text:
Limescale passivation by citric acid?
Cleaning products producers advise against using "home remedies" instead of their products. This advice against using citric acid to remove limescale was especially surprising for me:

Citric acid reacts unfavorably with limescale. In the case of very heavy deposits, the citric acid surrounds and seals the limescale like a protective coating. This results in insoluble deposits inside the appliances, which even descalers can no longer remove.

Surely the citric acid itself cannot "seal" anything as it is water-soluble, so I expect there to be some sort of "passivation layer" appearing on top of the limescale. Is there a known reaction that fits the description, and which one is it?

Calcium citrate is not very soluble in water : $0.9$ g in $100$ mL at room temperature. So it may happen that using citric acid on heavy limescale deposits may change this insoluble substance into another insoluble deposit. It is not a passivation effect.

Acetic acid is a much better bet - the solubility of calcium acetate is 34.7g/100ml

@Maurice I think 0.9g/100mL is still easily rinsed with water, and certainly not something I would call an "insoluble deposit".

@Waylander In the same page I have linked to they discuss acetic acid as well: it is said to dissolve plastifiers from plastics, making them brittle.

The low solubility of calcium citrate may result in precipitation of calcium citrate within crevices of the scale and reduce the surface available for dissolution, especially if the concentration  of citric acid is low and becomes depleted. On the other hand, high citric acid concentrations will certainly have a strong corrosive effect on the scale, and still not (very) corrosive toward the metal. Citrate also chelates calcium, so high concentrations of the acid will take away the scale nicely.

There is no cure-all solution for descaling and no "reaction" because scale is not a standard compound. Its composition will vary with water quality. So it is a bad idea by these companies to spread rumors that citric acid passivates limescale.
One has to know what the scale is and the construction material, to begin with descaling. I think it all depends on the application. For example, citric acid is ideal for descaling dishwashers and washing machines. It is not corrosive to steel and forms complexes with iron, if there are rust stains. Note that descaling is done with citric acid (citric acid is in excess) in these machines.


text:
True or false: "A used AA battery contains fewer moles of electrons than a new AA battery."
I recently took a Gen. Chem. 2 exam that contained this question. I answered false, but my professor said the answer is true. My reasoning was that any electrons that leave the anode end up at the cathode, so the number of electrons should be conserved. This was consistent with the way we studied redox reactions and electric cells: reactions were always broken down into half-reactions in which the electrons exchanged appeared on opposite sides of the half-reaction equations and canceled out. We never discussed whether electrons are lost in electrical circuits - a strict conservation of electrons was always implied.
I asked a physics professor who teaches a class on electromagnetism for his thoughts, and he sent me the following reply:

The number of electrons is conserved if there are no losses or leakage. Probably what was meant is whether these are free electrons or not. Clearly, in a used battery you have less free electrons, since there is no more energy to strip them from whatever compound is used in the battery."

I sent this (as well as a list of other sources that I won't quote here) to my chemistry professor and received the following reply:

The question states electrons. Period. There are no "free" electrons in a battery (there can be delocalized electrons, but that's not the question). Batteries are made of atoms. Atoms are made of protons, neutrons and electrons. As a battery is used, through the flow of electrons, electrons are lost to the environment (fyi - there is energy/electron loss, albeit small to "run" the voltmeter and even in the flow of electrons through a conducting wire). Those electrons are no longer in the battery. Thus, the battery has the same number of protons and neutrons, but less electrons. This also means more unreactive metal cations exist in a used battery.
I appreciate all your research to make a point, but hopefully you now see the answer is true. Even your physics professor agrees because there is loss/leakage. Thus, less electrons in the battery.
Story: I have a family friend, who is a full professor of electrical engineering at Caltech. She is clearly on the cutting-edge of this field. In one of our discussions, she shared displeasure in online information. She told me her grad students often cited sources that were not true. There is more to this story, but I think the point has been made. Keep it simple. Electrons are energy. They flow. That energy goes elsewhere, leaving the initial system with less energy/electrons.

I was a bit baffled by the "electrons are energy" remark - this seems at best poetic. He seems to be conflating electrical potential energy with electrons, but these aren't the same thing. If electrons actually "were" the electrical potential energy in a battery, wouldn't that imply that the compound at the cathode would never actually be reduced, since all energy, and therefore all electrons that "flowed" through the circuit, would be lost after depletion of the battery's potential energy? My understanding is that electrons can have energy but are not themselves energy. Although I have a very low level of knowledge regarding this topic, I've done a few hours of research and found that the common notion of electricity as a flow of electrons akin to a river is wrong, and that although electrons do move very slowly through a circuit, the flow of energy is due to electromagnetic fields associated with charged particles.
Unfortunately, I could not find any sources that directly answered this question, so I would greatly appreciate direct answers to this question from experts on this topic.

Your reasoning is right, and "electrons are energy" is sheer nonsense.

True or false: “The OP is ready to say goodbye to GenChem and hello to physical chemistry”,

It seems really simple.  Your instructor acknowledges that "the battery has the same number of protons and neutrons" before and after.  So if the number of electrons were to decrease, the battery would acquire a net postiive charge as it's used.  That's simply not the case. Indeed, if depleted batteries actually did have net positive charges, they could be used to create an electrical current!  Perhaps you could ask your teacher to run this one by their family friend at Caltech.

I guess technically a used battery does have less electrons than a new one, simply by wear and tear of the casing. Any object that's not new will lose molecules over time by usage, including their electrons. But of course that's not the intent of the question

Shocking truth: you have been in touch with a teacher of chemistry, a teacher of physics and a full professor of electrical engineering at Caltech and all of them said something wrong (even totally wrong, in one case). I am not surprised as for at the core of the question there is just a chemical fact, a reaction has occurred. (corrosion and physical losses as mentioned by @Ivo Beckers were of course not intended to be considered). You should really show this thread to them - once you are done with them!

Occasionally in academia, as in real life, you will meet someone who is drunk, stupid or fraulent. It is sad, but it is real. One of the important lessons to learn is to not answer with facts that match reality, but with the facts that your professor believes in. This is why one needs to attend classes, to learn what the teacher believes to be true, rather than the simpler truth of actual facts. This becomes more important as the academical rank of your teacher increases. A mere graduate can be corrected when wrong. A tenured multi-doctorate believes they are right, despite all evidence.

@Ivan Neretin it's not nonsense. Any form of matter is equivalent to energy.

@infinitezero it is nonsense context given. I also wanted to comment that the discharged battery has lost an incredibly tiny amount of mass. But the way that teacher drop it in the discussion is nonsense.

@infinitezero technically true but irrelevant to the point of nonsense unless your battery is generating its power with nuclear reactions rather than chemical ones. That's the line that made it clear to me as a physicist that this prof does not understand the basics of electricity.

He is wrong. Don't let him win here. This is really a physics question, not a chemistry one, so garner support from the physics department professors. Let us know if your professor eats humble pie and adjusts your test mark. Hopefully he can find a field of work in which he is competant and take that up instead of wasting people's time and money teaching falsehoods.

@Alchimista  The prof. of EE at CalTech wasn't wrong; it's just the Gen Chem prof's phrasing might make it look that way.  I.e., I thought the Gen Chem prof was recounting a prior discussion with the EE prof, in which the EE prof was merely discussing misinformation on the internet generally.  When the Gen Chem prof then continued with the following quote, he was speaking in his own voice:  "There is more to this story, but I think the point has been made. Keep it simple. Electrons are energy. They flow. That energy goes elsewhere, leaving the initial system with less energy/electrons."

@Alchimista - what was wrong in the physics professor's response?

@Logicus he refers to some free electrons which are in a new battery and gone in a depleted one. There is no sense at all to invent description like this. There are no free electrons in the charged battery. He just wrongly reword that a new battery delivers current to an external circuits. But I agree is the less wrong of the three (assuming I got the clear picture from your description).

@theorist I see. Better it is so. This bullshit I expect from teachers but I was really surprised to hear that from a caltech professor. Indeed he shouldn't have been mentioned, even.

UPDATE: I'm pretty shocked. I wrote two emails to my professor, one for this one, and one for the other thread I posted. Combined, I wrote about 3,000 words. I wanted to be thorough so we could conclude this quickly. I just received the following reply: "Why can't you admit that my response to both questions made sense and are most accurate? Is it really worth this much of your time to attempt to prove a point?" That's it. Pretty unbecoming of a professor, I'd say. Any suggestions as to what to do next? Go straight to the dean? I'd rather not alienate him, but this is ridiculous...

@Logicus, well, I'm two days late, and I suspect this follow-up question would be much better asked in Academia.SE, but I would personally do nothing at this point. Your instructor is apparently quite unreceptive to correction, so you have to decide what exactly you hope to gain by pushing matters further, and whether it's worth the opportunity cost for doing so. The time and hassle, as well as potential retaliation, would not be worth it to me. What makes it worth it to you?

@a-cyclohexane-molecule I decided to not respond to his email and let it go for the time being. After our final exam (which is, fortunately, drafted by the ACS!) next week, I might bring it to the attention of the dean of natural sciences. As for my motive, I think I just care a lot about truth, and I can't stand to see people who should know better refusing not only to correct their errors, but also to even engage in reasonable discussion. I'm definitely not doing it for any personal gain, though I must say that the responses I've gotten on SE have been very encouraging.

Very bad explanation in the email response.
The explanation reads..

"Thus, the battery has the same number of protons and neutrons, but
less electrons. This also means more unreactive metal cations exist in
a used battery."

