text:
Finding time t in a first order reaction (JEE Adv 2019)

The decomposition reaction $\ce{2N2O5(g) ->[\Delta] 2N2O4(g) + O2(g)}$ is started in a closed cylinder under isothermal conditions at an initial pressure of $\pu{1 atm}$. After $\pu{Y \times 10^{3} s}$, the pressure inside the cylinder is found to be $\pu{1.45 atm}$. If the rate constant of the reaction is $\pu{5 \times 10^{-4} s-1}$, assuming ideal gas behavior, the value of Y is?

I used the equation given here for a first order reaction, $[A]=[A]_oe^{-kt}$
Using $k = \pu{5 \times 10^{-4} s-1}$, I am getting 4.606 but the answer given is 2.3.
Why is this wrong?
The question is from JEE Advanced 2019 paper 2.

You can't directly use this formula.. the co-efficients of rate are different.. I would suggest that you read through the proof of this property and then try to solve it from the first principles. Concepts matter..

@Maurice I have done this same question before(and got the same answer).. So, I would assume that the OP has blindly copied the formula and not gone through the derivation which makes a difference in this case (as you may notice).

If the pressure has increased from 1 atm to 1.45  atm, it means that the number of moles of gas was n in the beginning, and 1.45 n at the end. It means that 2·0.45 · n
mole of N2O5  has been transformed into 2·0.45·n mole of N2O4 + 0.45·n mol of O2. So [A]/[Ao] = 0.1/1 = 0.1. Do try to finish this calculation.

You should your work to get help here. Just copy and paste a portion of a web page won't help. Also, do not post a question as an image.

@Safdar: A great job on editing! There was a typo on equilibrium constant what I fixed. Also, make sure to do complete editing such as including appropriate tags and editing the title according to question. The title should be attractive to the keywords used by future searchers. Other than that, excellent job.

The basic law of chemical kinetics states for $\ce{aR -> bP}$, that
$$A = kR = -\frac{1}{a}\frac{\mathrm dR}{\mathrm dt} = \frac{1}{b}\frac{\mathrm dp}{\mathrm dt}$$
(where $A$ is activity and $k$ is called rate constant1)
Note that, in our case, the stoichiometric coefficient of reactant (i.e., $\ce{N2O5}$) is 2, therefore the differential rate equation becomes,
$$-\frac{1}{2}\frac{\mathrm dR}{\mathrm dt} = kR$$
or,
$$-\frac{\mathrm dR}{\mathrm dt} = (2k)R$$
On integrating and re-arranging, we get,
$$R = R_0 \mathrm e^{(2k)t}$$
which means, $$t = \frac{2.303}{2k}\log_{10}{\frac{R_0}{R}}$$
This factor of $\frac{1}{2}$ was what you missed, I think.

1: As per my first equation, it's clear how rate constant is specific to a certain reaction. The following is an extract from Concentration and Rates (Differential Rate Laws) (under the heading "Reaction Orders"):

The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent.

I don't follow, even if we change the coefficient of the equations by which I mean we multiply throughout with 2, still the reaction will happen in the same time right. What I mean if we the concentration drops by a factor of ten in ten minutes it would drop by ten in ten minutes whatever the stoichiometric coefficient are.

Think of it this way, when the co-efficient is 1, every second a certain number of molecules react (this is an oversimplification), now if the co-efficient becomes 2, then that would mean that the number of molecules used up per second doubles. and so the time taken to reach a specific concentration halves.

@Wood: I've corrected some parts of my answer. Read the first bracket carefully. Further, I agree that time is independent of concentration or pressure, but it is dependent on rate constant, which is what I've edited. Thanks for pointing out!

The other answer is correct, but I'd like to try out a different way of explaining it.

We know that at fixed volume and temperature, the pressure is proportional to the number of particles.  From the formula given, we get 45% more particles when 90% of the original $\ce{N2O5}$ is broken down, since it takes two to make one $\ce{O2}$.

