woelen

Nature has decided..... I put a large spatula full of sodium bicarbonate in a test tube and added approximately 5 ml of distilled water. Next, I started heating the test tube. Well before the water starts boiling, the solution starts to bubble quite a lot. Many small bubbles escape from the liquid. This must be CO2. I continued heating but not boiling for several minutes, until bubbling ceased. After bubbling ceased, I boiled the solution for several minutes. It might be that more bubbles of CO2 were released, but of course, one cannot really see, because of the boiling of the water. In total, I have been heating for more than 10 minutes, and quite some water had evaporated. Next, the liquid was allowed to cool down. I measured the pH of the liquid, using a multi-color pH indicator. Using this indicator, I measured a pH around 10. So, the liquid definitely has become alkaline, but not nearly as much as a similar solution of NaOH. When a spatula full of NaOH is dissolved in 5 ml of water, then the pH will be around 14. Looking up the Kb value of Na2CO3, a pH around 10 can be explained very well. So, indeed, boiling a solution of NaHCO3 does not give NaOH, but Na2CO3. As I explained before this experiment, it can be understood simply from the fact that OH(-) and HCO3(-) react with each other giving CO3(2-) and H2O. Even at boiling temperatures, the CO2 is bound sufficiently strongly by the solution in order to prevent escape into the air. I think this issue has settled now.

I would just say, forget about making H2SO4 from NaHSO4. If you have read the thread on scimad, then I think you've seen that this is REALLY hard. I myself have quite some experience at a hobby level with chemistry, but this is one of the things which I do not even think of starting now, although SO3 is one of the reagents which I think is very interesting . Simply too dangerous and it requires too much apparatus. Don't do this. The other suggestion with bubbling SO2 through H2O2 also is a nice one. If you have access to sodium or potassium metabisulfite (you can buy that at stores for making wine, as a disinfectant and preservative) and you add that to conc. HCl and heat the solution, then SO2 will be bubbling out of the solution. Pass that gas through a solution of H2O2 (preferrably appr. 10% H2O2 by weight). This yields dilute H2SO4. Next boil until the liquid gives some white fumes. At that point the excess H2O2 is decomposed and what remains is H2SO4 of high concentration. As you see, this process also is quite complicated, but for the home chemist with some simple glassware and some rubber heat-resistant tubing it is feasible. Beware, SO2 is a very pungent gas! Try not to inhale much of the gas, it has a choking smell! Yet another time: please do not try the decomposition of NaHSO4, I mean this!. I would not like you to have a serious accident. Hassling around with SO3 only is for the most brave and most equipped of us (and I must admit, I'm not among them ).

No, phenol is a real pain to store and to my opinion it is worse on storage than benzene. I have 100 ml of 80% phenol solution in water and it is a real pain. It is extremely corrosive to the skin. I had a single drop on my skin and rinsed it away immediately, but it gave a really nasty tinging feeling and strong irritation. The skin became white. The smell also is a real pain and sickening if you smell it all the time. Now I store my phenol in a double container: Glass bottle with phenol, wrapped in a plastic food bag, put in a glass jar with a tight metal screw cap. Only with this type of storage, the smell is not coming out anymore. The phenol also is very prone to oxidation and subsequent formation of condensation products. Pure phenol is colorless, but it quickly turns red/brown on storage, due to oxidation. The oxidation products condense to large molecules of varying composition and give a kind of brown crap. I already have the phenol for several weeks now (I received it from a friend, with whom I exchanged some chemicals), but I still only did one experiment with it, due to its annoying properties. An interesting chemical though is hydroquinone. It is used as a photo developer and can be purchased easily as the pure compound at many places. This is a phenolic compound. It is benzene with two -OH groups attached to the benzene ring, in para-positions. Hydroquinone does store well, has many of the chemical properties of phenol, but not its highly irritating odour and corrosivity towards the skin. It is an off-white water-soluble solid, which can be oxidized easily and forms interesting complexes with many metals. It can also serve as the basis for all kinds of organic chemistry experiments and is relatively safe. Just an idea ??