No, not at all. An electrochemical cell is not ionized. It is always neutral overall. However, the electrodes are indeed electrostatically charged (just like your charged comb). If you had a charge sniffer (e.g., a charge sensor from Vernier) and if you touch the positive end of the battery with such a sensor, it will indeed show that this end was electrostatically charged with a positive sign.
The negative pole of the battery is equally charged but with a negative sign. Overall an isolated battery as a system is electrically neutral.
Why don't you feel the charge like a charged balloon or a comb near a battery? The reason is the voltage is very low!
Second, point: No, electrons are not energy. Electrons in a wire/electrode behave very much like negative charge (see Hall effect if you are interested)
While you cannot say True or False for "A used AA battery contains fewer moles of electrons than a new AA battery."
The number of electrons before and after are the same for the reasons that a battery is electrically neutral overall, assuming no mass loss during the battery usage. For a closed ciruit, no electrons were wasted or lost. The electrons were just travelling from one pole to the other pole while traveling with a direction in the circuit.
A used battery is reaching equilibrium which means that the battery is no longer able to do useful work (on the electrons in the external conductor). The voltage starts slowly to drop as you keep drawing current. The current also decreases.
For example, keep a flashlight running on battery overnight and you will initially see a bright bulb going dimmer, dimmer and finally no light at all.

(+1) Wow, even if batteries were just charged supercapacitors, and they are not, that “electrons are energy” nonsense is wacky! What are they teaching these students now? I hope that teacher/professor reads up on free energy before “teaching” it!

The OP should share this answer with his instructor. I am glad there are still a few students who can challenge their teachers for their wrong answers.

assuming no mass loss during the battery usage -- it doesn'e look like the fellow agrees with this assumption

This question and accepted answer illustrate how a question about batteries actually had nothing much to do with how batteries work: it was about electricity, etc. Replace battery with DC power supply and it would have been the same. And the answer, that was accepted, was about the Poynting vector! ;-(

M. Farooq - thank you for taking the time to write that answer. I just sent an email to my professor summarizing all of the arguments presented here. I quoted your answer (with the exception of the first line!). I'll let you know what he says.
Ed V - would you mind elaborating on what you meant re the answer about the Poynting vector?

As I said in a linked comment, “This comes up fairly often over at the chemistry stack exchange. You can benefit greatly by getting this and working through it: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior of this famous galvanic cell. Look up the famous Daniell cell.” Understanding how cells and batteries convert chemical potential energy into electrical energy (cont)

does not require any consideration of Poynting vectors. In fact, Poynting vectors and classical electromagnetism are to batteries as DC power sources as the details of solar nuclear fusion are to home solar cells as DC power sources. One last thing: a used up battery simply has run out of chemical reactant, so there is no more free energy to be had. The electrons just drifted around the external circuit, as always, but dead batteries can no longer “push them” as it were.

Your professor is wrong, and you can prove it.
According to your "professor":

As a battery is used, through the flow of electrons, electrons are lost to the environment ... Those electrons are no longer in the battery. Thus, the battery has the same number of protons and neutrons, but less electrons.

which implies the charge of the battery would become more positive during use, which it doesn't, but let's see if it would be large enough to measure.
An AA battery has about 10,000 J of stored energy, and a voltage of 1.5 V.
We know Joules = Volts * Coulombs, so if he's correct, the static charge of the battery would be 10000/1.5 = 6666 Coulombs!
Why the exclamation mark? For comparison, rubbing a balloon on certain fabrics typically creates a static charge of only about 1 millionth of a coulomb (no single credible source; consensus of lots of searching), yet such a charge produces macro effects like making a person's hair raise when held near and can allow the balloon to "stick" to the ceiling etc.
Another comparison: Lightning bolts have about 100 coulombs (albeit at colossal voltage).
The battery would have a charge billions of times that of a ballon and many times that of a lightning bolt. We wouldn't need sensitive instruments to detect such a static charge; it would be leathal.

Thank you for providing this argument, Bohemian. I just quoted it (except the very beginning!) in an email to my professor. I suspected that there would be an absurd result like the one you describe if he were right. However, I expect he'll respond that not all of the electrons are lost, but only a small number. How would you respond to that? I'm thinking of just shifting the burden to him and asking for sources to support his theory. Our textbook certainly never discussed this issue one way or the other!

@logicus even a small number of "lost" electrons would create a very notable charge on the battery. The (very obvious) static charge on a fabric rubbed balloon is caused by the transfer of a very small number of electrons. Get your prof to do some calculations and see what he says.

@Logicus come on, is teaching chemistry after all. Ask him if the same story would happen in a reaction flask. He cannot stay so blind forever, even assuming he has no idea about electricity and free energy of reaction, at least it should be familiar with the notion of balancing reactions. Obviously, I am not encouraging you to be disrespectful. The important point is that is clear to you.

The disagreement may have come from mutual misunderstanding.
Unless a cell has lost some matter, it contains the stable count of electrons, safe random fluctuations and eventual net charge caused by external electrostatic potential.
Aside of that, some of these electrons relate/belong to electro-active substances taking part in electrode half-reactions, releasing/absorbing electrons to/from ekectrodes. These electrons are able to leave the cell via anode and the equal count to come to the cell via cathode.
The total count of these migrating electrons decreases as the cell capacity decreases by static or cycling ageing.

Misunderstanding is always possible. However the answer to the questions as formulated in the title is Not. Except for some part of it flying off. But the same could be said for everything, it is certainly not the meaning of the story.

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Not able to match the number of components in a system with the "species-approach" to the Gibbs phase rule
The Gibbs' phase rule states that the number of degrees of freedom in a system $F = C - P + 2$
where $C$ are the number of components and $P$ are the number of phases. The term $C$ can be expanded to $C = s-r$, where $s$ are the number of species, and $r$ are the number of independent relations.
I'm trying to find number of components for two systems, but the two approaches doesn't add up. Where am I wrong?
System 1:
A closed, evacuated container with $NH_4Br(s)$ is heated, and the following equilibrium takes place: $NH_4Br(s) ⇌ NH_3(g) + HBr(g)$.
I believe the number of components C to be 1, $NH_4Br(s)$, since this substance alone can produce the gas mixture phase, and the solid phase. ($NH_3(g) + HBr(g)$ can also do this, but C should be the lowest number of components needed..).
Using the species approach, $s = NH_4Br, NH_3, HBr = 3$, and $r = $ 1 equilibrium reaction. Which gives that $C = s-r=3-1=$ 2 which doesn't match what I wrote above. Why?
System 2:
A saturated aqueous solution of $KCl(aq)$ with $KCl(s)$. There is also water vapor. I also assume this is an evacuated container, but the problem doesn't mention this.
I believe the number of components to be 2, $H_2O$ and $KCl$, since these substances can produce the solid phase $KCl(s)$, the gaseuous phase $H_2O(g)$, and the liquid phase $H_2O(l)+Cl^-(aq)+K^+(aq)$.
Using the species approach, $s = H_2O(aq/g), K^+(aq), Cl^-(aq), KCl(s)$, and $r =$ one phase with ion-balance + one equilibrium between water liquid and gas + one equlilibrium between $KCl(s)$ and the dissolved ions = 3. Which gives that $C = s-r = 4-3 = $ 1, which again doesn't match what I wrote above.

If the conclusion of a premise is wrong, compared to the observed F, check if the premise is right.

Your method for defining the # components seems a bit arbitrary (think how you would scale it for complex multiple reactions, or reactions with even product/reactant ratio at equilibrium - who is produced by who?). Normally you would include all species, reactions-aside.

@Poutnik Well, I don't know the F, nor do I know what the conclusion is, as I have two answers. I've tried to think about the premise of both of them, but I can't see where I'm wrong.. That's why I ask the question.

@ViníciusGodim I'm going by the following description "The number of components, C, is the minimum number of substances (...) from which we could in principle prepare each individual phase of an equilibrium state of the system, (...)." Taken from the Libretexts article "The Gibbs Phase Rule for Multicomponent Systems " by Howard DeVoe.

You do know F, at least for system 1: F=1 as you can choose T or p.

@Poutnik Right, that's true. In which case, there has to be 2 components, because F = C - P + 2 = 1 and P is 3. So what then is wrong with the C = s - r part? "s" is certainly 4, and as far as I understand, all of the above mentioned "r" has to factor in..?

@Quantonium I added answer following the methodology of the reference I cite at the end. It doesn't use this sort of "adapting" the $C$ value. There are probably other ways to do it, like the one you referenced suggests, but I'm not sure how. I hope my answer can clarify anyway.

@Quantonium P=2, not 3 ( solid NH4Cl and gas ), so C=1. There are 2 constrains - the reaction and the 1:1 gas ratio, so C=3-2

@Poutnik Sorry, I was looking at system 2. The last thing you said makes sense. The 1:1 gas relation was the missing link for me. Thanks

The Gibbs phase rule for reacting systems is
$$F = 2 - \pi + N - r - s$$
Where $F$ is number of degrees of freedom, $\pi$ is number of phases, $N$ is the number of components, $r$ the number of independent reactions and $s$ the number of extra constraints on the phase rule variables. These constraints originate by the way the system is created, for e.g. in the reaction
$$\ce{NH_4Cl(s) -> NH_3(g) + HCl(g)}$$
If the only source of products is the reaction, there is the special constraint $y_\ce{NH_3} = y_\ce{HCl}$ ($= 0.5$ if these are the sole components). This counts as a constraint in this case ($s = 1$). (This example appears in the reference)
In the reaction
$$\ce{NH_4Br(s) <=> NH_3(g) + HBr(g)}$$
$\pi = 2$, $C = 3$ (note that all components are considered), $r = 1$. Given the constraint $y_\ce{NH_3} = y_\ce{HBr}$, $s = 1$. So $F = 2 - 2 + 3 - 1 - 1 = 1$.
In the system $\ce{KCl(aq)}$ plus a vapor phase, I guess you model either $\ce{K^+}$ and $\ce{Cl^-}$ as independent components, or you could consider $\ce{KCl(aq)}$ as a component in equilibrium with $\ce{KCl(s)}$.
1st case: $\pi = 3$, $C = 4$, $r = 1$ (dissolution reaction), $s = 1$ (constraint given by mole balance), $F = 2 - 3 + 4 - 1 - 1 = 1$.
2nd case: $\pi = 3$, $C = 2$ ($\ce{KCl}$ treated as a single component), $s = 0$ (there is no extra constraint on $x_{\ce{KCl}}$), $r$ = 0 (no dissociation considerd, $\ce{KCl(s) -> KCl(aq)}$ is modeled as phase change). $F = 2 - 3 + 2 - 0 - 0 = 1$.
Reference: Smith, Joseph Mauk. "Introduction to chemical engineering thermodynamics." (1950): 584. (Chapter 14 - Chemical-Reaction Equilibria)

Thanks, but wouldn't $\pi$ = 3 in the KCl system (2)? KCl(s), an aqueous phase, and a gaseous phase. Also, shouldn't the phase change equilibria enter somewhere (water -> gas, and KCl(s) -> KCl(aq) or $K^+/Cl^-$)

@Quantonium Yes, you're right $\pi = 3$. The phase change equilibria is considered in the phase rule equation already (in its derivation), they should not be considered chemical reactions.

