text:
Do PEM Fuel Cells suffer CO poisoning at the oxidizer side?
Reading about PEM Fuel Cells, first on Wikipedia then on some other sites, I notice they talk a lot about the problem of CO poisoning. But almost always it's explicitly referred to as a problem in fuel supply, not oxidizer.
In fact, some reports show that govt agencies have actually tested these things in school busses with lifetimes of about 1200 hours. So if they get the oxygen from ambient air, and driving around normal urban environments, then it appears that CO through the air does not cause a problem.
My question is, how?
Here is a diagram from wikipedia too: https://upload.wikimedia.org/wikipedia/commons/a/a2/Membrane_Electrode_Assembly_-_Electro-Chemical_Reaction_Diagram.jpg
(Pic uploader not working on that link for some reason)
The diagram shows the catalyst on both sides. It is facing the oxygen input as well as the hydrogen input.
How is this possible? How is CO poisoning in the fuel side possible, but CO poisoning on the O side impossible? It is all using the same catalyst as far as I can tell, a Platinum-Rhuthenium alloy. Is the diagram incorrect here?

What's wrong with the question? Something obvious I'm not seeing?

What is the likely concentration of CO on the fuels side versus the oxidiser side? This might be the simple issue since CO is common in some fuel feeds but has a much lower concentration in normal air.


text:
Radicals, reaction intermediate, activated complex
I'm really confused about those three concepts: radicals, reaction intermediate, activated complex.

First of all, in my course we have seen radicals in the context of chain radical reactions. But does it have a wider meaning ? (in older exams of my course I am seeing the word "radical" for reactions that are simple elementary reactions, no propagation and stuff).
I have seen the use of an asterisk in the notation of radicals in radical chain reactions. Is it a specific notation for the particular case of radicals inside chain reactions? Not only ?
Is it true that all radicals are reaction intermediate but not all reaction intermediate are radicals? The contrary ? None of the 2, how could we describe it then ?
About activated complex now, is there any way to distinguish an activated complex (which I understand represent a maximum of energy) from a "classical" reaction intermediate (which I understand represent a local minimum of energy) just by the look at the shape of the chemical reaction(s) ? (so without calculations using specific data about the components).
Can we calculate a pseudo-equilibrium constant  (which is related to the fact that we have a steady state, correct me if I'm wrong) either in the case of complex activated and reaction intermediate ?

Thank you very much in advance !
EDIT: I understand your feeling that led you to close this topic. Tbh though, many students of our course were confused about those particular interactions and the point of the post was precisely to shed light on the links (or non links) between all those topics. I understand it may not be very meaningful to others who have had a different (and maybe clearer) learning structure (However, I suppose it would be possible that similar confusions could be happening elsewhere, not just with us).
Also, I think it's worth mentioning that Poutnik's answers were very helpful and helped many of my classmates to whom I passed on the information. I respect your choice to close this post but at least you know how it happened behind the scenes.

Always try to find answers on your own in offline and online resources before asking. The active searching will give you more than just passive waiting to receive an answer.  Write also notes about your prior effort results in the question when asking. As questions that look like it has not been done are frown upon on StackExchange site network and are often closed.  Respondents prefer not to tell you what you already know or what you can easily find. Also, it prevents cumulation of redundant Q/A on this site or internet.

I did search for it (for a while actually) but I can't find precise information where the distinction (or relation) between those concepts are clearly made. I can only find separate information about those things (but then I don't understand how and how much they are / aren't correlated to each other, hence my confusion).

Search for what each of it is, then you can make the distinction yourself. Vaque and implicit effort cannot be distinguished from no or negligible effort.

Radical is an atom or a molecule or a ion with an unpaired electron (Some of orbitals contains just a single electron so the electron spins are not paired).
$\ce{Cl^.}$ or $\ce{CH3^.}$ or $\ce{^.O2^.}$ are radicals. Oxygen molecule has 2 unpaired electrons.
Radicals are often also intermediate products in radical reactions.

In $\ce{H2 <=> 2 H^.}$, hydrogen atoms as radicals are not intermediate products, as they do not react to something else then $\ce{H2}$.
$\ce{[Al(H2O)5(OH)]^2+(aq)}$ is an intermediate product in $\ce{[Al(H2O)6)]^3+(aq)}$ hydrolysis, but it is not a radical.

Reaction intermediate/intermediate product is a product and at the same time a reagent in the context of a chain of reactions(including, but not limited to, chain reactions), which can be isolated, detected by analytical techniques or just assumed.
Being a radical is more about what an entity is.
Being an intermediate product is more about what an entity does.
so there is a large overlap.

Excited state is energetic state with higher than the base energy, e.g. with electrons excited in higher orbitals.
A molecule in excited state can be sometimes detected and undergoes further reaction. It is then also an intermediate product.
$$\ce{A ->[h \nu]A^{*}}$$

Activated complex is a transition state of atoms/molecules/ions happening during formation of products in context of an elementary reaction.
E.g. a reaction $\ce{A + B -> C}$  going like $\ce{A + B -> \ce{AB^{‡}} -> C}$

Activated complex has the maximum of potential energy of the reaction, it is kind of standing on the top of a energetic barrier. Intermediate products are existing  entities at the local minimum of energy.

In a children game with marble balls and holes, imagine (initial) reactants, (final) products and intermediate reactants/products like balls sitting in some of several adjacent holes. Some holes are deep, some shallow.
OTOH, the activated complexes are the balls in transition between two holes, that cannot stop there. They have to go forward or backward.

Thank you ! A few remaining points: 1. In my course, all intermediates that are radicals are denoted with * , should I understand that it is only because they are also excited state in that context of being an intermediate, but not because they are radicals? 2. I understand from your explanation that not all reaction intermediates are radicals and also not all radicals are reaction intermediates, sometimes they can be both, or just one, or just the other, right? (I'm asking because our teacher said one of the two, which confused all students).

Asterisks in superscript position are not used for radicals but excited states. For radicals, there are used "fat dots" in the middle position. ( MathJax cannot do so better AFAIK).

I see. Lastly, about the hypotheses we can use in different cases, in the case of activated complex: we can calculate a pseudo-equilibrium constant / In the case of reaction intermediate: we can equal their rate of appearance to their rate of disappearance. Correct? Those hypothesis are 2 different things, also correct?

IP are normal molecules/ions, that can have wide range of concentration, stability and reactivity.  AC are not  true molecules, rather transient state of close interaction of 2 or more entities, before one atomic arrangement transforms to another.

Thank you for the very nice edits. I said my previous question was the last one so I'm sorry to have another : is a reaction with a reaction intermediate automatically called a chain reaction? (even without propagation ?)

I have used the term incorrectly. I have meant a chain of reactions, not necessarily (but not excluding) chain reactions. For A -> B -> C,  B is a IP, reactions are chained, but they are not "chain reactions".

@Poutnik Still too soft after Christmas? ;)

@Mithoron Rather time by time. :)

@c.leblanc I have fexed few minor errors, check the current wording. Mainly Activated complex (was: Excited state) has the maximum of potential energy of the reaction, it is kind of standing on the top of a energetic barrier. Intermediate products are existing entities at the local minimum of energy.

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Comprehensive Functional-Group-Priority Table for IUPAC Nomenclature
I can't seem to find a source for a comprehensive list of FG-priorities. I also am not aware of any system to determine the priority for more uncommon FGs.
One of the two would be really helpful if one of you would know about something akin to that (I don't have access to the IUPAC blue book right now).
For more detail: I have to choose my core system; now I would naturally choose the heterocycle (in my case a oxazolidinone, with an alkyliden subst amongst others. Im not so sure about alkyliden/alkylidine subst. priorities too, that's why I mention it), but one sidechain features a guanidyl group.
According to IUPAC, the system featuring the FG with the highest priority determines the core system. I am at a loss here, since I do not know the priority of guanidyl-FGs.

For your detailed example, a drawing of the structure could be helpful. Note that the order of seniority is only used for the principal characteristic group expressed as a suffix, so the alkylidene group cannot be relevant.

Blue book would be indeed the reference to go to.  Chemistry.se's page resources for learning chemistry though includes a contribution about nomenclature in general, including the freely accessible online compilation by 2013 by G. P. Moss updated by 2022.  (For their brevity, the brief guide and the 4-letter abstract (open access) likely miss an entry about oxazolidinone.)  Chapter P-4, P-44 seniority order for parent structures may be of particular interest for you.
Last resort can be to sketch/import the structure in question for an assignment e.g., by ChemDraw, or ChemDoodle.  There is a JavaScript based test site of ChemDraw retaining the structure2name function.  Because the analogue test site for ChemDoodle JS is among the tutorial pages, there possibly is a limit of non-Hydrogen atoms to test the functions. (The program itself allows to adjust the nomenclature used, hence, there are occasions with different names assigned to one and the same (more complex) structure by ChemDraw and ChemDoodle.)

This is a very helpful resource, thank you! I will be looking into how oxazolidinones (as cyclic amides/heterocycles) differ from guanidine in seniority. Curious (at least to me) is, that e.g. chemdraw chooses 'guanidine' as the parent structure. This could mean that oxazolones might be classified as ketones rather than amides. I am just wondering how to determine the seniority of guanidine, since it is also used as prefix. There are many ways to deduce it (urea, amidine etc).

@Katjuscha P44-0, Introduction, provides an overview for the rules convened. Then P-44.1 Seniority order for parent structures, P-44.2 Seniority order only for rings and ring systems, P-44.3 Seniority of acyclic chains (the principal chain), and P-44.4 Seniority criteria applicable to rings, ring systems, or acyclic chains.  E.g., P-44.2.1.8 "The senior ring or ring system has the greater number of heteroatoms that occur earlier in the following sequence: F > Cl > Br > I > O > S > Se > Te > N > P > [cut for brevity] [criterion (g) in P-44.2.1]. Examples [...]: [illustrations follow]"

@Katjuscha To be precise, oxazolidinone is classified as pseudoketone in the seniority order.

