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how can a chemist detect the element boron in a sample of ordinary dust?
After the world trade center attack 22 years ago, the USGS sampled dust from the world trade center site and from lower Manhattan. They tested for every solid element except boron. I'm curious about this. I was told by a chemist that her electron microscope method (micro beam) was not a good way to search for boron. I'm trying to determine  what might be the best to go about this.

Mmm... probably some IR spectroscopy, depending on sample... But what's the point?

Perhaps boron was not considered a hazard? And I question if they checked for all solid elements -- can you provide a reference for that statement? Why would one test for iron, sodium, calcium or aluminum, common in dust from many sources?

I am trying to reconstruct why the USGS did what they did. I still don't know. That's why I posted here. Why boron? Good question. That cuts to the quick about the WTC collapse.

The USGS scientist I spoke with is a micro beam expert. She told me this method is not optional for identifying boron.

I don't think this is a bad question despite the down votes. However, the claim is indeed a bit fuzzy and could be substantiated with sources. For example where the claim is from what they tested about. For what it's worth, I don't even know what USGS stands for.

I have the US Geological Survey (USGS) dust study on order. When I have the data in hand I will answer your question.

I just hope eventual missing of boron analysis would not feed conspiration theorists.

you can find a reference for testing borons here

If you search on the web there are several papers on detecting Boron by flame Atomic Absorption Spectroscopy

As a light element, e-beam induced X-rays are indeed not a great technique. Mass spec, atomic emission, etc. are better.


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Stereoselectivity in catalytic hydrogenation
Why is cis stereo selectivity is observed in this case? See image below.

I can't think of any reason why cis addition is major.
In my opinion, the ratios should be opposite because of easier adsorption of alkene on metal from the front side. So $\ce{H2}$ should be added trans to the methyl group.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


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Avoid triethylphosphate as a side product during P-C-P coupling reaction with diethylchlorophosphate with LDA
I have a reaction in which I perform a deprotonation of an acidic proton at a phosphate and later do a P-C-P coupling. The coupling partner is diethylchlorophosphate. Deprotonation is performed with LDA in dry THF. Deprotonation is performed at -78 Degrees Celsius and argon. Unfortunately I also observe triethylphosphate (TEP) as a side product that can not easily be separated of from the desired product by silica flash coloumn chromatography. I am wondering what the mechanism could be, how diethylchlorophosphate forms triethylphosphate and how this formation can be avoided completely? Can LDA abstract an ethanol/ ethoxy unit that subsequently reacts with diethylchlorophosphate? Than less base should have an effect? Might the slower addition of the diethylchlorophosphate to the carben ion also reduce the formation?

Does the phosphate anion generated with LDA quench normally with other electrophiles e.g. MeI?

The ion is a carbanion. I would assume that it would react with a methylation agent like you suggested. I never tried that, but I imagine an experiment were you add ethanol and diethylchlorophosphate together with a base to see if TEP is formed.

Diethyl chlorophosphate + EtOH will certainly give triethylphosphate. I'm trying to explore possible causes of the triethylphosphate production as I've done this chemistry and it worked for me. How do you quench the reaction mixture?

Reaction mixture is quenched Bei KH2PHO4. I started to observe this species only when I increased batch size. Previous runs somehow were not contaminated.

Maybe the more interesting experiment would to check if diethylchlorophosphate forms triethylphosphate just in the presence of base. In this case it would make sense to use the base im slight molar deficiency.

Worth running the experiment

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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What mechanism do reaction of PCl5 follows, SNi or SN2?
I was taught that $\ce{PCl5}$ reacts with alcohols by $S_Ni$ mechanism as shown below:

Here it can be seen that $\ce{PCl5}$ exists as $\ce{[PCl4]+[PCl6]-}$ and so lone pair of the alcohol is donated to the electron deficient Phosphorus. Then nucleophilic Phosphorus deprotonates the good leaving +vely charged Oxygen. Now a Chlorine of the Oxygen-bonded Phosphorus attacks the alkyl group in a 4-centre cyclic transition state mechanism just like that in a reaction of an alcohol with $\ce{SOCl2}$. This yields an alkyl halide or two (if it was an ether) and $\ce{POCl3}$.

Once, I searched about this on internet and I couldn't find this mechanism but found written that $\ce{PCl3}$ and $\ce{PCl5}$ do $S_N2$ while $\ce{SOCl2}$ do $S_Ni$ (if there is no pyridine). However, the products were the same, leaving the stereochemistry.

I know the mechanism of $\ce{SOCl2}$ with alcohols and if I looked upon the above mechanism then I found no reason for why it should be wrong.
Also, the fact that $\ce{PCl5}$ is unstable and dissociates into ions to distribute its charge is true and can be validated by a simple Google search.
I am looking for some reasonable arguments to contradict or validate the above mechanism, if possible.

Have you seen https://chemistry.stackexchange.com/questions/33675/reaction-of-alcohols-with-pcl5-and-pcl3

@BuckThorn Yeah... Completely from top to bottom but, it discusses totally on reaction with PCl3 and it is disjoint to the mechanism above.


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Detection of contamination from salicylic acid via either single excitation fluorometer or via absorbance
My colleague detected fluorescent contamination likely from salicylic acid in some water samples. I have made a new batch and want to check whether my new samples also have this contamination.
1. Can I use a single excitation/emissions fluorometer to detect the presence of contamination from salicylic acid or a compound that fluoresces at the same wavelength as this. If so, what wavelength of emissions and excitation would be suitable?
2. Can I determine the presence of salicylic acid in my samples by measuring absorbance, if so, at what wavelength should I measure?

Not possible to assign "a random" fluorescence due to salicylic acid in an unlikely sample- seawater. There are millions of other natural and unnatural compounds which can fluoresce under UV. Point # 1: Not possible, 2. Again not possible. You first need to ensure your compound is salicylic acid from seawater from an another detector like mass spectrometer.

Also consider that a single molecule's fluorescence can be detected, but absorbance requires many to absorb a detectable amount.

(1) No, you will really need to measure the spectrum to be certain what is present. (2) fluorescence is a zero background technique so is orders of magnitude more sensitive that absorption which is a difference measurement. You can easily find the fluoresce and absorption spectra online.


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Bonding and symmetry
So I have been looking at different ways of symmetry and bonding
The so called counting method is something that I don't get
For each operation in the group:
For each object in the basis set:

    Apply the operation to the object to generate a resultant object
    Calculate the overlap between the resultant object and the object
Sum the overlaps to give the character for the operation.

You get +1 if the orbitals stays the same zero if it moves and - 1 if it changes I find it hard to imagine the AO not really sure if the method works perhaps someone here can show en example.

You should improve the quality of your question. To begin with, it is difficult to understand what your starting point is, and what you understand or misunderstand. Are you familiar with matrix representations, reducible representations and ireducible representations? Guess you are asking about symmetry-adapted linear combinations of molecular orbitals (SALC). You need to work on your basic concepts and improbe your question. You can find useful information and examples on Molecular Symmetry form the PhysChem or InorgChem point of wie at [Libretext.org Chemistry] (https://chem.libretexts.org)

Ive attached a figure that should help, in this case for a rotation. This only works for 90/180 rotation/reflection for any other angle you need to use sin/cosine. The procedure is the same for orbitals as for atoms as shown here.


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Why having a carbonyl group is not contributing to binding energy
I am carrying out a virtual screening project for a protein receptor to identify possible ligands (small molecules). After the virtual screening, molecular dynamics, and MMGBSA calculation, I identified two ligands that differ only by one carbonyl group. Because this has not been published, I cannot disclose the structure. The MD study shows that the ligand without the carbonyl group is more stable over time (0-100ns). And MMGBSA calculation also shows that the one without the carbonyl group binds better.
From my understanding, having a carbonyl group should increase the likelihood of interacting with amino acids, and thus contributing to binding. I appreciate it if anyone could offer some possible reasons why this is the case (no carbonyl group is better). I asked GPT and it tells me it might be the steric effect which makes sense. Are there any further analyses I can do to answer this question?
By the way, I meant it to be a fully computational study, and thus I do not plan to validate it by wet lab.

Well, there could be various reasons. For example carbonyl could bind to another part of the ligand.

@Mithoron, are you suggesting that the carbonyl is binding to another part of the ligand, like an intra-molecule interaction?

Yeah. That's one option.

Without more details, impossible to say (and I do molecular dynamics for a living, looking at ligand-receptor interactions). However, binding free energies via MM-PBSA are going to at best be +/- 2 kcal/mol; and you haven't told us how you prepped the system (dealt with missing residues, protonation states of certain amino acids, and so on and so forth). The net of it is that your question is not answerable in the present state. As a side comment, if you are skilled enough to be running 100 ns MD and MM-PBSA analysis on the trajectories, you really should know more basic chemistry than you do.

@ToddMinehardt, thanks so much for the insight. I should have asked a better question with more details. I was just seeking some perspectives to look at the question and therefore your comment is already helpful to me! I will consult with my chemistry professors for more chemistry knowledge.

"From my understanding, having a carbonyl group should increase the
likelihood of interacting with amino acids, and thus contributing to
binding."

No. This is a fairly foundational misunderstanding, too. Think about it... all binding pockets consist of "amino acids," are you suggesting that having a carbonyl will always increase binding strength? No, of course not.
Protein-ligand is a much, much more complex phenomenon involving the steric and electrostatic impacts of each particular amino acid in the local binding area. I think you would be wise to focus your studies on basic biochemistry prior to beginning any drug development project.


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How to clean naphtha of non-saponifiable fats
My dad has been using naphtha to degrease some animal bones, which based on the MSDS is probably some random jumble of hydrocarbons, mostly heptanes. He's trying to be environmentally friendly so he doesn't want to throw it out once it's saturated, and isn't comfortable trying to distill flammable liquids, so after some experimenting we started using sodium hydroxide to convert most of the dissolved fats into crude soap that separates out and can be discarded. This has worked well, but the naphtha is still accumulating some other impurities, most of which are (I suspect) other fats that are non-saponifiable like cholesterol.
I have a general interest in chemistry, but no formal education in it so I'm running out of ideas and don't have the correct terms to properly google my issue (or to understand most of what I've found). The closest I've come is finding that adding acetic acid causes something to precipitate out, but it hasn't even slightly settled out over 24 hours and when I run the resulting milky-white naphtha through a vacuum filter (10 micron) I don't see any change in clarity.
What are some ways I can try to clean my naphtha, using relatively little specialized labware and equipment?

You are doing good so far. Placing the naptha in a deep freeze above its freezing point might precipitate more impurities. try a small bottle and see. It might be possible to set up a solar powered [no flame or electricity] molecular still to slowly distill the naptha, [similar to the survival devices to allegedly distill water from the ground]

It would not be environmentally beneficial if you use chemicals or treatments that are more dangerous, or have a larger carbon footprint, than getting more naphtha. Perhaps there is a use for the partially-purified solvent, e.g. cleaning, fuel (yes, that releases CO2), or send it to an efficient commercial solvent recycler.

Throwing my solution in the freezer did the trick, a fair bit of solid settled out to the bottom. I'm going to do some testing to see how much has been removed, but it's looking promising!


