text:
Cell potential for redox titration of iron(II) with cerium(IV)

Calculate $E_\mathrm{cell}$ for an electrochemical cell consisting of SCE and a platinum wire indicator electrode dipped in a titration vessel containing $\pu{25 ml}$ $(\pu{0.01 M})$ solution of $\ce{Fe^2+}$ ions after addition of $\ce{Ce^4+}$ solution of concentration $\pu{0.01 M}$ of
A) $\pu{5 ml}$
B) $\pu{25 ml}$
C) $\pu{50 ml}$
SCE is attached to the negative terminal of voltmeter. Given data:
$$
\begin{align}
E^\circ(\ce{Fe^2+/Fe^3+}) &= \pu{0.77 V} \\
E^\circ(\ce{Ce^4+/Ce^3+}) &= \pu{1.61 V}\\
E(\text{SCE}) &= \pu{-0.24 V}
\end{align}
$$

I tried doing this, but one thing I don't get over with is that isn't iron ions and cerium ions both oxidising and reducing? Why is there another electrode? Please clarify as much as possible.

The initial solution is $0.01$ M and contains $250$ micromole $\ce{Fe^{2+}}$. If you add $5$ mL $\ce{Ce^{4+}}$ $0.01$ M, which is $ 50$ micromole, a chemical reaction will occur : $$\ce{Ce^{4+} + Fe^{2+} -> Ce^{3+} + Fe^{3+}}$$ The obtained mixture contains only $250 - 50 = 200$ micromoles of $\ce{Fe^{2+}}$ and $50$ micromoles of $\ce{Fe^{3+}}$. With this you can calculate the redox potential of the couple $\ce{Fe^{2+}/Fe^{3+}}$ using Nernst's law. Go on ! We will not make the whole calculations for you ! Hopefully you have understood my steps...

what is the role of Ce's Ecell here? also do you know any good books to get some basic practice for electrochemisry? i am struggling with some concepts and would appreciate any insight on how to approach this

Yes, cerium and iron ions are both oxidizing and reducing. Does it matter ? // Why the 2nd electrode ? Try it without it.

Consider it as the cerium redox system ( rather than being a cerium cell), existing aside of the iron redox system. Platinum electrode is inert, having redox potential of the dissolved redox systems ( which are not surprisingly equal ).

yes using cerium that way works in bit A but in bit B?  i feel like bit C is replacing iron reduction with cerium reduction. but i have no idea about bit B

Let's start at the beginning, with a solution where $\ce{[Fe^{2+}}] = 0.01$ M. In order to apply Nernst's law, we also need $\ce{[Fe^{3+}}]$, which is not known, but must be very very low, but cannot be zero. We will see that its exact value is without importance. Let's suppose it is  $\ce{[Fe^{3+}}] = 10^{-6}$ M. The redox potential of the solution is : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77 + 0.059 log(10^{-4}) = 0.53 V}$$ Now this potential cannot be directly measured. The only thing that can be measured is the difference of potentiel between this electrode and a reference, which is here SCE, at - $0.24$ V. So if you dip a platinum electrode and a SCE electrode in this solution, you will measure a voltage between them to be : $$\ce{E_{cell} = 0.53 V -(- 0.24) V = 0.77 V}$$ This not a reliable value, as it supposes that we know the amount of $\ce{Fe^{3+}}$ in the solution. Anyway, let's go on.
First addition of Cerium(IV).
Now you add $5$ mL $\ce{Ce^{4+} 0.001 M}$, that is $50~\mu$moles. This cerium ion will react with $\ce{Fe^{2+}}$ ions according to the equation $$\ce{Ce^{4+} + Fe^{2+} ->  Ce^{3+} + Fe^{3+}}$$ The initial solution contained $250~ \mu $moles $\ce{Fe^{2+}}$. Adding $~50~\mu$mole $\ce{Ce^{4+}}$ consumes $~50~\mu$mole $\ce{Fe^{2+}}$ and produces $~50~\mu$mole $\ce{Ce^{3+}}$ and $~50~\mu$mole $\ce{Fe^{^3+}}$. The final amount of $\ce{Fe^{2+}}$ is $250 - 50 = 200~ \mu$mol. The potential of this solution is, according to Nernst's law : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77V + 0.059 log\frac{50}{200} = 0.77V - 0.036 V = 0.734 V} $$ The potential of the whole cell (solution + SCE) is : $$\ce{E_{cell} = 0.41 V - (-0.734 V) = 1.144 V}$$ It is worth the trouble mentioning that here we could also have got the same result using the Cerium potential and Nernst's law. In principle. But, to apply Nernst's law to the Cerium couple we would need to know the residual concentration in $\ce{Ce^{4+}}$ which must be very small but not zero. So we have to use the couple $\ce{Fe^{3+}/Fe^{2+}}$.
Second addition of Cerium(IV).
Here you add $25$ mL $\ce{Ce^{4+} 0.001 M}$, that is $250~\mu$moles. The same reaction happens. But here, the amount of Cerium added is equal to the amount of original Iron in the solution. So at the end, there is in solution  $250~\mu$moles of $\ce{Ce^{3+}}$ and the same amount of $\ce{Fe^{3+}}$. The residual amount of $\ce{Ce^{4+}}$ and $\ce{Fe^{2+}}$ are very small, but unknown.
The potential of the solution must be the average of $\ce{E°(Fe^{3+}/Fe^{2+})}$ and $\ce{E°(Ce^{4+}/Ce^{3+})}$, and this is $(0.77$ V + $1.61$ V)/$2$ = $1.19$ V. The potential of the cell after this $2$nd addition is then $$\ce{E_{cell} = 1.19 V - (-0.24V) = 1.43 V}$$
Third edition of Cerium(IV).
Here there is a huge excess of Cerium(IV). There is exactly twice the amount of cerium(IV) necessary to oxidize all Iron(II). The final solution contains the same amount of both Cerium ions. The potential of the solution cannot be calculated with the Iron couple, as the amount of remaining $\ce{Fe^{2+}}$ is unknown and very low. So the potential of the cell must be calculated by Nernst's law applied to the Cerium couple. As the final amount of both ions $\ce{Ce^{3+}}$ and $\ce{Ce^{4+}}$ are equal, the potential of the solution is equal to the standard redox potential, which is $1.61$ V. And the measured difference of potential between the solution and the SCE is : $$\ce{E_{cell} = 1.61 V - (-0.24 V) = 1.85 V}$$
To summarize all these calculations, the measured potential in the cell (solution + SCE) increases steadily from about $0.77$ V, to $1.14$ V, then to $1.43$ V, then to $1.85$ V.


text:
Rust on stainless steel caused by salt solution
I am not a chemist and I could really use some help for a practical case.
I am fermenting vegetables (cabbage, onions, carrots, etc.). I recently switched from glass and ceramic vessels to stainless steel 304 ones (100L capacity). At the moment, I am only filling the vessels half-way, i.e. 50/60L and using a variable lid system. The vessel is airtight thanks to a plastic band around the lid, which is then placed right on top of the vegetables (it is a stainless steel lid and the plastic band around it is then pumped with air to compress it against the vessel). The lid also has an airlock system in the middle, allowing the co2 generated during the fermentation process to be pushed out. The brine created during the fermentation process also gets pushed out from that airlock (when the vegetables are put in the vessel, about 2% salt is added, so the brine contains salt). The brine ends up covering the lid and I removed it from time to time. It is not overflowing on the plastic band from the outside, but it seems that the brine is also pushed up against the plastic band and gets in between the plastic band and the vessel. After a few days, no more brine gets out.
At the end of the fermentation (after 4-5 weeks), when I removed the lid and emptied the fermentation vessel, I discovered traces of rust on the inside. It is right where the plastic sealed the vessel. It's not only rust, but the surface (or coating) of the stainless steel was even removed in some parts. There is no rust below or above that line (ie where the plastic sealed the vessel).
I am aware that stainless steel 304 is for sure not 100% rust-resistant but I was definitely not expecting such reaction after 1 usage! What a powerful brine...
My question is whether you believe that this reaction is due to the brine pushed up? Probably small amounts of brine end up getting stuck between the plastic and the vessel and thus potentially slightly exposed to air.
If that's the cause, I have two fixes in mind :

put the lid higher in the vessel so that the brine is not pushed up against the plastic/lid. I believe that the CO2 created should be sufficient to push out all the air below the lid.

continue to place the lid right above the vegetables. Once no more brine is pushed out, remove the lid to clean the hedges. Then put the lid back. I believe that the remaining fermentation process should push out the newly added air while not generating sufficient pressure to push new brine out.

I also came across another possible explanation while browsing full link

The same thing can happen at the water’s surface if the container is only half full. In this case, the steel above the waterline is exposed only to air, and the passive oxide layer is stable. Beneath the surface, the oxide layer is at a different potential and less stable because of the chloride ions. Now the crevice is represented by the waterline: stable area above, less stable but very large area below, and crevice corrosion occurs at the waterline. Usually this type of corrosion manifests as pitting or pinholes. The mechanism described is accelerated by localization, so a pit is most often the result.

Some pictures

the vessel
the rusted line / removed steel on the inside of the vessel
the lid on which the plastic band is then mounted

What do you think?

Chloride will pit/rust 304, not quite as much rust/pits on 316/317. Many things make it worse such as limited air access (under deposit or crevice corrosion). Oxygen can help rebuild the passive film. For your long term exposure ceramics and glass containers is the best fix. For metal containers, much higher alloy is needed such as Incoloy 825.

Here is a 2016 article that links the presence of ascorbic acid (A.A) to possible pit corrosion on stainless steel in the presence of tap water and chloride ions, to quote:

This study examined the effect of L-ascorbic acid (A.A) concentration on the pitting corrosion properties of 316L stainless steel (316L STS) of heat exchanger in synthetic tap water containing 400 ppm of Cl- ion...Above the 10^-4 M of A.A concentration, A.A generates soluble chelate rather than absorbs on the steel surface and it causes passive film deterioration and severe pitting corrosion.

I personally experienced pit corrosion on a stainless knife that was used to cut a mango (rich in Vitamin C), and which was carelessly left unwashed for days in a sink. On my part, an interesting, but unfortunate unintentional experiment.

source: Stack Exchange

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

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text:
Is hypervalency a real thing?
I saw a proper debate going on between answers to a question about whether the octet rule could be violated. Some people were pointing to hypervalency in period 3 elements, due to the available d-orbitals that are empty. Apparently, this interpretation has been pretty much disproven by real-life measurements of bond energy not corresponding with the bond energy that would exist in a hypothetical hypervalent molecule. But with the neat and simple explanation of hypervalency excluded, how does one explain molecules like SF6, or SO4? I suspect half-bonds are involved, but I don't even understand how a half bond can exist. One cannot split an electron in half (in this context at least), so how can a half bond exist?
Please explain to me as if I'm a High School student that's really learning more about the factory side of chemistry and is therefore not so acquainted with the more theoretical side of chemistry. Don't get me wrong, I love the theoretical side of chemistry, and anything mildly advanced within chemistry that I know is self-taught, but because of this, I have a lot of holes in knowledge and terminology. So, be painstakingly clear if you could :)

https://chemistry.stackexchange.com/questions/19433/how-can-iodine-bond-with-5-fluorines-in-iodine-pentafluoride https://chemistry.stackexchange.com/questions/105649/the-real-structure-of-difluorodioxoxenon-xeo2f2

https://chemistry.stackexchange.com/questions/4760/why-do-compounds-like-sf6-and-sf4-exist-but-sh6-and-sh4-dont

It was talked about far too many times and tends to attract answers that do not take into consideration like anything that was found out during last half a century.

