text:
Electrolysis of molten sodium iodide

Electrodes: Zinc electrodes
Electrolyte: molten sodium iodide

I know that at the cathode, sodium metal will be formed because only sodium ions are present for reduction.
At the anode, originally iodine should be formed with inert electrodes such as graphite. However, since zinc electrodes are used, and zinc metal is a stronger reducing agent than iodide, hence zinc should be oxidised into zinc ions.
My question is: what will be the reaction occurring at the anode? Will zinc ions or iodine be formed? Or is this case of electrolysis even possible?

Wow,  that's quite an exotic system. However, looks like it should work, and your reasoning is right. Zinc will be oxidized.

I'm curious why you would want to use Zn as the anode in this system if you are trying to electrolyze NaI.  You answered the question yourself -- use graphite if you going for electrolysis.


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Alkyl halide reaction
In the following the reaction ,
According to me the product should be compound  (A) due to substitution reaction.
But the product formed is the compound (B) , what could be the mechanism for the reaction for that product.

Part of product B is missing. I assume it is meant to be just a methyl group.

@bon yes you are correct , it is methyl group only .

To reflect on the comments below:
In literature this reaction seems to actually produce product A and synthesis of product B seems to require much more exotic conditions than described in the question. As such I have to conclude that the correct answer would have be A.

Old answer:
Due to the polar aprotic solvent (DMF) one would think Sn2 is favoured as the reaction mechanism, however if the product is indeed B it would suggest an Sn1 mechanism where the aromatic ring stabilises the cation. One of the resulting resonance structures would indeed react with fluoride. Due to the stability of aromatics the product is formed by re-aromatisation.

This is what I suspected must be the case but I can't give a good reason why the fluoride attacks at the ortho position rather than the benzyl position.

@bon If the primary carbocation is formed it will easily resonate to the secondary, much more stable, carbocation. Therefore the ortho position doesn't surprise me as much as the Sn1 mechanism does.

The secondary, non-aromatic resonance form, is more stable than the primary, aromatic resonance form...? Benzyl carbocations don't react on the ring

@orthocresol It struck me as weird the first time I saw similar mechanisms as well, but yes the aromaticity can be broken for such stabilization. However, I cannot find any mention of a synthesis ortho-fluorotoluene from benzyl bromide in literature. bon, are you sure B is the correct answer?

I have no idea what the correct answer is. I didn't ask the question. I too was unable to find any reference to this in the literature though so I suspect it may be wrong.

I just looked up the accepted ways of synthesising ortho-fluorotoluene and they are way more exotic conditions than these Finkelstein conditions. I'll edit my answer to reflect this.

Why the reaction proceeds like this ?

@koolman Which reaction are you asking about, to give A or to give B? To A would be due to the solvent (polar aprotic) promoting an Sn2 reaction, which occurs do to the higher stability of the fluoride (A) than the bromide (starting material).

Why B compond is formed ?

@koolman Well as you can read from my answer I actually highly doubt it would be in this reaction as none of the literature I was able to find supports that idea. I was wondering: who or what mentioned B as the correct answer to you?

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Drain cleaner on iodized salt reaction
I poured liquid heat brand drain opener onto iodized salt and didn't get a reaction? I think it's because the sulfuric acid is diluted. If so how can I remove the water or moisture from it?

And why do you think it wasn't basic cleaner?

According to the MSDS for Liquid Heat drain cleaner the chemical ingredients for the drain cleaner are $\ce{NaClO}$ & $\ce{NaOH}$.
Iodized Salt contains a simple mixture of KI and NaCl.
it makes sense that it doesn't look like much would happen when mixing those two together. $\ce{NaClO}$; however, would react with different substances.
See wikipedia on sodium hypochlorite reactions.
Sources:
https://www.quora.com/What-is-the-chemical-formula-of-Iodized-salt
https://www.menards.com/msds/113502_002.pdf

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Are certain substances in furniture truly toxic?
I'm looking to purchase a sofa, but have been bombarded with all kinds of internet info about the toxicity of certain materials used to make sofas.  For example, polyurathane foam is labeled as toxic--is this true?  Is something like a blend of cotton/polyester and polyurethane toxic?

I suppose being a couch potato is bad for you...

I wouldn't want to eat polyurethane foam, even less the ingredients it is made from. But that doesn't make sitting on it toxic.

@Jon Custer Truly.

The polyurethane would have to be absorbed through your pores when you're in contact with the sofa. That's not going to happen easily?
Theres a few tangent articles about the safety of these so called "cheap" sofas and mattress on livestrong and other websites.
I'd say it's not a concern. Sofa on bro.

Polyurethane is a polymer, how do you suppose absorbing such a huge molecule would work?

@caconyrn thats his whole point. It doesnt get absorbed.

Polyurethane is a polymer that's made up of toxic chemicals that react pretty completely with each other during the manufacturing process. While there's always potential that there could be residual unreacted materials, I'd say the risk of exposure to these is very minimal.
More important would be to look out for formaldehyde. It's a component in some of the adhesives that are commonly used in furniture making. New furniture can off-gas formaldehyde. If located in a room that doesn't get a lot of fresh air (either through the HVAC system or through building envelope leakage), then concentrations can build up. Sensitive individuals will experience upper respiratory tract and eye irritation. Long term exposure can cause sensitization - even small exposures would then lead to asthma like symptoms. It's also a recognized carcinogen, for good measure.
Problems mostly occur in office settings where they replace all the furnishings at once, and get a lot of off-gassing for the first several months. In a home environment, if you're just buying a sofa, it could still cause issues, but it's less likely.
The other broad category of chemicals to beware of are flame retardants. It's impossible to cover the full scope of these here, but suffice to say that different compounds are linked with health effects including atopic dermatitis, allergic rhinitis and asthma, and possibly endocrine disruption. Interesting to note that most Americans have measureable concentrations of various flame retardant chemicals in their blood. Related to this, there's a group of scientists that have come out strongly against incorporating flame retardants into furniture: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3002202/

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Mechanism of thyroid hormone synthesis: iodation and free radical formation

In the synthesis of thyroid hormone, what part of the thyroid peroxidase hormone is specifically responsible for abstracting the free radicals from the aromatic side chains in the iodated tyrosine residues on thyroglobulin?
Also, why are some of the tyrosines converted to monoiodotyrosine, and others diiodotyrosine? Does this have to do with steric hindrance, or the ring de/activating properties of the thyroglobulin backbone, or something else?


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Can ethyl acetate isolate cinnamaldehyde from cinnamon oil?
I recently watched a video on extracting cinnamaldehyde from cinnamon using dichloromethane as a solvent.  I would like to attempt the process.  However, I would prefer not to use haloalkanes because of the health risks.  Would ethyl acetate be a suitable replacement solvent for dichloromethane in extracting cinnamaldehyde from cinnamon oil?

You are wrong to generalise about the health risks of haloalkanes. dichloromethane is pretty safe by the standards of most laboratory chemicals.

Yes, ethyl acetate can be used to isolate cinnamaldehyde from cinnamon oil, though there are better non-haloalkane options. In a study that aimed to optimize the extraction of cinnamaldehyde from cinnamon powder, it was found that from methanol, ethanol, ethyl acetate, and water, methanol was the most efficient. The exact method used in the study was:

Collection of plant material: Bark of cinnamon was collected from a local market at Visakhapatnam, AP. The bark was cleaned and dried under sunlight for $24\ \mathrm{h}$. The dried bark was powdered and used as a raw material and stored in an air tight container. Cinnamon powder was sieved by using different particle sizes ranging from 354 to 125 microns.
Preparation of the extract: Cinnamon powder ($2\ \mathrm{g}$) was added with ethanol ($25\%$) and methanol ($25\%$) in different flasks and the volume was made $50\ \mathrm{mL}$. The solution was soaked for $1\ \mathrm{d}$ and $3\ \mathrm{d}$ respectively. After the soaking time, the solution was filtered using Whatman No.1 filter paper and the filtrate solution was heated to $78^\circ \mathrm{C}$ and $65^\circ \mathrm{C}$ and made up to $50\ \mathrm{mL}$ with distilled water and hexane and incubated for $2\ \mathrm{h}$.


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Would Al ions still be reducd to Al metal if it was aqueous soltion?
To my knowledge, Aluminium is mostly mined in the form of Aluminium oxide or Al2O3 and then the Bayer process is used to extract and refine the pure Aluminium.
Now, The Al2O3 must be in a molten state when the process occurs. Why is that? Why must it be molten?
Also, Would the Al(3+) ions still be reduced to the Al metal if it was in an aqueous solution instead of being molten?
Thanks,

That's the problem - they wouldn't. http://chemistry.stackexchange.com/questions/40199/why-cant-al3-be-reduced-to-solid-pure-aluminum

Ok. So, let me just clarify. Are you saying that Al3+ won't be reduced to pure Aluminium through the Bayer Process?

Al3+ can't be reduced in water, that's what I mean.

The Bayer process does not produce aluminium. It produces sodium aluminate, and finally aluminum oxide, which is later used in the Hall–Héroult process (without water) to produce aluminium metal.
As @Mithoron implied, aluminium will displace the hydrogen from water, so cannot be made in an aqueous environment.


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Why do sealed plastic bottles of pop collapse as the contents get old?
A number of bottles of pop (two varieties) in unopened plastic bottles have sat for a few years in a room-temperature environment. All of the bottles have now collapsed a certain amount (all to about the same degree).
They all started out under pressure from the carbonation, but now, not only is that gone, but the internal pressure has dropped below ambient. They all still have an "air" space above the liquid contents; not sure if it is or is not more than when originally purchased. If one of these bottles is opened, it draws in air and regains its original shape.
I have the following questions:

What happened to the carbonation? Did it leak out, or did it undergo some sort of chemical reaction, or perhaps both?
How could the pressure inside the bottle drop below ambient? What chemical or physical process is causing it, and what is driving the process? Why would it not simply reach equilibrium with ambient?
What would be the ultimate result if these bottles were left even longer?

The carbonate diffuses out until its partial pressure is the same in the bottle and outside. Strictly speaking the same is true for the water in the bottle.

@Karl Diffuses through the plastic? If the CO2 and H2O diffuse out, why doesn't O2/N2 diffuse in?

A net transport by diffusion depends on a gradient in partial pressure. Also the diffusion coefficients of the gases are very different.

comment: This happens to me with sealed plastic water bottles (non-carbonated) at work, where the indoor temperature is set by a thermostat to the same temperature (day and night, including weekends and holidays, summer winter). Only takes a few months for the sides of the bottle to cave in. I've noticed this for occurring for over a decade, in different buildings at work. Happens with many different brands of bottled water. (I keep the water bottles a year as part of my emergency supply). I wonder if the walls of the container weaken, and became stretched?

