text:
Are conformational isomers diasteromers?
I am currently struggling on seeing where exactly conformational isomers fit within stereochemistry & how (and whether) they are a subset of diasteromers. I've seen potentially conflicting information online on the definition of diasteromers (some sources mention that diasteromers are not mirror images of each other and not superimposable, while others don't explicitly mention the superimposable part).
I am convinced that enantiomers and geometric (cis-trans) isomers are a type of configurational isomer, the latter a subset of diasteromers (I hope this is correct), since they require bonds to break in order be superimposed with each other. However, due to some online sources differing in their mentioning of the superimposability requirement of diasteromers, I'm not sure whether diasteromers are a subset of configurational isomers (if this is the case, then it would seem incorrect to place conformational isomers within diasteromers, since it itself is a subset of configurational isomers).
For more of a visual explanation, see the below flowcharts:
(a) First flowchart; conformational isomers / conformers within diasteromers category. Doesn't mention configurational isomers & where they fit in.
(from Wikipedia)
(b) Second flowchart; conformational isomers is outside the stereoisomer category. I assume that optical isomers are a subset of configurational isomers?
(From LibreTexts)
(c) This flowchart separates configurational and conformational isomers and places diasteromers under configurational isomers. I feel that this third flowchart most closely resembles how I would personally organize all of the isomers (the only thing I'm unsure about is the mention of optical isomers, which I assume is a quality all non-cis-trans/geometric isomers exhibit).
(from the University of Cagalry)
Any help on this would be greatly appreciated! Thank you so much :)

The last flow chart should include identical molecules (they have the molecular formula, so they are part of the main branch).

Conformational isomers as in (boat vs chair for rings or s cis vs s trans for dienes or what the flowchart describes as rotamers) are not diastereomers, or any time of isomers strictly.
Conformational isomers are identical molecules, just in different momentary positions. Any two conformational isomers will be rapidly inter converting between each other and are impossible to isolate.
They only become significant when thinking about why a reaction can lead to different stereo chemical products. E.g. E2 eliminations that lead to cis vs trans isomers- here the difference in product is due to the reactant molecules being in different conformations when they reacted.


text:
Why isn't the n factor/valency factor in a disproportionation reaction = twice the number of electrons in a single half reaction?
I was reading the first answer to Equivalent weight in case of disproportionation reaction there it is said

Take a disproportionation reaction:- $\ce{nA -> xB + $(n - x)$C}$ where A is being oxidised to B and rest reduced to C. Also let $n_1$ and $n_2$ be the n factors of A to B and C respectively
We have 2 equations: one by conservation of electrons,
$$\begin{align*}
n_1 x &= n_2(n - x) \\[7pt]
\implies x &= \frac{n_2\cdot n}{n_1 + n_2}
\end{align*}$$
and the other from the definition of $n$-factor:
$$\begin{align*}
n_f &= \frac{\text{moles of electrons transferred}}{\text{moles of reactant}} \\
&= \frac{n_1 x}{n}    \\
&= \frac{n_1 n_2}{n_1 + n_2}
\end{align*}$$

In the second last line $n_f = \frac{n_1x}{n}$ shouldn't it be $\frac{n_1x+n_2(n-x)}{n}$ which is equal to $\frac{2n_1n_2}{n_1 + n_2}$ by expanding?
The reason for my thinking is because equivalent is defined as the weight of a substance that reacts with an arbitrary amount of another substance (eg:- 1g H). Hydrogen can both supply and take electrons therefore both the oxidation and reduction half reactions of A can take place with H and so I'm reasoning that n factor = adding up the electrons for both oxidation and reduction half reactions

Why would you use this concept here at all? One downside is apparent: it causes confusion. What are the upsides? When this reacts with that, you may use equivalents to find out how much "this" you need per certain amount of "that". Not the case here, though.

@IvanNeretin Can you explain why it cannot be useful? I'm learning this in school, and my teacher told me that we will eventually use these concepts to make solving titration and some mole concept problems easier. I've also been bugged by this for a week so i want to understand it.

I saw some comments on other posts by how it is rarely used in titration but I dont understand why

It is fine to use equivalents for titration problems. But your imaginary example has nothing to do with titration. It is about one compound reacting with itself. And that's where equivalents are useless.

Yes you're right. I was thinking what if a double titration was done with the one decomposing substance where the titration can help identify the amounts of the products of the decomposition, but then too you won't need the n factor of the one decomposing substance, you will only need the n factor of the products

Because it's the same as if you started out titrating the products.

If one insists on the formal usage of the n-factor and equivalent weight on disproportionation reactions (for which it is not intended), one must accept the fact the compound has simultaneously two n-factors (which can be incidentally equal):

one as the oxidant
one as the reductant.

Then the procedure is the same as for ordinary redox reactions where oxidants and reductants are different molecular entities.
But as Ivan and others have suggested, using n-factors for similar cases is rather useless and just complicates things. Be aware that the N-factor, equivalent, gram-equivalent and equivalent weight are obsolete concepts that should be taught only as what they are and how they were used.

The disproportionation redox reaction
$$\ce{n A -> x B + $(n - x)$C}$$
or alternatively written with small natural numbers $i,j$ as
$$\ce{$(i + j)$ A -> i B + j C}$$
is better to rewrite for easier analysis to 2 redox half-reactions:
\begin{align}
\ce{ i A &-> i B + n e-} \\
\ce{ j A + n e- &-> j C}
\end{align}
Then for n-factors of A as oxidant resp. reductant:
\begin{align}
n_\mathrm{f,A,ox} = \frac{n}{j} \tag{oxidant}\\
n_\mathrm{f,A,red} = \frac{n}{i} \tag{reductant}
\end{align}
It could be better understood if we reverse the reaction:
$$\ce{i B + j C -> $(i + j)$ A }$$

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text:
How to correct for acidic water when taking the pH of a soil/water solution?
I work in a soil science lab and not very much a chemist.
Recently I took the $\mathrm{pH}$ of a large number of soil samples using a glass electrode on a mixture of $\pu{10 g}$ soil and $\pu{20 mL}$ Millipore water. Partway through the assay, I discovered that the water I was using had a $\mathrm{pH}$ of $\pu{5.2}$.
Is there a way to correct my recorded $\mathrm{pH}$'s to values I would have measured had the water had a $\mathrm{pH} = 7$?

I also know the water concentration of each soil sample if that helps

There should be a standard procedure for this analysis. The procedure would include how to handle the distilled water. An educated guess is that $\ce{CO2}$ from the atmosphere is causing the low pH. (I remember boiling water and then putting a drying tube filled with $\ce{NaOH}$ on the container's headspace for some experiment that I did.)   Also the procedure should detail how to handle the soil-water mixture. If the soil has a pH above 5 or so, then the soil-water mixture too will eventually react with  $\ce{CO2}$ from the atmosphere.

The pH is not as important as the quantity of acid. When I have done similar tests , i assumed the several grams of soil had much more material to affect the pH than the water.

Regrettably, one cannot simply add or subtract pH to correct for that error. Correction would depend on knowing how much and what type of each contaminant was in the water.
For example, a very dilute, 0.00001 molar solution, of $\ce{HCl}$ would have pH about 5. It would take considerably more dissolved $\ce{CO2}$ to lower the pH to 5. The first would likely have less effect on your measurements, because the actual amount of $\ce{HCl}$ would have been small, but in the second case, far more $\ce{CO2}$ would be present to react with the soil sample.

Ah well, back to pH-measuring then. Thanks for answering!

We often acid-wash our containers in HCl in our lab. If the contaminant were enough HCl to lower the water pH to 5.2, how would I correct my pHs?

@Tillandsia123, How much water is left in the containers? If a 10 liter carboy were rinsed with HCl and shaken out, and if a 1 ml pipette were rinsed with HCl and shaken out, is that the same amount of remnant HCl? Answer: rinse with pure H2O, shake out, rinse, shake out, reiterate until the pH approaches that of the "pure" water. In other words, establish how much rinsing is needed, measuring pH each time, and establish a routine so that you need not repeat those tests. Even then, if the detergent or water source changes, you'd need you'd check again.

I'll make sure to do that. Thank you!

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text:
Why trans-dibromoethane is more favorable than cis-dibromoethane?
2 questions:

Is trans-dibromoethane more stable than cis-dibromoethane because it's more symmetrical (in terms of molecular weight distribution on either side)?

Is it always the case that trans would be more stable than cis?

It can be generalised that trans is more stable than cis form due to more symmetry

@Proscionexium I wouldn't say so. Center of symmetry isn't any worse than plane. Big substituents can lead to steric clash in cis, though.

Is the question about conformational isomers in ethAnes or geometrical isomers in ethEnes? In either case it depends on size, shape, intra [and inter] molecular forces and flexibility. Also Favorable to whom or what?

Yeah, you can hardly have a cis/trans in ethane derivative - it should be ethene. Better keep such "details" straight in the future...

I assume you mean 1,2 dibromoethene (not ethane as that doesn’t have a trans/ cis isomer as no double bond).
The bromine atom is demonstrably large from its position in the periodic table, much larger than carbon or hydrogen. The trans isomer evidently puts the bromine further apart which minimises interactions between them (known as torsional strain). Strain interaction would increase the energy of the molecule and so make a reaction with the cis isomer as its product have a higher (more positive) energy change, which is less favourable.


text:
Why does the oxygen in NaOH (sodium hydroxide) acquire an electron from the sodium if it's going to become unstable and give away the electron?
Mind you that the oxygen is already stable, having an even number of electrons and protons? Is the oxygen really neutral when it has an even number of electrons and protons?