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Measuring [A-]/[HA-] with Buffer and Indicator
I'm very confused on how to calculate the [A-]/[HA] value, given the fact that my dilution series uses buffers at various pH values with the same indicator (supplied at the same concentration to each diluent).
For example, I have been told that for the diluent composed of acidic buffer (pH=3, supplied at concentration of 0.2M, volume of 800µL) and indicator (concentration of 0.20µM, volume of 400µL), that the indicator is 100% protonated. As well, for the diluent composed of basic buffer (pH=9, supplied at concentration of 0.2M, volume of 800µL) and indicator (concentration of 0.20µM, volume of 400µL), the indicator is 100% deprotonated. I'm quite unsure what to even consider as the [A-] and [HA] (and or [B] and [BH+]) terms, since I'm working with both types of buffers but the same indicator.
As well, I have measured the absorbance value (using the same wavelength=480nm each measurement) across the dilution series (pH ranging from 3-9, using different buffers with varying pH values).
I'm just quite confused how to determine the pKa of my indicator, as well as the [A-]/[HA] values, based on the absorbance value at that given pH (as well as relating the differing concentrations of indicator and buffer). I would really appreciate if anyone could provide clarity on how to re-arrange the Henderson-Hasselbalch equation, and Beer-Lambert law, to determine these values and calculate the pKa, in this context.
Thank you, in advance.

This is a whole experiment based on Henderson-Hasselbach equation, but the explanation will take about 40 min lecture. Did your teacher/instructor explain it in the lab or what does the lab manual say?


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Is valency of an element equal to sum of covalency and electrovalence? Why do we calculate valency and oxidation number?
About valency books say following things:
1- Valency is combining capacity of an atom with number of hydrogen atoms or double of oxygen atoms with which the atom combines.
Doubt A: what is valency of C in C2H2 and CO?
2- Valency is number of electrons gained, lost or shared by atom to complete its octate.
Doubts B: in CO carbon shares 2 electrons and gains 2 electrons. There are triple bond between C and O. Should valency of carbon be 2 or 3 or 4?
Then
3- Electrovalency = number of charge on an ion.
And
4- Covalency = number of shared electrons.
Then books say that
5- For an element covalency and electrovalency are equal.
Then books say that
6- Valency of carbon is 4.
Now when I see carbon monoxide (CO), I am not able to figure out valency of carbon or oxygen.
Then comes oxidation number (hypothetical charge), which seems to be similar to electrovalency (actual charge), but still some online sources say these are different things.
Doubt C: Why do we calculate oxidation number? How is it different from electrovalency? Is the only difference hypothetical and actual?

There are Wikipedia pages that will answer most of your questions: https://en.wikipedia.org/wiki/Valence_(chemistry), https://en.wikipedia.org/wiki/Covalent_bond and https://en.wikipedia.org/wiki/Ionic_bonding

Carbon monoxide is admittedly a very strange molecule: https://en.wikipedia.org/wiki/Carbon_monoxide The usefulness of the concept of valency has its limitations...

I needed to know how to find valency from periodic table so that I can write correct formula of compounds. I have been reading links including similar questions on this website as well and I think that idea of valency to write chemical formula is much limited. I still don't understand why C in CO has valency 2. Is valency here implies covalency? What about 2 electrons borrowed from Oxygen? Why would oxygen give up 2 electrons when it is more electronegative than carbon?

The concepts of valency and Lewis structures are convenient but imperfect tools, and CO is tricky. Some explanations of bonding in CO involve resonance between various structures. This can be used to accommodate observations such as a bond order between 2 and 3 and a partial negative charge on C, but it is difficult to explain before the fact. CO valency would therefore lie between 2 and 3. Regarding selecting Lewis structures, Pauling suggest a rule of thumb called "the electroneutrality principle" which suggests that preferred structures keep the partial charges on the atoms to a minimum.

It is useful to think of atoms as separate characters being swapped and forming new associations. But it is truer to think that atoms and molecules are associations of electrons and nuclei. There are QM rules that ultimately decide where the electrons are mostly found around the nuclei. See for instance: https://chemistry.stackexchange.com/questions/32161/can-i-predict-the-preferred-resonance-structure-of-carbon-monoxide-from-its-mole

see also: https://chemistry.stackexchange.com/questions/65425/difference-between-valency-and-oxidation-state/65426#65426

And I almost skipped this one: https://chemistry.stackexchange.com/questions/25126/what-is-the-lewis-structure-for-co

Thank you for all useful links. When CO has triple bond what type of 3rd bond is other than the 2 covalent bonds.

If oxygen gave up 2 electrons there should be 2 ionic bonds along with 2 covalent bonds.

You should keep in mind that ionic bonds etc are a useful figment to represent partial charge transfer between the associated atoms. You can think of the triple bond (officially the bond order is closer to 2.6) as one sigma bond and two pi bonds. The Wikipedia page for CO discusses this to some extent.

Please have a look at the definition of valence in the gold book: https://doi.org/10.1351/goldbook.V06588 It clearly relates only to atoms, not to compounds. Therefore the valence of carbon is always 4. It doesn't change and has nothing to do with the compounds. As for the matter of CO: It is one of the molecules with the most interesting and complex bonding. That also means that that most of the traditional concepts are not useful to describe it.

@Martin-マーチン   and Buck Thorn thank you both of you. Some Wikipedia links contain technical terms which I need to read further and I will read them. The Gold book by IUPAC doesn't give any examples for laymans like me so thank you for your commentries.

For me, 1) Valency was a very useful value given by the number of doublets and the number of charges joining two atoms in the Lewis representation. It has been replaced today by the notions of covalency and ionic charges. 2) Electrovalency is an obsolete reality that is not used any more. 3) Oxidation number of an atom is a number that can be calculated using the following set of rules : a) Pure elements O.N. is zero; b) In compounds the sum of all O.N. is zero; c) In ions, the sum of the O.N. is equal to the charge of the ion; d) Fluor O.N. is $-1$ as O.N. e) Elements O.N. from the $1$st and $2$nd columns are the number of the column, f) Oxygen O.N. is $-2$ (except in compounds with fluor, in H2O2 and other similar peroxides), g) For other atoms in polyatomic molecules or ions, the O.N. is an unknown $x$ that can be obtained by applying the previous rules.

Why do we calculate valency if it can't help in writing formula for compounds?   Why do we calculate oxidation number?

In textbooks valency due to ions is referred to as electrovalency.

Oxidation numbers are necessary for writing redox chemical equations.


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Masses of atoms and molecules
This is my understanding so far:
Average atomic mass is the mass of an element considering isotopes and is measured in atomic mass units. However relative atomic mass is the average mass of an atom ( Considering isotopes ) relative to carbon 12. As a result, relative atomic mass possesses no units. The same is applied to atomic mass which is the mass of an atom ( Isotope ) measured in atomic mass units. Relative isotopic mass is the mass of an isotope relative to carbon 12 and also possesses no units. Is this understanding completly correct?
Furthermore, I was wondering about Atomic mass units. If this measures the actual mass of an atom for example then how is it that relative mass gives the same result just without the unit. Is atomic mass units somehow related to carbon 12 as well?

The atomic mass unit is no longer related to C-12 because Avogadro's number is defined as a base quantity since the 2019 update of the SI units (and not anymore as the number of C-12 atoms in 12 gr of pure C-12).

https://en.wikipedia.org/wiki/Dalton_(unit)#2019_redefinition_of_the_SI_base_units

It is a general topic of absolute and relative quantities.
Relative quantities are ratios of two quantities of the same unit, being unitless and independent on units.
Absolute quantities have numerical  values related to the scale unit, depend on the chosen unit, which is the unit part of the quantity value.
The absolute and relative quantities can have the same numerical value, if the relative quantity reference is at the same time used as the scale unit of the absolute quantity.
Hypothetically, we could always use just relative quantities for everything, if it was always clear and accepted, what is the implicit reference. But we could not then use dimensional analysis for formal checks of equations.
If I take $\pu{1 kg}$ mass etalon as the relative mass reference, my absolute mass is e.g. $\pu{75 kg}$ and my relative mass is 75.
The same for atomic masses, if $\pu{1 Da}$ (Dalton) as the atomic mass unit is at the same time used as the reference for the relative atomic mass.


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Naming of this organic compound
I am unable to share the image, so sorry; but can you tell me which is the correct name from these 2 of the same compound.

3-(1-chloromethyl)-2-(1-methylethyl)-pentan-1-ol

or

4-chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol

Shouldn't the first one be correct as it's the longest chain with a functional group? According to IUPAC we don't care about substitution groups while selecting the longest chain at all, right?

You are right to consider the pentane chain of higher seniority than the butane. Indeed, the two names differ in the perception of the primary chain, apparent if the names are converted into the corresponding structures (the upper line in the illustration below):

With this in mind, you may assign the names of the two formulae about the same molecular structure.
Note: In case of the chloromethyl group, there is no other place of attachement for chlorine than on carbon-1 (for this side chain); hence, 1-chloromethyl can be substituted by chloromethyl.  Second, 1-methylethyl is not wrong per sé, yet the motif frequently is faster to grasp (for the trained eye) as isopropyl (e.g., isopropyl alcohol [colloquial isopropanol]).
Side note: resources for learning chemistry includes references about nomenclature.

Note that the preferred IUPAC name for "3-(chloromethyl)-2-isopropylpentan-1-ol" is 3-(chloromethyl)-2-(propan-2-yl)pentan-1-ol. Also note that "isopropanol" is considered a bad name in nomenclature since there is no alkane "isopropane".

@Loong Moss' compilation, the full-pdf revised by 1 April 2022 states "P-29.6.2.2 The prefixes isopropyl, isopropylidene, and trityl are retained for use in general nomenclature but no substitution of any kind is allowed." with the example of the isopropylidene side chain. The term "isopropyl" equally is used by copy-paste of the structures into ChemDraw JS and twice by ChemDoodle .../...

@Loong, 2 In case of ChemDoodle, on can pick the rule set to assign a name to a structure (guide, though "Attempted PIN" (preferred IUPAC name, pdf for version 11/12, section 14.2.1 Types of Names, p. 338). Though neither isopropanol, nor isopropyl alcohol are good names, they are still alive in the catalogues of Sigma, Fisher, Merck next to propan-2-ol.

That's exactly what I said. "retained for use in general nomenclature" means "not the preferred IUPAC name"

Isn't there a C6-chain that should take precedence in naming? (cc @Loong)

@Martin-マーチン Since Rule P-44.1.1 (greater number of substituents corresponding to the principal characteristic group, which is expressed as suffix) comes before Rule P-44.3.2 (longest chain), a parent structure that includes the $\ce{-OH}$ group wins. In this case 3-(chloromethyl)-2-(propan-2-yl)pentan-1-ol.