I see. That's good, by this logic, my calculations in the OP also adds up.

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Equilvalence point in redox Titration of Fe(II) and Fe(III) with Reinhardt-Zimmermann solution
In the lab at university, we need to conduct a simultaneous determination of iron(II) and iron (III) ions using potassium permanganate solution.
In the first step, we just fill up our given sample solution to 100ml, and take 20 ml of it and fill it into an Erlenmeyer flask. Then we add 10ml of Reinhardt-Zimmermann-solution into the flask (my best guess is that our sample solution has hydrochloric acid in addition to iron (II) and iron (III)) and need to titrate with KMnO4-solution (1/50M) until the solution in the flask displays a very faint pink color. (determination of Fe(II))
The titration to determine Fe(III) is analogous, only that we need to reduce it beforehand.
But my problem is that I have a really difficult time recognizing red or pink colors (I have a condition), and when I did it the first time I didn't correctly identify the equivalence point. (Partly because the faint pink also discolours after a certain amount of time.)
My question is, if there is another indicator or extra chemical that I could add to the solution that makes the equivalence point easier to see and doesn't influence the reaction by a lot, the titration needs to be quantitative. I would appreciate any tips or help.

What about the ferroin redox indicator, provided commertialy AFAIK ?  Ferroin is the chemical compound with the formula [Fe(o-phen)3]SO4, where o-phen is an abbreviation for 1,10-phenanthroline, a bidentate ligand. The term "ferroin" is used loosely and includes salts of other anions such as chloride. // Reduced = intense red, oxidized = blue // Eventually, using o-phen itself, if available

The Zimmermann-Reinhardt solution has no chemical use in this titration. It is simply added in order to make a colorless complex with $\ce{Fe^{3+}}$ ions. This ferric ion is slightly colored brown - yellow. The end point is difficult to observe, without this Z-R solution, because it is not easy to see the first appearance of a pink color in a brownish solution. It is much better to observe this pink color if the solution is colorless.

Very good idea Poutnik. Indeed, when dissolved in 1M H2SO4 ferroin should have a redox potential of 1.07V which would be between Fe and permanganate, so after all the iron is oxidized it should oxidize the indicator


text:
Pressure dependence of the equilibrium constant
The equilibrium constant of a chemical reaction can be calculated if the $\Delta_\mathrm rG^\circ$ is known which is a function of only temperature. So if a reaction happens inside a container subjected to external pressure $p$, varying $p$ would have no effect on the equilibrium constant. I was reading this book which states:

$\mathrm d\Delta_\mathrm rG^\circ = -\Delta_\mathrm rS^\circ\,\mathrm dT + \Delta_\mathrm rV^\circ\,\mathrm dp$

It uses that relation to derive the pressure dependence of the equilibrium constant as:
$$\left( \frac{\partial \ln K}{\partial p} \right)_T = \frac{-\Delta_\mathrm rV^\circ}{RT} $$
What I can't understand is that the standard state subscript denotes a specific pressure. What is the meaning then differentiating standard change in free energy of a reaction with respect to pressure? As I said $\Delta_\mathrm rG^\circ = f(T)$
Can someone shed some light on what is going on the differential and why the equilibrium constant depends on pressure?

You need to define what type of $K$ you are using. By definition $\Delta G^\text{O}=-RT\ln(K_p)$ where $K_p$ is defined by the standard states of the gases, i.e. 1 atm. pressure so cannot depend on pressure.
If $K$ is measured wrt. mole fraction, for example, with partial pressure $P_i=x_iP$ for species $i$ then
$$K_x = K_pP^{-\Delta n}$$
where $\Delta n$ is the difference in moles product-reactants and $P$ total pressure. Taking the log of both sides of this last expression and differentiating (at const. $T$) wrt. $P$ gives
$$\left(\frac{\partial \ln(K_x)}{\partial P} \right)_T=\left(\frac{\partial \ln(K_p)}{\partial P} \right)_T-\Delta n\left(\frac{\partial \ln(P)}{\partial P} \right)_T$$
The first term on the right is zero because $K_p$ is independent of pressure, which leaves
$$\left(\frac{\partial \ln(K_x)}{\partial P} \right)_T=-\frac{\Delta n}{P}=-\frac{\Delta V}{RT}$$
assuming an ideal gas.
( When you use $K_p$ and the pressure is changed the degree of dissociation changes to compensate.)

Kp is the pressures at equilibrium not standard states.

@jmchmst,  No, it is defined wrt to free energies of species in the standard states of 1 atm. pressure.  Thermodynamics always examines the equilibrium  situation.

@porphyrin I am using the definition of K which includes activities. And also in the book as I said it differentiates the standard free energy change with respect to pressure. That is what it troubles me. How a derivative of a standard quantity with respect to pressure makes sense, if by definition, standard state refers to a fixed pressure?

It should all work out ok, the activity coeffs are constants so differentate wrt anything to zero.


text:
Problem understanding the rate of reaction of reversible reactions
Case 1:
$$\ce{CH_3COOH + NaOH<=>CH_3COONa + H_2O}$$
At equilibrium, there remain extremely small concentrations of the reactants (acetic acid and sodium hydroxide), and comparatively large concentrations of the products (sodium acetate and water). So, what are the rates of reaction at equilibrium and at other times?
My thought process:
At equilibrium, the rates of the forward and the backward reactions become equal. So, we can see that compared to the products concentrations, very small concentrations of the reactants are enough for the forward reaction rate to equal the backward reaction rate, when large amount of products are present. So, when similar amount of products and reactants are present, the rate of the forward reaction will be much higher than the rate of the backward reaction. Intuitively,
$$\text{forward reaction rate(small concentrations)=backward reaction rate(large concentrations)[at equilibrium]}$$
$$\text{forward reaction rate(similar concentrations)>>backward reaction rate(similar concentrations)[not at equilibrium]}$$
I think I'm correct here.
Case 2:
$$\ce{NH_3 + H_2O<=>NH_4^{+} + OH^{-}}$$
At equilibrium, there remain large concentrations of the reactants (ammonia and water), and much smaller concentrations of the products (ammonium and hydroxyl ions). So, what are the rates of reaction at equilibrium and at other times?
My thought process:
At equilibrium, the rates of the forward and the backward reactions become equal. So, we can see that compared to the products concentrations, very large concentrations of the reactants are necessary for the forward reaction rate to equal the backward reaction rate, when much smaller amount of products are present. So, when similar amount of products and reactants are present, the rate of the forward reaction will be much lower than the rate of the backward reaction. Intuitively,
$$\text{forward reaction rate(large concentrations)=backward reaction rate(small concentrations)[at equilibrium]}$$
$$\text{forward reaction rate(similar concentrations)<<backward reaction rate(similar concentrations)[not at equilibrium]}$$
My problem:
In the comment section of this post, @Poutnik generously pointed out that my explanation of case 2 is wrong.

The stronger the base and lower pKb, the weaker the acid and higher
pKa. And vice versa. Both pKa/b cannot be small. Ammonium dissociation
NH4+ <=> NH3 + H+ ( pKa ) is much weaker then ammonia dissociation NH3+
H2O <=> NH4+ + OH- (pKb )

@Poutnik Kind sir, if the ammonium dissociation is weaker than the
ammonia dissociation, then wouldn't that mean that the backward
reaction rate is lesser than the forward reaction rate in the
following reaction?  NH3(aq)+H2O(l)↽−−⇀NH4+(aq)+OH−(aq)  However,
that's not the case. Here the forward reaction rate is lesser than the
backward reaction rate. I'm a bit confused by your explanation kind
sir. If you have time, could you please explain it to me?

It is the case. Ka = [H+][NH3]/[NH4+] = Kw.[NH3]/([OH-][NH4+]) =
Kw/Kb.

I'm very confused now. All my life I thought that my explanation was correct. What do I do now?

I may somehow missed originally you talk about kinetics, not thermodynamics. // If you mean kinetic rates, then forward and backward reaction rates are equal at equilibrium ( reaction quotient is equal to equilibrium constant ). If reaction quotient is less than equilibrium constant, then the forward rate is greater than the backward rate. And vice versa.

But saying the above, I add that reaction rates being equal at equlibrium say nothing about the position of the equilibrium itself, which is defined by the respective dissociation constants.

@Poutnik In case 2, the reaction is shifted heavily to the left, right?

Do the math of that.

Please make sure your title uniquely and concisely reflects your question, or is the question itself. A complaint with a request makes a poor and clickbait'y header.

The core of the issue seems to be confusion about reaction of ammonium with H2O vs with HO-, which is the same as saying the effect of pH on the rate of dissociation of ammonium. In your case 2, ammonia is reacting with water, so the back reaction is reaction of hydroxide with ammonium.  But if the hydroxide concentration is very low (eg neutral or acidic pH), ammonium reacts primarily with water: $\ce{NH4+ + H2O <=> NH3 + H3O+}$ rather than $\ce{NH4+ + HO- <=> NH3 + H2O}$. The equilibrium of the  first reaction favors the left side. That of the second favors the right.

