6. QUANTITATIVE SALT-ANALYSIS 6.1 Volumetric Analysis: Definition and Principle 6.2 Use of chemical Balance 6.3 To prepare the standard solution 6.

Transcription

1 36 6. QUANTITATIVE SALT-ANALYSIS 6.1 Volumetric Analysis: Definition and Principle 6.2 Use of chemical Balance 6.3 To prepare the standard solution 6.4 Titration 6.5 Script of titration experiment 6.6 Calculation of titration

2 Volumetric Analysis : Definition and Principle In practical chemistry, based on chemical analysis the nature and amount of a substance ion can be determined. This chemical analysis is of two types quantitative analysis and qualitative analysis. Quantitative analysis deals with the volumetric and massive analysis of elements of a substance. Quantitative analysis is of two types. 1. Quantitative Analysis- In the quantitative analysis the substance or radicals, which has to be determined, is converted into the temporary compound. This temporary compound is either measure into its original form or it is burn to convert into the other compound, which is appropriate for measurement.now the weight of the substance or radicals is determined by the physical method. The ppt. method is the most useful method for the quantitative analysis. Note-Quantitative analysis is not included into the syllabus of Higher Secondary. 2. Volumetric Analysis- It is a sub-branch of quantitative analysis. It is a simple and accurate method of analysis. This method can determine amount of substance in terms of volume accurately the, hence the name volumetric analysis. In this type of analysis,a solution of a substance whose concentration is to be calculated is taken and a known volume of another substance whose concentration is known. The volume of the solution is measured when the chemical reaction is completed. The end point of such reaction is indicated by change in colour or precipitation etc.thus in this,method of calculation, known volume of two substances are taken and the concentration of one of them is known then the concentration of the other is easily calculated. This method is called Titration. The description of volumetric analysis of substance is given into this chapter. Solution:-The homogenous mixture of any substance is given into the water or any liquid in solution. Concentration or Strength-Amount of a substance in grams dissolved in a one-litre solution is called concentration or strength. Example-If the 2 grams of caustic soda is dissolved in 1 litre then the concentration of the solution will be 2gm/litre. Normality-Amount of a substance in grams equal to its 1 gm equivalent weight in 1 litre solution is called Normality of that solution. Or Normality can be express by the following formula, which is a ratio between concentration of that solution and gram equivalent weight of that substance. It is denoted by N. It is a number. It does not have any limit. Normality=Concentration /Equivalent weight Example-(1) Concentration of Caustic Soda solution is 4 gm/litre and equivalent weight of Caustic Soda is 40.Normality of this solution will be. =4/40=1/10=-N/10 =0.1N

3 38 Example-(2) 2N Caustic Soda solution will have 80 gm of it dissolved in 1 litre because: 2xN=2x40 i.e. 80 gm/litre. Molarity or Gram Molecular Weight :The amount of substance equal to its gram molecular weight dissolved in one litre solution,is called as Molar Solution. For example, 20 gm of caustic soda dissolved in one litre solution will form 0.5 M solution of Caustic Soda. Percentage : Amount of solute dissolved in 100 parts of the solution is called percentage of that solution. Example :10 percent solution of NaOH means 10 gram of NaOH dissolved in 100 gram of NaOH. Gram Molecular Weight : Amount of a substance in grams equal to molecular weight of that substance is called 1gram molecular weight of that substance. Example : Molecular weight of Sodium Carbonate is 106 then its 1 gram molecular weight is 106 gram and the molecular weight of 2 gram is 212 gram. Standard solution : Solution of known concentration (the solution in which the quantity of solvent is known in its known volume) is called a standard solution. Concentration of standard solution is expressed in terms of gram per litre or normality. Normal Solution : A solution containing 1 gram equivalent of a solute dissolved in 1 litre of solution is called a Normal Solution. Example : In the normal solution of Na 2 CO 3, 53 gram solvent substance is dissolved into the 1 litre (equivalent weight of Na 2 CO 3 is 53). Acid : Those substance which,when dissolved in water give Hydrogen ion (H+) as the only positive ion are acids. Example : HCL=H + + Cl - H 2 =2H + + Strength of Acid : The acid, which gives more H+ ions its strength, will be more. Two acids having same normality may have different degree of ionization and therefore the strengths will also different. Example : N/10 HCL is 92% ionized where as for acidic acid it is 1.3%.Therefore hydrochloric acid is a strong acid where as acidic acid is a weak acid. Relative strength of weak acids : HCL>HNO 3 >H 2 >H 3 PO 4 > CH 3 COOH Basisity of acids : Basisity of acids means can be express its strength to react with bases. Amount of Hydrogen ions given by one molecule of an acid is its basisity. It depends on the number of atoms of the H in acid. Therefore, number of atom of replaceable hydrogen of any molecule of acid is basisity of acid. HCI,HNO 3,CH 3 COOH are monobasic acids,h 2 is a dibasic acid and H 3 PO 4 is a tri-basic acid. Base : Those substances which when dissolved in water give Hydroxide ion (OH - ) as the only negative ion are called bases. Example : NaOH=Na + +OH - NH 4 OH=NH 4+ +OH - Strengths of Bases : The base those which give more OH - ions will have more strength. Two bases having same Normality may have different degree of ionization. Therefore, they have different strength.