Sodium hydroxide and any other compounds have a net charge of zero. The whole compound is neutral.

Sodium hydroxide and hydroxide ions in water solutions are very stable. They do not have tendency to decompose. OTOH, hydroxide ions are very reactive, meeting Broensted-Lawry acids.

You seem to assume a lot of things which are not there or not things at all.

First of all, welcome Phillip, to ChemistrySE :D.
I have noticed that you don't seem to have a solid ground on chemistry. It seems like you're struggling to understand how ions can be stable when they have an excess electric charge, compared to neutral atoms.
The reason for this is that you are equating the total charge of the atom with the electron-electron repulsion inside. You think that, because, say, $\mathrm{Cl^{-}}$ has an extra electron, compared to a neutral chlorine atom, it should be more unstable.
In order for an ion to be more unstable than a neutral atom, the inter-electronic repulsion between its electrons should go up more than the attraction between the extra electron and the nucleus. At first that idea seems to check out, after all, isn't the new electron facing the repulsion of $N$ electrons, but being attracted by a single nucleus?
This is a sensible way of thinking, if you are considering electron as small fragments of matter sitting till around an atom. But electrons are not classical particles of matter, they are not point like. Rather, electrons behave sort of like a fuzzy cloud. A very strange fuzzy cloud.
I'm not sure how familiar you are with quantum chemistry and the modern atomic model, but since you had difficulties with the basics, I'll refrain from using technical jargon and I'll simply focus on the concepts. If you feel like you're being patronized and would like a more accurate picture, just a leave a comment and I'll edit the answer to fit your tastes.
And because electrons are fuzzy clouds, ant not point particles, they repel each other using rules for fuzzy clouds, not point particles. And it turns out that there is a way for you to arrange fuzzy clouds around an atomic nucleus such that they do not repel each other much. We say that these clouds are orthogonal to one another.
As you go along adding electrons around an atom they take shapes that are, preferably, orthogonal to one another, so as to minimize their repulsion, such that each consecutive electron added to a shell increases little of the repulsion each electron in the shell feels, but it greatly increases the electron-nucleus attraction energy, because now there is a new electron being attracted by the nucleus with much the same intensity as the other electrons in the shell.
This can't go on forever, however, for electrons are quantum objects and, as such they have discreet states they can occupy. This means that rather than being an infinitely fluffy and flexible cloud that can deform however it needs, the electron must occupy one of a particular selection of distinct shapes that are mutually orthogonal. Thus, eventually, the electrons will run out of shapes on which to contort themselves and the next one will have to step on some toes. When that happens, you say that the electrons have closed a shell and any additional electron would rise the inter-electronic repulsion energy immensely.
Because electrons can't have any energy they want (this is a consequence of quantum mechanics), there are specific shells they can occupy, at specific distances from the nucleus. Each representing a particular energy level.
In order to add a new electron to an atom that has a filled shell, it has to enter an outer shell. This outer shell electron won't experience repulsion from neighbors in its shell, but it will feel an attenuated nuclear charge, as though it is being attracted by a weaker nucleus. This happens because it's feeling some repulsion from lower shells. The lower shells are shielding the outer ones from the nuclear charge.
Outer shells are larger and, as such, they allow the electrons to take more shapes than the smaller, tighter ones, so more electrons can fit inside them.  however, they are also further away from the nucleus, so adding more electrons doesn't usually contribute a great deal for the electrostatic stability of the atom.
Armed with this knowledge, let's take a trip through the periodic table to see how electrons are arranged on the various shells of the atoms.
We start with hydrogen ($\mathrm{H}$), one proton in its nucleus, atomic number 1 (atomic number is often detoned by $Z$). The neutral hydrogen atom has one electron. This electron enter the first shell, where one can fit up to two electrons. This is the only electron the hydrogen has, and it's positioned really, really close to the nucleus so it's very tight to take out. Hence, hydrogen don't often spontaneously loses its electron in orde to become a naked proton. However, one can fit another electron on the hydrogen first shell. Putting a new electron here would likely increase the electron-nucleus attraction energy of the entire system while increasing the electron-electron repulsion a tad less. Which is why you don't often see $\mathrm{H^+}$ as a stand-alone species (the $\mathrm{H^+}$ in acids and bases is short-hand for $\mathrm{H_3 O^+}$ so it doesn't count) but you do see $\mathrm{H^-}$, which is the hydride ion.
Alternatively, the hydrogen could share an electron with another atom that could also take a new electron on its shell in order to increase the electron-nucleus attraction energy. We'll see where these atoms are in a moment, but one obvious choice is... another hydrogen atom!
By sharing each of their electrons, two hydrogen atoms can fully occupy their outer shells for maximum electron-nucleus attraction while minimizing electron-electron repulsion provided the two atoms stay at such a distance from one another that their electrons don't overlap with one another and their nuclei don't start to repel one another. This effect can be counter-measured by the attraction between the nucleus on a hydrogen and the electrons around the other.
When you put it all together, there is a very particular distance between the two hydrogen atoms that optimizes all of these competing influences. Hence we have a molecule of $\mathrm{H_2}$ which is much more stable than isolated gas-phase hydrogen atoms and which has a specific $\mathrm{H-H}$ bond distance and strength.
Now, after hydrogen we have helium ($\mathrm{He}$), $Z=2$. Neutral helium needs two electrons in order to neutralize its nuclear charge. You put one electron in it, and it goes into the first shell, this first shell looks a lot like the one from hydrogen, only it is tighter because the nuclear charge is larger. Then the second electron goes also in the first shell and that's it.
Unlike hydrogen, helium doesn't want any more electrons. Any newer electron would have to occupy an outer shell, increasing the energy of the atom while providing very little electron-nucleus attraction since it would be shielded from the feeble helium nuclear charge by the inner first shell. One such electron would be very weakly bounded to the nucleus, and thus would very easily slip away, which is why we don't often see $\mathrm{He^-}$. For the same reason, helium isn't interested in sharing electrons either.
Alternatively, helium doesn't want to lose any electron either. Losing electrons only decrease electron-electron repulsion energy with lower shells, of which helium doesn't have any. In fact, helium is tighter than hydrogen, it's smaller, for it attracts its electrons with greater force than hydrogen. So losing electrons means losing a great amount of electron-nucleus attraction energy, the main stabilizer of the atoms.
Because of this lack of ways to increase its energy by interacting with the electrons of other atoms, helium atoms are chemically inert and all they form are monoatomic gases that do nothing all day long.
The next element is lithium ($\mathrm{Li}$), $Z=3$. Battery superstar and the only metal to have been formed from the primordial nucleosynthesis. Lithium needs three electrons. We've seen this movie before. Two of the electrons go into the first shell but now, unlike helium, there is one extra electron that is required to form a neutral lithium atom. Now, if there is only a lithium atom and the electron is passing by, there will be some electrostatic attraction between the positive charge in a semi-filled lithium atom and that electron, but if I explained things correctly, you're likely to anticipate that this outer electron isn't interacting as tightly with the lithium nucleus as the other two electrons. Not only is it further from the nucleus (because it is in an outer shell), it's also being shielded, by the inner shell, from the nuclear charge of the lithium nucleus to which it is bound.
At this point there is some repulsion between electrons, because electrons in different shells repel each other more strongly than those in the same shell. I should probably also mention that the extent to which a shell can shield the outer ones from the nuclear charge (as well as the inter-shell repulsion energies) are dependent upon the size of the neighboring shells and their spacing, two parameters that vary from atom to atom and which change for the same atom, with extra electrons or lacking electrons. So all of this we're talking about is an approximation, a tool to help you understand.
If the distance between the inner and outer shells is small enough and lithium nucleus is weak enough, this electron might be extraordinarily loose. Which, it is. Losing that electron would cause all of the inter-electronic repulsion between the two layers to go away, stabilizing the positive-charged lithium ion $\mathrm{Li^+}$. One would still have to pluck the electron away, however, because as high as the inter-electronic repulsion is in this case, it's still smaller than the electron-nucleus attraction. Which is why you can still see neutral lithium atoms in the form of the lithium metal and thin gases. If the repulsion between the shells was higher than the electron-nucleus attraction, it would have been impossible to have these gases and metallic pieces to exist in the mild conditions of a laboratory that can be inhabited by humans.
Now, not to extend myself further than I already have, let's talk about your particular problem. Here we have a poly-atomic anion ($\mathrm{OH^-}$) as well as a monoatomic sodium cation $\mathrm{Na^+}$. Sodium is from the family of lithium, only one period down so it has an extra layer between the inner and the outer ones. Much like lithium, sodium would love to lose its electron because then it would get rid of a whole bunch of electron-electron repulsion, so anything capable of taking that electron away from it could easily do it.
The problem is to understand $\mathrm{OH^-}$. Here we have an oxygen bound to a hydrogen and this somehow isn't unstable. Why doesn't this molecule give one electron back to the sodium?
Well... oxygen is on the same period as lithium, the second period of the periodic table. It's an exceedingly small atom because it only has two shells but they are wrapped tighter around its nucleus than the one on lithium, its a smaller atom. Though it has an electron-electron repulsion between its inner and outer shell, like lithium, it also has a lot of electron-electron attraction because it has two electrons in its inner shell (which is much like helium) and six electrons on its outer shell.  All of the electrons in the outer shell have a small repulsion to one another but they are attracted by a nucleus which is more charged than lithium. The shielding is not large enough to mask the intense electromagnetic attraction between these electrons and the oxygen atom.
When two atoms share electrons to fill each other's outer shell (we call that forming a bond), each of them participate with one of the electrons. A simple bond takes two electrons to form (one from each participant, in covalent bonds.), a double bond takes four and a triple takes six (and so on...).
In water $\mathrm{H_2 O}$, an oxygen is bound to two hydrogen bonds. Oxygen has space for two electrons in its outer shell, more than that and electron repulsion kicks and can't be balanced by the electron-nucleus attraction. hydrogen, as we have seen, has space for one. So, each hydrogen atom forms a bond with an oxygen and everyone is happy.
Except... Oxygen is six times more attractive than hydrogen and it is tight, small and with a lot of nuclear charge. Even at a distance, oxygen exerts an attractive force on hydrogen's electron that is larger than the attraction that hydrogen exerts over oxygen's electrons. This creates an imbalance. The cloud of electrons, considered collectively, is nudged towards oxygen (but you said electrons had definite shapes! Well they do, but the shapes are defined by the potential they are experiencing. Electrons under the influence of two nuclei instead of one have different shapes, one of which is this nudged one).
When one of the $\mathrm{H-O}$ of water is split, it is far more likely that the electron pair that forms the bond stays with the oxygen. One of those electrons was originally with the oxygen anyway, but the other one was originally in the hydrogen, so it adds an extra negative charge to the remaining $\mathrm{OH}$ bundel of atoms, and you have the following
$$
\mathrm{B\ +\ H_2 O\ \to\ BH⁺\ +\ OH^-}
$$
where $B$ is something that has extra pairs of electrons on its outer layers, such as a nitrogen atom bonded to two other hydrogen atoms. Nitrogen needs three more electrons to fill its outer shell, which it can easily get from forming bonds with eager hydrogens, but it does have a single pair o electrons that originally belonged to itself, so it can go this:
$$
\mathrm{N H_3(aq)\ +\ H_2 O(l)\ \to\ N H_4^+(aq)\ +\ OH^-(aq)}
$$
indeed, oxygen itself can provide electron pairs like this:
$$
\mathrm{H_2 O(l)\ +\ H_2 O(l)\ \to\  H_3 O^+(aq)\ +\ OH^-(aq)}
$$
a famous reaction called water autoprotolysis, often written like this, for stoichiometric friendliness:
$$
\mathrm{2 H_2 O(l)\ \to\  H_3 O^+(aq)\ +\ OH^-(aq)}
$$
The existence of species with available electron pairs to share with electron-depleted species (such as a hydrogen in a water molecule) is what drives the creation of negatively charged things. An amine ($\mathrm{N H_3}$) comes long and passes close to a water molecule minding its business. One of the hydrogens notice all of those loose and cushy electrons, nothing like the tight and depleted cloud that oxygen is offering it, and decides to change partners. The remaining hydroxyl ($\mathrm{OH^-}$) has a more evenly distributed electron cloud but still has atoms with all their shells filled.
The weakness of the hydrogen-oxygen bond is a result of the small size and high charge of the oxygen atom and the disproportional attraction that it has for hydrogen's shared electrons (this is a didactic statement, in reality electrons in a molecule aren't distinguishable, but speaking about that would open a can of worms that is better addressed in a different question).
When the $\mathrm{H=O}$ bond breaks, the oxygen is left with a negative charge but a filled outer shell, since the two electrons stais with it. That is what stabilizes the charged hydroxyl. In order to transfer the proton and lower the overall charge of the species to zero, the oxygen in the hydroxyl ($\mathrm{OH^-}$) would have to rip electrons from its outer shell... something that is likely not going to happen.
Incidentally, hydroxyl can still lose its last electron, provided you mix it with something with even better electrons or other conditions that allow for that bond breakage. The resulting ion is $\mathrm{O^{2-}}$ which forms oxides.
But one thing you can't do is have the hydroxyl exist as a neutral species an then it rips the electron away from sodium. That's not what is going on. In order to form a hydroxyl ion, you need to mix something with water that will be able to break that bond. Neutral hydroxyl has half-filled shells (thus making it a radical) and it is very much not stable. Hydroxyl ion is formed from things that have hydroxyl groups, the most abundant of which is water.
It's the opposite of sodium, which is a larger atom than oxygen with an appreciable electron-repulsion energy, eager to lose the outer shell.  Sodium, therefore, loses its outer electron in order to ditch the electron-repulsion energy whereas oyxgen retains hydrogen's electron when another species with better electrons come along to snatch its hydrogen away from it.
The two resulting species $\mathrm{Na^+}$ and $\mathrm{OH^-}$ are, therefore, very stable... Provided there are opposite-charged things around them to stabilize them. Although the microscopic sodium cation and hydroxyl molecule are very stable, if you bunch a mole of them together in order to get a macroscopic piece of them, the mutual electrostatic repulsion between each individual ion would make it impossible to counter. But if you mix the same amount of opposite charged things such that opposite charges are close together, then you can make a big thing like, say, a lump of table salt ($\mathrm{Na Cl}$) or, in your case, the white, soapy-like sodium hydroxide ($\mathrm{NaOH}$).
Water is a molecule with a dipole moment, which means that it works like a little triangle with one vertex negative and two others positive. When you put something charged into water, it may get in between the ions in the substance stuck together by electrostatic forces and its negative tip can form a tiny electrostatic interaction with the cation while its two other positive vertices form electrostatic interactions with the anions.
One molecule of water, individually, for an electrostatic interaction with the ions that is comparatively weaker than the ionic bond that the charged species would formed between themselves, but you surround them with a thick coat of water molecules (a process known as hydration) and then, depending on the properties of the ions and the lattice they were forming before water came in to break it, it might be better for the ions to interact with this combined water coat than with themselves, at which point the substance made from the charged species dissolves.
So you can have $\mathrm{Cl^-(aq)}$ or $\mathrm{N H_4^+(aq)}$ the $\mathrm{(aq)}$ indicating that the species are surrounded by a coat of water molecules, but you can never have $\mathrm{Cl^-(s)}$ or $\mathrm{N H_4^+(l)}$ that is a solid or liquid macroscopic substance made of charged parts of a single type.
We've come a long way. Hopefully by now you have understood why it is that some charge-excess species are more stable than neutral-charged alternative versions of themselves. It's all quantum mechanics' fault!
If the explanation is still murky and strange, I hope that, at least, you come out of this answer with better questions than the ones with which you walked into it. Chemistry is a fascinating subject, but also a deep and convoluted one, full of red herrings and surprising twists. Don't let the difficulty discourage you!


