# Last edited on 2017-04-11 23:56:17 by jstolfi SCRIPT Introduction When only one of the hydrogen ions is removed, what is left behind is the ion HC2O4-, with a single negative charge, called "hydrogenoxalate". When both hydrogens are removed, the resulting ion C2O4-- is called "oxalate". Note that oxalate (like carbonate CO3--) has no hydrogen, and is therefore one of the so called "oxocarbon anions", consisting only of carbon and oxygen. Goals and methods In this experiment we will make some sodium hydogenoxalate Na+ HC2O4-, also called "monosodium oxalate"; and sodium oxalate (Na+)2 C2O4--. We will obtain both substances by reacting solutions of oxalic acid and sodium hydroxide NaOH in water, by the reactions H2C2O4 + NaOH --> NaHC2O4 + H2O H2C2O4 + 2 NAOH --> Na2C2O4 + 2 H2O The reagents and the desired salt are all soluble in water. Therefore, mixing stoichometric solutions of H2C2O4 and NaOH should result in a solution of the desired salt. After the reaction, our plan is to remove all the water by evaporation, obtaining a solid that is the desired salt. If we are lucky, when we evaporate that solution, the first crystals that will form will be the desired salt. As that salt is removed from the solution, its composition will remain the same; and therefore continued evaporation will leave crystals of the desired salt only. If we are unlucky, the first substance to crystallyze would be some unwanted salt, e.g. the oxalate instead of the expected hydrogenoxalate; or a mixed salt containing both anions. In that case the composition of the solution will change as the water evaporates, and the final solid may be a mixture of various salts. In that case, we may have to look for another way to separate the solid salt. Let's hope we are lucky. The reagents and the other products Oxalic acid is a chemical compound with formula C2H2O4, that lies at the intersection of organic and inorganic chemistry. Its common uses include rust removal and marble polishing. Oxalic acid is responsible for the sour taste to some fruits, like carambola, and or certain leaves, like sorrel and rhubarb. Unilike other acids that serve that function (like citric, tartaric, maleic, ascorbic, and others), oxalic acid and its salts can be toxic if consumed in excessive amounts, especially to people with kidney problems. Oxalate removes calcium from the blood, which may impair neuron functions, and deposits it in the kidneys as insoluble calcium oxalate, which may form kidney stones. Moreover, oxalic acis is a strong acid: its dust is very harmful if inhaled, and solutions can be absorbed through the skin and cause burns. The acidity of the compound comes from the two carboxyl groups -COOH that form the molecule; specifically, from the hydrogen atom in each group. That atom is easily removed from the carboxyl, in the form of the positive ion H+ (a proton), leaving behind its electron that gives the remaining -COO group a negative charge. Indeed, oxalic acid is the simplest "dicarboxilic" acid, and organic acid with two carboxyl groups. ?? Oxalic acid and oxalates are fairly stable, but Sodium oxalate decomposes above 260 C releasing carbon monoxide. Hydrates To get the right amounts of the reagents, and check the mass of the result, we must take into account the fact that the solid forms of some of these substances contain some water, loosely bound to the metal ion and/or trapped in the crystal structure. ?? Note that hydration water contributes to the mass of the solid, and contributes some water to the solution. That is, when dissolving 10 grams of CaCl2(H2O)6 in 100 ml of water, we are actually dissolving a bit less than 10 grams of CaCl2 in a bit more than 100 ml of water. Thus, when weighting a desired amount of a solid substance, one must know which hydrate it is. ?? Reagent amounts We arbitrarily choose to make 50 g of each sodium salt. That will be enough for other experiments that we plan to make later on. From that we can compute how much of each reagent we need. First, we convert those 50 g to mols of the formulas NaHC2O4 and Na2C2O4: [ Recipe ] To make one mol of NaHC2O4 we need one mol of H2C2O4 and one mol of NaOH. Converting back to grams, we get: [ Recipe ] That will make also 1 mol of H2O. To make one mol of Na2C2O4 we need one mol of H2C2O4 and two mols of NaOH. Converting back to grams, we get: [ Recipe ] That will make also 2 mols of H2O. Water amount We will use the minimum amount of water needed to dissolve both the reagents and the salt produced. The