CHAPTER XXIX THE DETERMINATION OF LEAD IN ORES Lead occurs in ores chiefly as the minerals galena (PbS), cerrusite (PbCO3) and anglesite (PbSO4). It also occurs as molybdate, phosphate and arsenate. Lead ores are completely soluble in acids, hence require no fusion. Lead is determined satisfactorily either by gravimetric or volumetric methods. The gravimetric methods depend upon the precipitation of lead as lead sulfate or electrolytically as PbO2. Volumetric methods depend upon the precipitation of lead as lead molybdate or lead chromate. THE ELECTROLYTIC DETERMINATION OF LEAD Lead is precipitated on the anode from a nitric acid solution as PbO2. The precipitation is rapid, owing to the high atomic weight of lead and the small solution tension of PbO2 in nitric acid. The precipitated PbO2 does not adhere well to a smooth, flat electrode, but adheres well to a gauze electrode, so well, in fact, that it cannot be rubbed off with the finger. If bismuth or antimony are present, they will partially deposit with the lead, and manganese will also deposit as MnO2 with the lead unless the solution is very acid. The presence of arsenic, selenium or tellurium will prevent the precipitation of the PbO2 partially or wholly. If any of these are present they must be removed by chemical means, which is easily done. Phosphoric acid, if present, will prevent the precipitation of lead. Process of Analysis for Ores Free from Arsenic, Antimony or Bismuth. Weigh an amount of sample containing not more than 0.5 gram of lead. Put it in a 250-c.c. beaker, add 10 c.c. of strong HCl and heat to boiling for several minutes. Then, when most of the sulphur is driven off as H2S, add 15 c.c. of HNO3 and continue the heating until the ore is all decomposed and finally boil vigorously to expel all chlorine. There must be no chlorine present during the electrolysis. Add 20 c.c. of water and then NH4OH until there is considerable excess and heat until any lead sulfate is dissolved. Then add HNO3 until there is 15 c.c. excess present, dilute to 100 c.c. and electrolyze, using a gauze anode. With a cylindrical gauze anode 34 in. in diameter and 21⁄2 in. high a current of 3 amp. and a potential of about 3 volts will cause complete deposition in 15 minutes to a half hour. It is best to electrolyze in a hot solution. Break the current and set the anode in a beaker of pure water, then remove and dry at a temperature of 230° in an oven. Cool and weigh. Multiply the PbO2 by 0.8640 to get the weight of the lead. To remove the PbO2, set the electrode in a beaker containing warm 1:3 HNO3 and add a few cubic centimeters of alcohol or formic acid. nitric acid. The PbO2 is reduced to PbO, which is soluble in If the ore is soluble in nitric acid, the HCl may be dispensed with and the sample treated directly with the nitric acid. This will cause the formation of considerable lead sulfate, which will dissolve when the solution is made alkaline and heated. Method of Ores Containing Antimony or Bismuth.-Dissolve the ore as above directed and then add 50 c.c. of 10 per cent H2SO4 which has been saturated with lead sulfate by adding a few drops of lead acetate, shaking well and allowing to stand until lead sulfate settles clear. Evaporate to very copious fumes of H2SO4. Cool and add 30 c.c. of water containing 3 grams of tartaric acid to keep the antimony in solution. Heat to boiling to dissolve all soluble salts, cool, filter and wash three times with cold water. Wash the lead sulfate back into the beaker with a small amount of water and pour through the filter into the beaker hot ammonium nitrate solution, made by adding to 30 c.c. of 1:1 nitric acid 10 c.c. NH4OH. This will dissolve any lead sulfate left on the paper. Digest the solution in the beaker until the lead sulfate is all dissolved, then add nitric acid until there is 15 c.c. excess present, heat and electrolyze as above. To test the purity of the deposit the following method is good: Place the anode in a beaker containing 40 c.c. of 1:3 nitric acid, add a few cubic centimeters of formic acid and heat until all PbO2 is dissolved. Wash the anode, catching the washings in the beaker, add 5 c.c. of sulfuric acid and evaporate to copious fumes. Cool, add 5 c.c. of alcohol, Cool, add 30 c.c. of water and heat to boiling several minutes to change any lead bisulfate to sulfate. allow to settle and filter in a weighed several times with 10 per cent alcohol. a barely visible red over a Bunsen burner. the weight of lead sulfate by 0.6830. Ores may also be analyzed in this way if the analyst prefers to weigh lead sulfate instead of PbO2. That is, the ore is dissolved as in the method for ores free from bismuth and electrolyzed. Then the PbO2 (with oxides of bismuth and antimony and manganese, possibly) is dissolved with nitric and formic acids and evaporated to copious fumes. Then 30 c.c. of water are added and boiled, cooled, filtered and the lead sulfate washed with 10 per cent alcohol. The Gooch crucible and the precipitate are heated for five minutes over a Bunsen burner, cooled and weighed. No elements interfere and very good results are obtained. Permanganate