text:
Why does the melting point of hexafluorosilicic acid differ that much in these two solution concentrations?
From the Wikipedia article for hexafluorosilicic acid:

Melting point:
ca. 19 °C (66 °F; 292 K) (60–70% solution)
$<$ −30 °C (−22 °F; 243 K) (35% solution)

What's the reason for this huge difference in melting point while the solution concentration differs by about 30%?

if you compare it to e.g. the phase diagram of water - sulfuric acid, it is not a huge difference at all.

Yeah, it's nothing special. One could say something like (H3O)2SiF6 is essentially a compound that melts, not a solid solution.


text:
Electronic transition of iodine and non-bonding orbitals
As you all know, iodine has a purple color when in the vapor phase and when dissolved in non-coordinating solvents such as carbon tetrachloride. What electronic transition is responsible for this?
A quick google search (https://www.vedantu.com/question-answer/the-color-of-iodine-is-a-violet-b-green-c-white-class-11-chemistry-cbse-5f4a918f84fe9e109c4b5753) says that this electron transition is due to non-bonding electrons going into the antibonding sigma orbital.
This is certainly possible, but leads me to a more general question: Usually, we don’t draw the non-bonding orbitals into the MO-scheme of the molecule. Take a look at the fluorine molecule for example (as its MO scheme should be similar than that of iodine):

Fluorine also has lone pairs, and they also could transition electronically; why don’t we write them out specifically and how does one know when they will be involved in electronic transitions?
For example, in carbonyls, there is also the n-pi* transition for reference,in which the non-bonding electrons of oxygen can be promoted to an antibonding orbital. Why don’t they participate in the MO scheme?

Electrons from lower orbitals (higher bonding energy) will transition to higher orbitals (lower bonding energy) when excited (by whatever...). Even forbidden transitions occur because atoms (or molecules) in the real world aren't totally isolated. Thus microstate perturbations allow "forbidden transitions," but they are relatively low in probability. The rub here is not all of the transitions (excitation or emission) occur in the visible band.

I'm not sure I understand the question. Lone pairs and electrons in non-bonding orbitals are not the same thing

@IanBush how come it is not the same? For example in carbonyls, the oxygen electron pairs are the cause of the non-bonding-pi* transition: https://www.masterorganicchemistry.com/2016/09/26/uv-vis-spectroscopy-absorbance-of-carbonyls/

@MaxW true but where are non-bonding electrons located, exactly, in the MO? Should be somewhere, if they are available for a transition, like the n-pi* transition in carbonyls (see link in last comment)

In this source, they also mention non-bonding (n) electrons of the carbonyls for n-pi*-transitions: https://szerves.chem.elte.hu/oktatas/ea/Perczel/UV-VIS.pdf IMO, in MO schemes non-bonding electrons are not highlighted though.

