOH (ml) Step I Calculation of Strength of HCI Volume of HCl taken (V1) = 20 ml Strength of the HCl (N1) = ……………… Volume of NaOH (V2) = ……………. ml (from graph) Strength of NaOH (N2) = …………… According to the law of volumetric analysis, V1 N1 = V2 N2 V2 X N2 20 X N1 = = V1 ... Strength of HCl = ……………..N Calculation of amount of HCl The amount of HCl present In1000ml of the given solution = Eq.Wt of HCl x Normality of HCl = 35.45 X ……. N = ……………gms/lit The amounts of HCl present 250 in 250ml of the given solution =…………….. N X -------1000 = …………..gmsEngineering Chemistry Laboratory I & II 32 PROCEDURE: The burette is filled with std Sodium hydroxide solution upto the zero level. 20ml of the HCl is pipette out into a clean 100ml beaker. The conductivity cell is placed in it and then diluted to 50ml by adding conductivity water, so that the electrodes are well immersed in the solution. The two terminals of the cell are connected with a conductivity bridge. Now 1ml of NaOH from the burette is added to the solution, taken in the beaker, stirred for some time and then conductivity is measured. (The conductivity is going on decreasing upto the end point). The process is repeated until atleast five readings are taken after the endpoint has been reached. Now the graph is plotted by taking volume of NaOH in the X-axis and conductance in the Y-axis. The end point is noted from the graph and the amount of NaOH present in 250ml of the solution is calculated. RESULT: The amount of HCl present in 250ml of the given solution = ………..….. gmsV.S.Saravanamani Graph : (HCl and CH3COOH) Vs. NaOH Conductance Volume of NaOH TABLE – I Titration of mixture (HCl + CH3COOH) Vs NaOH S.No. Volume of NaOH added (ml) Conductance (mho) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 0123456789 10 11 12 Sudden decrease ● I end point Gradual increase ● II end point Sudden increase NaOH HCl Vs NaOH CH3COOH Vs NaOH A B Engineering Chemistry Laboratory I & II 34 8. CONDUCTOMETRIC TITRATION MIXTURE OF ACIDS (HCl & CH3COOH) AIM To determine the amount of a strong acid and a weak acid (HCl & CH3COOH) present in one litre of the given mixture of acid solution by conductometric titration using standard NaOH of 0.1 N. PRINCIPLE Solution of electrolytes conducts electricity due to the present of ions. Since specific conductance of a solution is proportional to the concentration of ions in it, conductance of the solution is measured during titration. When the Sodium hydroxide is added slowly from the burette to the solution, HCl (strong acid) gets neutralised first. Since the fast moving H+ ions are replaced by slow moving Na+ ions, decrease in conductance take place until the end point is reached. HCl + NaOH NaCl + H2O (I neutralisation). After the complete neutralisation of all HCl, the neutralisation of CH3COOH starts, CH3COOH + NaOH CH3COONa + H2O (II neutralisation) Since CH3COONa is stronger electrolyte than CH3COOH, conductivity slowly increases until all CH3COOH is completely neutralised. When the end point is reached, addition of NaOH will cause sudden increase in the conductance. This is due to the presence of fast moving OH ¯ ions. MATERIALS REQUIRED 1. Conductivity Bridge 2. Conductivity Cell 3. Beaker 4. Standard N/10 NaOH 5. Approximately N/10 HCl & CH3COOH 6. Burette, Pipette, Glass Rod Etc.,V.S.Saravanamani Step I Calculation of Strength of HCl Volume of the mixture (HCl), V1 = 20 ml Strength of the mixture (HCl), N1 =? Volume of NaOH V2 = ……… ml (I End point from graph) Strength of the NaOH N2 = 0.1N According to the law of volumetric analysis, V1N1 =V2N2 N1 = V2N2 V1 = x 0.1 20 Strength of HCl = ………..N Calculation of amount of HCl The amount of HCl present in 1000 ml of the given solution = Strength of HCl X Eq.Wt of HCl = …………….N x 35.45 = ……….gms Step I I Calculation of Strength of CH3COOH Volume of the mixture (CH3COOH), V1 = 20 ml Strength of the mixture (CH3COOH), N1 = ? Volume of NaOH V2 = ……… ml (II End point from graph) Strength of the NaOH N2 = 0.1N According to the law of volumetric analysis, V1N1 =V2N2 N1 = V2N2 V1 = x 0.1 20 Strength of CH3COOH = ………..N Calculation of amount of CH3COOH The amount of CH3COOH present in 1000 ml of the given solution = Strength of CH3COOH X Eq.Wt of CH3COOH = …………….N X 60 =…………….gms/litEngineering Chemistry Laboratory I & II 36 PROCEDURE: The burette is washed well with water and rinsed with the given standard NaOH solution.It is then filled with NaOH solution upto the zero level. 20ml of the given mixture of acids (HCl+ CH3COOH ) is pipetted out into a clean 100ml beaker. The conductivity cell is placed in it and then diluted to 50ml by adding conductivity water, So that the electrodes are well immersed in the solution. The two terminals of the cell are connected with a conductivity bridge. Now 1 ml of NaOH from the burette is added to the solution, taken in the beaker, stirred for some time and then conductivity is measured. (The conductivity is going on decreasing upto the end point). This process is repeated until atleast 5 readings are taken after the endpoint (A) has been reached. After the end point, again NaOH is gradually added, which causes increase in conductance. This increase in conductance is observed until the end point(B) is reached. After the second endpoint, sudden increase in conductance is observed on further addition of NaOH. The reading (conductivity) is continuously measured for each addition of NaOH and are tabulated. Now the graph is plotted between the volume of NaOH Vs conductivity. From the graph the first end point (A) and the second end point (B) are noted. From the end points the amount of HCl and CH3COOH present in 1 litre of the mixture of solution is calculated. RESULT (i) The amount of HCl present in 1 litre of the given solution =……….gms/lit (ii) The amount of CH3COOH present = …………….gms/lit. in 1 litre of the given solutionsV.S.Saravanamani Graph Bacl2 Vs. Na2SO4 End point Conductance Volume of Na2SO4 TABLE – I BaCl2 Vs Na2SO4 S.NO Volume of Na2SO4 added (ml) Conductance (mho) 1 0 Decrease ● End point Increases 2 1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 11 10 12 11 13 12 14 13 15 14 NaCl+Na2SO4 Bacl2 + Na2SO4+NaCl Engineering Chemistry Laboratory I & II 38 9. CONTUCTOMETRIC PRECIPITATION TITRATION USING BaCl2 – Na2SO4 AIM To determine the amount of BaCl2 present in one litre of the given solution by conductometric titration using standard Na2SO4 of 0.1 N. PRINCIPLE Solution of electrolytes conducts electricity due to the presence of ions. Since specific conductance of a solution is proportional to the concentration of ions in it, conductance of the solution is measured during titration. In the precipitation titration, the ions are converted into insoluble precipitate, which will not contribute to the conductance. When Na2SO4 is added slowly from the burette to the solution of BaCl2, BaSO4 gets