4 39 For example for same normality of N/10 NaOH IS 84% ionized where as N/10 NH4OH is only 1.3% ionized. Relative strength of few bases - KOH>NaOH>NH 4 OH>Ba(OH) 2. Acidity of Bases : Amount of Hydroxide ions (OH - ) present in one molecule of base, is called the acidity. Example : NH 4 OH,NaOH are mono acidic bases. Ba(OH) 2 is a di-acidic base. Equivalent Weight : Amount of substances which combines with or displaces parts of hydrogen or 8 part of oxygen or 35.5 parts of chlorine is called equivalent weight of that substances. Gram Equivalent Weight : Equivalent weight of a substance express in gram is called 1 gm equivalent weight of the substance. Example : Equivalent of H 2 is 49.So 1 gram equivalent of H 2 will weight 49 gram and 0.1 gm equivalent of H 2 acid will weight 4.9 gram. 1. Equivalent weight of acid : The equivalent weight of acid is that which has one position for replaceable hydrogen atom is equivalent weight of acid. For example, one molecule of HCL has (gram molecular weight) one replaceable hydrogen atom. Therefore the equivalent weight of HCL is 36.5.Similarly one molecule of H 2 has 2 replaceable hydrogen atoms (molecular weight 980. Therefore the equivalent weight of H 2 is 49(98/2). Equivalent weight of acid=molecular weight of acid/basisity of acid 2. Equivalent weight of base : The part of the base which has one portion for replaceable OH - radicals or the number of weights which completely neutralize the one equivalent weight of any acid. For example in the following reaction 36.5 gram, HCL (1 gm equivalent weight) completely neutralized 40 gm NaOH.Therefore the equivalent weight of NaOH is 40. NaOH + HCL=NaCl+H2O 40gm 36.5gm Equivalent weight of base=molecular weight of base/acidity of base 3. Equivalent weight of salt : The equivalent weight of salt is the sum of the equivalent weight of the radicals. Generally the equivalent weight of acid, base or salt can be determine by the ratio of its molecular weight to the total positive valancy. The equivalent weight of Na 2+ Co 3 is 23+60/ 2=53 or 106/2=53 Similarly the equivalent weight of HCL=36.5/ 1= The equivalent weight of oxidization and reduction agents : The equivalent weight of any oxidising agent is that weight which makes available the 8 weight oxygen for reaction with 1 weight of hydrogen. Similarly the equivalent weight of the reduction agents is that weight which makes available one part of hydrogen or reacts with 8 part of oxygen. 1. KMnO 4 (Potassium Permanganate)- This is react in acidic medium in following ways- 2KMnO 4 +3H 2 K 2 +2Mn +3H 2 O+5O Therefore, the 2 molecule of the KMnO 4 produces 5 atoms of oxygen in acidic medium i.e.

5 40 2[39+55=16x4]=5x16=10x1 Or 2[158]=10x8=10x1 Or 316/10=8=1 Or 31.6=8=1 Therefore, the equivalent weight of KMno4 is 31.6 into the acidic medium. In the similar way. The equivalent weight of oxidization agent=molecular weight of substance x2x Number of atom of the obtain oxidization agent. (Note- Hence 2 is multiplied in formula because the atomic weight of oxygen is 16 is eight times more then the equivalent weight of oxygen). Equivalent weight of KMnO4 in the neutral medium- 2KMnO 4 +H 2 O=2MnO 2 +2KOH+3O 2KMnO 4 =3O=6H 2x158=3x16=6H 316=6x8=6H Or 316/6=52.66=8=1 Therefore the equivalent weight of KMnO 4 in the neutral medium is Equivalent weight of KMnO4 in basic medium- 2KMnO 4 +2KOH=2K 2 MnO 4 +H 2 O+O 2KMnO 4 =O 2x158=16=2x8 Equivalent weight=2x158/2=158. In this way, the equivalent weight of KMnO 4 in the basic medium is 158. The example shows that the equivalent weight of any substance does not remain constant.it depend upon the reaction and the nature of the medium. Ferrous ammonium sulpheta : Fe (NH 4 )2.6H 2 O Or Ferrous Sulphate FeSO4.7H2O (i) Equivalent weight of iron : 2KMnO 4 +3H2O=K 2 +2Mn +5O+3H 2 O 10Fe +5H 2 +5O=5Fe ( ) 3 +5H 2 O 2KMn +10Fe +8H 2 =K 2 + 5Fe 2 ( ) 3 +2Mn +8H 2 O 2KMn =5O=10Fe.7H 2 O Or O=5Fe.7H 2 O 1/2O=Fe.7H 2 O Therefore the equivalent weight of Fe.7H 2 O is 278. Generally Ferrous sulphate is used in place of Ferrous ammonium sulphate Fe. (NH 4 ) 2.6H 2 O in titration because the ferrous sulphate oxidized into the ferric sulphate in the air 1.Therefore the equilibrium weight of ferrous ammonium sulphate is 399(the molecular weight of the ferrous ammonium sulphate. The equivalent weight of used compounds in the acid base titration Table Name of the compounds Molecular Basisity Acidity Equivalent Weight Weight Hydrochloric Acid(HCL) x 36.5 Nitric Acid(HMO 3 ) 63 1 x 63 Sulfuric Acid(H 2 ) 98 2 x 49 Acitic Acid(CH 3 COOH) 60 1 x 60 Oxalic Acid(H 2 C 2 O 4 2H 2 O) x 63 Potassium Hydroxide(KOH) 56 x 1 56 Sodium Hydroxide9NaOH) 40 x 1 40 Barium Hydroxide(Ba(OH) 2 2H 2 O) 315 x Amonium Hydroxide(NH 4 OH) 35 x 1 35 Sodium Carbonate(Na 2 CO 3 ) 106 x 2 53

