woelen

Can you mention any solvent, which shows the difference? As far as I know, NaOH already is 100% ionized, so how could CsOH be more ionized than NaOH?

No, not really. HF also attacks glass quickly, yet it is a weak acid. In fact, I intend to buy some CsI. It is affordable at a price of $16 per 10 grams. If I have that stuff, then I'll try whether a solution of KOH and CsI is more corrosive than a solution of KOH only of the same concentration. BTW, the CsI is going to serve as a place-holder for Cs in my collection of elements, until I have more $$$ for a decent Cs-sample But coming back on the experiment of thought, what do you think is more corrosive: CsI/KOH or CsOH/KI? Could you also give an explanation to this. I'm curious about your answer and maybe we can settle the issue on strength of alkalies.

Basically, yes... When you have HCl with some Cu(2+) in it, then when oxygen from the air is allowed to reach the mix, you can continue dissolving copper, turning that into copper (I), which is turned to copper (II) by the oxygen from the air. The limiting reagent is HCl. The net reaction is 2Cu + O2 + 4H(+) + 8Cl(-) ---> 2CuCl4(2-) + 2H2O As you can see, the reaction requires H(+) and Cl(-). Long before all HCl is used up, the reaction becomes so slow that in practice it hardly continues. But, the equation, given above, only is a net reaction. In reality, the reaction goes through the copper (I) complex, as you pointed out already. So, the liquid becomes very dark. When the copper metal is removed, though, the liquid finally will become green, because all copper (I) is converted to copper (II) by oxidation with oxygen from the air.

As far as I know, all Na-salts and K-salts are 100% ionic. No covalent bonding at all. So, the corresponding Cs-salt cannot be more ionic. Just an experiment of though. Suppose I have 1 mol of CsI and 1 mol of KOH. I also have 1 mol of CsOH and 1 mol of KI. I take 250 ml of water and I dissolve the CsI and KOH in it. I take another 250 ml of water and I dissolve the KI and CsOH in it. Which solution would be "stronger", i.e. more corrosive?

Have a look at page 8 of this document: http://www.cgcc.cc.or.us/Academics/dept/chemistry/RKovacich/notes/Notes%20Acids%20and%20Bases.doc The strong acids are mentioned separately all with their own Ka, which indeed differ by quite large factors. The strong bases, however, all are mentioned in a single group, with just one (uncertain) Kb value specified. So, there might be differences, but they are marginal at best. I still think that this is due to the fact that these strong bases are totally ionic and then the dissociation is approximately the same for all. When opening the file, please be prepared that it is quite large (>5 Mbyte). Of course, this single reference is not a proof of my theory, but it is at least some evidence supporting it .

No, if pure, it is white. In practice it may be very light grey. Just as with calcium phosphate. Magnesium salts usually are white or off-white, simply because the Mg(2+) ion is colorless. If color is imparted to magnesium salts then the anion gives the color to the salt.

As jdurg wrote, you have to memorize. In the case of copper, it forms copper (II) acetate. Copper (I) compounds can be formed, but the conditions for these must be quite specific. As a first step, just memorize that copper metal forms copper (II) ions in aquous solutions with almost all cations.

Indeed, the copper acts as reductor and it reduces the Cu(2+). So, in my test tube experiment you have green Cu(2+) (not as free ion, but complexed, also called coordinated, to chloride) and this reacts with the copper metal. A simplified equation is given below. Cu(2+) + Cu ---> 2Cu(+) The Cu(+) gives the dark compound with Cu(2+). When the test tube is not stoppered, then the liquid remains dark and finally becomes green again. The Cu(+) is VERY sensitive to oxidation by oxygen from the air, so any Cu(+) formed is oxidized to Cu(2+) by oxygen. When the test tube is stoppered, then no new oxygen can enter the tube and then finally it becomes (almost) colorless, because all Cu(2+) is reduced by the copper metal. As soon as the test tube is opened again, oxygen can go in and the Cu(+) is oxidized to Cu(2+) and the brown complex with copper (I) and copper (II) is formed. ---------------------------------------------------------- Now about the complexes. The story I told above is just the principle of the redox reaction between copper (II) and metallic copper (0). In reality, copper exists in the form of anionic species at high concentrations of HCl. Copper (II) exists as green CuCl4(2-) and copper (I) exists as colorless CuCl2(-). So, complex ions with chloride attached to copper are formed. These ions are not basically different than e.g. SO4(2-), where oxo-groups are attached to a central sulphur core. So, the real reaction when copper wire is added is the following: CuCl4(2-) + Cu ---> 2 CuCl2(-) The CuCl2(-) reacts with CuCl4(2-), forming the deep brown complex. The mechanism behind this is amazingly complex and not fully understood (at least not by me, but I'm afraid by no one). The chloride ions also have a very important role in this. If you have a solution of CuSO4 in 30% H2SO4 and you add copper metal, then nothing happens. You will not get copper (I). Chloride stabilizes the copper (I) complex.