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How does a reduction in potential energy in a chemical bond release energy
This question has to do with the idea of stability and energy. The premise is that systems will tend towards lower energy states, that’s why bonding happens and electrons prefer single orbitals. And that makes sense in all considering the second law of thermodynamics, but it still doesn’t answer what causes the release of energy when bonding. Clearly, the potential energy of each electron is lower when there is a greater charge and a stronger force of attraction between things—it makes sense, as when you have a stronger force of attraction, you have less “pathways” or degrees of freedom, hence less potential energy. It also checks out with the math of Coulomb’s law, as when the radius decreases from bonding, potential energy decreases as well. So by an atom exhibiting a magnetic pull on another and bonding, potential energy of each electron decreases. But the thing is I just don’t get where that loss of potential energy is going—in this scenario, it wouldn’t make sense for it to be randomly emitted. Or is that not even the reason bonds loose energy?
Consider this analogy, and correct me if I’m wrong:
Consider I have 2 magnets A and B that are oppositely charged. If I take magnet A and pull it away from magnet B, now I just did work on magnet A and increased the magnetic potential energy of it as well. Now lets say I released magnet A. The potential energy would be converted into kinetic energy and it would tend towards the other magnet. Now, let’s say I catch magnet A before it hits magnet B. The potential energy of Magnet A just decreased in the field because the kinetic energy transferred to my hand, right?
So instead in atoms, the “hand” that stops atoms from ramming into each other completely is other electrons and the repulsion of the nucleus. So it seems all energy is conserved. Each atom is just transferring energy between each other by both pulling on each other. So how does energy get released?

I don't know any thermodynamics but i would assume 2 atoms pulling on each other would lead to an increase in the activity of the system (aka temperature)

https://chemistry.stackexchange.com/questions/55914/how-is-heat-physically-released-in-an-exothermic-process https://chemistry.stackexchange.com/questions/119093/how-does-the-energy-released-during-a-bond-formation-typically-manifest-itself-o https://chemistry.stackexchange.com/questions/115888/what-actually-happens-when-energy-is-released

Energy is not released; it is transferred either as kinetic or potential energy of another particle or mass or the emission or absorption of a photon.

Consider $$\ce{2 H <=> H2^{*} ->[collision][emission] H2 + energy}.$$
Imagine a hydrogen atom figuratively rolling down to the potential pit of the bond with another hydrogen atom.
Without passing energy elsewhere, it will bounce and appear free again, keeping its initial energy. It more or less performs one molecule vibration at the top energy level which has just enough energy to break the bond.
Like a skateboarder on U-ramp in lossless scenario would appear on the other side of the ramp.
Now imagine the atom collides with another molecular entity during the time it has increased its kinetic energy and decreased the potential one. Or, it emits a photon. The atom then passes a part of its energy away and is then trapped in the potential pit of the bond.
Like if the skateboarder accidentally hits something/somebody during his ride and loses energy. He will not get in such a case to the other side and is "trapped" in the ramp.

In reactions with more than one product, the released potential energy is converted to kinetic energy of products as there is always something handy to push to release the energy, products flying away of each other.
In this way, there is not needed an inert object to pass momentum as in the prior case.

It's always interesting to see how most people completely take the presence of an inert third body for granted in standard conditions (condensed systems and relatively concentrated gasses), when in fact their absence reshapes chemistry (e.g. high-altitude atmospheric chemistry and astrochemistry)

@NicolauSakerNeto Yes, these chemistry domains are very interesting. Some exotic molecular entities are hard to detect elsewhere.

Our ambient conditions are highly unusual, but they are all we know, so we call everything else exotic.

As Poutnik has pointed out the reaction such as $\ce{H + H \rightleftharpoons H_2^\dagger \rightleftharpoons H + H }$ is entirely symmetrical which means that the bond is formed then breaks up again, i.e. energy is conserved as is always the case. The product $\ce{H_2}$ can only be formed if some energy is lost such as by collision with another inert atom or molecule during the small time interval that the nascent $H_2^\dagger$ exists, which is only a few femtoseconds.
If the reaction is atom+diatom, such as $\ce{Cl + H_2 \rightleftharpoons ClH_2^\dagger \rightleftharpoons HCl +H }$ then HCl will form as the H atom will take away some kinetic energy and HCL itself can have different amounts of kinetic, vibrational and rotational energy depending on the initial collision energy. We can picture the reaction as moving along a potential energy surface from reactants to products, see figure.
The picture shows the (LEPS) energy surface for $\ce{Cl + H_2}$, the numbers represent the potential energy and the black line the path the species take, the wiggle represents $\ce{H_2}$ vibrations (bottom) and $\ce{HCl}$ left side after reaction. (Red is high energy, blue low). You can see that $\ce{H2}$ initialy has only a small amount of vibrational energy but $\ce{HCl}$ has a lot more after reaction. The transition state is at the saddle point, the narrow part of the blue shaded contour.


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Reaction of eugenol with concentrated hydrobromic acid in the presence of gaseous hydrogen bromide
DISCLAIMER: I am not using any of this information to commit a crime, my university has a license from the relevant authorities to investigate clandestine chemical procedures.
We are investigating an amateur conversion eugenol to the 4-(2-bromopropyl)catechol compound you see there. In theory concentrated hydrogen bromide should be able to cleave the ether, yielding methyl bromide, which can be easily removed from the system, and also add to the double bond in markovnikov fashion. I have some questions for how this reaction can be performed. First of all, what solvent should be used? I am thinking of using acetic acid but i am slightly worried about acetate as a competing nucleophile, especially considering that eugenol acetate (alcohol replaced with acetic ester) is a naturally occurring component of clove oil meaning it must be somewhat stable. Perhaps I would need to take my impure product containing acetate derivatives and treat it again with HBr to fix it a little bit? Also, at what temperature should this reaction be performed? Please help answer me so we can stop the illicit drug trade.

You could provide your actual affiliation but you choose to remain anonymous, which makes bringing up the fact that what you are proposing could be used illicitly all the more curious. This is off topic for this site, but since you bring it up, how exactly is sharing this information with you and the wider world going to help stop illicit drug trade?

https://chemistry.meta.stackexchange.com/questions/3619/better-living-through-home-experiments-chapter-ii

Can you provide more context on this? Like, why you need this information and how this info will help you stop illicit drug trade, what is your action on this?

Since there is a lot of legality going on, I think you should read these posts: https://chemistry.meta.stackexchange.com/questions/3003/faq-update-how-to-ask-or-close-questions-that-involve-potentially-controlled, https://chemistry.meta.stackexchange.com/questions/5105/on-asking-a-question-that-may-inspire-unethical-or-illegal-activity

This suspicious question seems to me rather as an attempt to promote drug trade wrapped in a coat pretending the opposite. Any illegal lab worker could write it like this, if he stumbles upon some particular difficult synthesis step.

@Poutnik I suspect the same unless OP clarifies. Btw, who is closing the post as HW? Seems they are in hurry to close the question but can't find the appropriate closure reason.

I’m voting to close this question because it raises suspicious flags and the OP does not address it.

Consulting the oracle of wisdom that is Greene's "Protective Groups in Organic Synthesis" produces this reference that describes the cleavage of dialkyl and aryl alkyl ethers with 47% aq HBr at 115C in the presence of phase-transfer catalysts with yields >90%. These conditions would appear to meet your requirements.


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Color change in coordination compounds
When $\ce{FeCl3}$ is added in water it gives a pale yellow color, but on the addition of $\ce{KSCN}$ it changes to intense red color, so how would we explain the color change?
What I thought that the intensity would for sure change as when $\ce{FeCl3}$ is added with $\ce{KSCN}$ it would form a tetrahedral complex because $\ce{SCN^-}$ acts as a very weak field ligand, and as tetrahedral does not have center of symmetry La Porte Rule cannot be applied while in water it would form octahedral complex and on drawing the electronic configurations La Porte Rule will fail there, thus I came to the conclusion that color of the $\ce{KSCN}$ on would be more intense.
But how would we tell about the color change?

If someone is downvoting my post, would they at least tell me why are they doing so, I don't know all my previous posts were also heavily downvoted, and it only happens on chemistry stack exchange, on physics and math people do reply properly.

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.

@KutubkhanBhatiya "all my previous posts" -> there is only one other post and that was closed as HW. Perhaps, you can edit the question and clarify so it can get opened. Downvoters are not obligated to give an explanation. (I am not the downvoter btw)

@NilayGhosh I am not trying to offend someone, but I deleted my previous posts as they were all closed due to downvotes

@NilayGhosh And I also changed the questions by following the guidelines they attached about the homework question and I followed it as well but then too it didn't get open, so that's why

@KutubkhanBhatiya that also is not a right approach. Deleting questions frequently because you receive downvotes might get you a question ban. Instead edit the posts.

@NilayGhosh I think Kutubkhan was right to delete his downvoted threads. I never deleted my threads and eventually got permanently banned by low score. They shut down most of my threads just saying "not focused enough" yet they weren't homework questions. SE had to delete my questions in order to remove my permanent ban...

The choice of «intensity» is not a good one here, because the change from (pale) yellow to red relates to different positions of the absorption band(s)  in terms of an energy split between orbitals / eventually recorded absorption wavelengths.
Second, in aqueous solutions $\ce{Fe^{3+}}$ typically is said to be an hexaaqua cation.  The overall reaction in aqueous solution hence is better described as
$$\ce{[Fe(H2O)6]^{3+} + 3 SCN^- <=> [Fe(SCN)3(H2O)3] + 3 H2O}$$
However, the situation is a bit more complex (literally) because exchange of water by $\ce{SCN^-}$ is stepwise.  This is to say that upon addition of the aqueous solution of $\ce{KSCN}$ you simultaneously have the equilibria of
$$\ce{[Fe(H2O)6]^{3+} + SCN^- <=> [Fe(SCN)(H2O)5]^{2+} + H2O}$$
and
$$\ce{[Fe(SCN)(H2O)5]^{2+} + SCN^- <=> [Fe(SCN)2(H2O)4]^+ + H2O}$$
as well as
$$\ce{[Fe(SCN)2(H2O)4]^+ + SCN^- <=> [Fe(SCN)3(H2O)3] + H2O}$$
which each can be described by an individual thermodynamic constant $K$.
Third, tetrahedral complex refers to a cation in the centre with four bonds extending to ligands.  Neither the simplified description of $\ce{[Fe(SCN)3]}$, nor the better one of $\ce{[Fe(SCN)3(H2O)3]}$ fits this well.  (Here, both water and $\ce{SCN^-}$ are monodentate ligands. So the number of ligands equates to the number of «bonds» between the ligands and the central metal cation.)  A typical shape of complexes of formula $\ce{MA3B3}$ (with both $\ce{A}$ and $\ce{B}$ monodentates) is the octahedral one.

Oh got it, but just to make it clear I am a first year college student and this is my first chemistry course, and this was one of the previous year question of this course, and we barely even covered tetrahedral geometry, so like about the color change is it that that as ligands are changing the value of $\Delta$ changes, that is why there is a change in color?

@KutubkhanBhatiya How the MOs split (which are the orbitals lower, or higher in energy, relative to each other; the split of the ligand field to be square quadratic, tetrahedral, cubic, octahedral; how large is the energy gap between MOs) depends on the ligands, their number of bonds (monodentate/chelating ligand; hapticity of the central cation); initial electron configuration of the cation. For $\ce{ML6}$ complexes, depending if additional electrons «enter» orbitals antiparallel in spin to already present electrons, there are low-spin/high spin complexes. It certainly takes time to learn it.