Let say the kinetic rate of neutralization
$$\ce{CH3COOH + OH- -> CH3COO- + H2O}$$
is $k_\mathrm{f}[\ce{CH3COOH}][\ce{OH-}]$
And the kinetic rate of hydrolysis is
$$\ce{CH3COO- + H2O -> CH3COOH + OH-}$$
is $k_\mathrm{b}[\ce{CH3COO-}][\ce{H2O}]=k_\mathrm{b}^{*}[\ce{CH3COO-}]$
Then the equilibrium constant of the neutralization is $K = k_f/k_b^*$.


text:
Graph of equilibrium constant vs 1/T

I have a problem with the 1st option, I did
$\Delta G = -RTlog(Kp(RT)^{\Delta n})$
$\Delta H - T\Delta S = -RTlog(Kp(RT)^{\Delta n})$
$log(Kp) = \frac{\Delta S}{R} - \Delta nlog(RT) - \frac{\Delta H}{RT}$
Here, log(Kp) vs 1 / T isn't linear because of that log term that contains T, but the answer key states that that option is correct

This would be the case if it were Keq vs 1 / T instead(or $\Delta n = 0$, but nothing is stated). Did they intend for it to be Keq or am I doing something wrong?

Why are you converting to Kc? The equation with $\Delta G$ is only valid for Keq, which for a reaction with gases is the same as Kp, not Kc.

ohhhh, so for gases 'Keq' directly stands for Kp, I thought the equation always holds true with Kc regardless of reactants because 'nature doesn't know that the reactants are gases, so it wont just suddenly follow a different definition for Keq'

It's slightly more complicated but you have the idea. Keq is a unitless quantity that comes from the "activities" of each reactant, which are defined relative to a standard state of 1 bar for gases, 1 M for solutes and 1 for solids and bulk liquids. You have to reference those standard states because that is the reference point for $\Delta G$


text:
Why does isolated chlorophyl emits red light?
If we observe isolated chlorophyl solution with UV light, chlorophyl is seen as red. Explanation suggests that when a specific atom of chlorophyl absorbs UV light, the atom gets excited and in a short time electrons still goes from excited state to ground state. But why it is not emits same wavelength light, and instead emits lower energy light - red ?

Many materials do that. Search for fluorescence.

Besides generally searching for “fluorescence”, see here and perhaps search in this stack exchange for other answers.

It isn't just isolated chlorophyll that does this. The effect can be seen with cheap handheld near-UV laser pointers and leaves.

Very basic and quick - I think is what you are asking for. As molecules are constituted of different part linked and vibrating, the upwards path can be closed once absorption occurred. Part of the energy is dissipated in rearranging the molecule and by vibrations. Then from a certain already lowered down level, emission occurs. As a starting point for more look indeed at the thread linked by Ed V. Wikipedia on the subject is also nice. Note that "isolated"guarantee emission, it is not functional to the fact that UV in / red out. A molecule which is not isolated can have its emission quenched..

... because the surrounding can interact with it and remove the otherwise "to be emitted" energy.

Perhaps illustrating what @Alchimista and matt_black said, see how quinine in tonic water fluoresces when 405 nm light illuminates the solution. The main emission is a band around 460 nm, not just at 460 nm, and the emission can be quenched by adding ordinary table salt: the chloride ions effectively steal the energy from the excited quinine molecules before they can get rid of the energy by emitting light. The joke is that only 10% of all substances fluoresce and they are the wrong 10%!

@EdV indeed the "guarantee" I have used above is with the context given. OP: Isolated molecules can be not fluorescent, or doing it with a very small efficiency nevertheless.

There is nothing particularly unusual about using UV light to excite fluorescence of various colors: lots of minerals fluoresce when exposed to shortwave or long wave UV. Some people collect such minerals and the fluorescent colors are useful for identification purposes. And, of course, Stokes discovered fluorescence in fluorite (natural calcium fluoride), giving the phenomenon its name. Ruby fluoresces at 694.3 nm when excited with violet or yellow-green light.

Stunningly beautiful fluorescent mineral photos here: https://www.uvsystems.com/pages/tgms-fluorescent-displays.

There are three types of re-emission when photons interact with matter:

Emitting the same wavelength, e.g., simple reflection, resonance fluorescence and stimulated emission.
Emitting longer wavelength due to losses in absorption of initial photon, e.g., fluorescence and phosphorescence and Compton scattering.
Emitting shorter wavelength than that incident. This occurs in two photon or multiphoton fluorescence. It requires the molecule stay in a metastable state long enough for a second photon to raise it to a higher yet state, requiring intense light. AFAIK, there is some little literature on UV-incident multiphoton fluorescence, but I don't know if chlorophyll has been investigated.

I think it will be better to call "emit same wavelength" condition as resonance fluorescence rather than reflection. Atoms show this phenomenon in flames. If we excite Na atoms in flame by 589 nm, they will emit 589 nm.

Upvoted, but it is not clear to me what is puzzling the OP. Is it the matter of the “chlorophyl atom” not behaving like a hydrogen atom, i.e., it can absorb UV light, but fluoresces red? Or is the red fluorescence itself the issue? I think the former is what they are asking about, but unless they clarify or elaborate, not much more to say, I guess.


text:
Is there a way to calculate the rate constant of electrochemical cells?
I'm doing an experiment on temperature's influence on electrolysis of iron oxides and I need a way of measuring rate, preferably rate constant (k) or something similar. I've done some googling and there doesn't seem to be any calculations for rate constant for electrochemical cells or anything similar. The closest I've got was to use the Arrhenius equation and substitute the collision frequency/diffusion flux for the pre-exponential factor, but it doesn't consider voltage as an factor and I'm not sure if it's valid.
Can anyone tell me if electrochemical cells have a rate constant, and if so, how do I calculate it? If not, is there something similar?

Current times time equals coulombs of electricity. This must be related to the actual electrode reaction rates. You will have to do a voltage-rate- efficiency study and deconvolute the data.  This is done all the time in electroplating.

Thank you for your comment. I'm assuming you're referring to Faraday's laws of electrolysis. I did think of using that by just measuring the current, but I haven't heard of voltage rate efficiency studies so I need to do some research into that.

Electrode half-reactions have rates, not cells. See also https://en.wikipedia.org/wiki/Tafel_equation

Note also rates of fast electrode reactions are controlled by ion diffusion rate and rates of slow ones are controlled by the rate of electron transfer.

The reaction rate in a cell depends mainly on the outer resistance of the electric circuit. If the outer resistance is high, the current delivered by the cell is tiny, and so is the reaction rate. If the outer resistance is small, or even equal to zero, the current is maximum, the cell is not a  cell anymore, as the outer voltage is zero. Then the reaction rate is maximum and depends on a lot of factors like the area of the plates, the distance between them and the concentration of the solutes.


text:
Why this structure is diastereomers instead of enantiomers
In this diagram optical isomers has divided in to 2 categories, one is diastereomers and enantiomers. Is the example for diastereomers correct? For me its looks like enantiomers

What do you think?

Are the carbons bearing the Br the same?. What about the carbons bearing Cl?

I'm not sure why you think they are enantiomers. They are actuallually diastereomers as stated:

As depicted, the compound to the left is $(2R,3S)$- while the compound to the right is $(2R,3R)$-. Thus, they are pair of diastereomers.
To be enantiomers, either the compound to the left should be $(2S,3S)$- or the compound to the right should be $(2S,3R)$-.


text:
What does it mean for an electron to be in a bonding/antibonding orbital?
My confusion arises from the fact that bonding and antibonding orbitals are created simultaneously. I have read this answer that clarifies about the existence of the molecular orbitals itself.
But now I want to know what does it mean for an electron to be in a bonding or antibonding orbital. Are phases a property of the electrons or of the orbitals? If it is a property of the orbital itself, then how are they constructive and destructive at the same time? If it is a property of the electron, how can a single electron occupy an orbital and still be antibonding/bonding, even though it is not comparing phase difference with anything?
What if the electrons of the two atoms forming the bond have the same spin? Will one electron occupy bonding orbital and the other the antibonding orbital? Or will the covalent bond not be formed at all?
I believe I've understood the mathematical reason for this involving orbital functions (only the square of the function is physically significant), but I want it to understand the physical implications of that on the electrons

I admit is not a great piece but does this comment which I took from the thread you have mentioned says something to you? “While this does not explain why the orbitals combine that way, it might still be helpful if you take in mind that the molecule is hold by only one. In this sense the molecule is not formed by both constructive and destructive interference, that concerns the orbitals as functions, which include l a phase like in the sketch.“

@Alchimista I'm not sure if I understand it. Are you talking about how only the bonding orbital is usually occupied in the molecule?

yes, there is nothing simultaneously happening.

@Alchimista Well, that's the usual scenario. But I've read that it's even possible for scenarios like $\ce{H2-}$ ion to exist with 2 electrons in the bonding orbital and 1 in the antibonding orbital

how this relates to the question?

As of right now, I have concluded to myself that orbitals are a property of electrons. And when electrons form molecules, they have new orbitals to occupy. The formation of these new orbitals is through bases change as mentioned in an answer. And an electron can occupy only one molecular orbital at a time. I say, the electron behaves in a way favorable to both atoms in case of bonding orbital, and a way unfavorable to both atoms simultaneously in case of antibonding orbital

@Alchimista You spoke about how a molecule is held only by the bonding orbital. I misinterpreted that as only bonding orbital is filled, and gave that example