Please use normal grammar. What you are producing here is not even a normal sentence . I agree with YT that some respect for your audience would be very nice.

Well in fact, time to praise . Your reasoning is very good. Indeed, nothing in the mix wants the electron, so the Cu(+) is not oxidized. If you take red Cu2O and add HCl, then you see it turn white and CuCl (not CuCl2 !) is formed. In concentrated HCl you'll see it dissolve to the colorless complex ion CuCl2(-). ------------------------------------ HOWEVER, there is one big pitfall. Oxygen from the air is capable of oxidizing Cu(+) very well and it certainly does. Cu(+) is not present as a simple ion in hydrochloric acid, but it is present as the colorless CuCl2(-). When oxygen is present, then the Cu(+) in the CuCl2(-) complex ion is oxidized to Cu(2+). What happens next is incredibly complex and I believe even the world-leading chemists do not fully understand what happens then. The only thing I can say about this is that a mixed-valency complex is formed. A compound, containing copper (I), copper (II) and chloride is formed. This compound has a VERY intense color. A proposed formula for this compound is ClCu(μ-Cl)CuCl, where the copper atoms have oxidation state somewhere between +1 and +2. If you have concentrated HCl, H2O2 and Cu-metal and CuCl2, then you can do most of the experiments, I have performed a year ago and which I put on my website. I invite you to repeat some of these and to enter the world of mixed-valency copper chemistry . Be surprised, I mean it ! http://woelen.scheikunde.net/science/chem/riddles/copperI+copperII/index.html

Suppose that HCl would oxidize Cu(+) to Cu(2+), how do you think that would happen with hydrochloric acid as the oxidizer? Think of this and try to find a possible equation. By thinking about this and attempting to find a reaction-equation, you'll gain a lot of insight. When you come back with an answer, I'll praise you or I'll correct you , depending on your smartness .

I want to put a few things right. CsOH is not the strongest known base. It is not stronger than NaOH or KOH. If you dissolve CsOH, then you simply get Cs(+) ions and OH(-) ions. It might be that CsOH attacks glass more quickly than NaOH does, but this does not need to be due to stronger alkalinity. Stonger bases are the amides (e.g. NaNH2), alkoxides (e.g. the methoxide CH3OK), but also certain oxides (e.g. Na2O). All of these bases are so strong, that they cannot exist in water, but they can exist in more basic solvents. This strength of alkalinity (and also of acidity) can only be fully understood if the concept of acidity is generalized to non-aqueous solvents. For now, it suffices to say here, that each compound on its own has an intrinsic acidity or alkalinity and the observed alkalinity or acidity is limited to what a solvent can accomodate. In water, any compound more basic than hydroxide and any compound more acidic than hydronium seems equally strong. In liquid ammonia, any compound more basic than amide and any compound more acidic than ammonium seems equally strong. So, the observed acidity is limited by the solvent. As an example of this: HI is a stronger acid than HBr. In water, they look equally acidic, because both are fully ionized. In more acidic solvents (e.g. pure CH3COOH) the difference can be observed. HI ionizes more easily than HBr. The reason that HI and HBr differ, while NaOH and KOH and CsOH are not really different is that the latter already are fully ionized. E.g., liquid NH3 can discriminate between alkali strength of hydroxides and amides. In liquid NH3, all the above mentioned hydroxides are equally strong as well. In any solvent they are equally strong, simply because they always are fully ionized, regardless the solvent in which they are dissolved. --------------------------------------------- The compound NaF does attack glass (and I can say this, because I have experience with this:D ). The attack is not really fast, but it certainly happens. The reason of this is that NaF hydrolyses to a fairly large extent: NaF ---> Na(+) + F(-) F(-) + H2O <---> HF + OH(-) The HF, which exists with the F(-) in equilibrium, slowly attacks the glass. So, you can do test tube experiments with NaF without ruining the test tubes, but storing it in a glass bottle for days certainly results in etching of the glass.