Does this answer your question? Hypervalency and the octet rule

@SirArthur7 Not really, I didn't understand the explanations. And it wasn't conclusive either. Maybe there still isn't a conclusion, but that question was asked 5 years ago. Surely there has been some change in confidence in either the octet rule always being adherered to, or hypervalency being the answer.

See this answer https://chemistry.stackexchange.com/a/5242/9961 and this one https://chemistry.stackexchange.com/a/134402/9961

Hypervalency is real, yes. Hypervalency is the ability of an atom in a molecule to expand its valence shell beyond the limits of the Lewis octet rule. Hypervalent compounds are common for the second and subsequent row elements in groups 14–18 of the periodic table.
Half bonds don't exist. The closest thing to a half bond is e.g Li2+, H2+. Its a one-electron bond. One-electron bonds often have about half the bond energy of a 2-electron bond, and are therefore called "half bonds".

Seems you're not getting the point of the question at all...

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text:
Why does ethylamine have two infra-red spectra? Are both correct?
The one I have in my textbook is this:

Shouldn't the first trough at $3500$ to around $3000$ mean that it's an $\ce{O—H}$ group with a $\ce{C—H}$? And then the other trough at $1700$ be a $\ce{C=O}$? Why is this ethylamine $\ce{CH3CH2NH2}$?

The other one I found on Google:

Shouldn't the $\ce{NH2}$ group have double troughs? The only trough there is the $\ce{C—H}$ at $3000$. How is this ethylamine?

Can anyone explain those spectra for me? Are both correct? I'm very confused.

It would be nice if you added a citation for the book.

When looking at IR spectra of hydrogen-bonding groups, always check how IR spectra was recorded. It appears that NIST spectra is recorded using gas phase, while the first one used liquid film. In liquid film $\ce{NH2}$ groups form a network of hydrogen bonds, resulting in shape distortion and widening of $\ce{NH2}$-related bands. Same is true for carboxylic acids, alcohols, phenols, and many other group.  Comparison of water spectra in ice, liquid and gas is particularly enlightening.
If the spectrum is recorded in a solution, you also have overlapping spectrum of the solvent in your recording. If you look at water spectrum in liquid phase (nist webbook), you can see some wide peaks suspiciously similar to wide bands in the first spectrum you posted, so it might be that 'liquid film' is actually a water solution. In this case the situation is even worse, because in addition to ethyl amine spectrum you also have overlapping ethyl ammonium spectra.

Honestly I have doubt about your textbook version of IR spectrum. However, I can suggest by the experience I have gained that the shapes of the peaks depends on the method you have used to obtain the spectra such as ATR, smear, absorbed in inert surface, etc., thus cannot pinpoint which one is incorrect. Yet, you can compare an one with other related amines. The following shows the comparition of two spectra of ethyl amine (your google finding) and of butyl amine (from this site). I make the points $\pu{3000 cm-1}$ and $\pu{1000 cm-1}$ coincide for comparition purposes:

For instance, $\ce{NH2}$ group of primary amines absorb somewhere between $3100$ and $\pu{3500 cm-1}$ ($\ce{N-H}$ stretching). That double trough (typical of primary amines) can be seen clearly on the spectrum to the left of the $\ce{C-H}$ streaching absorptions (around $\pu{2900 cm-1}$). Thus, the spectrum you found in google is seemingly correct comparing aforementioned peaks and other overlapping peaks in fingerprint reagent.

I agree with Mathew about the textbook spectrum. A certain textbook had a mass spectrum that was interpreted in the text as propylmethyl ether. Unfortunately, the spectrum was that of diethyl ether.


text:
Cleaning silicone molds
I am a musician who wears custom-molded in-ear monitors made of silicone. They are rather expensive so I need to constantly clean them and carefully. I was told not to use isopropyl alcohol (96%) wipes because it will eat away at the silicone. I looked this up and somebody said it has to do with the molecular weight. Is it true that it would damage the silicone, or just make it bloat for a while until the alcohols are evaporated? What about carbamide peroxide, which is sold as an earwax remover?

Peroxide compounds might be too reactive to trust with polymers like silicone - just a general rule of thumb guess, I don't know specifically.

If the silicone swells in isoprop, it is in grave danger. Squeeze a jelly bear you put in water for an hour and you know what I mean. Otherwise, you only put them in your own ears, right? Wipe clean with a dry cloth, or use liquid soap / your shower gel if necessary. I would never use funny chemicals here.

Soap/water is your best bet. Do the IEM's have a silicone head that you can detatch and wash separately? Or are the drivers just embedded in the silicone. Asking because if you dunk it in water, you might damage the drivers.


text:
A question about the equality of heat and changes in enthalpy
I am wondering how the heat release or absorbed by a chemical reaction can be equal to the change in enthalpy of the reaction at constant pressure if there are changes in the amounts and potentials of each chemical in the reaction mixture. To be more specific, at constant pressure,
$$\mathrm dH = T\,\mathrm dS + \sum_{i} {\mu_i\,\mathrm dN_i}$$
How can we just ignore the summation for something like a chemical reaction?

In the absence of non-PV work (at constant pressure), the change in enthalpy of the system is equal to the heat exchanged between system and surrounding. This follows from the first law of thermodynamics and the definition of enthalpy.
The equation you are showing is based on the definition of the Gibbs energy (the chemical potential is defined via the Gibbs energy, and the Gibbs energy and the internal energy are different by TS). Just because you define some new thermodynamic functions and express the change in enthalpy with it does not invalidate the first law of thermodynamics.

How can we just ignore the summation for something like a chemical reaction?

Here is a silly calculation to model what is going on. Let's start with a variable $a$ known to be five:
$$a = 5$$
Now we add $b$ to both sides:
$$a + b = 5 + b$$
Let's define a new variable to simplify the right hand side:
$$c = 5 + b$$
and substitute that into the equation:
$$a + b = c$$
Now we solve for a:
$$a = c - b$$
It looks like $a$ depends on $b$ and $c$. However, we know $a$ is equal to 5, so it is constant. If you substitute the definition of $c$ back into the bottom equation, you get back to $a = 5$.

This is a very difficult question you have asked, and it has taken me a few days to figure out the answer.  This question is closely related to a Challenge Problem that I issued in PhysicsForums.com 3 years ago.  Even though I knew the answer to the challenge problem, it still took me significant time to figure out the connection to your question.  Here is the link to the challenge problem:  https://www.physicsforums.com/threads/thermochemistry-challenge-problem-chets-paradox.913567/
Consider the relationship between your question and the following part of the challenge problem:  If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?
The solution to this paradox posed by the challenge problem was masterfully provided by member Dr. Du in post #2.


text:
Why are there no alkaline foods?
It takes some work to find a list of the actual pH of various foods.  (Searches are confounded by a popular "alkaline diet" theory that the pH effect on a human body of consuming various foods is not a function of the pH of the food itself.)
It appears that almost all food has pH < 7.0 (most much lower), and I can't find any with pH > 7.5.
Is there a biochemical reason that there is no significantly alkaline food?

Apparently lutefisk is edible with a pH well above 8. (Preparation pH is in the 11 to 12 range.)

Counting lutefisk a food is quite a stretch, as to me. Now to the point. Our world is acidic. An acid will stay unchanged for pretty long, but any alkaline solution will quickly get neutralized by $\ce{CO2}$.

Proteins don't like alkalines. Our world is an oxgen world. Oxidation of organic substrates gives acids.

So no need to mention Icelandic Hákarl, which is reputedly high in ammonia content, but, properly prepared, is not poisonous! ;)

How about the common celery, which normally has a pH of 5.7 to 6.0 (see https://www.engineeringtoolbox.com/food-ph-d_403.html ). However, as it has a water content of 95% (per https://www.livescience.com/50640-celery-nutrition.html#:~:text=According%20to%20Ware%2C%20celery%20is,keep%20necessary%20fluid%20levels%20up. ), celery grown in alkaline soil and watered with alkaline water, may actually have a pH over 7.

I’m not sure what pH pretzels end up with but they are dipped in sodium hydroxide solution before baking. However, I offer red cabbage, an indicator, which in my home region is called blue cabbage because soil, water and sometimes additives give it a blue-ish colouring indicating slightly alkaline.

Most people find alkaline taste bitter and "bad tasting". There are some exceptions; shellfish, particularly conch, may have a pH of more than 8.0. This is also the approximate pH of bile, which tastes particularly foul when retching.
Crackers made with baking powder or baking soda also may be somewhat alkaline, as is egg yolk.
However, your question is valid... why are the vast majority of foods neutral or somewhat acidic? Why do most enjoy the taste of grapes, plums and citrus fruits, which may have pH below 3.0? Does it have to do with metabolism? With the preservative effects of low pH?


text:
Why has this molecule a S-chirality?
Can someone explain to me, why this molecule has an (S)-chirality? I don't get it. I tried to apply the rules, but I get only an (R)-chirality. Is there something special with cyclo-alkanes?

This is not cycloalkane. It is a cyclicether with ester grop in it at chiral carbon

Priority is $\ce{-O- \Rightarrow -CO2Et  \Rightarrow  -CH2-  \Rightarrow  -CH3}$

There is not much to be explained. This molecule is an S enantiomer according to the rules. There are no special rules for cyclic molecules. That's about the size of it.

@MathewMahindaratne Thank you for this correction.

@MathewMahindaratne I thought all the time, that whenever two carbons have the same priority, it depends on the priority of their substituents. So CH_3 was higher ranked than CH_2 for me.

Now why would CH3 be ranked higher than CH2, if the latter has some substitute instead of H, which surely has higher priority than H (because everything has higher priority than H)?

Working outward from the stereocenter, the methylene carbon has its 2 hydrogens and the next carbon in the ring attached to it. This is written as {C,H,H}. The methyl carbon has three hydrogens attached to it. It is designated {H,H,H}. The methylene carbon has priority because a one-to-one comparison has C>H based on atomic number.