I do not understand your question, so the best that I can do is show you how I think about orbitals. Broadly speaking, we want to minimize the energy of our total electronic ground state wavefunction $\Psi$. This condition leads ultimately to a set of molecular orbitals with fixed phases and the atomic orbitals serve as input for this procedure. I'll try to give you a rough idea of the procedure in the following.
The wavefunction itself can be expressed as a linear combination of products of molecular orbitals.
$$
\Psi(r_1,r_2) = \sum_{ij} c_{ij}\psi_i(r_1)\psi_j(r_2)
$$
The $\psi_i(r),\psi_j(r)$ are chosen from the set of molecular orbitals. This set includes the bonding and antibonding molecular orbitals. These molecular orbitals are themselves constructed by linear combination of atomic orbitals that we know from solving the hydrogen atom. We usually start with a set of atomic orbitals based on the atoms in our molecule and this atomic orbital basis must provide at least one orbital per electron. Lets limit ourselves to a minimal basis set of atomic orbitals. We have 2 Hydrogen atoms so only 2 electrons in total. We use the $s$-orbitals of both atoms as our atomic basis. Lets call the $s$-orbital which is centered on hydrogen atom A, $s_A(r)$ and the other $s$-orbital, centered on atom B, $s_B(r)$. Our atomic basis is thus the set with two elements, $\text{Atomic orbital basis} = \{s_A(r), s_B(r)\}$.
The problem that we want to solve is finding the electronic groundstate of our hydrogen molecule H2. This problem gives us equations and conditions for the optimal orbitals, the so called molecular orbitals. Covering the details would go to far. But we can use for starters the so called Hartree-Fock method to determine which orbitals would give us the lowest energy. When you solve the Hartree-Fock equations you obtain the set of molecular orbitals. Using our minimal atomic orbital basis as input to the HF-calculation, we would obtain the following spatial molecular orbitals,
$$
\psi_1(r) = \frac{1}{\sqrt 2} (s_A(r)+s_B(r))\\
\psi_2(r) = \frac{1}{\sqrt 2} (s_A(r)-s_B(r))\\
$$
These functions form the $\text{molecular orbital basis} = \{\psi_1,\psi_2\}$. Up to this point I have neglected spin. The orbitals that I have constructed so far do not include spin. Without going into detail we can again use these spatial molecular orbitals to construct spin-orbitals, that include spin. To do so, we simply multiply the spatial orbital with a spin up or spin down function,
$$
\phi_1(r,\omega) = \alpha(\omega)\psi_1(r)\\
\bar \phi_1(r,\omega) = \beta(\omega)\psi_1(r)\\
\phi_2(r,\omega) = \alpha(\omega)\psi_2(r)\\
\bar \phi_2(r,\omega) = \beta(\omega)\psi_2(r)\\
$$
and lets update my Ansatz from above to include these spin-orbitals
$$
\Psi(r_1,r_2) = \sum_{ij} c_{ij}\phi_i(r_1,\omega_1)\phi_j(r_2,\omega_2)
$$
Our set of molecular spin-orbital basis functions is thus $\{\phi_1(r,\omega), \bar\phi_1(r,\omega), \phi_2(r,\omega), \bar\phi_2(r,\omega) \}$ and the indices $i,j$ move over this set.
The electronic singlet groundstate is then obtained by taking the determinant of the product of the lowest energy molecular spin orbitals, where we have to include one orbital for each electron and each spin-orbital may only appear once.
$$\begin{aligned}
\Psi(r_1,\omega_1, r_2,\omega_2) &= \frac{1}{\sqrt 2}\det \left[\phi_1(r_1,\omega_1)\bar \phi_1(r_2,\omega_2) \right] = \frac{1}{\sqrt 2} \left[\phi_1(x_1)\bar\phi_1(x_2) - \bar\phi_1(x_1)\phi_1(x_2)\right]\\
&= \frac{1}{\sqrt 2}\psi_1(r_1)\psi_1(r_2)[\alpha(\omega_1)\beta(\omega_2)-\beta(\omega_1)\alpha(\omega_2)]\\
\Psi(r_1,r_2) &\propto \psi_1(r_1)\psi_1(r_2)
\end{aligned}$$
Lets neglect spin again for a second and look only at the spatial parts. You can see that this groundstate function only includes the spatial orbital $\psi_1\propto s_A+s_B$.
We can compare the energy of this wavefunction with a simple ansatz where we keep the electrons in the atom centered orbitals
$$
\Psi_{\text{non-bonding}}(r_1,r_2) \propto s_A(r_1)s_B(r_2)
$$
This would describe two Hydrogen atoms that are not interacting and each electron is always located on a single hydrogen atom. Unlike the actual ground state where each electron is in a binding spatial molecular orbital, a spatial superposition state with equal probability for an electron to be found on either hydrogen atom and
in-between both atoms, due to the positive sign/constructive interference of the underlying linear combination of atomic orbitals.
After this very lengthy exposition I'd like to address bonding and antibonding molecular orbitals. You saw that we obtained two spatial molecular orbitals, $\psi_1$ and $\psi_2$. The antibonding orbital $\psi_2$ is a mathematical necessity when we construct the set of spatial molecular orbitals from the set of atomic orbitals. The whole procedure is just a change of basis sets and such a change does not change the number of basis functions. We used the atomic orbital basis $\{s_A, s_B\}$ to construct our new molecular orbital basis $\{\phi_1, \phi_2\}$. You will always have the same number of functions in both sets. The coefficients that define the new molecular basis functions in terms of old atomic basis functions are fixed by the condition to minimize the energy of the final total groundstate wavefunction and the equations that follow from this condition.
If you truly want to understand orbitals I recommend that you try to get a good understanding of linear algebra, especially basis sets and how a change of basis works, then you need some basic quantum mechanics especially the hydrogen atom and what happens when you go from a single fermion system to systems with multiple fermions. Then you should take a closer look at the Hartree Fock method.
Qualitative hand wavy explanations will always leave some confusion and not a quantitative understanding. My understanding of orbitals improved by going through the math. All the qualitative explanations that I had gotten before typically only caused more confusion.

Thanks, this was helpful to some extent. My question was basically about what is the physical significance of the phases and the interactions between them. And I'll look into the things you've mentioned, I've already started reading a book on the basics of quantum mechanics

@Balu phase has no direct physical meaning (similar to the wavefunction itself). But it allows for interference, as observed. By the way I do not see how this answer above can be useful to you but at least its author said that at the very beginning.

@Alchimista Yeah, I meant the physical implications. Like wave function can be used for probability density. The above answer didn't answer my question, but it helped me understand the mathematical part better

@Balu and at least the phase account for interference without affecting spatially the distribution of probability for the base orbitals. I know it is not high level, but it is the implication of phase.


text:
what do the asterisks in the molecule stand for?
My 7 year old nephew found a chemistry text book and started copying some pictures and diagrams.  He drew this and asked me what do the asterisks mean ?  He said he normally sees letters which means atoms.  But he can't found an atom that called "asterisks" from the periodic table.

Asterisk some times indicates an excited molecule in spectroscopy. It is not an element. In this particular case, this asterisk appears to have a decorative purpose only.

Your nephew is apparently showing a polymer, with -CH2-CH2- repeating units.

We shall commit to enroll him in a couple of years

It is not an asterisk. It means "repeating the central unit -CH2-CH2- a lot of times, without any changes".

@Alchimista, he seems more prepared now that many other questioners. "Give me the child and I'll give you the chemist," as Aristotle stated (or was it Democritus? Dalton??).

M. Farooq gave answer in his comment above as: ```
Your nephew is apparently showing a polymer, with -CH2-CH2- repeating units.

Not sure you shall tell him, perhaps just a note for you. The way it is represented is somewhat technological, as it stress that the original  monomer had 2 C. Once formed, the polymer can be represented as - (CH2) -. The same happens with names, most accepted names refer to the starting material rather than the formal repeating unit. Another note, the terminals would be some other groups.

FWIW, the red dots are meant to represent the valence electrons.  In this structure, all the valence electrons are bonding  electrons (involved in bonds).  One pair of bonding electrons shared between two atoms gives a single covalent bond.  Two pairs of shared electrons gives a double bond, etc..  All the bonds in this molecule are single bonds.  You should ask him to double-check the picture in his book.  I believe he has an extra pair of bonding electrons between the two carbons.   This causes each carbon to be surrounded by 10 valence electrons, when it should be eight.

"I told my nephew, and then my nephew replied, "oooh!!! It's because you can't put infinite number of polyethene in this book! It will be TOO BIG! Bigger than the universe!"  I think your nephew misunderstood "repeating".  It sounds like he took it to mean infinitely repeating.  That's obviously not the case with a real molecule.  Thus the reason the entire polymer structure can't fit onto the page is because it is big, not because it is infinite.   Plus, conceptually,  once you've show that it's repeating, there's no reason to show more units, because they're just more of the same.

@theorist though theorists model conjugated polymers as infinite chains, it please them and makes everything more fancy. :)

@Alchimista  I work with biopolymers rather than conjugated synthetics, but I actually model mine as finite chains.  So maybe that means I'm at least partially grounded in the real world :).


text:
How can surface excess be negative?
If we have interphase of two phases $\alpha$ (water) and $\beta$ (air), surface excess is defined as
$$\Gamma_i = \frac{n_{i,\mathrm{tot}} - n_{i,\alpha} - n_{i,\beta}} A = \frac{n_{i,\mathrm{int}}} A,$$
where $n_{i,\mathrm{tot}}$ is total amount of some substance added in the system which can be some salt since in that case surface excess is defined as negative, and $n_{i,\mathrm{int}}$ is the amount of substance on the interphase.
Surface excess is the amount of some substance present on the interphase (salt or surfactant) divided by surface area $A$ of the interphase.
However, taking into account definition of surface excess, how can surface excess be negative when the amount of substance can't be negative number?

It may help if there is included the source and broader context of the equation.

http://www.jestr.org/downloads/volume1/fulltext1.pdf I must have forgotten this stuff because I did study adsorption phenomena at the university. After reading this I am still puzzled, but at least it seems that Surf Exc can be positive as well negative. Anyone able to explain the concept different than Wikipedia please do it.

You don't seem to have problems with positive surface excess, which by the same reasoning would violate the conservation of mass just as well, only with different sign. How so?

Conservation of mass does not play any role. Is the concept that we are missing (we I mean OP and I, at least). It is not clear to me what the n refer to, which volume. Perhaps n alpha and beta are those within the interface volume (across it, which has a thickness).

@Ivan Neretin Positive surface excess doesn't violate law of mass conservation. For example, if you add 1 mole of surfactant in water/hexane system, 0,2 mol go in water, 0,4 mol go in hexane and 0,4 go on water/hexane interface. Everything is okay with mass conservation.

It seems to be the surface excess is in the formula incorrectly formulated. It should be the difference to the expected n/A if the component was evenly distributed, which can be both positive and negative. E.g. the nonzero volume interface would have 1/100 volume of each phase,  expected to contain 1/100 of analyte in case of zero excess. Deviations from this 1/100, normalized to the unit surface, would be the surface excess.