I am working on an overview of the chemistry of many metals. Part of this work is finished by now (first row of transition metals except scandium), and I want to share it with others. It might be that the material is made into a book (or e-book), but for now, it first will be part of my website. I hope you like it. If there are errors or omissions, then I would be pleased to know about that. http://woelen.scheikunde.net/science/chem/solutions

o - ortho - next to each other m - meta - one C atom between the groups p - para - two C-atoms between the groups, at opposite sides of benzen ring

If you use cold reactants and the CuCl2 is not too acidic, then I expect you to get green copper hydroxochloride Cu(OH)Cl. Getting really pure Cu(OH)2 is hard from CuCl2, because chloride ions coprecipitate with the hydroxide. If you want Cu2O, then try the following: Crunch a tablet of vitamin C (not the multivitamin), dissolve in water and filter away the white crud. The clear solution contains vitamin C. Add this solution to a solution of NaOH. Now add a solution of CuCl2 to this NaOH/vitamin C solution. You will get a green precipitate, which turns orange or yellow. Assure that you have excess vitamin C. The Cu2O produced in this way is very flocculent and not that easy to separate from the liquid. Another way of getting Cu2O (the red stuff) is dissolving Cu(OH)2 in excess solution of NaOH, to which tartaric acid is added. This gives a deep blue solution (google Fehling's reagent for more info). Add some glycerin or dextrose to this solution and let stand. After a few hours you have a nice compact precipitate of Cu2O, which easily is separated from the liquid. Another nice experiment is making the dark blue tetrammine copper (II) complex, Cu(NH3)4(2+). This can be made by adding a solution of CuCl2 to a solution of ammonia. You need a large excess of ammonia.

Even when a solution is luke-warm, the Cu(OH)2 can decompose. It might be that you prepared the solution of NaOH from solid NaOH and that some heat of hydration was left in the liquid. Your CuCl2 also may contain a considerable amount of HCl as impurity, which with the hydroxide releases quite some heat. So, there are quite some possible explanations for your observations. So, indeed, your precipitate most likely is (impure) CuO.

When you use a very concentrated solution of NaOH, then part of the copper hydroxide redissolves again. Copper is somewhat amphoteric and in concentrated NaOH the so-called cuprate ion is formed: Cu(OH)2 + 2OH(-) ----> CuO2(2-) + 2H2O Here the copper hydroxide acts as acid! The cuprate ion is royal blue, it is not the sky-blue of copper (II) ions in water, but a deeper blue, more like indigo. Another effect that may plague you is that Cu(OH)2 is not very stable. Even at moderate heat, it decomposes and looses water, even when it is in water: Cu(OH)2 ---> CuO + H2O This may explain why you get a dark grey precipitate. Just try the following to see if my hypothesis is right: Add excess acid to a suspension of the dark grey stuff. If it dissolves and the liquid becomes light sky-blue on dilution, then it indeed is CuO. The following web-page may help you. Read the part about copper in oxidation state +2 and focus on the precipitates of copper hydroxide and the cuprate ion. Is the color of your solution similar to the dilute cuprate solution on the webpage? http://woelen.scheikunde.net/science/chem/solutions/cu.html

You don't want to have potassium at your house. It simply is too extreme to play with. It also is a REAL pain to store, unless you have it in a totally sealed glass tube. I was offered approximately 500 grams of (old) potassium a few weeks ago, but I did not take it, simply because I'm afraid to have that stuff around. It may be interesting as a sample for a collection of elements, but having it around is a fire hazard and cutting the metal into smaller pieces can be really dangerous, if the sample is somewhat older (peroxide layers). A friend of mine had 25 grams of severely oxidized/peroxidized potassium, but he has thrown it away in a river to get rid of it.