You really don't need the bold and dashed bonds as long as you realize that your ring implies that CH2 is in the foreground and the oxygen is in the rear. If not make the CH2 bond bold and the oxygen bond dashed. Once you establish priorities, don't worry about moving the structure but rather look here: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750

One subtlety:  when you rank the groups and look at the rotational order of ranks 1, 2, 3, you have to consider also whether group 4 is towards you or away from you.  A clockwise order for 1, 2, 3 (which the given structure shows) corresponds to $R$ when group 4 is directed away from you.  But here group 4 is directed towards you, so the chirality is reversed --- 1,2,3 clockwise now means $S$.


text:
Writing water as a reactant in acid/base dissociation (Brønsted Lowry)?
The dissociation of $\ce{HCl}$ in water is written as:
$$\ce{HCl (aq) -> H+ (aq) + Cl- (aq)} \tag{1}$$
Sometimes we also include $\ce{H2O}$ as a reactant to stress the fact that the protons are not isolated in water.
$$\ce{HCl (aq) + H2O (l)-> H3O+ (aq) + Cl- (aq)} \tag{2}$$
The last equation is quite useful to establish what is the conjugate acid and what is the conjugate base in Brønsted-Lowry's theory, as far as I understand.
Let's do the same thing for a strong base, say $\ce{NaOH}$. We write the dissociation in water:
$$\ce{NaOH (aq) -> Na+ (aq) + OH- (aq)} \tag{3}$$
What if I wanted to write equation $(3)$ in a form analogous to equation $(2)$? I tried including water as a reactant in the same way but then it yields a chemical equation where $\ce{H2O}$ is on both sides, so we can elide it.
In Brønsted-Lowry theory we generally write $\ce{H2O}$ in the dissociation for acids and for some basis, say $\ce{NH3}$. Why don't we do it for Arrhenius bases like $\ce{NaOH}$ or other hydroxides?

Use a different strong base, like $\ce{NaOCH3}$

Or write hydroxide hydrate as the product.

$NaOH+H_2O\rightarrow Na^+ +H_3O_2^-$

I omitted $(aq)$ here. Do you mean like this? If yes, why do we generally write the hydrsted form of $H^+$ explicitly and we don't for $OH^-$?

Yes, that's what I meant. I don't know why we do that. At least, hydroxide by itself is fairly common in ionic solids, so it is more of a thing. $\ce{H3O+}$ is not that accurate either, there is also $\ce{H5O2+}$ (Zundel cation) and $\ce{H9O4+}$ (Eigen cation).  See e.g. https://en.wikipedia.org/wiki/Hydrogen_ion

Yes, $H^+_(aq)$ can be found as you said as well. I guess we do that because of conjugated acid and bases.
$NH_3+H_2O\rightleftarrows NH_4^+ +OH^-$
In this case, $H_2O$ acts as a weak acid and its conjugate is $OH^-$.
While in the case of $NaOH$, the conjugate base of water will be the $OH^-$ released by sodium hydroxide. In terms of Arrhenius theory I'd say that sodium hydroxide releases its $OH^-$ while ammonia needs to "take" it from water.

I would add that $\ce{H+}$ is a simple proton. It is not an atom. It deserves a special treatment. All atoms are about the same size, between $\pu{5 · 10^{-11}}$ m to $\pu{5 ·  10^{-10}}$ m. A proton is at least thousand times smaller. It must react by addition on a water molecule yielding a "usual" ion, namely $\ce{H3O^+}$. The hydroxyl ion $\ce{OH-}$ does not have this property.

https://chemistry.stackexchange.com/questions/139922/writing-water-as-a-reactant-in-acid-base-dissociation-bronsted-lowry

Not all strong acids are like HCl
There are unusual strong acids that have ionic compounds with hydronium as the cation. Hydronium perchlorate is an example. It's dissolution in water would be written like this:
$$\ce{H3OClO4(s) -> H3O+(aq) + ClO4-(aq)}$$
This looks very similar to the dissolution of $\ce{NaOH}$.
Not all strong bases are like NaOH
Bases like $\ce{NaOCH3}$ lack hydroxide as an anion. The dissolution (and then reacting as a base) would be written like this:
$$\ce{NaOCH3(s) -> Na+(aq) + OCH3-(aq)}\tag{1}$$
$$\ce{OCH3-(aq) + H2O (l) -> CH3OH(aq) + OH-(aq)}\tag{2}$$
Leaving out the second part would not explain why the pH changes upon dissolution, and would not reflect the major products.
What actually happens to NaOH
If you label the oxygen of $\ce{NaOH}$, it would make sense to write the following:
$$\ce{Na^17OH(s) + H2O(l) -> Na+(aq) + ^17OH-(aq)}\tag{1}$$
$$\ce{^17OH-(aq) + H2O(aq) <=> H2^17O(aq) + OH-(aq)}\tag{2}$$
The equilibrium of the second reaction would be on the right hand side because the concentration of water is higher than that of hydroxide. So in that sense, the hydroxide in NaOH does react with water, yielding water and hydroxide.
Why are they different?

In Brønsted-Lowry theory we generally write H2O in the dissociation for acids and for some basis, say NH3. Why don't we do it for Arrhenius bases like NaOH or other hydroxides?

Acids and bases can change the concentration of hydroxide and hydronium ions in aqueous solution either by reacting with water, or by providing these ions directly. Metal hydroxides do the latter, while $\ce{HCl}$ and $\ce{NH3}$ and other acids and bases to the former.


text:
Calculating the number of volatile molecules in a container of known volume
Is it possible to calculate the number of volatile molecules (so only those in the gaseous phase) in a container of known volume, if the vapor pressure of the compound is known? So for example, if you have a container with a volume of $10\ \mathrm{ml}$ containing $3\ \mathrm{ml}$ of a Compound X with a vapor pressure of $10\ \mathrm{mmHg}$ at $\pu{25 ^\circ C}$, what is the number of gaseous molecules of X in the container?

Yes, this is a do-able calculation.

We have a policy which states that you should show your thoughts, effort and attempts to understand underlying principles and solve the question. It'll make us certain that we aren't doing your homework for you. ( As homework is considered literal homework, self-study questions, puzzles, worked examples etc.)  Next time,  provide your full reasoning or thoughts on this, otherwise, the  question may get closed. See Homework. This question may get closed  anyway.

Do you know about Raoult's law?


text:
Opposite effects of Le Chatelier's Principle
How does Le Chatelier's Principle deal with a change in reaction conditions which can simultaneously effect the equilibrium in opposite ways.
For example, here's the equation for the conversion of $\ce{N2}$ and $\ce{H2}$ into $\ce{NH3}$. It's an exothermic reaction, so I've added heat as a product on the right side.

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g) +\text{heat}}$$

If I increase the temperature (add heat) that would tend to push the reaction to the left because the reaction to the right is an exothermic reaction. But if I add heat (and the volume remains constant), then I'm also increasing the pressure, which should tend to push the reaction to the right, towards fewer moles of gaseous products.
Is there a simple explanation for how Le Chatelier's Principle works in such a situation?

It is important to know the externally caused change the system is dealing with.
If the primary change is increasing the pressure, then the equilibrium shifts to the lower volume, in case of gases the lower molar amount.
If the primary change is increasing the temperature, than the equilibrium shifts away from the direction that evolves thermal energy.
If external causes are combined, like pressurizing the system AND externally initiated heat transfer, the resulting direction of equilibrium shift would depend on what effect is stronger. As the Le Chatelier's rule is qualitative, it cannot provide answer here, but it must be quantitatively calculated via the equilibrium constant at particular temperature.

I removed "if both is changed at the same time" because that phrase suggests it matters whether the changes are done at the same time or in sequence, when in fact it doesn't matter whether you do #1 then #2, or #2 then #1, or both together.   I also changed, e.g., "If the primary change is increasing the pressure by lowering the volume, than the equilibrium shifts to the lower pressure.", because it doesn't matter how the pressure is changed, and because the new pressure could be a constant pressure,  in which case the shift in equilibrium isn't changing the pressure.

I guess it may be little overstepping the editor licence, by changing the meaning of the answer.. Better would-be  to comment it and let the author to rephrase it.

Yeah, in retrospect I can see I did overstep.  Sorry.  I thought I was making it clearer, but in fact I was actually changing the meaning.

Anyway, thanks for the attention, clarity attempts always count :-)

About the final constant pressure case, it would be like a continuous external cause, reacting on the system change.

I believe, this answer also removes the confusion to this question also!


text:
Carbonated Beverage Phenomenon
I've seen this phenomenon a couple of times--I actually saw it happen this night.
When a carbonated beverage is poured into a cup or glass, many tiny bubbles of carbon dioxide form on the bottom of the glass, then shoot up to the surface and pop when they get too big.
Today, as I was looking at a wine glass filled with a carbonated beverage, I noticed that there was something different--the carbon dioxide bubbles were rising up in a straight column from the very center of the cup; in other words, all the CO2 bubbles were coming from the point where the cup joined the stem.
Can someone explain how this phenomenon happens?

Look up "bubble nucleation sites". Hint: where in the interior of a wine glass is the roughest spot???

https://chemistry.stackexchange.com/questions/104595/why-do-bubbles-form-on-the-side-of-a-plastic-bottle


text:
Is this poor laboratory safety?
I'm a formulation chemist. My co-worker, is also a formulation scientist but has no previous formal training in chemical safety or lab chemistry settings.
I always work in the hood unless I’m just mixing fruit extracts or something non-volatile. My coworker doesn't ever work in the hood as they believe that since this goes in food it must be safe.
The lab is small and our work stations are pretty close. My coworker opens up reagent bottles and performs all other tasks on the bench top.
I don't think their reasoning is correct since in food items, the chemicals we work with are present in the ppm to ppb ranges whereas when a pure bottle of them is opened on a lab bench, they are in high concentrations. In addition we face prolonged exposure  ($8$ hr work days,$5$ days a week). The two scenarios are very different.
Also, I feel like I’ve had a sore throat ever since I’ve started working there.
I just don’t think these are prudent lab practices. Am I overreacting or does my coworker need to start working in the hood?

The lab should be shut down if there is no exhaust and your colleague and the higher up management do need to go through proper training.

Your coworker, like every chemist. absolutely must learn and remember the classic words of Paracelsus: “All things are poisons, for there is nothing without poisonous qualities. It is only the dose which makes a thing poison.”

"My coworker doesn't ever work in the hood as she believes that since this goes in food it must be safe." By that logic she would also have no problem with ingesting pure sodium and inhaling pure chlorine. After all, they are the ingredients for table salt, so they must be safe. Are you sure she has a chemistry education?

@Philipp she doesn’t. Studied nutrition in college

Or acetic acid aka vinegar. Can't you use that as an example for her?

@telcoM: I thought "dose makes the poison" comes from Galen (then again, Paraclesus could be quoting Galen)

Material Safety Data Sheets are available for all chemicals. It is sufficient to read those to see that most substances, indeed even those that are contained in food products, are quite harmful at high concentrations. Besides, I don't get it. It's not like the company would have to spend money to make you work safely: you already do have fume hoods, so why not use them?

This is more a comment than an answer, but too long for the former....
There are two things for you (and your company) to consider:

The SDS of the chemicals being handled - that will give you guidance for how best to handle a chemical. Flavouring chemicals like benzaldehyde, cinnamaldehyde, limonene all have irritation warnings, and need to be handled in well ventilated areas; others like diacetyl and allyl hexanoate are straight up nasty. And I don't mean to be patronising with this list of flavouring chemicals 101. I'm sure your list of chemicals is extensive, but they will all have SDS that will dictate how you should be handling them. Use that to base your arguments for change, not the opinions of an online discussion group.
Relevant building codes for laboratories that will specify the required number of air turnovers per hour. If your workplace is old, there is a good chance that ventilation is not quite what it should be. This is not an easy (cheap) problem to solve, but if the company is aware of any potential problems it will be easier to improve workplace practices of working in fumehoods.