It is most likely the CO2, H2O and air in the pop bottle headspace diffused (leaked) out of the pop bottle.
You are correct: the pressure has reached ambient (not dropped below ambient). Because the pressure in a freshly bottle pop bottle was originally higher than the pressure in the room (normally atmospheric pressure), and the bottle's plastic walls were smooth and strong, any decrease in pressure inside the pop bottle would cause the pop bottle walls to collapse.
If the bottles were left longer, the CO2 and water would continue to evaporate from the liquid phase until the headspace contained approximately the same pressure as the air in the room. If somehow all the water evaporated, you'd be left with sugar and a bunch of other solids (that were dissolved). I doubt full evaporation of water would occur.

Also, if the bottle is made from PETE (polyethylene terephthalate), prolonged contact with water (including pop) will result in many by-products. Two of them are:
a. UV exposure, air and water (together) promote peroxide formation on polymer end-groups (OH, COOH). Another way of saying this is that the bottle walls degrade.
b. The PETE plastic could leach some trace antimony (if that was used in the polymerization process) and/or phthalates, suspected to be endocrine disruptors.
IF water permeates through the bottle walls, it can allow for microoganisms to diffuse in/out of the bottle. You might get some mould in there!
Those are just some things that could happen.

Re: point #2: The pressure inside the bottles is slightly less than ambient - the partially-collapsed bottles "want" to return to their normal shape and do so as soon as they are opened. The diffusion process has evidently maintained a less-than ambient internal pressure.

re: micro-organisms... I was really only aiming to understand the limiting condition of the collapse process.

The weakest point in a bottle with a lid is usually the lid - note that differences in temperature can see the lid loosen and gas escape (I'm going to guess there's no wax or similar over the lid to prevent this).
Alternatively, some other form of slow leak exists.
I'm also going to guess that there's some temperature variation in the room they're stored in. In this case, when it's warmer, more gas will escape and when it's cooler the pressure will be lower than external ambient .
 There is a famous experiment where you take an empty bottle or similar, and place it in hot water with no lid, then place it cold water with a lid on, whereupon it collapses, illustrating this last idea.

I know the demonstration. Boil water in a can, tightly cap it and allow to cool. As the water condenses, atmospheric pressure crushes it. But in this case, temperature variation is just a few degrees, ranging from minimum of  maybe 19C to a maximum of 25C, and it is liquid volume which has been lost.  I haven't measured, but "guesstimate" that they have lost perhaps 25% or more of their  liquid volume.

Over a few years, across multiple change of seasons? Okay, that was a long way from my experience. In my place, it would be from 10 C night time winter to 30 C when we get home from work in summer (i.e. before we switch the A/C on), based on observing the thermostat.

CO2 diffused through the plastic or possibly leaked out the closure. PET is the most common "plastic" bottle in the USA but others are used. However, when PET was developed many things were stored for extensive periods in bottles to satisfy FDA rules and provide technical information. As I remember ,carbonated beverages went over 2 years with no change ( this may be searchable under Amoco Chemicals after about 1970.) I suspect you had a different plastic ; For example , H2S  totally diffuses through vinyl in 24 hours.


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Spin of electrons about it's own axis, release of energy & COLLAPSE of atom?
An electron spins about it's own axis. So it has rotational accln. Thus, according  to Maxwell's Electromagnetic Theory , Rotational Kinetic energy must be released. Thus, at some pt. of time , shouldn't the energy of an electron become 0 & then shouldn't the electron move out of  the atom, eventually leading to the atom only having the nucleus .

The electron isn't spinning on it's own axis

Why aren't you concerned with electron's "rotation" around the nucleus in the first place? (The explanation is pretty much the same: there is no actual rotation.) But wait, there is more to it: a free electron has spin as well! Had it lose energy through some mechanism, what would eventually become of it?

hey @IvanNeretin , I do know about the rotation of electrons around the nucleus(in fact it was a defect of Bohr's Theory). My question's  pretty similar(though it's about rotation of electrons about it's own axis). Hope u get my question.

You are confusing macroscopic spin with the spin angular momentum of electron here. Electron is a point entity and does not posses a macroscopic spin (i.e. it does not spin on its axis).
 It does have angular momentum which is expressed as 'spin angular momentum' which is of-course quantized.
This link   by  John Rennie will be useful for understanding : The electron isn't spinning on it's own axis


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Does the rate of movement of ions in a salt bridge affect the total voltage produced by a Galvanic cell?
I was thinking that different salts used in a salt bridge in a voltaic cell may have different rates of flow of ions in them (perhaps ionic radius or some other property affects the rate?)
I was wondering, if some ions in the salt bridge flow through the solution more quickly, will this have any effect on the overall voltage produced by the cell?

Voltage measurement is ideally a zero current process. Therefore the details of the ions in the salt bridge and their movements will not be an issue.
In reality, voltmeters do flow some current to measure the voltages. Yet, these currents are so small that in any self-respecting salt bridge, there will not be an issue. If your salt bridge was broken for whatever reason though, then your measurement will be completely incorrect. Most voltmeters will give you an overload in that case.

I'd point out that modern voltmeters use such high impedance solid state op amp circuits such that there is essentially no current drawn. However in the "olden" days this was not the case with analog voltmeters. I remember the headache of using a Wheatstone bridge to get more precise voltage measurements. So the "good old days" are now!

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Is there a such thing as a reburnable gas or can one be created?
I want to know is there a such thing as a gas that can be reburned multiple times?

What you essentially asking for is free energy. This is not possible.

In short: There are no compounds that can go through complete combustion more than once.
Incomplete combustion products, like carbon monoxide, can be burned to give the complete combustion product $CO_2$. This would however then not be the same gas as was (incompletely) burned to give the carbon monoxide.
As bon mentioned in his comment: a compound can never be burned twice, since if the product of the reaction would be the starting material and energy were to be expected to be released, as with a fuel, then energy would come from nothing, since no chemical change would have occurred.


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Do cyclic oxygen molecules exist and their theoretical stability?
Do oxygen molecules ever become cyclic shapes like carbon and other elements?

Why doesn't oxygen form cyclic compounds? How stable would they be if they do exist. Especially more than 3 oxygens. Why are there just 2 allotropes of oxygen known to mankind.

Possible duplicate of A different structure of Ozone?

see also http://scitation.aip.org/content/aip/journal/jcp/94/12/10.1063/1.460140 and http://www.sciencedirect.com/science/article/pii/S2210271X15003813 and https://en.wikipedia.org/wiki/Cyclic_ozone

I don't think this is a dupe. This question is more broad than the suggested dupe.

@bon yes, it's broader now; it was edited after I said possible duplicate

If it were me, I'd un-accept the currently accepted answer, and ask for supporting sources. Comments about lack of sources there are valid.

Unlike the likes of carbon, oxygen-oxygen single bonds are highly unstable. Because oxygen is such an electronegative compound, it can only become satisfied with bonds where it hoards a majority of the electrons. Because of this, compounds such as peroxides ($\ce{RO-OR}$) will readily decompose into two radicals. This is why hydrogen peroxide is used as a radical initiator for chemical reactions. Since a single $\ce{O-O}$ bond is so unstable, it would be extremely unlikely for a larger chain to exist. Peroxides by themselves are highly unstable. Not only do they decompose into radicals, but they also rapidly decompose into water. Peroxides have a single $\ce{O-O}$ bond, and are somewhat stabilized by donating groups on either side of the peroxide. Now, imagine a chain that looks like this: $\ce{O-O-O}$. The central oxygen is not only bonded one inductively withdrawing oxygen as in peroxides, but two of them. In addition, it is not stabilized by electron donation from any source. Despite this, it is predicted that small amounts of cyclic ozone might exist in the atmosphere. Cyclic ozone is so much less stable than regular ozone for the reasons I listed above, but it should exist. Any larger oxygen ring will most likely not form.
Carbon does not have this same problem. Since carbon has a mid range electro-negativity, it is stable forming bonds with itself. In addition, carbon can form four covalent bonds, normally two of which are with hydrogens (as in a cycloalkane). Carbon is perfectly stable with this electron density, leading to the stability of cyclic alkanes.

It's not that simple and you're overstating reactivity of those bonds. See http://chemistry.stackexchange.com/questions/35574/are-long-oxygen-molecules-possible

"Trioxidane has a half-life around 16 minutes in organic solvents, but only milliseconds in water."

But aren't unlikely to exist. You should use sources and write using NPOV, not try to make your point.


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Why does Devarda's alloy include copper?
Devarda's alloy is used in the detection of nitrates. However, judging by the reaction equation, only aluminium plays the role of reducer:
$$\ce{3NaNO3 + 8Al + 5NaOH + 18H2O -> 8Na[Al(OH)4] + 3NH3(g)}$$
Why does it contain copper? Is copper also capable of acting as a reducer here?
I know that zinc, the third component, is capable of that.

Galvanic couple, I guess.

Catalyst, I guess. ;-)

Blessed be the guessers, for they lead us to understanding.

Virtually every analytical chemistry textbook's chapter on anion determination mentions that Devarda's alloy is very brittle and can be easily pulverized in a mortar, unlike $\ce{Al}$ and $\ce{Zn}$ metals.
This is quite handy property as for qualitative analysis of nitrates in a heterogeneous media the reducing phase should be ground to fine powder in order to:

assure micro-scale precision of the added reactant, which is tricky to achieve with aluminium or zinc, as both metals are usually shipped in granules, and the fine metal powders are prone to oxidation, increasing the amount of unwanted oxides (plus, their powders are sticky);
increase the available reaction surface to speed up the analysis and increase its completeness (especially important for isotopic nitrate analysis).

On top of that, German Wikipedia
also supports both comments by Ivan Neretin

Galvanic couple, I guess.

and aventurin

Catalyst, I guess. ;-)

and suggests that copper facilitates catalytical formation of a local electrochemical cell.
A reduction with Devarda's alloy can be carried out in weakly alkaline solution or even in neutral solution, whereas reduction with aluminum often requires a strongly alkaline solution.
At the same pH, the reaction with Devarda's alloy is overall significantly faster than when using zinc or aluminum.
This was experimentally demonstrated by Mertens et al. [1]:

The superiority of Devarda alloy compared to aluminum dust as a reductant in alkaline medium is illustrated best with the following experiment (see Figure 3).
  Twenty ppm $\ce{NO3-}$ is subjected to reduction, first with aluminum dust and the potential of the ammonia probe is recorded continuously.
  One observes a slowly rising potential.
  On addition of Devarda alloy to this reaction mixture, a potential jump is obtained and the potential stabilizes rapidly (after 5 minutes).
  This shows clearly that the kinetics of the reduction with Devarda alloy are much more rapid than with aluminum dust.

Figure 3. Potential observed with the ammonia probe during the reduction of 20 ppm $\ce{NO3-}$ as a function of time
  From $t = 0$ to A: aluminum dust. At point A, Devarda alloy was added.

References

Mertens, John.; Van den Winkel, Pierre.; Massart, D. L. Determination of Nitrate in Water with an Ammonia Probe. Anal. Chem. 1975, 47 (3), 522–526. DOI: 10.1021/ac60353a009.