Probably there were tiny amounts of reducing contaminants already present in the air. My experience with solutions of selenium (IV) compounds is that they are reduced a very little bit quite easily (e.g. dust, organic matter in the air). That "tar" is iodine, with dissolved selenium and SeO2. Due to its hydrofobic nature it tends to float on the water. Most likely there also are some bubbles of NO and/or N2O in it as contaminant. There also is another name for this stuff: "useless crap" . What sulfur do you use or how do you light it? My sulfur burns very neatly. I can light it fairly easily, when it is heated somewhat, with a match.

No, not at all. A very well known ester is nitroglycerin. The name of this compound is very confusing, because it is not a nitro-compound (with -NO2 attached directly at a carbon atom), but a nitrate-ester. So a better name would be glycerine-trinitrate. The formula of this triple ester is CH2(ONO2)CH(ONO2)CH2(ONO2). It can be produced by an esterification reaction between glycerin and HNO3, with H2SO4 as added compound to take away any water formed in this reaction. To my opinion the smell of most esters is not that pleasant at all. They all have a very "chemical" smell. What makes natural flavors so attractive is the subtle mix. Pure ethylacetate does not smell fruity to me, it is the mix of ethylacetate with lots of other compounds, which make it fruity. You can have endless variations.... Citrates may work, but do not expect much smell from these. The molecule is so large that it is not sufficiently volatile anymore to perceive any smell.

Another easy to make (and isolate) ester is the "inorganic" ester methyl nitrite. Mix 1 part of methanol with 5 parts of dilute acid (e.g. 10% HCl or H2SO4) and add some solid NaNO2 or KNO2. You'll get a lot of bubbles of a colorless gas. This is mainly methyl nitrite, CH3-O-NO Do not breathe the gas mix too much. As a side reaction there may be some NO-gas and that it VERY toxic, you also will breathe some methanol vapors, which are not good as well. If you light the ester-gas, then it burns with a green/gray flame, really nice to see that. If you have a mix of the gas and air (I did not determine the precise ratio), then you can see the green/gray flame move slowly into the test tube, until it reaches the surface of the liquid. Really cool to see this. Higher nitrite esters can also be made in this way easily (e.g. from butyl- or allyl alcohol). These are liquids, which can be extracted with ether and then purified by letting the ether evaporate.

The properties of an OH group depend in the atom, to which it is connected. With compounds like NaOH and KOH, the OH group in fact is a separate ion OH(-). These are purely basic. In a compound like Al(OH)3, the OH-groups are not purely ionic, they are bound to the Al-atom somewhat covalently, there is an Al-O bond. When this compound is added to a strongly acidic solution, then it acts as base: Al(OH)3 + 3H(+) ---> Al(3+) + 3H2O When this compound is added to a strongly alkaline solution, then it acts as acid: Al(OH)3 + 3OH(-) ----> AlO3(3-) + 3H2O So, here you see that the OH-groups act as acid, by splitting off a H(+). The compound B(OH)3 is purely acidic, it cannot split off OH(-) ions, but it can split off H(+), albeit only in strong alkalies. A rule of thumb exists: The more electroposive a metal, the more basic the compound M(OH)n. If a metal exists in multiple oxidation states, then the metal hydroxide for the highest oxidation state is the least basic. At high oxidation states, the hydroxides are acidic, e.g. SO2(OH)2 is a very strong acid, NO2(OH) also is. These are sulphuric acid and nitric acid. For Xeluc: The metals you mentioned, these hydroxides can be dissolved without problem in HCl. The oxides may be a different story. E.g. Cr2O3 cannot be dissolved in any aqueous liquid, no matter how acidic, once it is calcined. Fe2O3 only dissolves with VERY great difficulty, once it is calcined.