6 Use of chemical balance The important task of volume analysis is to measure the volume the solution that is taking part in the given chemical reaction. Therefore the accuracy in the results of analysis depends on the pure measurement of the volumes of the reactants solutions.for measuring these solutions some special apparatus are used. In this chapter, an introduction of these apparatus and method of using them is described. Apparatus used to prepare the standard solutions are - 1. Chemical Balance 2. Weight Box 3. Weighting tube or bottle 4. Measuring Flask 5. Washing Bottle etc 1. Chemical Balance : The balance is enclosing in the glass tube. Its body is made of wood or aluminum alloy. Base is generally made of black glass or slate. Box can be open from the front by means of a sliding door. Two more small doors are present at the right and left side of the box. They can be open when required. Balance rod is made of metal. It is place on the knife-edge of agate stone in the center. The both end of the rod have knife-edge, each of them is made of agate stone. The pan of the balance are suspended from the terminal knife edges by means of shrimps pointer is attached to the center of the rod and its another end moves on the scale. When it reaches on the both side of scale until the equal number is achieve then the rod is horizontal. At the bottom of the balance, two horizontal or parallel screws are attached, so that to keep the balance in the horizontal position properly. A spirit level is present in the balance to gauge the plane of the balance. In some balance to check the horizontal position a plumbline is suspended with the base of the rod. A hook is attached by the rider and a rider is moved with the help of a rider carrier. Fine adjustment of screws are present on the two ends of the beam or rod to mark the rod horizontal.when the balance is not in use then the pans at rest on the pan barrier. Base plate at which the mirror is open having a knob at the front, after rotating it, the rod and the pans are becomes independent gram quantity can be measure (at the 4 places of decimal point) A labeled diagram of chemical balance is shown below- Marked Bar Adjustable Screw Hollow Stand Scale Key Indicator Fig. 6.1 Chemical balance Weight Box : A weight box containing weights of various denominations is used. A standard weight box contain following weights

7 42 Weight of Gram Weight of Mili Gram (Fractional Weight) Weight Number Weight Number 50 gm gm gm gm gm gm. 1 5 gm gm. 1 2 gm gm. 2 1 gm gm. 1 Fig. 6.2 Weight Box Figure: Fractional weights are of following shapes Rider The weight of 500 and 50 mg are have 5 sides, 200 mg. and 20 mg. weight are square or rectangular (each have 2-2 weight) and 100mg. and 10 mg. weight are triangular shape. 3. Precaution for the use of chemical balance: - 1. Clean the balance with soft cloth. 2. Adjust the balance plane with the help of adjustment screw. At the position of adjustment the plumbline is placed in the line of the points in the stand. 3. To see that the point goes to the equal point on the both side of the scale, rotate the handle or knob and see the horizontal rod. If it is not possible then use the balancing screw for this. 4. Pick the arrested knob of balance slowly and put it slowly. 5. The weight should place right pan and the substance, which has to be measure keep in to the left pan. 6. Do not touch the weight from the handle. Always use tongs for this. 7. Always starts the measurement with the big weight. 8. When the rod of the balance is upwards then do not touch the object. 9. The object should always measure by the watch glass or the weight it in the weighting tube. 10. Weight box should keep only when the right window is open. 11. Shut down the window at the time of weight. 12. At the time of weighting of these is a condition when the weight of 10ml is decreases or increases then use rider and place it always right side of rod. 13. Always write down the weight of the substance is into your practical notebook. 14. Do not weight the hot and the cold things. 4. The method of taking the reading by rider:- Use the rider for weighting below 10 ml. The zero is marked at the middle of the balance rod of chemical balance and both sides have blocks., which have 1 to 10 number are marked.

8 43 Weight Box : A weight box containing weights of various denominations is used. A standard weight box contain following weights. Weight of Gram Weight of Mili Gram (Fractional Weight) Weight Number Weight Number 50 gm gm gm gm gm gm. 1 5 gm gm. 1 2 gm gm. 2 1 gm gm. 1 Fig 6.2 Weight Box Fractional Weight or mg weight have following different shapes Fig : Rider The weight of 500 and 50 mg are have 5 sides, 200 mg. and 20 mg. weight are square or rectangular (each have 2-2 weight) and 100mg. and 10 mg. weight are triangular shape. 3. Precaution for the use of chemical balance: - 1. Clean the balance with soft cloth. 2. Adjust the balance plane with the help of adjustment screw. At the position of adjustment the plumbline is placed in the line of the points in the stand. 3. To see that the point goes to the equal point on the both side of the scale, rotate the handle or knob and see the horizontal rod. If it is not possible then use the balancing screw for this. 4 Pick the arrested knob of balance slowly and put it slowly. 5. The weight should place right pan and the substance, which has to be measure keep in to the left pan. 6. Do not touch the weight from the handle. Always use tongs for this. 7. Always starts the measurement with the big weight. 8. When the rod of the balance is upwards then do not touch the object. 9. The object should always measure by the watch glass or the weight it in the weighting tube. 10. Weight box should keep only when the right window is open. 11. Shut down the window at the time of weight. 12. At the time of weighting of these is a condition when the weight of 10ml is decreases or increases then use rider and place it always right side of rod. 13. Always write down the weight of the substance is into your practical notebook. 14. Do not weight the hot and the cold things. 4. The method of taking the reading by rider : Use the rider for weighting below 10 ml.

