Primarygun Posted October 27, 2006 Posted October 27, 2006 How does iodate ion react with ascorbic acid? How can I prevent it from occuring when I wanna to use iodine to oxide the acid instead so as to find out the percentage by mass of the acid in a vitamin C tablet.
woelen Posted October 27, 2006 Posted October 27, 2006 If you want to use iodine as oxidizer, then you could take iodine. But iodine is a bad compound on storage and it is very difficult to weigh out a precise amount of iodine (it emits vapor and its purity is questionable). Also making solutions of well-known concentration of iodine is hard because of this. Potassium iodate, on the other hand, can be weighed VERY accurately, and the chemical itself also can be obtained in a very pure state, without problems of uncertain amounts of water of crystallization, hygroscopic properties, which attract water and make the weight more uncertain, etc. Titrating for ascorbid acid, I would prepare a solution of potassium iodate of known concentration. I would add a large excess of potassium iodide to this solution, and a large excess of some acid (e.g. sulphuric acid). There is no need at all to precisely weigh the potassium iodide and the acid, as long as there is enough. KI also is very bad for precise weighing. It is hygroscopic and may have a strongly varying amount of water attached to it. So, for each mol of potassium iodate, add at least 8 mols of potassium iodide, but better is to add a clear excess, e.g. 15 mols of potassium iodide. Also add a generous amount of acid. Then add water, until the liquid has a precisely known volume. This liquid now contains a very-well known concentration of iodine, because the amount of iodate was very well known. Iodate reacts with iodide as follows, fast and quantitatively: IO3(-) + 5I(-) + 6H(+) ---> 3I2 + 3H2O And because iodate was your limiting reagent and this was added in precisely measured amount, you know also precisely how much iodine is present. This solution, now in turn is used to titrate the ascorbic acid. Starch can be used as indicator. Add a little starch to the solution, which contains the ascorbic acid.
Borek Posted October 27, 2006 Posted October 27, 2006 Add starch just before titration end, when the iodine color fades. Otherwise it may react with starch in an irreversible way. Borek -- General Chemistry Software www.pH-meter.info
Primarygun Posted October 28, 2006 Author Posted October 28, 2006 My teacher suggested me adding potassium iodide into a solution of acidified vitamin C solution and then later add potassium iodate solution into the mixture. Is it a bad to construct an experiment like this? Is it better if we produce iodine in another beaker and then later add it into vitamin C solution?
woelen Posted October 28, 2006 Posted October 28, 2006 I would make the iodine separately, but it really depends on the properties of iodate/ascorbic acid mixtures. If iodate simply reacts as oxidizer, then the method is equally well, but if there are other side reactions, then it is not suitable. Ask your teacher about that, I'm not sure about such side reactions. I cases like this, I take the certain way over the uncertain way, hence my suggestion for preparing the iodine in a separate beaker.
Borek Posted October 28, 2006 Posted October 28, 2006 hence my suggestion for preparing the iodine in a separate beaker In all analytical methods I am aware off you prepare reagents in the same beaker/flask it will be used, to avoid reagent losses in transfer. You are right about side reactions, but if there are side reactions possible it is better to look for other method. Borek -- General Chemistry Software www.pH-meter.info
woelen Posted October 28, 2006 Posted October 28, 2006 Borek, you make an important point over here. It can, however, be avoided almost completely by carefully thinking about how you prepare a solution. I'll explain with an example. Suppose I would want to make 100 ml of a 0.1 M solution of I2. I would weigh 1/3 of 0.01 mol of KIO3 on a glass watch glass. I would rinse this into a beaker of 100 ml. I would take somewhere around 60 ml of water, would add 10 ml of 20% H2SO4 and dissolve all. By rinsing the weighing glass, all iodate is taken into solution, and there is no real loss. To this, I would add 0.15 mol of KI (large excess, enough to make I2 and also keep this in solution as tri-iodide) Next, the 70 to 80 ml or so of the liquid is transferred to a volumetric flask of 100 ml. This is an important point, where there can be losses. I would take 10 ml of water, swirl this around in the original beaker and add this to the volumetric glass. A second swirl with 5 ml of water takes the final traces, and then the volumetric flask is filled up to 100 ml. In this way, the transfer losses really are neglicible, and still there is the advantage of not having to deal with (uncertain) side reactions. The starch, I then would not add to the iodine, but to the ascorbic acid solution, where there is no reaction. As soon as the iodine is in excess, the starch will indicate that, and no starch is in contact with iodine for a long time, causing an irreversible reaction with loss of iodine.