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


text:
Acids and Bases (Spectator Ions Identification)
In an acid reaction with water, or other solvents, how do we know which atom(s) from a compound reacted with that solvent are the spectator ions? Are they the ones soluble in the solvent?
Is the way in determining this the same as the way we usually do it in a precipitation reaction?

Spectator ions in any kind of reaction are those that do not take part in the interaction.
In precipitation reactions, they are those not forming the least soluble salt.
In acid-base reactions, they are those not manifesting in the given environment either acidic, either basic behaviours. Typically cations of alkali metal, or alkali earth metals, or anions of strong acids.
In redox reaction, they are those not being oxidized/reduced nor being produced by it.
For particular cases,  you must be familiar with chemistry of involved compounds.


text:
Comparing the stability of geminal diols

Arrange the given gemdiols in decreasing order of their stability:

I thought that high angle strain would correspond to low stability, and therefore I compared the stability of the rings, which yields the answer as C > B > A, but, the answer given in my textbook is the opposite i.e., A > B > C.
On googling, I found out that, in this case, when the corresponding ketone hydrolyses to form the gemdiol then the one which gets greater relief of angle strain will have higher stability. But why do we have to check that? And I don't  know the reliability of the source. Moreover, what is wrong in my reasoning?

Your reasoning is correct, but I think you'd mistakenly compared the ring strain of the rings, while I would compare the ring strain in the corresponding ketones.

@RahulVerma Yeah! That's my question. Why would you compare the ring strain of the ketones when the question is asking about the stability of the gemdiol?

@AniruddhaDeb The answer to that question tells how cyclopropane-1,1-diol is more stable than the ketone. My question is different

Your reasoning is correct, but I think you'd mistakenly compared the ring strain of the rings, while I would have compared the ring strain in the corresponding ketones.
Note that, you are asked about the stability of gem-diols, which could convert into the corresponding ketone. The point is, the ring isn't going to change, so the stability of rings don't matter.
Now, as you know the corresponding cyclic ketones would be $\ce{sp^2}$ hybridized. So, the ring angle should be $120^\circ$ in order to minimize the ring strain in ketone. The most closest is in C and least is in A.
Hence, the order.

If the angle is closest in C then should it not be most stable?

@RobinSingh, yeah, it means that the keto-form of C would be most stable. So, the gem-diol in C would be least stable.


text:
Does anyone have experience with polycaprolactone?
I am trying to create an emulsion from polycaprolactone. My efforts thus far have resulted in unstable emulsions which sediment and coalesce after some time. This is what I have tried:
Dissolve 5g of PCL in 100 mL of acetone.
Add the PCL solution drop by drop into a 1% Tween 80 solution agitated at 5000 rpm using a homogenizer.
Remove the solvent using a rotavap.


text:
How to determine the efficiency of intermolecular charge transfer?
Lets say we have 2 molecules: A and B. Both of these molecules associate with another molecule C. Upon association two complexes form A-C and B-C. We can experimentally see that upon association a new charge transfer (CT) absorption band appears, additionally intensive CT emission appears. Additional evidence for CT can be obtained by DFT calculations (where HOMO is located on A or B and LUMO is located on C). By determining photo luminescence quantum yields (PLQY) it can be seen that A-C has 4 times higher PLQY than B-C.
Are there any ways to determine whether specifically the efficiency of the CT is responsible for this drop of PLQY?

Do you mean that the fluorescence yield of C is reduced in BC vs AC ?

Indeed, if both complex A-C and complex B-C experience CT emission, the efficiency of CT character present in the complexes may be responsible for the difference in PLQY. This however, will be the case if there is no reverse in CT polarity (for example if in both complexes C is the acceptor (and emitter), this may also the reason for porphyrin's comment). Hence if the criteria above are met and You should generally observe the same (or highly similar) emission from both of the complexes. That would also indicate that one species is responsible for the observed emission.

Calculating efficiency of CT. Although I have not seen extensive literature about CT efficiency, there are few papers which try to quantify the CT. The best example I could find is available here: https://onlinelibrary.wiley.com/doi/full/10.1002/jcc.24822. In that paper the authors are trying to quantify CT within a single molecule. Technically the same methodology should also apply for the intermolecular CT in A-C and B-C. Complaring the values obtained should provide some insight in the CT character between both species. Importantly, to do this calculation You need to know exactly how the complex looks. One way to obtain their geometry is to obtain X-Ray single crystal diffraction representations of A-C and B-C in the solid state (this however works only if A-C and B-C do not dissociate upon crystallization).

Godspeed.


text:
Has a low spin tetrahedral Ni(II) complex been synthesised?
How can $\ce{Ni(II)}$ form a low spin tetrahedral compound? I was not able to find such compounds.
$\ce{Ni(II)}$ will have $\mathrm {4s^03d^8}$ and tetrahedral needs $\mathrm{sp^3}$. So $2$ electrons will always be unpaired. Is a rare case possible?
I am only aware of hybridisation theory and basic CFT theory for complexes.

Tetrahedral complexes are always high spin. Only exception I recall would be tetrakis-(1-norbonyl)cobalt(IV)

Don't use hybridisation for transition metal complexes...

@orthocresol your own answer should be relevant here: https://chemistry.stackexchange.com/questions/115051/why-are-low-spin-tetrahedral-complexes-so-rare

@NilayGhosh i want to know if a LOW SPIN tetrahedral Nickel complex has been made. I am aware of the square planar nickel compounds and high spin Nickel tetrahedral.

You asked two different questions in your title and the text body of the post. I wasn't sure which one is you meant to know. However, orthocresol has explained the question about low spin tetrahedral compounds: Why are low spin tetrahedral complexes so rare?. He also explain "Why it is wrong to use the concept of hybridization for transition metal complexes."

About your question in the title: "Has a tetrahedral $\ce{Ni(II)}$ complex been synthesized?"

Answer is "yes."  Orthocresol has given common example here, which is $\ce{NiCl4^2-}$. There are also non-common tetrahedral $\ce{Ni(II)}$ complexes have been synthesized. For instance, a series of $\ce{[(C2H5)3PO]2NiX2}$ where $\ce{X = Cl, Br, I}$ has been synthesized by Cotton and Goodgame (Ref.1; yes, that is the same Cotton in Cotton and Wilkinson):

The $\ce{Ni(II)}$ complexes $\ce{[(C2H5)3PO]2NiX2}$ where $\ce{X = Cl, Br,}$ and $\ce{I}$ have been prepared. Their physical properties, especially their electronic spectra and high magnetic moments, leave little doubt that the nickel ion in each case surrounded by two oxygen atom and two halogen atoms in an essentially tetrahedral array.

In addition, some distorted-tetrahedral nickel(II) complexes (Ref.2) and some pseudo-tetrahedral nickel(II) complexes (Ref.3) have been prepared as well.
References:

F. Albert Cotton, David M. L. Goodgame, "New Tetrahedral Complexes of Nickel(II)," J. Am. Chem. Soc. 1960, 82(22), 5771–5774 (https://doi.org/10.1021/ja01507a001).
S. Trofimenko, J. C. Calabrese, J. K. Kochi, S. Wolowiec, F. B. Hulsbergen, J. Reedijk, "Spectroscopic analysis, coordination geometry, and X-ray structures of nickel(II) compounds with sterically demanding tris(pyrazolyl)borate ligands and azide or (thio)cyanate anions. Crystal and molecular structures of bis[($\mu$-thiocyanato-N,S)(hydrotris(3-isopropyl-4-bromopyrazol-1-yl)borato)nickel(II)]-3-heptane and (thiocyanato-N)(hydrotris(3-tert-butyl-5-methylpyrazol-1-yl)borato)nickel(II)," Inorg. Chem. 1992, 31(19), 3943–3950 (https://doi.org/10.1021/ic00045a017).
John T. Donoghue, Russell S. Drago, "Non-aqueous Coördination Phenomena-Complexes of Hexamethyl-phosphoramide. II. Pseudo-tetrahedral Complexes of Nickel(II) and Cobalt(II)," Inorg. Chem. 1963, 2(3), 572–576 (https://doi.org/10.1021/ic50007a036).

I think OP is looking for "low-spin tetrahedral complex of nickel". Such complexes are already rare in itself and I don't know if such complex exist for nickel. The first such complex observed  is cobalt norboryl complex which Aniruddha pointed out. Other new examples is a new Co(II) complex which is a low-spin complex with a distorted tetrahedral geometry and a chromium complex. I have added those as a update in the comment section of ortho's previous answer.

I wanted to know if any LOW SPIN NICKEL TETRAHEDRAL complexes exist. Just specific for nickel.

I'm not sure I can find one for you. One of given references said it is very rare.


text:
Question about null space balancing method
I'm trying to use the null space method to balance the following equation:
.
I obtained the following composition matrix:
$$\begin{bmatrix}
3 & 8 & 1 & 12 & 4 & 2 \\
1 & 0 & 0 & 1 & 0 & 0 \\
4 & 4 & 3 & 40 & 3 & 1 \\
0 & 2 & 1 & 3 & 1 & 0 \\
0 & 1 & 0 & 12 & 0 & 0
\end{bmatrix}$$
where rows are in order H P O N Mo. I take the rref of this matrix, and augment it with one row of zeroes except the last element is 1. Taking the inverse of that matrix, I get $$\begin{bmatrix}
\frac{1}{51} & \frac{4}{17} & \frac{35}{51} & -\frac{1}{51} & -\frac{56}{51} & 1\\
\end{bmatrix}.$$
After scaling the elements by 51 I would get the end result of coefficients: 1, 12, 35, 1, 56, 51 which does not at all balance the equation. A correct solution would be 1, 12, 21, 1, 21, 12.
I have used the exact same steps to balance other equations. What am I doing wrong?

Fortunately this issue was just caused by a small oversight. In the midst of so many coefficients, you accidentally skipped the second nitrogen atom in $\ce{NH4NO3}$, which changes the fifth element in the fourth row from 1 to 2. The correct composition matrix is then:
$$\begin{bmatrix}
3 & 8 & 1 & 12 & 4 & 2 \\
1 & 0 & 0 & 1 & 0 & 0 \\
4 & 4 & 3 & 40 & 3 & 1 \\
0 & 2 & 1 & 3 & \color{red}{\textbf{2}} & 0 \\
0 & 1 & 0 & 12 & 0 & 0
\end{bmatrix}$$
After complementing with the bottom row $\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ and inverting the matrix, we obtain:
$$\frac{1}{96}\begin{bmatrix}
3 & 79 & 2 & -9 & -14 & -8 \\
36 & -204 & 24 & -108 & -72 & -96 \\
39 & -477 & 90 & -213 & -246 & -168 \\
-3 & 17 & -2 & 9 & 14 & 8 \\
-51 & 417 & -66 & 249 & 174 & 168 \\
0 & 0 & 0 & 0 & 0 & 96
\end{bmatrix}$$
Finally, transposing the rightmost column and dividing its coefficients by 8, the result is  $\begin{bmatrix} -1 & -12 & -21 & 1 & 21 & 12 \end{bmatrix}$, exactly as expected.
Thank you for the linked article! Hopefully others may find it useful. If you're into linear algebra applied to chemistry, I recommend you take a peek into chemical reaction network theory!