Hey @Buttonwood you seem pretty good, maybe you can attempt my question here https://chemistry.stackexchange.com/questions/172258/colored-inorganic-liquids-why-is-scl2-red-and-s2cl2-orange

@NealConroy I speculate someone deeply invested in quantum chemistry could provide an estimate which electronic transitions should be considered for $\ce{SCl2}$ and $\ce{S2Cl2}$, but it isn't me.


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Why does bond energy not vary with position?
Consider the formation of a hydrogen bond:
$$\ce{H + H -> H2}$$
The $\Delta H$ is $\pu{-436 kJ mol-1}$, so we can equally write the equation as: $$\ce{H + H -> H2 + \pu{436 kJ}}$$
But depending on the differences in position between the hydrogen atoms, there will be a different force of attraction between each other by Coulomb's law. So the potential energy will be larger the farther away the hydrogen atoms are from each other; thus, more energy would need to be released to form a covalent bond. So how does it always release $\pu{436 kJ}$ of energy when forming an $\ce{H2}$ bond if the energy must vary depending on initial position?

Of course while the strict definition of the bond energy is based on infinite separation, the shape of the potential energy surface gets very flat very quickly. So the difference in energy between say a separation of 10 bond lengths and 5 bond lengths as a starting point might be so small it is inside the error bars of the value.

It's a good point, and precisely because of this, bond dissociation energies (in fact, any kind of reaction energy or enthalpy) are defined such that different reactants are infinitely separated from each other. This includes, for example, the two hydrogen atoms on the left-hand side of
$$\ce{H + H -> H2}.$$
This means that there are no Coulombic forces between the hydrogen atoms (or, to be precise, no forces between the constituent particles of different hydrogen atoms).
You may have seen a diagram such as the one in this answer by porphyrin: https://chemistry.stackexchange.com/a/57157/16683
The infinite separation corresponds to the far right end of the diagram (where the energy $E = 0$), and the bond corresponds to the bottom of the well. That is why the bond dissociation energy is equal to the depth of the well.*
In general, if you have something like
$$\ce{A + B -> C + D}$$
what this means is that $\ce{A}$ and $\ce{B}$ are at infinite separation, and then you bring them together, they react to form $\ce{C}$ and $\ce{D}$, and you infinitely separate the products again.

* (Neglecting zero-point energy.)


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Strong acid definition
Strong acids are defined as being stronger than $\ce{H3O+}$ and thus dissociate completely in water. However, what does it mean exactly to be stronger than $\ce{H3O+}$? Is it the Ka value? Also, what makes $\ce{H3O+}$  the strongest acid in water that can exist in aqueous form? Does it have to do with water's ability as a proton receiver?

Yes, Ka value. It's not true that hydronium cation is strongest present in aq. solution, though. It's just that stronger ones are present in lower concentrations, at least in diluted solutions.

Water. That's what makes H3O+ the strongest acid that can exist in water.

It's called the solvent levelling effect

@Mithoron, could you please elaborate on your comment?

Everything is equilibrium. If an acid is just marginally stronger than H3O+ there will still be a measurable amount of undissociated acid except in very dilute solution. Think about how the equilibrium constant is defined. Nature doesn't possess an on/off switch, its more like a potentiometer.


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Using acetic acid to make vinegar
I'm aware that a very similar question has been answered, however that only deals with powdered form and there is no information about food safety in there.
From what I understand, white vinegar is just acetic acid diluted with water. Where I live I cannot find anything stronger than a 5% distilled vinegar, neither in shops nor online (not even Amazon). I can however get a 99% acetic acid solution in liquid form. If I dilute this 1:8 with water it would give me a roughly 12.4% vinegar solution. My question is

Is it safe to consume acetic acid diluted with water?
Are there other processes or agents required to make a distilled vinegar that mixing acetic acid with water does not provide?

Any help is appreciated.

12.5% is a very high concentration for food vinegar. But dilution of food grade acetic acid to produce vinegar is feasible and often done. The islamic restriction on brewing (which prohibits consumption of alcohol and naturally brewed vinegar) has led to the availability of "non-brewed condiment" which is, essentially, industrial acetic acid diluted to food-safe levels. (but needs food-grade acetic acid to start with).

12% is basically the standard for white vinegar/vinegar essence in Europe, it's only in the US that you will have trouble finding these concentrations. It's not particularly high, and almost all recipes using this will have some amount of water and sugar as well. Good point about verifying that it is "food grade".

Answering your first question,
I would recommend against trying to dilute 99% acetic acid to consume. First of all, the dilution of 99% acetic acid is dangerous without the proper precaution, as it will be highly exothermic and can cause hot acid to spray onto you. Second of all, your source of 99% acetic acid is most likely meant for use in a lab setting, not for human consumption. Although lab-grade materials tend to be purer than their food-grade counterparts, they may contain impurities that are not safe to be consumed and that would not be allowed in materials used for food. Additionally, consuming higher concentrations of acetic acid can be dangerous. According to this case report, "oral ingestion of greater than 12% has been reported to cause haemolysis, renal failure, shock and death".

Thanks for the thorough reply and link to the article. 12% is pretty bog standard in most of Europe and US so I wouldn't be too worried about accidentally consuming 12.4% or even 13%, but the point about impurities not approved for food grade is a good reason to avoid it. I'll have to keep looking around then :) thanks very much!

Vinegar can be fermented from wine to about 12%.


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Why does ethanol decrease viscosity of a high pH carbomer solution more compared to water
Carbomer 407 is made up of acrylic acid that undergoes polymerisation to form a long linear structure. At higher pH, the carboxylic acid functional groups will dissociate to form carboxylate ions. These carboxylate ions will have a strong electrostatic forces of repulsion that gives them its high viscosity.
When adding a water and ethanol to a high pH solution of carbomer, both the viscosity decreases but it is more significant in ethanol. Why?

Include a short summary of what you  already know or have found in offline and online resources. If nothing was found, write at least what a/o where you have searched.
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Water would create much stronger intermolecular hydrogen bond than ethanol. $\ce{\bond{-}CO-O^{-}\bond{...}H-O-H\bond{...}O^{-}-CO\bond{-}}$


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Adding colours to polymers
It is said that some polymers can be dyed, but I am not sure about the mechanism behind.
Let me take an example. In anodization, aluminium oxide can be coloured since there are tiny pores for the ink molecules to get in.
But then, for polymers, is it just so simple that the ink molecules form bonds with the polymers?

Polymer dying/colouring is several thousand years old.
It is done in several ways:

Forming covalent or hydrogen bonds
with polymer molecules, mostly natural hydrophilic polymers like plant polysaccharides (linen, cotton) or animal proteins (wool, silk)
Dissolving of dyes in melted thermoplastic polymers
Dissolving of dyes in reaction mixture producing thermoset polymers
The same as the latter two, but Disperging pigment particles instead.


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Are there common/trivial/archaic names for any of the periods of the periodic table?
There are "common names" for all of the s and p block groups of elements (alkali metals, alkaline earth metals, icosagens, crystallogens, pnictogens, chalcogens, halogens, noble gases) (although some of those aren't that common), "Trivial names" for some of those aswell (triels, tetrels, pentels), and up to d-block referring to a group as "the litest element in said group" group/family is accepted nomenclature.
What my question is is has there ever been a name used in any reasonable capacity or even proposed for any periods of the table beyond just period #? ie. is there any name I can use for the set of elements containing just H and He.
Period 6 and 7 f-block elements have commonly used names and this is kinda what I am looking for but not quite.

What would be the point in that? Also if you know triels, tetrels, or even some icosagens, I've likely never heard about, then I doubt we got much to add.

Though I don't know of terms for each row, there are a few that describe a horizontal section of the periodic table.

Astronomers refer to anything other than H and He (first period) as metals, so in that usage, the first period would be "nonmetals"... This hardly a useful chemistry definition, though ;-)

Lanthanides and Actinides are sections of a period, but certainly not the whole row.

Scandides are used for some period-4 elements, though again not a whole row. (courtesy Oscar Lanzi)

Transuranic element is a horizontal set, but also not a single period. The concept is more useful in nuclear physics rather than chemistry.

Heavy metal is far too broad a term, also, and omits nonmetals and alkalis (other than Ra). [BTW, I propose Malmsteenium for the next heavy metal to be created.]

From the pattern established by Lanthanides,  Actinides, and Scandides, for my purposes I am just gonna go with calling each period block combination by the first element in that set with the ending changed to -ides

Alkali-metal "-ides" would also be compounds containing anions of these metals.

The first extra row on the bottom of the table is known as rare earths.

Also: Coinage Metals

Coinage metals are not a horizontal period, but rather a group or column with similar properties, e.g., Cu, Ag, Au. The question was for, "a name used in any reasonable capacity or even proposed for any periods of the table."

The word "scandides" is sometimes used for the Period 4 transition elements from scandium (the namesake) through zinc, particularly in the label "scandide contraction" for what is more commonly called the d-block contraction. As with the lanthanides and actinides, the label does not cover the period as a whole.


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What is the difference between zero-point and total energy of a molecule in the output of an ORCA FREQ calculation?
I am trying to find the enthalpies of formation for a couple organometallics in ORCA and as a first time user I am a bit bogged down in the terminology. The Gaussian manual has a section on these calculations (4.3 on page 14) and I am trying to confirm whether the "total energy of the molecules/atoms" is the same as the "inner energy" or "total thermal energy" you can get from the ORCA FREQ calculations.

The ORCA manual is fairly explicit on most things. One thing to check (with a small molecule, Hartree-Fock, small well-defined basis set) is to set the temperature. Those numbers that change have a thermal component. You should be able to get them to match betwenn the two programs.

@Spencer. The title or your questions is a bit misleading. Your are asking about the meaning of the output of a computer program not about a general definition. Could you correct it?


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Improving the solubility of p-dibromobenzene in a reaction with water as solvent
I am trying to react p-dibromobenzene with Fenton's reagent (in an attempt to see if I can add alcohol groups to the ring and observe if there is any specificity in its addition), in this case using %30 H2O2 and FeSO4. The reaction will take place in water, since the oxidation of the Iron Sulfate and formation of hydroxyl ions/radicals is the major mechanism.
The main issue is that p-dibromobenzene is insoluble in water. (Clarification: p-dibromobenzene is solid at standard conditions.)
I have been adding acetonitrile to the solution and stirring hard in an attempt to improve the solubility of p-dibromobenzene, but had little luck in dissolving it. Literature tells me that it is very soluble in toluene, but this poses an obvious problem for the alcohol addition. (https://pubchem.ncbi.nlm.nih.gov/compound/1_4-Dibromobenzene). It seems like this would be the easiest way to get p-dibromobenzene into solution.
Perhaps I could double the amount of Iron sulfate and Peroxide, and hope that it reacts with both? It would still be reasonable to separate the dibromo compound from its solvent since the boiling points differ by about 100 degrees.
Any tips?

Would adding an emulsifier and/or using ultrasound to form micelles, greatly increasing surface in contact with aqueous solution, be useful? Both might react badly with H2O2, even at 30%, though... fizz, if not bang.

