Analytical Chemistry 1 CHM 241-1 PDF

Benha University Faculty of Science Chemistry Department ANALYTICAL CHEMISTRY Prepared by Staff members of Analytical Chemistry 1 GENERAL CONSIDERATIONS Classification of Analytical Chemistry: The two main divisions of analytical chemistry are qualitative analysis and quantitative analysis. The nature of the constituents of a given material is determined by the methods of qualitative analysis: the proportions in which these constituents are present are determined by the method of quantitative analysis. If the general composition of the material is unknown, a preliminary qualitative examinations is usually necessary before a satisfactory quantitative analysis can be made. This is because the method of determining accurately the percentage of a constituent depends to a great extent upon the nature and relative amounts of the other constituents present. Classification of quantitative analysis: A convenient classification of the methods of quantitative analysis is as follows:  Volumetric methods.  Gravimetric methods  Others physicochemical methods. Volumetric Methods: In a volumetric analysis, a well – defined reaction is caused to take place, where the reagent is added from a burret so that the volume of the solution employed to complete the reaction can be accurately measured. The concentration of this solution is accurately known, and the volume employed serves, therefore, as a measure of the amount of substance acted upon. For ex. The percentage, of silver in a sample of a soluble salt 2 can be determined by a volumetric process. The sample is dissolved in dil. HNO3, and a solution of potassium thiocytanate containing a known weight of the thiocyanate in each milliter of solution is slowly added from a burret until precipitation is just complete, as indicated by the color change of a suitable indicator (ferric ions, in this case). The amount of Ag can then be calculated from the concentration of the thiocyanate and the volume required. Volumetric methods are generally more rapid, require less apparatus, and are frequently capable of greater accuracy than gravimetric methods. Gravimetric Methods: In gravimetric processes the constituent to be determined is isolated, either as an element or as a compound of definite composition. This element or compound is weighed, and from its weight and its known chemical composition the amount of the desired constituent is determined. In most cases, the element or compound is separated by precipitation with chemical reagent. For ex. By one method the percentage of silver in a sample of a soluble salt is determined by dissolving a weighed amount of the salt in dil. HNO3 and precipitating the silver as AgC1. The precipitate is then separated by filtration and is dried and weighed. Since to quantitative composition of the precipitate is known (107.870 gms of Ag are contained in every 143.323 gms of AgCl) the amount of Ag, and hence its percentage in the sample, is easily calculated. 3 Physico chemical methods: Here quantities of constituents are determined from measurements of a physicochemical property as density, refractive index or color (colorimetry). In electrometric analysis, the process in which the progress and completion of the reaction is indicated by electrical means. It includes potentiometric, conductometric, amperometric, and coulometric methods. VOLUMETRIC ANALYSIS Methods of Expressing Concentration: Usually, the concentration of solution refers to the weight of solute dissolved in a unit volume of the solution. There are many ways of expressing concentration, and the following methods are the ones most commonly used. A) Gram per unit volume: By this method concentration is expressed in terms of the number of grams (or milligrams) of solute n each liter (or milliliter) of solution. This method is simple and direct, but it is not a convenient one, from a stoichiometric point of view, since solutions of the same concentration bear no simple volume relations to each other when they enter into reaction. Substances react on a molecule – to – molecule basis and not on a gram – to – gram basis. B) Percentage Composition: By this method concentration is expressed in terms of grams of solute per 100 gm of solution. A 10% solution of a given salt is made by dissolving 10 gm of the salt in 90 gm of water. 4 C) Specific gravity: The specific gravity of the solution of a single solute is a measure of the concentration of the solute in solution. Thus, a solution of sulphuric acid of specific gravity 1.14 at 15C contains 19.6% H2SO4 by weight. D) Volume ratios: Occasionally the concentration of a mineral acid or of ammonium hydroxide is given in terms of the volume ratio of the common concentrated reagent and water. Thus, H2SO4 (1 : 3) signifies a solution made by mixing one volume of the commonly used concentrated H2SO4 (sp. gr. 1.84) with three volumes of water. Non of the above methods provides a direct measure of the strength of the solution. For ex. a given volume of a 5% by weight solution of HC1 will react with a greater quantity of a base than will an equal volume of a 7% by weight HNO3. In order to make chemical calculations involving solutions, it is first necessary to transfer to concentrations into packaged units, such as the mole or equivalent. It is convenient, therefore, to state the concentration in packaged units in order to eliminate the necessity for the conversion. E) Molar and Formal Solution: A gram – molecular weight (or gram – mole, or simply mole of a substance is its molecular weight expressed in grams. Thus, a mole of K2SO4 (molecular weight = 174.3) is 174.3 gm. A mole of nitrogen gas (N2) is 28.016 gm of the element. A molar solution is simply one containing a gram – molecular weight of solute in a liter of solution (not of water). The molarity or molar concentration; M, is the number of moles of solute / liter of solution. 5 Since Moles = ( ) ( ) A formula weight is that weight in grams corresponding to the formula of the substance as ordinarily written in a chemical equation. It most cases it is identical to the mole, but occasionally the true molecular weight of a compound is a multiple of the weight expressed by the formula as ordinarily written in a chemical equation. F) Molal solution: The molality of a solution denotes the number of moles of solute per 1000 gm of solvent. A molal solution is that one which contains 1 mole of solute dissolved in 1000 gm of solvent. G) Normal solution: A normal solution is one that contains a gram-equivalent weight (or simply equivalent) of solute in a liter of solution or gram – milliequivalent weight (milliequivalent) in a milliliter of solution. Thus the normality or normal concentration, N: A 0.1 N (N/10) solution is read as one – tenth normal solution and denotes that a liter of this solution contains one-tenth equivalent of solution. 6 * Also the equivalent = Molarity and Formality: Molarity is unit of concentration expressed as the number of moles of dissolved solution. If the number of moles and the volume are divided by 1000, then molarity is expressed as the number of millimoles per milliliter of solution. To make a 1.000 M solution of KCl, exactly 1.000 mole of KCl (74.55 g) is dissolved in an amount of water or some other solvent to yield exactly 1.000 liter solution. Actually, this unit of concentration is often misused. For example, if water is used as the solvent the solution is exactly 1.000 M in K+ and M in Cl since KCl is completely dissociated. → K+ + Cl Therefore the molar concentration of KCl is zero. If the KCl were only partially dissociated, its molar concentration would be determined by the amount of dissociated in that taken place. Many compounds are only partially dissociated. For example, if 0.1000 mole or 5.995 g of acetic acid (HC2H3O2) is dissolved in *** liter of solution (H2O), the solution will be 0.00134 mole / liter in C2H3O2 and 0.0987 mole/liter in un-ionized HC2H3O2. Chemically, this partial dissociation can be represented by : HC2H3O2 + H2O H3O + C2H3O2. To avoid the misuse of molarity, the concentration term formality (F) was introduced and is defined as the number of gram-formula weights of the solute initially taken in the preparation of the solution. This unit of concentration is a summation of all the existing forms of the solute in the 7 solution. For the KCl solution the concentration can be expressed as 1.000 F KCl. Since the salt is completely dissociated the concentration can also be expressed as 1.000 FK+ and 1.000 F Cl. In the acetic acid solution there are three different species in solution, H3O+, C2H3O2, and HC2H3O2. However, dissolving 0.1000 mole of HC2H3O2 in 1.000 liter of solution results in a 0.1000 F HC2H3O2 solution. Molarity as a unit of concentration should be used to describe the actual concentration of the species in the solution. Frequently, this is called the equilibrium concentration. Formality, on the other hand, is a concentration which represents the total amount of a given species regardless of its state in the solution. Another term that **** used to describe this is the analytical (molecular) concentration (usually, if the concentration is clearly stated as analytical concentration, it will be stated in terms of molarity). Unit Symbol Definition Relationship Molarity M Number of moles of solution per liter of solution Formality F Number of gram-formula ( ) weights (GFW) of solute per liter of solution Normality N Number of equivalents of solute per liter of solution Molality m Moles of solute per kilogram of solvent Mole X Ratio of the moles of solute fraction to the total moles of solute plus solvent Percent by wt% Ratio of the weight of weight solute to the total weight of solute plus solvent. Percent by vol% Ratio of volume of solute volume to volumes of solute plus ( ) solvent needed to reach total volume 8 Equivalent Weights This quantity varies with the type of reaction. It often happens what the same compound possesses different equivalent weights in different chemical reactions. In neutralization reaction: a) The equivalent weight of an acid is that weight of it which contains one replaceable hydrogen i.e. 1.008 gm hydrogen. The equivalent weight of a monobasic acid, as HCl, HBr, HI, HNO3, CH3COOH is equal to its molecular weight. AN solution of a monobasic acid will therefore contain. 