@Poutnik It isn't. You can check Atkins:Physical Chemistry or Wikipedia. I think I've got what is actually going on and I've written the answer to my own question.

Negative surface excess is based upon idealized model of the interphase in which interphase has no volume. If we add 1 mol of NaCl in 1 L of water, concentration of NaCl is 1 mol/L.
However, if we tried to calculate number of moles of NaCl in water by measuring it's concentration, we would find that concentration measured is more than 1 mol/L because of which it would appear that number of moles in water is more than number of moles added.
This calculation of number of moles of NaCl in water is false because in reality interphase has volume and so volume of water is actually less than 1 L when interphase is formed.
If this is taken into account, surface excess isn't and can't be negative, but it appears to be when we want to calculate it based upon measuring concentration and idealized model of the interphase.
Gibbs adsorption isotherm is based upon idealized model of the interphase because of which negative surface excess has practical value, but it is important to note that in reality it isn't actually negative since number of moles can't be negative.


text:
Software to Draw Laboratory Apparatus
So I am preparing a blog post about chemistry for lay people and I have decided to start talking about the basics. It turns out that it would be super useful to create chemical apparatus on demand to illustrate the post. Such as this one:

I particularly like the clean vector image look. This is what I am looking for. A software that allows me to assemble these vector-graphics style 2D chemical apparatus as well as let me control the color (and texture, if possible) of the contents inside the flasks.
It is meaningful that the software contains correct laboratory glassware, preferably the classical stuff that is instantly recognizable like the bunsen burners and tripods rather than heat mantles, but that is me being picky. So far, anything goes.
Any ideas?

If you don't find an answer here in a few days, try posting in https://softwarerecs.stackexchange.com/

@DrMoishePippik Thank you, I will try that as soon as a few days pass. On the other hand, do you think it would be harmful if I had both posts going simultaneously?

though some resent cross-posting, see https://meta.stackexchange.com/questions/64068/is-cross-posting-a-question-on-multiple-stack-exchange-sites-permitted-if-the-qu

@urquiza: ChemDraw has a lot of templates for chemistry labs.

Likely you intend "clear" to apply to the graphic. Once you'll succeed on putting up the pics for your blog I would skip the rack and especially its joints as I am pretty sure they can be mixed up with pipes and valves.

Bitonal illustrations e.g., with ChemSketch (demo video, or the colorful templates from Chemdraw (sample page by the stamp icon (before last, top row)). Subscription based chemix follows a little bit different approach (liquids may have adjusted colors; the meniscus adjusts if the container is tilted; export as jpeg/png/svg). But the more modern the lab is, the less often you encounter open flames (e.g., Bunsen burner) for extended heating below a flask ...

@Buttonwood That is very close to what I was thinking. The only thing is that it can't show liquids boiling, for instance. Anyway, would mind publishing it as an answer? I am thinking on accepting your answer but publishing a new question at softwarerecs.stackexchange.com.

@urquiza In the past days, there were other priorities than chemistry.se; now I added an illustrated brief overview as an answer.

The short answer to this topic is that the typically used programs for this
tasks vary in their coverage of of lab ware to prepare such illustrations and
focus on drawing molecules, exporting molecules in machine readable (chemical)
formats understood e.g., by databases, and performing some computations (e.g.,
averaged molecular and isotopic weight).  Beside popularity of the programs if
you want to share/collaborate with colleagues intermediate files, an additional
point to consider are the graphic formats these programs offer for file I/O.
Thus, a native export of the .svg formats understood e.g., by
inkscape may be an advantage.
Without aim to provide an exhaustive description, the following examples may
illustrate this with templates and building a distillation.

ACD ChemSketch contains quite a number of lab utensils in the template library.  The free (as in free beer) version disables some functions which however are not relevant to drawing the beakers, flasks, etc.  The representation of them did not change for decades, however the Windows program offers a native export as .png and .pdf.  Recent releases improved interaction with   wine to equally work well enough in Linux, too (this includes the current version 2021.1.3).  It may take some tinkering to adjust the the individual pieces' orientation to build a setup.
Overview of the chemical lab utensils:

A distillation:

ChemDraw
offers templates which may be bitonal, or in color (see, e.g., here. Primarily written for Windows and Mac, with only varying success to be deployed in Linux, the program is widely used in academia and industry (definitively not for free as in free beer, often accessed within a campus license).  The templates include parts aligned to fit better into the round bottom flasks.  Among the export formats are .png and .svg.  The later allows you e.g., to adjust the fill and stroke of the  paths, or to remove the ace label (which actually is a trade mark of Ace Glass, NJ).
With many chemistry-relevant functions removed, the ChemDraw JS page allows to get familiar with them (stamp button opens a pull-down menu), to save the   drawings in the native format (Structure -> Get .cdxml), as .png (-> Get   image) or vector file (-> get .svg).
Some of the templates:

(image credit to a Russian blog post)
A distillation (color adjustments with Inkscape):

ChemDoodle is the youngest of these three   sketchers with the largest number of lab utensils in the template library.    Capable to interact with many chemistry-relevant file formats (including the
public .cdxml of ChemDraw), the export of the graphics includes many options
for round-trip edits, and export e.g., as either .png, or (optionally   layered) vector format (.svg, .ps, .pdf) and anticipate their use in web pages and services like twitter.  The purchase of one of their licenses offers the user to choose between a program for Windows, or Mac, or Linux; this includes the option to shuttle the license key among the operating systems.
An overview of the chemistry templates:

With light retouches in inkscape, an illustration of a short-path distillation:

Contrasting to the programs above, chemix's focus is about drawing a lab setup exported either as bitmap or vector file.  (Maybe drawing organic structures will be added.) By number, the inventory of lab utensils (still) is smaller than e.g., the one offered in ChemDoodle, though it contains material absent in the other collections (e.g., a waterless condenser, or the GHS symbols).
In addition to standard options to move and scale the objects, there are interesting details in handling the objects like (incomplete list):

joining the elements is guided by snap-points like magnets
both color and height of liquids in the containers may be adjusted within the interface, including boiling-like bubbles
a tilt of the container automatically affects the meniscus of the liquid
changing the height of the lab boy affects the scissoring

The green arc sign in the illustration below mark utensils you access when entering a paid subscription.  Based on their twitter feed, there is continuing development and addition of utensils for this application running remotely in your web browser.
An illustration:

The comparison with the utensils in the lab may reveal differences between the sets offered (e.g., Chemix' missing pressure release for a distillation present in Chemsketch and ChemDraw's sets/how you should mount safely a distillation) may be seen; thus, design with care for detail.

@Buttonwood That was a stupendous answer. However, I tried Chemix and wasn't able to put the boiling liquid mask to my fluids inside the glassware. Is that a premium feature?

@urquiza The seventh category «Accessories» contains bubbles, which equally are available in the free section. As colour, «default» seems to blend nicely if this rectangle is stashed behind the beaker/round bottom flask in question. This may require some adjustment of rows (i.e., height) and bubbles per row (i.e., width) of the «bubble rectangle» to fit for the (adjustable height) of liquid in the container (which may be moved as usual by the mouse, too).


text:
Characteristics of metals compatible with semiconductor grade Ultra Pure Water (UPW)?
I am looking to build a water conductivity sensor for semiconductor grade Ultra Pure Water (UPW). UPW conductivity sensors already exist, but because of price and low water volume I need to fabricate my own.
The electrode material that the commercial product link above cites is Titanium. I do not know the alloy or the surface treatment. Typically in mechanical components constructed of titanium are anodized (like aluminum) to protect the relatively reactive metal. Not sure if this is the case for these electrodes however, as thick oxide layers would greatly reduce the conductivity.
These sensors are also likely used for high volume processes and any contamination from this small area may be considered inconsequential. So even though they are "industrial grade" they may not be satisfactory for my lab application.
The Ultrapure Water Wikipedia page only mentions compatible plastics.
Questions:

Is it possible to calculate the contamination rate of polished titanium given the electrode
areas (with assumptions of course)? With a titanium dioxide layer?
Are other metals that are lower on the reactivity series, such as Gold or Platinum, better low-contamination electrode choices?

It is counterintuitive, but "pure" water is relatively aggressive to metals. Normally is is contained in rubber lined vessels or plastic vessels.

I understand, hence the question of how to select an electrode. Any thoughts on how to calculate the rate ions are lost for different metals?

Perhaps use an alternating current conductivity tester would be more suitable: see https://earthscience.stackexchange.com/questions/13363/are-there-any-diy-conductivity-sensors-that-will-hold-calibration-for-long-perio At a high frequency, the electrodes could be completely insulated by anodizing, coupled through capacitance to the water.

@DrMoishe Pippik, Thanks for the information and the DIY link! I think the problem with the alternating current solution is that the conductivity of ultrapure water is so low that the influence of ambient RF noise and metal across the room would possibly be larger than the signal itself. Might be able to improve it with some shielding but this would be a difficult route. Any thoughts on electrode selection for direct measurement?

@blacksmith37 Please stop perpetuating this myth. The reason why ultrapure water is stored in plastic containers is not because of high corrosion levels for metals. It is because even a very tiny amount of contamination by metals renders the water no longer ultrapure.

In order to replicate a commercial sensor, you are going to need to copy what they do or replicate the R&D commercial suppliers put into their sensors. That sounds like a lot of work and might cost you more than buying a sensor. Plus, creating ultrapure water is also expensive and reliant on a great deal of experience. Doing it yourself on the cheap doesn't sound like a good idea.

@matt_black, How can I better phrase my question to help me select an electrode material? I'm a mechanical PE, and fabricating the sensor is not a problem. Generating a small amount of ultra pure water is easily achieved by polishing with mixed bed ion exchange resin. The part missing in my understanding is electrode material selection and contamination rate for ultra pure water.  I find it unlikely that only the commercial sector understands this.  I am probably just searching the wrong vocabulary. Is there better terminology to describe this process?

Titanium grades 1 to 4 are unalloyed titanium; Grade 1 has the lowest level of impurities. However, I must agree with other comments , the lowest final total cost will be a commercial unit.