Blessed be the commenters for they lead us to an answer after 3.5 years ;)

More recent science (as compared to 1975), the action of Al with NaOH is believed to produce surface absorbed hydrogen atoms on the aluminum. Related chemistry is commonly employed via the vigorous action of HCl on Zn (so one of the roles of zinc is to provide added surface area, which may be acted upon by NaOH also, to answer a posited question).
More precisely on the referenced surface chemistry, let me quote from this 2008 doctoral thesis " Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena" by Saikat Adhikari, as an example:

Hydrogen embrittlement, a process in which brittlement and subsequent cracking of metals is caused by ingress of atomic hydrogen into the metal, is also sometimes thought of as a type of stress corrosion cracking. Due to the detrimental effects of such corrosion processes, it is important to have a fundamental understanding of associated alkaline or acidic dissolution mechanisms.

And more directly:

Apart from activation of the surface, dissolution of aluminum in alkaline solutions at open-circuit also leads extremely high rates of H-absorption into the metal, [9-14] and also formation of nanoscale voids or hydrogen bubbles [15, 16] near the aluminum surface.

H-absorption on the Aluminum surface is a source of hydrogen radicals, •H, which could engage in a stepwise breakdown of nitrate to ammonia.
For example, with the nitrate ion a radical reaction:
$\ce{•H + NO3- ⇌ •OH + NO2-}$
$\ce{•H + •OH -> H2O}$
As: e- + H+ ⇌ •H, the net of the above two reactions can be written as (source, see Section 8.28.2.1 Summary of Properties) the following cited reaction for the first stage creation of nitrite:
$\ce{NO3- + 2 H+ + 2 e- -> NO2- + H2O}$
My experience with just alkaline Al is that it is apparently difficult to stop the reaction at the point of just nitrite creation only, by say limiting the amount of aluminum.
So we have a complex reaction system involving surface chemistry, and in the presence of other metals (like copper), a galvanic cell (which implies two different metals, namely Zn and Cu, to answer a second question) introducing solvated electrons, which may push the equilibrium below to the right:
$$
e-(aq) +  H3O+(aq) ⇌ •H (aq)
$$
promoting hydrogen radicals. The latter could explain the acceleration in the breakdown of nitrate (as reported).
Interestingly, one may assume that the •H radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = •H ) as apparently a (e-, H+) pair acting on ions. For an example from 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive leaching application citing the reaction (based on surface-treated Aluminum) to quote:
" PbS + 2 •H = Pb + H2S (5) "
Source reference for the above.
which I view functionally as follows:
Pb(+2)S(2-) + 2 (e-, H+) = Pb + H2S (g)
Further, I do NOT recommend experimenting with Al/NaOH in the presence of impurities in a chemistry class. To quote from Bretherick, Volume 1, page 23-24:

A student mixed aluminum foil and drain cleaner in a soft drink bottle, which started emitting gas. Another student carried the bottle outside and was claimed to have been overcome by the toxic fumes [1]. Most drain cleaners are alkalis, so that aluminum will dissolve to produce hydrogen. The bleach that it is is suggested may have been present will produce no toxic fumes in alkali, and one would be surprised to find arsenic or antimony compounds present. If the collapse was not purely hysterical, the remaining, though remote, possibility would be phosphine. The soft drink the bottle had contained was one of the many perhaps best described as impure dilute phosphoric acid [2]. "

I suspect that the fumes coming from the Al foil and drain cleaner were a mist of H2O and NaOH, possibly with some NH3. Drano consists of NaOH plus aluminum and other stuff, including NaNO3 (to reduce H2 gas production). The nitrate can be reduced to NH3. A soft drink bottle (1-2 L?) with ~200 mL H2O and a lot of Al foil and Drano will create a lot of heat and H2 and mist, and NH3 would be scary - but it is all much safer if done on a small scale to see how fast the reaction goes. And in an open space, with ventilation.

No, per the dialogue, possible mention of PH3 from the phosphate-rich drink, which parallels the breakdown of nitrate to NH3 !

I am referring to the unintentional formation of lethal compounds from miscellaneous salts interacting with the hydrogen radical. Per the hydrometallurgy example, sulfides can introduce H2S. More generally and problematically, salts of Hydroarsenic acid (arsenical analogue of phosphoric acid), for example,  could produce deadly Arsine (AsH3),...


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What techniques can determine the concentration of ethyl acetate in a solution containing ethyl acetate, acetic acid, ethanol and water?
I need to monitor the concentration of ethyl acetate in an Fischer esterification reaction. What techniques or spectroscopies can detect the concentration of the compound? And which one do you recommend based on precision and less sample preparation?

How much ethyl acetate is present in the mixture ?

Hold on, why EA? I assume that the esterification you are doing is between the citric acid and ethanol, so shouldn't you be looking to monitor the ethyl citrate product? I can't imagine where EA would come from unless you added it from the beginning.

@ITTsoi I'm sorry, you're right. I meant acetic acid. It was a typo.

@ankit7540 The concentration of ethyl acetate is considerable (>10% ww)

You might consider taking Aliquots of the reaction (or if you have the equipment directly sampling via a cannula) into a GC-FID or quantitative GC-MS. This way you could monitor the concentrations of all species in solution given the right GC method. Areas under each peak will give you relative amounts, you would need to use an appropriate known amount of an internal standard that remains constant to get absolute numbers.

You don't mention the scale of the reaction, however 1H NMR will be the easiest and most effective method for monitoring the progress of the reaction, and give very accurate results.
If done on a small scale, you could take 50uL aliquots, and dilute in deuterated solvent. If you are doing this on a large scale, you simply take 0.4mL aliquots, and either add a small amount of deuterated solvent for lock, or simply run it unlocked. If running it neat, you can return your sample to the reaction vessel to continue the reaction.
Monitor the reaction via the concentration of the methylene groups for ethanol and ethyl acetate. The ethanol CH2 comes around 3.7ppm, and the ethyl acetate CH2 comes around 4.1ppm. Easy peasy.

Great answers above using specific techniques.
I would use in-situ Raman spectroscopy.
Sample preparation : No sample preparation needed. The whole reaction vessel can be moved to the sample point of Raman spectrometer (if the Raman spectrometer allows it). The incoming laser beam will not affect the reaction since none of the chemicals mentioned are light sensitive.
Advantage : No need to open the reaction vessel and take aliquot after some time duration. You can continuously monitor the change (~ every 1-5 minutes depending upon spectrometer throughput).
Quantitative information:
2 approaches:

Using an internal standard:
Raman peak from some species whose concentration stays same throughout can be used as a relative internal standard. Concentration of other species can be found relative to this standard.
Using abolute Raman cross-section: Absolute Raman cross-section of any species if known can be used and concentration can be found absolutely in the reaction vessel. If concentration of one species in known accurately others can be found relative to it.

*If you need more info send a comment.

Since custom Raman spectrometer is hard to find, it is more practical to use an instrument which is available in your department and is free to use, which is more likely to be an NMR instrument.


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Is UV-Vis absorption polarization dependent ? Why?
Consider molecule of interest be dissolved in a solvent (and this molecule has absorption in the visible region). We are using a broadband visible light source say tungsten lamp.  This light passed through a polarizer and then split into two beams, one passing through sample cell, and other through reference cell having solvent.
Would there be difference in the electronic absorption spectra when measured keeping the polarizer parallel or perpendicular with respect to the incident polarization ?  Also, no external electric field is present and we have randomly oriented molecules.
(Consider the transmission of optics and detector are corrected for this polarization difference)

Only if there is a good reason for the solute to have a preferred orientation in the liquid (external electric field, say).

@Jon - Thanks for the point about electric field.

With the experimental setup you describe, you would not observe any intentional polarization dependence, because the light passes only through a beam splitter, not a polarizer.
There can be unintentional polarization due to the nature of optical materials and manufacturing, caused by

the diffraction grating,
the polarization contribution of mirror reflections, and
the diffraction grating slits.

With that out of the way, UV/Vis absorption is polarization dependent. In theory, polarization can occur in any spectroscopic technique where the polarization of the incident EM radiation can be controlled. To quote the Wikipedia article,

In general, any kind of anisotropy in the media results in some sort of change in polarization. Such an anisotropy can be either inherent to the media (e.g., in the case of a crystal substance), or imposed externally (e.g., in the presence of magnetic field in plasma).

Assuming that no other external fields are applied other than your probe beam and your bulk phase has no anisotropy, any differential absorption of polarized light must come from the molecules themselves, namely those that are "optically active". Linear dichroism and particularly circular dichroism are used to describe the differential absorption of linearly and circularly polarized light, respectively. Electronic (UV/Vis) CD is widely applied in the determination of protein secondary structure and the absolute configuration of stereocenters in chiral molecules. An ECD spectrum is usually presented as a difference spectrum between the left- and right-polarized spectra, which themselves are no different from a typical UV/Vis spectrum.
The parent dichroism article is a good starting point for learning about these spectroscopies.

Thinking about the first sentence, since in the setup I describe I do keep a polarizer to get linearly polarized excitation light. Do you mean that the beam splitter may induce polarization aberration for the two split beams ?

I'm not sure why I thought that you didn't have a polarizer present. Are you sure you don't have a monochromator instead of a polarizer? I didn't intend to say that the beam splitter would cause an aberration, but it might be possible. I will ask my 2D spectroscopy friend. I'm thinking a real answer to your question with your setup is "yes", with an explanation  of orientation dependence (like here).


text:
How the equation that describes evolution of alkyl nitrates has been derived?
I was reading this paper, Bertman, S. B., J. M. Roberts, D. D. Parrish, M. P. Buhr, P. D. Goldan, W. C. Kuster, F. C. Fehsenfeld, S. A. Montzka, and H. Westberg (1995), Evolution of alkyl nitrates with air mass age, J. Geophys. Res., 100(D11), 22805–22813, doi:10.1029/95JD02030 (link attached), and had a hard time trying to derive equation (12) from equation (11). If anyone knows how to do that, help would be highly appreciated.