1 gm molecular weight (or 1 mole) in a liter of solution. The equivaleat weight of a dibasic acid as H2SO4 and (COOH)2, or of a tribasic acid e.g., H3PO4 is likewise 1/2 and 1/3 respectively of its molecular weight. It is known that molecules of di- and tribasic acids may react so that not all or the ionisable hydrogen atoms are involved. Evidently in such cases the values of their gram equivalents are different. For ex. since each H3PO4 molecule yields two hydrogen ions in the following reaction. H3PO4 + 2NaOH = Na2HPO4 + 2H2O Its equivalent weight in this reaction is not 1/3 but 1/2 of the molecular weight. Similarly, in the reaction H3PO4 + NaOH = Na2HPO4 + H2O the equivalent weight of the acid is equal its molecular weight. b) The equivalent weight of a base is that weight of it which contains one replaceable hydroxyl group, i.e., 17.008 gm of hydroxyl are equivalent to 1.008 gm of hydrogen. The equivalent weights of NaOH, 9 KOH < and NH4OH are the molecular weight of each. The equivalent weight of Ca(OH)2 and Ba(OH)2 is 1/2 its molecular weight. c) Salts of strong bases and weak acids possess alkaline reactions in equeous solution because of hydrolysis. Sodium carbonate, with methyl organ as indicator, reacts with 2 moles of HCl to form 2 moles of NaCl; hence its equivalent weight is 1/2. Borax, under similar conditions, also react with 2 moles of HCl and its equivalent weight is 1/2 mole. II- In precipitation reactions: Here the equivalent weight is the weight of the substance which contains or reacts with 1 gm atom of a univalent metal, 1/2 gm atom of a bivalent metal…etc. For a metal, the equivalent weight is the atomic weight divided by its valency. For a reagent which reacts with this metal, the equivalent weight is the weight of it which reacts with one equivalent of the metal. The reactive constituents of salt are its ions. The equivalent weight of a salt in a precipitation reaction is the gram molecular weight of the salt divided by the total valency of the reaction ions. Thus the equivalent weight of Ag NO3 in the titration of chloride is its molecular weight. Also in the reaction FeSO4 + 2NaOH = Fe(OH)2 + Na2SO4 Or FeSO4 + BaCl2 = FeCl2 + BaSO4 the equivalent weight of FeSO4 = 1/2 its molecular weight. In the same way find that the equivalent of Al2(SO4)3 and BaCl2 in the reaction. Al2(SO4)3 + 3 BaCl2 = 3 BaSO4 + 2AlCl3 are respectively 1/6 and 1/2 of the molecular weight of the corresponding substance. 11 In a complex formation reaction the equivalent weight is most simply deduced by writing down the ionic equation of the reaction. For ex. the equivalent weight of KCN in the titration with silver ions is 2 mols. Since the reaction is 2CN + Ag+ [Ag (CN)2] III- In oxidation – reduction reaction electrons are transferred from the reducing agent to the oxidizing agent. This leads to the following definitions. Oxidation is the process which results in the loss of one or more electrons by atoms or ions. Reduction is the process which results in the gain, of one or more electrons by atoms or ions. An oxidizing agent is one that gains electrons and is related to a lower valency condition. A reducing agent is one that losses electrons and is oxidized to a higher valency condition. The following are examples of oxidizing agents which are of importance in quantitative analysis: potassium permanga –nate, potassium dischromate, ferric sulphate, iodine potassium bromated and potassium iodate. Examples of reducing agents are : ferrous sulphate, metallic iron, sodium thiosulphate, oxalic acid and aresenious oxide. In all oxidation – reduction process there will be a reaction undergoing oxidation and one undergoing reduction, since the two reactions are complementary to one another and occur simultaneously – one can not take place without the other. The equivalent weight in redox reaction can be calculated as follows: a) By the number of electron lost or gained: To calculate the equivalent weights of oxidizing and reducing agents, which we meet in oxidimetry, let us consider the following ex: 11 10 FeSO4 + 2KMnO4 + 8H2SO4 = 5Fe2(SO4)3 + 2MNSO4 + K2SO4 + 8H2O. The reducing agent is FeSO4 or, the Fe++ ion, which loses one electron and is oxidized to an Fe+++ ion. Fe++ Fe+++ The oxidizing agent is KMnO4 or MnO4 ion which is reduced to Mn++ as follows: MnO4 + 8H+ + 5e = Mn++ + 4H2O The calculation of the equivalent weights of oxidizing and reducing agents must be based on the number of electrons gained or lost by one molecule of substance in a given reaction. The equivalent weight of an oxidizing agent is equal to its molecular weight divided by the number of electrons gained by one molecule of it in the given reaction. The equivalent weight of a reducing agent is equal to its molecular weight divided by the number electrons lost by the molecule in the given reaction. In the above example, the equivalent weight of KMnO4 (in acid solution) and the equivalent weight of FeSO4 = its molecular weight. The other procedure which is of value in the calculation of the equivalent weight of substances is b) by the oxidation number method. The equivalent weight of an oxidizing agent is determined by the change in oxidation number which the reduced element exerts. It is that quantity of oxidant which involves a 12 change of one unit in the oxidation number. Thus in the normal reduction of KMnO4 in acid medium to manganous salt. The change in the oxidation number of the manganese is from +7 to +2. The equivalent weight of KMnO4 is therefore is 1/5 its molecular weight. Similarly for the reduction of K2Cr2O7 in acid solution. ( ) the change in oxidation number of two atoms of chromium is from + 12 to +6, or by 6 units of reduction. The equivalent weight of dischromate is 1/6 its molecular weight. For reducing agent as FcSO4 ( ) ( ) Here the change in oxidation number per atom of iron is from +2 to +3 or by 1 unit of oxidation, hence the equivalent weight = molecular weight. A useful summary of the common oxidizing and reducing agents, together with the various transformations which they undergo is given in the following: Oxidants KMnO4 (in acid) MnO4 + 8H+ + 5e Mn++ + 4H2O KMnO4 (neutral) MnO4 + 2H2O + 3e MnO2 + 4OH KMnO4(alkaline) MnO4 + e MnO4 K2Dr2O7 Cr2O4 + 14H+ + 6e 2Cr+++ + 7H2O Iodine I2 + 2e 2I 13 KIO3 IO3 + 6H+ + 6e I  + 3H2O H2O2 H2O2 + 2H+ 2e 2H2O Reductants Zinc zn–2e zn++ HI 2HI – 2e I2 + 2H+ Oxalic acid C2O4– – 2e 2CO2 FeSO4 Fe++  e Fe+++ Na2S2O3 2S2O3– – – e S4O9– – Calculation of the Results of Volumetric Determination: The most important advantage of the equivalent system is that the calculation of volumetric analysis are reduced very simple, since at the end point the number of equivalents of the substance titrated is equal to the number of equivalents of the standard solution employed. We may write: Hence: number of equivalents = N x number of litres. If the volumes of solutions of two different substances (1) and (2) which exactly react with one another are V1 ml and V2 ml respectively, then these volumes orally contain the same number of gm equivalents of (1) and (2). Thus V1 x N1 = V2 x N2 In other words the product of the volume V of a solution used in a titration and its normality N is constant value for both reacting substances. 14 Ex: 1- How many ml of 0.2N HCl are required to neutralize 25ml of 0.1 N NaOH? N1 x V1 = N2 x V2 0.2 x V1 = 0.1 x 25 V1 = 12.5 ml Ex: 2-How many ml of N HCl are required to precipitate completely 1 gm of AgNO3. The equivalent weight of AgNO3 in a precipitation reaction = molecular weight = 169.89 gm. 1 gm of AgNO3 = 1 x mgm equ. Now number of mg equivalents of HCl = number of mg equivalents of AgNO3. N x V of the acid = number of mgm equivalent of AgNO3 V x 1 = 5.886 V = 5.9 ml Ex: 3 – 25 ml of acidic FeSO4 solution react completely with 30 ml of 0.125 N KMnO4. Calculate the strength of FeSO4 solution in gms of FeSO4 per litre. The eq. wt of FeSO4 as reducing agent = its molecular wt. = 151.9. N1 x V1 = N2 x V2 N1 x 25 = 0.125 x 30 N1 = 0.15 The strength of FeSO4 = 0.15 x 151.9 = 22.785 gm/l. 15 Ex: 4- Find the amount of acetic aid in a solution if 20.5 ml of 0.1145 N NLOH solution was required to neutralize it. Solution: Since N x V of NaOH = number of m.grm equivalent of acetic acid. the mgm eq. of acetic acid = 20.5 x 0.1145 the gm eq. acetic acid = since the gm eq. of any substance = and the eq. wt of acetic acid = 60.05 the amount of acetic acid = = 0.1410 gm Ex: 5-calculate the molarity of a 0.3 N Al2(SO4)3 solution? Solution: For this salt the gm. Eq. is 1/6 of its gm. Molecule Therefore, in order to find how many moles are present in 0.3 by 1/6, thus M=Nx = 0.3 x = 0.05 i.e. the molarity of the solution is 0.05 Ex: 6 - Calculate the normality of 0.2 N Bi (NO3)3. Solution: Since a gm. Molecule of Bi (NO3)3 corresponds to 3 gm molecules of HNO3, i.e. to 3 gm. Ions of H+, the gm. Eq. of this salt is equal to 1/3 16 of a gm molecule. Therefore, a 1M solution is 3N, and a 0.2M solution is 0.2 x 3 = 0.6 N. Ex: 7- Calculate the normality and molarity of a 20% sulphuric acid solution its specific gravity is 1.14. Solution: It is known that the weight (w) of a substance, its density (d), and its volume (V) are connected as follows: 20% H2SO4 solution means that each 100 gm of the solution contains 20 gm H2SO4 and 80 gm of water. At first we calculate the volume of 100 gm of 20% H2SO4 solution. Now calculate the grams of H2SO4 in litre (1000 ml) of 20% H2SO4 solution. 87.7 ml contains 20gm H2SO4 1000 ml contains x gm H2SO4 Since the equivalent weight of H2SO4 is 49.04. Then a 20% H2SO4 acid solution is 4.65 N The molarity of the same solution = 4.65 / 2 = 2.32 M 17 Ex: 8- Find how many milliliters of concentrated H2SO4, sp. gr. 1.84, containing 96% H2SO4, must be taken for preparation of 5 litres of approximately 0.1 N solution. Solution: First calculate how many grams of H2SO4 is required for the given volume (here 5 litres) of 0.1 N solution. Since the gm. Eq. of H 2SO4 is M/2 = 49.04, and 1 litre of 0.1 N solution contains 0.1 gm. Eq., the total amount of H2SO4 required is X. X = N x eq. wt x number of litre = 0.1 x 49.04 x 5 = 25 gm Now the weight of 96% sulphuric acid which contains this weight of anhydrous H2SO4. 