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Does temperature affect ionic strength?
Hello I apologize if this is a dumb question but if I have a phosphate buffer made up of Na2HPO4 and NaH2PO4 and I gradually increase the temperature (lets say 10,20,30,40,50) what would happen to the ionic strength. Would it remain the same or change?

As solution density and acid dissociation constants depend on T - what would you say ? another question is - Can you afford to consider it constant ?

I was wondering how would it change. Would it increase as the temperature increases? Im rather confused so if my question doesn't make a lot of sense thats why

also thank you for your response

Consider that even pure water increasingly dissociates with higher temperature, so the pH of plain water at 25°C is 7.0, but at 50°C is 6.63 (remember pH is a negative logarithmic value, so "less is more" ionized.). See https://www.westlab.com/blog/how-does-temperature-affect-ph


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What is meant when the solubility for an ionic compound is listed as "reacts with water"? Does this mean that it cannot form an aqueous solution?
I think my main question/concern really relates to the formation of a product that is listed as "reacting with water" after the combination of two aqueous solutions.
For example, both sodium iodide (NaI) and iron(III) chloride (FeCl3) form aqueous solutions. When combined, the expected products from double replacement would be sodium chloride (NaCl) and iron(III) iodide (FeI3). Equation something like this (Where the ? indicates the root of my question):
$$\ce{3 NaI (aq) + FeCl3 (aq) -> 3 NaCl (aq) + FeI3 (?) }$$
Now, FeI3 is listed as "reacting with water" for its solubility. Does this mean that FeI3 is, in fact, not produced, but rather begins decomposing into iron(II) iodide (FeI2) and elemental iodine (I2)? Or do we get some form of iron compound with an iodine oxyanion like Fe(IO3)3 and hydrogen gas (H2)?
And in either case, would this reaction be said to take place or not? I'm not really sure what my end goal is here, I just don't understand how an ionic compound that might be formed from aqueous solutions will behave in "reacting with water" considering its surrounded by water from the two aqueous solutions.
Thanks in advance for the help!

$$\ce{2Fe^3+(aq) + 3I-(aq)-> 2Fe^2+(aq) + I3-(aq)}$$ $\ce{I2}$ is not very soluble in water unless it combines with an $\ce{I-}$ to form $\ce{I3-(aq)}$. The reaction is not with water but rather in aqueous solution.

Related: https://chemistry.stackexchange.com/q/51610/17175. Note that FeI3 is known only in nonpolar environments or as a soft-base complex. Forming a salt by double displacement in water solution does not offer either stabilizing condition.

@OscarLanzi So then it would be simplest to identify this specific situation, in terms of double displacement for these solutions, as being a situation where "no reaction" takes place?

I would say it is a reaction, but not double displacement. Instead, with sodium and chloride as spectators, we are looking at a redox reactions between iron(III) and iodide.

Addressing the general topic of the title itself: There is a wide range of reactivity, since "so reactive the solution just does not make sense" to "reaction does not allow to reach dissolution equilibrium".

It seems that your textbook or teacher is highlighting the fact that some products formed in water-based reactions can decompose into other substances. It is important to note that many salts are not stable in water. For example, anhydrous aluminum chloride will immediatly react with water and iron (III) chloride, although soluble in water, will begin to hydrolyze and form solid, brown-colored hydrated iron oxides within hours at neutral pH.
In the specific reaction mentioned in the question, it's important to keep in mind that it is likely to be a pretty complex reaction with multiple side products. For example, when the two solutions are mixed, free iodine will be released due to the mild oxidizing properties of iron (III) chloride. This free iodine can potentially react with NaI to form a triiodide ion. Additionally, the iron (III) chloride can hydrolyze in water to form other insoluble side products. Overall, it can be difficult to predict the outcome of this reaction at neutral pH. Furthermore, if ferrous iodide is formed, it will quickly react with oxygen in the air. This complexity is a characteristic of many chemical reactions in the real world. This is why chemistry is still an experimental science. Nobody should ask this question in the exam.


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Mechanism for a reaction of reduced fluorophenol with BF3?
In the given question, $\ce{BF3}$ will act as an electrophile. Will it form $\ce{BF4-}$ and leave a positive charge on the carbocation to which $\ce F$ was attached or react with $\ce{OH}$. It will form phenol but what will be the mechanism for the reaction?

It will react with the phenol to form a complex

In my textbook the major product formed was phenol. Can you pls provide the mechanism for the formation of phenol?

The boron trifluoride will indeed remove the fluoride ion to form tetrafluoborate. Fluoride removal is favored by two related factors:

The carbon-fluorine bond us more polarized than the carbon-oxygen bond, making the fluorine a stronger negative charge center for the partially-positive boron.

Because fluorine is more electron-withdrawing than hydroxyl, the carbocation is more stable and more easily formed by retaining the hydroxyl function.

The carbocation does not simply retain its positive charge in the originally fluoridated carbon. That carbon is conjugated with double bonds, and the $\ce{C-H}$ hydrogen in the $\ce{CHOH}$ group can tautomerize to form an aromatic cation (protonated phenol):

Then the fluoborate can be protonated by this cation, giving phenol, hydrogen fluoride, and regenerated boron trifluoride. The boron trifluoride is thus actually a catalyst for the thermal decomposition.

So after carbonation is formed it will rearrange and hydrogen will shift on OH to form aromatic ring thus making it more stable and the H from H2O+ will form Hydrogen fluoride after reacting with BF4-. Am I right?

I would regard the carbocation as a superposition ofcall four structures. The same proton is either the polarized C-H proton on the CHOH group or the excess proton on oxygen in the peotonated phenol structure; because that is the same proton we can't distinguish that experimentally.

Alright got it thanks a lot.


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Why do hydrogen bonds between atoms of the polypeptide backbone, form both helix and pleated structures, instead of only one structural type?
Proteins have segments of their polypeptide chain/chains that can be repeatedly coiled or folded into helix and pleated structures, respectively. This is due to hydrogen bonds between partially charged oxygen and hydrogen atoms in the repetitive polypeptide backbone (which excludes the amino acid side chains).
My guess is that one structure (either a helix or pleated structure) is lower in potential energy compared to the other (because one structure likely maximises the number of hydrogen bonds that can form, compared to the other), which begs the question, why do both structures form, if they are largely a result of hydrogen bonds in the polypeptide backbone, rather than only one structure forming?
Is it a result of both hydrogen bonds in the polypeptide backbone, and interactions between amino acid side chains?
My own attempt at answering the question:
Intuitively, I believe the terms 'secondary' and 'tertiary' structure are misleading. My reasoning is that, it does not make sense to say that hydrogen bonds between the atoms of different amino acids in a polypeptide backbone, bring separate amino acids closer together, before interactions between amino acid side chains (R groups) occur (especially since some of these interactions between R groups are hydrogen bonds themselves).
I assume, the previous statement can be valid stated vice versa, i.e., that interactions between amino acid side chains bring separate amino acids closer together, before hydrogen bonding can occur between the atoms of different amino acids.
Ultimately, if this is the case in reality, I would expect both H bonds and AA side chain bonds to occur, effectively, simultaneously, blurring the distinctions that have been established by humans.

Your very line of inquiry is misled. Pretty much all organic chemistry sits pretty far from its absolute potential energy minimum. The same goes twice for biochemistry. So what if there are two local minima? It could have been three just as well.

@IvanNeretin I appreciate the response, albeit, it only raises more confusion (for me), since it makes me wonder why all matter is not found at the lowest state of potential energy (if such a thing is conceivable), since it allegedly has a tendency towards it, and why matter found in organic chemistry (and biochemistry) is unique in that it tends to be far from the 'absolute potential energy minimum' (if such a thing is even conceivable, quantifiable, measurable etc). This is especially confusing because the distinctions between organic chemistry and biochemistry are entirely arbitrary.

The conformation is highly coupled to the amino acid side chains. Some like to form helices, some like to form strands, and some don’t like either.

Proteins don't have some configurations carved in stone, but "conformational ensembles". They tend to be pretty close to a minimum, usually even global one, but wobble all the time, especially in vivo.

The stable state for a protein is not the absolute energy minima for all of it's components taken separately. There might be cases where some interactions might be unfavorable in some configuration, but there might be other interactions nearby which stabilize the entire structure. This is something we regularly run into in MD simulations.

"Why do hydrogen bonds between atoms of the polypeptide backbone, form both helix and pleated structures, instead of only one structural type?" I assume it wasn't what you intended to say because not the hydrogen bonds form helix/pleated structures, but the proteins.

It sounds like you are getting philosophical. Buildings (or bridges or...) are different, and buildings can be classified into particular types of structures (houses, highrises, mosques etc), without needing to consider how they are built. They may  also share elements in their structures, and they may all be regarded as stable despite these similarities.

Your comment " it makes me wonder why all matter is not found at the lowest state of potential energy (if such a thing is conceivable)" is very interesting and an important biological (or general) consideration. You may want to search the site for similar questions and if you can't find one try to formulate this into a question for a new post.

A simplistic answer to "why all matter is not found at the lowest state of potential energy " is that barriers exist between local minima and the global minimum. For example, a boulder can remain at rest in a depression on the side of a mountain for many millenia before an event occurs that allows it to move to a position of lower gravitational potential energy. And the barriers may be so high that it is never able to move.

@Andrew I appreciate the response and I am aware of that interesting point you have raised. However, it does not answer the question of why not all matter in the universe is at the lowest potential energy state, because it does not present an answer as to why the matter you mentioned (and all matter in the universe, that is found higher than the lowest potential energy state) was raised to that higher energy potential state in the first place.

Thank you for all the comments everyone, seriously, a diverse array of knowledgeable people have responded. @BuckThorn I think I will ask that question elsewhere, as per your advice, although I suspect it is is unanswerable.

Proteins have segments of their polypeptide chain/chains that can be repeatedly coiled or folded into helix and pleated structures, respectively. This is due to hydrogen bonds between partially charged oxygen and hydrogen atoms in the repetitive polypeptide backbone (which excludes the amino acid side chains).

Hydrogen bonds alone can't explain the structure of proteins because making hydrogen bonds with water is just as good. When you look at the overall picture (entropy and enthalpy of folding in the presence of water), it turns out the helices and sheets do occur frequently in folded structures.