You have to quote a bit more of the paper for context, namely the reaction scheme. In the paper, it is given as
$$\ce{RH ->[\beta k_\mathrm{A}] RONO2 ->[k_\mathrm{B}] P}$$
The first thing to notice is that the concentration of $\ce{RH}$ as a function of time is given by the first-order integrated rate law
$$[\ce{RH}]_t = [\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)$$
and therefore equation (11) becomes
$$\frac{\mathrm{d}[\ce{RONO2}]}{\mathrm{d}t} = \beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)- k_\mathrm{B}[\ce{RONO2}]$$
This is a ordinary differential equation and there are many ways of solving them (this is hardly the place to discuss ODEs; consult a maths book if you are interested). In this case, the way I did it is to multiply throughout by $\exp(k_\mathrm{B}t)$. I am going to use $c$ to represent $[\ce{RONO2}]$ to simplify the equation a little bit:
$$\begin{align}
\frac{\mathrm{d}c}{\mathrm{d}t} &= \beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)- k_\mathrm{B}c \\
\exp(k_\mathrm{B}t)\frac{\mathrm{d}c}{\mathrm{d}t} + \exp(k_\mathrm{B}t)k_\mathrm{B}c &= \exp(k_\mathrm{B}t)\beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t) \\
&= \beta k_\mathrm{A}[\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t]
\end{align}$$
Now if you consider the derivative of $\exp(k_\mathrm{B}t)c$ with respect to $t$, we have (by the product rule, and bearing in mind that $c$ itself is a function of $t$)
$$\frac{\mathrm{d}}{\mathrm{d}t}[c\exp(k_\mathrm{B}t)] = \frac{\mathrm{d}c}{\mathrm{d}t}\exp(k_\mathrm{B}t) + ck_\mathrm{B}\exp(k_\mathrm{B}t)$$
which is the same as the left-hand side of the equation above. The thing we multiplied by throughout, $\exp(k_\mathrm{B}t)$, is called an integrating factor. So, we can now integrate both sides:
$$\begin{align}
\frac{\mathrm{d}[c\exp(k_\mathrm{B}t)]}{\mathrm{d}t} &= \beta k_\mathrm{A}[\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] \\
\int_0^t \frac{\mathrm{d}[c\exp(k_\mathrm{B}t)]}{\mathrm{d}t}\mathrm{d}t &= \beta k_\mathrm{A}[\ce{RH}]_0 \int_0^t \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] \mathrm{d}t \\
\left[c\exp(k_\mathrm{B}t)\right]_{t=0}^{t} &= \beta k_\mathrm{A}[\ce{RH}]_0 \left(\frac{\exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - 1}{k_\mathrm{B}-\beta k_\mathrm{A}}\right) \\
c_t\exp(k_\mathrm{B}t) - c_0 &= \beta k_\mathrm{A}[\ce{RH}]_0 \left(\frac{\exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - 1}{k_\mathrm{B}-\beta k_\mathrm{A}}\right) \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - [\ce{RH}]_0) \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_t \exp(k_\mathrm{B}t) - [\ce{RH}]_t\exp(\beta k_\mathrm{A}t)) \\
c_t\exp(k_\mathrm{B}t) &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_t \exp(k_\mathrm{B}t) - [\ce{RH}]_t\exp(\beta k_\mathrm{A}t)) + c_0 \\
\frac{c_t}{[\ce{RH}]_t}\exp(k_\mathrm{B}t) &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (\exp(k_\mathrm{B}t) - \exp(\beta k_\mathrm{A}t)) + \frac{c_0}{[\ce{RH}]_t} \\
\frac{c_t}{[\ce{RH}]_t} &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{\exp(-k_\mathrm{B}t)c_0}{[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)} \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{c_0}{[\ce{RH}]_0}\exp[(\beta k_\mathrm{A} - k_\mathrm{B})t] \\
\end{align}$$
I have shown quite a lot of the intermediate steps, if anything seems puzzling just work through it, it is really just algebraic manipulation.
Replacing $c$ with $[\ce{RONO2}]$ gets you to nearly your desired equation
$$\frac{[\ce{RONO2}]_t}{[\ce{RH}]_t} = \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{[\ce{RONO2}]_0}{[\ce{RH}]_0}\exp[(\beta k_\mathrm{A} - k_\mathrm{B})t]$$
The only difference in the paper is that $\beta$ was sometimes dropped (i.e. some of the $\beta k_\mathrm{A}$'s simply became $k_\mathrm{A}$). I glanced through the paper and can't find a rationale for it, but I will admit I did not look at it in great detail.


text:
What is the difference between the equilibrium position and the equilibrium constant?
I used to think that the equilibrium constant was a quantitative measure of the equilibrium. However, changing concentration and pressure in a system at equilibrium that shifts the equilibrium doesn't affect the equilibrium constant, which is solely temperature dependent. This means that my definition was wrong. What is the real relationship between the two, or what do they actually mean? (I haven't been able to find any suitable definition for "equilibrium position," and the only definition I know how "equilibrium constant" is the value of the reaction quotient when a system has reached equilibrium).

Here's a more simplistic response.
Equilibrium position: a particular set of concentrations of reactant and product species.
Equilibrium constant: a number that describes how far "forward" equilibrium lies for a particular system (as described by a particular equation).
Example: for the system $\ce{A <=> B}$, the mass action expression is $\ce{\frac{[B]}{[A]}}$.  Let's say $K = 200$.
This means, at equilibrium, $\ce{[B]}$ will be 200 times greater than $\ce{[A]}$. That's what the equilibrium constant tells you - equilibrium for this system lies to the right, and the K value is greater than 1.
There are infinitely many different equilibrium positions that satisfy K.  We could have $\ce{[B]} = 20~\mathrm{M}$ and $\ce{[A]} = 0.1~\mathrm{M}$.  Or we could have $\ce{[B]} = 1~\mathrm{M}$ and $\ce{[A]} = 0.005~\mathrm{M}$.  Those are two different equilibrium positions that are both at equilibrium.  In both cases, $Q = K$ and the system is at equilibrium.
Some notes:

If we describe the exact same system with a different equation, $\ce{B <=> A}$, then the mass action expression is $\ce{\frac{[A]}{[B]}}$. We would find that $K = 0.005$ (that is, $1/200$) and we would say that equilibrium lies to the left.  Of course, the molecules don't behave any differently, but the equation we use to describe the system is different, and that difference is reflected in a different $K$.
If we have a system with more than one product or reactant or with stoichiometric coefficients other than 1, then it's a bit more complicated than just saying something like "product concentrations are 200x greater than reactant concentrations". Instead, equilibrium may be achieved at many different equilibrium positions. You have to look at the mass action expression to interpret the meaning of $K$.
The equilibrium constant $K$ is just one way of expressing how far "forward" a particular reaction will proceed.  Gibbs free energy change ($\Delta G$) is another way. And, for redox systems, cell potential is yet another way.

Equilibrium position:
A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .
This happens when the reaction Gibbs energy becomes zero viz. $\Delta_r G= 0\;.$
Equilibrium constant:
The reaction quotient $Q$ is given by
$$\Delta_r G= \Delta_r G^{\Theta} + RT\ln Q\;\tag 1 $$
where $^\Theta$ denotes the term is measured under standard conditions.
Now, in equilibrium, $\Delta_r G= 0$ which implies
$$\Delta_r G^\Theta = -RT\ln Q_\textrm{equilibrium}\tag 2$$
From $(2)$ we can define
$$Q_\textrm{equilibrium}= e^{-\frac{\Delta_rG^\Theta}{RT}}=K\equiv \textrm{equilibrium constant}\;.$$
Response of Equilibrium constant to the shifting of Equilibrium position:
Shifting of equilibrium position implies a change in the conditions the reaction is carried under. This may include change in temperature or pressure.
That would definitely influence the standard reaction Gibbs energy $\Delta_r G^\Theta$ and hence the value of $K\;.$
Note that the equilibrium position is still the condition  when $\Delta_r G$ becomes zero; but the value of $K,$ the equilibrium constant, does not necessarily need to remain the same (for instance, $K$ doesn't change for a change in pressure since $\Delta_rG^\Theta$ remains the same irrespective of the change in pressure; nevertheless the individual components can change without altering the value of $K$ in whole).

I will try to give a simpler answer!
$Q,$ the reaction quotient, basically tells you the ratio of products:reactants right now. When you add $\ce{HF}$ to pure water $Q,$ is $0$ at the start because in the reaction $$\ce{H2O + HF <=> H3O+ + F-}$$ the ratio of products to reactants is $0$. $$\ce{Q=\dfrac{[H3O+][F^{-}]}{[HF]}=0}$$
However, the equilibrium constant of HF is $$K=2.7\times 10^{-2}~\rm M$$
$K$ is the value of $Q$ at equilibrium. The expression for $Q$ and $K$ are the same however $Q$ describes the state of a system and $K$ describes the system at equilibrium. Thus, as the reaction proceeds, more and more $\ce{HF}$ dissociates. This causes a larger $\ce{[H3O+][F^{-}]}$ and hence $Q$ increases and approaches K. Le Chatelier's principle tells us that a system that has $Q\neq K$ will approach $K.$ When $Q$ is not $K,$ the system is not at equilibrium.
Hope this helps!

Q is the reaction quotient not the equilibrium constant. K is the equilibrium constant and it is equal to the reaction quotient at equilibrium.


text:
Why did the alpha-particles in Rutherford's experiment not collide with the electrons?
In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.
But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.

The body of your question has nothing in particular to do with neutrons - perhaps you mean nucleus? (Yes, very similar names). As for scattering off of electrons, the alpha particles most certainly were interacting with the electrons, and as a result lost energy as they traversed the material. This so-called electronic stopping is very small compared with a nucleus-nucleus collision event primarily because the mass of the electron is so much smaller than even a proton - you just can't have much energy transfer in such a collision.

Protons and neutrons have a mass approximately 1836 times greater than an electron, ignoring relativistic effects and nuclear binding energies. There are 4 of these baryons in an $\mathrm{\alpha}$-particle. There is no way a puny electron could do much to deflect an $\mathrm{\alpha}$-particle that is almost 7500 times more massive. This is like if a 150-lb human tried to block a 747 (the 747-100's maximum takeoff weight is only about 6400 times greater than the person's).

Your answer is essentially on the right track but you should point out that the "target size" isn't actually measured in mass but in terms of a quantity named the cross-section.

@MaxW You're absolutely correct. I was trying to aim more for a more intuitive answer, but yes, a cross section would be the technically correct way to approach this problem.


text:
Is it possible to quantify alkane branching?
Is it possible to express the extent to which an alkane has branching (branch count)?

Your question is a bit unclear as to what exactly you are wondering. However, for a simple saturated hydrocarbon what's IUPAC name is given, its base name is given by the largest chain. Knowing this, all branching is indicated by the named substituents that branch off of this main chain.
For example, 2,3,5-trimethylhexane will be a 6-carbon saturated chain with 3 branches. Hopefully this helps answer your question, maybe you can add some clarification?


text:
Naming a tri-substituted amine
Just came across this couldn't decide how to name it, anyone think they can give it a go?
$\hspace{7cm}$

Related: Amine Nomenclature – When to use “N” and Nomenclature of Amines

This is N-ethyl-N-methylpropan-2-amine. The base of this amine is propan-2-amine, and the nitrogen is substituted with an ethyl (N-ethyl) and a methyl (N-methyl) group.


text:
pH of Strong Acid (and why it seems like it should be zero)
I understand the Henderson-Hasselbalch equation for acids that do not completely dissociate but I am having trouble calculating the pH of a strong acid using the equation: for example a .25 m solution.
pH = p(Ka) + log([A-]/[HA])

which is
pH = p([A-][H+]/[HA]) + log(.25/0)

which is approximately
pH = p(.25^2/0) + infinity

and therefore
pH = -infinity + infinity = 0

So why isn't the pH of any strong acid 0?