There is a lot of literature on Fentons hydroxylation of arenes, have you looked at some of those papers to see how it was done?

Hi Waylander, one paper I found mentions using a 0.05 M Sulfuric acid solution https://pubs.acs.org/doi/pdf/10.1021/ja00279a057. I already have been adding a few drops of 12M H2SO4 just so I have acidic conditions, but this might be worth trying. The main issue is that I am working with a homogenous (solid/liquid) mixture.


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Positive and negative terminal in an electrolytic cell
In the figure below, which electrodes are negative terminals? I have worked out that E: Cu2+ -> Cu, F: OH- -> O2, G: H+ -> H2, H: OH- -> O2. However, I am still confused about the polarity of the electrodes and I have thought of two explanations

By definition, electron flows out of the negative terminal, so E and F are negative terminals
In an electrolytic cell, the negative terminal is the cathode. So E and G are negative terminals

Which one is correct?

At electrolysis, the cathode (absorbing electrons) is usually more negative electrode. But if for some reasons you reverse the voltage and current direction, the cathode is more positive electrode.

Note that for F and H case, it would be H2O -> O2

If you remember a couple of simple things most of the confusions in electrochemistry will vanish.

Positive and negative in electrochemistry mean that a particular electrode has an excess electrostatic charge.

Anodes and cathodes are defined with respect to the chemical processes. Electrode where oxidation occurs is the anode, electrode where reduction occurs is the cathode.

The bulk of the solution has to remain electrically neutral, i.e., the charge balance condition must hold all the time. If an electron enters the solution from one electrode, another electron from the solution must leave from another electrode.

Using the three points, we see that there is a battery, and electrode E is connected to its negative terminal.
(i) We can conclude that E is negatively charged because it has excess of electrons. If electrolysis is occurring, electronic current in the external circuit is flowing. If an electron left the electrode E, in order to balance the charge, another electron from electrode F must leave, so F is positive (cond. 3).
(ii) The electron which left the electrode F, has to enter the solution via electrode G. G has excess of electrons, and it will have a negative charge.
(iii) Electrode H is connected to the positive terminal of the battery. It must be positive.

I have heard from physics that positive terminal means a higher potential and a negative terminal means a lower potential. Does this apply to electrochemistry? If yes, could I also understand it as F having a higher potential than G?
Edit: The potential refers to the electrical potential (as in Volts)

(iii) It implies the external power source is harder voltage source than eventual galvanic cell. If - as extreme case - I attach 9V compact battery to 2 V cell of the car acid-lead battery to perform electrolysis, the lead-acid cell is much harder voltage source and would not care to which its contact is attached + contact of the 9V battery.

@JellyQwerty, Yes by convention, a positive positive (relative) is higher than negative potential. F is at higher potential than G.

Note that the potential of galvanic/electrolytic cell electrodes is relative to each other. They can be both positive or both negative wrt the potential of the standard hydrogen electrode (SHE), taken conventionally as 0.00 V potential reference. And the SHE has potential $\pu{+4.44 \pm 0.02 V}$ wrt the potential of a free electron.
Therefore, the "positive" electrode, contact or potential means being more positive than the "negative" one.
The electrolytic cell can be considered as a power source with the open circuit voltage $U_\text{cell}$ and internal resistance $R_\text{cell}$. The external source similarly is the combination of a voltage source $U_\text{src}$ with the internal resistance $R_\text{src}$.
Case 1 - The "Positive" contact of the voltage source is connected to the "positive" contact of the electrolytic cell. The source has higher voltage than the cell.
The current flows to the more positive electrode contact, electrons flow from it and this more positive electrode is the anode. The current $$I = \frac{U_\text{src} - U_\text{cell} }{ R_\text{src} + R_\text{cell}}$$
The voltage on the contacts is then
$$U_\text{cont} = \frac{ U_\text{src} R_\text{cell} + U_\text{cell}  R_\text{src} } { R_\text{src} + R_\text{cell}} $$
The contact with the more positive potential is then obviously at the more positive contacts of both the source and the cell.
Case 2 - The "Positive" contact of the voltage source is connected to the "positive" contact of the electrolytic cell. The source has lower voltage than the cell.
The formulas above still stand, but there is the opposite direction of the current. The measured voltage on contacts is lower than the voltage of the cell at the open circuit, but higher than the voltage of the source at the open circuit. The cell works in the galvanic mode instead and the positive electrode is the cathode now.
Case 3 - The "Positive" contact of the voltage source is connected to the "negative" contact of the electrolytic cell.
In this case, we have two power sources, connected in a series, the cell working in the galvanic mode.
The current $$I = \frac{U_\text{src} + U_\text{cell} }{ R_\text{src} + R_\text{cell}}$$
The potential of the nominally "positive" electrode of the cell wrt to its nominally "negative" is:
$$U_\text{cont} = \frac{ U_\text{cell}  R_\text{src} - U_\text{src} R_\text{cell}} { R_\text{src} + R_\text{cell}} ,$$
and can be both positive and negative. As in the case 2, the cell works in the galvanic mode and the nominally "positive" electrode is the cathode now.
Note that this mode may lead to very high currents and potentially to damage or destruction of the source or the cell.


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Distillation of free fatty acid from soya acid oil
I have soya acid oil with FFA percentage of about 40. After degumming and glycerolysis I am able to bring down the FFA percentage to 5. It does go down to about 2 percentage but it takes an enormous amount of time.
At this stage I want to distill the oil and extract two fractions, first being FFA and other being TGs.
Assuming this would also improve the color of the oil and will settle pigment and unsaponifiables as pitch in the distillation flask.
I have following available:

Distillation flask 1 litre
Packed column 12 inches (ratchet rings)
Thermometer head along with thermometer.
Condenser Collection flask
with vaccum adaptor
Vaccum pump
Heating mantle with temperature dial
but no visual temperature display.

Is it possible to achieve this?

If I understand you post correctly. you want to distill TGs too. Such a distillation may be a challenge even for vacuum distillation, having much lover vapor tension than FFAs.

You do not say what is the purpose of the product.

To use it as feedstock for FAME

To readers, FAME =  Fatty Acid Methyl Esters // Avoid using unexplained less common abbreviations. Even if many readers may not be familiar with the topic and topic specific abbreviations, they may still be expert in topics you ask about or you need to know more about.  Better if 1 writer explains them than if many readers have to find their meaning.

Will take care of it next time. Thank you.

What is the vapor pressure and at what temperature does decomposition become significant? Note that the longer you heat the mix, the more time it has to degrade.

Distill the methyl esters.


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If E° of a Half Cell is based on the chosen reference Electrode (Generally SHE) then how is the equation ΔG°= -nFE° be valid for a Half Cell
For example let us take the reaction:
Zn(s) ---> Zn^2+(aq) + 2e^-(aq) ;ΔG°, which we are considering as our half cell
And now suppose, by taking the value of E° of SHE(Standard Hydrogen Electrode) as zero and using it as reference, we have measured the E° of the zinc electrode. But clearly this value is relative to SHE and not the absolute value.
Now we have the equation ΔG°= -nFE° ,
relating the change in Gibbs energy of the half cell reaction and the value of E°.
The change in standard Gibbs energy of a reaction is constant at a constant temperature and has nothing to do with the chosen reference electrode so how can we relate both the values?
My another doubt is why even bother defining a reference electrode if we can find the electrode potentials of a half cell with its change in gibbs energy.
Or is it the other way around, that we calculate the value of ΔG based on the measured value of E.

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If we use the tabelated reduction potential $E°$ of a half-reaction in the equation $ΔG° = -nFE°$ the we implicitly assume the reaction $\ce{Ox(aq) + $\frac n2$ H2(g) -> Red(aq) + n H+(aq)}$.
If other than SHE half reaction had been chosen as the conventional potential reference, the respective implicit reaction would have been different, with different $ΔG°$ and different $E°$ for both the reaction and the zinc half reaction.
If we consider two general half-reactions then $Δ_\text{r}G{^\circ} = -nF(E{^\circ}_\text{cathode} - E{^\circ}_\text{anode} )$

So does that mean when i refer to the value of ΔG° of the reaction that i mentioned above , what is actually being implied is that its the ΔG° value for the reaction Zn(s) + 2H+(aq) ---> Zn2+(aq) + H2(g)   ?

It is, if the E° used in the equation is the E° for Zn^2+ + 2- -> Zn(s) half-reaction.

Thank you! that fully cleared my doubt

Oops, I have overlooks the direction you have used. It is the opposite direction,  H2 + Zn^2+ -> 2 H+ + Zn.  E°Zn/Zn^2+ < 0,  ΔG > 0 so your directions is not spontaneous reaction. The opposite one is.

You can’t calculate the Gibbs energy for a half cell, just like you can’t do a redox reaction with a half cell, only. In the formula $$ΔG°= -n F E°$$ the $E°$ refers to the cell potential, not the half cell reduction potential. The former is independent of a reference electrode, while the latter is not.


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Why we apply Pauli Exclusion Principle to Molecular Orbital Theory?
Can we say that an electron with a specific set of quantum numbers of a fluorine atom is exactly the same as that of another electron with the same set of quantum numbers of another fluorine atom.
If no (which I don't think is the case), then on what basis will we differentiate them?
If yes, then how are we able to apply "Pauli's Exclusion Principle" to the Fluorine molecule. What happens when the two atoms combine? Do the atomic quantum numbers change into molecular quantum numbers? Even if so, is there any way to figure it out?

https://en.wikipedia.org/wiki/One-electron_universe may be of interest

You may want also review the related topic of Electron degeneracy pressure in context of  white dwarfs and solid metals.

Quantum numbers are for the whole system. In the first case the system is a totally isolated F atom, and you have two entirely disconnected systems, in different universes if you like. So You can distinguish the two electrons by being godlike and seeing which universe they are in. In the second case the quantum numbers are those from solving for the entire F2 molecule - those for the F atom are irrelevant as you can't simply factorize the solution into two separate contributions, partially because of the Pauli principle. Thus you apply the principle for the molecule, not two atoms.

@IanBush it's agreeable what you explained, but only before and after the formation of the molecule. What was happening during the formation of the molecule is not clear from the argument. What happened to the spin of the electrons and things like that?

You (try to) solve the appropriate equations for the full system, get the quantum numbers and use the exclusion principle. The fundamental equations that describe Life, The Universe and Everything don't just apply at the beginning and end points, they are Universal.

There are several forms of the 'Pauli exclusion principle', and you have described one that is a special case only applying to atoms.
More generally speaking, one should say that no two electrons can occupy the same state.*
In atoms, states happen to be defined by the four quantum numbers $(n, l, m_l, m_s)$, which means that no two electrons can have the same four quantum numbers.
In molecules, the states are labelled in a different way, no longer using the same quantum numbers as before. But the same principle holds: no two electrons can have the same state (i.e., be in the same orbital, with the same spin).
The opening section of the Wikipedia page goes into more detail about this: https://en.wikipedia.org/wiki/Pauli_exclusion_principle

Do the atomic quantum numbers change into molecular quantum numbers?

Well, the atomic states do slowly morph into molecular states. The atomic states are described using quantum numbers, but that is just a short way of writing the full states out (which are wavefunctions).

* Fermions, if you want to be really general.