I2 has the ground state configuration $(\sigma_g5p)^2(\pi_u5p)^4(\pi^_g5p)^4(\sigma^_u5p)^0$ and the first excited state $(\sigma_g5p)^2(\pi_u5p)^4(\pi^_g5p)^3(\sigma^_u5p)^1$, so $\pi^* \to \sigma^*$. The transition is between states $^1\Sigma^+_g ,\to,^3\Pi_u$ . As this molecule contains heavy atoms spin-orbit coupling is very important and the singlet - triplet spin transition prohibition is broken.

@Mäßige Your MO diagram for F2 does include the electrons which in another view of the world would be the lone pairs. Count the electrons, you have all 7 2s & 2p electrons for each F, so where are the electrons for the lone pairs? They must be already in that diagram! The problem is that you are mixing 2 different views of the world. In MO theory orbitals are across the whole molecule; MO theory is a totally delocalised view. Lone pairs are a localised concept and as such don't occur in MO theory. You can make the connection, but you have to a a lot more careful than you have in the above.

True @IanBush they have to be inside of there somewhere. So in actuality, iodine in an MO scheme doesn't have lone pairs per se, rather non bonding orbitals somewhere in it (namely the pi/pi* and sigma/sigma* (2s) orbitals). So the transition is due to pi* to sigma*, would you agree. Furthermore why would a carbonyl have nonbonding electrons then? Or is it a semi-quantitative MO scheme?

To be sure I would have to read up on this more, but my first guess would be as you say, pi* -> sigma* But the point @porphyrin makes about spin orbit coupling is important as well

Hybrid orbitals and molecular orbitals are different ways of looking at the same thing
Let’s take your fluorine example first.
It doesn’t quite make sense to ask why we “don’t write out the lone pairs specifically” in $\ce{F2}$; we’re using a different model. Once we’ve decided that the $1s$ electrons are core and won’t participate significantly in bonding, we have 14 electrons to worry about, all of which are shown in the MO diagram; there are no extra lone pairs that we’ve chosen to ignore.
Alternatively (but worse from the perspective of understanding electronic transitions), we could say that these 14 electrons are arranged in three lone pairs on each of two F atoms (3 pairs $\times$ 2 atoms $\times$ 2 electrons per pair = 12 electrons total) plus one $\sigma$ bond between the atoms (the final two electrons).
So it’s not that we’re choosing to ignore lone pairs, it’s that they don’t exist in this model.
Non-bonding orbitals are typically included in MO diagrams
The link you gave to illustrate your carbonyl example only shows three orbitals. For a fuller picture, let’s have a look at a qualitative MO diagram for the simplest carbonyl compound, formaldehyde:

(This diagram has some flaws but it was the best I could find quickly online.)
The LUMO, here given the symmetry label $2b_1$, is an example of an orbital conventionally labelled “non-bonding”, though in fact you can see that it is mildly antibonding between C and O and bonding between C and H. There’s a more quantitative picture here. The point is that we don’t have to decide whether or not to include this; it turns up naturally when we make the diagram.
The visible transition in $\ce{I2}$ is better thought of as being between two antibonding orbitals
Returning to your initial question, as your own MO diagram and several comments have indicated, the HOMO-LUMO transition in $\ce{I2}$ is best described as $\pi^*\rightarrow\sigma^*$, not $n\rightarrow\sigma^*$. The source you link to for the latter description is not very scholarly! As the comments have pointed out, the reason for iodine’s strong colour includes both that this transition occurs in the visible spectrum and that spin-orbit coupling breaks the “selection rule” prohibiting a singlet-triplet transition.


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Alleged temperature dependance of activation energy?
According to Master Organic Chemistry (https://www.masterorganicchemistry.com/2012/09/10/elimination-reactions-are-favored-by-heat/), in point number four, the activation energy for a reaction is temperature dependant ("So you can see by analyzing this term that activation energy can change with temperature!"). This is in contradiction with what multiple threads on here on this topic have claimed. Could someone clarify this confusion.

The original Arrhenius model considers Ea independent on T. But later, more sophisticated models consider its T dependence, involving thermal energy of molecules. //  So what is stated depends on the implicit context.


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Are these reactions in synthesis of melamine correct?
I found the Wikipedia article for melamine to be incorrect in "synthesis and reaction" section; so I decided to edit it. But on the other hand I'm not sure my assumption to be correct that the entry is incorrect. So I decided to ask here to seek assist.

Today most industrial manufacturers use urea in the following reaction to produce melamine:
$$\ce{6(NH2)2CO -> C3H6N6 + 6NH3 + 3CO2}$$
In the first step, urea decomposes into cyanic acid and ammonia:
$$\ce{(NH2)2CO -> HNCO + NH3}$$
Cyanic acid polymerizes to cyanuric acid, which condenses with the liberated ammonia forming melamine. The released water reacts with cyanic acid, which helps to drive the reaction:
$$\ce{6HNCO + 3NH3 -> C3H6N6 + 3CO2 + 3NH3}$$

Questions:

So are there two methods of obtaining melamine; one by decomposing $6$ moles of urea and another by mixing $6$ moles of cyanic acid with $3$ moles of ammonia?

Why is $\ce{3NH3}$ both in reactors and products parts of the last reaction?

So where are the cyanuric acid and water mentioned in the last reaction? They are in the text but not in the reaction equation.

I combined the latter two reactions and I got this: $$\ce{(NH2)2CO + 5HNCO -> C3H6N6 + 3CO2 + NH3}$$, which differs from the first reaction.

Think it like this way: Urea decomposing to cyanic acid which then polymerizes to melamine.

And a species can be at both sides of the reaction. It is the net moles that matter. See for example: https://chemistry.stackexchange.com/questions/34138/preparation-of-sulfuric-acid-through-contact-process

In the first step, urea decomposes into cyanic acid and ammonia:
$$\ce{(NH2)2CO -> HNCO + NH3}$$
Cyanic acid polymerizes to cyanuric acid,

$$\ce{3 HCNO -> (HO)3C3N3}$$

which condenses with the liberated ammonia forming melamine.

$$\ce{(HO)3C3N3 + 3 NH3 -> (NH2)3C3N3 + 3 H2O}$$

The released water reacts with cyanic acid,

$$\ce{HCNO + H2O -> NH3 + CO2}$$

which helps to drive the reaction:
$$\ce{6HNCO + 3NH3 -> C3H6N6 + 3CO2 + 3NH3}$$

It gives the summary reaction:

$$\ce{6(NH2)2CO -> C3H6N6 + 6NH3 + 3CO2}$$

So are there two methods of obtaining melamine; one by decomposing 6 moles of urea and another by mixing 6 moles of cyanic acid with 3
moles of ammonia?

That's not what the article is saying.
The first equation it presents describes the stoichiometry of an overall process for converting urea into melamine, ammonia, and carbon dioxide.
The second two equations each describe contributing reactions: first urea decomposes into cyanic acid and ammonia; then, some of those intermediate products react further to produce the final products.
In the industrial processes described, both of these reactions happen in the same reaction vessel, and they are going on at the same time. And in the real world, probably neither runs all the way to completion, and probably there are some side reactions, too, diverting some of the input material to different products.

Why is 3NH3 both in reactors and products parts of the last reaction?

Because ammonia is both consumed and produced.  The reaction does not happen without ammonia: "cyanuric acid [...] condenses with [...] ammonia", but overall, the ammonia that is consumed is also replaced.  Putting it on both
sides of the reaction conveys that.

So where are the cyanuric acid and water mentioned in the last reaction? They are in the text but not in the reaction equation.

Yes, the last equation presented is also a summary scheme.  It does not show the cyanuric acid and water being both produced and consumed (as opposed to consumed and (re)produced), possibly because they are short-lived intermediates, possibly because the author considered those details less important, possibly just for brevity.  The prose description sketches in those details.  But if you preferred, you could say
$$\ce{3HNCO -> C3N3(OH)3}$$
$$\ce{C3N3(OH)3 + 3NH3 -> C3N3(NH2)3 + 3H2O}$$
$$\ce{3HNCO + 3H2O -> 3NH3 + 3CO2}$$
The last reaction drives this sub-process forward by consuming the water produced during the condensation and replenishing the ammonia consumed in that step.  But note well that this can't happen without the condensation happening first, and the condensation requires ammonia.
As for

I found the Wikipedia article for melamine to be incorrect in "synthesis and reaction" section

, I don't see any basis for that conclusion.

I don't see any basis for that conclusion. Yes, and I myself told everyone that I'm not sure about my assumption.