My guess is that one structure (either a helix or pleated structure) is lower in potential energy compared to the other (because one structure likely maximises the number of hydrogen bonds that can form, compared to the other), which begs the question, why do both structures form, if they are largely a result of hydrogen bonds in the polypeptide backbone, rather than only one structure forming?

As stated above, potential energy alone is not sufficient to rationalize protein conformation. In the end, it is Gibbs energy that decides, so enthalpy and entropy. Also, you have to consider effects on the protein and the solvent, not just the protein. This becomes clear when you add other solvents, or place a protein in vacuum, where it often denatures.

Is it a result of both hydrogen bonds in the polypeptide backbone, and interactions between amino acid side chains?

The side chains are important, also for the energy of the main chain conformations. For example, the alpha helical conformation is unusual for amino acids with a branched side chain at the beta position (e.g. valine or threonine).

Intuitively, I believe the terms 'secondary' and 'tertiary' structure are misleading.

If you think of them as hierarchy of structure (from local to larger and larger scale), they make more sense than if you interpret them as an order of folding. In the biological context, the N-terminal end of a protein sometimes folds before the C-terminal part is even synthesized (so parts of the tertiary structure form before the primary structure is fully formed).

Ultimately, if this is the case in reality, I would expect both H bonds and AA side chain bonds to occur, effectively, simultaneously, blurring the distinctions that have been established by humans.

Yes, secondary and tertiary structure form mostly concurrently after formation of a molten globule (i.e. hydrophobic side chains starting to cluster). It is thought that the last feature that forms are the disulfide bridges, as shown in the famous Anfinsen set of experiments.

Particular protein structures arise because of a potentially very complex interplay of interactions within a chain, with other chains (from other subunits in a multimeric protein or from chaperones or other biomolecules) and with the solvent which includes water, lipids, ions, small molecules. All of these contribute to the adoption of a final fold but also to that of intermediates on the folding pathway. Sometimes the final fold is not "triggered" until an apo structure is further processed or associates with a target, for instance by cleavage (insulin), cross-linking (insulin), binding of a cofactor (azurin) or self-association (amyloid).
Generally the most important factors determining the secondary structure are the balance of hydrogen bonds, interactions between the hydrophilic sidechains of polar residues with the solvent, and hydrophobic interactions that result in burial of hydrophobic sidechains in the protein interior. The balance between specific interactions is generally very finely tuned so that only a desired structure is adopted, such as a helix, a beta sheet or a turn between secondary structure elements. Sometimes folding is hierarchical, with secondary structure elements forming more rapidly (for instance helices) before associating into higher-order tertiary structures (motifs). Sometimes cooperative interactions between secondary structure elements is required (for instance in sheets). Some aminoacids are bulky (tryptophan) and cannot be accommodated except in particular structures because of steric clashes. Some aminoacids have structural constraints and are only found in particular sites such as turns or unusual secondary structures (proline). Some residues have a low propensity to be found in helices (glycine, proline) because of entropic penalties or constrained backbone dihedral angles.
Because the many individual interactions can be weak, proteins sometimes are assisted during folding by other proteins (chaperones) that encourage desired interactions and discourage unwanted ones. The folding history and tertiary interactions can decide the final fold, such that the protein undersamples its possible conformational realm before acquiring its functional form.

Thank you Buck, as a second year Biology student I tried my best to understand all the responses. May I ask what your academic background is?
Regardless, you reaffirmed some assumptions I made about my own question and were able to understand 'where I was coming from' or my intuition, for lack of better wording, thus, you answered basically all my questions. Which is impressive because I wanted to rewrite my entire question in more detail as soon as I posted it, realising that it was a bit misleading of my own understanding.

Nonetheless, from your response, it does make me support the idea that the terms 'secondary' and 'tertiary' structure are a misleading distinction, because secondary structure (which is stated in my Biology textbook, as the result of hydrogen bonding between atoms of the polypeptide backbone) is a result of tertiary structure as well and vice versa.

I should edit into my answer that there are definitions for what constitutes a particular type of secondary structure in terms of backbone dihedral angles (geometry) and H-bonding patterns. These definitions are independent of the mechanism (path) that led to the structure. Folding mechanisms are of great interest but ultimately irrelevant for classifying the type of secondary structure (although you might find that particular aminoacids constrain how a protein folds and are therefore correlated with particular pathways). PS I have a doctoral degree in physical chemistry.

I imagine this is largely determined by function. In a protein such as bacteriorhodopsin there are several parallel helices which hold the protein as a fairly rigid structure with a channel down the middle, between the helices, where ions are transported. The rhodopsin chromophore acts as a switch for ion transport and sits in the middle of this channel. To function the structure has the be help rigidly because the isomerising chromophore pushes against a helix to open the ion channel. The LHII protein of photosynthetic bacteria has many helices arranges in a ring with each holding chlorophylls and a xanthophyll. These are held exactly in place so that efficient energy transfer around the ring of pigments so formed is efficient. By contrast, the protein I27 is part of the muscle protein titin and has beta sheets that can extend and contract under the action of force, and can extend reversibly to a certain limit without unfolding, so that the muscle can itself work. (btw All these structure are known from x-ray crystallography and easily found using Wikipedia).
These examples given do give some clues as to why there are two types of structures.  However, things are never so simple, the protein in the eye's lens, crystallin, is almost all beta sheet and serves to prevent uv damage, such as formation of cataracts, by dumping the uv energy via tryptophan and cysteine residues. So there seems to be no clear cut answer after all.


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What is the difference between distillation temperature and boiling point?
Is there a difference between the distillation temperature and the boiling point of a fluid, in my case kerosene? And if there is, what is the difference?

Describe in details your search or understanding failure, what may help against the question closure. As vaque general description of effort is frequently a placeholder for no effort.

Kerosene is a mixture of many compounds. It doesn't have a single well-defined boiling point, but a wide range over which different components will distill off. In this case there will be no well defined temperature for distillation nor a single boiling point.

Both depend on the composition of your sample (keyword Raoult's law), as well as the external pressure.
One may argue the term boiling point leans more toward the characterization of a compound (e.g., water) with a denomination like $bp = \pu{100 ^\circ{}C} (\pu{1 atm})$.
On the other hand, distillation temperature leans more toward the (technical)  application of distillations in synthesis and chemical engineering (large scale distillation, steam cracker, etc) as a unit operation.  Here, temperature and pressure are set as process parameters, e.g. solvent x was removed from the reaction mixture by distillation at a pressure of $\approx \pu{10 mbar}$ and $\pu{80 ^\circ{}C}$, or to yield blends of a distillate (e.g., kerosene) for which only a range of boiling points is provided (e.g., petrol ether of $40\ldots\pu{60 ^\circ{}C}$).

Boiling point and distillation point are similar if we take a glance but the significant difference among them..

Boiling point

When a liquid starts converting into its gaseous state and the constant transition temperature is boiling point.

Distillation temperature

On the other hand distillation temperature is used while distillation process where only the temperature of a substance that you want to yield is considered in a mixture with other substances.
in plain words boiling point is a characteristic property of a substance


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Height of light beam when measuring fluorescence or absorbance
Beer's law relates absorbance to concentration.  The "width" of the sample is part of the equation.  I'm wondering why the height or diameter of the beam of light is not included as a factor.  Or perhaps the height/diameter does affect the absorbance but can be factored out later.  In other words would the absorbance change if the only thing varied is the height or diameter of the light beam; say if the height or diameter is doubled.

Absorbance is a relative measure, Beer's law is $I_{trans}=I_0e^{-\epsilon [c]\ell}$ so that you have to measure both the transmitted light $I_{trans}$ and the amount when there is no solute in the sample cell $I_0$. Thus is does not matter how big or small your beam is assuming always that it all passes properly through the sample cell and the solution, i.e. not clipped by the cell or messed up by the miniscus etc.  The beam size etc does not matter (unless you are using a high powered laser and I assume you are not) because all this is taken into account when you measure $I_0$.

So, if I understand you correctly; if the beam height is changed, then $I_trans$ and $I_0$ will change as well.  However, they will change in the same proportion, so when $I_trans$ is divided by $I_0$ the result is the same as if the beam height hadn't been changed.

@rdemyan, Yes you are right. As long as the beam is still passing through the solution, your understanding is correct. We cannot use a beam which is wider than the container width. In fact if you have a spectrophotometer, insert a piece of white paper and see how the beam shape looks like, by setting the spectrophotometer to blue or green wavelength (450 nm or 540 nm)

I should add that fluorescence is not measure via Beers law, it is just the number of photons counted at each wavelength. The fluorescence yield is tricky to get accurately but can be found most easily by comparison with a standard compound whose yield is known.

A larger beam diameter might help increase the SNR under some circumstances.

Beer's law relates absorbance to concentration. The "width" of the sample is part of the equation. I'm wondering why the height or diameter of the beam of light is not included as a factor.

It is not the width but the depth, i.e. the distance a beam of light travels through the sample. The height and width, or taken together the cross-sectional area or shape, are arbitrary as long as they same geometry is used for the blank.

In other words would the absorbance change if the only thing varied is the height or diameter of the light beam; say if the height or diameter is doubled.

You can take the same cuvette and place it in a cuvette-holder with a smaller aperture. The absorption will be the same. There will be more noise because it is harder to measure the smaller intensity of light, however. Here is an example of an adapter that changes the cross-section of the beam that arrives at the detector:

height of light beam when measuring fluorescence or absorbance

For fluorescence, the signal is in arbitrary units. Again, if you compare two samples, the cuvette and cuvette-holder should be the same, otherwise you get a systematic error.

Thank you for the correction.  I did say "width" but I meant "depth".


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Air/Acetonitrile Interfaces
Say I drop some acetonitrile on a grounded surface and it forms a meniscus. The acetonitrile is in air. Lets forget about evaporation for the moment.

Will there be an interfacial dipole formed by the orientation of acetonitrile at the surface with the air?
If so, what would be the strength of that dipole be? What temperature/diffusion would be enough to overcome that orientaiton.

Add more context to the question. What is purpose of dropping ACN on a grounded Al plate?

The Al was just a distraction (removed now). I'm interested in the orientation of the dipole layer (if it forms) at the surface of ACN. I'm not sure how to go about understanding that.

