From: [J--b--a] at [hotmail.com] Newsgroups: alt.drugs.chemistry,alt.drugs,rec.drugs,rec.drugs.chemistry Subject: Hydriodic Acid FAQ 1.0 by Iyemtetum Date: Sun, 05 Oct 1997 19:39:28 GMT The Hydriodic Acid FAQ 1.0 By: Iyemtetum This FAQ covers only one of many methods for the production of Hydriodic acid. No information regarding its use is included. This procedure requires the use of iodine, red phosphorus, and distilled water. Iodine has the chemical symbol I. It occurs in its stable form as a diatomic molecule, I2. Iodine has a formula weight of 253.809. 1 mol of iodine has a mass of 253.809 grams. Red phosphorus has the chemical symbol P. In its stable form, phosphorus occurs as four molecules bound together, P4. It has a formula weight of 30.973. 1 mol of phosphorus has a mass of 30.973 grams. Water has a chemical formula of H2O. Its formula weight is 18.015. 1 mol of water has a mass of 18.015 grams, or 18.015 milliliters. Hydrogen iodide gas has a chemical formula of HI. Its formula weight is 127.912. 1 mol of HI gas has a mass of 127.912 grams. Phosphoric acid has a chemical formula of H3PO4 its formula weight is 99.995. 1 mol of phosphoric acid has a mass of 99.995 grams. Hydrogen iodide gas can be formed through the reaction between iodine, water, and red phosphorus. This is shown by the chemical equation below. 16(H20) + 10(I2) + 2(P4) ---> 20(HI) + 4(H3PO4) +P4 16 mol of water added to a mixture of 10 mol iodine and 2 mol red phosphorus yields 20 mol hydrogen iodide gas, 4 mol phosphoric acid, and 1 mol red phosphorus. The extra mol of red phosphorus in the first half of the equation is to ensure that the reaction will be carried out. As you can see, the additional mol remains unchanged, and can be filtered out for later use. You can NOT add 16 grams of water to 10 grams of iodine and 2 grams of phosphorus to produce 20 grams of hydrogen iodide gas, 4 grams of phosphoric acid, and one grams of phosphorus. In order to find the quantities of reagents necessary to carry out this reaction, the mol’s must be converted to their mass equivalent. For example: 16 mol H2O 18.015 grams -------------------- * ---------------- = 288.82 milliliters H2O 1 1 mol The mol units cancel out leaving grams. 16 mol H2O = 288.82 milliliters of water 10 mol I2 = 2538.09 grams 2 mol P4 = 61.95 grams 20 mol HI = 2558.24 grams 4 mol H3PO4 = 399.98 milliliters 1 mol P4 = 30.97 grams This give you: 288.82 milliliters of water plus 2538.09 grams of iodine plus 61.95 grams of red phosphorus will produce 2558.24 grams of hydrogen iodide gas plus 399.98 milliliters of phosphoric acid plus 30.97 grams of red phosphorus. A 57% solution of hydriodic acid can be produced by bubbling 57 grams of HI gas through 100 milliliters of water. To ensure the required concentration is obtained, 60 grams of HI gas for every 100 milliliters of water will provide a sufficient safety net. The above quantities of reagents will produce enough HI gas to make approximately 4 Liters of hydriodic acid. How To Make 300 ml of Hydriodic Acid Materials: 198.42 grams Iodine 4.84 grams red phosphorus 250ml Erlenmeyer flask 500ml Erlenmeyer flask filled with 300 ml distilled water one hole stopper two hole stopper addition funnel filled with 22.52 ml distilled water long stem medicine dropper inch section of glass tubing 6 inches of rubber tubing hot plate Procedure: Step 1. In a 250 ml flask, mix 198.42 grams of iodine and 4.84 grams of red phosphorus. Step 2. Prepare the two hole stopper by inserting the addition funnel through one hole, and the 2 inch section of glass tubing in the other. This stopper will go into the 250 ml flask containing the red phosphorus and iodine. Make sure that both the stem of the addition funnel and a portion of the glass tubing are exposed on the side of the stopper that will be inside the flask. Step 3. To the exposed end of the glass tubing, attach the 6 inch section of rubber tubing. At the other end of the rubber tubing insert the medicine dropper. Step 4. Insert the medicine dropper into the one hole stopper and LOOSELY insert the stopper into the 500 ml flask filled with 300 ml water. Step 5. Put some goggles on and slowly open the addition funnel valve to allow water into the 250 ml flask. A significant amount of heat will evolve. HI gas will begin bubbling through the rubber tubing into the water in the 500 ml flask. If you are adventuresome, VERY VERY gently heat the 500 ml flask with the hot plate. Take precautions to ensure that the pressure inside the flask does not become too great. The increased pressure will increase the amount of HI gas that can be absorbed by the water, however this does not benefit very much because as the pressure is equalized when you remove the stopper, any HI gas that was absorbed over 57 grams will become gaseous once again. Congratulations, you have succeeded in producing approximately 300 ml of 57% hydriodic acid. If the quantities of the chemicals are changed at all, do it proportionally!! If they are not changed proportionally, another reaction could occur involving the production of a phosphoric acid, and much less HI than the method detailed. For those of you interested in the phosphoric acid.... Maybe later. Have fun, be careful, Don't take any of this as law, research, enjoy. Special thanks to Phiber Optik for all of his help! Also, Scott, Thanks for for the constructive criticism. I hope you agree with all this!