precipitated while the chloride ions remain unchanged. BaCl2 + Na2SO4 BaSO4 + 2NaCl unionized The Ba2+ ions in the solution are replaced by free Na+ ions. Since the mobility of Na+ ions are less that of Ba2+ ions the conductance of the solution decreases. After the end point, when all the Ba2+ ions are replaced, further addition of Na2SO4 increases the conductance. This is due to the increase of Na+ and SO42-ions in the solution. MATERIALS REQUIRED 1. conductivity bridge 2. conductivity cell 3. beaker 4. standard N/10 Na2SO4 5. approximately N/10 BaCl2 solution 6. burette, pipette, glass rod etc., CalculationV.S.Saravanamani Calculation of Strength of BaCl2 Volume of BaCl2 mixture, V1 = 20ml Strength of the BaCl2 solution, N1 = ………..? Volume of the Na2SO4, V2 = ……………ml (from graph) Strength of the Na2SO4, N2 = 0.1N According to the law of volumetric analysis, V1N1 =V2N2 N1 = V2N2 V1 ………ml X 0.1N = 20 Strength of BaCl2 = ………..N Calculation of amount of BaCl2 The amount of BaCl2 present in 1 litre of the given solution. = Strength of BaCl2 X Eq.Wt of BaCl2 = …………….N X 122.14 = ……….gms/lit Engineering Chemistry Laboratory I & II 40 PROCEDUREThe burette is filled with Na2SO4 solution upto the zero level. 20ml of the given BaCl2 solution is pipetted out into a clean 100ml beaker. The conductivity cell is placed in it and then diluted to 50 ml by adding conductivity water. The two terminals of the cell are connected with a conductivity bridge. Now 1 ml of Na2SO4 from the burette is added to the solution, taken in the beaker, stirred, and then conductivity is measured. This is continued upto the end point. (The conductivity is going on decreasing upto the end point). After the end point, again Na2SO4 is gradually added and few more readings are noted. Thus the conductivity is continuously measured for each addition of Na2SO4and are tabulated. Now the graph is plotted between the volume of Na2SO4 and conductivity. From the graph, end point is noted and hence the amount of BaCl2 present in 1 litre is calculated. RESULT The amount of BaCl2 present in 1 litre of the given solution =……………….gms/lit.V.S.Saravanamani TABLE – I K2Cr2O7 Vs Iron Solution Volume of Ferrous Iron Solution = 20 ml S.No Volume of K2Cr2O7 (ml) Emf (Volts) 1. 0 2. 1 3. 2 4. 3 5. 4 6. 5 7. 6 8. 7 9. 8 10. 9 11. 10 12. 11 13. 12 14. 13 15. 14Engineering Chemistry Laboratory I & II 42 10. POTENTIOMETRIC TITRATION – ESTIMATION OF FERROUS IRON AIM: To estimate the amount of ferrous iron (Fe2+) present in Whole of the given solution potentiometrically. A standard solution of potassium dichromate of strength 0.1N is provided. PRINCIPLE Potentiometric titrations depend on measurement of emf between reference electrode and an indicator electrode. When a solution of ferrous iron is titrated with a solution of potassium dichromate, the following redox reaction takes place. 6Fe2+ + Cr2O72¯ +14H+ 6Fe3+ + 2Cr3+ + 7H2O During this titration Fe2+ is converted into Fe3+, whose concentration increases. At the end point, there will be a sharpe change due to sudden removal of all Fe2+ ions. The cell is set up by connecting this redox electrode with a calomel electrode as shown below Hg/Hg2Cl2(S), KCl //Fe2+, Fe3+/Pt A graph between emf measured against the volume of K2Cr2O7 added is drawn and the end point is noted from the graph. Materials Required: 1. Potentiometer 2. Pt electrode 3. Saturated Calomel electrode 4. Standard K2Cr2O7 solution 5. Given ferrous iron solutionV.S.Saravanamani Graph: I Graph: II emf ∆E ∆V Volume of K2Cr2O7 (ml) Volume of K2Cr2O7 (ml) TABLE – II K2Cr2O7 Vs Iron Solution Volume of Ferrous Iron Solution = 20 ml S.No Volume of K2Cr2O7 (ml) Emf (Volts) ∆E (Volts) ∆V (ml) ∆E/∆V Volts/mlEngineering Chemistry Laboratory I & II 44 PROCEDURE: The given ferrous ion solution is made up to 100ml in a standard flask. 