100 gm of 96% acid contains 96 gm H2SO4 y gm of 96% acid contains 25 gm H2SO4 and hence We now convert the weight of 96% sulphuric acid into the corresponding volume. Therefore to prepare 5 litres of approximately 0.1 N sulphuric acid solution we must measure out about 14 ml of conc. H2SO4, sp. Gr. 1.84 and dilute it with water (the acid must be poured into water and not water into the acid) to 5 litres. Ex: 9 – Find the volume to which 50 ml of 2N HCl must be diluted to convert it into 0.3N solution. 18 Solution: Since N x V = constant. N1 x V1 before dilution = N2 x V2 after dilution 0.3 x V1 = 50 x 2 V1 = Therefore to convert a 2 N HC1 solution into a 0.3 N solution, 50 ml of 2N solution must be diluted with water to 333 ml. Ex: 10 – What volume of a 1 N solution contains the same amount of a given dissolved substance as 30 ml of a 0.2N solution? Solution: Since the amount of substance is the same in both solution, the products of N1 x V1 = N2 x V2 Therefore, V1 x 1 = 30.0 x 0.2 and V = 6 ml Ex: 11- calculate the proportions by weight and by volume which a 54% solution of HNO3 (sp. Gr. 1.39) must be mixed with a 14% solution of HNO3 (sp. Gr. 1.08) to give a 0% solution. Solution: This can be calculated by means of very convening graphical device. 54 6 (i.e., 20 – 14) 20 (i.e., 54 – 20) 14 34 On the left we write, one above the other, the percentage concentrations of the two original solutions and the final concentration of 19 the mixture is written in the center. On the right, at the opposite ends of the diagonals we write the differences between the final and each of the initial concentrations. The smaller quantity being subtracted from the larger. Each of the difference represents the weight of the solution the percentage concentration of which is written in the same horizontal lines. Thus 6 parts by weight of 54% acid must be taken with 34 parts by weight of 14% acid to obtain a 20% solution of HNO3. The volume proportions can be easily found from these weight proportions. The volume of 6 gm of 54% acid is 6/1.33 = 4.5 ml and the volume of 34 gm of 14% acid is 34/1.08 = 31.5 ml. Therefore 4.5 ml of 54% HNO3 must be added to 31.5 ml of 14% HNO3. Ex.: 12- Calculate the weight of water to be added to 100 ml of 72% sulphuric acid (sp. Gr. 1.63) to convert it into 25% acid. Solution: 72 26 25 0 46 Therefore we must take 46 parts of water to26 parts of 72% acid solution by weight. We now convert to volume proportions. Volume of water = = 46 ml Volume of acid = = 16 ml To 16 ml H2SO4 we must add 46 ml H2O. To 100 ml H2SO4 we must add x ml H2O 21 X= = 290 ml Ex: 13- (a) What is the normality of a solution containing 0.008292 gm of K2CO3 per milliliter. (b) What is the normality of a solution made by diluting 50 ml of H2SO4 (sp. Gr. 1.06) containing 11.60% H2SO4 by weight to 1 liter with water. (c) How many (milliliters of the carbonate solution will be completely neutralized by 7 ml of the acid. Solution: (a) Eq. wt of K2CO3 = Then 1 milliliter of N K2CO3 solution contains = = 0.0691 gm Then 0.0691 K2CO3 give 1 ml of 1 N solution and 0.008292 gm K2CO3 give 1 ml of X N olution. The normality of the solution X = (b) The weight of 50ml of acid solution is given V= w = V x d = 50 x 1.08 gm of solution. 100 gm of acid solution contains 11.6 gm H2SO4 then 50 x 1.08 gm of acid solution contains x gm H2SO4. x gm of H2SO4 in 50 ml= = 6.264 gm H2SO4 since 50 ml of the solution diluted to 1 liter. Then the liter contain the same amount of H2SO4 = 6.264 g/ml and its normality = = 0.1277 21 (c) N x V = N1 x V1 0.12 x V = 0.1277 x 7 V = 7.45 ml Preparation of Standard solution If a reagent is available in the pure state, a solution of definite normality is prepared simply by weighing out an equivalent weight, or a definite fraction or multiple therefore, dissolving it in the solvent, usually water and making up the solution to a known volume. When the reagent is not available in the pure form as in the cases of most alkali hydroxides, some inorganic acids and various deliquescent substances, solutions of approximate normality required are first prepared. These are then standardized by titration against a solution or a pure substance of known normality. This indirect method is employed for the preparation of solutions of acids, sodium, potassium and barium hydroxides, potassium permanganate, potassium thiocyanate and sodium thiosulphate. These are known as secondary standard. Primary standard substances: A primary standard substance should satisfy the following requirements. 1) It must be easy to obtain, to purify, todry, and to preserve in a pure state. 2) The substance should be unaltered in air during weighing, this conditions implies that it should not be hygroscopic, nor oxidized by air nor affected by CO2. 3) It should have a high equivalent weight so that the weighing errors may be negligible. 22 4) The substance should be readily soluble under the conditions in which it is employed. 5) The reaction with the standard solution should be stoichiometric and practically instantaneous. The titration error should be negligible. The following substances are commonly employed as primary standards: sodium carbonate Na2CO3, borax Na2B4O7, potassium hydrogen phthalate KHC8H4O4, silver nitrate AgNO3, sodium chloride NaCl, potassium bromide KBr, potassium dichromate K2Cr2O7, potassium iodate KIO3, iodine I2, sodium oxalate Na2C2O4, and arsenious oxide As2O3. Classification of reactions in volumetric analysis: Volumetric methods of analysis are generally more rapid than gravimetric methods. The process where by a standard solution is brought into reaction is called a titration, and the point at which the reaction is exactly completed is called the equivalence point. An indicator shows the end point of the titration, usually by a change of color. The volume of the standard solution used and its concentration furnish a measure of the substance being determined. The mathematics of this kind of measurement is called stoichiometry. The processes of volumetric analysis are classified into three main classes. 1. Neutralization reactions: These include the titration of free bases, or these formed from salts of weak acids by hydrolysis, with a standard acid (acidimetry) and the titration of free acid, or those formed by the hydrolysis of salts of weak 23 bases, with a standard base (alkalimetry). The reactions involve the combination of H+ and OH ions to from water. 2. Oxidation – reduction (redox) reactions: These include all reactions involving change in oxidation number of transfer of electrons. A reducing, substance is titrated with a standard solution of an oxidizing agent, or an oxidizing substance is titrated with a standard solution of a reducing agent. 3. Precipitation and complex – formation reactions: Those depend upon the combination of ions, other than H+ and OH- ions, to form either a simple precipitate, as in the titration of silver with a solution of chloride, or a complex ion, as in the titration of a cyanide with silver nitrate solution. Problems 1. 250 ml of NaOH solution contains 10 gm NaOH that is the normality of this solution? 2. Find the normality of an HCl solution if its concentration is 0.003592 g/ml. 3. What is the weight of KOH in 200 ml of a 0.092 N solution? 4. What is the normality and strength of an HNO3 solution if 20 ml of it takes 15ml of 0.12 N NaOH solutions? 5. How many gm. Eq. are contained in (a) 1.8909 gm of chemically pure oxalic acid H2C2O4.2H2O; (b) 20ml of 0.12 N NaOH solutions? 6. What are the normality and strength of a KOH solution if 26 ml of it was required for titration of 0.1560 gm of chemically pure (dibasic) succinic acid H2C4H4O4? 24 7. How many milliliters of 0.02 N KMnO4 solution would be required for titration of 20 ml of 0.03N FeSO4 solution? 8. How many milliliters of 0.02N KMnO4 solution would be required for titration of an FeSO4 solution containing 0.02 gm of iron? 9. Calculate the normality of a 40% CaCl2 solution of sp. Gr. 1.395. 10. Calculate the molar concentration of 10% NH3 solution (sp. Gr. 0.958). 11. How many milliliters of 2.0 N HNO3 solution should be taken for preparation of 3 litres of 0.1 N solutions? 12. What volume of 1 N HCl solution is equivalent to 23.8 ml of 0.2 N HCl solution? 13. How many milliliters of 20% HCl solution of sp. Gr. 1.098 should be taken for preparation of 5 litres of 0.1 N solution? 14. How many milliliters of 10% HCl solution (sp. Gr. 1.047) should be added to 50ml of 37.23% solution of sp. Gr. 1.19 to give a 25% HCl solution. 15. How much water must be added to 200 ml of 46% HNO3 solution (sp. Gr. 1.285) to convert it into 10% solution? 16. What is the normality of an HCl solution in a 100 ml of it yields 0.3075 gm of AgCl? The Neutralization Method Ionization of acids and bases: An acid may be defined as a substance which when dissolved in water, undergoes dissociation with the formation of hydrogen ions as the only positive ion: HCl H+ + Cl 25 Actually the free hydrogen ions H+ (or protons) do not exist in the free state in aqueous solution, each H+ ion combines with one molecule of water to form the hydroxnium ion H3O+. Thus HCl + H2O H3O+ + Cl HCl, HNO3 are almost completely dissociated in aqueous solution. Polybasic acids ionize in stages. In H2SO4 the first hydrogen atom is almost completely ionized. H2SO4 + H2O H3O+ + H2SO4 Primary ionization the second hydrogen atom is partially ionized, except in very dilute solution. H2SO4 + H2O H3O+ + SO4 Secondary ionization Acid of the type CH3COOH, the degree of dissociation is accordingly small CH3COOH + H2O CH3COOH + H3O+. The former are termed strong acids (ex: HCl, HBr, HNO3 and primary ionization of H2O4) and the latter are called weak acids (ex: CH3COOH, H2CO3, boric acid). A base may be defined as a substance which when dissolved in water, undergoes dissociatic: with the formation of OH- ions s the only negative ions. Thus NaOH, KOH< and he hydroxides of the alkaline earta metals are almost completely dissociated in equeous solution. NaOH Na+ + OH- 26 These are strong bases. NH4OH solution, however, is a weak base, its degree of dissociation is small NH4OH NH4+ + OH. Modern views regarding acids and bases: The simple concepts given above suffice for most of the requirements of quantitative inorganic analysis. According to Bronsted theory: an acid is a donor of protons, and a base is an accepter of protons. When an acid (HA) loses its proton, a base (A-) is formed. HA and A- are called a conjugate pair: A- is the conjugate base of the acid HA, and vice versa. Acid Base HA H+ + A CH3COOH H+ + CH3COO NH4+ H+ + NH3 H2PO4 H+ + HPO4 Acid may be neutral molecules, positive or negative ions. Ionization of water: Water is a very weak electrolyte. It is amphiprotic, which means that it can act either as an acid or as a base. The ionization reaction is H2O + H2O H3O+ + OH Applying the law of mass action. ( )( ) KW = ( ) where KW is the ionic product of water. 