This is a question on math, not chemistry. In short, you just can't operate with infinities as if they were normal numbers, because they are not.

so we just arbitrarily define it to be the log of molar concentration of the strong acid?

We do not have to define anything special for this case, arbitrarily or not. The ordinary definition of pH still applies without problems.

People make the mistake that anything divided by 0 is infinity. In fact it is not defined.  Besides that the Henderson-Hasselbalch equation comes from the equilibrium constant equation for the dissociation of a weak acid not a strong acid, so don't use it. $pH = -\log[H^+]$. So assuming you have a strong acid, you have complete dissociation of acidic Hs. So you now know the $[H^+]$ so simply get $-\log$ of it. Don't over complicate things :)

https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation especially "limitations"

You are confusing infinitely strong acid with simply a strong acid.
For any real acid, there will be some undissociated acid (HA) present.  While countless books say strong acids completely dissociate, this is just an approximation.
See Table of Acid and Base Strength for the Ka values of some common strong acids.

The ${K_{a}}$ of a strong acid is not infinity. It is a defined number depending on solvent. Thus the assumptions you make do not hold and you don't have to worry about dealing with infinities.

Indeed, that's true.

Yeah that is true but then let's look at the the definition of ka = h a- / ha ....multiply both sides by ha and take the log we get log(ka) + log(ha) = log(h^2) since h and a- are equal... This still doesn't give the rift andwer

So it seems we just break from the normal rules

We very much do follow the normal rules. Broken, if anything, is your perception thereof.

One factor here is that of temperature. As the temperature rıses, ionızation ıncreases. pH ıs measured at room temperature. Heating the solutıon of a strong acıs wıll ıncrease ıonizatıon and drop the pH.
Melek

One principle from mathematics is that you cannot say that two infinities are equal to one another. For example, while there are an infinite number of natural and real numbers, these two infinities are not the same. Here you can find a derivation of the Henderson–Hasselbalch equation. Since the Henderson–Hasselbalch equation bundles up the math in a way that is difficult to analyze, lets look at the original equilibrium equation, which you can derive from the Henderson–Hasselbalch equation simply by undoing all of the steps:
$$k_a=\frac{[H^+][A^-]}{[HA]}$$
Since a strong acid dissociates completely, $k_a$ must be equal to $\frac{1}{0}$. In other words, the concentration of HA must be zero at equilibrium. Since we are doing no nuclear reactions here, this rule that my friends like to call "conservation of stuff" takes effect. In other-words, all of the strong acid that you started with must have gone somewhere (hint: it dissociated into $\ce{H+}$ and $\ce{A-}$). Now, to calculate the final pH of a strong acid solution, we use the number of protons that dissociated to find the concentration of $\ce{H+}$ and from there, the actual pH.

The statement "the concentration of HA must be zero at equilibrium" is incorrect.  Ka is 24 for nitric acid.  For Ka of other strong acids see http://depts.washington.edu/eooptic/links/acidstrength.html  The undissociated fraction depends upon Ka and concentration.

I was using the assumption that a strong acid dissociates completely, which I know is not actually the case. If it dissociates completely, then the concentration of HA would be zero.

So you are basically repeating what the OP did to get infinities, rather than explaining that there are no infinities involved.

I am explaining why the OP's infinity's would not imply that the pH is always zero.


text:
Choose electrolyte for electroplating
I am planning to try electroplating a steel ring with copper, than nickel, and than finally gold.  It seems to me that this would be as simple as using a bath of HCl for electrolytes and attaching the object to be plated to the cathode and the metal to be plated to the anode.  I though I would then simply run a current through the setup.  However, I have seen several "recipes" for electroplating using other electrolytes such as copper and nickel sulfate or dissolving the gold in aqua regia or cyanide before the plating.  Is this necessary, as the metal to be plated will dissolve into solution when electricity is run through the circuit anyway.  I would prefer to buy fewer chemicals if possible and reduce the complexity of the process, while obtaining the best results.

There are two issues here:
First, you will be oxidizing water. You would end up making oxygen in most of those cases, before oxidizing your metal and making the cation you need.
Second, even in cases where your metal will oxidize before water, the concentration of the metal cation in the solution will be low and variable. This will create problems with controlling rates of deposition and uniformity of deposition.
That is why the electroplating baths rely on the already dissolved cation and not the metal as the source. This way you have lots of the cation you need in the solution to work with and you can put some inert counter electrode.


text:
How can proteins reduce surface tension?
How can the protein lactalbumin reduce the surface tension of water and work like an emulsifying agent in milk?

I dont know why lactalbumin reduces the surface tension but If you know the concepts of surface tension and surfactants then it becomes easier to extrapolate on that concepts to reach the conclusion why lactalbumin reduces surface tension?
According to HyperPhysics surface tension is the

The cohesive forces between liquid molecules are responsible for the
  phenomenon known as surface tension. The molecules at the surface do
  not have other like molecules on all sides of them and consequently
  they cohere more strongly to those directly associated with them on
  the surface. This forms a surface "film" which makes it more difficult
  to move an object through the surface than to move it when it is
  completely submersed.

And the choesive force is caused by the hydrogen bonds due to the  strong interactions between water molecules

And according to wikipedia

Surfactants are compounds that lower the surface tension (or
  interfacial tension) between two liquids or between a liquid and a
  solid.

This image shows how a surfact decreases the surface tension of a liquid by causing interference in the hydrogen bonds between water molecules

Similarly lactalbumin which has a core hydrophobic group and other functional group such as thiol(Beta lactalbumin) which may interfere in water-water hydrogen bond causing decrease in surface tension.


text:
Why is 2-methyl-2,3-pentadiene achiral (allene)?

2-Methyl-2,3-pentadiene is achiral. Why?

Centre of symmetry
Plane of symmetry
C2 axis of symmetry
Centre and Plane of symmetry.

I've drawn the structure:

So far I've dealt with planes of symmetry in determining whether a compound is chiral or achiral and nothing more. So I'm relatively new to axes and centers of symmetry.  I can't visualise the rotations in the axis of symmetry.
What I don't understand is: How can you say a compound is chiral/achiral based on it's axis of symmetry?

I'll start by linking to Symmetry @ Otterbein, as it's a very useful tool for looking at various structures and seeing planes an axes of symmetry.  If you're having issues visualising rotations and such, then this website should certainly help.
I suppose the first step would be to make sure that you're always, always, always visualising the structure in 3D. I'm not saying that you haven't been, but I just want to make sure.
A general explanation for testing for chirality would be to take your two structures and check to see if they are superposable. If they are, then they aren't chiral, but if they're not superposable, then chirality isn't ruled out.
If you were to take models of (R)- and (S)-bromochlorofluoromethane, you'd find that under no conditions are they superposable. They are indeed chiral.
Now consider your structure:

There is a mirror plane going right through the middle of the molecule (cutting carbons 2, 3, 4, and 5 in half, as well as one of the hydrogens on carbon 5).
Let's view it from above, with the mirror plane going through the centre of the image:

Now let's consider one of the core concepts of symmetry. If you were to only see what's on one half of the mirror plane, could you draw the entire molecule, knowing with complete confidence that you didn't draw any kind of isomer?

The answer should be yes, because that's what a mirror plane implies: it's the same on both sides of the plane, so if you know what's on one side, then you know what's on the other.
If you were to take the mirror image of this molecule through that same plane, you'd end up getting the identical molecule. If you mirror this molecule through a perpendicular axis, you'd end up getting the same thing, but with the methyl and hydrogen on carbon 5 switched around.  With the symmetry of the molecule, though, you could just rotate the molecule 180° to get the original one back.  That would make them superposable, and therefore not chiral.
When you recognise any symmetry, you'll know (or eventually intuitively see without thinking) what you need to do to generate the superposable conformation.  A reasonable analogy would be that if you write a message on white paper using black permanent marker and hold it up in the mirror, you'll see a mirrored image; BUT, realising that the paper itself becomes a mirror plane when the ink soaks through to the other side, you'll know that all you need to do is flip the paper over, showing the mirrored writing to the mirror, and see forward writing in the mirror.  This is very related to what you'd do with molecules.
I think that your best bet is to play with as many 3D models (physical or computational) as possible, doing rotations and mirror operations whenever possible.  At some point, it should click and be completely intuitive when just thinking about the molecule.  Try it on all sorts of things.  Try making a pizza with C6v symmetry.  Think about what your pizza would look like if it had D6h symmetry instead (note: very messy to make and to bake).

An axis of symmetry will never by itself generate an achiral compound. For a compound to be achiral you need any kind of improper rotation, denoted as $S_n$ or $\bar n$ for a rotation around $\left (1 \over n \right )^\circ$. The former notation implies a rotation followed by mirroring through a plane perpendicular to the rotation axis while the latter implies rotation followed by an inversion. The special cases of planes of symmetry or simple inversion can be written as $S_1$ or $\bar 2$ (plane) and $S_2$ or $\bar 1$ (inversion). Higher-order symmetry such as $S_4$ or $\bar 4$ that do not appear alongside of simple planes or inversions are rare but exist.
A centre of symmetry is inversion symmetry. To visualise it, rotate by $180^\circ$ and then apply a $\sigma$ plane orthogonal to the rotation axis.

I would not call the $\bar 4$ axis rare. It is present in $\ce{CH4}$, for example (along with some other axes and planes). Well, if you want it to be the only symmetry element of a molecule, then it calls for something more exotic.

@IvanNeretin Thanks, I wanted $\bar 4$ to be the only chirality-removing symmetry present; clarified.


text:
Superoxide dismutase of Mn (SOD)
I have a little problem with my paper about Mn superoxide dismutase. On this picture we have "B" behind the 3 OH ions, I really don't understand what is this. Where is this "B" from? Please, help me, my lecture will not accept my work today.

That ain't MnSOD, it's arginase.
B = boron
The molecule in red is 2(S)-amino-6-boronohexanoic acid.
See figure 2 of Arginase−boronic acid complex highlights a physiological role in erectile function Nature Structural Biology 6, 1043 - 1047 (1999)


text:
Is these a pair of diastereomer or identical compound?
I attempted on this question (e) and found out that they are diastereomers, but the answer from the book is that they are identical. I worked out the configuration at each of the centres for the molecules then compare the results. If you find that one is RS and the other is SS then you'll know that they differ at only one stereo-centre and hence I think the book is wrong. I've got this question from Solomons: Organic Chemistry, 11th edition..

You are right, the left one is SS the right one is SR (if we keep Br-C to the left and F-C to the right in both isomers).