Are you sure? Esterification is between an alcohol and an acid. So I can imagine esterification between phenol and benzoic acid, but not between benzene and benzoic acid. R-OH + HO-(O)C-R' <---> R-O-(O)C-R' + H2O The problem is that benzoic acid has a -COOH group attached to the benzene ring. You can reduce the -COOH group with a strong reductor like LiAlH4, but that still leaves the C atom on the benzene-ring. I can imagine that making phenol from benzoic acid can best be done via the benzene route, by first taking off the -COOH group, then substituting one of the H's on the benzene ring (e.g. with Cl) and then replacing that with -OH. But I'm not an organiker, so there may be easier routes to phenol. @YT: You know that benzene is a proven carcinogen? Please do not breathe the vapors from your hot NaOH/benzoic acid mix.

Indeed, this evening I'll put this question in front of our final jury, being Nature itself

Well, now I hope that you see the gap in your reasoning in case of the solution of the bicarbonate. Suppose some hydroxide is formed, according to your proposed mechanism, that hydroxide will react with other bicarbonate in solution immediately, forming carbonate: HCO3(-) + OH(-) <----> H2O + CO3(2-) This equilibrium is very much to the right, CO3(2-) only is a weak base. Now another gap in your reasoning: A solution of a bicarbonate is weakly acidic. A solution of a carbonate is fairly basic. A solution of hydroxide is very basic. You expect the hydroxide to form fast from the somewhat acidic solution of bicarbonate, while it is formed much slower from the more alkaline carbonate. Think of the bicarbonate being converted to hydroxide, then alkalinity of the liquid moves through the medium situation with the carbonate and the process comes to a halt. You see this reasoning? So, making hydroxide from bicarbonate is harder than making hydroxide from carbonate. In fact, the carbonate ion is quite stable. Think of a cettle with hard water in it. Hard water contains calcium and magnesium bicarbonate in solution. When you boil the water, then the bicarbonate decomposes to water, carbon dioxide and carbonate ion. When the liquid cools down, then the solubility of the calcium/magnesium carbonate drops to a low point, such that it precipitates out of solution. This is the main problem with hard water. You don't get calcium hydroxide. If the latter would be formed, then we would not have problems with hard water, because calcium hydroxide is sufficiently soluble and does not precipitate.

In fact, it is +7. ClO4(-) has a charge equal to -1. You have four oxygens, each having an oxidation number equal to -2, that makes up a total of -8. In order to have a net charge of the ion, the chlorine must be +7. Remember, oxidation state and ionic charge at an atom-basis are two completely different things, but the following is true: The sum of all oxidation numbers of all atoms in an ion or a molecule must be equal to the total charge of the ion or molecule.

In fact they do, but just a little bit. Sulfate ion is a little bit basic. The reason that it is only a little bit is that HSO4(-) is a fairly strong acid. Take a concentrated solution of e.g. sodium sulfate and you'll see that it is very weakly basic: SO4(2-) + H2O <<<---> HSO4(-) + OH(-)

Wow, how skilled you are ! If you really are 13 then I would say that for your age (compared to the great majority of 13-year old people) you indeed are very skilled (this is a compliment, really). But looking at it from an absolute point of view I'm inclined to say that you still have to learn a lot. This is not meant as an insult, but please let's put things in the right perspective. People's real strength often is in their ability to recognize their own limitations. I did not mean to crank you with my remarks. With the words 'playing around' I meant 'doing experiments'. Forget about it. Even with a propane torch at full blast you'll not be able to decompose the NaHSO4 further than Na2S2O7, when you heat the stuff in a crucible. This is REALLY difficult. At just over 100 C you get the following: 2NaHSO4 ---> Na2S2O7 + H2O At a MUCH higher temperature (near 1000 C) the following happens: Na2S2O7 ---> Na2SO4 + SO3 This SO3 then needs to be added to water, but this is increadibly dangerous. Even a single drop of cold SO3, added to a small amount of water will result in detonation and shattering of the container. How do you think you will collect the SO3? No, making H2SO4 (or SO3) from NaHSO4 simply is not feasible for the home chemist. This has eluded even the real brave ones from a site like this But now the other thing. What experiments do you want to do with H2SO4? A solution of NaHSO4 acts as a solution of H2SO4 in water, with an equimolar amount of Na2SO4 dissolved as well. For almost all aqueous reactions, requiring dilute H2SO4, NaHSO4 works equally well. In fact, I use this stuff myself very often at places, where H2SO4 is demanded, simply because it is much easier and safer to handle. Only experiments, requiring concentrated H2SO4 cannot be conducted with NaHSO4.