27 In pure water or in dilute aqueous solutions, the concentration of the undissociated water may be considered constant, hence. KW = (H3O+) (OH) = 10-14 pKW = pH + pOH = 14 The hydrogen – ion exponent pH defined by the relationships pH =  log (H3O+) = log ( ) All the states of acidity and alkalinity between these solution with respect to hydroxonium and hydroxyl ions can be expressed by a series of positive number between 0 and 1. Thus a neutral solution with (H3O)+ = 10-7 has a pH = 7. A solution with molar (or normal) concentration of H3O+ ions has a pH of O (= 10+) and a solution molar with respect to OH ions has (H3O)+ = / 10+ = 10-14 and ( ) posscess a pH of 14. A neutral solution is therefore one 'in which pH = 7, an acid solution one in which pH < 7, and alkaline solution one in which pH > 7. Ex: 1- Find the pH of a solution of which (H3O+) = 4.0 x 10-5. Solution: log 4.0 = 0.602, hence. log 4.0 x 10-5 = 5.602 = 0.602  5 =  4.398 pH =  log (H3O+) = 0 ( 4.398) = 4.398 28 Ex: 2- Find (H3O+) corresponding to pH = 5.643 Solution: pH =  log (H3O+) = 5.643 log (H3O+) =  5.643 log (H3O+) = 6.347. Soluble, the antilogarithms of.357 = 2.28 (H3O+) = 2.28 x 106 Ex: 3- Calculation the pH of a 0.01 M solution of acetic acid (The degree of dissociation is 12.5 per cent). Solution: (H+) = 0.125 x 0.01 = 1.25 x 103 log 1.25 = 0.097 pH =  (3 + 0.097) = 2.903 Ex: 4- Calculate (H3O+), (OH), pH and pOH in 0.0080 M HCl. If Kw = 1.01 x 10-14. Solution: HCl is a strong electrolyte and is completely ionized Hence (H3O+) = 0.0080 M pH =  log (H3O+) =  log 0.008 =  log 8 – log 103 = 0.90 + 3 = 2.10 (OH) = ( ) pOH =  log (OH) =  log (1.26 x 1012) = 11.9 pH + pOH = 14, pH = 14 – 11.9 = 2.1 29 Ex: 5- What is (H3O+) in a solution where pH = 4.72 ? Solution: (H3O+) = 10pH = 104.72 = 10–5 x 100.28 = 1.91 x 10-5 M Acid-Base properties of salts. Hydrolysis: Salts may be divided into four main groups: i) Those derived from strong acids and strong bases, e.g., KCl. ii) Those derived from weak acids and strong bases, e.g., CH3COONa. iii) Those derived from strong acids and weak bases, e.g. NH4Cl. iv) Those derived from weak acids and weak bases, e.g. CH3COONH4. i) The aqueous solution of a salt of group (i) neither the anions (Cl) have any tendency to combine with H+ ions nor the cations (K+) with the OH ions of water since the related acids and bases are strong electrolytes. The equilibrium between H+ and OH ions in water: H2O H+ + OH Is therefore not disturbed and the solution remains neutral. ii) if a salt (NaA) of weak acid (HA) and a strong base (NaOH) is put into water, the solution does not remain neutral: A  has basic properties, and the solution becames basic. A + H2O HA + OH. This interaction between the ion (ions) of salt and the ions of water is called hydrolysis. 31 iii) if a salt of group (iii) of a strong acid and a weak base is put in water (MH), the M+ ion will be reduced by combinations with OH ions of water to form the little dissociated base MOH M+ + H2O MOH + H+ the (H+) of the solution will thus be increased an the solution will react, as acid. iv) For salt of group (iv), in which both the acid and the base are weak, two reactions will occur simulataneously M+ + H2O MOH + H+ A + H2O HA + OH The reaction of the solution will clearly depend upon the relative dissociation constants of the acid and the base. If they are equal in strength, the solution will be neutral, if Ka > Kb it will be acid, and if Kb < Ka, it will be alkaline. Neutralisation indicators: Suitable indicator must be added to acid base titration in order to determine the equivalent point. If both the acid and base are strong electrolytes, the resultant solution will be neutral and have a pH of 7 at the equivalence point. But if either the acid or base I a weak electrolyte, the salt will be hydrolysed to a certain degree, and the solution, at the equivalence point will be either slightly alkaline or slightly acidic. For any actual titration the correct end point will be characterized by a definite value of (H+) of the solution, the value depending upon the nature at the acid and the base and the concentration of the solution. A large number of substances are available, called neutralization or acid-base indicators which possess different colors according to the (H+) 31 of the solution. The chief characteristic of those indicators is that the change of color from the predominantly acid color to a predominantly alkaline color is not sudden and abrupt, but takes place within a small interval of pH (about 2 pH units) called the color – change interval of the indicator. Certain indicators changes color when the solution is quite strongly acidic; others, when the solution is quite strongly basic. The following table shows the most important acid base indicators and their color – change intervals. Figure (1): Acid –base indicator ranges at 25C. 32 Ostwald's Theory of Indicator: All indicators in general use are very weak organic acids or bases. The undissociated indicator acid (HIn) or base (InOH) has a different from that of its ions. The equilibria in aqueous solution may be HIn H+ + In and InOH OH + In+ unionized ionized color color Let us consider an acid indicator. In acid solution; i.e., in the presence of excess of H+ ions, the ionization of the indicator will be depressed (common ion effect) and the concentration of In – will be very small, the color will therefore be that of the unionized from. If the medium is alkaline the decrease of (H+) will result in the further ionization of the indicator. (In–) increases and the color of the ionized form becomes apparent. By applying the law of mass action and if the activity coefficients are assumed to be unity, then ( ) ( ) Kin a = ( ) ( ) (H+) = ( ) ( ) = ( ) where (Kin a) is the ionisation constant of the acid indicator. 33 The actual color of the indicator, which depends upon the ratio of the concentrations of the ionized and unionized forms, is thus related to (H+) or pH ( ) pH = log ( ) For a base indicator ( ) (OH–) = ( ) where Kin b is the base dissociation constant. This may be written ( ) (H+) = ( ) since Kw = (H+) (OH–) The Ostwald's simple theory of the color charge of indicators is not quite correct. According to the newer theory the color change is no longer being due to ionization alone, because it has been shown that an indicator consists of two or more different tautomeric forms possessing different constitutional forms and different colors. One of the forms is a non – electrolyte (a pseudo – acid or pseudo base). The other form an acid or base. The ions of the true acid or true base possess the same color as the undissociated form of the same structure. We can illustrate this by reference to phenolphthalein. 34 Phthalein indicators (III) Colorless Red Red Pseudo acid acid ion HIn' HIn In– + H+ Colorless red red In the solid state or in acid solution, the indicator exists as the colorless pseudo acid (I). This is in equilibrium in solution with a trace of the tautomeric form (II) which is an acid since it possesses the –COOH group and dissociates to yield the H+ ion and indicator ion In– (III). Two equilibria must be considered in aqueous solution HIn' HIn HIn H+ + In– Applying the law of mass action to both 35 ( ) KT = ( ) ( )( ) and KI = ( ) assuming that the activity coefficients equal unity. The ion In– has the same color (red) as the weak acid HIn. In acid solution, i.e., in presence of excess H+ ions, the indicator exist largely s HIn' and now these two forms have different colors, and HIn has the same color as In–. The indicator will only be useful if equilibrium constant for the tautomeric equilibrium KT is very small, then practically the whole of the undissociated indicator will be in the form HIn'. Combining the above two equations. ( )( ) Kin a = KT x K1 = ( ) ( ) (H+) = ( ) Kin a ( ) = ( ) Or ( ) ( ) The above equation can write as follows ( ) pH = pKin a + log ( ) The color of an indicator is determined by the ratio of the concentrations of In– and HIn'. Both forms are present at any hydrogen ion concentration. It must be realized that the human eye has a limited ability to detect either of the two colors when one of them predominates. Experience shows that the solution will appear to have the acid color, i.e., 36 of HIn' when the ratio of (HIn') to (In–) is above approximately 10, and the alkaline color, i.e., of In–, above a similar ratio of (In–) to (HIn') of approximately 10. Thus only the acid color will be visible when (HIn')/(In–) > 10, the corresponding limit of pH given by the equation pH = pKin a-1 The alkaline color only will be visible when (In) / (HIn') > 10 and the pH pH = pKin a +1 The color change interval is accordingly pH = Kin a 1 i.e., over approximately two pH units. Within this range the indicator will appear to change from one color to the other. The change will be gradual, since it depends upon the ratio of the concentration of the two colored forms (pseudo acid and ionized). For an indicator base we have, according to the modified theory the two equilibria. In'OH InOH and InOH In+ + OH– The following equation may be deduced along similar lines to that described for an indicator acid. ( ) (OH) = ( ) ( ) ( ) 37 where Kin b is the apparent dissociation constant of the base. Remembering that Kw = (H+) (OH–) ( ) (H+) = ( ) ( ) ( ) ( ) The above theory applies only to a monobasic aid or to mono acid base. Universal or multiple range indicator: By suitable mixing certain indicators the color change may be made to extend over a considerable portion of the pH range. Such mixtures are usually called universal indicators. They are not suitable for quatitative titrations, but may be employed for the determination of the approximate pH of a solution. The following example is given by Bogen. Dissolve 0.1 gm phenolphthalein, 0.2 gm of methyl rad, 0.3 gm of methyl yellow, 0.4 gm of brom thymol blue, and 0.5 gm of thymol blue in 500 ml of absolute alcohol, and add sufficient NaOH solution until the color in yellow. The color changes are as follows : pH2, red, pH4 orange, pH6 yellow, pH 8, green, pH 10, blue. Several indicators are available commercially as solutions and as test papers. Neutralisation (Titration) Curves: In order to select the suitable indicator for an acid base titration, it is necessary to understand the changes in the (H+) or pH value during the course of the titration. The curve obtained by plotting pH as ordinates against the ml of alkali added as abscissa is known as the titration curve. 