Ah, yes. The classic one. That's easy to fix. Thank you very much


text:
Finding the Freundlich adsorption isotherm from the graph of log(x/m) v/s log(p)
The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

The mass of gas adsorbed, $x$, per unit mass of adsorbate, $m$, was measured at various pressures, $p$. A graph between $\log\frac xm$ and $\log p$ gives a straight line with slope equal to $2$ and the intercept equal to $0.4771$. The value of $\frac xm$ at a pressure of $\pu{4 atm}$ is: (Given $\log 3 = 0.4771$)

I know the Freundlich adsorption isotherm formula, but I decided to write the equation of given line $( y = c + mx)$ as,
$$ \log \frac xm = \log 3 + 2\log p $$
So, the relation comes out to be,
$$ \frac xm = 3p^2 $$
On plugging $p = 4$, we get,
$$ \frac xm = 3(16) = 48 $$
But, the answer was wrong as per the official key, which mostly has genuine answers. Where am I wrong?

The answer given is,

 6

$1/n$ lies between 0 and 1..

@Safdar: Yes, I'd doubt over this during the exam. But, if I take $n=2$ and then, draw for $\log p$ v/s $\log x/m$, then it goes very complex.

I think the question has a mistake. You should take $n=2$ then equation is: $$\log \frac{x}{m} = \log 3 + \frac{1}{n} \log p$$ and you got the given answer.

@MathewMahindaratne, Great, I never thought that. Actually, I have to challenge the given key, that's why I asked. Thanks for pointing this thing :)

One version of Freundlich adsorption isotherm equation is:
$$ \frac xm = Kp^{\frac12}, $$
which can also be written as:
$$ \log \frac xm = \log K + \frac12\log p $$
This is a straight line equation of type $( y = c + mx)$ as given in the question. However, I think the question has made a mistake saying the slope is $2$, but instead it should be $\frac12$. Accordingly, the equation with given numeric values should be:
$$ \log \frac xm = \log 3 + \frac12\log p $$
Or without logarithm:
$$ \frac xm = 3p^\frac12 $$
On plugging $p = 4$, You'll get:
$$ \frac xm = 3 \times 4^\frac12 = 3 \times 2 = 6 $$
Hence, you get the given answer.


text:
Detailed kinetic explanation for vapor pressure reduction by dissolved solute?
The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

An open beaker of water in equilibrium with water vapor is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules:
(A) leaves the solution increases
(B) leaves the solution decreases
(C) leaves the vapor increases
(D) leaves the vapor decreases

According to the NCERT for Class XII, Part I, pg. 46, para 3,

In  a  pure  liquid  the  entire  surface  is occupied by the molecules of the liquid. If anon-volatile solute is added to a solvent to give  a  solution  [Fig.  2.4.(b)],  the  vapor pressure  of  the  solution  is  solely  from  the solvent  alone.  This  vapor  pressure  of  the solution at a given temperature is found to be  lower  than  the  vapor  pressure  of  the pure  solvent  at  the  same  temperature.  In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced.  Consequently,  the  number  of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.

So, after reading the last lines of above quoted text, I think that the vapor pressure is reduced, because the rate at which solvent molecules leaves the solution decreases, due to a decrease in exposed surface area. So, acc. to me option (B) should be correct. But, it was incorrect, acc. to the key.
Is there anything else, I'm missing?

The answer given is,

 (C) leaves the vapor increases

It seems to me it is less about principles and more about the formulation of the task. In some ways both B and C are correct. B is correct as the rate of evaporation decreases. C is correct, as the net rate of the phase exchange increases in favor of vapour condensation. But I consider this interpretation as tricky, supporting B.

B is correct. So there is too much vapor the immediate rate does not change! as the partial pressure of water is lowered the rate of condensation lowers until it reaches the rate of evaporation in the new solution so D is also correct.

I'd like to specifically commment on this:
According to the NCERT for Class XII, Part I, [pg. 46, para 3][1],

In  a  pure  liquid  the  entire  surface  is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give  a  solution  [Fig.  2.4.(b)],  the  vapor pressure  of  the  solution  is  solely  from  the solvent  alone.  This  vapor  pressure  of  the solution at a given temperature is found to be  lower  than  the  vapor  pressure  of  the pure  solvent  at  the  same  temperature.  In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced.  Consequently,  the  number  of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.
[emphasis mine]

NCERT's explanation for why vapor pressure is lowered due to the presence of a dissolved solute is incorrect, because vapor pressure is independent of the surface area accesible to the solvent (assuming it's non-zero).  Consider these two examples:

You have two sealed containers, both of which contain an open beaker of identical solvent.  In container A, the beaker is low and wide.  In container B, the beaker is tall and narrow.  The surface area of solvent in container B is thus smaller.  Yet, assuming the conditions in the respective containers are identical, the equilibrium vapor pressures will be the same.

You have two sealed containers, both of which contain an open beaker of identical solvent.  The beakers are identical, so the surface area of liquid is the same. The conditions in the containers are also identical.  Glucose is dissolved into the solvent in container A, while sucrose is dissolved into the solvent in container B.  The final concentrations are the same.  Sucrose is larger than glucose.  Consequently, the fraction of the surface area covered by solvent in container A is larger than in container B.  Yet, ignoring non-ideality, the vapor pressures are the same; they depend only on the concentration of solute, not its nature.

The surface area only matters for kinetics, i.e., for how quickly the solvent can escape the container to reach equilibrium.  Clearly, with a larger surface area, the equilibrium vapor pressure will be reached more rapidly.
Feel free to send an email to NCERT with a link to this comment.
Here's the thermodynamic explanation: The addition of solute lowers the chemical potential of the solvent, due to the entropy of mixing.  [There can also be energetic effects, in either direction, but with a soluble solute the entropy term is dominant.]
At equilibrium, the chemical potential of the solvent in the liquid and vapor phases must be equal.  Thus, since the chemical potential of the solvent in the liquid phase has lowered, that of the the solvent in the vapor phase must lower as well.
And since the chemical potential of a gas increases with its partial pressure, the vapor pressure of the solvent in the gas phase will decrease until its chemical potential reaches the new (lowered) chemical potential of the solvent in the liquid phase.
As far as the question itself goes, I'm uncertain, because the question is asking for a kinetic, rather than a thermodynamic, explanation.  But I can offer two alternative pictures:
Let's represent the movement between the liquid and vapor phases of the solvent ("X"), as follows:
$$\ce{X_{(l)}<=>X_{(g)}}$$
Picture (I): Both "B" and "C" are correct.
Let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase.  This shifts the reaction to the left, which means that the rate of the forward reaction initially decreases (which is answer "B") and the rate of the backward reaction initially increases (which is answer "C").  Eventually the system reaches its new equilibrium point, at which point the rates of the forward and reverse reactions are again equal.
Picture (II): "B" is correct.
Again, let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase. The rate at which gas enters the liquid phase depends only on the concentration of gas, so the backwards reaction will be unaffected by the presence of the solute in the liquid.  However, the rate of the forward reaction will be reduced, since it is now more favorable to for the solvent to remain in the liquid state.  Hence the answer is "B".
As I mention in the Comments:
In Picture (I), the change in relative chemical potentials would raise the energetic barrier in the forward direction and lower it in the reverse direction, which would in turn increase the % of collisions that are succesful in the rev. direction, thus increasing the rate of the rev. reaction even though the collision frequency of the vapor with the liquid woudn't be changed by the presence of the dissolved solute.
But: If there isn't any barrier in the rev. direction, such that 100% of vapor molecules that collide with the liquid phase become liquid, then Picture (I) would not apply.
I'm afraid I don't know enough about the microscopic kinetics to have certainty on this.  Perhaps someone that has done simulation work on this could provide an answer.  [But to get this, you would want to vote to reopen the question.]

If I've understood you clearly, then you're saying that my answer seems reasonable, as decrease in exposed SA takes it longer to reach the equilibrium vapor pressure. But, the reduction in vapor pressure isn't a consequence of decrease in exposed SA. Am I right?

NCERT's explanation is supported here and here. I can't think of a way to refute your explantion. What do you think is the reason for lowering of vappour pressure?

@RobinSingh  See the explanation I added.  BTW, this is not my explanation, this is the standard thermodynamic explanation.

@RahulVerma  Your second sentence is correct.  You can read what I added to the end of my answer.

This doesn't sound right: "the rate of the backward reaction initially increases". Following your own logic there is no reason why the rate of collision of the gas molecules with the surface of the solution should increase simply because something was added to the solution.

@BuckThorn  Yeah, I was wondering along those same lines, which is why I doubly-qualified my answer with "...unless I'm missing something subtle about the kinetics..." and "...I believe..."  But then I thought in terms of a simple kinetic model, where the rate of reaction is proportional to the rate of collisions times the percent of those that are sucessful. Given this, my picture was that the change in relative chemical potentials would raise the energetic barrier in the forward direction and lower it in the reverse direction, which would in turn increase the % of collisions that are....

....succesful in the rev. direction, thus increasing the rate of the rev. reaction even though the collision frequency of the vapor with the liquid woudn't be changed by the presence of the dissolved solute. But: If there isn't any barrier in the rev. direction, such that 100% of vapor molecules that collide with the liquid phase become liquid, then my thinking wouldn't apply. But, again, I don't know enough about the kinetics to have a solid picture on this, hence the explicit uncertainty with which I characterized the "answer" part of my post (that's also why I placed it at the end).

I just edited my answer to mention both of these pictures.

Water is in equilibrium with vapor. The simple interpretation is that rates of condensation and evaporation are equal, but that is not really true in the description of an open beaker in a closed space with respect to the beaker eventually the liquid would be dispersed thruout the container droplets here, there, and everywhere. Addition of a solute to the liquid definitely lowers the vapor pressure of that liquid . It does NOTHING to the vapor at the moment of dissolution. A simple experiment to show this would be to have two identical beakers in the enclosed space one with pure water the other with the solution; a careful observation will show that water will leave the pure water and accumulate in the solution. the rate of condensation is not affected by the solution. It is affected by the partial pressure of the water vapor in the container.
If the only source of water vapor is the beaker with the solute[!] the rate of evaporation will decrease; the initial rate of condensation will be the same and the partial pressure of water vapor will then slowly decrease and the rate of condensation will slowly decrease until equilibrium at the new vapor pressure is established. The question is simply inane.