Which gas do you light? At the anode or at the cathode? I only can imagine formation of H2 at the cathode. Are you sure that the soot is not from the flame of the lighter? Just by keeping it upside down or in a confined space with limited ventilation you may have a sooty flame. I hardly can imagine that you get a hydrocarbon from this electrolysis process. I would expect formation of H2 gas at the cathode and formation of copper (II) carbonate or some copper (I) species at the anode. The green stuff almost certainly is a copper (II) species, but not pure copper (II) oxide. Copper (II) oxide is black. Copper (I) oxide is red, orange or even yellow, exact color depending on how finely it is divided and on the amount of water, incorporated into the compound. I think that the green stuff is a mix of copper (II) carbonate, oxide and hydroxide. Try heating the stuff. It probably turns black also.

It is one of the chemicals I use most (together with HCl and H2SO4). Almost every aqueous experiment I do either involves an acid or a base. If I need a base I take NaOH, if I need an acid I take HCl when the complexing of chloride does not disturb the experiment, otherwise I take the more expensive and harder to obtain H2SO4. Many redox reactions either consume H(+) or OH(-), that's why I need acid or base. Many organic acids also can only be brought in solution as their sodium salt or potassium salt. For this reason I also use NaOH (sometimes KOH, but the latter is MUCH more expensive and harder to get). NaOH also is very interesting for dissolving elements like sulphur and silicon. These solutions have really interesting properties. In fact, there are too many uses of NaOH to name them all. Have a look at my site and look at the experiments I do. Then you'll see a lot of experiments with one of the chemicals being NaOH: http://woelen.scheikunde.net/science/chem/exps/exppatt.cgi?compound=sodiumhydroxide

I still do not agree, but let's give a turn to this discussion: A counter question: What would you expect if you boil a solution of Na2CO3 to dryness, instead of NaHCO3? You also expect NaOH in that case or do you expect Na2CO3 as residue?

The fact that the volume seems a little larger indeed is quite possible, because now you have a free powder, while first you had a compressed tablet form. Another reason may be difference in density, but for that I would suggest to look up a table. As far as I know, benzoic acid is odorless and (more or less) tasteless. It is used in small quantities in foods as a preservative and of course for that purpose you don't want added smell or taste.

Don't worry about the sulphur-hydrogen rocket. It won't work. Sulphur and hydrogen can be made to react, but only with great difficulty. H2S certainly does not form spontaneously and you have to heat a gaseous mix of H2 and S8 and slowly (very slowly) an equilibrium is obtained with H2S, H2 and S8. In fact, this method of making H2S is not used anywhere. H2S indeed is VERY toxic and one should not regard this as a nice playing thing. As Jdurg pointed out, at higher concentration it kills your sense of smell and you can be in a high concetration of H2S without you knowing it. It will kill you in that case. The stink bombs I know from my youth were a dilute solution of ammonium sulfide. When exposed to the air, they give a mix of H2S and NH3. The amounts formed, however, were so low that no serious risks were introduced.

I strongly doubt that, actually, I do not believe this. If you heat a solution of NaHCO3, then indeed bubbles of CO2 will be driven out, until carbonate remains: 2HCO3(-) ---> H2O + CO2(driven off) + CO3(2-) The carbonate will be in equilibrium with a very small amount of hydroxide: CO3(2-) + H2O <----> HCO3(-) + OH(-) So, continued boiling may indeed remove a little more CO2 than one would expect for pure carbontate to remain, but this will only be a very weak effect (carbonate is only a weak base and OH(-) is a very potent obserber of CO2). If you continue heating, then all water will boil away and a dry residue of sodium carbonate remains with at most a few percents of NaOH, but I even doubt that, my bet would be just a few tenths of percents of NaOH. Have you actually tried this way of making NaOH?