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If I spilled mercury in a carpeted room 28 years ago will it have evaporated by now?
I made a really stupid mistake as a kid 28 years ago and spilled about one teaspoon of mercury onto a carpeted floor of a room about 12x12 feet. I’m assuming it was organic mercury only because it was taken from a junior high school science department and I’ve read that in this context it is typically organic mercury.
Is it likely there still could be mercury in the house at toxic levels after 28 years?
People still live in the house and I would like to get the air tested, however if it would have evaporated by now I would prefer not to worry people unnecessarily. My understanding is that there is no way to do this sort of test discreetly myself.
I lived in the room for 4 years after the spill and there was very little ventilation, the door was always closed so the only air coming in was from one air conditioning vent. During that time it probably was vacuumed a few times. After 4 years I moved out and for the past 24 years the room has had the door open and the air from the vent coming in. At one point the carpet was replaced.
I feel really terrible that I probably have caused my family harm. Later in life when I learned of the harmful effects, I told my dad about it who was a chemistry major and a
medical doctor. This was about 20 years ago and he told me not to worry about it then. I assume that he must have had it checked and dealt with (later they changed the carpet) but I didn’t get a chance to ask him about it and confirm this before he died last year.
Now it is gnawing at me and I would like to get it tested but if it is safe by now, I’d rather not make everyone worry. They are very anxiety prone people and I don’t want them to worry the rest of their lives that they may have some long term subtle effects from that past exposure, considering there’s nothing that can be done about that past exposure now.
Thanks for any advice.

So, was it metallic mercury or organic mercury ? There is huge difference. What particular compound if not the metal?

I don't know what you mean by "organic mercury".  If what you spilled was a silvery, metallic liquid, it was NOT "organic mercury" but rather inorganic mercury.  Metallic mercury is less toxic than mercury salts.  And more likely to evaporate.

According to a paper about dental amalgam and mercury exposure:

The evaporation rate of elemental mercury at room temperature (20 °C) is approximately 50 μg/cm²/h (range of 40–60 μg/cm²/hr).

If your teaspoon of mercury was about 5 mL in volume, and if it was metallic mercury, it would have amounted to about 67 g, based on mercury's density of 13.5 g/mL.
We need to know the surface area of the mercury to use the figure for the evaporation rate.  In all probability, the surface area likely became very very large after the mercury was divided into many small droplets after it was spilled on the carpet.
But let's say the surface area remained very, very low, about 5 cm² only.  That means about 250 μg, or about 0.25 mg, of mercury would evaporate each hour.  And therefore the 67 g, or 67,000 mg, of mercury would evaporate in 268,000 hours, or about 30 years.
But a surface area of 5 cm² seems very low.  If the mercury did not stay coalesced in one giant droplet, it was probably much higher than this, meaning the mercury is probably gone.
However, at these very long time scales, mercury may also oxidize to salts and other non-metallic forms of mercury, which would have effectively zero volatility.  So it's probably a safe bet that a large fraction of the mercury you spilled is evaporated, with a smaller fraction that is still contaminating the carpet, not evaporating, perhaps removed by vacuuming, but perhaps with some low level of detectable mercury still there.
Only testing the carpet for mercury would say for sure, I guess.

Just a question. Do you remember the beginning of the problem ? Was it metallic mercury, which is a bright liquid often present in small droplets? Or was it a powder ? Mercury compounds like chloride, sulfate, or organic compounds (mercurochrome) are in general powders, like salt, sugar or talk powders. These powders will not get evaporated. So their vapors do not exist and are not toxic.

According to Curt F. 5 hours ago, a droplet of metallic mercury about 5 mL in volume would be evaporated in 30 years. This happens if no chemistry, no oxidation for example, is occurring during this time. If such a chemical reaction happens, the metallic mercury will be transformed into a new stuff, which is probably less volatil. So the carpet does not produce any toxic vapor any more today. And if the new stuff is more volatil than metallic mercury, mercury has already disappeared a long time ago from the carpet.


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Which is better ligand, PH3 or PF3?
It confuses me that many researches pointed out different results about phosphine ligands properties (σ-donor and π-acceptor). Since both phosphine $\ce{PH3}$ and phosphorus trifluoride $\ce{PF3}$ possess similar properties, which one is the better ligand in terms of forming a stronger bond with the center metal atom?

It may depend on the metal ion. PF3 would seem to be better wuth a soft metal ion that can donate pi electrons to it. (And won't blow up if you breath on it wrong.)

It would very much help your question if you provide references or better links to at least some of the papers you are referring to in the question.

What kind of metal center are you talking about? While all phosphines are good σ-donors and poor π-acceptors, $\ce{PF3}$ is more π-acidic than $\ce{PH3}$ because of the electron withdrawing effects of the $\ce{F-}$ substituents. The more electron-rich metal centers would then be more stabilized, theoretically, by $\ce{PF3}$ than $\ce{PH3}$ due to their ability to back-donate into the phosphine.
Hard-soft acid base theory would also suggest that the softer $\ce{PF3}$ ligand will coordinate to softer metal centers, while the $\ce{PH3}$ ligand would coordinate more easily to harder metal centers.
This is all to say, in reality, I would not want to work with either phosphine due to their dangers, but would rather work with alkyl and aryl phosphines instead. Tunability of these would be easier as well.


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What is the name of the bond between oxygen atoms and Na+ ions when NaCl dissolves in water?
When NaCl (sodium chloride, table salt) dissolves in water, the NaCl molecules break apart. The hydrogen atoms form hydrogen bonds (I think?) with the Cl- ions. What is the name of the bond between the oxygen atoms and the Na+ ions?

There is no molecules in solid sodium chloride. There is a series of alternating $\ce{Na+}$ and $\ce{Cl-}$ ions alined on the three axes Ox, Oy and Oz. When dissolved in water, these ions are simply separated without any molecular breaking.

Hydrogen bonding is mainly used for polar covalent bonds and not for free ions.

Both the sodium cation and chloride anion will be surrounded by polar water molecules forming a "coordination sphere." However such + and - groups can be attracted to one another in concentrated solutions and form an ion pair. Ion pairing is one reason that activities differ from concentrations.


text:
What is correct sign convention for Helmholtz free energy change?
I have been studying about the Helmholtz free energy function ($A$) that
$$A=U-TS$$
Evidently the change in $A$ is given by (generally for isothermal reversible processes)
$$\Delta A = \Delta U - T\Delta S \tag1$$
and we know that  $$\Delta S =  \frac{Q}{T} \tag2$$
Now this Wikipedia page describes two different sign conventions informally named Clausius's sign convention and Planck's sign convention.
If I follow Clausius's convention then
$$\Delta U = Q-W \tag3$$
Using (1),(2) and (3),
$$W= -\Delta A \tag4$$
which makes complete sense to me that the decrease in work function ($A$) should be equal to work done ($W$) by the system.
But confusion occurs when I use Planck's convention for the same thing. According to it
$$\Delta U= Q+W \tag5$$
Using (1),(2) and (5),
$$W=+\Delta A \tag6$$
Now I know that obviously both the sign conventions are correct and paint the same picture. But Clausius' convention seems to be  correct and Planck's convention do not both intuitively as well as logically. I am not calling any of these conventions wrong. I am asking why (4) and (6) gives different interpretations of the same thing?

Is not it obvious ? If W is work done to system then W= +ΔA. If by the system then W=-ΔA. It is not about intuition but about historical context. Work done by the system was used by engineers/physicists, more interested what the system can do. Chemists/scientists were interested more in what happens to the system.

So can I say (4) and (6) are fundamentally same in meaning? And sign is a matter of preference?

It is not a matter of preference. It is a matter of defining the internal energy of a system. Either add the work done on the system or subtract the work done by the system.  The work must result in random change in molecular momentum and change the heat content of the system.

The Helmholtz energy change does not have a sign convention. Work does.
If there is provided reversible work 10 J to an isothermic, isochoric, closed system, then

in the Clausius convention (focusing on work done by thermal engines):

$\Delta U = Q - W$
$W=\pu{−10 J}$
$\Delta A=\Delta U - T\Delta S=\\
T\Delta S - W - T\Delta S =\\
−W =\pu{+10 J}$

in the Planck convention (focusing on changes of the system state):

$\Delta U = Q + W$
$W=\pu{+10 J}$
$\Delta A=\Delta U - T\Delta S=\\
T\Delta S + W -T\Delta S =\\
+W =\pu{+10 J}$

Note that while both conventions are correct, if not mutually confused, physical chemistry prefers the Planck convention (work done on the system). It is a good habit to explicitly mention which convention is used.

Helmholtz energy does not have a sign convention. Work does. This sentence was the key to end the confusion. Thanks.


text:
Exceptions in ionization energies of Oxygen and Nitrogen
We all know that it is difficult to remove an electron from atoms that are half-filled or fully filled. Thus they have higher ionization energy.
But today my teacher said that $N^{-3}$ has a lower ionisation potential than $O^{-}$. Can anyone help me ? Sorry if the question is dumb.

Well, the title isn't doing it any favors...

Why is nitrogen happy with 3 extra electrons flying around. In fact, no isolated -2 ion is stable. Many -1 ions are happy in isolation.

Atoms have ionisation potentials, and bare nitride anion doesn't even exist.

https://en.wikipedia.org/wiki/Ionization_energy#Exceptions_in_ionization_energies

Please edit your title to reflect what is being asked.

Be aware N is aside of few others the exception, having negative electron affinity ( and more negative ones for the 2nd and 3rd electrons). $\ce{N^3-}$ is stable only in ionic nitrides, stabilized by strong energy of ionic lattice, like for $\ce{Mg3N2}$.

I am not an inorganic chemistry specialist. But it seems like $\ce{N^3-}$ has a filled valence shell whereas $\ce{O^-}$ is missing one electron to close the shell. which is why this oxidation state is often found bound to hydrogen in the form of hydroxyl. So it might be possible that taking the extra electron from the oxygen mono anion might be easier.