I am also a JEE Aspirant and this question intrigued me. This is also my first answer on Stack Exchange :)
First of all we need to understand that Vapour Pressure decreases when any solute is added to a solvent, i.e $$VP_solution < VP_solvent$$
A very simple explanation for this is that the solute particles introduce new forces of attraction between themselves and solvent molecules and these forces help to prevent vaporisation of solvent.
Another thing to know that Vapour Pressure is an equilibrium phenomenon. In the question as soon as the glucose is added, the already existing equilibrium was disturbed.
There was already an equilibrium stage attained where as many molecules were leaving the solution, were also leaving the vapour state.
Now the addition of glucose has lowered the amount of molecules leaving the solution. So when the new equilibrium is attained, the Vapour Pressure will be less than what it was when no glucose was added.
Now it is understood that Vapour Pressure must decrease, i.e, more molecules should leave the vapour state. Just lowering the amount of molecules leaving the solution wouldn't decrease the Vapour Pressure. So, the rate at which the molecules leave the vapour must increase and this is why Vapour Pressure decreases.
Mind you, when the new equilibrium is attained the amount and rate of molecules leaving and entering the solution become equal but they are less than what they were initially.
I hope this helped!

If you remove vapor the rate of removal must slow;  as you state in your penultimate paragraph. [thanks for the chance to use that word]

Please can someone provide constructive criticism on why my answer was downvoted?

Because your explanation was not correct. You explain it yourself in the next to last paragraph. At equilibrium the rates are equal if the forward reaction is slower the reverse reaction must also be slower. The condensation reaction monotonically slows.

@jimchmst I see. Thanks for clarifying it! I'll try to understand this better :)

Guri It took me years and I have a long way to go. the best way is to assume equilibrium then study carefully what happens if it is perturbed by various methods. In this case the sticky sugar solution is similar to the vacuum sucking mechanism.


text:
Can Na2S + H2SO4 react to give Na2SO4 + H2O?
Can $\ce{Na2S + H2SO4}$ react to give $\ce{Na2SO4 + H2O}$ or $\ce{Na2O + H2O + SO2}$?
In my textbook, it has been given that the product will be $\ce{NaOH + H2O + SO2}$. Can also the other two products created? If not, then why?

Your textbook might be incorrect. AFAIR $$\ce{Na2S + H2SO4 -> Na2SO4 + H2S}$$ Were any other reaction conditions mentioned?

Concentrated acid would have oxidative properties. $\ce{Na2S + 5 H2SO4 -> 2 NaHSO4 + 4 SO2 + 4 H2O}$

Elemental sulfur could also be a product.

At least, what is sure is that $\ce{Na2O}$ or $\ce{NaOH}$ will never be produced by action of an acid like $\ce{H2SO4}$ on any sort of substance.

OP should at least enumerate the considered chemical reactions.

As Poutnik said, concentrated sulphuric acid is oxidising, dilute sulphuric acid is not. So whether Aniruddha is correct or whether your textbook is accurate here depends on the conditions involved. (The reaction with concentrated sulphuric acid may or may not be complete.)

Well, first of all, it is important to understand the nature of this reaction. The reaction here is a redox reaction where one species is oxidised and another species is reduced. In this case, $S^{2-}$ is oxidised to $\ce{SO2}$ and $\ce{SO4^{2-}}$ is reduced to $\ce{SO2}$. Hence, even though $\ce{Na2SO4}$ does form (and it usually does since the Sulfide anion is quite basic and would readily accept $\ce{H+}$ from $\ce{H2SO4}$ to form $\ce{HS-}$ or even $\ce{H_2S}$), I believe your textbook is trying to illustrate the redox nature of this reaction.
For the second set, $\ce{Na_2O}$ usually never forms because it is quite reactive and would readily dissolve in water to produce $\ce{NaOH}$. (Usually we would just label this as $\ce{OH- (aq)}$ rather than $\ce{NaOH (aq)}$ because $\ce{Na+}$ is a spectator ion and does not really react.)

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

In the case of excess of $\ce{H2SO4}$, bases like $\ce{Na2O}$ or $\ce{NaOH}$ cannot be produced.
The first mentioned option ( $\ce{Na2SO4 + H2O}$ ) is not possible, as there would be oxidation without reduction. Note that in excess of $\ce{H2SO4}$, $\ce{NaHSO4}$ is formed, not  $\ce{Na2SO4}$
Diluted $\ce{H2SO4}$ would produce $\ce{H2S}$, concentrated one would oxidize sulphides/sulphane, producing $\ce{SO2}$ and/or elementary sulphur.
$$\ce{Na2S + 2 H2SO4 -> H2S + 2 NaHSO4}\tag{1}$$
$$\ce{Na2S + 4 H2SO4 -> S + SO2 + 2 NaHSO4 + 3 H2O }\tag{2}$$
Sulphane reduces sulphur dioxide tu sulphur:
$$\ce{2 H2S + SO2 -> 3 S + 2 H2O}\tag{3}$$
Sulphur can be oxidized as well:
$$\ce{S + 2 H2SO4 -> 3 SO2 + 2 H2O}\tag{4}$$


text:
Reaction of glucose acetal with acetic anhydride
The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

Consider the following reactions:
$\ce{ (i) Glucose + ROH ->[dry HCl] Acetal ->[$x$ eq. of (CH3CO)2O] acetyl derivative }$
$\ce{ (ii) Glucose ->[Ni/H2] A ->[$y$ eq. of (CH3CO)2O] acetyl derivative }$
$\ce{ (iii) Glucose ->[$z$ eq. of (CH3CO)2O] acetyl derivative }$
What are $x, y,$ and $z$ in these reactions, respectively?

I know that the reaction in (i) is acetal formation (used to protect aldehyde), and that alcohol reacts with anhydrides to form acids. As, there are only 5 hydroxyl group in acetyl deriv. of (i), so $x=5$.
In (ii), carbonyl part of glucose reduces to hydroxyl group, so $y=6$. And, in (iii), glucose directly reacts with anhydride, so $z=5$.
But, my answer was wrong (x only), as per the key. What am I doing wrong?

The answer given is,

 x = 4, y = 6, z = 5

Look at the cyclic form of glucose - how many OH groups are there?

@Mithoron wouldn't there still be 5 and then ROH/HCl breaks the ring?

What does methyl glucoside look like? Go from there.

@Mithoron, so that means y should be 5, if we are considering the cyclic form?

@user55119, does the cyclic structure justify for y's value? That's the reason I didn't started with cyclic one.

If you meant x's value, it's a mixed acetal between ROH and the C5-OH. That leaves C2, C3, C4 and C6-OH's available to acetylate. x=4. If you meant y's value, there isn't any cyclic form. It's hexaol. y=6.

@user55119, I meant value of "y". If I start with cyclic glucose, then how will I get hexaol?

The cyclic form is in equilibrium with the open form. The aldehyde is reduced to form glucitol, a hexaol.

Let us continue this discussion in chat.

I think there's nothing to discussed about this further as all comments directing to the correct answer: $x = 4, \ y = 6,$ and $ z = 5.$ I think it's better show in the structures:

Glucose is naturally occurring in pyranose hemiacetal form. When subjected to react with $\ce{H+/R-OH}$, the remaining acetal hydroxy group (anomeric hydroxyl group) will undergoes alkylation. Thus, the first scheme is showing alkylation of this anomeric hydroxyl group by methanol (Ref.1). The resultant methylglucose has four free hydroxy group, hence, $x = 4$.
When subjected to reduction with suitable conditions (e.g., $\ce{Ni/H2}$, $\ce{Ru/C/H2}$, etc.), the hemiacetal group would be reduced to a $\ce{-CH2OH}$ group (second scheme in the image). The resultant product is sorbitol (Ref.2), which has six free hydroxy groups, hence, $y = 6$.
Finally, pyranose glucose structure has five free hydroxy groups, hence, $z = 5$.

References:

B. Helferich, W. Schäfer (Checked by H. T. Clarke and M. R. Brethen), "$\alpha$-methyl d-Glucoside," Org. Synth. 1926, 6, 64 (DOI: 10.15227/orgsyn.006.0064).
J. Wildschut, J. Arentz, C.B. Rasrendra, R.H. Venderbosch, H.J. Heeres, "Catalytic Hydrotreatment of Fast Pyrolysis Oil: Model Studies on Reaction Pathways for the Carbohydrate Fraction," Environmental Progress & Sustainable Energy 2009, 28(3), 450-460 (https://doi.org/10.1002/ep.10390).

Starting from any cyclic form of glucose, there are two key ways to form the acetal in sequence (i). Taking glucopyranose as a starting point, here is a reaction scheme:

Initially, you could argue that you can protonate either the free hydroxy group (left-hand path) or the cyclic oxygen (right-hand path). The two paths give you either of the two aldehydes in the third row. The left-hand path is immediately complete once an alcohol molecule attacks and gives the O,O-acetal in the box. The right-hand path must continue (proton eliminations were forgotten in this row). A second protonation and elimination of water leads to the structure at the bottom left corner. Why did I write it in that way? Well, no matter how you do it, the hydroxy group on either C4 (not shown) or C5 are ready to attack the double bond and form another cyclic acetal. This is an intramolecular reaction and thus much faster than any intermolecular attack with a second alcohol molecule – even if you conduct the reaction using the alcohol as a solvent. Arguably, you will have a hard time going down the right-hand path at all, since the first two steps are highly reversible: the hydroxy group is, after all, still poised to re-attack the newly-formed carbonyl.
The take-home message is that it is very hard to access linear acetals from glucose. Under the standard reaction conditions, a cyclic acetal is the main product by far. Thus, only 4 hydroxy groups are available for acetylation.


text:
Will these K-alum crystals combine into a "cohesive/well-glued" mass?
The attempt to make a single solid mass of K-alum, using over 125 grams of McCormick K-alum powder, has resulted, in less than 2 hours, in:

So besides the growth around the loop of the seed "crystal", there are 3 other regions where crystals are forming.
Context: I am seeking a solid mass of K-alum to use as an astringent and antiseptic after shave (starting at 30 seconds into this video).
Questions: (1) Assuming that the over 125 grams of powder is large enough, will the bottom two or three masses combine into a big "cohesive/well-glued" mass? (2) Or should I dismantle the setup -- keep the seed-block, but heat the water to dissolve the other three masses to become very slightly under-saturated and put the seed back in?
Update: It has been under 3 hours, and the bottom three masses have combined into a single mass. I do not know how "cohesive/well-glued" that single mass will be when it is eventually removed and dried with a paper-towel.
The comment by @tstone-1 suggests this simpler approach for creating a single mass of K-alum: Heat about 200 ml of water to about 88C; saturate this with K-alum powder (about 100 grams); filter this solution into a paper coffee cup; cover cup and wait till most of the water evaporates; tear off paper cup to get solid mass. Question: how cohesive/well-glued will this mass be?
Update 2: Did end up with two consolidated masses, one from the starter near the rim and the other from the bottom three chunks;  both are suitable for use after-shave as an astringent and antiseptic:

There is a good chance they combine if there is plenty material dissolved. If you are trying to grow a single crystal, you likely want to avoid that. I can't see it well enough whether your seed is actually good enough for a single crystal attempt. The question depends very much on your intention with the final crystal, but if you want to grow a crystal, you will want to use supersaturated solutions.