Maybe the answer of the book is wrong? For the fisher projection, it is Diastereomer definitely.


text:
Why thermal decomposition of iron(II) nitrate and sodium nitrate yields different products?
\begin{align}
\ce{4Fe(NO3)2 &-> 2Fe2O3 + 8NO2 + O2}\tag{1} \\
\ce{2NaNO3 &-> 2NaNO2 + O2}\tag{2}
\end{align}
Why do nitrates of different metals decompose in different ways?
Why do we get a nitrite with sodium, but an oxide with iron(II)?

iron (II) -> iron (III); sodium oxides are not really very stable; not comparable reaction conditions. I am quite certain, that you are not looking for reaction mechanisms in any of your recent questions. A reaction mechanism is a detailed step-by-step description in terms of elementary reactions (also including transition states).

@Martin-マーチン - then I'm not sure which tag to pick. I'm looking for something to help me memorize, something logically explaining the differences. I wonder why can't we have Fe(NO2)2, for instance. My latest questions are from USE Exam Sample Question No.37, considered one of the hardest in the exam. The question contains a textual description of a chain of reactions which you should write down, guessing the products. I've already memorized some interesting reactions, like SO2 + H2S (occurs in volcanoes) and am plowing further.

For the thermal decomposition of sodium nitrate see Eli S. Freeman,
J. Phys. Chem. 1956, 60 (11), 1487–1493.

Iron(II) cation is moderately reducing- an iron(II) cation has a valence electron configuration of $\mathrm{3d^5 4s^1}$, and by giving up an electron in its $\mathrm{4s}$ orbital the cation will transform into an iron(III) cation with $\mathrm{3d^5}$ valence electron configuration, which, with every $\mathrm{d}$ orbital half filled, has a higher degree of symmetry over $\mathrm{3d^5 4s^1}$ electron configuration and hence is more stable (thermodynamically favored).
Sodium(I) cation, on the contrary, has a highly stable valence electron configuration of $\mathrm{2s^2 2p^6}$, with all its $\mathrm{2s}$ and $\mathrm{2p}$ orbitals filled, identical to the electron configuration of the chemically inert noble gas neon, and either removing or adding one electron to the sodium(I) ion will lead to an electron configuration much less stable and hence thermodynamically unfavored.
You may notice that in the decomposition reaction, the iron(II) cation is oxidized into iron (III), which is favored, and the sodium cation remain intact.
Put it in another way, we may imagine a first step thermal decomposition reaction of iron(II) nitrate resembling decomposition of sodium nitrate and yields iron(II) nitrite and oxygen gas as products. Then we easily notice that iron(II) nitrite is reducing, and oxygen gas is a strongly oxidant, hence the two products of the first step decomposition can further react with each other to produce iron(III) oxide and nitrogen dioxide.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I disagree that $\mathrm{[Ar],3d^5,4s^1}$ has a lower symmetry than $\mathrm{[Ar],3d^5}$.


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How soluble is uranium in molten iron?
As I know, uranium and iron simply aren't mixable, which means there are no U-Fe alloys. The Google says, uranium is insoluble in molten iron.
But, if something is "insoluble", it actually means only that there is a very low maximal concentration.
What if we want to mix only 1ppm of U in molten Fe? What if 0.01 ppm? What is the maximal concentration?

uranium / iron alloys do exist, they have been reported in the inorganic crystallograpic database

According to Uranium partitioning between liquid iron and silicate melt at
high pressures: implications for uranium solubility in
planetary cores

The solubility of U in
  liquid Fe is in the range of 0.6 to 800 ppm and increases with temperature (T) and
  pressure (P).

See also U solubility in Earth’s core which finds a solubility of 2 ppm at zero pressure (run number 198).

Hehh, you know the reason, why I asked it! Respect :-)

There's a difference between solubility and partitioning, which is not clear from the papers you linked to (which weren't peer reviewed I guess). The solubility of U in liquid Fe is almost complete. Partitioning experiments (including the ones you linked to) show that it preferentially goes to the silicate, not the core. Read this related question in ESSE.

At any temperature where the iron is liquid, uranium is also liquid and 100% miscible with it. See Chen et al.[1], from which the diagram below is taken. As with many metal pairs, the liquid phase solubility does not carry over to the solid state, where several intermetallic compounds form.

Reference
1.
Chen, Tianyi;  Smith, Travis; Gigax, Jonathan; Chen, Di; Balerio, Robert; Shao, Lin; Sencer, Bulent; Kennedy, J. (2015). "Intermetallic formation and interdiffusion in diffusion couples made of uranium and single crystal iron." Journal of Nuclear Materials 467, 82-88. https://doi.org/10.1016/j.jnucmat.2015.05.026.


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Chemical processes and chemical reactions difference
Whats the difference between chemical processes and chemical reactions?
Ive heard that the chemical industries only uses chemical processes to produce new substances, but isnt that also what happens when a chemical reaction occurs? I couldn't find a lot on this topic online except for the wikipedia aricles that only made me more confused.

According to IUPAC:

Chemical reaction
A process that results in the interconversion of chemical species.
Chemical reactions may be elementary reactions or stepwise reactions
(It should be noted that this definition includes experimentally
observable interconversions of conformers.) Detectable chemical
reactions normally involve sets of molecular entities as indicated by
this definition, but it is often conceptually convenient to use the
term also for changes involving single molecular entities (i.e.
'microscopic chemical events').

A chemical reaction is a kind of process in which interconversion of chemical species takes place. Because of this, we can say that a chemical reaction is a chemical process.
Chemical process can be a broad term, which describe any process that is commonly studied by chemists because they are closely related to chemical reactions. It is somewhat diffuse, as boundaries of the field of chemistry.
Industries use a variety of procedures to obtain substances. Chemical processes/reactions take place. If your interest is realising what is done in industries, do not get stuck in these subtleties. Most of the times they are not intentionally distinguished, and their usage is customary in the field.

While a reaction is a kind of process in the broader sense of that word, in the chemical industry the term process usually refers to an engineered system for producing a chemical product, including one or more reactions, the equipment, the feedstocks, the separation methods, etc. An example of how a process is not the same as a reaction is that a given substance may be produced from the same precursors using either a batch process or a continuous flow process.

In other words, the chemical industry uses the term "process" primarily to refer an engineering process.


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What is the use/importance of the equilibrium constant Kc
I understand what the equilibrium constant is, however when reading in my textbook it has occurred to me that I do not understand what it's actual use/importance is. One specification in my syllabus is "Understand the importance of the numerical value of Kc".
Thank you for any help.
I do not understand why this question has been put on hold. How can I demonstrate that I have put in an effort into understanding this question when I literally cannot find the answer on the internet or in any of my four textbooks? I have familiarised myself with Kc, however nowhere can I find it's importance, hence the reason that I ask the question.

$K$ is the value of $Q$ at equilibrium. The expression for $Q$ and $K$ are the same however $Q$ describes the state of a system and $K$ describes the system at equilibrium. Le Chatelier's principle tells us that a system that has $Q\neq K$ will approach $K$. When $Q$ is not $K,$ the system is not at equilibrium.
K tells you where an equilibrium lays. For a given reaction, if $K$ is larger than 1 than the equilibrium is towards the right. If $K$ is less than one then the equilibrium is towards the left.
   For a simple equilibrium such as acid dissociation, the forward reaction's equilibrium constant is
$$K = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
so you can compare the strengths of acids by comparing the equilibrium constant. Since the $K$ of nitric acid is larger than the $K$ of hydrofluoric acid we know that it is a stronger acid.
Hope this helps!


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Which is more amphoteric: zinc or copper hydroxide?
$\ce{K_{a} ~Cu(H2O)_6^2+}{=~5*10^{-7}} $
$\ce{K_{a} ~Zn(H2O)_6^2+}{=~2.5*10^{-10}} $
Given this information, which forms a more amphoteric hydroxide?
My reasoning is that copper hydroxide is more amphoteric than zinc hydroxide. My reasoning is that because $\ce{K_{a} ~Cu(H2O)_6^2+}$ > $\ce{K_{a} ~Zn(H2O)_6^2+} $, $\ce{Cu(OH)_2}$ should be more amphoteric because it will probably have the bigger $\ce{K_{a}}$. This fulfills half of the definition of amphoteric - behaving as both an acid and base.
But how do I compare the base strengths of $\ce{Zn(OH)_2}$ and $\ce{Cu(OH)_2}$? I need to, because being amphoteric means the substance acts as both an acid and a base!

Consider the following:
$1. ~{K_{a1}:} ~\ce{Cu(H2O)_6^2+ + H_2O -> Cu(H2O)_5(OH)^+ + H_3O^+}$
$2. ~{K_{a2}:} ~\ce{Cu(H2O)_5(OH)^+  + H_2O -> Cu(H2O)_4(OH)_2 + H_3O^+}$
$3. ~{K_{a3}:} ~\ce{Cu(H2O)_4(OH)_2  + H_2O -> Cu(H2O)_3(OH)_3^- + H_3O^+}$
Basic character of copper hydroxide is illustrated by the below equation:
$4. ~\ce{Cu(H2O)_4(OH)_2  + H_2O -> Cu(H2O)_5(OH)^+ + HO^-}$
The above equation, reversed, is:
$5. ~\ce{Cu(H2O)_5(OH)^+ + HO^- -> H_2O + Cu(H2O)_4(OH)_2 }$
This equation is very similar to $~{K_{a2}}$ except that the base is hydroxide ion instead of water.
We can write a similar equation for zinc hydroxide:
$6. ~\ce{Zn(H2O)_5(OH)^+ + HO^- -> H_2O + Zn(H2O)_4(OH)_2}$
Let's compare equations $5$ and $6$. We can expect $5$ to be a larger extent reaction than $6$ because hydrated copper ion is a stronger acid than hydrated zinc ion to start with. So:
${K_{rxn5} > K_{rxn6}}$
And therefore,
${K_{rxn4} < K_{rxn7}}$
$7. ~\ce{H_2O + Zn(H2O)_4(OH)_2 -> Zn(H2O)_5(OH)^+ + HO^-}$

This info isn't enough to tell, it's even misleading I'm afraid.

@Mithoron - could you expand your comment into an answer perhaps? :D. I'm very curious as to what your thoughts are.

@Mithoron - what do you mean?

Can you define "more amphoteric"? More acidic and more basic are unambiguous, but this is not.

@Zhe - more amphoteric: reacts more readily as an acid and a base.

OK. Apply your definition to this scenario: Compounds A and B are amphoteric. A is a better acid, B is a better base. Which is more amphoteric?

I think you should come to chat to discuss questions. It may be cheaper then massive bounty.

@Zhe - I would say the better acid. Case in point: consider the hydroxides of Mg2+ and Al3+. The Al3+ based hydroxide is amphoteric. The Mg2+ based hydroxide is not amphoteric for all practical purposes (i.e. 6 M NaOH won't touch it).

Zinc hydroxide is more amphoteric as quantitatively investigated in Effect of pH, Concentration and Temperature on Copper and
Zinc Hydroxide Formation/Precipitation in Solution
In summary, in the case of Cu(OH)2, only at extremely low concentrations and very high pH can a third OH- be coordinated.