The units of our rate constant include no concentration, so we know the reaction is first order, meaning it progresses at the same rate without regard to pressure. It means rarely, something just "goes wrong" with a molecule of $\ce{N2O5}$, at random, with no obvious cause.

Our rate constant can be rewritten as $\pu{\frac{1}{2000} s-1}$. The $\pu{2000 s}$ is the mean lifetime - like a half-life but the time for the overall amount of a starting reactant to be reduced to not 1/2 but to $1/e ≈ 1/2.718 ≈ 0.368$ of what it was.  (The reason for the odd number is that some of the remaining sample will survive multiple mean lifetimes, and when you average it all out, the mean lifetime is the average time any one molecule will survive)

We want the amount of $\ce{N2O5}$ remaining to be $10\% = 1/10$, so we have to figure out an exponent $n$ so $(1/e)^n = 1/10$.  In other words, we need $\ln(10) ≈ 2.30$.

At this point I'm just about ready to make the same mistake -- I get $2.30 \times \pu{2000 s}$.  But ... we need to remember that 2 molecules of $\ce{N2O5}$ are used per reaction, so the rate of its depletion is twice the rate of the reaction. So we divide by 2 to get $2.30 \times \pu{1000 s}$.

The computation is similar to half-life computations with radioisotopes, but in those we never have a stoichiometric coefficient to worry about because there's no way that each atom of, say, $\ce{^{32}P}$ can reliably "take out" another such atom when it decays.  Notice this time is also different from the time it would take 45% of $\ce{N2O5}$ to react if it did not affect a second molecule.  Even though it is not rate limiting, using up the second molecule of $\ce{N2O5}$ in each reaction reduces the amount available to continue reacting.

by e-life, did you mean mean-life ?

The 'e-life' is called the lifetime $\tau=1/k$


text:
How were urolithins discovered and named?
Urolithin is a product of the degradation of ellagic acid by gut flora and has significant biological activity. (Wikipedia)
I am curious as to how it was discovered but cannot find the story. Although I can imagine how it might have been discovered through the study of fecal flora, I cannot imagine why it was named with reference to kidney stones.

I strongly suspect that it's just derived from the word urolith, literally a urinary stone.

I just updated my answer with some links to free pdfs describing the very first isolation and naming.

I meant ‘fecal’

The identification of urolithins as products of human metabolism of ellagic acid was reported by Cerda et al. in the European Journal of Nutrition in 2004. The full reference is
Cerda, B., Espin, J.C., Parra, S., Martinez, P. and Tomas-Barberan, F.A. (2004) The potent in vitro antioxidant ellagitannins from pomegranate juice are metabolised into bioavailable but poor antioxidant hydroxy-6H-dibenzopyran-6-one derivatives by the colonic microflora of healthy humans. Eur J Nut 43:205-220.
However, these researchers did not name the compounds as urolithins. Rather, they note that the compounds

have also been found as constituents of ‘clover stone’, a type
of renal calculus found in sheep which can also ingest large
amounts of ellagitannins[26]. In fact, the metabolite d
(3,8-dihydroxy-6H-dibenzo[b,d]pyran-6-one) has been previously named
as ‘urolithin  B’ [22, 26].

That explanation seems sufficient to understand the source of the name, but if you are interested in the specifics of the original discovery, reference 26 is to the Dictionary of Natural Products published in 1992 by Chapman & Hall, which is available only by subscription. Perhaps someone with access to that Dictionary can look up the entry on urolithin and provide more info.
UPDATE: Since this answer has received a decent amount of attention, I dug a bit deeper to find the references for the first isolation and naming of the compounds (thereby actually answering the question).
In 1963 at the 430th meeting of the Biochemical Society in Oxford, England, Nottle and Pope first reported isolating two compounds that they named "urolithin A" and "urolithin B". Their report was published among the proceedings of the meeting in the Biochemical Journal, which are available for free from the journal here. The full reference is
Nottle, M.C. and Pope, G.S. Two Constituents of 'CloverStone', a Type of Urinary Calculus Found in Sheep in Proceedings of The Biochemical Society. Biochem J (1963) 89 (1): 1P–67P.
The Nottle and Pope report is on p. 67P. In that report, however, the structures of the urolithins were not described, as they had not yet been determined. In a subsequent paper, Pope reported the structures of the urolithins, noting that they were actually previously known compounds (specifically urolithin A is identical to pigment I of castoreum, isolated from beaver scent glands), but despite the compounds already being known, Pope and others continued to use the new name, which seems to have stuck. That report is also freely available from the Biochemical Journal here. The full reference is
Pope, G.S. (1964) Isolation of Two Benzocoumarins from 'Clover Stone', a Type of Renal Calculus Found in Sheep. Biochem J 93:474.