Growing potassium alum single crystals is a typical experiment in our undergrad laboratories. From my very personal view and experience: While one can guess some larger single crystalline domains from your photograph, the stuff looks rather polycrystalline and aggregated. As you already noticed, this compound crystallizes very readily within hours (without seeding) and it is quite easy to obtain translucent and clear single crystals with edge lengths in the range of centimeters. Try to cool down the solution as slowly as possible, e.g. with the vessel placed within a larger bath.

Thanks -- have corrected my misuse of the term "crystal" and have made explicit the goal of the activity.

Perhaps the proper question for me to ask is "How best to make a block of K-alum from powder?" with the post made above being my attempt at doing so.

According to the Wikipedia article Potassium alum, the dodecahydrate melts at 92 to 95 °C. So if I wanted a large lump, I'd put the salt in a test tube (or some other container), dunk the container in boiling water to melt the salt, then pour the melted salt into some type of a mold to cool and solidify.

I tried the suggestion with a piece of Ammonium alum (which has a similar melting point as K-alum) that I had handy. On melting, the liquid was first colorless and then became light-green-blue in color. This liquid solidified as small clumps that adhered with each other -- but there is still some light-green-blue liquid left over. (So it is not like melting candle wax and re-solidifying it.) Is it known that melting K-alum and re-solidifying it will bring back K-alum?

Not sure exactly how to explain that. // Another option would be to press the powder into a pellet. In the XRF lab I had dies that made 1/2 inch and 1.25 inch pellets.


text:
Is acetylene IR active?
I've a very straight forward question, I'm confused because I've been told that a atom which has a oscillating transient/permanent dipole is IR active. But there are few asymmetric stretching/bending modes in acetylene? So, why does my teacher say it's IR inactive?

IR activity is not a property of an atom, nor that of a molecule. It is a property of a particular oscillation.

@IvanNeretin This should have been the answer. And its clear, concise, to the point, too.

Clear as it may be, my comment is way too short for an answer.

You may find some vibration frequencies of acetylene here.

Also, vibration patterns (active and inactive) here with animations.

Looks like it has an IR spectra to me:

Source: https://webbook.nist.gov/cgi/cbook.cgi?ID=C74862&Type=IR-SPEC&Index=1


text:
How can CO2 exist in two different phases?
Given a container of $\ce{CO2}$ at a pressure of 100 bar and 295 K, I check the chart below and find:

I'm within the liquid region
I'm before the critical point (304 K)

My question then is, why can I go purchase a tank filled at 100 bar that contains both liquid and gas?  What stops the gas from condensing, since according to this chart, the pressure and temperature map to a liquid phase.

Why is there water vapour in the atmosphere despite liquid water bing the most stable form an normal conditions?

Using a better phase diagram, it does not correspond to a liquid: https://chemistry.stackexchange.com/a/128112/79678.

A CO2 bottle at 20°C does not have 100 bar, but ~57! Or, if it has 100bar, there is zero gas in it, and it is going to explode if you leave it in the sun for a few minutes!

The $\ce{CO2}$ exists in two phases when you buy a partially-filled $\ce{CO2}$ cylinder because there is extra volume (which is needed to allow expansion as the container warms). Therefore, the pressure in the container is not 100 bar, but closer to 80 bar at 293K.
If the container had only enough room for the $\ce{CO2}$(l), then, indeed, there could be no gas phase. But if space is allowed above the liquid, then $\ce{CO2}$ evaporates, filling that space, and the $\ce{CO2}$(g) and $\ce{CO2}$(l) reaches equilibrium pressure.
On the molecular level, consider it an ongoing exchange between the molecules of liquid and gas. For the same reason, there is humidity in the air and oceans full of water, a 1 bar at sea level.

"But if space is allowed above the liquid, then CO2 evaporates, filling that space, and the CO2(g) and CO2(l) reaches equilibrium pressure." --
But after everything settle sand the bottle comes back up to room temperature, what will that equilibrium pressure be?  Is it the vapor pressure of liquid CO2 at that temp?

Yes. Vapor is compressible, liquid not much.

Thanks DrMoishe.  Would it then be to correct to say, if I had 3 bottles at room temperature filled with pure CO2 and nothing else (one where 25% of the volume is liquid, another at 50%, and another at 75%), the pressure of all three is the vapor pressure of liquid CO2 at this temperature?

Yes. The pressure would not go above that until it's 100% liquid (and you apply pressure, e.g., from a piston and spring); and would not go below that until it's 0% liquid, i.e., you're drawing gas from the container.


text:
Modelling pH as function of time and position
I already posted this on the Engineering Stack Exchange under Chemical Engineering as I don't know where it's a better fit.
I have a system whose initial state consists of three different sections with different chemical composition. All sections are aqueous solutions of different $\mathrm{pH}$ that are brought into contact for $t \gt 0$. The system is isothermal in each section and is of equal temperature in all sections and is assumed to remain isothermal, as heat generation is not considered. Section 1 (the leftmost section) is basic and contains reagent 1, section 2, the middle section, contains just water at a $\textrm{pH}$ of 7 and section 3, the rightmost section, is acidic. Section 1 and 3 are of equal size, whereas section 2 is larger. The system is three-dimensional, but can be reduced to one dimension due to symmetry (it is a tube in which angular and radial coordinates are assumed to not play a role). All sections can be considered equal in terms of physical properties (viscosity, density, etc.). I would like to model the $\mathrm{pH}$ as a function of time and position as there are reactions occurring whose reaction rate is a function of $\mathrm{pH}$. There is no convection, only diffusion. I already tried setting up a PDE system using
$$
\frac{\partial c_i}{\partial t}= D\nabla^2 c_i + r_i(\mathrm{pH},T)
$$
and solving it in Matlab, while setting a very high rate constant for the recombination of $\ce{H+}$ and $\ce{OH-}$. The Matlab solver fails as the recombination kinetics are too fast, but lowering them is not an option, as this wouldn't represent reality anymore (acid-base reactions are thought of the be instantaneous). How could I rewrite my PDE system to model the evolution of the $\mathrm{pH}$ at every position as well as calculate the concentrations of products and educts of reactions that depend on $\mathrm{pH}$ (but are only catalysed by it), without using reaction kinetics for the recombination of $\ce{H+}$ and $\ce{OH-}$ to make use of the fact that this reaction is instantaneous?

First I doubt that there is a closed solution. Second you have not given key details to even begin to work on the problem. All sections start at the same temperature? What is the physical shape of the three sections? Viscosity of the three sections the same? and so on...

Just updated the question to contain more information

When reaction is instantaneous, there's no reason to consider it at all.

Another thing "lowering them is not an option, as this wouldn't represent reality anymore" really? And const temp. and density are realistic? Not to say that given how slow diffusion is, you could make neutralisation billion times slower then it is and it wouldn't change a thing...

Often in cases like this its is necessary to use scaled units, say time in microsec, distance in mm etc to whatever suits your system. This may need some fiddling to get right.  Sometimes dimensionless units  are used but these are often hard to workout. Black-box integrators can be 'too smart for their own good ' and coding up, say, a simple modified Euler method will work, where the integrator fails,  and can give essentially identical results.

continues..The true H+OH reaction rate constant is so large that it will be diffusion limited (or controlled) ($k_{effective}=4\pi N RD$, R sum of radii, N Avogadro const, D diff coeff.) and so you may be able to use this to simplify the scheme.

@Mithoron The system is very small and has a large surface/volume ratio, so I don't consider temperature fluctuations right now, but will in the end. As the reaction is instantaneous, I used large rate constants to represent it that way, as the reaction I'm interested in is affected by the pH and thus artificially low recombination kinetics would presumably alter the outcome a lot. Because the reaction is instantaneous, I want to completely omit the neutralisation reaction and instead reformulate the PDE system while still accounting for a time- and position dependent pH (original question).

@porphyrin thank you, I will try that

To answer my own question:
Using either the diffusion limited or the literature values of the acid-base neutralisation (Baldyga et al., Non-isothermal Micromixing in Turbulent Liquids: Theory and Experiment), I was able to simulate the $\mathrm{pH}$ as a function of time by making the simplifying assumption of equal diffusion coefficients of proton and hydroxide ions. I re-expressed the proton concentration in terms of $\mathrm{pH}$:
$$
\frac{\textrm{d} c_{\textrm{H$^+$}}}{\textrm{d} \mathrm{pH}} = -\ln (10) \cdot 10^{-\mathrm{pH}}
$$
plugged this into my original equation and made everything dimensionless.


text:
Why and how is rust forming on moon?
note: @Mithron's proposed duplicate Why can't rust form without water? does not have anything about the conditions on the Moon, so no, it's not a duplicate.

Several popular news articles mention that "rust" has been found on the Moon, and this is surprising because rust requires the presence of both oxygen and liquid water.

nasaspaceflight.com: Rust on the Moon. How is that possible without oxygen and liquid water? "But how can rust form far from water ice deposits on a barren oasis devoid of oxygen?"
Axios: Researchers find rust on the Moon "Scientists were surprised by the findings because rust requires oxygen and water to form on Earth."

When I open a plastic-wrapped cake or muffin there is often an "O-buster" packet with finely powdered iron. When I open the packet (chemistry is more interesting than cake) and watch the powder, indeed it rusts. Of course air has plenty of both oxygen and water, so this is no surprise.
Question: But why and how is rust forming on moon? Why does it need both oxygen and water there, and more importantly, how is it getting them?

Hmm,but rust is not the same as hematite, that can be reportedly formed without water, e.g. as result of volcanic activity.

@Mithoron no it doesn't, because it doesn't seem to address what's happening on the Moon. Luckily though the answer posted here does! btw it's certainly to simply ask "Does this answer your question" in a comment without simultaneously voting to close, which has the primary effect of preventing other users from posting Moon-based answers.

@Poutnik I have not followed it closely, but on Mars the presence of hematite, or at least Martian blueberries is often cited as evidence of the previous presence of bodies of water, but I guess it's the shape that's key in this case, not simply the presence of mineral itself.

Does this answer your question? Why can't rust form without water?

UPDATE:
Today I got a chance to read the abstract the original paper which made news everywhere that moon has rust on the side which faces the Earth.
The authors propose that upper atmosphere oxygen from the Earth reaches the moon. In their words
"Oxygen delivered from Earth’s upper atmosphere could be the major oxidant that forms lunar hematite. Hematite at craters of different ages may have preserved the oxygen isotopes of Earth’s atmosphere in the past billions of years. Future oxygen isotope measurements can test our hypothesis and may help reveal the evolution of Earth’s atmosphere."
Widespread hematite at high latitudes of the Moon in Science Advances (open access

Let us split your query into three parts:
a) Will iron oxidize without water and form oxides?
Iron would happily burn in oxygen to form iron oxides. But we cannot call it iron rust. So iron can oxidize without moisture. In fact, iron powder may spontaneously burn in air.
b) Will iron form rust without water?
Rust is more of a semantic issue.  Iron rust may not have constant composition. As per the dictionary definition (OED) "

A red, orange, or yellowish-brown substance which forms progressively
as a flaking, permeable coating on the surface of iron and its alloys
as a result of oxidation, esp. through exposure to air and moisture.