9 44 The zero is marked at the middle of the balance rod of chemical balance and both sides have blocks., which have 1 to 10 number are marked.between each two number there is 5 or 10 blocks are present. Such type of scale has 0 to 50 or 100 blocks on one side. i.e. the half portion of the rod of balance is divided into 50 or 100 small blocks. The weight of the rider is generally 10 mg therefore if 10 mg weight from the left pan of balance and the rider is placed at the right end of the scale then according to the principle of balance, after picking up the rod of the balance the points go to the equal mark on the both side of the scale i.e. The value of 50 small blocks = 10mg The value of one blocks = 10/50 =.2mg If the balance has 100 small part then the value of small part is = 10/100 =.0001 gm. Example : The position of rider is displayed on the balance rod in the following diagram. The rider is placed at the third small mark after the fifth mark point of main scale of the balance rod. Precaution : And every bigger part is divided into 5 small parts. Therefore reading will be as follows. 5 big parts = 5 mg. 3 small parts = mg. = 0.6 mg. Total weight = 5 mg mg. = 5.6 mg. = gm This weight will be added in the weight of the pan. Rider (a) (b) Fig : 6.3 Use of Rider Procedure of weighing : Before taking weight pan of balance should be cleaned first and see whether the balance is at equal level. To bring it in equal level screws then screws on the edges of the marked rod can be set by rotator if needed. Rotate the handle and see whether the indicator goes to and from zero equally or not. If it does not move equally and screws on the edges of the marked rod can be set by rotating. The direction in which indicator covers the more distance that pan is the lighter one. To make it right, rotate the screw inside from the heavier pan or rotate the screw outside from the light is pan. After checking whether the balance working properly keep substance in the measuring tube (clean & dry) in the left pan from the left window. Keep weight in the right pan from the right window. First start weighing with larger weight, then smaller weight are used in place of larger weight sequentially. After putting each weight it is cheeked that which pan heavier. The direction in which indicated covers more distance that pan is lighter one. If there is the need for the weight of less than one gm then start putting weight of mg. If the weight less than 10 mg is needed then rider is used. At first rider is kept on one end point of the right side of the marked bar and it moves to and fro from the zero. Observe this situation carefully this is the exact measurement. Now take out the weight from the pan & note the reading. 6. Calculation of weight : Suppose the following weights are kept on the right pan for weighing tube & weight of substance. Gram weight = 5 gm + 2 gm + 1 gm = 8 gm. Milligram = 500 mg mg mg + 20 mg + 10 mg. = 950 mg = gm. Rider = 3 main parts on the right side + 2 subparts = gm gm.

10 45 = gm gm Total weight = 8 gm gm gm = gm weighing tube + weight of substance = gm Similarly, the weight of weighing tube = gm. weight of the substance = ( ) gm = gm 7. Measuring flask : Measuring flask is a flat bottom pot of glass, which has a long neck and there is a cork of glass. which fixed on the neck. Neck of the flask has the circular mark. Fill the liquid to the mark, and liquid will become to the capacity of each flask is written on it at certain temperature. Use flask of 250 ml. or 100ml. to prepare a standard solution in lab. Do not heat the flask and do not pour any hot liquid in it. 8. Necessary tube, watch glass and washing bottle : These are shown with measuring flask in the following figure (i) Measuring tube (ii) watch glass (iii) washing bottle (iv) measuring flask Fig Preparation of Standard Solution Important standard substance : Some important substance are obtained in pure condition. There are substance that react with the components of atmosphere. These substance are important standard substances for example : Oxalic acid, Sodium carbonate. Characteristic of important standard Substances : 1. The substance should hundred present pure. 2. The substance should not affect with air and light. 3. The substance should not give vapour into the air. 4. The substance should not be porus otherwise there will be difference into the weight of substance. 5. Substance should not be soluble in distilled water. The following substance do not prepare standard solution.

11 46 Acid : HCl, HNO 3, H 2 etc. Base : NaOH because it absorb the air moisture and CO 2. Salt : KMnO 4 because it decomposed in air & light. Fe because it oxidized to ferric sulphate from air. Secondary Standard Substance : Some substance are such type which can not be obtained easily in pure state like KMnO 4, H 2 or NaOH etc, then substance are called secondary standard substance. The standard solution of such substance cannot prepare easily. In such a condition the solution of substance is prepare whose concentration is near to the required solution. To find the acurate. conc. of solution,it is titrated with standard solution. For Example : Before making the standard solution of NaOH its solution of approximate solution is prepare and to find its concentration,it is titrated with standard solution of oxalic acid determine by adding required water in NaOH solution. To calculate the weight of the substance required for the preparation of standard solution : In lab 250ml standard solution is prepare. For the 250ml standard flask in used. The required weight can be calculate with following formula. E N V W = gm 1000 When W = Weight of solote (substance) E = Equivalent weight of substance N = Normality V = Required Volume From the above formula the required weight of substance in calculated to prepare standard solution of derived volume. Example : To find the required weight of acid for the preparation of 250 ml. N/10 solution of oxalic acid (Equivalent weight of acid = 63) Quantity of substance = Equivalent weight Normality Volume /1000 gm = / = gm Method to Prepare Standard Solution. The method of preparing standard solution, is stated by an example. Objective : To prepare 250ml solution of oxalic acid of N/10 concentration (Equivalent weight = 63) Method 1. Weight the dry, clean watch glass Now weight it by adding 1.6 gm oxalic acid. 2. Cast the substance of watch glass in a clean beaker according to figure. Fig Weight the substance in watch glass. Pour the attached substance of watch glass in beaker by washing bottle. Fig Pour the attached substance in beaker.