Thank you. I tried looking up some references but only abstracts were available in front of pay walls. Some of the prices for a single article are astonishing. I feel like we are going back to 18th century France where all the wealth was concentrated in the top 1% of the population.

Could you double check the first link, I get the message "Your session has timed out. Please go back to the article page and click the PDF link again."

@M.Farooq - I think I fixed it. Is it working now? Thanks for letting me know.

@Andrew, Yes, it works now. Thanks for demystifying the reference. The pagination as 67P was important. SciFinder needs to correct it.

As @Zhe says, the name urolithin A seems to have derived from the word urolith which is basically kidney stone. Quoting from here1:

The name urolithin was first given to two metabolites isolated from
the renal calculus of sheep (Trifolium subterraneum has been
reported as the cause of clover stone and might be a relevant source
of ellagitannins) that were named urolithin A and urolithin B. These
two molecules coincided with pigments I and II previously described
from castoreum and beaver glands

Also, a strain helps metabolizing urolithin to isourolithin-A and was named accordingly2:

Recently, a specific strain also isolated from human microbiota was
able to further metabolize EA up to isourolithin-A and was named as a
consequence Ellagibacter isourolithinifaciens. Nonetheless,
until now, no other bacterial strains that could metabolize EA to
urolithins B have been reported.

References

Biological Significance of Urolithins, the Gut Microbial Ellagic Acid-Derived Metabolites: The Evidence So Far by Juan Carlos Espín, Mar Larrosa, María Teresa García-Conesa and Francisco Tomás-Barberán, 2013, Article ID 270418, 15 pages,  DOI:10.1155/2013/270418
Food Ellagitannins: Structure, Metabolomic Fate, and Biological Properties
By Karen Johana Ortega Villalba, Fabrice Vaillant Barka, Carlos Vélez Pasos and Pablo Emilio Rodríguez, 2019, DOI: 10.5772/intechopen.86420

Looks like we were writing at the same time. . .

@Andrew Good answer complementing mine. +1

Thank you nilay, i gave Andrew the checkmark but your answer was also good. I imagine that sheep must have a chronic urinary infection with the necessary organism to synthesize it in their urine.

OP's request: I am curious as to how it (urolithin) was discovered but cannot find the story.

3,8-Dihydroxybenzo[c]chromen-6-one or 3,8-dihydroxy-6H-dibenzo[b,d]pyran-6-one (CAS #: 1143-70-0), now known as urolithin-A is first discovered as a metabolite of ellagic acid from mice in 2003 (Ref.1) by the same research group who also discovered that in as a metabolite in humans later as Andrew noted elsewhere:

However, given that urolithin-A has a CAS number of [1143-70-0] (and 3-Hydroxybenzo[c]chromen-6-one or urolithin-B; CAS #: 1139-83-9), it should have been known for an awfully long time. However, the important biological activity of this group of compounds is not known until 2003, yet keep in mind that pomegranate (Punica granatum) fruits have been used for centuries in folk medicine. The antioxident activity of this specific fruit was scientifically analyzed in 2000 and compared to that of red wine and green tea (Ref.2). However, urolithins are not identified, but the presence of ellagic acid ($301 \ m/z$; $\mathrm{M-1}$ peak) was confirmed among others with similar structural features such as gallagic acid ($601 \ m/z$; $\mathrm{M-1}$ peak). The presence of punicalagin ($1083 \ m/z$; $\mathrm{M-1}$ peak) was also confirmed. Following is the proposed metabolic path of urolithins (Ref.3):