You need to enclose \ce{...} inside math mode, i.e. $\ce{...}$ (this differs from original LaTeX)


text:
Titration of butyllithium methodology
I am a physical chemist with limited experience in organic work. I am going to be using butyllithium for insertion of lithium into a metal oxide structure (Cava, Solid State Ionics, 1983). One goal is to determine the amount of butyllithium consumed during the reaction. Cava uses "acid-base titration", but does not go deeper into the detail of what that entails. My assumption is that I quench the BuLi with 2-propanol, and then titrate the new organolithium compound outside of the glovebox with just a typical HCl or something. But my experience in this department is very limited.
Do any of you organic folks know the best way to go about doing this? This work will be performed inside of an argon glovebox.
Thank you.


text:
Why is folic acid very slightly soluble in water while it's a B vitamin?
The solubility of folic acid is just 1.6 mg/L (25 °C).
Why is it so insoluble in water while it is one of B vitamins which are called water-soluble vitamins beside Vitamin C?

You don't need much of it - 400 micrograms could still be dissolved in a glass of water.

But when we say a compound is water-soluble we are saying it has a significance solubility in water.

What's the solubility of the so-called water-insoluble vitamins ?

The structure with two acid groups and at least one amine suggests that the solubility is pH dependent. Water soluble or fat soluble is determined by the ability of the body to store vitamins in fat cells or possibly the liver and to transport them from the intestines. The concentration of folate in the blood is much lower than the solubility limit. It is hard to make definitive conclusions on one data snippet.

Folate is a supplement and vitamer, see e. g. here. More important than absolute solubility is the ratio of solubilities in water and octanol, or log POW.

Of course folate/folic acid (vitamin B9) is just one of the examples. There are other B vitamins too that are very slightly soluble in water; such as riboflavin (vitamin B2).

Well, that's what you get if one tries to categorise stuff arbitrarily. Still, as I said you could live on  folic acid dissolved in water, while I seriously doubt the same can be told about vit. A, for example.

Consider that sodium folate  is more soluble that folic acid. similar as sodium palmitate (soap) is more soluble than palmitic acid.

But they don't mention salts of vitamins when speaking about the solubility of vitamins. Also, let's compare solubilities of ascorbic acid and sodium ascorbate. Vitamin C has a solubility of $330$ g/L, while its sodium salt has a solubility of $62$ g/L. See? Its sodium salt is even less soluble. Also, we get several solubilities with different cations. So it's not very good to measure the solubility of vitamins as salts. Some of them aren't even acids to produce salts.

Oh, by the way, solubility of sodium palmitate is also less than that of palmitic acid! At least $4.6$ mg/L for the acid (in 0 °C), and $120$ μg/L for the salt. (So palmitic acid is even more soluble than folic acid!)

Ascorbic acid is called a water-soluble vitamin (vitamin C) and its solubility in water is $330$ g/L.

You better have some real good data on this "sodium palmitate is also less than that of palmitic acid" because it's nonsense. You ever seen insoluble sodium soap? I doubt that.


text:
What factors cause a few sodium salts to be water-insoluble?
We know that almost all salts with sodium as cation are water-soluble. But there are a few examples where the salt is water-insoluble; such as sodium bismuthate, sodium fluorosilicate, and much more.
I think that's because water can't separate the ions in these examples. But if true, are there other factors that cause this strange solubility in some sodium salts?

I guess that's kinda follow up of https://chemistry.stackexchange.com/questions/82328/are-there-any-sodium-salts-that-are-insoluble-in-water

related https://chemistry.stackexchange.com/questions/75989/why-is-there-an-increasing-solubility-in-water-for-chlorides-chlorates-perchlo

The first question is just about mentioning some sodium salts which are water-insoluble. But the second question (Nilay's answer) has a good point.

But how are those few anions capable to make so strong bonds with sodium that it's very hard to break?

It's not a matter strong bonds, more like fitting into crystal lattice and low enthalpy of solvation.

Several insoluble sodium salts can be obtained by adding some sodium salt solution to a mixture of uranyl acetate and magnesium (or zinc or manganese) acetate, in the presence of acetic acid. The precipitate is a triple acetate, namely sodium-magnesium-uranyl acetate. See Earl E. Caley, A New Quantitative Reagent for Sodium, J. Am. Chem. Soc. July 1929, p. 1965.

Having fluorine as the element ionically bonded to sodium is one factor that inhibits solubility. Like all other pretransition metal ions (which are hard acids), bonding to fluorine bases is typically stronger than bonding to similarly charged bases containing other elements, making the lattice energy/solvation balance less favorable.
We can see this in the binary compounds, which means halides (other ionic binary compounds of sodium react with water). Usually we do not list sodium fluoride as an "insoluble" sodium compound. But the respective Wikipedia articles show that it has seven times less solubility in water than sodium chloride (molar basis), and sodium bromide is similar to the chloride rather than the fluoride.


text:
What common plastic types would be most compatible with pool water?
What thin-film plastic types are most compatible with typical pool water (pH 7.2, normally 5ppm chlorine but up to 15ppm on occasion) in full sun?
I'm finding that most standard film materials degrade quickly.  How does polyethylene faire?  Mylar?  Teflon?  Polyetherimide?  A table of such material compatibility would be wonderful.

It's more of an issue with in being film and with sunlight.

Note that pool ropes and buoys are often polyethylene or polypropylene because they're more resistant than nylon (and float), but they have UV inhibitors to extend life.


text:
Difference between allosteric and chelate cooperativity
As the title suggests, i'm trying to wrap my head around what differentiates an allosteric cooperativity from a chelate cooperativity by looking into this article. However i'm not able to grasp the core idea behind them.

What i understand so far, by looking in this other article, is that this two types of cooperativity give rise to the same effect at macroscopic level. The main difference is at molecular level, where in the case of chelate cooperativity the single ligands are from the same system while in allosteric cooperativity they are not.

But other than that, i'm not able to understands what this implies and why we need to differenciate between them. I would appreciate if some of you could give me some more information about this aspects and underlying their main differences.

Thanks in advance.

You should include in questions a short summary of what you already know or found (with eventual quotes/links), so others will not write what you are already aware of, or what you should know if you searched about it. If nothing was found, write at least what (a/o where) you have searched for. This way responders avoid duplication of your effort and redundancy of internet content.

@Poutnik Hope is easer to understand now, thanks!

Here seems to be the free access links to your references on sci-hub.se: What is Cooperativity?   and  Allosteric, Chelate, and Interannular Cooperativity:
A Mise au Point

@Poutnik sorry, maybe i didn't express myself well enough. I've already read the full articles when i've wrote the question, however i don't understand the difference between the two.

It was just a side comment for readers who do not have access to the paywalled articles. (admiting usage of sci-hub.se may be controversial).

The cooperativity in the context of enzymes or substrates with multiple active centers is generally mutual affecting activities of these centers, by various mechanisms.
Allosteric cooperativity (AC) means the bonding of one of centers to a ligand leads to sterical reconfiguration of the enzyme, increasing or decreasing the activity (rectivity, availability) of remining centers. The typical case is hemoglobin, consisting of four enzymatic units. When one unit binds a molecule of oxygen, other units reconfigure themselves to expose their hem centers for easier oxygen bonding. This leads to steeper absorption of oxygen wit raising oxygen partial pressure, compared to a case without the AC effect.
Chelate cooperativity (CC) is typically the case when the active center is a bound atom/ion of a transition metal, intended to bound itself to a bidentate ligand, forming a chelate. In this case, after binding the first binding point of the ligand, the second bonding point of the chelate will much more probably bind as well, compared to independent ligand. another option could be two adjacent substrate active centers, binding each to one of two centers of the ligand.
In summary, CC assumes formation of chelate structure, while AC assumes reconfiguration of the structure of the substrate.


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Why does p-aminophenol oxidize in acid solution
I'm reading Bard's Electrochemical Methods. He cites the oxidation of p-aminophenol in acid solution. I haven't taken organic chemistry in a while and don't understand why oxidation takes place in acid. The pKa's of the amine and alcohol are 5 and 10, respectively. If you have a solution, say pH 1, wouldn't both moieties remain protonated and nothing happen? Thank you.

The reaction (1.5.15) is not spontaneous. It only happens if an oxidant is in contact with the p-aminophenol

Non-negligible amounts of double protonation will almost certainly require a superacidic medium, like pure triflic acid or carborane superacids in non-protic solvents. You need to check the pKb values of anilines and phenols (or equivalently the pKa values of aniliniums and phenyloxoniums). The pKa of phenol and aniline are irrelevant. Note that your reported "aniline pKa" is actually the anilinium pKa.

There is quite a difference in p$K_a$ between an (alkyl) amine, and a phenyl amine. Even more so the difference between an (alkyl) alcohol, and a phenol (a reference), to an extent phenol is a weak acid willing to offer protons and is labelled with a GHS05/corrosive sign (e.g. to skin and eye) while e.g., ethanol does not get it.  This is why I suggest to call the starting material in 1.5.15 a phenol (when describing its chemistry around the OH group).