12 47 3. Dissolve the substance of beaker by stirring and fill the measuring flask by funnel according to figure Fig 6.7 Pour the solution in measing flask. 4. Now stirr the funnel gradually to drop quantity of substance in the flask. At this situation neither take out the funnel from flask nor move it. When the substance falls in the flask by funnel then wash the edges of funnel with distilled water, wash bottle gradually so that the substance of funnnel reaches into the flask Now remove the funnel from flask. Now wash the funnel 2 or 3 times. Now try to drop the substance attached with the wall of flask neck in the flask with the help of distilled water. Now pour more distilled water in flask to fill it 1/3. Now cover the flask with cork and stir it properly so that the substance is completely dissolved. During the observation it is necessary to observe that the solution should not comes up to the neck of flask. 5. Now pour water drop by drop in flask with the help of a clean and washed pipette. When the lower level of semi circular and solution touches the circular mark then stop pouring water. Remember that the water should not be more than the circular surface otherwise the concentration will be disturbed. It should be in such a manner that the liquid should not come out Now press the dot with hand and stirr the solution so to prepare homogenous solution. Now standard solution of oxalic acid has been prepared. Fig 6.8 To pour the solution in the flask upto last drop. Fig 6.9 To wash the funnel and fill the measuring flask Fig 6.10 To fill the mearuring flask with the help of pipette

13 Titration Titration A known volume and concentration of one substance is reacted with a known volume of another substance whose concentration is to be determined this process is called. titration. Titration is classified into 3 categories depending upon the nature of the reaction which happen in the solutions. (i) (ii) (iii) Acid - Base titration or Acidimetry and Alkalimetry. Oxidation and Reduction titration. Precipitation Titration. (i) Acid - Base Titration : In this method acid and base are reactants. They react with each other and neutralize each other. This titration is called Neutralization Titration. Acidimetry : In this method concentration of an acid in a solution is determined with the help of an alkali of known concentration. Alkalimetry : In this method concentration of an alkali in a solution is determined with the help of an acid of known concentration. (ii) Oxidation and Reduction Titration : In this method, an oxidising agent and a reducing agent are reacted. In this, one solution is of oxidation substance and other is of reducing substance. In this method one substance is oxidized and another is reduced. Ex : (i) Titration of KMnO4 with Oxalic Acid. (ii) Titration of KMnO4 with FeSO4 4 KMnO acts as an oxidising agent in a acidic medium. It Oxidizes Oxalic acid & itself gets reduced. 2 KMnO4 + 3H2SO4 K2 SO4 + 2MnSO4 + 3H2O [ COOH] CO2 5H2O (iii) Precipitation Titration : In this method of titration concentration of a substance is determined by its complete precipitation with the help of another substance of known concentration Indicator Those chemical substance, which indicate end point in the titration by changing their colour are called indicators. Success of Titration experiment depends upon the correct measurement of reacting solutions. For this exact end point is required. An indicator changes its colour at the end point. This indicates the physical - chemical phase or state of completing the reaction and do not affect the concentration of actual solution. Indicators are of 3 types : (i) Internal Indicator (ii) External Indicator (iii) Self Indicator (i) Internal Indicator : Such indicators are added to the actual solution in the titrating flask during titration. The other solution is added from the burette till end point is reached; are called internal Indicator. In this type of indicator methyl orange and phenolphthalein are mostly used. They are used in acid - Base Titration.

14 49 (ii) External Indicator : Indicators that are not added in the titration solution are called external indicator. Few drops of such indicators are kept on a tile and after the titration, the reacting solution mixture is treated with external indicator used in the titration to reach the end point. For example : Potassium ferrocyanide is an external indicator used in the titration between dichromate and Ferrous Ammonium Sulphate. (iii) Self Indicator : When one of the reacting solutions during titration changes its colour at end point then such a solution is called Self Indicator. For example: During titration between Potassium Permangnate and Ferrous Ammonium Salphate the Potassium Permangnate acts as a self indicator. Theory of indicators : Indicators used in acid base titration are generally coloured organic compounds. They are weak acid or base and they have one ionization constant. In titration Methyl orange and Phenolphthalein indicators are used. They change their colour at a specific ph. They exhibit different colour in ionised and unionized state. Methyl Orange : It is orange colour weak base; it is expressed by the formula MeOH. It ionized in water as follows. MeOH Me + + OH Atom (Orange) (Pink) (Colour less) In basic medium the above equilibrium shifts in the backward direction ( ) and Methyl orange remains wnionized state. Therefore in Basic medium the methyl orange do not change any colour as a result colour of the solution remain yellow (MeOH) (During the presence of colour atom of Methyl Orange). In the acetic medium the OH ion acid add with + H ion to make atom of water and reaction shift towards right side. More + Me ions are formed due to which solution becomes pink. Phenolphtholein : It is a week organic acid it is denoted by formula HPh. It is ionises as follows. HpH = H + + Ph (Colour less) (Pink) + In base solution the H ion of phenolphthalein + react with H ion of base to make water. Therefore concentration of Ph ions increases due to which solution (due to the pink colour of Ph ion) turns pink. In acidic solution the ionization of (HpH) is affected therefore the Ph ions do not present in solution therefore phenolphthalein does not give any colour. Choice of indicator : The choice of indicator depends upon the presence of concentration of hydrogen in present in solution during the titration. At least one of the two solutions should be strong because suitable indicator is not available for titration between weak acid and weak base. Suitable indicator can be chosen on the basis of the following table. S.No Solution Suitable Indicator used in Titration for Titration 1. Strong Acid and Methyl Orange or strong Base Phenophthelim one of them 2. Strong Acid and Methyl Orange Weak Base 3. Weak Acid and Phenolphthalein Strong Base 4. Weak Base and None of them gives right weak Acid result