References:

Begoña Cerdá, Rafael Llorach, José J. Cerón, Juan Carlos Espín & Francisco A. Tomás-Barberán, “Evaluation of the bioavailability and metabolism in the rat of punicalagin, an antioxidant polyphenol from pomegranate juice,” European Journal of Nutrition 2003, 42, 18–28 (https://doi.org/10.1007/s00394-003-0396-4)(PDF).
María I. Gil, Francisco A. Tomás-Barberán, Betty Hess-Pierce, Deirdre M. Holcroft, Adel A. Kader, “Antioxidant Activity of Pomegranate Juice and Its Relationship with Phenolic Composition and Processing,” J. Agric. Food Chem. 2000, 48(10), 4581−4589 (https://doi.org/10.1021/jf000404a)(PDF).
Juan Carlos Espín, Mar Larrosa, María Teresa García-Conesa, Francisco Tomás-Barberán, “Biological Significance of Urolithins, the Gut Microbial Ellagic Acid-Derived Metabolites: The Evidence So Far,” Evidence-Based Complementary and Alternative Medicine 2013, Volume 2013,  Article ID 270418, 15 pages (http://dx.doi.org/10.1155/2013/270418)(PDF).

Thank you Matthew. Indeed the story is peculiar. Looking at all those urolithins, I get the feeling they would like to intercalate between the bases of a DNA molecule. I will continue my research but sometimes it is hard as many articles are behind paywalls.

I understand your frustration. That's why I attached PDFs whenever I found them with answers. Good luck on your search and please post your finding as an answer to your own question here.

It is interesting that the very first modern reference to urolithin and its identification is in 1963 in the article cited as  Two constituents of clover stone, a type of urinary calculus found in sheep by Nittle, M. C.; Pope, G. S., in Biochemical Journal, 89(1), pg 67 is wrong even in SciFinder and in Google Scholar. If we check the journal website, there is no such paper with this title. I am surprised. Can anyone double check?
Anyway, looking up German sources from Google books from 1836 brings up a Homeopathy Pharmacopea, here. It shows the Greek etymology of urolithin, which is obvious "der Harn = urine" and "Stein = stone".
However, this "Urolithin hominum" indicates that is referring to human urine stones, but at least the etymology is confirmed :-)

source: Stack Exchange

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

sort: votes


text:
Finding the pH of the resultant solution when a pH1 solution mixed with a pH2 solution
I want to know that is it possible that a $\mathrm{pH_1}$ solution mixed with a $\mathrm{pH_2}$ solution and the mixture would be less than $\mathrm{pH_1}$?
Is it possible that the result has a lower $\mathrm{pH}$ value than the original solutions?

Strictly speaking no, you can write a free-energy diagram which would show that this is impossible. However, if the addition of the second solution somehow reacted to create a new compound that was less acidic then this could be possible.

And what is "PH2 solution "?

And it spells pH, not PH or ph.

If you have equal volumes of Solution A containing 0.01 M $\ce{HCl}$ and 0.1 M $\ce{BaCl2}$ and solution B containing 0.1 M apiece $\ce{Na2SO4}$ and $\ce{NaHSO4}$, each one has a pH of 2 or nearly so (based on the Henderson-Hasselbach Equation for B).  If you now mix them the barium ions from A selectively precipitate the sulfate ions rather than the bisulfate ions from B, forcing the latter to dissociate and the pH will drop below its initial value of 2.

Never answer a question with a negative score, maybe?

I upvoted both answers. Unfortunately, I seem to fall into the “answer questions with negative scores” pit on a regular basis. I am frequently surprised at what does not find favor.

I did so with the other answer.  @ed if you know me on this site you probably recognize that when I detect downvotes and no constructive criticism, I squawk.  Yup this is one of those squawks.