These seems relevant (second behind paywall sadly) - but doesn't really solve the problem. https://pubs.acs.org/doi/pdf/10.1021/j100541a015 and http://www.columbia.edu/cu/chemistry/fac-bios/eisenthal/group/pdf-files/JPCC_2009.pdf

Have you discussed this with your supervisor? This is why I was asking for a context. Is this a PhD or MS or BS project or just a classroom curiousity? In short, this is not something which can be answered on a web-post (and I don't know much about it). The first paper is behind pay-wall not the second one.

Just a curiosity. I naively thought dipoles would be quite easy to calculate from the charge distribution of the solvent molecule, or at least an idea if the dipole even exists! Thought the community would have a general idea of the surface of common solvents!....


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Are lipid membranes destabilized by aligning so many negatively charged phosphates in close proximity?
From my understanding, lipid membranes in biological systems are made mostly of phospholipids, which have a hydrophobic chain region and hydrophilic head group containing a phosphate. In the membranes, these head groups then all come close together to prevent water from reaching the bilayer's interior, where the hydrophobic chains meet.
However, wouldn't doing this require moving numerous negatively charged head group phosphates in very close proximity to each other? In my mind, this would destabilize the membrane structure. Is there more going on here?

note that many of the common phospholipid head groups contain a positively charge ammonium group so the net molecular charge is zero.

See https://doi.org/10.1021/acs.langmuir.6b04073 for single charge detergent.

See https://doi.org/10.1016/0095-8522(66)90012-2 for micelle size limited by electrostatic repulsion

In a bilayer the charged head groups are in contact with the water and the fatty acid chains are in the middle of the bilayer. In water, bilayer formation is largely an entropic effect. In isolated phospholipids in water the fatty acid chain has to be bridged by the H-bonded water molecules as it is hydrophobic. This reduces the water molecules entropy vs that in pure water. Forming bilayers, or micelles, allows many individual lipids to free up many waters as the water is excluded from the centre of the bilayer. This is entropically favourable, i.e. entropy increases vs isolated lipids, and this gain is so large that it exceeds that entropy reduction of forming the bilayer. (The charged head groups are mainly just the same whether as individual lipids or as a bilayer.).
The water around the head groups has a high dielectric constant (relative permittivity = 80) which means that the electric field of any charge is reduced v rapidly vs distance away from the charge, approx as $q/80R$ for charge $q$ and distance $R$. Thus two charges can be close without  any large repulsion. In a non-polar solvent such as hexane the dielectric constant is $\sim 2$ so v slow reduction in electric field and charges cannot come close to one another as repulsion soon becomes greater than thermal energy.


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How would I draw orbitals for specific atoms (with shapes, not diagrams)?
I'm relatively new to chem and I'm learning about orbitals.  I know that they have distinct shapes (s is a sphere, p is a dumbbell, etc.). However, apparently their shapes change when rotated (s changes into a half sphere or is invisible when rotated about the XYZ plane). Why do they do this; and how would you combine the cloud diagrams for specific atoms? For example, Sulfur (1s22s22p63s23p4)'s cloud diagram, with orbitals combined. If it's possible, please explain in beginner terms; the internet isn't helping. Thank you!

These orbitals shapes are good for hydrogen only. So focus on hydrogen for the time beings. Introductory chemistry textbooks do an injustice by drawing orbitals for other heavier atoms. It is their imagination. May be paste a picture so that readers can provide you with the search terms.

Thanks for the answer! Is there a way to mark this as correct?

This is not an answer, but it was a suggestion. Still you can post a picture? What do you mean by saying that s orbital becomes a half sphere?

Added! I assume that that's what it looks like, but I may be misinterpreting the diagram

It is a sphere, they are showing a half slice to show nodes, regions where the probability of finding the electron is zero.

The Orbitron site has some very good visualizations: https://winter.group.shef.ac.uk/orbitron/atomic_orbitals/1s/index.html


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What do we use Rare Earth Elements/ lanthanides for in geochemistry?
What kind of things are lanthanides used for in geochemical studies in terms of high temperature igneous systems and dating rocks from the earth, moon, and solar system?

What particular info did not you understand while reviewing offline and online resources about this topic?

Just can't seem to find anything related to high temperature igneous systems or dating

Put then enough relevant details, purpose, background and found partial confusing info in the question. Otherwise responders would be confused  what exactly you want to know and especially need to know.

What gets used in gechemistry is not really the rare earth elements as such, but minerals containing them. The preference of these elements to for higher oxidation states (+3 or more) and relatively large size of their atoms/ions causes rare earths to be incorporated into different minerals from those that make up most of our felsic or mafic igneous rocks. An example is provided by zircons, which are used in dating rock formations by measuring their lead content versus their uranium content.
Zircons are based on zirconium orthosilicate ($\ce{ZrSiO4}$; note the +4 oxidation state of zirconium). As described in this question, the chemical composition of zircons favors inclusion of rare earth and early transition metals as they are formed. Actinide elements, also having large atoms/ions and sufficiently high oxidation states, also join in, including uranium. Lead, by contrast, is disfavored. Yet zircons are commonly found with lead, which must have emerged after the fact from radioactive decay of heavier elements, principally uranium. Given this mechanism zircons, with their rare-earth-loving formation chemistry, can be used to date the rock formations in which they appear from the amount of heavier elements (assumed to be principally uranium) that have decayed to lead since the formation of the zircon.


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Solubility of amines
I ran across something saying that n-butanamine is miscible in water while butanol is only partially soluble. Is this correct? Can anyone offer an explanation? I would have guessed that the decreased polarity of the amine relative to the alcohol would have resulted in lower solubility for n-butanamine.

Tendency to form n-butylamonium cation may play a role too.

$n$- butanamineis only a weak base. We would not expect much of it to dissociate into ions.

The different solubility of certain alcohols versuses corresponding amines cannot explain simply by considering the electronegativity advantage of oxygen over nitrogen alone to make hydrogen bonding. For example, intermolecular forces between neat amine is weaker (e.g., boiling point of 1-aminobutane is $\pu{78 ^\circ C}$) than that in alcohols (e.g., boiling point of 1-butanol is $\pu{118 ^\circ C}$). To dissolve either compound in water, those corresponding intermolecular forces should be broken before making new intermolecular forces with water molecules, thus, amine has the advantage over that for having weaker intermolecular bonds. My speculation can be supported by following data, which shows the different solubilities of triethylamine in water (boiling point: $\pu{90 ^\circ C}$; solubility in water $\pu{14 g}$ in $\pu{100 mL}$ water) compared to that of dipropylamine (boiling point: $\pu{111 ^\circ C}$; solubility in water $\pu{4 g}$ in $\pu{100 mL}$ water).
Also, primary amine such as 1-aminobutane has two $\ce{N-H}$ bonds to make hydrogen bonding compared to one $\ce{O-H}$ bond in 1-butanol. Again, advantage: amine.
It's worth noting that, as Poutnik pointed out in the comment section, the tendency to form n-butylammonium cation may play a role too:
$$\ce{n-Bu-NH2 + H2O <=> n-Bu-NH3+ + OH-}$$


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Why does the reaction of dissolution stop at an equilibrium point?
The formula for Gibbs free energy is $\Delta G=\Delta H-T\Delta S$.
If Gibbs free energy is negative, the reaction is spontaneous. This also applies to dissolution reactions.
However, we know that every soluble element has a dissolution constant. At a certain equilibrium point, the reaction stops. Which term in the Gibbs free energy indicates that the reaction should stop?
Moreover, according to the Le Chatelier's principle, the dissolution reaction can go in either way if we change some concentrations at the equilibrium point. But only changing concentrations does not seem to affect neither the enthalpy nor the entropy. Why does the value of $\Delta G$ vary?
Finally, is there any way to predict the dissolution constant based on $\Delta H$ and $\Delta S$?

The entropy of mixing depends on concentrations, which change during the dissolution process. That is the basis of the law of mass action (equilibrium when Q = K) as well.

"Why does the reaction of dissolution stop at an equilibrium point?" It does not stop! There are two sides to a state of equilibrium:

The forward reaction, e.g. a salt dissolving in water, $\ce{NaCl -> Na+ + Cl-}$.  Simplified for illustration.
The backwards reaction, e.g. a salt precipitating from solution in water, $\ce{Na+ + Cl- -> NaCl}$.

At the point of equilibrium, they are proceeding equally rapidly in both directions. That means salt is dissolving and precipitating constantly. If you place a crystal of salt in a just saturated solution, it may gradually change shape, as parts dissolve and precipitate elsewhere.

Equilibrium is the state at which the forward and reverse reactions have the same rate, not the point at which the reaction stops.


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How to get Raman shift frequency values from neutron scattering measurements?
I have read some papers that say that a Kohn anomaly mode is a particular feature of 2H- transition metal dichalcogenides and that the existence of such a mode can be seen in neutron scattering data where an anomaly (frequency decrease) in the phonon dispersion plot can indicate a Kohn anomaly. An example of the neutron scattering data for 2H-TaSe2 can be seen below (DOI link to paper:
https://doi.org/10.1103/PhysRevLett.34.734):

I've seen papers (such as this one: https://doi.org/10.1038/srep02593) say that this anomaly corresponds with a broad peak seen at 140 cm$^{-1}$ but I don't understand how that has been calculated. Does anyone know how the phonon dispersion data like that from neturon scattering experiments can give values corresponding to Raman shift frequencies?

Photons scattering off of phonons is Raman scattering. Neutrons also scatter off phonons.

@JonCuster Indeed. I've had another look at the paper and have seen that I had misunderstood it. I believe there is a correspondance between neutron scattering exerpiments and Raman experiments in that both can say something about the phonon energy. I found this link to be useful: https://physics.stackexchange.com/questions/423969/how-to-convert-cm-1-to-ev


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Why does electrochemically formed CuO not adhere to the bulk Cu unlike oxide formed through heating?
It is my understanding that copper forms an oxide layer upon heating. The colour of the copper depends on the thickness of this layer and can range from reds to purple to green etc.
I have confirmed this by heating a sample and noticed that the oxide layer is robust is 'adhered' to the bulk copper
Separately, I have attempted to form an oxide layer through electrolysis, passing a current from one copper electrode to another via a CuSO4 solution. A black coating (assumed to be CuO) formed around one electrode. However, the CuO does not adhere at all to the bulk material, rubbing away extremely easily, leaving only shiny metal.
Assuming I am indeed forming CuO, why does this not adhere to the bulk in the same way CuO produced by heating does?

Red/purple colour is due Cu2O, not CuO. Green is CuCO3 . Cu(OH)2.