15 Principle of Titration Acid base titration is based on the principle of neutratlization and law of Equivalence according to the ionc theory. When the base is added in the conical flask then the OH H 2 O of base in produced, in the solution. But practically it is not seen. When the one drop of acid is add into the neutral solution the + H ions are produced. So that the colour of the solution changes by the indicator, this condition arises when acid in taken in the conical flask and base in added by burette. According to the law of Equivalence the reaction between base and an acid takes place in the ratio of their equivalent weight. HCl + NAOH = NaCl + H2O Acid base Salt Water (36.5) (40) (58.5) (18) It in clear from the equation that for complete neutralizes 40 gram of NaOH must react with 36.5 gram of HCl. i.e one gram of equivalent of NaOH Completely neutratizes one gram equivalent of HCl. If Normal Solution of both reagents are used then one litre of each solution or any same volume of both solutions will be required for complete netralisation i.e. end point 36.5 gm HCl = 40 gm. NaOH 1000 ml N. HCl = 1000 ml. NaOH V ml. N. HCl = V. ml N. NaOH Therefore it is clear that according to these law for complete reaction if required quantity of the substance is expressed in their gram equivalents then they must be equal. On this basis, calculation of titration is calculated as under. Suppose solution A of N, normality and V, ml. Volume for complete reaction requires solution β of N 2 normality and V 2 ml. Volume. Then for solution A 1000 ml = N 1 gm. equivalent V ml, = N 1 V gm equivalent N 1 Therefore during titration for solution A, 1 V 1000 gm equivalent substance is used. Similarly Solution B, N 2 V gm. equivalent substance must be used. (because Normality N 2 and used volume = V 2 ml.) Therefore according to law of equivalent equal gm. equivalent of both must react completely. N1 V1 N2 V2 = N 1 V 1 = N 2 V 2 Normality of solution A volume of solution A = Normality of solution B Volume of solution B Apparatus of Titration 1. Burette and its use : It is long cylindrical glass tube of uniform diameter Fig 6.11 (a) Stopper of burette. (b) pinch cock burette.

16 51 Divide its length into equal parts and make many marks Q on it. After every 10 mark, the no, 0, 1, or 100 are marked from upto bottom repecivety, which include the inner volume of burette in cubic cm or ml. In the same way the volume is read upto 1 no. of decimal. (0.1 qubic cm). The tube get the jet shape into its bottom. The stop cork or pinckcork in fixed on the jet. So that the stop cork in rotate as required. To drop the solution of burette drop - by - drop with help of jet. Method to use burette : 1. Wash the burette with water. If the burette in dirty then wash it with uramil acid and then wash it with water. If the inner surface of burette in flat then wash it with NaOH solution, wash it with water. At last wash it with distilled water. Then since it with the solution, which to fill in it. The rinsing process is shown in figure. Fig 6.12: Rinse the burette. 2. Rinse the burette and make it empty, and fill the used solution upto the zero mark. The bubble of jet of burette is remove by pressing pitch rightly upside and then law it. Now again fill the burette upto zero after removing air bubbles. Now burette is ready for titration. Fig. 6.3 Fill the solution in the burette and measure it.

17 52 3. Now take the known volume of the solution into the beaker in the conical flask with the help of a pipette & put it below the burette & note the meniscus of the surface the burette s solution. Now press the pinch cock to pour the burette solution into the beaker by the left hand finger & hold the beaker into the right hand shape is slowly. Now again note down the meniscus of the surface of burette s solution when the neutral point is obtained & determine the difference between first and the 2 nd reading. This is the volume of the liquid of burette in titration. 4. For colourless solution lower meniscus of half moon the solutions read. At the time of reading the position eye placed at the point of the lower surface of half moon Neither above nor below. Correct way of reading of burette must be observed in the following diagram. It is a glass tube having cylindrical part is the middle. It is provided with a jet at its lower end, It is used for correct measurement of definite volumes of solution. The volume of pipette is marked in its middle cylindrical part. It is 10ml, Pipette and its use 5. It must be checked that the burette is not leaking. In case of leakage some grease must be applied at the glass stopper and replaced the hard pinch cock of the burrete to another. 6. Before the experiment if any drop solution is hanging at the tip of the end point of jet, then it should be removed by using a filter paper. But if the drop in hanging after the addition of solution into the titration flask then it should be removed by touching it with the wall of the titration flask. 20ml, 25ml, 50ml. Its upper side tube has a mark. On filling the solution upto that mark the liquid becomes equal to the marked volume on the pipette. Fig : 6.15 Use of Pipette.

18 53 Method of using pipette : 1. Wash pipette with water. 2. Rinse the pipette with the given solution. Fill the liquid into the pipette, put it horizontal & rotate, & take out the liquid outside. This whole all procedure is shown in figure Now put the given solution in pipette according to fig 6.14, close the upper end of the pipette by the index figure. To allow the solution to its fall solwly upto the marked point in pipette. At the marked point put the surface of water in pipette constant by increasing pressure on the index finger. Now place the pipette in the titration flask vertically and remove the index finger and pour the complete solution into the titration flask. 4. For removing the left solution in jet of pipette, touch the end of the jet of pipette to the surface of the wall of titration flask, & close the upper end by index finger. In case some drops are left into the jet, don t try to remove them by any other method. Because the marked pipette is calibrated taking this liquid in to account. Fig : 6.16 (i) To fill the solution in the pipette (ii) To stop the solution with finger (iii) To bring the solution upto the marked point (iv) To pour the solution in the conical flask (v) To remove the last drop of the pipette. (Record of Titration-Experiment) Correct result is obtained after completing Volumetric analysis with accuracy. It is to be carefully recorded in the practical note book. Here the method are given for recording the practical work related with titration. 1. Single titration : In this titration the unknown concentration of solution of substance is determined with the help of standard solution of another substance. This kind of titration in performed.