It will be useful to check the molecular orbital diagram of $\ce{F2}$, it is not difficult to understand the possible spatial states. Since the exclusion implies that two electrons should not have similar states, each molecular orbital is the state of two electrons with different spin up to the highest occupied molecular orbital. In the first quantization scheme one should not think about the electrons but about the states, the states are labeled with the quantum numbers for the atom, but when they overlap and form a molecule, the potential does not have the perfect spherical symmetry, it especially affects the angular momenta, while the first quantum number $n$ is specific to all discrete level systems. So to summarize, the orbitals overlap and form new molecular orbital states, what you now call electrons are simply indistinguishable and each has a specific molecular state.
There is no need to decipher the exclusion principle any further than the assumption that each spatial molecular orbital is the state of two electrons with different spins.

Glad to know what is indeed necessary and what not in MOT.


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Formation of hypofluorite from F2 and hydroxide
Why $\ce{FO-}$ can't form in reaction of $\ce{F2}$ with $\ce{OH-}$ but $\ce{ClO-}$ can form on reacting $\ce{Cl2}$ with $\ce{OH-}$?
Is $\ce{FO-}$ very unstable?
I know that fluorine can't have positive oxidation state that's why it don't behave like chlorine but isn't it good thing as now it will attract electron cloud from oxygen making it partially positive charged while chlorine would make oxygen partially negative charged because oxygen can pull electrons from chlorine.
Positively charged oxygen in case of $\ce{FO-}$ would have higher attraction for electron as compared to negatively charged oxygen in $\ce{ClO-}$ and hence there should be no reason for $\ce{FO-}$ to not form in reaction of $\ce{F2}$ with $\ce{OH-}$.  $$\ce{F2 + 2 OH-(aq) -> 2F- + OF2 + H2O}$$ $$\ce{Cl2 + 2 OH-(aq) -> Cl- + ClO- + H2O}$$

Consider the strength of the O-F bond

Your last paragraph answers it you just draw a wrong conclusion. F is more electronegative than O; Cl is similar or less than O. Hence different behavior. Explanations help understand observations not vice versa .  It is reasonable that a group has similar properties but a Leap to assume the same.


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How do you determine point groups of more complex molecules when there are 2 or more central atoms?
I am trying to determine the point group of Fe2(CO)8 and am having difficulty with the bond angles and geometry. How would one go about this in a logical sense?

Find the center of mass (which must be preserved by any symmetry operation) and go through the process as if it were the central atom, even if there is no atom at that exact place.

As an aside note that you can pick any point as an origin, its just that you will also in general have to include translators over and above the normal operations that are taught. But for molecular systems there always exists a point such that if chosen as the origin the number of required translators is zero, resulting in what is usually taught for chemists;  @IvanNeretin has given the recipe. Also note that  for periodic system, such as crystals, this is not the case. I doubt this is a big surprise.

Have a look at this site https://molecule-viewer.com/ where you can add symmmetry elements (rotation axes, reflection planes etc.) and test your skill at finding the point group.

The determination of the point group of a molecule does not mandate the presence of one atom in the centre of the molecule.  Indeed you can assign point group symmetries to objects other than molecules, e.g. the Platonic solids, polyhedra in general (see Kai Landskron's synopsis on chem.libretexts.org, for example), and objects of the daily life (coffee mug, pair of scissors, Allen key, the barrel of revolving doors, propellers, tires of cars by different manufactures, etc.)
In addition to building molecular models, the compilation of materials on Symmetry@Otterbein includes a gallery which not only allows you to toggle on/off highlights of the symmetry operators, but to see their application which can be halted to «walk around» the molecule.  As an example about naphthalene ($D_\text{2h}$) and one of its $C_2$ axes (note, there is no atom at the mass centre of the molecule):

If you happen to have the geometry of your compound in a format accessible e.g. to Jmol, the assignment of the point group symmetry and the subsequent display of the operators can be relayed (while retaining a critical eye on it) to a program, too.  As an example, $\ce{Fe2(CO)9}$ (a molecule different to the one of your question) from the program's console (file -> console) by the commands compute pointgroup followed by draw pointgroup:

(The relevant thresholds are adjustable, see calculate pointgroup in Jmol's documentation.)
Perhaps Bertini's paper hints to a reference file about $\ce{Fe2(CO)8}$ the abstracts describes as «dibriged $C_{2v}$ isomer for the singlet and the unbridged $D_{2d}$ isomer for the triplet.»
Bertini, L.; Bruschi, M.; De Gioia, L.; Fantucci, P. Structure and Energetics of $\ce{Fe2(CO)8}$ Singlet and Triplet Electronic States. J. Phys. Chem. A 2007, 111, 12152–12162, doi 10.1021/jp075343z.


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How to calculate the pH of a solution with addition of a complex
The scenario is this. I have $50$ mL of $0.1$ M $\ce{NH4^+}$ at a certain temperature which gives it a $K_a=5.2\times 10^{-8}$. To this solution, I add $0.02$ moles of $\ce{Cd(NO_3)_2}$. It is known that cadmium ions undergo the following reaction: $\ce{Cd^2+ + 4NH3 <=> [Cd(NH_3)_4]^2+}$ which has an equilibrium constant $K_f=7.3\times 10^{17}$ at this temperature. Assuming no other reactions, I want to calculate the pH of the resulting solution.
The solution presented in the textbook goes as follows: They first find the equilibrium constant for the equilibrium $\ce{Cd^2+ + 4NH_4^+ <=> 4H3O^+ + [Cd(NH_3)_4]^2+}$ which is $K_{eq} = (K_a)^4\times K_f = 5.3375\times10^{-12}$. They have the initial amount of $[\ce{Cd^2+}] = 0.4$ M and $[\ce{NH_4^+}] = 0.1$ M. Then they let $x$ be the change in concentration and retrieved the following equilibrium expression: $$\frac{\left[\ce{[Cd(NH_3)_4]^2+}\right]\left[\ce{H3O^+}\right]^4}{\left[\ce{NH_4^+}\right]^4\left[\ce{Cd^2+}\right]} = \frac{x(4x)^4}{(0.4-x)(0.1-4x)^4}= 5.3375\times10^{-12}$$
Then by applying the small change assumption – where $0.4-x = 0.4$ and $0.1-4x=0.1$ – we get x = $0.000556$ M. This results in a pH = $-\log{4x}= 2.65$
My qualm is what happens if we do it this way: instead we try to find the equilibrium constant to $$\ce{Cd^2+ + 3NH3 + NH_4^+ <=> H_3O^+ +[Cd(NH_3)_4]^2+}$$ which is just the addition of the two reactions. This intuitively seems like more of an accurate depiction of what is going on but I'm not sure. This time the equilibrium constant is $K_{eq}=K_a\times K_f = 3.796\times10^{10}$. There's now a couple more steps. Using the $K_a$ we find the equilibrium concentration of $\ce{NH_3}$ in just the acid solution which will give us an initial value of $[\ce{NH_3}] = 7.21\times 10^{-5}$ M. We then take the the same initial for $\ce{Cd^2+}$ and decrease the ammonium one by the amount we just calculated for the sake of accuracy and we get the following expression: $$\frac{\left[\ce{[Cd(NH_3)_4]^2+}\right]\left[\ce{H3O^+}\right]}{\left[\ce{NH_3}\right]^3\left[\ce{Cd^2+}\right]\left[\ce{NH_4^+}\right]} = \frac{x\left(7.21\times10^{-5}+x\right)}{\left(7.21\times10^{-5}-3x\right)^3(0.4-x)(0.0999-x)}= 3.796\times10^{10}$$
This one isn't so easily solved on paper but when I plugged it into wolfram-alpha I got $x=0.0000236$ M as the only solution which results in no negative concentrations, giving a pH $= -\log\left(x+7.21\times10^{-5}\right) = 4.02$
These answers are two orders of magnitude apart from each other. Came anyone help clear up what is going on; which one (or maybe neither) is the more accurate way and why? Thank you!

@Poutnik Yep, thank you for that, I'll edit it!

Perhaps, apply some formatting for easier reading and orientation. // For plain text formatting by StackExchange markdown, see as inspiration SE meta - formatting // SE meta -  tables  // Advanced formatting

You can always apply the universal method of solving the set of nonlinear equations for all equilibrii and mass and charge balances.  :-) // The above solutions do not count with the initial H3O+ nor NH3. In my ballpark from head estimation the [NH3]init would be very roughly around 10^-5 mol/L. (0.1 M NH4+ and pH 5.25), comparable with $x$.

For such equation set solutions, it is advised to formally substitute variables by single-letter symbols for easier manipulation.

How did you get the initial [NH3]=7.21E-5 M? When I have calculated pH=0.5(pKb - log c) = 0.5(9.25 - log 0.1) = 5.12, so [H+]=[NH3] = 7.5E-6 M

@Poutnik the ammonia concentration I got just from using an ice table with $\ce{NH_4^+}$ initial as $0.1$ M. You then have the equation $\frac{x^2}{0.1-x} = K_a$ which I then solved to get the ammonia concentration

The Ka seems wrong. pKa of NH4+ is about 9.25 at 25 deg C. what is 5.62e-10, not 5.2e-8. I doubt a lot it would change value by 2 orders within reasonable T range.

@Poutnik the problem did not specify 25°C. The ammonium pKa could be for a warmer solution.

@OscarLanzi I am aware of that. But I am afraid it would need such high temperatures, for which complex stability constants are not known. It is possibly a made up task. Also, strong water autoionization may play a role.

@Poutnik The numbers are made up. Can you explain theoretically just what is not working

Why do you think increase of [H3O+] is equal to increase the complex concentration? Complex formation itself does not change [H3O+] directly. That is controlled by ammonium dissociation constant.  // By other words combining 2 reactions and their equilibrium constants, we get valid equilibrium constant, but we do not get a new reaction, that is ongoing instead of the original two ones. Therefore H3O+ honors the new constant, but not the new assumed reaction kinetics.