38 1) Neutralisation of a strong acid and a strong base: In order to calculate the change of pH during titration, we shall start with 100ml. of say N HCl and add N NaOH solution. (The pH of N HCl is 0). When 50 ml of N NaOH have been added, 50 ml of unneutralised N acid will be present in a total volume of 150 ml. From N x V = N` X V` and since N = 1 (H+) = 50 x = 3.33 x 10-1, or pH 0.48 For 75 ml of base. (H+) = 35 x = 1.43 x 10-1, pH = 0.94 For 90 ml of base (H+) = 10 x = 5.27 x 10-2, pH = 1.3 For 98 m of the base. (H+) = 2 x = 1.01 x 10-2, pH = 2.0 For 99 ml of base (H+) = 1 x = 5.03 x 10-3, pH = 2.3 For 99.9 ml of base (H+) = 0.1 x = 5.01 x 10-4, pH = 3.3 Up on the addition of 100 ml of base, the pH will be change sharply to 7, the equivalence point with 100.1 ml of base (OH) = = 5 x 10-4, p OH = 3.3 and pH = 10.7. 39 with 101 ml of base. (OH) = x 10-3, p OH = 2.3, pH = 11.7. The above results show that as the titration proceeds, the pH rises slowly. But between the addition of 99.9 and 100.1 ml of base, the pH of the solution rises from 3.3 to 10.7, i.e., in the vicinity of equivalence point the rate change of pH the solution is very rapid. Alkali added (0.1 N N2OH) From the curve, it is evident that any indicator with a color change range between pH 3.5 and pH 10.5 can be used without much error. Phenolphthalein and methyl orange are most often used in such titrations. The titration curve is influenced by the concentration of the titrating agent. The more dilute the acid or base used in the titration, the less abrupt is the inflection at the equivalence point, the more carefully must the indicator be chosen. It is seen from the curve that in dilute solution (0.01 N), the titration error for methyl orange will be 1-2%. Neutralisation of a weak acid with a strong base: 41 For ex: acetic acid and NaOH solution. Let us study the neutralization of 100ml of 0.1 N acetic acid with 0.1 N NaOH solution. a) At the beginning of titration. In this case (H+) at the start is less than the above example, since acetic acid is less ionized than strong acid. The pH can be calculated from the ionization constant Ka of acetic acid which at 25C is 1.06 x 10-5, Thus CH3COOH H+ + CH3COO. ( )( ) = Ka = 1.82 x 10-5. ( ) Since (H+) = (CH3COO-) and dissociation of the acid is relatively so small that it may be neglected in expressing the concentration of acetic acid. ( ) - 1.82 x 10-5 (H+) = √ = 1.35 x 10-3. pH = 2.87 b) When 50ml of 0.1 N NaOH have been added. The addition of NaOH to acetic acid solution continually forms sodium acetate. The approximate (H+) at any point prior to the equivalence point can be found by substituting the relative concentrations of acetate ion and undissociated acetic acid in ( ) ( ) = 1.82 x 10-5. ( ) (CH3COOH) at any point during titration before the end point is taken as equal to the concentration of the acid that has not yet been 41 neutralized, and the concentration of the salt formed (CH3COO) is assumed equal to the concentration of the alkali added. (Salt) = 50 x = 3.33 x 10-2 (acid) = 50 x = 3.33 x 10-2 and from ( )( ) ( ) ( ) ( ) pH = log (3.33 x 10-2 / 3.33 x 10-2) + 4.74 = 4.74 The pH values at other points on the titration curve are similarly calculated. c) At the equivalence point the solution would not be neutral but would be neutral but would be slightly alkaline due to the hydrolysis of the salt. 42 (CH3COO) + H2O = OH + CH3COOH. The pH can be obtained from the equation. (H+) = √ Where Kw is the ionic product (i.e., 10-14), Ka is the ionization constant of the acid, and C is the molar concentration of the salt produced at the equivalence point. Hence. 43 pH = 1/2 pKw + 1/2 pKa – 1/2 log C. (pKw =  log Kw and pKa = log Ka) pKw = 14, pKa = 4.73, C = pH = 7 + 2.37 + ⁄ (-1.30) = 8.72. d) After the equivalence point. Beyond the equivalence point, NaOH is added and the ions in the solution have no appreciable effect on the (H+) concentration. The pH values and the titration curve are therefore the same as those in the proceeding titration of HCl. It is clear from the titration curve that methyl organge can not be used as indicator. The equivalence point is at pH 8.7, and it is necessary to use an indicator with a pH range on the slightly alkaline side, such as phenolphthalein. terilization of a weak base and a strong acid. We my illustrate this case by the titration of 100 ml 0.1 NH4OH (Kb = 1.8 x 10-5) with 100 ml 0.1 N HCl. Neutralisation of a weak base with a strong Acid: The pH value at the beginning of the titration from the ionization constants of NH4OH. ( )( ) = Kb = 1.8 x 10-5 ( ) Since (NH4+) = (OH) and the salt is weak base the degree of hydrolysis is very small. ( ) (OH) = √ 44 OH can be calculated and therefore, pH can be calculated and is equal 11.2. Similarly from this ionization constant, the pH can be culated during the progress of the titration. At the equivalence point in the titration is reached a 100 ml of 0.1 N acid is added and the solution would be htly acidic due to hydrolysis: (NH4+ + H2O H+ + NH4OH) The (H+) and the pH value is given by ( ) Since Kw = 10-14, Kb = 1.8 x 10-5 45 and C = pH = 1/2 Kw – 1/2 pKb – 1/2 log C = 7 – 2.37 – 1/2 (-1.3) = 5/28 For other concentrations, the pH may be calculated from the equation. ( ) pH = pKw – pKb – log ( ) where the (salt) is the (acid) added. It is clear from the curve that phenolphthalein is not suitable in the titration of 0.1 N NH4OH. The equivalence point is at pH 5.2, and it is necessary to use an indicator with a pH range on the slightly acid side such as methyl orange. Neutralization of a weak acid with a weak base: Ex: the titration of 0.1 N CH3COOH and (Ka = 1.8 x 10-5) with 0.1 N NH4OH (Kb = 1.8 x 10-5). The pH at the equivalence point is given by √ ( ) PH = 1/2 PHw + 1/2 pKa – 1/2 pKb = 7 + 2.37 – 2.37 = 7 It is clear from the curve that the change of pH near the equivalence point is very gradual. There is no sudden change in pH, and hence no sharp end point with simple indicator. 46 Neutratisation of a polybasic acid with a strong base: The shape of the titration curve will depend upon the relative magnitudes of the various dissociation constants. For many dibasic acids, the two dissociation constants are too close together and it is not possible to differentiate between the two stages. If the second ionization constant is not very small, all the replaceable hydrogen may be titrated e.g. sulphuric, oxalic and malonic acid. Similar remarks apply to tribasic acids. Ex: orthophospheric acid, for which K1 = 7.5 x 10-3 K2= 6.2 x 10-6 and K3 = 5 x 10-13. 47 Hence the three constants are not close to each other so that the acid will behave as a mixture of three mono basic acids with dissociation constants given above. H3PO4 H+ + H2PO4 H2PO4 H+ + HPO4 HPO4 H+ + PO4 Netralisation proceeds almost completely to the end of primary sage before the secondary stages is affected, and the secondary stage completed before the tertiary stage is apparent. The pH of the first equivalence point is approximately = 4.6, and the second is = 9.7, in the very weak third stage, the curve is very flat and no indicator is available for direct titration. 48 For the primary stage (H3PO4) as mono basic acid, methyl orange may be used as indicator. The second stage mixed indicator of phenolphthalein and – naphthol – phthalein may be used as indicator. Titration of sodium carbonate with a strong acid: The solution of Na2CO3 may be titrated to the bicarbonate stage (i.e. with one equivalent of acid), when the net reaction is Na2CO3 + HCl NaHCO3 + NaCI The pH at the equivalence point at his stage = 8.3. Therefore phenolphthalein may be used as indicator to detect the end point. Na2CO3 solution may also be titrated until all the carbonic acid is displaced (two equivalents of acid). The net reaction is then Na2CO3 + 2HCl = 2NaCl + H2CO3 The same end point is reached by titrating NaHCO3 solution with HCl. NaHCO3 + HCl = NaCl + H2CO3 49 The pH at the equivalence pint is normally 3.8. Suitable indicator is methyl orange. Ex: 1- What is the pH value of a dill. Acetic acid having (H+) of 2.5 x 10-4 mol. Per lites ? What color would be given to the solution by a drop of phenolphthalein? Solution: Ph. Ph = colorless Ex: 2- Calculate the pH at the equivalence point in the titration of 0.1 N weak acid HA (Ka = 1.0 x 10-5) with 0.1 N NaOH. Solution: At the equivalence point ( ) √ Where C is the concentration of the salt produced NaA at the end point pH = 1/2 pKw + 1/2 pKa – 1/2 log C pH = 8.85 Ex.: 3- How many grams of H2SO4 are required to prepare 1 N, 2N, 2M, 1/2 M solution of H2SO4. Solution: The eq. wt of H2SO4 = 98 / 2 = 49. 1 N solution, we need 1 x 49 = 49 gm/l 2 N solution, we need 2 x 49 = 98 gm / l To propare 1M, 98 gm/l are required. For 1/2 M, 98 / 2 gm / 1 are required. 