20ml of this made up solution is pipetted out into a clean 100ml beaker. About 10ml of dil.H2SO4 and 20ml of distilled water are added in it. A platinum electrode is dipped into the solution. This electrode is then coupled with a saturated calomel electrode and the cell is introduced into potentiometric circuit. The std. K2Cr2O7 solution is taken in the burette and is added. TITRATION-I First a preliminary titration is carried out by adding std. K2Cr2O7 solution in portion of 1 ml and the emf of the cell is measured after each addition. The addition of K2Cr2O7 is continued even after the end point and the range at which endpoint lies is found out by plotting volume K2Cr2O7 added against emf (graph-I )V.S.Saravanamani Step – III Calculation of strength of ferrous iron solution Volume of ferrous iron solution, (V1) = 20 ml Strength of the ferrous ion solution, (N1) = ………………? Volume of the K2Cr2O7, (V2) = ……………. (from graph) Strength of K2Cr2O7, (N2) = 0.1 N According to the law of volumetric analysis, V1 N1 = V2 N2 V2 X N2 X 0.1 N1 = = V1 20 ... Strength of ferrous ion solution N1 = ……………..N Calculation of ferrous iron The amount of ferrous ion present in1000ml of the given solution = Eq.Wt of Fe x Normality of Fe2+ =………………. N X 55.85 The amount of ferrous ion present 100 in100ml of the given solution =…………….. N X 55.85 X -------gms 1000 = ………………….gmsEngineering Chemistry Laboratory I & II 46 TITRATION-II Another titration is carried out by adding std. K2Cr2O7 solution in portion s of 0.1ml near the end point and the emf of the cell is measured after each addition. The addition of K2Cr2O7 is continued even after the endpoint for further 1ml. The accurate endpoint is determined by plotting ∆E /∆V Vs Volume of K2Cr2O7 added (graph-II).From the end point, the strength of ferrous iron solution and hence its amount can be calculated. RESULT: The amount of ferrous ion present in 100ml of the given solution is = ------------------mg/litV.S.Saravanamani TABLE – I Titration I HCl (20ml) Vs std NaOH Model Graph pH ΔpH/ΔV Volume of NaOH Volume of NaOH S.No. Volume of NaOH pH 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 123456789 10 11 12 13 14 Gradual increase ● end point Sudden increaseEngineering Chemistry Laboratory I & II 48 11. PH TITRATION – DETERMINATION OF STRENGTH OF HCl by NaOH AIM To determine the strength of given HCl by pH metry, a standard solution of NaOH of 0.1 N is provided. PRINCIPLE Since the pH of the solution is related to the H+ ion concentration by the following formula. pH = – log [H+] Measurement of the solution gives the concentration of H+ ions in the solution. When NaOH is added slowly from the burette to the solution of HCl, the fast moving H+ ions are progressively replaced by slow moving Na+ ions. As a result pH of the solution increases. HCl + NaOH NaCl + H2O The increase in pH takes place until all the H+ ions are completely neutralised (upto the end point). After the end point, further addition of NaOH increase pH sharply as there is an excess of fast moving OH¯ ions. MATERIALS REQUIRED 1. pH meter, 2. Glass electrode, 3. Beaker, 4. Standard N/10 NaOH, 5. Approximately N/10 HCl, 6. Burette, pipette, Glass rod etc. V.S.Saravanamani TABLE – II Titration of HCl (20 ml) Vs NaOH S.No. Volume of NaOH (ml) pH ∆pH ∆V ∆pH/∆V 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.Engineering Chemistry Laboratory I & II 50 PROCEDURE The burette is filled with std. NaOH solution. Exactly 20ml of the given HCl solution is pipetted out into a clean beaker. It is then diluted to 20ml by adding distilled water. The glass electrode is dipped in it and connected with a pH meter. TITRATION – I First a preliminary titration is carried out by adding std. NaOH solution in portions of 1ml from the burette to the HCl solution taken in the beaker and pH of the solution is noted for each addition. This process is continued until atleast 5 readings are taken after the end point, and the range at which the end point lies is found out by plotting volume of NaOH added against pH (graph I).V.S.Saravanamani Step I Calculation of Strength of HCI Volume of HCl taken (V1) = 20 ml Strength of the HCl (N1) = ……………… Volume of NaOH (V2) = ……………. ml (from graph) Strength of NaOH (N2) = …………… According to the law of volumetric analysis, V1 N1 = V2 N2 V2 X N2 20 X N1 = = V1 ... Strength of HCl = ……………..N Calculation of amount of HCl The amount of HCl present In1000ml of the given solution = Eq.Wt of HCl x Normality of HCl = 35.45 X ……. N = ……………gms/lit The amounts of HCl present 250 in 250ml of the given solution =…………….. N X -------1000 = …………..gmsEngineering Chemistry Laboratory I & II 52 TITRATION -II Another titration is carried out by adding std. NaOH solution in portion of 0.1ml near the end point and PH of the solution is noted after each addition. The addition of NaOH is continued even after the end point for further 1 ml. The accurate end point is found out by plotting ∆pH/∆V against volume of NaOH added (Graph II).From the end point, the strength of HCl solution and the amount of HCl can be calculated. RESULT: 1. Strength of the given HCl Solution = ---------------------N 2. Amount of HCl Present in 1 litre of the solution =--------------------------gms.V.S.Saravanamani Calculations Weight of silica crucible, W1 = .…………… gms Weight of silica crucible + residue (CuSO4), W2 = ……………. gms Weight of residue, (W2 __ W1) = ………………. __ …………….. i.e., Weight of the CuSO4 = ………………. gms Thus, 2 gms of CuSO4 · 5H2O contains, = 2 – (W2 __ W1) gms of water. = ……………….. gms of water. 2 (W2 __ W1) X 100 ... 100 gms of CuSO4 · 5H2O contains = _______________________ mgs 2 Water of crystallization of crystalline salt CuSO4 · 5H2O = …………….. gms% Engineering Chemistry Laboratory I & II 54 12. DETERMINATION OF WATER OF CRYSTALLIZATION OF A CRYSTALLINE SALT (CuSO4 · 5H2O). AIM To Determine the quality of water of hydration present in CuSO4 · 5H2O, you are provided with a sample of A.R. CuSO4 · 5H2O. PRINCIPLE Crystalline copper sulphate is blue in colour, which is due to the hydration of water. Its molecular formula is CuSO4 · 5H2O. When CuSO4 · 5H2O is roasted, it losses its water molecule and becomes white in colour. CuSO4 · 5H2O CuSO4 + 5H2O (Blue) (White) From the weight loss, the amount of water molecule present in the CuSO4 · 5H2O can be calculated. According to stoichiometric equation CuSO4 · 5H2O CuSO4 + 5H2O Mol. wts = 249.68 159.68 90 i.e., 249.68 gms of CuSO4 · 5H2O contain 90 gms of water χ X 90 .