Believe it or not ! If a saturated NaCl solution is added to an equal volume of $\ce{HCl}$ $0.01$ M, the pH goes from $2$ to $1.4$ ! I know : it seems contradictory, and even incredible. Diluting an acidic solution with a non-acidic solution produces a more concentrated acidic solution. Personally when I read about this experiment a long time ago, I would not believe it. So I tried it at home, with a glass electrode, or an indicator. And I had to believe what I saw. The pH decreases by adding $\ce{NaCl}$. It even decreases more by adding a concentrated $\ce{CaCl_2}$ solution.
This cannot be explained by usual application of any Henderson-Hasselbach equation. It is only a question of activity. Remember that the pH is not minus the logarithm of the concentration of $\ce{H^+}$. No ! It is the logarithm of the activity of $\ce{H+}$. And the activity depends on the ionic strength, which depends on the concentration of all substances in solution.
It looks as if the activity could be considered as the number of moles divided by the volume of the "free water", if you admit that the ions $\ce{Na+}$ and $\ce{Cl-}$ can "catch" or "capture" a lot of water molecules in their Debye layers, so that there is practically no more water as solvant in the solution. I have said "as if". As if !

The glass electrode is sensitive to sodium ion as well as hydrogen ion. Under the conditions of this experiment, there is a large excess of sodium ion, so the glass electrode does not accurately measure the actual pH.

So all that is necessary is to change the ionic strength by using something that does not significantly interfere with the pH electrode’s performance.

"[The pH] decreases even more by adding $\ce{CaCl2}$ solution".  This would seem to point to a real water-activity related effect.

@ iad22agp. Of course the glass electrode is sensitive to sodium ions. But this change of pH could also be checked by using indicators like thymol blue, whose color changes between 1,2 and 2.8. And the effect happens with CaCl2.

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


text:
Why is the boiling point of trichlorofluoromethane lower than that of chloroform?
If $\ce{CFCl3}$ has a higher molar mass, how come $\ce{CHCl3}$ has a higher boiling point?
$$
\begin{align}
\mathrm{bp}(\ce{CFCl3}) &= \pu{23.77 °C} \\
\mathrm{bp}(\ce{CHCl3}) &= \pu{61.15 °C}
\end{align}
$$

related https://chemistry.stackexchange.com/questions/21613/melting-and-boiling-points-of-benzene-and-fluorobenzene

https://chemistry.stackexchange.com/questions/80695/why-are-fluoroalkyl-chains-hydrophobic-oleophobic https://chemistry.stackexchange.com/questions/43334/why-do-chlorinated-silanes-have-lower-boiling-points-than-their-methane-analogs

Does this answer your question? Reason for decreased boiling point of hexafluoroisoproponal?

@User 95897. Your question is contradictory. You say that CFCl3 has a lower boiling point than CHCl3. But it is wrong. CFCl3 boils at -23.77°C. This is higher than -61.2°C, where CHCl3 is boiling. Don't forget the minus sign.

@Maurice :-) I think those were meant as bullet points. The boiling point of $\ce {CFCl3}$ is ~296.9K (23.8°C) not -23.7°C

The main reason acting here is Dipole Moment. The higher the dipole moment of a molecule, the greater will be the intermolecular attractive forces and the higher will its boiling point be.
Dipole moment is the displacement of electron density in a molecule and it is a vector quantity. The net dipole moment of a molecule is the vector summation of all the individual dipole moment of 2 bonded atoms created by the electronegativity difference between them.
So, at first, let's get the structure of the molecules. Both of them are sp3 hybridised and have a tetrahedral structure.

For CHCl3, if you consider the bond between carbon and chlorine, its highly polar. Due to the high electronegativity of chlorine, there will be a net negative pole on the bottom part caused by the three Carbon chlorine bonds. And Hydrogen being less electronegative,however, will not attract the electrons to a great extent. Hence, there will be a pretty significant net dipole moment in CHCl3.

But in the case of CFCl3, yes, there will be a charge separation by the chlorine atoms, but on the upper portion lies the F atom which is very electro-negative. As a result, it will also pull its bond pair of electrons towards itself. Since, the net dipole moment is the vector summation of the individual dipoles, the charge separation of C-F, being in the opposite direction of the charge separation by the 3 chlorine atoms, will reduce the net dipole moment of the molecule.