You can read about the quinhydrone electrode here: https://en.wikipedia.org/wiki/Quinhydrone_electrode
This electrode reaction is similar to the oxidation of Para-aminophenol and is measurably reversible over a pH range of about 2-10. It would be reasonable that the aminophenol coupled with the acid requirements of the oxidant would have an optimum pH [range] for the redox reaction.
Standard electrode potentials are when all activities are at unit levels. Some reactions can actually be measured under those conditions but sometimes the standard potentials must be calculated from other thermodynamic data. [Imagine trying to measure the standard potential of sodium metal immersed in one normal acid!] Research is needed to determine what is going on here.
About " and nothing happen.", Acid-base reactions in water involving proton transfers are very fast and frequent; something is always happening [sometimes catalysis is helpful].

You are right in your assumption that in a solution with an acidic pH nothing would happen as is. But the example is considering a redox reaction.
If a suitable oxidant oxidises the p-aminophenol first, the resulting positively charged p-aminophenol will have drastically different pKa values because of the positive charge. The pKa might be low enough, that the deprotonation can even happen in acidic solution.
Alternatively, the reaction might also proceed via a concerted mechanism, where both a proton and an electron are transfered at the same time. In that case, pKa values don't apply anymore.


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Using sodium metasilicate as a non-reactive reusable electrolyte in water electrolysis
I'm trying to increase the transport of electricity in the tap water I use in my dry cell electrolyser. Table salt is out, as it rusts the stainless steel electrodes and produces toxic gasses. Lye is caustic to skin and mucous membranes but the next best option as the products of the reaction seem to be harmless. Then I thought of water glass (sodium metasilicate) as a solution. It dissolves in water and acts well enough as an electrolyte but less extreme than lye and perhaps more stable. I got an appreciable increase in the production of hydrogen and oxygen in a simple test setup.
So what happens chemically when we use a solution of water and sodium metasilicate and say run 12 volts through it using stainless steel electrodes? Was I creating toxic silanes? What really happened?

There could be formed silicagel at the anode due local acidification.

If you're concerned about the safety of caustic $\ce{NaOH}$, sodium metasilicate itself is quite alkaline. From the SDS: "Causes severe skin burns and eye damage." (Though my experience is, it may be so viscous in solution that skin damage does not propagate deeply from brief contact.)
In addition, if you're passing DC current through it, there will be $\ce{NaOH}$ buildup at the negative electrode, anyway.
As far as silanes, no worry; electrolysis just yields $\ce{SiO3^2-}$ on the positive end, which likely would precipitate silica, eventually.
BTW, you might find this Study on the Electric Conductivity of Sodium Silicate interesting,

Looking back and considering your knowledge. The anode and cathode were both producing gasses. There seemed to be a fairly decent production of hydrogen gas (very vigorous bubbling, small bubble size) on one terminal, and very little oxygen production. (large bubbles, blurp... blurp...) Probably due to some oxygen being spent in the generation of silicates as you mentioned. These silicates might create a clogging problem in the device itself eventually. The production of lye seems negligable not detrimental. Not sure if this is a superior, equal or inferior electrolyte to be honest.

In the beginning, the electrolysis will be all right. But, after a while, the anodic zone becomes acidic, and non-conducting silica $\ce{SiO2}$ will be precipitated around the anode.  The will increase the resistance of the solution. The current will slowly decrease.

Also, the viscosity of the solution might hinder ion flow. Still, an interesting experiment.

12 volts is excessive. current flow should be measured and optimized. My first thoughts are that metasilicate would increase viscosity, hinder ion flow and the increased voltage would destroy the anode. Water glass is oxidized silicia it is not oxidized by O2


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Why is it that in a buffer solution the equilibrium concentrations may be assumed to be the initial concentrations?
Considering the Henderson–Hasselbalch equation,
$$\text{pH} = \text{p}K_a + \lg \frac{[\ce{AcO⁻}]} {[\ce{AcOH}]}$$
$$\text{p}K_a = \lg \frac{[\ce{AcO⁻}][\ce{H⁺}]} {[\ce{AcOH}]}$$
Why are the values for the concentrations often the initial concentration added?
$$\text{pH} = \text{p}K_a + \lg \frac{[\ce{AcO⁻}]_0} {[\ce{AcOH}]_0}$$

Because concentration shift to reach equilibrium is negligible.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

Remember this isn't pure math where $\pi$ is known to billions of digits. Equilibrium constants are only known to 2 or 3 significant figures generally. Book problems will in almost all cases require nothing more than a quadratic equation to solve by making the right chemical assumptions. Solving 5 equations with 5 unknowns isn't problematic for a computer but it is miserable when doing so by hand.

Related: https://chemistry.stackexchange.com/a/172402


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Fischer Projection for Different Compounds

Which of the following Fischer projections is different from the other?
Just by looking at the R-S configuration, figure C is the odd one out. But when I look at D I see that both methyl and hydrogen are below the plane while in other diagrams it hydroxyl group and hydrogen, meaning howsoever you orient both C and D by rotation you will never be able to get the same spatial arrangement as A and B. I am confused by this, my TA says the answer should be opt C because of the configuration and I haven't talked to him about this approach.

If you try to redraw the structure of compound D you will obtain a spatial arrangement that is only in apparence different from structure of A and B.

With a rotation of 120° around the C-H bond you will obtain a spatial arrangement that is identical to the others.

Which of the following Fischer projections is different from the
other?

The final answer is C because it has a opposite configuration from the others:

C = R

A,B,D = S

Any pairwise swap changes the stereochemistry, so an even number of swaps retains the stereochemistry. This is the rule to figure it out on paper. Of course building the models in 3D is much better for a deeper understanding and believing the answer.


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calculation of enthalpy change
So for for the calculation of $\Delta{H}$,we have two formulaes in our book:
$\Delta{H}=E_{2}-E_{1}$
where $E_{2},E_{1}$ are the internal energies of the product and reactants respectively.
Now this makes sense to me,But we've another formula for calculating $\Delta{H}$ which is
$\Delta{H}=B_{1}-B_{2}$
Here $B_1$ is the total bond energy of the reactant and $B_2$ is the total bond energy of the products
I don't understand where this formula comes from,in this formula we are not taking into account the individual energies of the atoms.Dont they differ according to their orientation?And the orientation of them becomes different when they react to become products.
Here are my questions
1.So does this formula really gives a close approximation of $\Delta{H}$?
2.And,where does this formula comes from?

Both formulas are wrong, as enthalpy is not the same as internal energy and the difference in the bond energies is not the same as the enthalpy change.

so what are those formulas for exactly?

Please cite the source of the these equations and provide more context. It is really hard to just have an equation without the actual definitions of the variables within. And for your own benefit, you might want to consider not starting sentences with 'so', and pay more attention to spelling and punctuation.

About the equation $ΔH = E_2 - E_1$:
Note that by definition: $ΔH = ΔU + Δ(pV) =  E_2 - E_1 + Δ(pV) $
For an ideal gas: $ΔH = E_2 - E_1 + Δn \cdot RT$, with the last term being typically about $\pu{2.2 kJ mol-1}$ with total amount change by $\pu{1 mol}$.
For isobaric changes in condensed phases, the term $Δ(pV)$ is often negligible.
For isochoric changes in condensed phases, the term $Δ(pV)$ can be enormous due big changes of pressure due very limited compressibility.
About the equation $ΔH = B_2 - B_1$:
Note that $ΔH = ΔU + Δ(pV) = ΔU_\text{bond} + ΔU_\text{kin, transl} + ΔU_\text{kin, vibr} + ΔU_\text{kin, rot} \\
+ ΔU_\text{pot, vibr} + ΔU_\text{pot, other} + ΔU_\text{electronic} + ... + Δ(pV)$

U form
U legend

$U_\text{bond}$
potential energy of bonds(negative)

$U_\text{kin, transl}$
translational part of thermal kinetic energy

$U_\text{kin, vibr}$
vibrational part of thermal kinetic energy

$U_\text{kin, rot}$
rotational part of thermal kinetic energy

$U_\text{pot, vibr}$
vibrational part of thermal potential energy

$U_\text{pot, other}$
collision part of thermal potential energy

$U_\text{electronic}$
electron excitation part of thermal potential energy


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Term for one chromatography sample affecting the next?
Suppose I am running a collection of samples on a chromatograph (gas or liquid) in some order. Some of the sample molecules have large retention times due to their size and/or interactions with each other or the stationary phase to such an extreme extent that molecules from the previous sample are still arriving at the detector while molecules from the current sample are arriving at the detector.
What is this problem called? Is it an example of carryover, or would some other term better apply? Or would it be termed "contamination"? Or something else?

Yes, it is termed "carryover".

Carryover is recognized as the presence of a small analyte peak that appears when a blank is injected following the injection of a sample that produces a large peak of the same analyte. When it occurs, peaks attributed to the previously analyzed sample may be observed in the subsequent chromatogram(s) which may co-elute or interfere with desired analytes.

Solving Carryover Problems in HPLC

In a high-performance liquid chromatography (HPLC)-based analytical method, carryover denotes one type of systematic error that is derived from a preceding sample and introduced into the next sample.

Zeng et al 2006

+1 from me, "carryover" in its most general sense will work.  However in its more specific senses, it can mean specifically the "non-ideal" retention of analytes into the next sample, e.g. by contamination of the autosampler needle.  This question is about analytes that interact with the column, which is ideally what's supposed to happen in chromatography, but they're just doing it too strongly.  Again, I think "carryover" in its general sense will work OK, but sometimes people do have more specific meanings for this term in mind.