·. χ gms of CuSO4 ∙ 5H2O contains = ______________ gms of water. 249.68 V.S.Saravanamani Engineering Chemistry Laboratory I & II 56 PROCEDURE 1. Determination of weight of empty silica crucible A silica crucible is washed and heated to about 200˚C in a hot air oven and cooled in a desiccator and then weighed. This process of heating, cooling and weighing is repeated until a constant weight is obtained. Let the weight of empty silica crucible = W1 gms. 2. Determination of weight of silica crucible + residue (CuSO4) Exactly 2.0 gms of the given A.R.CuSO4 · 5H2O sample is taken in a weighed silica crucible and roasted (strongly heated with stirring) until the blue colour changes into white. Then it is cooled in a desiccator and weighed. This process of roasting, cooling and weighing is repeated until a constant weight is obtained. Let the weight of silica crucible + residue (CuSO4) = W2 gms. 3. Determination of weight of the residue (CuSO4) The weight of the residue (CuSO4) is obtained from the difference in weights of W1 and W2. i.e., weight of the residue (CuSO4) = (W1 – W2) gms. RESULT Water of crystallization of crystalline salt (CuSO4 · 5H2O) = ……………. gmsV.S.Saravanamani λmax = 480 nm S.No Concentration of Fe3+ (mg/L) Absorbance 1. 0(Blank) 0 2. 3. 4. 5. 6. 7. 8. UnknownEngineering Chemistry Laboratory I & II 58 13. ESTIMATION OF IRON BY SPECTROPHOTOMETRIC METHOD (USING THIOCYANATE) AIM: Estimate the amount of Fe3+ present in the given water sample by spectrophotometric method using NH4SCN reagent. PRINCIPLE: When monochromatic light falls upon a (colored) homogeneous medium a portion of the monochromatic light is absorbed and a portion is reflected and the remaining is transmitted. Io = Ir +Ia + It Where Io = intensity of radiations Ir = intensity of reflected radiation Ia = intensity of absorbed radiation It = intensity of transmitted radiation Ir is usually eliminated, hence Io = Ir +Ia + It Beer-Lambert‘s Law Intensity of a transmitted beam of monochromatic light decreases exponentially, as the concentration of the absorbing substance in the solution and the path length of the light through the solution increases arithmetically, both being considered independently. This can be mathematically represented as, It =Io 10-ξ cl Where, ξ is molar absorption co-efficient C is concentration of solution in moles /litre L is thickness of the cell (or) path length of the solution through Which light passes = 1 cm. Io Log _______ = ξCl = A It Where, A = absorbance V.S.Saravanamani Engineering Chemistry Laboratory I & II 60 Keeping the path length constant say 1 cm, the variation is only in concentrations. Solutions which are coloured can be studied by this method, or solutions which can be coloured by the addition of a reagent can also be studied. Example: Ferric forms a blood red colour when thiocynate is added. This colour is due to the formation of ferric thiocynate, an octahedral complex. Fe3+ + 6KSCN → [Fe (SCN) 6]3-+ 6K+ This solution absorbs white light in the blue region, λ = 480nm which is complementary to red colour. In the spectrophotometer the radiation is allowed to pass through the solution and the transmitted radiation is measured. To measure Io, a blank solution (distilled water) and transmitted light is adjusted for hundred percent transmissions. The instrument then converts transmittance in terms of absorbance. A calibration graph can be drawn with absorbance