The problem with solution 2
First, a quick calculation checking solution 2. If you plug in the claimed equilibrium concentrations for the second solution and calculate $Q$ for the ammonium dissociation, you will see that:
$$[\ce{H3O+}]=\pu{9.57E-5 M}$$
$$[\ce{NH3}]=\pu{1.3E-6 M}$$
$$[\ce{Cd^2+}]=\pu{0.4 M}$$
$$[\ce{NH4+}]=\pu{0.0999 M}$$
$$Q = \frac{\pu{1.3E-6} \cdot \pu{9.57E-5} }{0.0999} = \pu{1.2E-9} $$
Because $Q$ is smaller than the equilibrium constant, more ammonium will dissociate (which will mess up the other equilibrium). This means you have to solve the two equations simultaneously if you go down this path.
Why solution 1 is the better strategy
There are three equilibria you have to satisfy. The complexation, the acid dissociation of ammonium, and the autodissociation of water.
Solving this with a single equation (instead of more than three, if you include mass balance) comes down to knowing major and minor species. As the correct solution shows you after the fact, you have a lot of free ammonium and cadmium, some hydronium and complex, and much less of ammonia and hydroxide. The first solution path uses an equilibrium equation that does not contain the minor species.
$$\ce{Cd^2+ +4NH4+ + 4H2O<=> 4H3O+ +[Cd(NH3)4]^2+}\tag{1}$$
Ignoring minor species (setting them to zero) does not introduce a large error (and you could iteratively fix that small error later - see answers from Poutnik for similar questions).
Let's look at the second solution path. The equilibrium you are focusing on is:
$$\ce{Cd^2+ +3NH3 +NH4+ + H2O<=> H3O+ +[Cd(NH3)4]^2+}\tag{2}$$
You calculate the concentration of the species before adding cadmium, ensuring that all but reaction (2) is at equilibrium. However, as reaction (2) goes forward, it consumes ammonia and ammonium, and produces hydronium, so that the acid dissociation equilibrium of ammonium is now out of equilibrium. As shown above, more ammonium would dissociate into ammonia, which would then form more complex.
How would you know before you try both?
We know that free cadmium will be a major species (in excess). We know that the solution will be acidic (some ammonia will dissociate), so there will be more (free) ammonium than ammonia. In fact, the pH would have to be higher than 7.3 for there to be more ammonia than ammonium. The reason that adding the cadmium lowers the pH is that it depletes ammonia, allowing more ammonium to dissociate (Le Chatelier's principle, if you will). The two equilibria that show this nicely are:
$$\ce{NH4+ + H2O<=> NH3 + H3O+}\tag{3}$$
$$\ce{Cd^2+ +4NH3  <=> [Cd(NH3)4]^2+}\tag{4}$$
However, they both contain minor species, so you want to combine them to eliminate the minor species.
If you are experienced at solving equilibria with approximations, you might realize that you can get away with ignoring ammonia and hydroxide at first, and that setting up your equation with initial concentrations of complex and hydronium that are zero makes things easier.  Also, if you find this problem in the textbook, you would expect that the solution does not require you to solve a cubic equation. If your first attempt at a path to a solution gets that complicated, you can take a step back and try a different approach.

The second approach fails, the first one works.
In the second approach you initially assume that the ammonium ion hydrolyzes to give equal amounts of $\ce{H3O^+}$ and $\ce{NH3}$, so you implicitly assumed that the $\ce{H3O^+}$ concentration must have matched the $7.21×10^{-5}$ molar you found for the $\ce{NH3}$. But then your calculation gave a much higher value for $\ce{H3O^+}$, indicating a contradiction.
The contradiction came because in fact you do not have equal concentrations of $\ce{H3O^+}$ and $\ce{NH3}$. The reaction of the cadmium ion with the $\ce{NH3}$ inteferes, drawing off the $\ce{NH3}$ and thus forcing the hydrolysis of $\ce{NH4^+}$ to accumulate more $\ce{H3O^+}$. So in reality $\ce{H3O^+}$ should have a much greater concentration than $\ce{NH3}$.
You can add reactions, but you should take a linear combination that balances out suppressed species like $\ce{NH3}$ in this case (or like free electrons in redox reactions). Here the component reactions are, as you describe,
$\ce{NH4^+ + H2O <=> NH3 + H3O^+}$
$\ce{Cd^{2+} + 4 NH3 <=> Cd(NH3)4^{2+}}$
and balancing out the $\ce{NH3}$ which is suppressed in the acidic solution requires taking four times the first reaction and then adding the second. This gives
$\ce{Cd^{2+} + 4 NH4^+ + 4 H2O <=> Cd(NH3)4^{2+} + 4 H3O^+}$
which is then the reaction used in the first, correct method.

When $\ce{NH_4^+}$ initially disassociates, it will produce equal amounts of ammonia and hydronium.  That's why their initial concentrations are the same. Their equilibrium concentrations are not in the second reaction :/

The first method is correct only if concentration of NH3 is constant during the complex formation and there is no conversion between NH3 and NH4+.

One thing should catch you eyes immediately.
The first way creates $\ce{4 H+}$ for $\ce{1 [Cd(NH3)4]^2+}$, your way creates $\ce{1 H+}$ for $\ce{1 [Cd(NH3)4]^2+}$. And there can be formulated three other versions with $\ce{0 H+}$, $\ce{2 H+}$ and $\ce{3 H+}$  for $\ce{1 [Cd(NH3)4]^2+}$.
It is obvious maximally one of five versions can be true. And, in fact, not even one.
All five possible equilibrium constants $K_\text{eq} = K_\text{f} \cdot K_\text{a}^N$, $N=0..4$ for
$$\ce{Cd^2+(aq) + $(4-N)$ NH3(aq) + $N$ NH4+(aq) <=> [Cd(NH3)4]^2+(aq) + $N$ H+(aq)}$$
are formally correct and equilibrium concentrations will honor them in contexts of concentration = activity simplification.
But the equilibrium equations do not say there is formed $\ce{$N$ H+}$ ions for each $\ce{[Cd(NH3)4]^2+}$ ion.
Before introduction of the cadmium salt, there is established the equilibrium ammonia concentration due ammonium hydrolysis. This concentration shifts when the cadmium complex is formed.

Let perform this formal substitution for easier expression manipulation:
$\ce{h = [\ce{H+}]}$, $\ce{n1 = [\ce{NH4+}]}$, $\ce{n2 = [\ce{NH3}]}$, $\ce{c1 = [\ce{Cd^2+}]}$, $\ce{c2 = [\ce{[Cd(NH3)4]^2+}]}$
and we can now formulate visually compact equations for equilibrii,
$$K_\text{a} = \frac{h \cdot n_2}{n_1}$$
$$K_\text{f} =  \frac{c_2}{c_1 \cdot n_2^4}$$
for mass balances:
$$n_1 +n_2 + 4 c_2 = c_\text{N,tot}$$
$$c_1 + c_2 = c_\text{Cd,tot}$$
and for charge balance:
$$h + n_1 + 2 c_1 + 2 c_2 = c_\text{chrg}$$
$\ce{OH-}$ is neglected and furthermore due unknown temperature we do not know $K_\text{w}$.
Substituting $c_2 = c_\text{Cd,tot} - c_1$:
$$K_\text{a} = \frac{h \cdot n_2}{n_1}$$
$$K_\text{f} =  \frac{c_\text{Cd,tot} - c_1}{c_1 \cdot n_2^4}$$
$$n_1 +n_2 + 4 (c_\text{Cd,tot} - c_1) = c_\text{N,tot}$$
$$h + n_1 + 2 c_1 + 2 (c_\text{Cd,tot} - c_1) = c_\text{chrg}$$
Substituting  $n_1 = c_\text{chrg} - 2c_1 - 2(c_\text{Cd,tot} - c_1) - h$:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_1 - 2 (c_\text{Cd,tot} - c_1) - h}$$
$$K_\text{f} =  \frac{c_\text{Cd,tot} - c_1}{c_1 \cdot n_2^4}$$
$$(c_\text{chrg} - 2c_1 - 2(c_\text{Cd,tot} - c_1) - h) +n_2 + 4 (c_\text{Cd,tot} - c_1) = c_\text{N,tot}$$
Simplifying:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_\text{Cd,tot}- h}$$
$$K_\text{f} =  \frac{c_\text{Cd,tot} - c_1}{c_1 \cdot n_2^4}$$
$$c_\text{chrg} + 2c_\text{Cd,tot}  - h +n_2 - 4c_1 = c_\text{N,tot}$$
Substituting  $c_1 = \frac 14(c_\text{chrg} + 2c_\text{Cd,tot}  - h +n_2 - c_\text{N,tot})$
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_\text{Cd,tot}- h}$$
$$K_\text{f} =  \frac{c_\text{Cd,tot} - (\frac 14(c_\text{chrg} + 2c_\text{Cd,tot}  - h +n_2 - c_\text{N,tot}))}{(\frac 14(c_\text{chrg} + 2c_\text{Cd,tot}  - h +n_2 - c_\text{N,tot})) \cdot n_2^4}$$
Simplifying:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{chrg} - 2 c_\text{Cd,tot}- h}$$
$$K_\text{f} =  \frac{2c_\text{Cd,tot} - (c_\text{chrg} - h +n_2 - c_\text{N,tot})}
{(c_\text{chrg} + 2c_\text{Cd,tot}  - h +n_2 - c_\text{N,tot}) \cdot n_2^4}$$
Simplifying: As $c_\text{chrg} = 2c_\text{Cd,tot} +c_\text{N,tot}$:
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{N,tot}- h}$$
$$K_\text{f} =  \frac{ h - n_2 }
{(c_\text{chrg} + 2c_\text{Cd,tot}  - h +n_2 - c_\text{N,tot}) \cdot n_2^4}$$
Approximating:  $c_\text{N,tot} \gg h$ and $c_\text{chrg} + 2c_\text{Cd,tot}  - c_\text{N,tot} \gg |- h + n_2| $
$$K_\text{a} = \frac{h \cdot n_2}{c_\text{N,tot}}$$
$$K_\text{f} =  \frac{ h - n_2}
{(c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot}) \cdot n_2^4}$$
Substituting:    $n_2 = \frac{K_\text{a} \cdot c_\text{N,tot} }{ h}$
$$K_\text{f} =  \frac{ h - (\frac{K_\text{a} \cdot c_\text{N,tot} }{ h})}
{(c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot}) \cdot {(\frac{K_\text{a} \cdot c_\text{N,tot} }{ h})}^4}$$
$$K_\text{f} =  h^3\frac{ h^2 - ({K_\text{a} \cdot c_\text{N,tot} })}
{(c_\text{chrg} + 2c_\text{Cd,tot} - c_\text{N,tot}) \cdot {({K_\text{a} \cdot c_\text{N,tot} })}^4}$$
The Excel based iteration $h=\mathrm{f}(h)$ provided
$$h=[\ce{H+}]=\pu{9.73E-4} \implies \text{pH} = -\log{h} = 3.01$$
Comparatively, the initial $[\ce{H+}] \approx \sqrt{K_\mathrm{a}.[\ce{NH4+}]}=\sqrt{\pu{5.2e-8}\cdot 0.1} \approx \pu{7.21E-5 mol L-1} \\
\implies \mathrm{pH} \approx 4.14$


text:
OOM Error Calculating ESP Charges with Psi4
I'm trying to create a script to generate partial charges for a given .dat file.
As an example of my process, I'm currently using gromacs to run minimization on a box of propane. It outputs a min.gro file, which i wrote a script to convert to min.dat so I can use Psi4. Here is the script:
def gro_to_dat(input_file, output_file):
    with open(input_file, 'r') as gro, open(output_file, 'w') as dat:
        # Skip the title and total atoms lines in the .gro file
        gro.readline()
        num_atoms = int(gro.readline().strip())

        # Write the initial lines to the .dat file
        dat.write("molecule mol {\n")
        dat.write("   0 1\n")

        # Process each line in the .gro file
        for _ in range(num_atoms):
            line = gro.readline()
            atom_symbol = line[12:15].strip()[0]  # Extract the first character of atom name
            x, y, z = line[20:28].strip(), line[28:36].strip(), line[36:44].strip()
            dat.write(f"   {atom_symbol}   {x}   {y}   {z}\n")

        dat.write("}\n")

gro_to_dat("min.gro", "min.dat")

And it creates a .dat file, which looks like this:
molecule mol {
   0 1
   C   3.384   2.059   2.720
   C   3.419   2.209   2.725
   C   3.308   2.297   2.664
   H   3.465   2.000   2.761
   H   3.368   2.026   2.617
   H   3.294   2.038   2.777
   ...
}