19 54 between the solution of opposite nature (such as acid & base), there solutions mutually naturalise one another during titration. In such case the titration is performed only once, therefore it is called single titration. 2. Double Titration : It is also based on titration between two solutions of opposite nature. If the solution A and B are of same type and concentration or normality of one solution (suppose A) is known then the concentration of solution B can not be determined directly after titration. In such a case a solution having opposite nature is used, is called intermediate solution & the titration of both solutions is performed by this. At first the titration is performed between A, E (standard solution) and C (intermediate solution). So that the concentration of solution C in determined. Now the titration is performed between C and B and the concentration of B is determined by calculation. This kind of titration is called Double titration. Record of titration experiment in practical note book : In practical note book on both side of the page titration experiment in recorded in following method is popular, acceptable and systematically. Left page Right page 1. Experiment No.1 1. Experiment no & dt. 2. Object 2. Object 3. Formula 3. Apparatus & Reagent 4. Calculation 4. Principle or equation 5. Method (not necessary to units) 6. Observation table 7. Result 8. Precaution Experiment No 1. Object : To find out the concentration of hydroxide. solution by titrating it against the N/10 standard solution of oxalic acid. Apparaturs: Burette, pipette, conical flask, burette stand funnel. Reagent : Standard solution of Oxalic acid, Solution of NaOH Phenolphthalein (indicator) Principle : This titration is performed between strong base and weak acid they are Phenolphthalein is. a indicator & NaOH is taken into the burette. The neutralization process in completed in following way: COOH + 2NaOH COONa + 2H O phenol-ph+nalein COOH COONa as an indicator Method : 1. Clean all the apparatus with water. Wash the burette with given NaOH and fix it to the stand, fill it with acid. In case, if there is a air bubbles in the rubber of pinch cock then remove them and put the surface of base at zero. 2. Wash the pipette with standard oxalic acid, take 25 ml. acid unit and pour it into the conical flask and. add 1-2 drop of phenethicillin indicator in this solution. 3. Now drop the base from the burette into the flask slowly and carefully and determined the end point. Note this point of the burette. 4. Repeat this experiment 3-4 times and find the two same reading of burate for the above value of oxalic acid. Observation Table : Table number 1 S.No. Volume Reading of Burate Volume of acid Starting End of base ml 0.01 ml 20.3 ml ml 0.01 ml 20.3 ml 20.0ml ml 0.01 ml 20.3 ml

20 55 Formula : N 1 V1 = N2 V2 Standard solution = Solution of unknown concentration. Calculation N 1 V1 = N2 V2 (acid) (base) N = N N 25 N N 2 = = Normality of given base = N Concentration of base = Normality Gram equivalent weight base N 40 = 5 Gram / liter Result : The normality of given NaOH is N & the concentration of given NAOH is 5gm / litre. Precaution : 1. Wash the instrument with water before use. 2. Rinse the burette and pipette with respective solution. 3. Solution in the burette should be filled with the help of a funnel. Remove the funnel. 4. Air bubble from the pinch cock must be removed. 5. The indicator should be used in very little quantity (1, or 2 drop) Experiment No. 2 Object : Make the standard solution of sodium carbonate (about N/8) and determine the concentration of Unknown H 2 with the above solution by using titration method. Principle : This titration is performed between conc. acid and weak base. Therefore Methyl orange indicator is used. The base solution is taken into the beaker & acid solution is taken into the burette. Therefore the red colour is produced at the end point. Suppose that a standard solution has to prepare into the 250 ml titration flask therefore weight of Na 2 CO 3 is 53. Therefore weight gm Na 2 CO 3 for preparing the solution of N/P normality. Now take a weighing tube in which a cock is fixed & note down its weight. Take gm. Na 2 CO 3 and note down its weight. Now take Na 2 CO 3 into the beaker and wash it with some water. Fit the cock into the weightube weight it with the left Na 2 CO 3 and note its weight. Pour the solution of beaker is the 250ml of titration flask with the help of funnel. Wash the beaker many times with water in the titration flask. Now fill the water into the titration flask upto the marked point carefully and make a homogeneous solution by shaking the flask. Observation : 1. Weight of the weighting tube = gm. 2. Weight tube + weight of Na 2 CO3 = gm. 3. Other weight of the weighing tube = gm. Weight of Na 2CO3 = ( ) gm = gm. Concentration of Na 2CO3 = ( gm / littre) Concentration of Na 2CO3 = ( gm / littre) Precaution : Now wash the burette with the solution of H 2 & fill it with H 2, and fixed to the burette stand. Take 20ml solution of Na 2 CO 3 with the help of pipette. into the titration beaker or flask Now add 2-3 drop methyle orange indicator to perform the titration. The colour of the solution becomes light pink at the last drop of solution. Repeat the titration process until the Volume of H 2 becomes 2 times.