verses concentration. PROCEDURE The spectrophotometer is switched on and warmed up for 15 minutes. The monochromatic is adjusted for λ = 480 nm. Blank solution is kept in spectrophotometer and transmittance is adjusted for 100 percent, for which absorbance is 0. The solutions of known concentrations are prepared by developing the red colour using NH4SCN and 4 mL of dilute HNO3 and making upto 100 mL in a series of 100 mL standard flasks (containing1, 2,3,4,5 and 6 mg of iron per litre). The solutions are kept in the spectrophotometer one by one and absorbance for unknown concentration is measured. The cuvet (sample cell) should be handled properly and cautiously. A calibration graph is drawn between absorbance and concentration which passes through the origin. From this the unknown concentration is determined. RESULT Amount of Fe3+ present in the given water sample = ___________ ppm V.S.Saravanamani Weighing and Preparation of some Important Reagents Dilute Acids (1N) Reagent Normality Volume required to make 1 litre of 1N solutions Hydrochloric acid (HCl) 11.30 89ml Nitric acid (HNO3) 16.0 63ml Sulphuric acid (H2SO4) 36.0 28ml Acidic acid (CH3COOH) 17.4 58ml Ammonium hydroxide (NH4OH) 14.3 70ml Sodium hydroxide (NaOH) 40gm Ortho phosphoric Acid 41.1 23ml Liquid Ammonia (NH3) 14.3 71ml Preparation for other reagents (For 30 students) Expt.No: 1 -Hardness of Water S.No Name of the reagent Quantity M or N Preparation 1. Standard EDTA. (Mol.Wt = 372.24) 1 litre 0.01 M 3.72gms of EDTA + 1 litre distilled water 2. Std.Hard water (Mol.Wt =100) 1 litre 0.01 M 1.0 gm of CaCO3 + min.HCl + 1 litre distilled water 3. Sample Hard water 1 litre 05 gm of NaHCO3 or Ca(HCO3)2+ 1 litre distilled water 4. Buffer solution 1 litre 70 gms NH4Cl+568ml liq.NH3 make upto 1 litre using distilled water 5. Indicator Eriochrome Black -T 100ml 1 gm of EBT + 100ml Methanol or Ethanol Engineering Chemistry Laboratory I & II 62 Expt.No: 2 -Estimation of Copper in Brass S.No Name of the reagent Quantity M or N Preparation 1. 0.01 M ZnSO4.7H2O. (Mol.Wt = 287) 1 litre 0.01 M 2.87gms of ZnSO4.7H2O + 1 litre distilled water 2. Buffer solution 1 litre 70 gms NH4Cl+568ml liq.NH3 make upto 1 litre using distilled water 3. Indicator Eriochrome Black -T 100ml 1 gm of EBT + 100ml Methanol or Ethanol 4. Standard EDTA. (Mol.Wt = 372.24) 1 litre 0.01 M 3.72gms of EDTA + 1 litre distilled water 5. Brass solution 1 litre 20 gms brass powder + 80ml con.HNO3+150ml dil.H2SO4 in1 litre distilled water 6. 0.1 M CuSO4 (Mol.Wt = 249.68) 1 litre 0.1 M 24.968 gms of CuSO4 + 1 litre distilled water 7. 0.01 M CuSO4 100ml 0.01 M 10ml of 0.1 M CuSO4 solution in 100ml distilled water Expt.No: 3 – Dissolved Oxygen S.No Name of the reagent Quantity M or N Preparation 1. 0.01 N Sodium thiosulpahte. (Mol.Wt = 248.17) 1 litre 0.01 N 2.48gms of Na2S2O3 + 1 litre distilled water. 2. 0.01 N Potassium dichromate. (Eq.Wt = 49.02) 1 litre 0.01 N 0.49gms of K2Cr2O7 + 1 litre distilled water. 3. MnSO4.H2O. 100ml 5% 0.49gms of MnSO4.H2O + 100ml distilled water. 4. KI Solution. 100ml 10% 10gms of KI + 100ml distilled water. 5. Alkali-Iodide mixture. 500ml 100 gms KI (20%) + 40gms of NaOH (2N) + 500ml distilled water 6. Starch Solution 100ml 1% 1 gm Starch powder + 100ml boiled distilled water 7. Con. H2SO4 8. Dil. H2SO4. (4 N) 1 litre 4 N 112ml of Con. H2SO4 +900ml of distilled water.V.S.Saravanamani Expt.No: 4 – Chloride content S.No Name of the reagent Quantity M or N Preparation 1. Standard NaCl. (Mol.Wt = 58.44) 1 litre 0.01 N 0.584 gms of NaCl + 1 litre distilled water 2. Standard AgNO3. (Mol.Wt =169.87) 1 litre 0.01 N 1.698 gm of AgNO3 + 1 litre distilled water 3. Water Sample. 