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Interpretation of the Gibbs free energy
Gibbs free energy is defined as the total non-expansion work (i.e. electrical work) that a system can perform. However, it also indicates whether or not a reaction is spontaneous. I am therefore confused on the interpretation of the Gibbs free energy when it comes to reactions such as the combustion of methane, which is evidently spontaneous - meaning that its $\Delta G$ is negative. By the above logic, the combustion of methane should therefore be capable of producing non-expansion work, but I can't see how this is the case (i.e. what types of non-expansion work can it do?).
Can someone explain, or if I'm misunderstanding something?

You can make an engine based on methane burning.

Thank you for your comment. I am assuming that you are talking about rocket boosters? If so, could you please clarify what type of work this would be?

I mean a classical thermal engine, similar to one in petrol or LPG powered cars,  converting (part of) chemical energy to mechanical work via the torsion of rotating crank. But rocket boosters would be possible too, if liquid methane would be used. Another way could be fuel cells.

The best non-expansion work that a system can perform is the electric work (or electric energy). This is what you wrote in the first line of the text, To get an electric work, the reaction must be carried out in a galvanic cell. If the reagents (methane and oxygen for example) are used in an electric cell, the reaction proceeds if the reaction is spontaneous (if Delta G is negative). If Delta G is negative, the reaction will produce a tension and a current, whatever the value of Delta H. In a spontaneous reaction, Delta G is negative, and Delta H can be either positive or negative.

@Poutnik please correct me if I am wrong, but aren't heat/thermal engines devices converting heat energy into work? The heat given off by the combustion of methane itself is not work. Furthermore, dG correlates to the maximum non-expansion work that can be performed, while a heat engine, doesn't it produce mechanical/expansion work anyway?

Have a look here https://chemistry.stackexchange.com/questions/2881/difference-between-thermodynamic-and-kinetic-stability and also https://chemistry.stackexchange.com/questions/141155/about-spontaneous-combustion-of-methane

The (standard) reaction Gibbs energy change is the top limit of nonexpansion work, that can be provided by a closed system at isothermal and isobaric conditions, where 1 mol of the referred chemical reactions happens (at standard conditions).
It does not say that the limit work can be practically achieved, nor it say by what way such a work could be achieved.
Note that the expansion work is already involved in the enthalpy change $ΔH = ΔU + pΔV$.
Heat engines transform the Gibbs energy of fuel to Gibbs energy of heat reservoirs via creation of temperature difference, which is then used for nonexpansion work  $\delta W = T \cdot \omega \cdot \mathrm{d}t$, where $T$ is torsion and $\omega$ is angular speed.
Similarly, human metabolism transforms part of the Gibbs energy of food and oxygen to nonexpansion work of body muscles.
The schema generally is not
$$\text{Reaction} \rightarrow \text{nonexpansion work}$$
but rather
$$\text{Reaction} \rightarrow \overset{ΔG \rightarrow W}{\text{"black box"}} \rightarrow \text{nonexpansion work}$$

So if a reaction happens and the number of moles changes, would that be PV work or non-PV work? P∆V = ∆nRT

@Proscionexium See the 3rd paragraph.

I would say you have your definitions backwards, in a sense. The Gibbs free energy is the Legendre transform of the enthalpy. When you do this and derive the differential form of $G$ from the fundamental thermodynamic relation, you find that $G$ is a function of $P$, $T$, and the number of moles of each type of species in your system. The special thing about $G$ is that, because of the second law, you will always have $dG<0$ for a spontaneous process $\textit{under constant temperature and pressure}$. That conditional is critical. Without it, there is nothing special about $G$ at all, and it tells you nothing about spontaneity. It just so happens that chemists often consider processes where this happens to hold, and so $G$ is a very useful function. This is sort of like how $H$ is quite useful under conditions of constant pressure, as it gives you the amount of heat transferred in a process. But in general the enthalpy is just a Legendre transform of the internal energy: no more, no less.
Through a different set of arguments, you get that $G$ allows you to determine the maximum amount of non $PV$ work obtainable from your system. It is not obvious at all how you would actually obtain this work, but you know that it is theoretically possible to get it, as suggested in another answer. Unfortunately, thermodynamics has a nasty habit of telling us interesting things about nature without giving us a guidebook on how to actually capitalize on them.

Chemists are not usually aware of the Legendre transform (its TD application). I have met the orthogonal Legendre polynomials few times, but not the transform.

That’s fair. I’m actually teaching physical chemistry this fall, and my students will be very familiar with them!


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Dipole Moment in P4O8 molecule

So from the structure, which seems to be quite symmetrical, according to me the dipole moment should be zero.

It's difficult to comment on dipole moments of such extended oxides as they are usually isolated in the crystal phase and will require QC calculations. However even on a basic observational basis, we can perhaps comment that not all the P-O bond lengths are same. Also, you called P4O8 "symmetric", but a cursory glance at a 3-D model will show that the two P=O vectors don't exactly cancel each other. So qualitatively we may suspect this to have a dipole moment. But proper bonding analysis and confirmation will require computation.

You may check the conformer here: https://pubchem.ncbi.nlm.nih.gov/compound/Tetraphosphorus-octoxide#section=3D-Conformer

Try this one link.springer.com  but paywalled.  sci-hub.se may be blocked in some countries.

@Poutnik Ok i will check it out..

@YusufHasan yes the 3d model shows the P=O aren't symmetric, yes I understood, that there is no such symmetry...Thank you!

Try to work out the point group, its possible it is C2v .

http://csj.cumhuriyet.edu.tr/tr/download/article-file/1055622 says P4O8 is C2V, but it's a computational study so I'd like to study the method more carefully before I believe it, on first glance the basis set doesn't look that big. (note I am a computationalist). https://pubs.aip.org/aip/jcp/article/5/8/626/209772/The-Molecular-Structure-of-P4O6-P4O8-P4O10-and is an experimental study from 1937 (!) which says they can't tell.

This is an example of the danger of reading too much into two-dimensional paper representations of three-dimensional molecules.
Imagine the two pendant oxygen atoms were not there and you had just $\ce{P4O6}$. The phosphorus atoms would then be equivalent. So to make  $\ce{P4O8}$, instead of putting the two additional oxygen atoms on top and bottom you might have put one on top and one on the right. Now the molecule looks (di)polar.
To resolve this paradox, you must use the third dimension properly. In the $\ce{P4O6}$ core each phosphorus atom is the vertex of a polyhedron whose edges lie along the $\ce{P–O–P}$ linkages. Thus four vertices and six edges, meaning a tetrahedron (not a square). With a tetrahedron the two additional oxygen atoms are not on opposite vertices because a tetrahedron does not have any such pair of vertices, so those bond polarities do not cancel and $\ce{P4O8}$ ends up having a nonzero dipole moment.


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How can I dissolve my Copper(II)sulfate in my MOPS buffer pH 7.2?
For my research project I need to analyze a copper-ligand complexes in a basic buffer through UV-Vis spectroscopy. After testing several buffers, I currently use the MOPS buffer which I want to adjust to a pH of around 7.3
The problem is: my copper(II)-ion reacts with almost everything by becoming a hard soluable solid! I have now prepared the MOPS Buffer and adjusted the pH with NaOH (1M) to 7.2 and when I wanted to dissolve my $\ce{CuSO4}$ in it, and it once again resulted in a blue precipitation. I am now at a loss of options since I also get blue precipitation with carbonates, phosphates, ammonia and citrate. Does anybody have insight into what i can do?
structure of MOPS / 3-(N-morpholino)propanesulfonic acid:

(image credit Wikipedia)
PS:

adding a chelation agent is also not a useful option since it will compete with the ligand i need to research
The concentration of the copper(II)-solution is about 40 µM

Whatever the nature of the buffer, it will react with copper ions

Did you exhaust [Good's list of buffers](https://en.wikipedia.org/wiki/Good%27s_buffers?

The point is, copper ions form insoluble copper hydroxide or many insoluble salts at neutral/alkalic pH, UNLESS there is already present a strong ligand binding itself to copper ions.

I choose the MOPS because it doesn't interact with any metal ions, especially not copper.
Yes, i did exhaust Good's list of buffers.
There is a lot of literature with the use of the MOPS Buffer for Copper-Ion research, but (and that is the main problem right now) my Buffer has to be adjustet by adding a Base. In the Literature it says to add NaOH, but this just doesn't work (the blue precipitation)... now i am at a loss of options and ideas with what i can adjust the pH with. NaOH, KOH and Ammonium certainly do not work.

Ammonium might work, since it forms a strong bond with copper: https://en.wikipedia.org/wiki/Tetraamminecopper(II)_sulfate . If that interferes, raising the temp might work, for just 40 µM of Cu++.

Thank you for the suggestion! in my head i had the notion, that the tetraamminecopper(II)-complex was unsolvable in water, so i guess i will conduct a test tomorrow. A milestone would be reached if i at least get a clear solution in alkalic pH.

How did you try to dissolve the $\ce{CuSO4}$? It might be the case that precipitation occurred due to high local concentrations.

UPDATE: ok guys so, thanks for all your help, it really helped get a clear thought about this problem. I have now, at least for this moment, found a solution to my problem. I have set up the MOPS Buffer and adjustet the pH with L-Arginin. It also forms a complex with copper according to literature, but it is fully watersoluable. When analysing the copper solution in the UV-Vis, the copper solution only showed absorbance around 680 nm, which is totally fine since i dont need that wavelength anyway :)
My next alternative would have been to try ammonium acetate to adjust the pH.