21 56 Table No. 1 S.No Volume of Reading of Burrete Volume Na 2CO3 Start End of H 2 SO ml 1.7 ml 22.9 ml 21.2 ml ml 3.2 ml 24.3 ml 21.1 ml ml 6.2 ml 24.3 ml 21.1 ml Calculation : N 1 V1 = N2 V2 Normality=Concentration equivalent weight = N1 = N = V1 = 20 ml N 2 =? V2 = 2101 ml N 20 = N N2 = N 2.50 = = N Concentration=Normality equivalent weight = 4. 9 = Experiment no. 3 : = 6.09 litre. Objective:To find out the concentration of HCl solution with the help of standard solution of N/ 10. Sodium Carbonate, Principle : When HCl is added into the sodium carbonate then the reaction is completed into two steps : (i) Na 2CO3 + HCl NaHCO3 + NaCl (ii) Na 2CO3 + HCl NaCl + H2O + CO2 In the first step the Carbonate is converted in to the bicarbonate. In the 2 nd step the bi - carbonate reacts with acid and gives carbondie-oxide. In this titration methyle orange is used as a indicator and acid solution is be taken in to the burette. Apparatus : As per practical No. 1 Method : 1. Rinse the burette with the HCl acid and then fill the burrete with acid upto 01 ml mark. 2. Take 25 ml solution of Sodium Carbonate into a conical flask or beaker with the help of pippete. 3. Pour 2-3 drop of Methyl orange into the solution. And pour the acid solution from the burette into the conical flask slowly. Shake the solution. Note the reading of the burette at which the pink colour is obtained of the drop. 4. Repeat this experiment.3 to 4 times, So as to obtain two consecutive readings. Observation table S.No. Volume of Reading Burette Used the base of Start End Volume of acid ml 0.0 ml 26.7 ml ml 0.0 ml 26.4 ml 26.4 ml ml 0.0 ml 26.4 ml Calculation : Sodium Carbonate Hydrochloric acid N 1 V1 = N2 V2 N 10 N 23 = N N 25 = = Normality of HCl solution = N N Concentration=Normality equivalent weight = = gm/litre. Result : Concentration of HCl = gm/ litre Precaution : As per the experiment no. 1

22 Calculations of Single Titration Example 1: 1.2 gram of a solution is dissolved in 200 ml solution then calculate the strength of a solution Solution 1: When 200 ml of solution contains 1.2 grams solute ml of solution will contain 1.2 = 1000 = 6 gm 200 Strength of solution 6 gm / litre Example 2: 53 gm of sodium carbonate is dissolved in 100 ml solution find strength and normality of the solution. Equivalent weight of sodium carbonate is 53 Solution 2: 100 ml of solution contains.53 grams solution 100 ml solution will contain = = 5.3 gm 100 Strength = 5.3 gram / litre Example 3: Calculate amount of caustic soda in gram in a deci normal solution of it prepared in one litre Solution. 3: Normality of Deci normal solution = 10 N Strength = equivalent weight X Normality = N 40 = 4 gram /litre 10 2 litre solution will contain 4 2 = 8 gram Result -2 litre solution of caustic soda contains 8 gram solvent Example 4 : Find the amount of KOH dissolved in 500 ml of N KOH solution Solution 4 : Strength = Normality Equivalent weight = = 4.76 gram/litre 1000 ml of solution contain 4.76 gram solution 500 ml solution contain will contain Normality = Strength Equivalent weight = 5.3 = = = 2.38 gm N = Result- (1) Strength of solution of sodium carbonate is 5.3 gram/litre (2) Normality of solution of sodium carbonate = N/10 Result : 500 ml solution contain 2.38 gm. KOH. Example 5 : To find the quantity substance in gm. to prepare 500 ml N/10 Sodium Carbonate. Solution 5 : Strength = Normality equivalent weight

23 58 N or 25 = N = N = = N Result : Normality of HCl in N Example 7 : 2.65 gram Na 2 CO 3 is dissolved in 250 ml water to prepare standard solution. It neutrilize 25 ml of this solution with H 2, 27.3 ml H 2 was required. Find the normality and strength of H 2. Solution : 500 ml solution contain 2.65 gm. Solvent 1000ml solution will contain = 5.3 gm. Normality = Strength Equivalent weight = = = N = N 1 V1 = N2V2 = base acid N = = N N = N 10 2 or = N Strength = Normality Equivalent weight = = gram / litre = 4.46 gram / litre Result : (1) Normality of H SO N 2 4 = (2) Strength of H. 46 gram/ litre 2 4 = Q.1 What do you understand by titration? Q.2 Define Normality, Molarity and Strength? Exercise Question Q.3 Explain the function of indicators in Acid Base titeration? Q.4 What is the difference between dinormal & decimolar solution of Sulphuric acid? Q.5 Which indicator is suitable for titration between oxalic acid and caustic soda solution? Q.6 Explain giving reasons, Acidimetry & Alkalinimetry with example? Q.7 What is the normality of Na 2 CO 3 solution prepared by dissolving 10.6 gram in 500 ml flask? Q.8 How much NaOH is required to prepare 0.01N solution in 500 ml solution? Q.9 How much Na 2 CO 3 is required to prepare 1 N solution in 500 ml. Q.10 How much water must be added in 6 3 NHNO acid solution so that it becomes 2 N concentration?

24 59 Q.11 How much water must be added in 1 litre N NaOH solution so as to make on N solution? Q.12 Find the amount of KOH dissolved in it was completely neutralized by 25 ml of N/10 HCl Solution? 58 Q N HCl Solution completely neutralized by 22.5 ml Caustic Soda Solution, find the amount of Caustic Soda dissolved in the Solution? Q.14 To neutralized 35 ml of H 2 of unknown strength 28.6 ml. N/10 NaOH was used. Find the normality of H 2. Q ml of N/20 was completely neutralized by N/10. Calculate the used quantity of acid? Q ml of 1 base was neutralized by 25 ml of N/10 HCl solution. If strength of base is 4.5 gm/litre then find the equivalent weight of base. Answer (7) 0.4 N (8) 2 gm (9) 26.5 gm (10) 1 : 2 (11) 250 ml (12) 0.14 N (13) 0.45 gm (14) N (15) 100 ml (16) 55.8