At higher pH values, copper
  hydroxide, Cu(OH)2, is the dominant species up to pH 12.3 where the copper ion
  Cu(OH)3
  - forms according to Equation 3. At higher copper concentrations, solid
  Cu(OH)2 is formed and precipitates out of solution at copper concentrations above the
  solubility product of copper hydroxide at 1×10-8 M. It is important to note that the
  domain of stability of solid Cu(OH)2 is expanding to lower and higher pH values with
  increasing copper concentration.

On the other hand:

Above pH 11.4, Zn(OH)3
  - forms according to Equation 7. At
  higher zinc concentrations, solid Zn(OH)2 forms and precipitates out of solution at zinc
  concentrations above the solubility product of zinc hydroxide of 1×10-5 M.

So qualitatively both display similar amphoteric behavior in the limit of infinite dilution, but at a appreciably non-zero concentration zinc is more amphoteric.
The article says it is getting the equilibrium constant values from Mineral Equilibria, Low Temperature and Pressure.
An alternative source of this information is Hydrolysis of cations. Formation constants and standard free energies of formation of hydroxy complexes Inorg. Chem., 1983, 22 (16), pp 2297–2305, which gives quantitative values for OP equations 1, 2 and 3 as well as the corresponding equations for zinc, in terms of Gibbs energy of formation and in terms of cumulative equilibrium constants.
Simplifying the general equation in the paper to only mononuclear (one metal atom) complexes:
$$K_y$$
$$\ce{M^{n+} + yH2O <=> M(OH)_y^{n-y} + yH+}$$
$\ce{Cu^{2+}}$ :
$pK_1 = 7.96$
$pK_2 = 16.26$
$pK_3 = 26.7$
$pK_4 = 39.6$
$\ce{Zn^{2+}}$ :
$pK_1 = 8.96$
$pK_2 = 16.9$
$pK_3 = 28.4$
$pK_4 = 41.2$
(These are all experimental values, calculated values are also given)
So really, without considering the solid phase (solubility), Cu2+ and Zn2+ seem very similar, with Cu2+ showing slightly more acidity.
Another study, which does consider the solid phases, is Zinc Hydroxide: Solubility Product and Hydroxy-complex Stability
Constants from 12.5-75 [degrees] C Can. J. Chem. 53, 3841.
This study approaches the problem with 5 equilibria involving the solid phase. (These authors use the symbol "c" for the solid phase.)
$\ce{Zn(OH)2(c) <=> Zn(OH)+ + OH-}$
$K_1 = [\ce{Zn(OH)+}][\ce{OH-}] = 2.54 \times 10^{-11}$
$\ce{Zn(OH)2(c) <=> Zn(OH)2(aq)}$
$K_2 = [\ce{Zn(OH)2}] = 2.62 \times 10^{-6}$
$\ce{Zn(OH)2(c) +OH- <=> Zn(OH)3-}$
$K_3 = [\ce{Zn(OH)3-}]/[\ce{OH-}] = 1.32 \times 10^{-3}$
$\ce{Zn(OH)2(c) +2OH- <=> Zn(OH)4^{2-}}$
$K_4 = [\ce{Zn(OH)4^{2-}}]/[\ce{OH-}]^2 = 6.47 \times 10^{-2}$
$\ce{Zn(OH)2(c) <=> Zn^{2+}} + \ce{2OH-}$
$K_{sp} = [\ce{Zn^{2+}}][\ce{OH-}]^2 = 1.74 \times 10^{-17}$
This study (see Fig. 1) shows that aqueous $\ce{Zn(OH)2}$ is only the major aqueous species in the pH range 9-11, having acted as an acid or base outside this range.

Instead, if I were just given the information in the OP, meaning the two supposed Ka values, as some kind of test or homework question,  I would say copper is more amphoteric.  My reasoning would be, for zinc, just to remove one proton, you need to get up to almost pH 10.  For two more protons to be removed, as required to get to Zn(OH)2 and lose yet another proton to act as an acid, seems impossible because I would expect pKas to be spaced apart and if you would need to approach or go beyond pH 14 to observe the acidic behavior, that isn't reasonably considered amphoteric.
Note that EXAFS Investigations of Zn(II) in Concentrated Aqueous Hydroxide Solutions J. Phys. Chem. 1995, 99, 11967-11973 finds that rather than being 6-coordinate as indicated in the OP, 4 hydroxides coordinate Zn2+ tetrahedrally with no water ligands.


text:
Will sulphuric acid react with water produced in an esterification process?
Sulphuric acid acts as a catalyst to catalyze the reactions of alcohol and carboxylic acid to form ester. Alcohol reacts with carboxylic acid to produce ester and water.
However, as sulphuric acid have strong afinity towards water, will sulphuric acid react with the water produced in an esterification process?
If so, will it mean that the equilibrium is shifted to the right as water is reacted with sulphuric acid? And will this mean that there would be less sulphuric acid present as a catalyst to catalyse the reaction?

Sulfuric acid acts as catalyst meaning it is regenerated at the end of the reaction. In the presence of water or other bases, sulfuric acid will primarily be deprotonated. $$\ce{H2O + H2SO4->H3O+ + HSO4-}$$
It is occasionally more correct to consider the water acting as a catalyst. Any time that you have a strong acid acting as a catalyst, because its deprotonation is much more likely than the organic reaction, it is nearly completely deprotonated at a given time. Thus the protonated $\ce{H3O+}$ will act as your catalyst. protonating the carboxylic acid and deprotonating the ester.

Sulfuric acid is a dehydrating agent. This should the water produced in Fischer Esterification. This will stop the ester from hydrolyzing. The ions produced from the deprotonation of sulfuric acid should also help in catalyzing the reaction. This will push the reaction to the right.


text:
Reactor mass balance with solid feed and overflow
I am modeling a lime slaker which is basically a stirred tank reactor with two separate feeds:
1. CaO feed screw (powder)
2. H2O binary valve (liquid)
These react according to the reaction:
$\ce{CaO(s) + H2O(l) -> Ca(OH)2(s)}$
The slaker is operated on a semibatch fashion where water and lime are fed until the desired amounts have been added and when the reaction has finished a new batch can be started. Water is added in excess with a mass ratio of approximately 1:4. The addition of the reactants takes about six minutes and after that some minutes are required for the temperature to rise due to the reaction. The reactor is not emptied between the batches but the contents exit to a secondary tank through internal piping on an overflow basis. An example of the development of the reactor temperature is given below.

Because of the overflow the volume of the reactor (and the contents) stays constant. The output feed is not measured. In general, the material balance of a model is:
the rate of accumulation of mass in the system = rate of mass entering the system - rate of mass leaving the system
or
$\frac{dM}{dt}=\dot{m}_{in}-\dot{m}_{out}$
or
$\frac{dV\rho}{dt}=\dot{F}_{in}\rho_{in}-\dot{F}_{out}\rho$.
For the lime addition weighing data is available to determine the mass of the added quicklime. The lime feed screw is operated at a constant speed so the feed rate can be determined. The water feed has a flow measurement.
What would be the best way to form the material balances as the reaction happens between a liquid and a solid? How to determine how much material is displaced by the addition of lime? How does the formation of calcium hydroxide affect the balance?


text:
Zwitterions/IEP of glycine at pH 6? (Paradox?)
I am trying to understand an experiment: Some crystalline, solid glycine was solved in water and a pH=6 was measured (also calculable with $\mathrm{pH} =\frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2} = \frac{1}{2}(2.34 + 9.6) = 5.97)$
I have tried to explain the acidic pH-value: Solid glycine is in a crystalline structure — all molecules are in a zwitterionic state. If they get solved in neutral water, the crystal-structure is lost and $100~\%$ of zwitterions are dissolved in the neutral solution. The carboxylic and amino group now act as an acid and base in aqueous solution. The acid reaction of the carboxylic group predominates ($\mathrm{p}K_\mathrm{a1} < \mathrm{p}K_\mathrm{a2}$). Some amino acids are now negative ions, some are still zwitterions.
But now is my question (the contradiction?): pH = 6 is also the isoelectric point IP with the definition, that all amino acids are in zwitterionic state. But according to my theory above, some amino acids have already reacted "away" from the zwitterionic state.
Are there flaws in my reasoning — is my explained "paradox" not true?

Your definition of pI is wrong, it only refers to the entire collection of molecules having overall no net charge, it does not mean EVERY molecule has to have no charge

@orthocresol Thanks for your reply, but also my whole collection has a charge, because the whole collection lost some protons?

Are you German? You capitalise and hyphenise a lot ;)

@Jan And yes, I am German. (-:

You need to separate the two concepts.
When dissolving a neutral amino acid in water, a buffered acid-base reaction will take place as you highlighted, giving a certain pH value. In this case, the following equilibrium:
$$\ce{HOOC-CH2-NH3+ <=> ^{-}OOC-CH2-NH3+ <=> ^{-}OOC-CH2-NH2}$$
will be somewhere between the pure second and the pure third species. (The neutral, non-zwitterionic species is neglectable in aquaeous solution.) Superfluous protons liberated into the solution will end up as $\ce{H3O+}$ explaining the overall reduction in pH value when compared to a neutral solution. You won’t notice it unless you perform really accurate experiments, but the solution’s pH value will be a tad higher than the isoelectric point.
If you go on and add external protons, i.e. by carefully adding an acid, you can adjust the overall solution’s pH. Since the glycinide anions are the strongest base present, they will take up the protons first, followed by the zwitterions. At a certain pH value (corresponding to a certain external proton concentration added), the concentrations of glycinide anions and glycinium cations will be identical and we have reached the isoelectric point of glycine. It is, and has to be, slightly different from the pH value obtained by dissolving glycine.

@user2365846 The Stack Exchange-preferred way to say ‘thanks’ is upvoting and/or accepting ;)

After it dissolves, there will be four species, the zwitter-ion, the positive species, the negative species, and the neutral non-zwitter-ion species.
In the limit that the concentration of amino acid is much greater than the effect of water auto-ionizing to H+ and OH-, there will be equal positive and negative species.  The concentration of each of the negative and positive species will be $10^{5.97-9.6}$ times the concentration of the neutral (zwitter-ion + non-zwitterion) species.
If the concentration of amino acid is very low, the pH will be closer to 7 because some of the OH- of water will react with the amino acid to shift the negative species to be greater concentration than the positive species (and [H+] greater than [OH-]).
Overall, for quantitative prediction, you would need to solve for 5 concentrations: H+, OH-, the positive and negative amino acid species, and the sum of the two neutral amino acid species, considering Kw, the 2 "Ka"s, conservation of charge, and the fact that the sum of the amino acid species is a constant.  5 unknowns and 5 equations.

Thanks for your answer! Ok, maybe my calculation above is false - but I measured the pH=6.

@user2365846 That sounds right.  It has to be somewhere between 5.97 and 7.00, and it would only get closer to 7 as concentration becomes extremely low.