Therefore, rust is an product of ambient environmental factors that affect iron at (approx.) typical room temperature. Rusting in sea water might different than rusting on a mountain. Acidic gases such as carbon dioxide, sulfur dioxide, nitrogen oxides certainly accelerate this process. Iron rust can be green and it can be common brown version. Since, chemically rust is a hydrated oxide of iron plus its hydroxide, water's presence is a must. More importantly water is a necessary ingredient which works as mediator of the electrochemical cell that forms between oxygen and iron surface. If you search Google Scholar you would find tons of articles on the electrochemical mechanism of iron rusting.
Coming to the more interesting question of rust on the moon:

nasaspaceflight.com: Rust on the Moon. How is that possible without
oxygen and liquid water? "But how can rust form far from water ice
deposits on a barren oasis devoid of oxygen?" Axios: Researchers find
rust on the Moon "Scientists were surprised by the findings because
rust requires oxygen and water to form on Earth."

Since you would know more astronomy than I do, wasn't moon a part of the Earth? I recall this from a documentary. I do remember seeing layers of "rust" in moutainous paths, so the oceans had a lot of iron which settled as a layer in mountains. If moon were a part of Earth, why is that surprising?
a) The question is what was the age of this rust?
Secondly, Mars is also "rusty", its surface is highly oxidizing with a huge amount of perchlorate in the soil (there is a long story how that was discovered by the folly of analytical chemists). Who knows what is the moon soil like. What about the shower of radiation on the moon. That may promote oxidation as well.

Just a small correction: there is almost surely no evidence of life on the Moon. Its creation has been geologically dated to well before any kind of life on Earth, much less multicellular life - Earth was still a molten ball of rock. Furthermore, its genesis was so cataclysmic that any evidence of life would have been destroyed in the first place (as well as resterilising Earth...). The impactor was roughly the size of Mars! That's no small pebble!

Good point. I do recall some mention moon fossils from some documentary that must be wrong.

@uhoh, You might have seen the original paper.

@M.Farooq oh, Widespread hematite at high latitudes of the Moon I haven't seen it yet, I need to be in certain places to access paywalled articles, but I'll make a point of getting it today! I've had my morning coffee but it hasn't really kicked in yet, so I'm a little slow right now :-)  update: Oh, this one is not paywalled! I can see it right now!

What is rust?

According to Wikipedia, rust is an iron oxide, a usually reddish brown oxide formed by the reaction of iron and oxygen in the presence of water or air moisture.

So, how does rust is forming on Moon although it has no Earth-like atmosphere, devoid of oxygen/moisture?

According to nasa.gov: moon has trace amounts of oxygen. The source of that oxygen is speculated to be Earth. Earth's magnetic field trails behind the planet like a windsock. In 2007, Japan's Kaguya orbiter discovered that oxygen from Earth's upper atmosphere can hitch a ride on this trailing magnetotail traveling 239,000 miles (385,00 kilometers) to the Moon.
More on space.com: Earth's magnetotail can reach all the way to the near side of the moon, where most of the hematite was found. Moreover, it was found out that at every full moon, the magnetotail blocks 99% of solar wind** from blasting the moon, drawing a temporary curtain over the lunar surface, allowing periods of time for rust to form. It is also possible that more rust were formed when earth and moon were close together in the ancient past

Where does water comes from?

Water ice has been found in shadowed lunar craters on Moon's surface. But those ice deposits are far from where rust is detected. So, scientist proposed that fast-moving dust particles that regularly pelt on the Moon could release water molecules from ice-water deposits. These dust particles carried the water molecules, gets implanted onto the lunar soil which contains iron, mixes with it and formed rust. Heat from these impacts could also increase the oxidation rate.

**What's with the solar wind?

The solar wind is a a stream of charged particles that flows out from the Sun, bombarding Earth and the Moon with hydrogen. Hydrogen makes it harder for hematite to form since, it's a reducer and while the Earth has a magnetic field and an ozone layer like atmosphere, shielding from hydrogen is possible, but not for the Moon.

Why it hasn't been detected earlier?

This rust formation is not sudden and took years to form, also taking into account the inhibition due to hydrogen from solar wind, the process was very, very slow. Also, the rust was detected on the far-side of the moon, where it is difficult for probes to reach, or proper satellite image was difficult to take, this is a mystery and it will take quite time to solve.


text:
Kinetics of Decomposition of Hydrogen Peroxide
Consider the decomposition of Hydrogen Peroxide:
$$\ce{H2O2 -> H2O + 1/2O2}$$
The  rate law for this reaction is :
$$\text{Rate} = k[\ce{H2O2}]$$
Further
$$\text{Rate} = -\frac{\mathrm d[\ce{H2O2}]}{\mathrm dt} = \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = 2\frac{\mathrm d[\ce{O2}]}{\mathrm dt}\tag1$$

Now consider :
$$\ce{2H2O2 -> 2H2O + O2}$$
The  rate law for this reaction is :
$$\text{Rate} = k[\ce{H2O2}]$$
Further
$$\text{Rate} = -\frac {1}{2}\frac{\mathrm d[\ce{H2O2}]}{\mathrm dt} = \frac {1}{2}\frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\tag {2}$$

My question:

Which one is correct - eq $(1)$ or eq $(2)$?

What is the general principle for deciding this?

Update:
As I have been informed through comments that both the equations are equally true, I have another doubt:
Consider half-life time of the reaction (first order kinetics):
$$t_{1/2} = \frac {\ln 2}{k}$$
As stated in the comments rate constant of both the reactions denoted by $k_1$ for $eq 1$ and $k_2$ for $eq 2$ are different. This means that the half life time of these reactions are different.
But the half-life time for first order kinetics is independent of initial concentration and no matter what equation we use , the actual reaction happening in the jar is the same. Provided this, how can there two different half-life times?

Both are correct. Why not?

As @IvanNeretin said, both eq. are correct. But, both of the reactions can't have the same rate constants. I think this is what you want to ask. ?

Sure, they are not the same. They are different; they differ by a factor of 2. That's not much of a surprise, since the reactions themselves are different.

@IvanNeretin So which is more generally accepted or is used in questions?

There does not seem to be an obvious preference.

@IvanNeretin Sir please see the updated part.

Try fully solving the whole rate equation.. You should get your mistake.. the effective k here would be same..

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details


text:
Denticity of bridging ligands
How would we assign denticity to bridging ligands?
For eg, consider $\ce{NH^-_2}$, it has 2 lone pairs and can hence act as a bridging ligand, making 2 metal-ligand bonds. So I thought it should be bidentate. However, my professor says it is monodentate, why so?

My professor told us bridging ligands are monodentate, including mu-hydroxo ligands

I was wrong. Jan's answer clears that up. Bridging ligands with a single donor atom are monodentate.

Does this answer your question? Why Cl⁻ can't act as bidentate ligand?

I think these two questions are sufficiently different to not warrant a dupe-closure. This one asks ‘What is the denticity of bridging ligands’ while the other asks ‘can an atom with multiple lone pair donate twice (I assume, although it isn’t implicit, to the same atom)?’. Full disclosure: I have a horse in the race as I have answered here.

It is important to clearly separate the terms denticity, hapticity and bridging.

An atom or a ligand is bridging is one single atom bonds to two different acceptors.

A ligand (but never a single atom) can be multidentate or multihaptic if two atoms of that ligand can bond to the same acceptor.

A ligand such as the amide anion cannot be multidentate or multihaptic as it only has one coordinating atom: nitrogen. This nitrogen can coordinate to two different metals (in a μ-amido complex) but it still remains a monodentate ligand.
A ligand such as ethylenediamine or en for short ($\ce{H2N-CH2-CH2-NH2}$) can coordinate to the same central atom via each of the two nitrogen atoms. This is a bidentate ligand as it has two ‘teeth’ in the central atom.
A ligand such as ethene ($\ce{C2H4}$) can coordinate to a single central atom in a bihaptic (η2) fashion. Drawing shortest lines from nucleus to nucleus, both carbon atoms would be the same distance from the acceptor atom (in reality, it is better thought of the π system donating). Hapticity is reserved for such cases in which a continuous set of atoms coordinate to a central metal.
A ligand can be both multihaptic and multidentate. An example would be dibenzylideneacetone or dba ($\ce{Ph-CH=CH-C(=O)-CH=CH-Ph}$), in which both of the $\ce{C=C}$ double bonds act as a bihaptic ligand. (Due to geometry constraints, dba usually coordinates to two different acceptors rather than to a single one in a chelate fashion.)
Finally, it is possible for a ligand such as en to be both bidentate and briding, if either of the nitrogen atoms coordinates to two different metals. This requires deprotonation of at least one nitrogen.
The bottom line is: briding ligands can be monodentate or multidentate; they can be monohaptic or multihaptic. Only bridging ligands based on a single donor atom must be monodentate.


text:
Comparing basic strength of phosphorus trichloride and phosphorus trifluoride
I need to compare the Lewis basic strength of trihalides of phosphorus
My thoughts :
Since fluorine is more electronegative, the percentage of $\mathrm s$-character in $\ce{P-F}$ bonds will be lower, which will in turn increase the percentage $\mathrm s$-character of the orbital which contains the lone pair of phosphorus atom .
So the lone pair will be closer to the nucleus and so its donation will be more difficult, and thus I came to the conclusion that the order should be
$$\ce{PF3}<\ce{PCl3}<\ce{PBr3}<\ce{PI3}$$
But I have read that $\ce{PF3}$ behaves as a better ligand in coordinate compounds.
Also, I have seen different orders at different places.
Could someone help me figure out the correct order and also the actual reason for that order, and also about whether lewis basic strength and ligand strength go parallel to each other or whether there can be discrepancies in order?

You need to distinguish between $\sigma$-donating and $\pi$-accepting properties. As the electronegativity of halide substituents increases, the $\sigma$ donor strength of the phosphorus atom will decrease. However, the $\pi$-accepting character will increase. The amount of back-donation from M will also depend on the $\pi$ acidity of the metal. It is often said that PF$_3$ behaves like CO in terms of ligand strength. In order to assess the donating properties of PX$_3$, compare one particular series of complexes [L$_n$M(PX$_3$)] and alter just X. Have a look QALE or Tolman's parameters.

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Indeed, ligand binding of PX3 is heavily influenced by pi backbonding.

@tstone So the basic strength order I've mentioned is correct , right ?

@ARROW If the question is solely about $\sigma$ donor strength: yes. However, if a ligand's strength (= Lewis basicity?) is to be classified according to the sum of $\sigma$ and $\pi$ donor properties, ask the question: At which system are we looking at? As I said, you need to look at one specific complex [L$_n$M(PX$_3$)] and not compare across systems. For X$_3$P$\cdot$BBr$_3$, the order given above is correct. In a textbook case, the poor $\sigma$ donor character of PF$_3$ will be overcompensated by its $\pi$-accepting properties, thus rendering PF$_3$ the strongest ligand.