@ericnutsch I think the title already asks a better question. But I suspect that, even if you get better theoretical answers, they won't be of practical use.


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Solubility of nitrogen dioxide in sodium hydroxide
I know $\ce{NaOH}$ can absorb $\ce{NO2}$, but I could not find any data of solubility of $\ce{NO2}$ in $\ce{NaOH}$ at room temperature and atmospheric pressure.
I am trying to figure out how much NaOH solution I will need to absorb the $\ce{NO2}$ gas being produced from a result of another reaction. My setup is having $\ce{NaOH}$ solution in a beaker (how much is what I want to calculate) and feed in the $\ce{NO2}$ gas into the solution. Is there any way to theoretically calculate the quantity of $\ce{NaOH}$ I will need to absorb the $\ce{NO2}$ gas without solubility data?

NO2 reacts with NaOH, forming NaNO2  + NaNO3, so solubility does not make sense.

You may find useful these links for text formatting ( not to be applied to titles ):
notation ,  formulas/expressions/equations  and
upright vs italic

@Poutnik Relevant line of argument: https://chemistry.stackexchange.com/questions/42696/should-bromine-water-be-called-a-solution

@NilayGhosh Well, bromine water is in very majority a true solution, with the equilibrium strongly shifted to the left. For NO2/NaOH, it is the opposite.

By other words, practically, it is possible to achieve saturated solution Br2(l)<=>Br2(g)<=>Br2(aq), but not NO2(g)<=>[NaOH]NO2(aq).  NO2 or NaOH will be spent.

A numerous studies have shown that the absorbtion of $\ce{NO2}$ in alkaline solutions is due to the reaction (For example, Ref.1):
$$\ce{2NO2 + 2OH- -> NO2- + NO3- + H2O} \tag1$$
Thus, as Poutnik pointed out, you may use dilute solution of $\ce{NaOH}$ solution, if you preferred. For insterd, Ref.2 has used sodalime, mainly $\ce{Ca(OH)2}$, in its solid form (following is the method in the paper for your benefit):

The test gas containing $\ce{NO2}$ $\pu{57 ppm}$ in high purity nitrogen and a trace of oxygen was passed at a flow rate of $\pu{1 litre min-1}$ through the column filled with soda lime $\pu{120 g}$ (Soda sorb) and the $\ce{NO2}$ concentration was measured continuously at the outlet of the column by an electrochemical fuel cell technique (TM-100, Taiyo-Sanso, Co., Ltd, Osaka). Break point was determined as the time when $5\%$ of the initial concentration of $\ce{NO2}$ was detected.

The authors have used two different varieties of soda limes, Soda sorb (which has the chemical composition of $73\% \ \ce{Ca(OH)2}, 5\% \ \ce{KOH}, 3\% \ \ce{NaOH}$, and less than $19\% \ \ce{H2O}$  with ethyl violet colour indicator) and Wako lime-A (which has the chemical composition of $81\% \ \ce{Ca(OH)2}, 4\% \ \ce{NaOH}$, and $15\% \ \ce{H2O}$ with ethyl violet colour indicator). The difference between these two types is the absence of $\ce{KOH}$ in Wako lime-A.  The authors have claimed that they have shown that soda lime completely absorbed $\ce{NO2}$ regardless of the type of soda lime used. Thus, it is safe to suggest that the only necessity is the presence of $\ce{OH-}$ in the system.
Further, according to Ref.2, following equilibrium would establishe in aqeous medium:
$$\ce{2NO2(g)  + H2O (l) <=> HNO2 (aq) + NO3- (aq) + H+ (aq)} \tag2$$
The following is the best result of particular experiment measuring the absorption of $\ce{NO2}$ (for example, flue gas used: $\pu{800 ppm}$ $\ce{NO2}$ in $\ce{N2}$ with flowrate of $\pu{1 L/min}$) in caustic aqueous scrubbing solution, which, for example, is $\pu{1 L}$ of $\pu{0.5 M}$ $\ce{NaOH}$ for the best result (all experiments have been conducted at room temperature and pressure). The average percent absorption $(n = 3)$ is reported as $98.1\%$ in $\pu{0.5 M}$ $\ce{NaOH}$ solution. For compartion, it was reported as $89.9\%$ in $\pu{0.1 M}$ $\ce{NaOH}$ solution and $81.8\%$ in $\pu{0.01 M}$ $\ce{NaOH}$ solution. Thus, even though change in absorption varies from different concentration of $\ce{NaOH}$ used, the difference is close enough to say concentration would not be effective if you have increased the time of exposure of gas to the scrubbing solution.
References:

T. Ishibe, T. Sato, T. Hayashi, N. Kato, and T. Hata, "Absorption of nitrogen dioxide and nitric oxide by soda lime," British Journal of Anaesthesia 1995, 75(3), 330-333 (DOI: https://doi.org/10.1093/bja/75.3.330)(PDF).
Chen-Lu Yang, "Aqueous Absorption of $\ce{NO_x}$ Induced by Sodium Chlorite Oxidation," Master's Thesis 1989, New Jersey Institute of Technology, NJ.


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Question regarding Entropy
I had some doubts regarding the concept of entropy. My first doubt was

Entropy is defined as $ dS = dQ _{rev}/ T $ yet my textbook says that entropy is a state function. Isn't it the definition of a state function that we can define its value uniquely depending on the current system with no regards to the previous states of the system, yet in the formula for entropy we need to find the change in its heat energy which is not uniquely determined by the system and we need the previous state of the system to define it properly.

My course says that entropy is strictly defined only for reversible paths, that is all processes done on the system must be strictly reversible, however there are no such restrictions for that on the surrounding. What is the reason for such a definition of entropy and why we can't define it for irreversible processes on the system? The only reason my course if providing is that because the concept of entropy is derived from the working of a Carnot Engine, where the system has reversible processes done on it, we confine the definition of entropy as such as well.

Another doubt that I had was if we do want to calculate the change in entropy of a system if we  provide it $ dQ$ heat reversibly, how do we ensure that the temperature of the system does not change, since for calculation of $ \Delta S$ they only consider the change in heat energy and not that of temperature.

If you consider the alternate definition, S = k ln W, it is easier to see that entropy is a state function.

For 1, the essence is this: A state function depends on states. However, accessible states could depend on path, so state function appears to be dependent on path.

To get the change in entropy between equilibrium states of a system, you do not need to know the absolute value of the entropy in either state.  All you need to know is the integral of dQ/T for a reversible path between the states.  Entropy is a state function even though Q depends on path because the integral of dQ/T is the same for all reversible paths between the same two end states.
Does your book say that entropy is defined only for reversible paths, or does it say that the integral of dQ/T is equal to the entropy change only for a reversible path?  I think the latter.  Entropy change is defined for an irreversible process between two thermodynamic equilibrium states.  But to determine the entropy change of the system for such a process, you need to devise an alternate reversible path for the change, and calculate dQ/T for that reversible path.
The integral of dQ/T includes the changes in T over a reversible path, and does not require that T be constant for the reversible path.


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Is it okay to use indicator as a powder instead as a solution in titration experiments?
My experiment involves titrating iodine solution against some sample solutions to find the amount of ascorbic acid in it. The indicator I'm using is 0.5% starch solution which will turn the solution to blue-black signaling the end point of titration.
However, the last time I conducted the experiment, I found that even 10 drops of the indicator was not enough to signal the endpoint of titration (no color change was observed). After I added the entire 100ml solution (containing 0.5g of starch + 95.5ml water), the color change could be observed.
But, I don't want to add 100ml of indicator for titration experiment in future. I was wondering if I could just add 0.5g (entire amount of starch in 100ml of starch solution) of starch powder directly instead of making a starch solution and then using it while titrating? If not, what can I do to make starch be able to detect the endpoint of titration without using too much of starch solution?
Thank you,

In principle, you could perhaps use solid indicator. In practice, starch powder can take a long time to dissolve in water, so this approach might not be best in your specific case. How old is your indicator solution? In our teaching labs, we find starch indicator solution has a useful shelf life of 1-2 weeks. Water slowly hydrolyzes the starch.

They sell spray starch in the laundry section of grocery stores. It worked well for the iodine indicator reaction and was convenient to use: just a short spritz was enough. Worth a try.

Concentration of the indicator is one thing.  But if the dilution of your analyte is too high (or: its concentration too low) then you do no recognize the moment when colours change. 1) Perform a series of check runs with solutions known in advance to contain $\pu{xy g/L}$ of analyte to learn how it should look like, and to check how well the results are accurate and precise (two different, though related concepts). 2) If your your eyes are your recorder, place a white paper (better: a white glazed tile) below the Erlenmeyer flask during the titration, this increases contrast.

If still not successful enough, you might resort to perform the titration with a UV-Vis spectrometer.  Depending on the spectra of blank and reaction mixture, you might consider an interval of wavelengths instead of a single wavelength for higher sensitivity.

Do you have hands on a classical quantitative analysis book?

Starch solution cannot be kept more than a couple of days. After this tlme, starch will be hydrolized, transformed into glucose and not be able to react in blue with iodine.
On the other hand, if you put solid starch into the solution to be titrated, the reaction with iodine will only happen on the surface of the grains. It will be difficult to notice the very beginning of this blue color, if it only occurs on the surface of grains which are not in solution. They may fall on the ground of the container, and take time to react with iodine produced in the solution. The equivalence point is much more visible when starch is regularly distributed in the whole solution.


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Continuous variable for graph of boiling point of different chain lengths
Looking at the boiling points of alkanes (for examples), many graphs plot boiling point against chain length :

However, there is an issue with the line of best fit- since chain length is a discrete variable (the line of best fit suggests you can have non-integer carbon chains). What is a more suitable alternative ?

I have seen examples which use wiener indices (not familiar with these), but it appears the indices should be integer values.
Relative formula mass is another possible alternative (since there could be decimal Mr from isotopes).  However, is there a more suitable alternative ?

It's fine to have a fit equation where some of the values in the domain (nonintegers in this case) are invalid. 2. Your plot doesn't look like a fit line, it looks like it simply connects each data point with a straight line.

