CHAPTER XVI OXIDATION-REDUCTION CALCULATIONS NORMAL SOLUTIONS. FACTOR VALUES 258. There is not much difficulty in making the calculations which are necessary in connection with oxidation-reduction titrations if we base them upon the changes of charge that are involved. For simple ions the change of charge in going from one state of oxidation to the other is directly obvious, thus, For complex ions the change of charge is not so directly obvious and must be calculated after the manner described in § 248, namely, first construct the ionic equation which represents the transformation, by writing down as the first term of its left-hand member the complex ion under consideration, and as the first term of the right-hand member, the ion (simple or complex) which results from reduction; as the second term of the left-hand member, the appropriate number of hydrogen ions necessary to form water with the number of oxygen atoms which the complex ion loses in the reduction, and as the second term of the right-hand member the number of moles of water formed. The change of charge attendant upon the transformation will be given by the balancing of the charges, thus, 1. permanganate ion → manganous ion MnO4+8 H+ → Mn+++ 4 H2O 3. thiosulphate ion → tetrathionate ion 2 S203S40☎ 4 Ꮎ = = 20 +20 the change of charge is one per mole of thiosulphate ion 4. oxalate ion → carbon dioxide C2O→ 2 CO2 2 Ꮎ = zero +2 → For the halogens, we have, taking iodine as an illustration, 2 II2 2 Ꮎ = zero + 2e the change of charge is one per mole of iodide ion. Illustration 1. Suppose that in titrating a solution of stannous chloride by means of 0.1034 molar ferric chloride, it required 46.54 c.c. of the ferric chloride solution, and we want to know how much stannous chloride this represents. We have for the changes of charge that are concerned Fe+++ →Fe++ gives up 1 Sn++ → Sn++++ requires 2 → whence 1 mole Sn++ = 2 moles Fe+++ Consequently, 46.54 c.c. of 0.1034 M FeCl3 = 23.27 c.c. 0.1034 M SnCl2. Since the molecular weight of SnCl2 is 189.9, we have that 1 c.c. of 1 M SnCl2 = 0.1899 g. SnCl2, and therefore 23.27 c.c. 0.1034 M SnCl = 23.27 X 0.1034 X 0.1899 g. SnCl = 0.4569 g. SnCl2. Illustration 2. Suppose that in standardizing a solution of approximately 0.03 molar potassium permanganate by means of sodium oxalate we have found as a result of three determinations that 1 c.c. of permanganate solution is the equivalent of 0.01033 g. sodium oxalate and that we wish to express the value of the permanganate solution in terms of iron. We have for the changes of charge that MnO→ Mn++ gives up 5 → C2O2 CO2 requires 2 Fe++ Fe+++ requires 1 whence 2 moles MnO4 = 5 moles C2O4 = 10 moles Fe+++ or 2 KMnO4 = 5 Na2C2O4 = 10 Fe From the relationship between the sodium oxalate and the iron we can construct our stoichiometrical equation, viz., Na2C2O4 2 Fe: 0.01033 g. x which upon solving gives x = 0.008608 g., that is, 1 c.c. of the permanganate solution corresponds to 0.008608 g. Fe. n moles 259. Normal Solutions. For oxidizing and reducing agents a normal solution is one which contains per liter of solution that weight of reagent which would, in accordance with the proper stoichiometrical equation, oxidize 126.9 grams of iodide ion or reduce 126.9 grams of iodine. Another way of defining normality as applied to oxidizing and reducing agents is to say that a nor1 mal solution is one which contains per liter of solution of reagent where n represents the change of charge which the reagent undergoes in the reaction. The criticism put forward in § 124 that the scheme of normality as a basis of definition is inadequate and should be abandoned applies a fortiori here. Thus a normal solution of potassium permanganate (mol. wt. 158.03) may contain either 31.61 g., or 39.51 g., or 52.68 g. KMnO4 per liter of solution according to the particular reaction in which it is employed (see § 250). 