Many organic acids, especially, when the organic group is large, compared to the acidic group, have only slight solubility in water, because of the fact that they are bulky mostly apolar compounds. The sodium salts, ammonium salts or potassium salts usually are soluble (as almost all salts of these three cations), but when a strong acid like HCl or H2SO4 is added, then the cation is protonated and the free organic is produced, which in turn precipitates So, in the case of benzoic acid: C6H5COO(-) + H(+) ---> C6H5COOH(s) Here the C6H5- group is bulky compared to the COO(-) group.

Indeed, you're right, it decomposes to Na2CO3. In electrolysis you'll get the same, but there the process continues and you'll end up with a solution of NaOH with still quite some Na2CO3 in it. Heating of NaHCO3 will first give you H2O, CO2 and Na2CO3 at fairly low temperature (around 100 C). In order to drive off the other molecule of CO2 requires very strong heating. I'm afraid that even pointing the flame of a propane torch directly on the Na2CO3 still does not drive off the CO2, so that is not a useful way of getting NaOH. But now for something else, is this just you want to play around and see if you can make NaOH yourself, just out of curiousity, or are you really interested in the NaOH itself? If you are after NaOH, then I would suggest buying it. It is sold as drain cleaner in many places (hardware stores, drugstores and some super markets) in the form of white pellets. It also is cheap (around EUR 2 per pound, or $2 per pound). Another thing you could try is to use a salt bridge instead of a membrane. For this purpose, take two beakers with NaCl solution. Take a rubber tube and fill this with cotton and completely soak this with salt-solution. Now do the electrolysis with the anode in one beaker and the cathode in the other beaker. The salt bridge assures that there is no mixing of liquids, but a disadvantage is the larger resistance of the total circuit, you'll loose more electrical energy as heat.

The first is simple. Dissolve some Na-benzoate in water and add some acid (e.g. dilute H2SO4). The benzoic acid precipitates as a coarse flaky precipitate. The second one I do not know.

At the anode you'll get CO2 and O2 simultaneously: 4HCO3(-) --> 4CO2 + O2 + 2H2O + 4e At the cathode you'll get H2 and OH(-) 2H2O + 2e --> 2OH(-) + H2 So, theoretically you'll end up with NaOH solution only, in practice it will be REALLY hard to get it completely free of carbonate. When you are close to the end, then any CO2 formed at the anode will be absorbed by the hydroxide at once and you get carbonate again: CO2 + 2OH(-) --> CO3(2-) + H2O The solution also will pick up CO2 from the air. In practice you'll never succeed in making it free of carbonate, not even with rigorous exclusion of air. Use of NaCl does not work very well, unless you use some method of separating the anode liquid from the cathode liquid and use some special membrane, which is permeable for sodium ions and chloride ions, but not for hydroxide. When the liquid is allowed to mix, then you get hypochlorites and chlorates in solution as well.

Well, I have the idea that you still do not really grasp the concept. Another time, oxidation state is just bookkeeping, it does not reflect real structure. I'll try to explain with two very simple compounds: CO (carbon monoxide) CaO (calcium oxide) The first is a purely covalent compound. This means that the C atom and O atom are bound tightly and that both atoms share electrons. Two electrons from the O and two electrons from the C are shared between the C and O atoms. Of each of these 4 electrons you cannot say to which atom they belong, all 4 of them belong to both atoms. Now, if you look at CaO, this is a purely ionic compound. It consists of Ca(2+) ions and O(2-) ions, packed in a crystal lattice, but nevertheless, these ions exist as real entities. In the last example, oxidation state really matches the real physical structure. In the first example, there is no ionic thing at all. However, in both compounds, the oxidation state of the oxygen atom is said to be -2. It is just a matter of appointment to say this. Both compounds are neutral species, so for the other element the oxidation state must be +2. Again, for CaO this is the real charge on the calcium ion, but for CO again, it the oxidation state of carbon is said to be +2. Now back to CrO3. There is no actual +6 charge on the chromium, we simple say it has oxidation state +6, because we say that oxygen has oxidation state -2 in its compounds (except the few I mentioned earlier). Now, the link with real chemistry is just that we order elements from most electronegative to most electropositive. Electronegative elements usually have negative oxidation states. Fluorine is the most electronegative element and it is said to have oxidation state -1 always, except in F2. Oxygen has oxidation state -2, except in the compounds I mentioned earlier and in OF2, where it has oxidation state +2 (this must be the case, because of the rule for fluorine). Chlorine is the next electronegative elements and it has oxidation state -1 usually, but in its compounds and ions with fluorine and oxygen it has a positive oxidation state. The oxidation state, hence, tells something about how much an element is oxidized. The higher the number, the stronger it is oxidized. So, Cr2O3 contains a much less oxidized form of chromium than CrO3. When you know the rules and have some experience, you can see at once, whether a compound is a strongly oxidized compound or not. You understand? Just plain arithmetic.