Primarygun Posted October 29, 2006 Author Posted October 29, 2006 If iodate simply reacts as oxidizer, then the method is equally well, With excess iodide added, one mole of iodate can generate 3 moles of iodine. What's equation of oxidation of thiosulphate by iodate? I can't figure it out. However, I guess that it cannot yield the same effect compared to iodine which can be produced by the iodate.
Borek Posted October 30, 2006 Posted October 30, 2006 S4O62- + I- Borek -- General Chemistry Software www.pH-meter.info
Borek Posted October 30, 2006 Posted October 30, 2006 Suppose I would want to make 100 ml of a 0.1 M solution of I2. I would weigh 1/3 of 0.01 mol of KIO3 on a glass watch glass. I would rinse this into a beaker of 100 ml. I would take somewhere around 60 ml of water, would add 10 ml of 20% H2SO4 and dissolve all. By rinsing the weighing glass, all iodate is taken into solution, and there is no real loss. To this, I would add 0.15 mol of KI (large excess, enough to make I2 and also keep this in solution as tri-iodide) Next, the 70 to 80 ml or so of the liquid is transferred to a volumetric flask of 100 ml. This is an important point, where there can be losses. I would take 10 ml of water, swirl this around in the original beaker and add this to the volumetric glass. A second swirl with 5 ml of water takes the final traces, and then the volumetric flask is filled up to 100 ml. In this way, the transfer losses really are neglicible, and still there is the advantage of not having to deal with (uncertain) side reactions. I would rather go with iodate solution prepared from solid directly in the volumetric flask, filling up to the mark. Then I would transfer exactly known volume of iodate to baker and I would add iodide and acid there, avoiding all steps that can introduce error to the iodine amount When titrating iodine you usually start with yellowish solution - and you don't add starch till solution loses almost all color. Borek -- General Chemistry Software www.pH-meter.info
woelen Posted October 30, 2006 Posted October 30, 2006 Borek, we are talking about different things with the starch. In the case of titrating a vitamin C tablet, I would drip in the solution of iodine into the solution of vitamin C. You are talking about the vitamin C dripped into the iodine? Even better of course would be to use thiosulfate, but in the case of vitamin C and iodine, I'm afraid that will lead to extra errors, because besides the fast reaction between iodine and vitamin C, there will be another much slower reaction between iodine and the clean oxidation product of vitamin C. A reaction, very similar to the reaction, you mention with the starch. So, in this particular case, I would go for crushing the tablet of vitamin C, and dissolving this (any insoluble solid just leaving in the beaker). To this, I would add starch and then I would drip in the iodine. I know, that the iodine solution is somewhat messy in a burette, but when a glass burette is taken (and they usually are), then I see no real problem. As soon as excess iodine is added, the solution in the beaker will turn dark blue. The other option would be to add a well-known amount of iodine to the vitamin C, such that an excess of iodine is added. Then this excess can be titrated with thiosulfate. In that case, I indeed would not add the starch from the beginning. But, this approach has the disadvantage, that the excess iodine may react a little with the oxidized vitamin C (which still is an hydrocarbon, which certainly can be oxidized further). The latter problem does not exist with the former approach, because the iodine only exists for a very short time in solution and quickly is consumed by still unoxidized vitamin C. Iodate, reacts with thiosulfate, forming iodide and tetrathionate, as Borek pointed out. But iodate is quite a strong oxidizer and adding this directly to thiosulfate, without first converting it to iodine with excess iodide, may also lead to oxidation of part of the thiosulfate, all the way up to sulfate. This side reaction makes the titration inaccurate, so it really is important to first make iodine.
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