Finally, my partial charge generation script:
import psi4

# Set Psi4 memory and number of threads
psi4.set_memory('8 GB')
psi4.set_num_threads(8)

# Read the min.dat file for molecular geometry
with open("min.dat", 'r') as f:
    lines = f.readlines()
    molecule_data = "".join(lines[1:-1])  # skip the first and last lines

# Create the molecule object using the data from the file
molecule = psi4.geometry(molecule_data)

# Set computation options
psi4.set_options({
    'basis': 'STO-3G',
    'scf_type': 'df',
    'e_convergence': 1e-6,
})

# Perform a single-point energy computation to obtain wavefunction
energy, wfn = psi4.energy("SCF", return_wfn=True, molecule=molecule)

# Compute the ESP charges
esp_charges = psi4.core.MoleculeESP.compute_esp_charges(wfn.molecule())

# Write the ESP charges to a file
with open("esp_charges.txt", 'w') as f:
    for charge in esp_charges:
        f.write(f"{charge}\n")

print("ESP charges written to esp_charges.txt")

I actually want to use 6-31G*, but even using STO-3G, the computation does not finish.
My output is very long, but it basically is this:
 ==> Geometry <==

    Molecular point group: c1
    Full point group: C1

    Geometry (in Angstrom), charge = 0, multiplicity = 1:

       Center              X                  Y                   Z               Mass
    ------------   -----------------  -----------------  -----------------  -----------------
         C            0.901238549949    -0.441467579029     0.193243059784    12.000000000000
         C            0.936238549949    -0.291467579029     0.198243059784    12.000000000000
         C            0.825238549949    -0.203467579029     0.137243059784    12.000000000000
         H            0.982238549949    -0.500467579029     0.234243059784     1.007825032230
                               ...
Running in c1 symmetry.

  Rotational constants: A =      0.00022  B =      0.00020  C =      0.00020 [cm^-1]
  Rotational constants: A =      6.53586  B =      6.11387  C =      6.09670 [MHz]
  Nuclear repulsion = 18130296.238725978881121

  Charge       = 0
  Multiplicity = 1
  Electrons    = 13000
  Nalpha       = 6500
  Nbeta        = 6500

  ==> Algorithm <==

  SCF Algorithm Type is DF.
  DIIS enabled.
  MOM disabled.
  Fractional occupation disabled.
  Guess Type is SAD.
  Energy threshold   = 1.00e-06
  Density threshold  = 1.00e-06
  Integral threshold = 1.00e-12

  ==> Primary Basis <==

  Basis Set: STO-3G
    Blend: STO-3G
    Number of shells: 8500
    Number of basis functions: 11500
    Number of Cartesian functions: 11500
    Spherical Harmonics?: true
    Max angular momentum: 1

Killed

I've tried increasing memory, and have already downgraded my basis set, but nothing seems to be working.

Suggestion: because the question is more about the setup of an eventually successful computation, move it to mattermodelling.  To increase visibility of your question on the sibling page, there is the specific tag psi4 (examples of previous questions).


text:
31P-13C coupling in bis(diphenylphosphino)methane
In the 13C NMR spectrum of bis(diphenylphosphino)methane (dppm), some of the phenyl carbon signals appear to be split into triplets. It makes sense to me that the center methylene is split into a triplet, but why would any of the phenyl C signals be triplets? I expected to see doublets.
You can see the 13C NMR of dppm here: https://www.sigmaaldrich.com/deepweb/assets/sigmaaldrich/quality/spectra/293/067/FNMR000746.pdf or
In contrast, the 13C NMR of diphenylmethylphosphine only exhibits doublets in: https://www.sigmaaldrich.com/deepweb/assets/sigmaaldrich/quality/spectra/313/648/FNMR004111.pdf

Suggest to edit the question: 1) put chemical structure and corresponding spectrum either in one illustration altogether, or at least next to each other.  2) Seek 13C NMR spectra recorded at a higher Larmour frequency (e.g. 13C 100 MHz, or better) with Reaxys, or Scifinder, or a spectroscopy database.  For 13C-31P coupling, the Aldrich spectra from the catalogue of chemicals sold is not well suitable (fix interval, no zoom into the regions of interest). 3) Shapes of theoretical spectra need not manifest in experimental ones (e.g., lack of recording resolution, applied field strength, etc.).


text:
Sodium borohydride and lithium aluminium hydride
Both sodium borohydride and lithium aluminium hydride are common reducing agents in organic chemistry. Then can these reducing agents be applied in inorganic chemistry? I mean, for example, reducing acidified potassium dichromate, an inorganic compound.

NaBH4 and LiAlH4 are themselves both inorganic compounds, so yes, they can react with inorganic compounds.

In the case of sodium borohydride (which is more compatible with aqueous solutions), applications are known involving metal reduction. A few examples:
Precious metal recovery/recycle
Silver and platinum group metals are reducible from water-soluble salts using sodium borohydride. Initially the method was used for recovering silver from thiosulfate solutions used for photographic fixing; the application has now spread to recovery of "most precious metals". See Medding and Lander[1].
Wastewater treatment/Metal recovery
Sodium borohydride reduction can be used to remove certain heavy metals from wastewater, including copper. See Sithole et al. [2].
Ceramic nanomaterials
The reduction of cobalt chloride solution with sodium borohydride under controlled conditions yields $\ce{Co2B}$ powder with particle sizes 20-100 nm. See Jianming et al.[3].
References

G.L. Medding, J.A. Lander,
"Applications for sodium borohtdride in precious metal recovery and recycle", Editor(s): E.D. Zysk,
Precious Metals 1981, Pergamon, 1982, Pages 3-10,
ISBN 9780080253923,
https://doi.org/10.1016/B978-0-08-025392-3.50007-7.

N T Sithole, F Ntuli and T Mashifana (2018). "The Removal of Cu (II) from Aqueous Solution using Sodium Borohydride as a Reducing Agent". IOP Conf. Ser.: Earth Environ. Sci. 120 012022.
DOI 10.1088/1755-1315/120/1/012022.

Jianming Lu, D.B. Dreisinger, W.C. Cooper (1997). "Cobalt precipitation by reduction with sodium borohydride",
Hydrometallurgy 45(3), 305-322,
ISSN 0304-386X,
https://doi.org/10.1016/S0304-386X(96)00086-2.


text:
Weight Ratio vs Molar Ratio - Chloramines
I'm working with a delicate experiment whilst trying to test corrosion of stainless-steels in a chloramine environment. I've run into various literature that discusses the formation of different types of chloramines relative to pH and ratios of chlorine to nitrogen (or ammonia). In order to create this environment without being a complete f-bomb and accidentally creating concentrated Nitrogen Trichloride. I'd love some clarification for my non-chemical-based knowledge.

I've read that monochloramine dominates when the chlorine to nitrogen weight ratio in <5:1 (or 1:1 molar ratio). Separate ratios exist for other chloramines but let's focus on this for the purpose of my question.
Firstly, I'm a little confused how to convert between these units. I imagine the 'weight ratio' is measure as PPM or mg/L, while molar ratio is g/mol? I feel like this should be easy but my head's in a twist.
Secondly, separate literature use similar figures for 'chlorine to nitrogen ratio [Cl:N]' whilst other reference 'chlorine to ammonia ratio [Cl:NH3]' (or even chlorine to ammonia-nitrogen). Are these synonymous or should weight of the included hydrogen (albeit minimal) be included?
Tertiary , I'd assume the ratios mentioned above refer to 'free chlorine residual' not total chlorine, which includes chloramines (chorine already bonded to ammonia). Sounds obvious but just checking.

molar ratio [mol/mol] = weight ratio
[g/g] /  molar mass ratio[(g/mol)/(g/mol)]. The ratios can be expressed in any compatible form, like molar PPM= mass PPM / molar mass ratio. Important thing is not to ignore proper unit arithmetics. // Nitrogen is synonym to ammonia nitrogen if there is not other nitrogen. Nitrogen is obviously not a synonym for ammonia. Molar ratios are the same for nitrogen and ammonia, but not weight ratios. N has molar mass approx. 14 g/mol, ammonia 17 g/mol.

@Poutnik, thanks working with units directly makes life simpler. I thought it may have been that simple so appreciate the clarification. As a simple Mech. Eng I'll stick with the familiar grams and litres.


text:
Williamson Ether synthesis use of particular solvent
In Williamson ether synthesis conjugate acid of corresponding alkoxide ion is taken as solvent instead of any other alcohol. Why it is like that?

What could happen if you were using methoxide base and ethanol as solvent?

The $\text{p}K_a$ of alkyl alcohols* are close to each other; the public compilation by the Bordwell table for example lists MeOH (29.0), EtOH (29.8), iPrOH (30.3).  If one assumes the exhaustive deprotonation e.g., of MeOH by sodium to yield $\ce{MeO^-}$ and $\ce{Na+}$, this newly generated based is capable to move the equilibria of
$$\ce{EtOH + MeO^- <=> EtO^- + MeOH}$$
$$\ce{iPrOH + MeO^- <=> iPrO^- + MeOH}$$
sufficiently to the right hand side that -- if the molar ratio of e.g. $\ce{MeO^-}$ and EtOH is big enough -- these additional nucleophiles interfere with the Willamson reaction.  Instead of only
$$\ce{RCH2-Cl + MeO^- -> RCH2-OMe + Cl^-}$$
you are going to observe substantial amounts of side product along
$$\ce{RCH2-Cl + EtO^- -> RCH2-OEt + Cl^-}$$
which can alter the both physical properties of the product (crystallinity, melting point, etc) as well as chemical and ADMET characteristics.
* The reason to highlight alkyl alcohols is that phenols are substantially more acidic because of an easier stabilization of the negative charge of the anion.  In both cases, though, electron withdrawing substitution can increase the acidity / lower the relevant $\text{p}K_a$ value.


text:
What can be inferred from overlap matrix S if coefficient matrix C is unitary in Roothaan equation?
I was going through the Hartree-Fock algorithm, transforming the spatial molecular orbitals {$\psi _i$} to be a linear combination of Roothaan basis set {$\phi_i$}. From my knowledge {$\psi_i$} is set to be orthonormal, and there is a linear transformation $[\psi_1\ ...\psi_K] = [\phi_1\ ...\phi_K]\begin{bmatrix}C_{11}&C_{12}&...&C_{1K}\\C_{21}&C_{22}&...&C_{2K}\\...\\C_{K1}&C_{K2}&...&C_{KK}\end{bmatrix}$ Where {$\phi_i$} is not necessarily orthogonal, albeit being normalized.
Let the coefficient matrix be $C$. With that in mind, and that matrix representation of Fock operator (transformation) under the new basis set is related to the old one by a similarity transform $F = C^{-1}F'C$, I went on calculating the element of Fock matrix under each basis: $$\langle \psi_i|f|\psi_j\rangle = \Sigma_{\mu}C_{\mu i}^*\langle \phi_\mu |f\ |\Sigma_{\nu}C_{\nu j}\phi_\nu\rangle$$ $$= \Sigma_\mu\Sigma_\nu (C)^\dagger_{i\mu}\langle\mu|f|\nu\rangle C_{\nu j}$$ $$= (C^\dagger F'C)_{ij}$$ So basically I have got $F = C^\dagger F'C$ and $F = C^{-1}F'C$. I am not an expert on maths but $C$ looks suspiciously unitary. Let $C^\dagger C = L$ where $L\in M_{K\times K}(\mathbb{C})$ Since $C$ is non-singular there can be $C^\dagger = LC^{-1}$ substituting this to $F = C^\dagger FC = LC^{-1}F'C = LF$ so $L$ can be none other than $I$.
This causes a problem. The overlap matrix $S:=\langle \mu|\nu\rangle$ has a special equality:$$(C^\dagger SC)_{ij} = \Sigma_{k,l}C^\dagger _{ik}\langle k|l\rangle C_{lj} = \Sigma_{k,l}C^* _{ki}\langle k|l\rangle C_{lj} =\langle \psi_i|\psi_j\rangle = \delta_{ij}$$ This means $$C^\dagger SC = I$$ but $C^\dagger = C^{-1}$ so $S$ is similar to $I$. Similar matrices have the same eigenvalue. $S$ clearly does not have all 1 eigenvalues. What went wrong here?