less water we use, the less we will need to evaporate at the end. It turns out that the water from the two solutions that remains in liquid form will always be sufficient to dissolve all the CuCl2 formed. Therefore, we will use as little water as possible in the two starting solutions. Namely, we will use saturated solution of the two salts. Preparing the calcium chloride solution Since we do not know which hydrate of CaCl2 we have, we do not know how much of the solid to use. We will get around this problem by exploting the fact that all hydrates give the same solution, and the saturated solution contains a know concentration of CaCl2. Namely, we choose some temperature {Ts}, say 20 C, and look up the solubility of CaCl2 at that temperature. That is, how many grams of anhydrous CaCl2 can be dissolved in 100 ml (or 100 g) of water at that temperature: [ Recipe ] (Note that the solubility value depends on which hydrate was used by whoever measured it. If he did not use the anhydrous form, the number must be adjusted.) Then, from that solubility value, we work out how many grams of saturated CaCl2 solution we need to use, in order to get the desired amount of CaCl2: [ Recipe ] Then we put a large enough amount of water in a beaker, at that chosen temperature {Ts}, and we add CaCl2, until some of it will remain solid even after lengthy stirring. Then we will decant the right amount (by mass) of the clear solution into a separate beaker, carefully leaving the remaining solid behind. By this procedure, we know that the first beaker will contain the right amount of CaCl2, no matter what hydrate we have. [ Recipe ] Preparing the copper sulfate solution The copper sulfate one can get in stores is almost always the pentahydrate CuSO4(H2O)5, and the other hydrates do not form spontaneously. Thus, the problem that we had with CaCl2 does not arise here. We could prepare the CuSO4 solution by weighting the desired amount of CuSO4(H2O)5. and adding as much water as needed to dissolve it. However, for this video we have chosen to use the same method we used for the calcium chloride solution. Namely, we prepare a saturated solution of CuSO4 in water at the chosen temperature {Ts}, and we separate the right amount (by mass) of the liquid. Redoing the same computations we did for CaCl2, we get [ Recipe ] We will use a little excess of this solution, for reasons to be explained later. Reaction After weighting the desired amounts of the two saturated solutions, we mix them by pouring in a large beaker. To prevent the salts from crystallizing out before the mixing, the solutions can be heated at some higher temperature, and/or diluted with some water. Obviously, that would not affect the amount of CaCl2 and CuSO4 in the mix. On mixing, a white precipitate of CaSO4(H2O)2 is formed immediately. Depending on the amount of water present, the result may be a paste. Heating the mixture to near boiling and letting it cool may help increase the sizes of the particles (by dissolving and re-crystallizing a bit of the calcium sulfate). That will speed up filtering. Filtering The mixture is then filtered through some coarse paper, such as a coffee filter (or even a couple of sheets of paper towels, if they don't break down when wet). Most of the precipitate should remain on the paper, and a relatively clear solution (the "main filtrate") should go through, containing mainly CuCl2. (A bit of precipitate may get through, but that is no problem; we will take care of it later.) The solution should be some color between green and blue, depending on the concentration of the solutions (that is, on the tepmerature {Ts} at which they were prepared). The precipitate on the filter should be washed with a bit of water squirted over it and deep into it. The filtrate of this washing, that contains significant amounts of CuCl2, should be added to the "main filtrate" of the original mixture. This washing should be repeated a couple of times, until the wash water is a very light blue-green and the precipitate is mostly white all through. We save the precipitate for later. The main filtrate (including the water from the first washes) should then be run again through a real filter paper, in order to remove the small amounts of precipitate that it still contains. Again we wash the precipitate a couple of times, with a bit of water, adding the wash water to the "main filtrate". We should then have a few hundred ml of a perfecly clear blue-green solution. The precipitate While our goal is the