I think there is more than simple statistics treatment. I am even not sure that a fit should be there. In so saying, I assume the BP of alkanes as data with basically no fluctuation. But I know it depends on the context as well as that you just gave an example.

I refer to a good response on the general question "Discrete variables in regression model? found on a sister site.
The quote in part:

General confusion appears when you mix in ordinal data, such as those 5-point "how satisfied are you?" questions. They are expressed in whole number, very easily to be confused with discrete data. However, each jump in the scale does not necessarily mean the same thing. E.g. a jump from "4: happy" to "5: very happy" is not necessarily the same as a jump from "1: very unhappy" to "2: unhappy." In that case, the variable should not be put into the regression as is, but treated differently (search "dummy variable in regression" to learn more.)

Now, in the current context of chain length as the explanatory variable per a standard regression analysis, one could argue for a solution composed of a mixture with known varying chain lengths, a standard regression analysis could still present meaningful information in such a heterogeneous composition.
However, with respect to a homogeneous single chemical composition, I believe a more difficult argument to make in general. So, do consider a dummy variable approach especially with a large number of data points. Be mindful, however, of the impact of a loss in degrees of freedom (from employing an explanatory 'dummy' variable for each chain length). This expanded regression model generally results in higher expected errors on parameter estimates and also wider associated confidence intervals as well.
You may also want to consider weighing (inversely by the ascribed regression variance estimate for each model) the fitted value especially when working with smaller data sets and extended chain lengths.


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Is there any Iron solution that can reduce to Iron metal upon heating etc?
The chemical should be water soluble and then should reduce to form Iron metal when heated up or upon adding some other chemical etc. The final product does not need to be solid metal, it can be in the form of powder (particles of nano micro etc) or just layer of deposits of Iron particles. Is there any chemical that is in the liquid form or water soluble, that can ultimately reduce to Iron metal?

The people who give negative votes should explain if the question is not up to standard, without leaving me wondering.

While I agree downvoting deserves explanation (unless it is already provided before), it does not necesserily mean the question is not up to standard. It just expresses subjective opinions, illustrated by mouse hints ( UP=This question shows research effort; it is useful and clear / DOWN=This question does not show any research effort; it is unclear or not useful ). // Be aware many SE sites  generally frown upon questions without explicit own trying to find the answer or to understand the topic. Questions being just the questions are frequently downvoted or voted to be closed.

Minus one. Also because a preliminary search would have answered your question or helped you writing a better one. Ex: https://en.m.wikipedia.org/wiki/Iron_nanoparticle

Since iron can displace hydrogen from water, leaving you with rust, you might try another solvent.

@Alchimista, "In the presence of oxygen and water, [iron nanoparticles] rapidly oxidize to form free iron ions. "

@DrMoishePippik well iron is iron,

Actually, boiling many concentrated solutions containing a soluble Iron salts in an aluminum pot (which is the "or upon adding some other chemical etc.") will liberate Fe.
However, be mindful to limit the presence of air (else just common rust) and even an excess of water (see "The Reaction Between Iron and Water in the Absence of Oxygen").
Enjoy experimenting!

thanks a lot, so aluminum reacts with Fe solution to produce aluminum oxide + Iron, am I correct? So, I should be able to put aluminum pieces to a pot and Fe solution and still get the same result as using the aluminum pot isn't it. And about excess of water, to prepare the Fe solution, we need water to be added, so is it necessary to have no water in the solution? or I just have to limit it. I'm asking this because I'm thinking to use FeCl3 and it needs water to become a solution. Any other suggestions?

@PrettyGirl this thread got more attention than what it deserve. In addition to the lack of search, you do not give context. Why? A trick to display? A homemade recipe for something? Etc. Equipment? Why not iron filings?

With dilute conditions, some so called hydrolysis reactions may occur, With significant air contract, oxidation reactions also (especially in the presence of Al+++ which forms a semi-valence complex, and note the presence of Al and Fe gives a galvanic cell aspect,). Perhaps run parallel reactions in limited air to a plate (significant surface area with interface) with both concentrated and dilute conditions and report back on observed results. Varying results may support why this experiment is a actually a good teaching exercise for both elementary and advanced students.

References: Article ‘Oxidative Stress Gated by Fenton and Haber Weiss Reactions and Its Association with Alzheimer’s Disease’ by Tushar Kanti Das, et al, published in Archives of Neuroscience, July 2014 2(3): e20078, DOI: 10.5812/archneurosci.20078. Key reaction formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex: Al(H2O)4 + O2•− <-> Al(O2•−)(H2O)4.

Reference: On the galvanic aspects (source of e- and with O2, the superoxide radical anion), see for example, https://www.sciencedirect.com/science/article/abs/pii/S1876107020303230  . Quoting;  "The present paper constitutes a conjugate numerical and experimental study of a novel process patented as Galvano-Fenton, based on the association of galvanic corrosion and Fenton reaction. The study investigates the pathways of the in-situ formation and regeneration of iron species involved in electrochemical and chemical mechanisms of the Galvano-Fenton process...."

Reference: The hydrolysis of iron(III) chloride is the cationic reaction of the salt with water. Cationic hydrolysis is possible because the salt, iron(III) chloride, formed by a weak base (iron(III) hydroxide Fe(OH)₃) and a strong acid (hydrochloric acid HCl).  Link: https://melscience.com/US-en/articles/hydrolysis-reaction-iron-iii-chloride/

@AJKOER thanks a lot I'll go through one by one.


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Comparing acidity between 4-hydroxybenzaldehyde and 2-hydroxybenzaldehyde

https://hmdb.ca/metabolites/HMDB0011718
https://hmdb.ca/metabolites/HMDB0034170
The above two links provide the pKa values for the 4-hydroxybenzaldehyde (pKa = 7.32) and 2-hydroxybenzaldehyde (pKa = 8.4) ...
(not sure as to how reliable the source is)

How do I compare their acidic nature? In either cases there's an electron withdrawing group (aldehyde) at ortho and para position, conjugate base is stabilised either way.
How would the meta- isomer differ from the acidity comparison?

Consider interaction of groups in ortho position. // Note that questions should be as self-contained as reasonably possible. Links should merely be just data references or source of further study, if the relevant part of link target content can be written as quoted text, eventually formatted by SE Markdown or MathJax/mhchem.


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How is relative interaction energy defined and quantified?
I understand the interaction energy is the energy associated or caused by the interaction between the objects. So, for atoms it has to be the sum of van der Waals interaction and Coulomb interaction.
If there are a bunch of atoms in the system, what does the term relative interaction energy mean? How to conceptually understand this term, and how do we measure or calculate it?

See also wikipedia.org Interaction_energy

Yeah, I have already looked into it. But, I confusion is regarding conceptually understanding it, and the term relative.

It was meant rather just as a general additional reference, as  you should provide in the question the explicit compact result of your preliminary effort before asking. You should e.g. provide the context using the term "relative interaction energy".

Based on the general concept "relative X = absolute X / absolute X0", the contextless guess says that relative interaction energy is ratio of 2 interaction energy values, e.g. during comparison of 2 related cases.


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Contribution, and the way thereof, of the core 6p orbitals of compounds of plutonium(VIII)
According to the NIST database, the ground-state configuration of plutonium(VIII) is [Hg]5f³6p³, when intuitively it should be the same as neutral radon([Hg]6p⁶). Since plutonium(VIII) compounds are already known, there are two problems:

the plutonium(VIII) cation is paramagnetic;
the core 6p shell of plutonium must, according to this electronic configuration, take part in bonding in compounds of plutonium(VIII).

Are there any plutonium(VIII) complexes with diamagnetic ligands known to be (experimentally or theoretically) paramagnetic or with meaningful contributions (of the central plutonium's 6p shell) in its bonding?

N.B. the same problem occurs for neptunium(VII); https://physics.nist.gov/cgi-bin/ASD/ie.pl?spectra=Np&submit=Retrieve+Data&units=1&format=0&order=0&at_num_out=on&ion_charge_out=on&el_name_out=on&shells_out=on&conf_out=on&ion_conf_out=on&e_out=0&unc_out=on&biblio=on

Cation and compound are two different things - such configuration, even if correct (probably not) isn't enough to say that 6p orbitals would have significant contribution to bonds.

It's rather 6d orbitals that do have https://en.wikipedia.org/wiki/Plutonium#Electronic_structure

There is actually some question about the existence of plutonium(VIII). Huang et al. [1] investigate the electronic structure and stability of nominal Pu(VIII) oxide and fluoride complexes. They find that in all cases studied, either the proposed compound decomposes with loss of fluorine or oxygen, or in more oxide-rich cases the most stable conformation has two of the oxygen atoms forming a superoxo ligand and the plutonium thereby only in oxidation state +5. Pu(VIII) conformations of $\ce{PuO4, PuO3F2, PuO2F4, PuO2F5^-}$ are metastable with the corresponding Pu(V) superoxide-containing conformations being fully stable.
Regardless of oxidation state, all molecules and conformations show the $6p$ orbitals completely filled and bonding from the normal plutonium valence orbitals instead; no significant bonding with the $6p$ orbitals is indicated. The Pu(VIII) conformations are rendered most stable as as spin singlets, consistent with the conventional radon-core model.
One reason for the noted gas-phase electronic structure is that multiple subshells in plutonium have energy levels close to one another and thus what are normally secondary factors may come into play. With bare Pu(VIII), $5f^36p^3$ could be favored over the expected $6p^6$ by exchange energy of the unpaired electrons in the $5f$ and $6p$ orbitals. However, loss of unpaired electrons in the complexes could be expected because the bonding to ligands may drive electron pairing, as it does in many other examples involving the metal in high oxidation states.
Reference

Wei Huang, Pekka Pyykkö, and Jun Li (2015). "Is Octavalent Pu(VIII) Possible? Mapping the Plutonium Oxyfluoride Series PuOnF8–2n (n = 0–4)". Inorg. Chem. 54(17), 8825–8831. https://pubs.acs.org/doi/10.1021/acs.inorgchem.5b01540.