Method. The PbO2 obtained by electrolysis may be impure from the presence of arsenic, antimony or manganese. The error due to the presence of arsenic and antimony, which may have precipitated because of using the direct method of electrolysis without previous removal of arsenic and antimony, may be corrected as follows: Put the anode with the deposited PbO2 in a tall beaker, cover with water and add 5 c.c. of HNO3 free from nitrous acid, add standard H2O2 solution until the PbO2 is completely dissolved. Now titrate the excess H2O2 promptly with standard permanganate. Add the same amount of H2O2 solution as used and titrate again with permanganate. The difference between the amounts of permanganate used in the two titrations gives the permanganate equivalent to PbO2. The H2O2 solution is made by dissolving 20 c.c. of U.S.P. H2O2 in a liter of water and adding 50 c.c. of HNO3. The reactions are: PbO2 + H2O2 + 2HNO3 = Pb(NO3)2 + 2H2O + 02. 5H2O2 + 2KMnO4 + 6HNO3 = 2KNO3 + 2Mn(NO3)2 + 8H2O + 502. Hence 2KMnO4 = 5Pb = 10Fe and, theoretically, 5Pb 710Fe = 1.851, but actually the ratio is found to be 1.92. REFERENCES: SMITH, "Electroanalysis." PERKIN, "Practical Methods of Electrochemistry." BENNER and Ross, "Electrolytic Determination of Lead in Ores," LIST, Metal. Chem. Eng., 10, 135. WOICIECHOWSKI, Met. Chem. Eng., 10, 108. For the determination of lead by the molybdate method of Alexander, see Low, "Technical Methods of Ore Analysis." This method is less reliable, and no more rapid, than the dichromate method, and so is not given here. For the permanganate method for lead, see BOLLENBACH, Chem. Ztg., 33, 1142. THE VOLUMETRIC CHROMATE METHOD FOR LEAD The following method is the one devised by Guess and modified by Low and Waddell. Low says, "I find it more generally satisfactory than any other." It depends upon the precipitation of lead chromate from an acetic acid solution which is then dissolved in hydrochloric acid, potassium iodide added and the liberated iodine titrated with standard thiosulfate solution. Solutions Required. Extraction Solution.-Make a cold saturated solution of sodium acetate and filter it. Dilute it with two volumes of water and add 30 c.c. of 80 per cent acetic acid per litre. Hydrochloric Acid Mixture.-Make a cold saturated solution of NaCl and filter it. To 1 liter of the salt solution add 250 c.c. of water and 100 c.c. of hydrochloric acid (sp. gr. 1.2). Potassium Dichromate.-Make a cold saturated solution of the commercial salt and filter it. Starch Solution.-Make as directed on page 97. Process of Analysis.-Weigh 0.5 gram of the ore and put it into a 150-c.c. flask. Add 10c.c. of strong HCl and heat until no more H2S comes off. Add 5 c.c. of nitric acid and boil until the ore is completely decomposed. Now add 10 c.c. of 1:1 sulfuric acid and boil until copious white fumes come off. Cool, add 50 c.c. of water and heat to boiling until all the soluble salts are dis solved, cool and add 5 c.c. of ethyl alcohol, allow to settle, then filter through a 9-cm. filter and wash with cold 10 per cent sulfuric acid solution five times and then twice with water. Have the extraction solution nearly boiling, and with a fine jet wash the lead sulfate back into the flask, then wash the filter thoroughly with the hot solution until all lead sulfate remaining on it is dissolved, catching the washings in the flask containing the rest of the lead sulfate, and taking care to wash under the folds of the filter paper. Heat the filtrate to boiling and add more of the acetate solution if necessary to dissolve all of the lead sulfate. Finally, dilute to 150 c.c., heat to boiling, add 10 c.c. of the dichromate solution and boil for seven minutes. It is necessary to boil about this length of time to insure always the same constitution of the lead chromate. Now filter through a large filter paper and wash the flask and precipitate 10 times, with a hot solution of sodium acetate made by diluting 50 c.c. of a cold saturated solution to 1 liter. Place the clean flask under the funnel, and with a jet of the cold HCl mixture dissolve the precipitate on the filter. Continue the washing and stirring up of the precipitate with the HCl mixture until all the residue and all color are removed from the filter. Use at least 50 c.c. of the mixture. Now add 4 c.c. of a 25 per cent solution of KI and titrate at once with standard sodium thiosulfate solution. Continue adding the thiosulfate until the brown color of the liberated iodine becomes faint; then add enough starch solution to produce a strong blue color and continue the titration until the solution becomes a pale green with no tinge of blue. The end point is very sharp. Standardize the thiosulfate solution on pure lead. The reactions in the determination are: 2Pb(C2H3O2)2 + K2Cr2O7 + H2O = 2PbCrO4 + 2KC2H3O2 + 2HC2H3O2. 2H2CrO4 + 6KI + 12HCl=6KCl + 2CrCl3 + 8H2O + 61. 2Na2S2O +21=2NaI + Na2S406. Therefore, 1Pb equals 1H2CrO4, which liberates 31, which equals 1Na2S2O3. That is, to titrate the iodine liberated by the H2CrO4 from 1Pb requires 3Na2S2O3 5H2O. Then to make a solution of thiosulfate, 1 c.c. of which will equal 0.005 gram of |