For this reason, the dipole moment of CHCl3 is more than that of CFCl3 and so, CHCl3 has a higher boiling point.

In addition to the poor polarizability of $\ce{C-F}$ bonds described here, we may consider hydrogen bonding. $\ce{CHCl3}$ has this to a greater extent than most chlorinated organic compounds because the electronegative $\ce{CCl3}$ function draws off the electron density in the bonding orbitals, leaving more of the unoccupied antibonding orbitals on hydrogen.  We then have relatively good molecular orbital overlap between hydrogen in one molecule and chlorine in another.  $\ce{CFCl3}$ does not offer this possibility.


text:
sp2 hybridisation of alkyl radicals causing formation of racemic mixture

What I know about $\ce{^{.}CR3}$ radical is that it has both $\ce{sp^2}$ and  $\ce{sp^3}$ character but the $\ce{sp^2}$ character dominates and so the radical is $\ce{sp^2}$ in nature.
So, suppose we have the above situation. In this, a radical is formed  at  the carbon of the 6 membered ring at which one $\ce{-CH3}$ is attached since the bromination is selective, but will it form enantiomers? That is, can the $\ce{^{.}Br}$ radical attack from both sides?
In this question, the answer given is (c), but I think that enantiomers can not form, as the electron of the radical would be present only on one side of the radical carbon, and so cannot migrate to other side.

In the first step, due to the selectivity of bromine in a free radical reaction, the tertiary carbon which has a $\ce{-CH3}$ group attached to it forms the radical.
Now, since the radical is a $\ce{^{.}CR3}$ radical, the carbon radical in the intermediate becomes $\ce{sp^2}$. This means that $\ce{^{.}Br}$ can attack from both the top and the bottom since this lone electron in the $\ce{^{.}CR3}$ radical is in a p-orbital. This is because the electron is equally likely to be found in both lobes of the orbital.
Now after the attack of the $\ce{^{.}Br}$ radical, you get two products. The first one shows the product formed when $\ce{^{.}Br}$ attacks from the bottom and the second one shows the product formed after the $\ce{^{.}Br}$ attacks the intermediate from the top.
You may have noticed that both products are enantiomers of each other and hence the answer would be (c).
Reference
Chem Libretexts


text:
What is the origin of leaving group ability?
For example, during an SN2 attack:

In the previous organic chemistry book that I am self-studying with, it says the fluorine atom is a better leaving group (become fluoride ion). This is because HF (having a conjugate base of F-) has a pKa of 3.1 compared to CH4 (having a conjugate base of CH3-) - pKa = 50. Therefore, F- is more stable when it leaves the molecule and remains in the solution, making a better leaving group.

In this lecture that I am now self-studying, at 17:50, I learn the choice of the leaving group of SN2 in a very different view. F is a better leaving group is because antibonding sigma R-F has lower orbital energy compared to antibonding sigma R-CH3, so much lower activation energy is required to break the bond and as a leaving group.

From my understanding, the downward trend of antibonding energies from C-C to C-F is because the corresponding bonding sigma energies have an upward trend. As the atomic radius decreases from F to C in the second period, the bond length of C-X (X = C / N / O / F) becomes shorter, hence bond energy increase. (with C-N bond somehow as a weird exception for me, what caused that?)
In the lecture, the professor says about the leaving group abilities (which I think is about pKa) is totally 'fictitious' and the antibonding orbital energy is the reason behind.
What is the true reason for F in C-F a better leaving group or CH3 in C-CH3? (thermodynamic or kinetic) I am extremely confused.
Also, if there is anything/concepts that I say it wrong in my deductions above, please kindly let me know. I really want the right concept. Thank you.

A good paper to read is this by Jeremy Harvey https://doi.org/10.1002/anie.200501526 and references therein. They write that leaving group ability is a kinetic property. I think maybe we shouldn't say that it is because of pKa; instead that it is correlated with pKa (which as far as I am aware, is true up to a certain point).

