Back titration is also titration. It is called back titration because it is not carried out with the solution whose concentration is required to be known (analyte) as in the case of normal or forward titration, but with the excess volume of reactant which has been left over after completing reaction with the analyte. Back titration works in the following manner (with an example) : 1: The substance or solution of unknown concentration (4 gm of contaminated chalk, C a C O 3 ) is made to react with known volume and concentration of intermediate reactant solution (200 ml, 0.5N HCl). The reaction goes past the equivalence point. The amount of intermediate reactant is in excess of that required for completing reaction with analyte. 2: After completing the reaction with analyte, the resulting solution containing excess of intermediate reactant is titrated with known volume and concentration of titrant (50 ml of 0.5N NaOH). If subscripts 1 and 2 denotes intermediate reactant and titrant, then N 1 V 1 = N 2 V 2 ⇒ 0.5 V = 0.5 X 50 ⇒ volume of excess HCl, V = 50 m l Also, meq of execss HCl = meq of titrant, NaOH = 0.5 X 50 = 25 3: Determination of excess volume or meq of intermediate reactant allows us to determine the volume or meq of intermediate reactant which reacted with analyte. This, in turn, lets us determine the amount of analyte. ⇒ meq of chalk = total meq of HCl - meq of excess HCl ⇒ meq of chalk = 0.5 X 200 - 25 ⇒ meq of chalk = meq of HCl used for chalk = 75 Applying gram equivalent concept to chalk and HCl, ⇒ meq of chalk = meq of HCl used for chalk = 75 g E X 1000 = 2 X 1000 g 40 + 12 + 3 X 16 = 75 ⇒ g = 75 20 = 15 4 = 3.75 g m This means chalk contained 0.25 gm of impurities in it.

Purpose of back titration Back titration is designed to resolve some of the problems encountered with forward or direct titration. Possible reasons for devising back titration technique are : 1: The analyte may be in solid form like chalk in the example given above. 2: The analyte may contain impurities which may interfere with direct titration. Consider the case of contaminated chalk. We can filter out the impurities before the excess reactant is titrated and thus avoid this situation. 3: The analyte reacts slowly with titrant in direct or forward titration. The reaction with the intermediate reactant can be speeded up and reaction can be completed say by heating. 4: Weak acid – weak base reactions can be subjected to back titration for analysis of solution of unknown concentration. Recall that weak acid-weak weak titration does not yield a well defined change in pH, which can be detected using an indicator. Problem : 50 litres of air at STP is slowly bubbled through 100 ml of 0.03N B a O H 2 solution. The B a C O 3 formed due to reaction is filtered and few drops of Phenolphthalein is added to the solution rendering it pink. The solution required 25 ml of 0.1 N HCl solution when indicator turned colorless. Calculate percentage by volume of C O 2 in air. Solution : Carbon dioxide reacts with B a O H 2 to form B a C O 3 . The excess B a O H 2 is back titrated with HCl. total meq of B a O H 2 solution = 0.03 X 100 = 3 meq of excess B a O H 2 solution = 0.1 X 25 = 2.5 meq of B a O H 2 solution used for C O 2 = 20 − 2.5 = 0.5 g = m e q X M O 1000 x = 0.5 X 12 + 2 X 16 1000 X 2 = 0.01 g m Using Avogadro’s hypothesis, 44 gm of C O 2 occpies 22.4 litres at STP. This mass of CO2 corresponds to : volume of C O 2 = 22.4 X 0.01 44 = 0.01 l i t r e s ⇒ % of C O 2 in air = 0.01 X 100 50 = 0.02

Back titration and dilution While dealing dilution in titration calculation, we need only to remember that dilution does not change geq or meq of a solution. Problem : 2.75 gram of a sample of dolomite containing C a C O 3 and M g C O 3 is dissolved in 80 ml 1N HCl solution. The solution is then diluted to 250 ml. 25 ml of this solution requires 20 ml of 0.1 N NaOH solution for complete neutralization. Calculate percentage composition of the sample. Solution : We shall first determine the amount of HCl required to react with the sample by using back titration data. Here, we make use the fact that dilution does not change meq of a solution. ⇒ meq of 25 ml of diluted excess HCl solution = meq of NaOH solution = 0.1 X 20 = 2 ⇒ meq of 250 ml of diluted excess HCl solution = 20 ⇒ total meq of HCl = N V = 80 X 1 = 80 meq of HCl used to react with dolomite = 80 − 20 = 60 Mass of HCl used in reaction with dolomite is calculated using expression of meq : m e q = g E X 1000 = x g M O X 1000 ⇒ g = m e q X M O 1000 x = 60 X 36.5 1000 X 1 = 2.19 g m In order to find the composition, we apply mole concept. The reactions involved are : C a C O 3 + 2 H C l → C a C l 2 + C O 2 + H 2 O M g C O 3 + 2 H C l → M g C l 2 + C O 2 + H 2 O Let mass of C a C O 3 in the sample is x gm. Then mass of M g C O 3 is 2.75-x gm. x M C a C O 3 moles of C a C O 3 = x 100 moles of C a C O 3 = 2 x 100 moles of HCl ⇒ mass of HCl required for C a C O 3 = 2 x X M H C l 100 = 2 x X 36.5 100 = 73 x 100 = 0.73 x g m Similarly for M g C O 3 , ⇒ mass of HCl required for M g C O 3 = 2 2.75 − x M H C l 84 = 2 X 2.75 − x 36.5 84 = 73 2.75 − x 84 = 2.39 − 0.87 x g m ⇒ total HCl required = 0.73 x + 2.39 − 0.87 x = 2.19 ⇒ 0.14 x = 0.2 x = 1.43 g m ⇒ % mass of C a C O 3 = 1.43 2.75 X 100 = 52 ⇒ % mass of M g C O 3 = 100 − 52 = 48