51 Ex: 4- Calculate the molality of solution which contains 10 gm K2CO3 per 1000 gm of water. Solution: Moles of K2CO3 = = Molality = Molality = ⁄ Ex.: 5- Find the normality, molarity and molality of 10% solution of KOH by weight. The resulting solution of KOH as density 1.20 gm/l. Solution: Wt of water in 100 gm solution = 100 – 10 = 90 gm. Since the density of the solution is 1.2, then the volume of 100 gm of solution = 100 / 1.2 = 83.33 ml. Normality = equivalent eight of KOH = 56. N = (10/56) / (83.33 / 1000) = 2.1 N Molarity = = (10 / 56) / (83.33 / 1000) = 2.1 M Since the eq. wt = mol wt of KOH, N = M. Molarity = 51 = (10 / 56) / (90 / 1000) = 1.984 molal Ex.: 6- How many ml volume of 0.1 N solution can be prepared from 2.86 gm of Na2CO310 H2O. Solution: Eq wt. of Na2CO310 H2O = 286/2 = 143. Now 143 gm of Na2CO3.10H2O give solution = 1000 ml IN or 10000 ml. 0.1 N 2.86 gm of Na2CO310H2O will give = Ex. 7- Calculate the number of ml of 0.1 N H2O4 solution necessary to react completely with solution that contains 0.106 gm of pure Na2CO3. Solution: Gm eq. of Na2CO3 = 0.106/53 = 0.002 gm. Eq. since the gm. Eq. of Na2CO3 = gm. Of H2SO4 at the end point. 0.002 = N x V 0.002 = 0.1 x V V = 20 ml Ex: 8- How many ml of 12 N HCl solution is required for the preparation of 250 ml of 0.1 N HCl? Solution: N1 x V1 before dilution = N2 x V2 after dilution 12 x V1 = 0.1 x 250 V1 = 2.083 ml. 52 Ex: 9- What volume of N/3 HCl and N/12 HCl be mixed to get 1 litre of N/6 HCl? Solution: Let the volume of N/3 HCl required = X litres. The volume of N/12 HCl required = 1- X litres. Since for (N / 3 HCl)…………. 1/3 = gm eq of N/3 HCl = x/3 gm eq. similarly gm. eq. of N/12 HCl = 1-X / 12 and the gm q. of N/6 HCl = 1 x 1/6 = 1/6 Thus x = 1/3 litre the volume of N/3 HCl required = 0.333 liter and the volume of N/12 HCl required = 1 – 0.333 = 0.667 liter Ex.: 10 – 0.07 gm of a monobasic acid required 10ml of N/12 NaOH solution for complete neutralization. Calculate the molecular weight of the acid. Solution: Sine number of gm. Eq. NaOH = N x V = gm.eq gm eq. of NaOH at the end point = gm eq. of acid gm eq. of acid = 1/1200 gm eq. = eq.wt = 90. 53 Ex.: 11- Exactly 20 ml of 0.8 N acid was required neutralize completely 1.12 gm of an impure sample of CaO. The molecular weight of CaO = 56. Calculate the % of purity of CaO? Solution: eq. wt of CaO = 56/2 = 28 gm eq. of acid = 0.8 x = 0.016 at complete neutralization gm. Eq. of aci = gm eq. of oxide and hence gm eq of oxide = the weight of CaO = 0.16 x 28 = 0.448 gm. The 1.12 gm of impure CaO contains 0.448 gm of pue CaO % of purity = Ex: 12- What is the purity of conc. H2SO4 (sp. Gr. 1.8) if 5 ml is neutralized by 0.084 of 2N NaOH? Solution: gm.eq. of NaOH = 2 x = 0.168. gm. eq. of NaOH at the end point = gm. eq. of H2SO4 gm. eq. of H2SO4 = 0.168 gm. eq. = The weight of sulphuric in the solution = 0.168 x 49 = 8.2323 gm. The weight of sulphuric in the solution = 5 x 1.8 = 9 gm H2SO4. % of purity = x 100 = 91.46 54 Ex.: 13 – 3 gm of mixture of NaCl and NaOH were dissolved in distilled water to make 250 ml of the solution. 20 ml of this solution required 24 ml HCl for neutralization, 18 ml of this HCl required 20 ml of 0.1 N KOH solution for complete neutralization. Find the percentage composition of the mixture. Solution: N1 x V1 for HCl = N2 x V2 for KOH N of the acid = also N1 x V1 for HCl = N2 x V2 for the mixture N for the mixture = = The normality of the mixture is the normality of NaOH equal the strength of NaCH per liter since NaCl does not reat with HCl. N x eq. wt = x 40 = 5.33 gm/litre The % of NaOH = = 44.44% The % of NaCl = 100 – 44.44 = 55.56% Ex.: 14 – 12.6 gm of hydrated oxalic acid was dissolved per liter of the solution. 10 ml of the solution required 20 ml of N/10 NaOH for complete neutralization. Find the number of molecules of water of crystalisation. Solution: N1 X V1 for the acid = N2 x V2 for the base N1 = 55 The eq. wt of anhydrous oxalic acid = 90 / 2 = 45 The strength of the acid = 1/5 x 45 = 9 gm Applying the relation: 9 (90 + 18x) = 90 x 12.6 x=2 The number of water molecules = 2 Ex: 15-A 1.2 gm sample containing Na2CO3, NaHCO3 and inert impurities is titrated with 0.5 N HCl. With ph. Phth as the indicator the solution turns colorless after the addition of 15 ml of the acid. Methyl orange is then added, and 22 ml more of the acid are required to change to color of this indicator. What is the % Na2CO3 and of NaHCO3 in the sample? Solution: Na2CO3 is first convented into NaHCO3 required 15 ml of the acid. From the 22 ml of additional acid added, 15 ml must have been required to complete the reaction with this NaHCO3 formed and 7 ml to react with NaHCO3 originally present. Na2CO3 required 30 ml of the acid and NaHCO3 required 7 ml of the acid. gm. eq; of Na2CO3 = N x V = (30/1000) x 0.5 = 0.015 and the wt of Na2CO3 = gm eq. x eq. wt. = 0.015 x 35 = 0.795 gm 56 % Na2CO3 = 0.795 / 1.2 = 66.25% % of NaHCO3 = x 100 = 24.5% % of impurities = 100 – (66.25 + 24.5) = 9.25% Ex.: 16 – A 1.2 gm sample of a mixture of NaOH and Na2CO3 with inert impurity is dissolved and titrated with 0.5 N HCl. With pH.ph.th as indicator, the solution, turns colorless after the addition of 30 ml, of the acid. Methyl orange is then added, and 5 ml more of the acid are required for the color to change to pink. What is the % NaOH and of N2CO3 in the sample? Solution: With phenolphthalein 30 ml of the acid is required to neutralize NaOH an to convert the Na2CO3 to NaHCO3. With methyl orange 5 ml of acid is required to neutralize the NaHCO3. The total acid used = 30 + 5 = 35 ml. 10 ml of the acid are required react with Na2CO3 and therefore 25 ml with the NaOH. NaOH = x 100 = 41.67% % Na2CO3 = x 100 = 22.08% PROBLEMS 1. It is found that 20 ml of H2SO4 solution reacted completely with 0.2 of pure NaOH. What was the normality of the acid solution? (2.5 N). 57 2. 0.2 gm of a diacid base required 25 ml of N/10 HCl for complete neutralization. Calculate the molecular weight of the base (160). 3. Find the number of ml of conc H2SO4 (sp. Gr. 1.83 and 93.2% by weight) required to make 500 ml of 1.5 N acid (21.5 ml). 4. How much volume of N. H2SO4 is needed to convert it to 32.8 ml of 0.1452 N H2SO4? (4.76 ml) 5. 25 ml of HCl solution containing 7.3 gm of the acid/l neutralized 30 ml of NaOH solution. A 20 ml of this alkali solution was neutralized with 24 ml of H2SO4 solution, calculate the normality and strength of H2SO4 acid (0.1388 N H2SO4 and 6.8 gm/l). 6. 250ml of conc HCl sp. Gr. 1.1 and containing 20.2% HCl by weight is required to neutralize 30.4 ml of a KOH solution. Determine the normality of the KOH solution (5N). 7. 5 CC of strong H2SO4 are diluted to 500 cc by adding water. On titration 10.2 cc of this diluted acid neutralized 227 cc of N/10 Na2CO3 solution. What volume of water must be added to 400 cc of the diluted acid to make it decinormal. (490.2 cc). 8. 2.5 gm of a monobasic acid was dissolved in water and the foluem made up to 500 ml, 10 ml of this acid solution required 15 ml of N/24 NaOH solution for complete neutralization. Calculate the equivalent weight and molecular weight of the cid (eq. wt = mol. Wt = 96). 9. 0.225 gm of a dibasic acid required 10.84 ml of 0.25 N NaOH for complete neutralization. What is the molecular weight of the acid ? (165.04). 58 10. The density of a commercial sample of H2SO4 is 1.8/ml. 10ml of this acid was made up to liter with water. 20 ml of this dilute acid required 60 ml of N/10 NaOH for neutralization. Calculate the % purity (by weight) of the commercial sample. (81.67%). 11. 0.1 gm of anhydrous organic acid required 19.84 ml of 0.112 N NaOH for neutralization. 0.25 gm of the hydrated acid required 35.4 ml of the same alkali. Calculate the number of molecules of water of crystallization per aquivalent of the anhydrous acid (1). 12. 2.325 gm of a mixture of Na2CO3 and NaCl was dissolved in water and volume raised to 250 ml. 10ml of this solution was completely neutralized by 15 ml of N/15 HCl. Find the % composition of Na2CO3 and NaCl in the given mixture (Na2CO3 = 66.25%, NaCl = 33.75%). 13. 10 cc of a mixture of H2SO4 and HCl solutions required for complete neutralization of 18 cc of N/8 NaOH. 20 cc of the same mixture when treated with excess BaCl2 gave 0.3531 gm of BaSO4. Calculate the amount of HCl in one liter of the mixture. (1.825). 14. If 10 cc of N/2 HCl, 30 cc of N/10 H2SO4 and 40 cc of N/20 HNO3 are mixed together, what will be the normality of the mixture. 15. 25 cc of diluted HCl liberate 10 cc of CO2 at N.T.P. when treated with excess of pure CaCO3, calculate the normality of the acid. 16. A mixt. of Na2CO3 and BaCO3 weighing 0.2 gm requires 30 ml of 0.1 N acid for complete neutralization. What is the % of each compound in the mixt? (55.7% NaaCO3, 44.3% BaCO3). 59 17. A sample of pure (NH4)2SO4 weighing 0.6607 gm is dissolved in water and treated with excess NaOH. The liberated NH3 is passed into 50 ml of HCl. The excess acid then requires 15 ml of KOH solution for titration. The HCl has a normality exactly 10% greater than that of the KOH. Calculte the normality of the HCl. (0.275 N). 18. A sample consisting of Na2CO3, NaOH, and **** weights 1.179 gm. It is titrated with **** ph. Ph. Th as the indicator, and the solution becomes colorless after the addition of48.16ml Mehyl orange is then added and 24.06ml mere of the acid are needed for the dolor change. What is the % of NaOH and Ma2CO5? (24.51% NaOH, 64.95% Na2CO3). 19. If 30ml of H2SO4 are required to neutralize 25 ml of 0.66 N KOH solution: to what volume should 200 ml of 0.66 N.KOH solution; to what volume should 200 ml of the acid be diluted with water in order for the resulting solution be 0.5N? What volume of 1.0 N NaOH should be added to a litre of the KOH in order for the resulting solution to be 0.7 N as a base? What volume of the diluted acid will be neutralized by 25.1 ml of the alkali mixture? (220 ml, 133 ml, 351 ml). 61 Oxidation – Reductions Reaction The nature of redox reactions: A redox (i.e., oxidation – reduction) reaction is on is which the reacting substances undergo changes in examination number. The oxidizing agent becomes reduced, draining electrons and decreasing in oxidation number. The reducing agent becomes oxidized. Losing electrons and increasing in oxidation number. It is important to note that: 1. An oxidizing agent is an electron acceptor, and a reducing agent is an electron donor. 