I just see that the C-X bond cleavage of substitution reaction is heterolytic but not homolytic, so is it still plausible to use bond strength (based on homolytic bond cleavage)to determine the leaving group ability?


text:
Cleavage of C-C bond in 1,2-diketo compounds with periodic acid
Today we learned the Malaprade reaction in class. My teacher told me that a similar reaction can take place when the substrate is a 1,2-diketo or 1-keto-2-hydroxy or even 1-hydroxy-2-amino(in presence of periodic acid):-

(A similar one with 1-hydroxy-2-amino)
I wanted to know the mechanism of this reaction. This is the closest thing I could find.

Google is your friend!  A search of "periodate cleavage mechanism of diols" gives several well explained examples including https://www.masterorganicchemistry.com/2011/10/21/reagent-friday-sodium-periodate/

Yes, but I want to know that of "diketones"

The mechanism is the same as the reactive species is the enol

@waylander, is it thru the enol or the hydrate? Is it known?

I can find very little literature on the periodate cleavage of diketones to give a definitive answer. It could proceed through the hydrate depending on the ketone and solvent conditions.


text:
Comparing heat of hydrogenation among 1-methylcycloprop-1-ene, 3-methylcycloprop-1-ene and methylenecyclopropane
Problem

Which of the following orders is correct for heat of hydrogenation of these compounds?

A) 1 > 3 > 2
B) 3 > 2 > 1
C) 3 > 1 > 2
D) 2 > 1 > 3

Answer

 D) 2 > 1 > 3

My Attempt
Heat of hydrogenation is inversely proportional to stability.
Stability is directly proportional to the number of α-hydrogen.
The heat of hydrogenation is inversely proportional to the number of α-hydrogens:
1: one α-hydrogen;
2: four α-hydrogens;
3: five α-hydrogens.
Also, here in the 3, the intermediate formed is more stable due to dancing resonance.
So, the order of heat of hydrogenation must be 1 > 2 > 3, which contradicts the answer.
Edit:The given solution to this question contidicts our argument

Consider the angles of the double bonds. Stability is in this case principally determined by ring strain.

can you please elaborate it as an answer @Waylander?

why is the order heat of hydrogenation is failing when we compare with the $alpha$
hydrogens? @Waylander

You also have to consider factors like ring strain, resonance. Stability can't be decide just be based on one thing.

The number of alpha hydrogens is only a determining factor when considering unstrained alkenes or alkenes of identical strain.

ok, I understand,
but how can we conclude the order with the ring strain, please explain in detail.
(as i am a student)
@Safdar and waylender

I think the given answer is incorrect, as 2 is the most stable! Also, can you explain how do you see dancing resonance in 3?

same question. The answer in your key is wrong..

So the answer is correct.. If you find it satisfactory consider upvoting and accepting.. It helps.

In the given problem, we have three compounds and have been asked to compare their heat of hydrogenation.
Starting Premise
We start from the fact that the heat of hydrogenation of a compound is inversely proportional to the stability of the double bond in the system. Using this we can say that the opposite of the order of stability would be the answer
Comparing stability of the compounds
Here, the three compounds given contain three-member rings. Cyclopropane rings with all carbons being $\ce{sp^3}$ have a very high ring strain since the angle between the carbons are $\mathrm{60 ^\circ}$ when the expected $\ce{C-C}$ bond angle is $\mathrm{109.28^\circ}$.
Therefore, the higher the total strain, the higher the heat of hydrogenation, since the the final product is same in all cases. Finding the total strain in the three cases using WebMO, the total strain energy is as follows.
$$ \small
\begin{array} {lcc}
\hline
\text{Compound} &\text{Total strain energy(kcal/mol)} \\ \hline
\text{methylenecyclopropane} &\text{224.836} \\ \hline
\text{3-methylcyclopropene} &\text{389.564} \\ \hline
\text{1-methylcyclopropene} &\text{395.751} \\ \hline
\text{1-methylcyclopropane} &\text{90.490} \\ \hline
\end{array}
$$
Now, from this table, we can see that methylenecyclopropane (2) has the highest stability, then $\ce{3-methylcyclopropene}$ (1) and finally $\ce{1-methylcyclopropene}$ (3)
Edit: After Rahul Verma's comment, taking hyperconjugation(due to similar total strain) for (3) and (1), due to better hyperconjugation on (3), (3) would be more stable.
Therefore the order of stability would be (2) > (3) > (1)
Order of Heat of Hydrogenation
With the above data, we can say that the order of Heat of Hydrogenation would be (1) > (3) > (2)
Source:Relation between total strain energy and stability