Back titration and decomposition Back titration is analyzed using geq or meq concept. On the other hand, decomposition is analyzed using mole concept. We combine two concepts by converting final geq or meq data of back titration into either mass or moles. Problem : 4.08 gm of a mixture of BaO and an unknown carbonate M C O 3 is heated strongly. The residue weighs 3.64 gram. The residue is then dissolved in 100 ml of 1 N HCl. The excess acid requires 16 ml of 2.5 NaOH for complete neutralization. Identify the metal M, if atomic weight of Ba is 138. Solution : We need to determine atomic weight of M to identify it. From the question, we see that first part involves decomposition, which can be analyzed using mole concept. Note that difference of mass of mixture and residue is mass of C O 2 . On the other hand, analysis of back titration gives us meq of HCl required for neutralization of residue. To combine meq data with decomposition data, we shall convert required meq in milli-moles. ⇒ total meq of HCl = N V = 1 X 100 = 100 ⇒ meq of excess HCl = N V = 2.5 X 16 = 40 ⇒ meq of HCl used for neutralization of residue = 100 − 40 = 60 ⇒ milli-moles of HCl for neutralization of residue = x X m e q = 1 X 60 = 60 Now, we analyze the data which is given for decomposition of mixture. On heating, BaO is not decoposed. Only M C O 3 is decomposed as : M C O 3 = M O + C O 2 Carbon dioxide is released and is not part of residue. The amount of C O 2 is equal to difference of mass of mixture and residue. Hence, moles of C O 2 released are : ⇒ n C O 2 = 4.08 − 3.64 12 + 2 X 16 = 0.44 44 = 0.01 m o l e Applying mole concept, moles of MO in the residue is 0.01 mole i.e. 10 milli-moles. Now, MO reacts with HCl as : M O + 2 H C l → M C l 2 + H 2 O Moles of HCl required for MO is 2X0.01 moles = 20 milli-moles of HCl. But total milli-moles that reacted with residue is 60 milli-moles. Hence, milli-moles of HCl that reacted with BaO is 60-20 = 40 milli-moles of HCl. Applying mole concept, B a O + 2 H C l → B a C l 2 + H 2 O The milli-moles of BaO is half of that of HCl. Hence, millimoles of BaO in the mixture is 20 milli-moles. ⇒ amount of BaO in the mixture = milli-moles X M B a O 1000 = 20 X 138 + 16 1000 = 3.08 g m ⇒ amount of MO in the residue = 3.64 − 3.08 = 0.56 g m Using formula : ⇒ g = milli-moles X M M O 1000 = 10 X M M O 1000 ⇒ M M O = 100 g = 100 X 0.56 = 56 Clearly, atomic weight of M is 56-16 = 40. The element, therefore, is calcium.

Back titration and multiple neutralizations The concentration of titrant used to determine the excess reactant is not directly given. We make use of subsequent neutralization data to ultimately determine the concentration of titrant. Problem : A sample of M n S O 4, 4 H 2 O (molecular wt : 223) is heated in air. The residue is Mn 3 O 4 , whose valency and molecular weight are 2 and 229 respectively. The residue is dissolved in 100 ml of 0.1 N F e S O 4 solution containing dilute sulphuric acid. The resulting solution reacts completely with 50 ml of K M n O 4 solution. 25 ml of the K M n O 4 solution requires 30 ml of 0.1N F e S O 4 solution for complete oxidation. Calculate the weight of M n S O 4, 4 H 2 O in the given sample. Solution : The excess of F e S O 4 completely reacts with 50 ml of K M n O 4 solution of unknown concentration. We need to know this concentration. From the data of complete oxidation, however, we can know concentration. Let 1 and 2 subscripts denote K M n O 4 and F e S O 4 , then N 1 V 1 = N 2 V 2 ⇒ N 1 X 25 = 0.1 X 30 ⇒ N 1 = 3 25 = 0.12 N Now, considering back titration, ⇒ total meq of F e S O 4 solution = 0.1 X 100 = 10 ⇒ meq of excess F e S O 4 solution = 0.12 X 50 = 6 ⇒ meq of F e S O 4 solution used for residue = 10 − 6 = 4 ⇒ meq of M n 2 O 3 in the residue = 4 Mass of Mn 3 O 4 is determined using meq expression : m e q = g E X 1000 = x g M O X 1000 ⇒ g = m e q X M O 1000 x = 4 X 229 1000 X 2 = 0.458 g m The decomposition of M n S O 4, 4 H 2 O takes place as : 3 M n S O 4, 10 H 2 O → M n 3 O 4 + 3 S O 2 + 10 H 2 O Applying mole concept, the amount of M n S O 4, 4 H 2 O is : ⇒ Mass of M n S O 4, 4 H 2 O = 3 X 223 229 X 0.458 = 1.338 g m