260. Factor Values. As in § 125, we can apply this system to oxidation-reduction reactions in a similar manner. Thus suppose that we have a solution of potassium permanganate 1 c.c. of which is the equivalent of 0.006024 gram of iron, and we wish to take such a weight of iron ore for analysis that each c.c. of permanganate solution used in titration shall correspond to 2% of iron in the ore. The formula is W = 100 f where W = desired weight of sample in grams = Ρ % constituent to be represented by 1 c.c. of standard solution f = number of grams of constituent corresponding to 1 c.c. of standard solution. namely, we take a weight of sample equal to 0.3012 g. 261. Examples 1. A solution of potassium permanganate contains 3.156 g. KMnO4 per liter. Find its value in terms of iron, ferrous oxide, and ferric oxide. 2. The tin from 0.2800 g. of a sample of block tin, when in hydrochloric acid solution and in the stannous condition, was titrated with 0.0997 M FeCl solution, 46.54 c.c. being required. Calculate the percentage of tin in the sample. Ans. 98.35% 3. To 60.0 c.c. of a stannous chloride solution an excess of approx. 0.1 M FeCl3 solution was added, and the resultant solution then titrated with 0.1600 M K2Cr2O7 solution, 40.0 c.c. of the latter being required. How much SnCl; per c.c. did the original solution contain? Ans. 0.0608 g. SnCl2 per c.c. 4. For the titration of ferrous ion, how many g. of potassium dichromate per liter of solution will give a solution equivalent to a solution of potassium permanganate containing 2.500 g. per liter? Ans. 3.878 g. 5. In standardizing an approx. 0.1 M KMO4 solution by means of sodium oxalate the following results were obtained: What was the average deviation? What was the molarity of the KMnO4 solution, and what was its value in terms of iron? 6. The calcium oxalate precipitate from 1.1450 g. of a sample of rock was dissolved in dilute sulphuric acid and titrated with 0.01995 M KMnO4, requiring 11.30 c.c. Find the percentage of CaO in the rock. Ans. 2.76% 7. 100 c.c. of a sample of commercial hydrogen peroxide solution were diluted to 1,000 c.c., and 10 c.c. of this diluted solution were titrated with titanous chloride solution (1 c.c. of which = 0.005156 g. Fe), the titration requiring 23.1 c.c. How many g. of H2O2 were present per 100 c.c. of original H2O2 solution? For method, see Knecht & Hibbert, Ber. 38, 3324 (1905). Hint, H2O2 → 20H ̄ + 20 Ans. 3.63 g. H2O2 per 100 c.c. 8. 1.000 g. of a sample of chrome-iron ore FeO Cr2O3 was oxidized by sodium peroxide, and one-fifth of the resulting chromate was treated with 35.24 c.c. of a ferrous ammonium sulphate solution, which proved to be in slight excess. The excess of ferrous ion was titrated back with 2.72 c.c. of a potassium dichromate solution. The values of the solutions were as follows: 1 c.c. Fe(NH4)2(SO4)2 solution = 0.005823 g. Fe; 1 c.c. K2Cr2O7 solution = 0.006257 g. Fe. Find the percentage of chromium in the ore. Ans. = 29.20% 9. The phosphorus in 5.000 g. of a sample of iron ore was precipitated as ammonium phospho-molybdate. The precipitate was dissolved, etc., and the solution passed through a Jones reductor. It then required 22.68 c.c. KMnO4 solution (1 c.c. = 0.005988 g. Fe) for the titration. Find the percentage of phosphorus. Note. After passing through the reductor the molybdenum is in the form of Mo+++; the permanganate oxidizes this to MoO ̄. Hint, see § 188. Ans. 0.042% 10. Given a solution of KMnO4, 1 c.c. = 0.008637 g. of Na2C2O4. What weight of iron ore must be taken so that each c.c. of KMnO4 solution used in titration shall correspond to 2% of iron in the ore? Ans. 0.3600 g. 11. How many g. of K2Cr2O7 per liter of solution will be required in order that 1 c.c. of solution shall correspond to 1% of iron when 0.500 g. of iron ore is taken for analysis? Ans. 4.391 g. 12. A solution of sodium thiosulphate is freshly prepared and 48.00 c.c. are required to titrate the iodine liberated from an excess of KI solution by 0.3000 g. of pure KIO. What is the normality of the thiosulphate and its value in terms of iodine? L. F. Hamilton and S. G. Simpson, loc. cit., § 13, p. 121. Ans. 0.1752 normal. 0.02224 g. |