2. The redox process involves a transfer of electrons from the reducer (donor) to the oxidizer (acceptor) A reducer cannot give electrons unless an oxidizer is present to accept them. Thus reduction is always accompanied by an equivalent amount of oxidation. The reagents react in such proportions that the number of electrons furnished by the reducer is exactly accepted by the oxidizer. 3. The oxidized and reduced forms of a substance may be regarded as interconvertible, and are called a rodox conjugate pair. The chemical equation for the interconversion is called a half – reaction. 4. An element in an intermediate oxidation state is theoretically capatic of acting either as an oxidizer or a reducer depending on the relative oxidizing or during ** the other reacting substained for ex : **** iodine may act as an oxidizer if mixed with a strong reducer such as H2SO3. 61 Theory of oxidation reductions: Radox process involves a transfer of electrons (i.e., electricity) from a reducing agent to an oxidisng agent. Thus in the reduction of FeCl3 by SnCl2: 2 FeCl3 + SnCl2 = 2 FeCl2 + SnCl4 Or 2 Fe+++ + Sn++ = 2 Fe+++ + Sn++++ For energy gm atom of Fe+++ iron (56 gm) reduced 96500 coulombs or one Faraday of electricity is lost by the iron and for every gm atom of Sn++ thin (119 gm) oxidized, the latter gains 2 x 96500 coulombs. According to modern theory, an electric current is essentially a transfer of electrons. It should be possible to proof of the transfer of electricity in the oxidation – reduction experimentally. For ex: solutions of SnCl2 and of FaCl3, each acidified with dil. HCl, are placed in separate beaker A and B. The two solutions are connected by means of salt bridge containing KCl. Platinum foil electrodes are introduced into each of the solutions, and the two electrodes are 62 connected to a voltammeter V. When the circuit is closed it will be found that the (negative) current in the external circuit passes from the SnCl2 solution to the FeCl3 solution. After a time, Sn++++ ions can detected in A and Fe++ ions in B. In the above ex., the flow of current is due to E.M.F. (electro- motive force) produced by the chemical change in the redox reaction. Electrode potential: When a metal is immersed in a solution containing its own ions, say, Zn in ZnSO4 solution, a potential difference (e) is established between the metal and the solution. The potential difference (electrode potential) for an electrode reaction M Mz+ + Ze, is given by e = eo + ln CMZt assuming that activity coefficient equal unity. R is the gas constant. T is the absolute temperature, F the Faraday, z the valency of the ions and e is the standard electrode potential e = e when the solution at 25C and the activity equal unity. The above equation can be simplified by introducing the known values of R and F. e = e + log CM2+ For non – motals which yields negative ions the equation becomes. e = e + log CM2- The E.M.F. of the galvanic cell is the algebra sum of the two electrode potentials. Thus 63 E.M.F. of cell = C1 – C2 If the standard electrode potentials e of the elements are calculated and arranged in order give what is known as the electrochemical series of the elements. The standard electrode potential e of hydrogen is arbitray taken to equal zero. The more important standard electrode potentials e are collected in the following table. K / K+ - 2.924 H/H 0.00 Na/Na+ - 2.715 Cu/Cu++ + 0.344 Al/Al+++ - 1.33 Hg/Hg++ + 0.799 Fe/Fe++ - 0.441 Ag/Ag+ + 0.799 Sn/Sn++ - 0.136 Au/Au+++ + 1.36 A metal with a more negative potential will displace any other metal below it in the series from solutions of its salts. Thus Al, Fe, Sn will displace Cu from solutions of its salts. Also C will displace Ag. The standard electrode potential is quantitative measure of the tendency the element to lose electrons. It is therefore, a measure of the strength of the element as a reducing agent, the more negative the potential of the element, the more powerful is its action as a reducing agent. Oxidation – Reduction cells: Reduction is accompanied by a gain of electrons, and oxidation by loss of electrons. In a system containing both an oxidizing agent and its reduction product, there will be an equilibrium between them and electrons. If an inert electrode, such as platinum, is placed in a redox system, for ex : system containing Fe+++ and Fe++ ion, it will assume a definite potential indicative of the position of equilibrium. This potential I 64 known as redox electrode potential. The magnitude of the redox potential will be a measure of the oxidizing or reducing properties of the system. For a Fe++ - Fe+++ chloride electrode PT : Fe++ and Fe++ The redox potential. eredox = e + ln Where Z is the number of electron transfer in the above ex. Z = 1 (Fe+++ + e Fe++) and generally eredox = e + ln where e is known as standard oxidation potential. Credox = e when the ratio of concentration of oxidizing agent Cox to that of the reducing agent Cred. Examples of standard oxidation potentials C is given below. Electrode Electrode reaction e MnO4-, Mn++/Pt MnO4+ 8H+ + 5e Mn++n +4H4O +1.52 Ce++++, Ce+++/Pt Cc++++ + e Ce++++ + 1.45 Cr2O7--, Cr+++/Pt Cr2O7--+14H+ + 6e 2Cr+++ + 7H2O +1.36 Fe+++, Fe++ / Pt Fe+++ + e Fe++ + 0.77 I2; I- / Pt I2 + 2e 2I- + 0.53 Sn++++, Sn++/Pt Sn++++ + 2e Sn++ + 0.15 H+ , H2/Pt 2H+ + 2e H2 0.00 Cr+++, Cr++ / Pt Cr+++ + e Cr++ -0.41 The standard oxidation potentials e help us to know which ions will oxidize or reduce other ions at molar concentrations. The most powerful oxidizing agent are those at the upper end of the table and the 65 most powerful reducing agent at the lower end. Thus permanganate ions can oxidize I and Fe++ ions. Fe+++ ions can oxidize I- ions but not Cr2O7 ions. Equilibrium constants of oxidation – reduction reactions The general equation for an oxidation – reduction electrode may be written: pA + qB + rD … + He sx + ty + uz + … The potential is given by C = e + log Now consider the following reaction MnO4 + 8H+ + 5e Mn++ + 4H2O ( ) ( ) e = e  log ( ) The concentration (or activity) of the water is taken as constant. We are now in a position to calculate the equilibrium constants K of oxidation – reduction reactions, and thus to determine whether such reaction can find application in quantitative analysis. If the value of the equilibrium constant is large, this indicates that the reaction will go from left to right almost to complete. Consider the following reaction. MnO4 + 5Fe++ + 8H Mn++ + 5Fe+++ + 4H2O the equilibrium constant K is given by 66 ( )( ) K= ( )( ) ( ) the water concentration may be assumed constant. The hydrogen – ion concentration is taken as molar. The complete reaction may be divided into two half – cell reactions corresponding to the partial equations. MnO4 + 8H+ + 5e Mn++ + 5Fe+++ + 4H2O (1) and Fe++ Fe+++ + e. (2) For (1) as an oxidation reduction electrode ( )( ) ( ) Since ( )( ) ( ) The partial equation (2) may be multiplied by 5 in order to balance (1) electrically. 5 Fe++ 5Fe+++ + 5e as an redox electrode ( ) ( ) since for the electrode = 0.77 volts. ( ) ( ) 67 Combining the two electrodes into a cell, the E.M.F. will be zero when equilibrium is reached, i.e., e1 = e2 ( )( ) ( ) ( ) ( ) Or ( )( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( ) This results clearly indicates that the reaction proceeds virtually to completion. Fro the general oxidation – reduction reaction a Ox1 + bRedII bOXII + a RedI It can be shown that the concentrations at the stoichiometric point, when equivalent quantities of the two substances OXI and PedII are allowed to react, are given by √ This expression helps as to calculation the exam *** at the equivalence point in at redox *** given above *** the reactive *** solution and a 0.1 N ferrous salt solution in *** of molar sulphuric acid. 68 Then equilibrium constant **** reaction is 3 x 1063. At the equivalent point. ( ) ( ) √ = √ It is evident that for all practical purpose to ** is a quantitative one. Change of the electrode potential during the titration, of a oxidant and a reductant. Oxidation – reduction curves: A suitable is the titration of 100 ml of 0.1 N ferrous iron with 0.1 N ceric cerium. Ce++++ + Fe++ Ce+++ + Fe+++ We are concerned here with two systems, the ferrous – ferric ion electrode (1), and cerous ceric ion electrode (2). For (1) ( ) ( ) ( ) ( ) For (2) ( ) ( ) ( ) ( ) The equilibrium constant of the reaction is ( )( ) Log K = ( )( ) (1.45 = 0.77) = 11.84 69 Or K = 7 x 1011. The reaction is therefore virtually complete. During the addition of the ceric solution up to the equivalence point, its only effect will be to oxidise the ferrous ions and consequently change the ratio (Fe+++) /(Fe++). When 10 ml of the oxidizing agent have been added, (Fe +++), (Fe++) =10/90 approximately. and e1 = 0.77 + 0.059 log 10/90 = 0.69 volt. with 90 ml, e1 = 0.77 + 0.059 log 90/10 = 0.81 volt. with 99 ml, e1 = 0.77 + 0.059 log 99/1 = 0.87 volt. with 99.9 ml, e1 = 0.77 + 0.059 log 99.9 / 0.1 = 0.93 volt. At the equivalence point (100 ml) (Fe+++) = (Ce+++) and (Ce++++) = (Fe++) and the electrode potential is given by 71 The subsequent addition of the ceric solution will increase the ratio (Ce++++) / (Ce+++) thus with 100.1 ml, e2 = 1.45 + 0.059 log = 1.27 volts. with 190 ml, e2 = 1.45 + 0.059 log = 1.45 volts. The results are plotted graphically in the above figure. It is clear that (Fe+++) / (Fe++) and e change abruptly in the region of the equivalence point. Indicators used in oxidation – reduction methods: A. Internal oxidation – reduction indicators An oxidation – reaction indicator should mark the sudden change in the oxidation potential in the neighbourhood of the equivalence point in oxidation – reduction titration. The ideal oxidation – reduction indicator will be one with an oxidation potential intermediate between that of the solution titrated and that of the titrant, and which exhibits a sharp, readily detectable color change. An oxidation – reduction indicator is a compound which has different colors in the