"C looks suspiciously unitary" - it isn't, at least in general,that's your mistake. Look at your last equation which clearly shows the Hermitian conjugate of C is not the inverse of C unless S is the Identity matrix. More generally C in the AO basis is not unitary, but in the MO basis it is - this is almost the point of the basis transformation.

In fact to answer your question in the title what can be inferred about the overlap matrix if the coefficient matrix in the AO basis is unitary is that the overlap matrix is the identity matrix.

Thanks Ian! Although I am still stumbling against what went wrong in my reasoning in the beginning.

I'll write a proper answer in a bit, but v. Busy at moment. But I think you may be a little confused between the atomic orbital and molecular orbital bases. The latter is unitary, the former is not.


text:
Closed pressurized water vessel
Suppose we have a closed container not isolated from the external environment so that heat exchanges can occur on the lateral wall (mainly for convection). The water inside is pressurized by a gas (nitrogen for example) starting from a pressure that is not exceeding 20 bar.
As water is drained from the tank I'm expecting that pressure will decrease as a result of the expansion and so will temperature that will lead to the onset of heat exchanges between liquid and gas and internal lateral walls.
At this point a question comes: If a thermal control is inserted in the tank so that liquid temperature stays always in the range of 10 - 30 degree Celsius, boiling will occur only if vapor pressure of liquid will equate the pressure exerted by nitrogen on the liquid free surface. Can evaporation take place before reaching this condition? In this case, it is possible to consider that phase equilibrium is present?
I'm saying this since I can expect that some water molecules will have enough energy to escape liquid but will eventually come back due to volume constraint (even in the case of small quantities of vapor).
The same can be said also when boiling happens in such a way that its onset is always accompanied by condensation?

For plain text formatting by StackExchange markdown, see as inspiration SE meta - formatting // SE meta -  tables  // Advanced formatting

If there is an inert gas as nitrogen, the liquid would never boil if there is enough time for evaporation of liquid vapor. Because external pressure would be always bigger than  vapor pressure.

Thanks @Poutnik. Suppose that pressure of nitrogen will lower up to equate vapor pressure of water (around 0.03 bar for a temperature of 25 degree Celsius). Boiling will occur in these conditions?

Imagine a pressure cooker that contains water and  air, that is sealed and slowly heated, within safe margins. The the p = p_vapor + 1 atm (of air). Pressure is  always higher than vapor pressure. Boiling is possible only temporary and locally, if water is heated fast enough so gas phase is not about  saturated.

Ok thanks, so if heat exchanges are controlled we could assume phase equilibrium in the tank. An equal amount of water will evaporate while the exact amount will condense.

Boiling can only occur if the equilibrium vapor pressure of water at the liquid phase temperature exceed the the gas phase pressure.  But, of course, evaporation can occur at the upper liquid surface if the equilibrium vapor pressure at the liquid phase temperature exceeds the partial pressure of water vapor in the gas phase.
You are controlling the temperature within the tank, so that it stays about constant.  In that case, the partial pressure of the water vapor in the gas phase will about match the equilibrium vapor pressure of the liquid.  Since the volume of the gas phase is increasing, some of the liquid water will have to evaporate to maintain vapor-liquid equilibrium.
A more interesting question would be what would happen if no heat was added to maintain the temperature.

Thanks so much @Chet Miller, so I'm expecting that as liquid is drained outside the tank a small amount of water will evaporate for compensate the additional volume so that water vapor will become saturated. At that point, since no more vapor is allowed, there will be a mass exchange due to condensation towards the bulk liquid.
As you said, boiling can only occur if vapor pressure is sufficiently higher than external pressure.
At this point, can I consider properties unchanged during evaporation? Equal amount of mass and heat will be transferred during the processes until boiling occurs.

I don't understand these question at all.

Sorry @Chet Miller, it's mainly because I'm trying to understand the physics behind the process.  We both agree that equilibrium exists since water is evaporating as liquid is drained. My question was mainly regarding what happens when vapor pressure of liquid equates pressure above its surface but as you said boiling will occur in that condition.

You can model it and see how it plays out.  Would you be able to handle that.

Yep, I will work on it these days


text:
Can any compound form into an amorphous solid?
I am curious whether it is possible or not to make any solution of a specific compound into a crystal lattice or into an amorphous solid? I know that crystals form under specific conditions but seem to be very wide spread throughout nature and chemistry labs. Therefore, I am curious if it is possible to turn something that usually is a crystalline structure into an amorphous solid. For example, can sodium pyruvate be naturally found as an amorphous solid or can it be made synthetically into such by some means?

While Maurice's answer isn't fully correct, something like sodium pyruvate would be difficult to make amorphous .

Very simple crystalline substances like water can still form amorphous glasses, but you need astounding cooling rates on the order of millions of degrees per second.

Whether a substance crystallizes or not largely depends on the rate of cooling or loss of solvent.

K. Gibson states, "Cooling materials super-quickly... prevents the normal crystalline structures of materials from forming."

Beginning glass blowers soon learn that extended heating of glass causes devitrification, i.e., partial crystallization.

Compare slowly-cooled, crystalline granite with rapidly chilled vitreous obsidian.

I just want to clarify that the first link to Phys.org states "Splat quenching can cool the material as high as 108 Kelvins per second." when the correct text is "Splat quenching can cool the material as high as 10⁸ Kelvins per second.", which is a very different thing. Phys.org has a tendency to botch science articles one way or another...

@NicolauSakerNeto, thanks for that note. You are truly an exponent of proper mathematical formatting. Oops... I'd better log off before I make another pointless comment.

I see how rapidly freezing a solution would create an amorphous solid. However, this method is only possible if you have an equipment capable of freezing at astounding rates. Is there not some other way of creating and amorphous solid, for example, by mixing with another compound or such?

@BrianBlumberg, certainly! Mixing pure chemicals may produce a mix that crystallizes slowly, or not at all. Pure SiO2 forms quartz crystals, but mix silica with potash, washing soda and other substances, and you've a recipe for glass. https://www.wikihow.com/Make-Glass

Solids made of covalent bonds like $\ce{SiO2}$ can be made into an amorphous solid. Solids made of ions like $\ce{NaCl}$ cannot be made into an amorphous solid. They can be made of piles of rather tiny crystals, but this is not a glass.

That's not true: https://chemistry.stackexchange.com/questions/46/are-there-any-ionic-amorphous-solids

Depends on how small you consider a speck to still be a crystal. Spin-cooling, vapor deposition onto a material with different atomic spacing, etc. can  create effectively amorphous ionic solids.

I think it's possible to make very heavily radiation-damaged ionic salts which you could argue are amorphous on a microscopic scale, though the damage tends to create byproducts.

Usually it is possible for a large range of solids by using different techniques, for example, a lot of vacancies can generate an amorphous phases. In small scales, technique of deposition can be used to generate amorphous materials such as amorphous silicon which is used for solar panels, while the common structure is a solid with a diamond lattice, in the same group carbon also has amorphous phases.


text:
Parentheses use in non stoichiometric compounds
When should one use (or not) parentheses in writing the formula for non stoichiometric compounds.
I mean, the mineral sphalerite is often represented by the formula $\ce{(Zn,Fe)S}$ and, all though I know that this really means $\ce{Zn_xFe_{1-x}S}$, I would like to know when the use of the first formula is recommended over the second.
That said, I have done a brief search on the IUPAC's red book about the use of parentheses and found nothing.

Looks like a duplicate of What does a comma signify in inorganic chemistry?

Rule IR-2.2.3 / Parentheses, sub rule IR-2.2.3.1 / Use in formulae illustrates the use of parenthesis :

(d) In solid-state chemistry, to enclose symbols of atoms occupying the same type of site in
a random fashion. The symbols themselves are separated by a comma, with no space.
Example:
9. $\ce{K(Br,Cl)}$

Note there is a subsequent, contrasting entry stating

(f) To indicate the composition of a non-stoichiometric compound.
Examples:
11. $\ce{Fe_{3x}Li_{4-x}Ti_{2(1-x)}O6}$ ($x = 0.35$)

$\ce{LaNi5H_x}$ ($0 < x < 66.7$)

source: p. 21 of IUPAC's Red Book by 2005 the resources page (section nomenclature) links as freely available online release (pdf).

Certainly the example 11 is not one that could be denoted by parenthesis, since it does not represent atoms on the same site(s).

@JonCuster  My interpretation of the OP's question was as the notations of $\ce{(Zn,Fe)S}$ and $\ce{Zn_xFe_{1 - x}S}$ were mutual synonyms of and fully equivalent to each other. So far, I speculate this is not the case and hence included example f).

I indeed agree, just pointing out that parenthesis could not be used for example 11. Coming from materials science, the implication of (Zn, Fe)S is you don't care what the ratio is, while for Zn$_{x}$Fe${1-x}$S you do care and there will be plots of something vs $x$.

(Zn,Fe)S and ZnxFe1−xS have different meanings, as stated by Jon Custer. In the 1st case Zn and Fe form a solid solution (Fe and Zn are located at the same crystallographic site). In the 2nd case, the formula implies a not-stoichiometric compound with Fe and Zn at different sites.  (Zn,Fe)S and (ZnxFe1−x)S are equivalent; in the 1st case, the relative amounts of Fe and Zn are not specified (even though the global amount corresponds to a stoichiometric coefficient equal to 1). In the 2nd case, you can specify the specific amounts of Fe and Zn by putting, for example, x=0.1, 0.25, and so on


text:
Vapour pressure and Raoults law
Suppose we take a binary ideal solution having components A and B ....  let's say that their mole fraction is 1/3 and 2/3 respectively (for simplicity)
Now ... We know that this solution will start evaporating untill a dynamic equilibrium is achieved ... And some molecules of A and some molecules of B from solution will break apart from solution and go into vapour phase .... Also the molecules breaking apart will be different for A and B(depending on which is more volatile)
Now **my **
In the expression of raoults law
what value we have to put in for  Xa(mole fraction of A) .... The initial one before dynamic equilibrium (1/3) or the final one after equilibrium is achieved (which will be different)
And if both the mole fractions are same then how come that is true??

The prior site search about the topic is the basic prerequisite for asking on most sites.