@Poutnik - It is my understanding that the blue/red/purple/greenish colours that come from heating are due to the thin film interference effect, and are related to  the thickness of the oxide layer, and not due to the colour of the oxide (or carbonate) itself.

It is well known copper is covered by cuprous and not cupric oxide. Aside of that, dry thermal formation and wet electrolytic formation are way different circumstances without justification of the same result.


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Why did industry develop mainly to fix nitrogen in ammonia (Haber-Bosch) instead of in nitrate?
Since approximately WWI, humanity has been able to fix nitrogen gas from the atmosphere ($\ce{N2}$) into compounds that are much more bio-available, like ammonia and nitrate. The first (hard) step is the Haber-Bosch process $\ce{N2 + 3H2 -> 2NH3}$, which has to be carried out under elevated pressure, temperature and highly anoxic conditions.
Given that our atmosphere is 80% $\ce{N2}$, 20%  $\ce{O2}$ and therefore naturally quite oxic, given that nitrate seems to have a much lower enthalpy of fusion(*), given also that typical fertilizer prescriptions are overweight on $\ce{NO3}$ compared to $\ce{NH3}$ (in ratios like 4/1)... why did humanity not develop an industrial process emulating e.g. fixation in lightning and thus synthesizing useful $\ce{NO_x}$ for its fertilizer needs as opposed to $\ce{NH3}$?
(*) In effect, if the Haber-Bosch process is followed by the Ostwald process to make nitric acid, it seems there's going to be a massive loss of exergy (not least because the extraction of $\ce{NH3}$ from the reaction mixture in the Haber-Bosch plant seems to involve a massively wasteful cooling-down step). As an illustration I looked up the standard enthalpy of formation of a mixture (1 mol $\ce{NH3}$ + 1 mol $\ce{KHCO3}$ + 2 mol $\ce{O2}$) vs a stoichiometrically equivalent mixture (1 mol $KNO_3$ + 2 mol $H_2O$ + 1 mol $\ce{CO2}$) and the former contains $\pu{448.5 kJ}$ more than the latter mixture!

Comments are not for extended discussion; this conversation has been moved to chat.

I do not address agriculture application, but extreme energetic demand and inefficiency of direct synthesis of nitrogen oxides by the Birkeland–Eyde process, or the Frank–Caro process of thermal capture using $\ce{CaC2}$. Synthesis of nitric acid via ammonia by H-B process is the most cost and energy efficient process,
There is always long way from labs to industry. Consider how many revolutionary new designs of rechargable cells have been announced during decades.
That quote of the abstract of your linked resource;

This analysis shows that the energy consumption for NOX synthesis with plasma technology is almost competitive with the commercial process with its current best value of 2.4 MJ mol N−1, which is required to decrease further to about 0.7 MJ mol N−1 in order to become fully competitive.

Industry has to first see the decisive advantage to change the technology, as aside energy demand there would be new technology challenges and extra investments.
Much more energy than needed for endothermic reaction $\ce{N2+O2 -> 2NO}$ is wasted, because of creating condition where such reaction can provide sufficient yield by sufficient rate.

Because the capital investment and energy consumption if the Haber + Ostwald process is still far lower than the alternatives
The reason why specific industrial processes are chosen are usually economics and practicality. While some chemical reactions look, in principle, to be better alternatives (surely direct routes to nitrate are better than making ammonia first and then oxidising it?) they may not be practical or economic.
We also know that some plants can fix nitrogen using enzymatic reactions at normal temperatures and pressures. Every chemist wishes there were a way to industrialise that. It is clearly possible but nobody has ever demonstrated it on an industrial scale. And industrial routes to nitrate were known before the Haber process was developed. But those direct processes (using plasma reactions) were far more expensive in both capital and energy inputs than the, apparently, indirect, combination of Haber and Ostwald processes. 100 years later this is still true. The linked paper shows that the energy cost of plasma processes to nitrate is still about five times higher than the the Haber  Ostwald process (per mol of fixed nitrogen). Plus, the capital and maintenance costs are higher.
In short, the choice of which processes are used is based on accessible known reactions when the processes is adopted and the overall economics of the competing processes. Plasma processes were and are still far from competitive despite the theoretical advantages of a direct route to nitrate.


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How can we make seacrete / biorock grow faster?
Seacrete is an interesting building material (precipitated limestone in seawater) but is quite slow to grow.
https://naturalbuildingblog.com/seacreteseamentbiorock/
https://en.wikipedia.org/wiki/Biorock
With the purpose of making it grow as fast as possible, for the purpose of building structures, are the variables known (salinity, temperature, exact electric current, any others) that affect its rate of accretion/'growth' to boost the process, i.e., faster in cold or warmer water, salinity, etc.
I.e, have any known studies been made of the optimal conditions for fast seacrete accretion.

You might find some tips here and here...

It is a calcarious deposit that builds up on steel with cathodic protection in seawater. Similar to limestone. It is desirable to reduce current needed for cathodic protection otherwise of no particular value that I have heard of.


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How can I determine a stereogenic centre?
I have a problem understanding stereogenic centres. A stereogenic centre is specified (in the case of organic chemistry that I am interested to) as a carbon atom which is bounded to four different atoms or groups. However in my teacher's notes in order to make clear the way we find stereogenic centres he gives as an example a cholesterol molecule with its stereogenic centres that are shown in the image:

What I cannot understand is why the 18th carbon atom is said to be stereogenic since it is bounded to four carbon atoms?
Why is the third carbon atom said to be stereogenic since it is bounded to 2nd and 4th carbon atoms which are identical?
I have some confusion regarding the definition of an assymetric carbon atom. Even though it seems to me quite straight forward the way it is implemented seems to not obey to my theoretical expectations.Can someone please justify the way stereogenic centres are specified in the case of cholesterol? Could someone give me some examples of chemistry molecules and the stereogenic centres that we meet there ? Any help is really needed and appreciated!

Note an error recognizing the assignment of numbers here. Sterogenic carbon atoms are number 3, 8, 9, 10 (not 19!), 13 (not 18!), 14, 17, and 20 (not 21!).

As your teacher's notes says, a carbon atom in a molecule which is bounded to four different atoms or groups can be considered as stereogenic center. An example for a molecule with a carbon atom which is bounded to four different atoms is 1-bromo-1-chloro-1-fluoroethane:

The stereogenic center is the $\ce{C}$1, which is bounded to another $\ce{C}$, and three halogens, $\ce{Br, Cl,}$ and $\ce{F}$, respectively.
An example for a molecule with a carbon atom which is bounded to four different groups is 3-methylhexane:

I choose this molecule to clear up your confusion on stereogenic centers in cholesterol. The stereogenic center in 3-methylhexane is $\ce{C}$3, which attached to 3 different carbon groups, mamely methyl, ethyl, and propyl. The forth group is hydrogen.
Now, let's see one of your concerns:

Why is the third carbon atom said to be stereogenic since it is bounded to 2nd and 4th carbon atoms which are identical?

Do they identical? Not really. What is the next atom to $\ce{C}$2? It is $\ce{C}$1 which is $\mathrm{sp^3}$ hybridized. What is the next atom to $\ce{C}$4? It is $\ce{C}$5 which is $\mathrm{sp^2}$ hybridized. Thus, $\ce{C}$2 and $\ce{C}$4 are not really identical such as methyl and ethyl groups are not identical in my second example.
And, if you really looks carefully there is no steocenter on $\ce{C}$18 (which is a methyl group). I believe the stereocenter you really talking about is $\ce{C}$13 (also, if you have problem with $\ce{C}$13, you should have problem with $\ce{C}$8 as well). On $\ce{C}$13, the three other groups are $\ce{C}$12 ($\ce{CH2}$ group), $\ce{C}$14, and $\ce{C}$17 ($\ce{C}$18 methyl is the forth). You misunderstood as $\ce{C}$14 and $\ce{C}$17 are identicle because both of them are $\ce{CH}$ groups. It is also true that $\ce{C}$14 attached to two other groups with $\ce{CH2}$ ($\ce{C}$15) and $\ce{CH}$ ($\ce{C}$8), as well as $\ce{C}$17 attached to two similar groups with $\ce{CH2}$ ($\ce{C}$16) and $\ce{CH}$ ($\ce{C}$20). Thus, to determine whether they are different or not, you need to go one step more. For $\ce{C}$14, $\ce{C}$8 attached to $\ce{CH}$ ($\ce{C}$9) and $\ce{CH2}$ ($\ce{C}$7).
For $\ce{C}$17, $\ce{C}$20 attached to $\ce{CH2}$ ($\ce{C}$22) and $\ce{CH3}$ ($\ce{C}$21). That observation exclusively shows $\ce{C}$14 and $\ce{C}$17 are not identicle.

Consider the third carbon for instance, As you mentioned, it is connected to three carbon atoms, which are its immediate neighbors. Since the three atoms are carbons, we now go to the next-nearest neighbors, and now we see that the next-nearest neighbors for atom 3 are different. Thus, the local environment for the atom number three has four types of atoms (one O from -OH group, one H, and two distinct carbon atoms). This same reasoning can be extended to atom number 13, where we see that the neighbors of the four carbon atoms connected to atom 13 are in different chemical environment, rendering the atom number 13 to be a chiral center.

So far, user55119 provided 19 answers how to apply the CIP rules and I recommend to work through these e.g., this one, or this. The results are developed gradually and amended by useful illustrations.

I was always puzzled by the definition of stereocenter based on how many 'different groups' it has. Perhaps it is more useful to learn what stereoisomers are; then a 'safer' definition may be 'an atom is a stereocenter if, by swapping any two different groups bound to it, a new stereoisomer is generated'. See https://en.wikipedia.org/wiki/Stereocenter . This covers cases like 4-methyl-cyclohexanol, which has two stereocenters despite being achiral and having two identical groups on each of them. But yes, the OP's had simply misinterpreted the term 'different atoms/groups', not this aspect.


text:
When NaOH ionizes in water, does the OH– react with the water molecules or with the hydronium ions from the dissociation of water?
NaOH in water becomes  $\ce{Na+}$ and $\ce{OH-}$ ions. If it's a Bronsted Lowry base then the $\ce{OH-}$ ions will take $\ce{H+}$. But FROM where exactly - from the water molecules or from the hydronium ions (from the dissociation)?
Also do all the $\ce{OH-}$ ions from $\ce{NaOH}$ combine with the $\ce{H3O+}$ to form 2 molecules of water or some of them remain,as there is an excess of  $\ce{OH-}$ ?
If  $\ce{OH-}$ reacts with the water then the reaction is: $$\ce{OH- + H2O -> H2O + OH-}.$$ So from this reaction we just conclude that $\ce{OH-}$ exist freely in our solution.