oxidished and reduced forms. Such indicators include diphenylamine NH(C6H5)2, which is used in qualitative analysis as a reagent for the NO3- ion. The nitro ion oxidizes diphenylamine (which is colorless in solution) to another compound (diphenyl-benzidine violet), which has a blue – violet color. Diphenyl – amine is also oxidized by many other oxidizing agents with high oxidation potentials, such as KMnO4, K2Cr2O7…etc. 71 It is clear from all this that redox indicators are substances which can be reversibly oxidized or reduced, with different colors in the oxidized and reduced froms. Inox + ne Inred thus it is a redox system, thus. ( ) ( ) Here eo is the standard oxidation potential of the system, i.e., the potential corresponding to the case when (Inox) = (Inred). If 1-2 drops of a solution of some redox indicator is added to a solution of a reducing (or oxidizing) agent, the ion concentration of the oxidized and reduced forms of the indicator will be in a ratio corresponding to that ratio. If this solution is titrated with an oxidizing (or reducing) agent, the oxidiation potential e changes. The (Inox)/(Inred) ratio alters accordingly. However, with acid base indicator, not energy change of this ratio corresponds to a color change which can be detected by the eye. Usually the color change for a redox indicator will be over the potential range (eoIn + 0.05) to (eoIn – 0.05) volts. For a sharp color change at the end point, eoIn should differ by at least 0.15 volt from the other systems involved in the reaction. During the titration of ferrous salt potassium dichromate solution, in order to gain a satisfactory end point (an intense blue – violet coloration), phosphoric acid must be present, this forms a complex with the ferric ions, thereby reducing the concentration of the latter. The action of diphenylamine (I) as an indicator depends upon its oxidation first into colorless dipheylbenzidine (II), which is the real indicator and is reversibly further oxidized to dipheneybenzidine violet (III). 72 Examples of some oxidation indicators are: Inox Inred eo Neutral red Red Colorless 0.24 Methylene blue Greenish blue Colorless 0.53 Diphenylamine Blue violet Colorless 0.76 Phenylanthranilic Acid Red violet Colorless 1.08 Ferroin Pale blue Red 1.14 B. The reagent may serve as its own indicator: Titration without an indicator is possible for ex: when various reducing agents are oxidized by permanganate in acid solution. The purple – violet color of the MnO4- ion disappears owing to reduction to the almost colorless Mn++ ion. When all the reducing agent has been titrated a single excess drop of permanganate gives a visible pink coloration to the solution even in the presence of slightly colored ions such as ferric ions. Similarly, reducing agents can be titrated with iodine solution without the use of indicators, because the dark brown color of the iodine disappears as the result of reduction of I2 to I ions. However, since the color of I2 solution is not very deep, it is convenient in such cases to use 73 an indicator – starch solution, which gives an intense blue color even with very small amounts of free iodine. The use of starch is based on its ability to form a blue adsorption compound with iodine. C. External indicators: The best – known example of an external indicator in a redox process is the spot test method for the titration of ferrous iron with standard potassium dichromate solution. Near the equivalence point, drops of the solution are removed and brought into contact with dilute freshly prepared potassium ferricyanide solution on a spot plate. The end point is reached when the drop first to give a blue coloration. Permanganate titration: KMnO4 is strong oxidizing agent. Oxidation may proceed in acid or neutral or alkaline medium. When KMnO4 acts as an oxidizing agent in acid solution the Mn7+ in it is reduced to Mn++ ions and the manganous salt of the acid used is formed, thus in presence of H2SO4. 10FeSO4 + 2KMnO4 + 8H2SO4 = 5Fe(SO4)3 + 2MnSO4 + K2SO4 + 8H2O Thus the eq. wt = During oxidation in neutral or slightly alkaline, the Mn7+ ion is reduced to Mn4+ with the formation of MnO2 in the form of a brown precipitate. In this case eq. wt = In strongly alkaline solution, the Mn7+ ion is reduced to Mn6+ or in other words the permanganate ion MnO4 is reduced to manganate ion MnO4. In this case the eq.wt= mol. wt = 158.04. 74 Dichromate titration: K2Cr2O7 is not so powerful an oxidizing agent as KMnO4, but it has several advantage over KMnO4 for ex: K2Cr2O7 is an excellent primary standard. K2Cr2O7 is used only in acid solution, and reduced rapidly to a green chromic salt. Thus the chromium ion in the dichromate reduces from Cr6+ to Cr+++ is Cr2O7 + 14H+ + 6e = 2Cr+++ + 7H2O This equation shows that if K2Cr2O7 is used for oxidation then the eq.wt = The usual indicator in the dichromate method is diphenylamine. It is possible to use phenylanthranilic acid. Iodimetry and iodometry: Iodimetry: covers titration with a standard solution of iodine. Iodometry deals with the titration of iodine liberated in chemical reactions. Iodine is a much weaker oxidizing agent than KMnO4 and K2Cr2O7. Strong reducing agents such as stannous chloride, H2SO3, and sodium thiosuphate, react completely and rapidly with iodine even in acid solution. The equations of some reactions are: Sn++ + I2 = Sn++++ + 2I SO3 + I2 + H2O = SO4 + 2H+ + 2I 2SO2O3 + I2 = S4O6 + 2I If a strong oxidizing agent is treated in neutral or acid solution with a large excess of iodine ion I, the latter I reacts as a reducing agent and 75 the oxidizing agent will reduced. In such cases, an equivalent amount of iodine is liberated, as is then titrated with a standard solution of a reducing agent which is usually Na2S2O3. For ex: 2MnO4 + 16H+ + 10I = 2Mn++ + 5 I2 + 8H2O Cr2O7 + 2H+ + 2 I = 2H2O + I2 IO3 + 6H+ + 5 I = 3 I2 + 3H2O 2HNO2 + 2H+ + 2 I = 2NO + I2 + 2H2O Detection of the end point: A solution of iodine in aqueous iodide has an intense yellow to brown color, so that iodine can serve as its own indicator. The test is made much more sensitive by the use of a solution of starch as indicator, starch reacts with iodine in the presence of iodide to form an intensely blue coloured adsorption complex containing both iodine and iodide which is visible at very low iodine concentrations: Ex.: 1- What is the normality of a solution of KMnO4 if 40 ml will oxidize 0.3 gm of Na2C2O4? What is the value of 1.0 ml of the permanganate in terms of grams of FeSO47H2O. Solution: milieq. wt. of oxalate = = 0.067 milli equivalent = = 4.476 at the end point milligram eq. of oxalate = milligram eq. of KMnO 4 = 4.4776 N x V = millieq. 76 N x 40 = 4.4776 N = 0.1119 The expression the value of 1.0 ml of the permanganate in terms of grams of FeSO47H2O means the number of grams of FeSO47H2O capable of being oxidized by 1.0 ml of KMnO4 solution. 1.0 x 0.1119 x = 0.3108 gm of FeSO47H2O Ex: 2- What is the normality as an oxidizing agent of a solution of K2Cr2O7 containing 9.806 gm/l, and what volume of 0.1 N ferrous ammonium sulphate will 30 ml K2Cr2O7 oxidise in the presence of acid? How many gms of FeSO4(NH4)2. SO4.6H2O are contained in each milliliter of the ferrous solutions? Solution: The eq. wt of K2Cr2O7 = 49.03 N= = 0.2 N N x V for reducing agent = N1 x V1 or oxidizing 0.1 x V = 30 x 0.2 V = 60 ml For ml of N solution of ferrous salt = ( ) = 0.3922 gm/ml For 0.1 N ferrous solution needed 0.03922 gm/ml Ex.: 3- What is the % of Fe and of Fe2O3 in an iron ore if 0.5 gm of the ore, after solution in acid and complete reduction of the iron, requires 25.5 ml of KMnO4 (1 ml of it equivalent 0.0126 gm 77 H2C2O4.2H2O) for oxidations? What volume of this KMnO4 would be required for a 10 sample of 3% (by weigh) H2O2 solution. Solution: the normality of KMnO4 N x V for KMnO4 = gm eq of oxalic acid Nx1 = N of KMnO4 = 0.2 N ( ) x 100 = 56.96% Fe and x 100 = 81.45% Fe2O3. x 100 = 3 V = 88.211 of KMnO4. Ex.: 4- A sample of pyrolusite weight 0.5 gm. To this is added 0.6674 gm of As2O3 and dil acid. After solvent action has ceased, the excess three – valet arsenic is titrated with 45 ml of 0.1 N KMnO4. Calculate the oxidising power of the pyrolusite in terms of % of MnO2. Solution: Number of gm millieq. of As2O3 added is = 13.5 Number of gm. millieq. of KMnO4 = 45 x 0.1 = 4.5 number of gm milli equivalent of MnO2 78 = 13.5 – 4.5 = 9 gm milli eq. of MnO2. The % of MnO2 = x 100 = 78.25% Ex.: 5- Given a solution of I2 (1 ml equivalent to 0.004946 gm, As2O3) and a solution of Na2S2O3 (20 ml equivalent to 0.09351 gm KBrO3). What is the normality of each solution, and what volume of 0.1 N KMnO4 would liberate an amount of I2 from excess KI to required 40 ml of the above thiosulphate for reduction? Solution: N x V for I2 = gm millieq. of I2 = gm. millieq. of AS2O3. N of I2 = = 0.1 N Also the normality of Na2S2O3 = = 0.15 N NxV=NxV 0.1 x V = 0.15 x 40 V = 60 ml of KMnO4. Problems 1. What is the percentage of iron in a sample of iron wire, if the titration FeSO4 solution, formed by dissolving 0.14 gm of the wire in H 2SO4 without excess of air, took 24.85 ml of 0.1 N KMnO4 solution? 79 2. For standardization of Na2S2O3 solution, 0.1125 gm of chemically pure copper was weighed out, dissolved and treated with KI as in iodometric determination of copper. The iodine liberated was titrated with the thiosulphate solution, 18.99 ml was required. Calculate the normality of N2S2O3 solution? 3. For determination of H2S in a solution, 50 ml of 0.0196 N iodine solution was added to 25 ml of the H2S solution, the excess of iodine took 11 ml of 0.0204 N Na2S2O3 solution. Find the H2S content of the solution in gm/l. (0.515 gm/l). 4. How many gms/ml does a solution KNO2 contain if it is 0.1 N as a reducing agent? How many gms of SO2 is contained in a liter of a solution of H2SO3 which is 0.0586 N as a reducing agent? (0.00425 gm and 1.877 gm). 5. If 100 ml of a solution contain a mixture of 0.158 gm of KMnO 4 and 0.49 gm of K2Cr2O7, what is its normality as an oxidizing agent in the presence of acid ? How many ml of 0.1 N FeSO4 solution will 40 ml of the above solution o

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