copper chloride, we will try to save the calcium sulfate too, for future experiments. It should be more pure than the gypsum that one can buy at hardware store, that may contain the cheaper calcium carbonate, sand and silcates from rocks, etc. So, we put all the precipitate (from both filtering steps) into a single beaker, scraping and washing as much as we can off the filter papers. We add about 100 ml of water, mix thoroughly, and repeat the coarse and fine filtering steps. In both steps we wash the precipitate several times while on the filter paper. This time, however, we keep all the light blue filtrate in a separate flask of "wash solution". If we added it to the main filtrate, it add to it very little CuCl2 but a lot of water that would have to be boiled off. Once the precipitate is throughly washed snow-white, we scrape and wash it off the filter paper. To remove the last amounts of CuCl2 from the the precipitate, we spread it over a filter paper and several sheets of towel paper. When it is firm we squirt a bit of water on top of it, to wash it even more completey, changing the towels (which must be discarded). finally we cover it with another disk of filter paper, more paper towels, and some non-metallic waterproof weight, such as a plastic container or a ceramic tile. After an hour or so we remove the paper towels, leaving the two filter papers, and we let the precipitate dry out completely in air for a day or two. In the end we weight the precipitate and compare to the predicted amount of CaSo4(H2O)2: [ Recipe ] Not bad! Crystallization and purification To extract the copper chloride from the main filtrate, we basically boil off the water until only dry solid remain, and heat the solids above 110 C to turn any CuCl2(H2O)2 into anhydrous CuCl2. Then we will redissolve the dry residue in ethanol and evaporate the ethanol to crystallize the anhydrous CuCl2. When the recipe is executed at 20 C, he filtrate will contain a few grams of calcium sulfate in solution. To reduce that impurity in the residue, we will do the evaporation in two steps. First we boil the filtrate until just enough water is left to dissolve the CuCl2 completely at 20 C (step 1). Then we cool the solution to 20 C. That will cause most of the CaSO4 to come out of solution. We filter again the liquid and proceed to complete evaporation (step 2). The brown residue should contain mostly the CuCl2, plus some excess CuSO4, possibly hydrated, and some CaSO4 that remained in solution. Recall that, in the basic recipe, used a small excess (5%) of copper sulfate. One advantage of doing so is that the extra SO4-- ions will drive a little more of the CaSO4 out of solution. However we will end up with some CuSO4 in the solid residue of the evaporation. Another advantage of using some excess of CuSO4 is that CuSO4 and CaSO4 are quite insoluble in ethanol, whereas CaCl2 is very soluble. Therefore, for the purification step, it is useful to ensure that there is no CaCl2 in the product that results from evaporation and dehydrating. False starts In my first attempt, I tried to prepare the reagent solutions at 100 C and carry out the reaction at that temperature, in order to minimize the amount of water in the resulting CuCl2 solution. At that temperature, only 46 and 21 ml of water are needed to dissolve the CuSO4(H2O)5 and the CaCl2, respectively, and the final CuCl2 solution has only 94 ml of water. As in the recipe above, I prepared each solution separately at 100C, with a little excess of the salt, and decanted the right mass of the saturated liquid to a separate beaker. Since these solutions cooled a bit after pouring, some of the salt crystallized out. I then re-heated the two solutions until the salts dissolved again, before mixing them. The reaction worked, but the amount of calcium sulfate was so large compared to the amount of liquid that they formed a thick paste of a beautful lime green color. I had to add more than 200 ml of water to make the mixture fluid enough for the calcium sulfate to precipitate out. I also made a technical mistake that could have resulted in burns. When re-heating the two solutions before mixing, I overheated the calcium chloride one to about 120 C. (A solution of CaCl2 that is concentrated at 100 C will boil only at 120 C or more.) Impatient, I added it to the CuSO4 solution while still at 110 C. But a saturated solution of CuSO4 still boils at 100 C, so there was vigiorous boiling that could easily have splattered the paste out of the beaker.