Note the mouseover hint at down pointing grey triangle as the default reason for the question downvoting:  "This question does not show any research effort; it is unclear or not useful." Just in case you would wondering why the question got downvotes or was even closed.

@Poutnik is this a sarcastic remark?

Don't consider water ionization as a static process. It is highly dynamic at a molecular level and no doubt water is indeed a mystery solvent.
It is an interesting question and that can perhaps only be answered by isotope labeling studies. Instead of $\ce{NaOH}$ start with $\ce{NaOD}$ (deuterated base) and alternatively, take $\ce{NaOH}$ and D2O.

Search in this direction using Google Scholar or Google.

Almost a duplicate of https://chemistry.stackexchange.com/q/88323

@Poutnik cool, i wasn't sure if you meant something else that's why i asked.

Even pure water contains $\ce{H2O, OH- and H3O+}$. The two ions can either react with water, or with each other:
$$\ce{OH-(aq) + H2O(l) <=> H2O(l) + OH-(aq)}\tag{1}$$
$$\ce{H3O+(aq) + H2O(l) <=> H2O(l) + H3O+(aq)}\tag{2}$$
$$\ce{OH-(aq) + H3O+(aq) <=> H2O(l) + H2O(l)}\tag{3}$$
All three are fast acid/base reactions. In neutral water, reaction (3) in the forward direction is most unlikely because both reactants are present at very low concentration.

If it's a Bronsted Lowry base then the OH− ions will take H+. But FROM where exactly?

Both reaction (1) and (3) will happen. Because of reaction (3), the $\ce{H3O+}$ concentration will rapidly decrease, making the rate of this reaction very slow. Both reactions (1) and (3) have hydroxide as one of the reactants, but the other reactant is either water (present at high concentration) or hydronium (present at very low concentration in alkaline solution). This makes reaction (1) more likely.
As the OP mentioned, reaction (1) does not use up any water, so that reaction will continue to proceed at a high rate. In fact, some argue that $\ce{OH-}$ is not the best description, and it should rather be $\ce{H2O.OH-}$ or $\ce{H3O2-}$. For the positive ion, $\ce{H3O+}$ is a compromise, acknowledging that there is no "naked" hydrogen ion in water, but not trying to describe higher aggregates.

does the OH– react with the water molecules or with the hydronium ions from the dissociation of water?

The short answer is both.

I don't understand this part: "Because of reaction (3), the H3O+ concentration will rapidly decrease, making the rate of this reaction very slow."  We know that increasing the concentration of one or more reactants, in this case the OH-, will increase the rate of a reaction. So which one is it?

@CaptainAmericaWhyso see my edit to the answer.


text:
Interpretation of Helmholtz energy and Gibbs energy
I know that this question has many other variations on this site, but I'm trying to see if I understood Helmholtz and Gibbs energy properly or not. The material I'm reading from is Physical chemistry by Thomas Engel and Philip Reid, third edition, and An introduction to Thermal Physics  by Daniel Schroeder.
In Schroeder's book, Chapter $5$, the author defined Helmholtz energy, $A$, as $A= U-TS$, where,
$U$ is internal energy of the system and
$S$ is the system's final entropy.
The author says that, It is the total energy needed to create the system minus the heat you can get free from the environment at temperature $T$. He further states that it is the available or "free" energy.
Then, for a system in constant pressure $(P)$ and temperature $(T)$ environment, he defines Gibbs energy, $G$, as $G= U-TS+PV$, where $PV$ is the atmospheric work term that's in enthalpy,
$H$ $=$$U+PV$.
Also, from Engel and Reid, chapter $6$, we have, for isothermal process,
$dA$ $\le$ $đw_{total}$ $...$$(1)$,
including expansion and non-expansion work, where the equality is satisfied for reversible process. Equation (1) allows us a way to calculate maximum work that a system can do on the surroundings.
And in a similar manner $dG$ $\le$ $đw_{non-expansion}$ $....$$(2)$, where the equality is satisfied for reversible process.
It is stated that equation $(2)$ allows one to calculate maximum non-expansion work  that can be produced.

Now, this is my understanding:
For reversible process, $A$ represents total available internal energy, and a part of $A$, which is $G$, is the available energy to do non-expansion work.For irreversible processes the inequalities (1) and (2) gives the lower bound for expansion and non-expansion work respectively. Is it the correct way of understanding or am I way off?
Also, if $A$ does represent total available internal energy for reversible process then what does it represent for irreversible process? Thanks in advance

pV is an atmospheric work term only at isobaric conditions.

@Poutnik, I did write that the system is in constant temperature and pressure environment.

OK, but it is misleading formulation there, making impression  G is defined for constant pressure (P) and temperature (T), what is not true. It is defined generally, having just special advantages at those conditions, like delta pV being the volume work term. The similar for A and constant T and V.

@Poutnik Yeah, you are right. It does seem that way. What do you think about the last paragraph, though? Am I thinking correctly? Thank you for replying!

I will not probably pay enough attention today, so you may want to wait until US wakes up, there are some TD experts.  // Yeat another interpretation of A and G is $\Delta S_\mathrm{tot} \ge -\Delta G_\mathrm{sys}/T$ for T,p const, resp.  $\Delta S_\mathrm{tot} \ge -\Delta A_\mathrm{sys}/T$, for  T,V const, where $\Delta S_\mathrm{tot}=\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}$

For my students,  $\Delta H$ is the amount of thermic energy accompanying a reaction at constant pressure, and $\Delta G$ is the amount of electric energy that accompany a chemical reaction occurring in a system (or redox cell) at constant pressure.

@Poutnik, thanks for the hint! I'll work with what you have written and let you know my thoughts about it.

@Maurice, thanks for replying! I've read that interpretation in Schroeder, and it is good way to think about it that way. I'd still like to know, if what I understood is correct or not.

Note that $-\Delta G$ represents the maximal amount of work the system may provide during the respective system state change. For isothermal change of ideal gas state, $\Delta G = W=nRT\ln{\frac{p_2}{p_1}}$


text:
Is there a good synonym for "a group connected to a N-atom"?
I'm writing a publication in which I compare the effect of alkyl, alkene and similar chains of different length connected to a N-atom of a heterocyclic organic compound. Often I have to refer to these chains, without the N-atom.
Right now I'm using something like "the ethyl/propyl/etc. group connected to the N-atom", which feels awkward. Is there a better chemical term I could use? Based on my online research, amine group or amine substituent appears to be the closest, but it includes the nitrogen itself.
Some examples (not from my work):

Is there a term that properly describes the encircled parts?

Though I asked the question in general, the molecule I'm concerned with has the nitrogen in a ring structure, like my examples. If there's no general name, one applying for this specific case would also suffice.

o.O Looks like you have a fundamental misunderstanding. Substituents are named for what they are, not what they are connected to.

@Mithoron I know how it looks like, but (I think) it's not a misunderstanding. I'm working with molecules with different length of CH-groups attached to a the N-atom of a pyrinide ring. I'm comparing certain (density-like) quantities on the ring (including the N) and on the connected group (excluding the N). It's kind of awkward to refer to this "tail" as "the ethyl/propyl/etc. group connected to the N-atom", so I was looking for a general way to refer them at.

Amyl is an old name for pentyl, so you at least have a misunderstanding. Substituents are substituents, in particular alkyl, or aryl groups.

@Mithoron That was a mental typo on my part. fixed.

See nomenclature in resources learning chemistry, especially the Blue Book (or chapter 4, P-44 of Moss' compilation): first you define the senior chain/cycle/motif of structure; second the substitutents/side chains.  But the name of an "ethyl", "acetyl", or "3-methoxyphenyl" side chain is not determined by what atom of the more senior motif binds to this side chain.

Where is the problem ? Your first molecule is N-ethylpyrrole; the second is N-acetylpyrrole.

@Maurice I need a general word to refer to the circled groups, ot the name of the molecules.

If general, what about "N-substituent", "N-substituent in X"  or "R-N" in a formula?

@Poutnik wouldn't that still include the N-atom? I arrived to a phrase like "N-bound chain" but it feels a bit clumsy.

As I assume N may be bound by various ways, I doubt there could be a general name covering N implicitly. Try a general name without the substituent.

I think "N-substituent" is just fine.
A substituent refers to whatever is sticking off something. The prefix "N-" doesn't change that; rather, it qualifies what it is sticking off. So the N-substituent doesn't mean "the nitrogen plus the substituent", but rather "the substituent sticking off the nitrogen".
(As a parallel example: in, say, aromatic rings it is common to refer to a C-2 substituent or an ortho-substituent or something. This doesn't include the carbon on the ring, it just refers to whatever's sticking off it.)

Inspired by the locant N- and by biochemical nomenclature for glycolisation, I could see use of the term “N-linked moiety”.


text:
Safe disposal of hydrochloric acid
We recently bought a home and found about 2 litres of hydrochloric acid 20% (also known as muriatic acid) that was used to manage acidity in a swimming pool that no longer exists. I would like to dispose of it but our local recycling centre is unfortunately not accepting this substance.
Is there a safe way to dispose of it? I have done limited research and here are the 3 ways I could think of:

Diluting the acid in water and dispose in a sewer when pH is above something like 5-6. I have testing bands so I can ensure that the water solution is not too acid.
Neutralize with a basis, like baking soda or garden lime. Basically, it's similar to solution 1 except that it uses less water but produces dangerous gaseous chlorine as a side product.
Use a solid metal (like iron or aluminum) to produce a precipitate and gaseous hydrogen. This might be safer (no toxic chlorine is produced as gas) but I have the problem of getting rid of the precipitate. I am not sure if that's simpler.

I am not a chemistry expert but I do remember some basic concepts. I can find online that the pH of HCl is 0 for a concentration of 3.647%. From there, I compute that the theoretical pH of my solution is -0.74. This means that, to reach a pH of 5, I believe that I need to dilute by solution by more than 500'000 times, assuming that the pH of the water in our pipes is around 7 (which is the case). That means that I would need to use 1 million litres of water, which is insane!
That leaves me with the last 2 solutions. Is it reasonable to consider using a metal to form a precipitate or is really the only option to use a basis? Also, what amount of garden lime or baking soda should I expect to need to bring my pH to around 5?