Burn some magnesium metal in a pure CO2 atmosphere, you get pure C and MgO. As a secundary side reaction you may also get some magnesium carbide. This separates the C, but not the O.

Dissolve in HCl (forms a soluble chloro-stibnate (III) complex) and slowly add an excess of sodium sulfide while stirring constantly. The nice orange/red precipitate of Sb2S3 separates from the liquid. Do this experiment in a good fumehood or outside. A lot of free H2S is formed as well!

As I wrote before, it just is a matter of bookkeeping and does not reflect the real structure of this covalent compound. There are some bookkeeping rules and for this discussion I repeat the one, which is important here: Oxygen has oxidation state -2 in almost all of its compounds. Exceptions are peroxo-compounds, superoxo-compounds, ozonide-compounds. Now back to CrO3. This is a neutral species. Oxygen has oxidation state -2, there are three oxygens, so chromium has oxidation state +6 in order to obtain neutrality. Another example: NO2(-) ion, nitrite. Oxygen has oxidation state -2, there are 2 of these. The total charge equals -1 and there is one N-atom. Make it +3 and you end up with the charge equal to -1. So, N in nitrite has oxidation state +3. The N in the brown nitrogen dioxide (NO2 without charge) has oxidation state +4. The N in the nitronium ion (NO2(+)) has oxidation state +5. Things become even more interesting, such as in a salt like nitronium perchlorate (NO2 ClO4, with the NO2(+) ion and the ClO4(-) ion). I'll leave it as an exercise for you to compute the oxidation state of the chlorine atom in this salt. The actual charge on that atom, however, is much lower, because it is covalently bonded to the oxygens.

Oxidation state is just a book-keeping device. For example in the compound CrO3, the metal chromium is in the +6 oxidation state, but this does not mean that it gas lost 6 electrons. The bonds in CrO3 are mostly covalent, with just a small charge-bias towards the oxygen atoms. Only for purely simple ionic compounds, the oxidation state equals the charge. E.g. in CaO, the calcium really lost two electrons and exists as Ca(2+) ion and the oxygen gained two electrons and exists as O(2-) ion. In a compound like HgO, things are quite different. The bond in this compound is not fully ionic, but something between purely covalent and purely ionic. So, comparing these two compounds: (2+)Ca O(2-) (+)Hg....O(-) In CaO, the ions are completely disconnected, in HgO, there is still some bond between the Hg-atom and O-atom, but charge is distributed with a positive end at the Hg-atom and a negative end at the O-atom. However, the Hg-atom does not have a +2 charge. This effect becomes stronger with increasing oxidation state. So, for a compound like CrO3, the Cr-atom carries a slight positive charge and the O-atoms carry a slight negative charge. Why is the concept of oxidation state introduced? It provides a good way of bookkeeping the level of oxidation (or reduction) of an element and it allows one to see whether an element is in a common oxidation state or a very special one. With some experience, you can see at once, whether a compound is as expected or something very special.

Look around on the same site I posted here. It covers the process of making chlorate from NaCl in great detail, including instruction for how to construct an electrolysis cell.