1 litre 0.05 N 2.92 gms NaCl + 1 litre distilled water 4. Indicator K2CrO4 100ml 2% 2 gms of K2CrO4 + 100ml distilled water Expt.No: 5 – Alkalinity S.No Name of the reagent Quantity M or N Preparation 1. 0.1 N HCl. (Con.HCl = 11.30 N) 1 litre 0.1 N 8.9ml of Con. HCl +991ml of distilled water. 2. Sample Water. NaOH (0.1 N) + Na2CO3 (anhyd) (0.1 N) 1 litre 0.1 N 4 gms of NaOH +5.3 gms of Na2CO3 (anhyd) + 1 litre distilled water 3. Phenolphthalein indicator. 100ml 1% 1 gm of Phenolphthalein + 100ml ethanol. 4. Methyl orange indicator 100ml 1% 1 gm of Methyl orange + 100ml ethanol. 5. Sample Water. Na2CO3 (anhyd) (0.1 N)+ NaHCO3 (0.1N). 1 litre 0.1 N 5.3 gms of Na2CO3 (anhyd) + 8.4 gms of NaHCO3 +1 litre distilled waterEngineering Chemistry Laboratory I & II 64 Expt.No: 6 – Determination of molecular weight S.No Name of the reagent Quantity M or N Preparation 1. Stock PVA Solution 500ml 1% 5 gms of PVA is dissolved in 500ml of hot water with stirring and then cool. Expt.No: 7 – Conductometric titration of Strong acid with Strong base S.No Name of the reagent Quantity M or N Preparation 1. 0.1 N Std.HCl. (Con.HCl = 11.30 N) 1 litre 0.1 N 8.9ml of Con. HCl +1 litre of distilled water. 2. 0.1 N Std NaOH. (Eq.Wt = 40) 1 litre 0.1 N 4 gms of NaOH +1 litre distilled water Expt.No: 8 – Conductometric titration of mixture of acids S.No Name of the reagent Quantity M or N Preparation 1. Mixture of acids. (HCl + CH3COOH) 1 litre 0.1 N 8.9ml of Con. HCl + 5.8ml of glacial CH3COOH +1 litre of distilled water. 2. 0.1 N Std NaOH. (Eq.Wt = 40) 1 litre 0.1 N 4 gms of NaOH +1 litre distilled water Expt.No: 9 – Conductometric precipitation titration S.No Name of the reagent Quantity M or N Preparation 1. 0.1 N BaCl2. (Eq.Wt = 122.2) 1 litre 0.1 N 12.22 gms of BaCl2 +1 litre of distilled water. 2. 0.1 N Na2SO4. (Eq.Wt = 71.0) 1 litre 0.1 N 7.1 gms of anhyd.Na2SO4 +1 litre distilled waterV.S.Saravanamani Expt.No: 10 – Potentiometric titration S.No Name of the reagent Quantity M or N Preparation 1. 0.1 N Potassium dichromate. (Eq.Wt = 49.02) 1 litre 0.1 N 4.9 gms of K2Cr2O7 +1 litre of distilled water. 2. 0.1 N FAS. [ FeSO4(NH4)2SO4.6H2O ] (Eq.Wt = 392) 1 litre 0.1 N 39.2 gms of FAS + 15-20ml H2SO4 in 1 litre distilled water 3. 2.5 N H2SO4 250ml 2.5 N 17.36ml of con.H2SO4 in 250ml of distilled water. Expt.No: 11– pH metry S.No Name of the reagent Quantity M or N Preparation 1. 0.1 N HCl. (Con.HCl = 11.30 N) 1 litre 0.1 N 8.9ml of con. HCl +991ml of distilled water. 2. 0.1 N NaOH. (Eq.Wt = 40) 1 litre 0.1 N 4 gms of NaOH +1 litre distilled waterEngineering Chemistry Laboratory I & II 66 Expt.No: 13 – Spectrophotometry S.No Name of the reagent Quantity M or N Preparation 1. Stock Fe3+ Solution. 1 litre 1ml= 10ppm 0.0838 gms of FAS + 2-5ml HNO3 +1 litre distilled water 2. Std Solutions of Fe3+ 1 litre 0.1 N 39 gms of FAS + 2-5ml HNO3 +1 litre distilled water Volume of Fe2+ ion Solution. Volume of HNO3 Volume of NH4SCN Volume of distilled water Concentration of Iron (ppm) 1 1 1 7 1 ppm 2 1 1 6 2 ppm 3 1 1 5 3 ppm 4 1 1 4 4 ppm 5 1 1 3 5 ppm 6 1 1 2 6 ppm 3. 4 N HNO3. 100ml 4 N 25ml of con.HNO3 + 75ml of distilled water 4. NH4SCN 500ml 10% 50 gms of NH4SCN + 500ml of distilled water