Concerning your second point: Why do you think you will get chlorine when neutralizing the acid? Baking soda and hydrochloric acid will give you a sodium chloride solution.

That's a fair question. The advice I got from my recycling centre was: "While wearing protective gear for your eyes and mouth, slowly add the acid to the bucket, keeping your face away while pouring.  Caution that the reaction of lime and acid can release chlorine gas and/or hydrogen gas, so wearing a chlorine gas respirator is recommended (or at the least do this process in a well-ventilated area and keep the air moving away from you)." So I assumed that it could release some chlorine gas.

That being said, if I only use sodium bicarbonate only, should I expect to have (almost) no chlorine gas production? The advice seems to be related to garden lime specifically.

NO ! No chlorine ! Adding lime or chalk or baking soda will neutralize hydrochloric acid and produce some sodium chloride or calcium chloride. But it will not produce any chlorine gas. No !

OK this is reassuring then! I was worried when I read the advice to wear a "chlorine gas respirator". In any case, I plan to neutralize the substance outside (in open air) and measure the solution acidity (with test papers) before releasing it in a sewer.

Pour 200 ml in your toilet and flush, 10 times.

Outdoors or well-ventilated area is a good idea just on general principle, and any other chemicals present might do something.  Maybe wait for next spring if it's cold out, and you can't find anywhere to give it away.  Protective gear is a good idea because the reaction could be energetic and fling drops of acid around.  The chlorine gas respirator advice makes sense if the person giving it isn't sure about the chemistry and already had easy access to a respirator; better to wear a respirator you didn't need than to breathe poison gas! Chlorine likes to form compounds, hard to free it.

Oh, but as Robert points out, there can be HCL gas, some of which will escape instead of further reacting.  So that might be a reason for a respirator if you're indoors or can't keep your distance upwind.

Yes I plan to neutralize it outdoor. I'll do small quantities at once so it doesn't get out of control! Will try to find a respirator too, just in case.

Give it to someone who has a pool and can use it.

Just pour it in sewerage, don't neutralize with soda to avoid drops of HCl everywhere.

$1$ liter $\ce{HCl}$ $20$% weighs $1.101$ kg, and contains $0.20 · 1101$ g/($36.45$ g/mol) = $6.041$ mol $\ce{HCl}$. $2$ liters $\ce{HCl}$ $20$%, contain $12.082$ mol $\ce{HCl}$. To neutralize this acid with baking soda ($\ce{NaHCO3}$), you will need $12.082$ mol baking soda, which weighs $12.082 · 84.02 g = 1015$ g $\ce{NaHCO3}$. To neutralise the same amount of acid with lime (impure $\ce{CaCO3}$), you will need $6.041$ mol $\ce{CaCO3}$, which weighs $604.1$ g lime.
The reaction will be quickly finished. The obtained final solution may be thrown away with the waste waters. And I repeat, no toxic chlorine gas will be produced.
You may also replace baking soda or lime by a metal like iron or aluminium. But the reaction will be slow, especially at the end, when $90$% of the acid will have reacted. It may last more than one hour. Just to get an idea, you will need $6.041$ mol of metal, which weighs $338.3$ g with iron, and $163.1$ g with aluminium.

I would just add a notice of caution: While the math is simple enough, hydrochloric acid at that concentration is hazardous. It will release some hydrogen chloride gas, which OP will notice in their nose instantaneously. And adding the carbonate or bicarbonate base too rapidly will result in a terrible mess due to the carbon dioxide produced. Be prepared. If you don't know what I mean don't do it.

Thank you for the answer and the comment. I will pour the acid very slowly, mix and monitor the situation for any smell. Just a last comment, I think that I should expect some heat coming from the reaction? I find +28.5kJ/mol for the reaction so I should expect about 340kJ of net energy (12 mols), which is around the energy to raise 1 litre of water from 20C to 100C. To be on the safe side, I will have at least 10 litres of water for the reaction and monitor the temperature.

The reaction with aluminum may make a sludge that isn't really nice to dispose of in waste water. See the answers to my question https://chemistry.stackexchange.com/questions/112616/safe-and-responsible-disposal-of-aluminum-chloride

In the end, we neutralized with baking soda. I used slightly more than the theoretical value but that ensured us that we have a slightly basic solution, which is fine for our sewer.

Thankfully, 2 liters is not alot.  Muriatic acid is used to clean bricks and concrete.  A local mason or contractor might happily take it for free.
But, if you love a good science experiment, start by being safe with googles and have a baking soda solution ready in case there is a spill.  Use a large plastic tub for containment and you are ready to work.  Make it outdoors or in a well ventilated area.  Outdoors is far better.
Start by diluting a portion of the 20% muriatic 1 to 10 with water.
Add acid to water. Concerns about heat evolution are well founded.  You won't get Cl2 gas, but you will get HCl gas during neutralization, which is only slightly less worse if inhaled.  The chemistry is fairly simple:
$$\ce{NaHCO3 84g + HCl 36 g -> NaCl + H2O + CO2 fizz}$$
You can make a dilute baking soda solution as well and add slowly with stirring.  Just keep adding until it no longer fizzles, then test with pH paper.  Repeat.

Chemical equations contain atomic symbols and molecular or ionic formula. This is all. Please don't introduce masses in the equation !

I understand your intention but "fizz" is not a state of matter. You can denote with "^" to indicate gas.

You are so lucky as to have found 2 liters of toilet bowl cleaner (concentrated). There used to be a variety in my grocery store that was 20% HCl, but now it is only 10%, so now I have to use twice as much. The commercial materials contain a little surfactant to make some foam, and usually some blue color.
However, you can make do with the clear stuff, and don't really need the surfactant; if you want to, you can add that to the toilet water separately.
As a first try, pour about 10-20 mL of the 20% acid into your toilet bowl and scrub around with the usual toilet brush. Then flush. Depending on how often your toilet needs cleaning, your hydrochloric acid will last about 2-3 years. Consider it a blessing. When it's gone, you may consider buying a gallon from the hardware store.
Now, if it really gets on your nerves to have a generic toilet bowl cleaner in your house, you could use twice as much, twice as often, and make it go away four times faster.
BTW, I have a gallon of 31% HCl that I use - but I dilute it to 15% for toilet bowl cleaning. I keep the jug (tightly stoppered) in a plastic bucket, in a place where there is plenty of ventilation (just in case of leakage). So, you do need to be careful, but aggressive chemicals can be useful if well contained.

Some of the answers above go into a lot of good detail as to the chemistry behind the answer, but I want to get to the point and avoid extraneous detail.

Gradually pour chalk or baking soda into it, stir and repeat until it stops reacting.
Congratulations, you've neutralised your acid and now have a salty solution of water and sodium or calcium chloride (depending on whether you went with chalk or baking soda respectively). This can be poured down the drain no bother.
Just make sure you do it outside so the hydrogen doesn't build up and you do it slowly enough that the heat doesn't cause a problem.


text:
Protonation in acid-catalyzed nucleophilic attack of alcohol on carbonyl
When studying acid-catalyzed nucleophilic attack of alcohol on carbonyl, I got puzzled by the first protonation step. In textbooks, they often show that the first step is to protonate the carbonyl oxygen by the acid to make the carbonyl carbon into a better electrophile, so that even a weak nucleophile such as the alcohol's oxygen could attack it. But why isn't that alcohol's oxygen getting protonaded by the acid instead? After all, it has a lone pair too, it is capable then of grabbing a proton if someone wants to get rid of it (and the acid certainly does). But if the alcohol's oxygen got protonated, it would get positively charged, which would undermine its ability to act as a nucleophile! So why is it that the acid protonates the carbonyl oxygen instead of the alcohol oxygen?
My guess is that this might have something to do with pKas, or perhaps with the fact that the alcohol's oxygen atom already has one proton on it. But I haven't seen any notions of pKa involved when these reaction mechanisms are described in textbooks. Am I on the right track?

Bot the carbonyl and -OH [or -OR] are basic and will accept protons from a stronger acid. Both protonations are probably very rapid. The question is what can happen to the intermediate in either case. In this case either can give appropriate product from nucleophilic attack at the carbonyl carbon. The gem-diol intermediate is esthetically more pleasing and possibly of lower energy.

I assume OP is wondering why always showing carbonyl carbon protonation instead hydroxyl carbon protonation in the mechanism of Fischer esterification. As I have mentioned this in another answer, those reactions are reversible and protonation happens in both oxygens. yet reaction would go through to give the thermodynamically most stable product, which in this case the final ester.
Then, why protonation go through carbonyl carbon? Carbonyl carbon of acid group is $\mathrm{sp^2}$ hybridized and hence, when oxygen is protonated, carbon is capable of stabilizing the positive charge on oxygen by resonance, although this act may migrate the positive charge to unfavorable carbon atom. Yet, since that same carbon (previously carbonyl carbon) already has another oxygen attached to it, which can stabilize the unfavorable positive charge on carbon by mesomeric effect (giving one of the lone pairs to make the third resonance). The higher the participation of the third resonance form (the higher the ability of $\ce{OH}$ group to release electrons), the more stable the protonated form of a carbonyl group (the larger the basicity of carbonyl group).
If the initial protonation has gone to the hydroxyl oxygen (instead of carbonyl oxygen), there is no way to stabilize the positive charge by any mesomeric effects (neither by first and second resonance). Thus, the protonated form is very unstable and reversed reaction (deprotonation) occurs very quickly. On the other hand, in a protonated carbonyl oxygen of a carboxylic acid, the first and third resonance forms are identical (both having $\ce{HO^+=C}$), which gives an extra stabilization ($\mathrm{p}K_\mathrm{a}$ of protonated acid is about $-6$).

Your mentioned points are somewhat correct. Look in acidic medium electrophilicity of carbonyl increase, secondly nucleophile(like alcohol) get consume due to protonation. For the reaction to happen we maintain a particular particular pH(4.5-5) at which both factors are maintained so that not much consumption of electrophile takes place along with increase in electrophilicity of carbonyl.

This answer is poorly worded and not comprehensible.

I agree. All that I understood from it is that yes, the alcohol indeed gets protonated as well, just not in the same pH range, which is what I suspected. Still don't know the details.