But according to my "theory", every IEP should have a pH=7,  no matter which concentration, ...


text:
How can I clean up from electrolysis?
So, I was doing a little experimental electrolysis in my basement. I used 2 (presumably) galvanized steel screws as my electrodes and ran 5A through tap water with a lot of NaCl dissolved in it (to decrease resistance). It worked great: I got a lot of bubbles from the - side. Problem is, my cup started filling up with this black crud.
Now, I have a little plastic disposable cup with really salt water, 2 screws (black with what I presume is some corrosion), and a lot of black and red insoluble crud. How do I get rid of it?
Also, what it that crud (most likely)?
EDIT: I figured out that putting NaCl in an electrolytic cell means that I was actually generating chlorine gas instead of oxygen gas. That means that I have a bunch of iron and zinc chlorides in that water. It's also probably pretty basic (I saw no bubbling at that electrode). Any disposal problems?

most likely iron oxides, pretty safe to dispose.

@permeakra I update the question to reflect my current research: it turns out that I was generating Cl2 which reacted with electrodes (I think).

The black one is probably $\ce{ZnFe2O4}$; insoluble compound of very weakly toxic metal ions, or $\ce{Fe3O4}$. The red one is $\ce{Fe2O3}$ In terms of disposal all above counts as non-toxic mineral materials, similar to sand.
It is also possible to have some carbon in this setup, but I doubt it.
Anyway, you have nothing too toxic in this setup, you can safely dump it into ground without any meaningful lasting environmental consequences.
The problem may arise, if the steel used was stainless steel or another iron alloy, but it doesn't seem to be the case from your description (and they are rarely galvanized anyway)


text:
Sodium fluoride and varying pH - HF risk
I'm looking into conducting an experiment on the adsorption of fluoride (for my International Baccalaureate Extended Essay). As I will be investigating the role of $\mathrm{pH}$ in adsorption capacity, I will be dealing with $\ce{NaF}$ solutions of varying $\mathrm{pH}$. Could hydrofluoric acid be created by this process, and could it be dangerous in the dilute concentrations I am dealing with (<$0.5\ \mathrm{g\ L^{-1}}$)?
My reasoning is that added $\ce{H+}$ will move the equilibrium of the buffer $\ce{F^{−}(aq) +H3O+(aq) <=> HF(aq) + H2O(l)}$ to the right, and the $\mathrm{pH}$ will hardly change until the buffer capacity is exceeded. I'm still not sure whether this poses  a health risk, though.

related http://chemistry.stackexchange.com/questions/51223/can-hydrofluoric-acid-etch-glass-at-a-ph-of-7-0

(Soluble) fluorides are always poisonous, only the undissociated, relatively non-polar HF (i.e. fluorides under acidic conditions) quickly penetrates the skin.

Your reasoning is correct.  Solutions containing dissolved fluoride salts can form HF under acidic conditions.  I personally would not take any chances with exposure, because even dilute concentrations of HF can cause severe burns if enough of your skin area is exposed to it.  Standard PPE would include goggles, faceshield and neoprene gloves.  PVC sleeves and lab apron are recommended as well.

HF does NOT cause visible or even tangible burns on the skin, unless highly concentrated, and even then much less than ordinary hydrochloric acid. The effect is one of intoxication of the body parts below the affected skin part, followed by systemic intoxication.


text:
Haloalkanes and E1, E2 reactions
In an E1 reaction, I understand that the base reacts with the hydrogen of the carbon which is next to the positive charged carbon of the cation (of an haloalkane), because this hydrogen is supposed to be acidic. Why is that? And again, in an E2 reaction, why is the hydrogen that is attacked by the base, acidic? I am not sure how the polarity of the halogen affects the situtation.

After the first step of the reäction (i.e., loss of the halogen), you have the carbocation.  You could draw a resonance structure of this: push the electrons from the C–H bond into making a new C–C bond, leaving you with an alkene and a nearby proton (note the conservation of charge, moving the + from a carbon to the hydrogen).  This is like the reverse of an addition of a hydrohalic acid to an alkene, whereïn you'd have the double bond attack a proton, then have the halogen attack the carbocation.  Either way, you've drawn a very reasonable* resonance structure showing an alkene and H+.  That H+ is pretty much the definition of a Brønsted acid, so that would be a hand-waving explanation of how to see this hydrogen as being particularly acidic.
For the E2 example, we can't use the same method because we can't show this resonance structure (well, we kind of can if you want to get into a really tedious discussion that would be highly debated).  You'd probably want to use the argument about halogens being inductively electron withdrawing, which pulls electron density away from nearby bonds.  This results in less electron density existing in the C–H bond, making the bond longer, and slightly increasing the positive charge on the hydrogen.  Since Brønsted acid theory defines the acid as being H+, the more we can show a partial positive charge on the hydrogen, the more Brønsted-acidic we can argue it to be.
* Note that I mentioned drawing a "reasonable" resonance structure.  That's important, because we could technically show a resonance structure of benzene as having a carbon with a negative charge and a hydrogen with a positive charge, but that wouldn't be an argument for that hydrogen being particularly acidic, as it's a horrible resonance structure to draw.
It certainly is more complicated than this, but I feel like this highlights some of the most important effects without oversimplifying to the point of making anything explicitly wrong or creating a bunch of exceptions.

For E1: So we can say that the hydrogen of the carbon next to the positive charged carbon is acidic because the electrons of the C–H bond tend to balance the positive charge of the cation? For E2: I still don't understand why the C–H bond next to the halogen bonded carbon sholud be more affected than the hydrogens attached to halogen bonded carbon...

For E1:  That's a good way of describing it, yes.  Since C+ is far less stable than H+, it would be favourable to move the two electrons over from the C–H bond to make a new C–C bond, since doing so would transfer that positive charge from the C to the H.  While we may draw the intermediate as a carbocation, there is already quite a bit of electron density that has moved from C–H over to make a new C–C bond, making that H quite a bit closer to H+.

For E2:  The halogen certainly does affect the hydrogens on the same (geminal) carbon.  There are a few ways of explaining why the base wouldn't get involved with one of those hydrogens, so I'll choose one.  If you did reprotonate there, you'd end up removing both a proton and the halide from the same carbon.  This would make a carbene, which is quite reactive, both nucleophilically and electrophilically.  The process would likely just reverse.  This can be done, though!  For example, you can use potassium tert-butoxide on CHCl3 to make a :CCl2 carbene.

Ok I think I got it. So in E2, the halogen affects the bonds nearby making the hydrogens bonded to the carbons next to the geminal carbon little, yet enough acidic for a strong base to come and get them?

That sounds right to me.  There are other factors that complicate things, but that's true for anything in science.  For example, depending on your molecule, you may have other hydrogens that are more acidic, so a base would want to deprotonate at the more acidic sites first, BUT if there's no good leaving group next to that carbon, then an elimination won't happen, and the process would likely reverse.  You also need to use a base that is small enough and strong enough to deprotonate where you want it to.  But what you said is still correct (I'm just stressing that other factors exist).

Thank you very much for the information and your patience.

authors: Αντώνιος Κελεσίδης


text:
What is the difference between the chemical potential μ and the molar Gibbs free energy Gm?
I'm not too sure if I understand this correctly:
Is the $G_\mathrm{m}$, the Gibbs free energy of the entire system, divided by the amount of substance making up the system, while $\mu$ is the Gibbs free energy of a component of the system?
If so, what does ${\partial G_\mathrm{m}}/{\partial n}=\mu$ signify? As in, what's the difference between this and $\mathrm{d}G/\mathrm{d}n=\mu$?

@orthocresol Any particular reason you removed the subscripted unicode m?

@Martin http://i.imgur.com/0GueIzJ.png

@orthocresol O.ô Guess you have some missing fonts then. Can you read any of the wikipedia page on unicode sub/superscript? (Also you just triggered a rollback war auto flag, lol.)

@Martin oops, sorry! Well, about two-thirds of the symbols are fine, the others don't appear. The subscripted numbers work (so chemical formulae in titles tend to work)

@orthocresol don't worry about it, but I thought you might want to look into it on your computer. It does not really make a difference here...

$\partial G_m / \partial n = \mu$ is wrong. $\partial G_m / \partial n =  0$ as  $G_m$ is an intensive property it can not  variate by solely change the amount of substance.

And $dG/dn = \mu$ is also wrong by definition: $\partial G/\partial n = \mu$.

@user1420303 ok I now get why ${\partial G_m}/{\partial n}=\mu$ is wrong, but I'm not quite clear on why $dG/dn=\mu$ is wrong. Would it still be wrong if p and T were kept constant, and assuming that it's a pure substance (a single component system)?

You are thinking in the right direction, but it has no mathematical sense. Now there is a misunderstanding of derivatives :-) We can fix $T$ and $p$ in a system, but $\partial G / \partial T$ (or $p$) won't be null, because they represent how $G$ would change if in infinitesimal change in $T$ (or $p$) is performed.

The chemical potential of a species is the partial derivative of the Gibbs free energy of the mixture of components with respect to the number of moles of that particular species, holding the temperature, pressure, and number of moles of all the other species constant.  For a single component system, this is reduces to the total free energy divided by the number of moles.

I request that the answer be made more clear by giving an example of the difference between molar Gibbs Free Energy  and Chemical Potential.

I don't know how I can say it any more clearly.  The Chemical Potential is defined as the partial molar free energy.


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Are there biodegradable pigments that are soluble in oil, but not in water?
I'm hoping to find a purple pigment that preferably, but not necessarily, can be used at room temperature. The goal is a pigment that will break down in septic systems.

"Biodegradable" is a relative concept. Does it need to break down in hours, months or centuries? Does it matter if the degradation products are also colored, or perhaps toxic? More information is needed on the intended use.

I never considered time frames! Sometimes disposal will be in septic systems. Maybe a better stated goal is an oil soluble pigment that can be consumed by the typical bacteria found in septic systems.

There are several such pigments which are bio-degradable by some microbial action (mostly bacterial action) and are oil soluble (lyophilic) and water insoluble (hydrophobic).
Some are:

chlorophyll : certain bacteria can decompose chlorophyll.
A paper citing zoo-plankton activity on chlorophyll is here.
Xanthophylls : This is a general category of several pigments whose degradation by marine bacteria has been widely  studied. (You can search for many relevant papers online).
Carotenes : Certain carotenes are also reported to be degraded by fungi and mushrooms. For example here

Anthocyanins (and related polyphenols) which have dark red to purple color are mostly water soluble.

Thanks much. A purple color and lyophilic  are absolutely essential properties. You said that purple anthocyanins are "mostly water soluble". Would it be possible for you to cite any exceptions?

@steveorg : I'll check on internet and if i find something  I'll add in the post.

I came up empty for 100% of my requirements. However, I did come up with second best, which is an emulsified purple that mixes well with oil. After mixing, it doesn't leach into any water that contacts the oil.
While it should be possible for anyone to create this, it requires the proper equipment and it is ineffective in a commercial situation when trying to control costs. I found a company that has been very helpful that provides custom samples and commercial quantities at affordable prices. The company is Color Maker.