I guess you forgot about hyperconjugation. If you'll take account of both the factors, then answer should be 1 > 3 > 2 (as I got). Btw, you can take ring-strain of 1 & 3 to be nearly same.


text:
Stability of radicals
I recently came across a question about the stability of radicals and this option was given as correct.

$\ce {F2C^.H}>\ce {C^.H3}  $

And the solution was given that $\ce{FC^.H2}$ and $\ce{F2CH^.}$ are more stable than $\ce {C^.H3}$ but $\ce{C^.F3}$  is less stable. Can anyone explain why?
If it has got something to do with inductive effect then all three $\ce {F2C^.H},\ce{FC^.H2},\ce{C^.F3}$ should be less stable than $\ce {C^.H3}  $. Right?

you might also want to consider lone pair repulsion between F atoms.

https://chemistry.stackexchange.com/questions/49754/what-is-the-hybridisation-of-trifluoromethyl-free-radical

@RahulVerma Do you have anything to back up that claim? Normal radicals are electron-deficient.

@Zhe: No. I don't have. As the claim was wrong.


text:
Substitution reactions haloalkanes confusion
"In haloalkanes, the large electronegativity difference between halogen and carbon atoms means that the carbon-halogen bond is polar. This makes the bond weaker than one without a dipole" - excerpt from my textbook
I don't particularly understand the last sentence. Can someone elaborate and explain what does "this makes the bond weaker than one without a dipole" means?
the strength of the inter molecular force goes the following: Dispersion, Dipole–Dipole and Hydrogen Bonding.
From my interpretation: Without dipole is equivalent to dispersion and dispersion is subsequently weaker than dipole dipole. But why does dipole dipole weaker than one without a dipole? Shouldn't it be stronger?
That sentence continues with "Species such as another halogen ion, a cyanide ion, ammonia can be substituted for the halogen in the polar bond."
Any help will be great. Thanks.

Polar bond is not weak, but can easily be breakdown in polar and ionic solvents due to electrostatic interaction. In such solvent the negative and positive part of the solvent attracts the positive and negative part of the haloalkanes, thus facilitates the bond breaking of haloalkanes.

Please provide a reference to the textbook you are citing.

@Manu I am a novice. So in a polar solvent, C-F bond is the weakest (highest electronegativity difference, hence the highest electrostatic attraction). In a non-polar solvent, C-I bond is the weakest (longest bond length, hence strongest bond strength)? Thank you!

Yes, due to high electrostatic interaction with the polar solvent component.


text:
What are the most potent hydrogen peroxide decomposition catalysts available to an amateur?
Some months ago I was playing with $60\,\%$ $\ce{H2O2}$ and unknowingly I used a silver spoon to transfer the liquid from a container, and the peroxide immediately started bubbling (it did look and sound like violent boiling) around the spoon.
Are there solids or thick liquids available to an amateur that are even more potent in catalysing the decomposition of $\ce{H2O2}?$

I did not downvote, but I am a bit concerned about the “playing with” 60% hydrogen peroxide. It is dangerous, causes nasty painful burns (worse than the 30% grade) and practically everything catalyzes its highly exothermic decomposition. Silver, including sterling silver teaspoons, is one of the many substances that act as efficient catalysts. You may well be lucky nothing really bad happened. Please read up on hydrogen peroxide and be safe: use the appropriate PPE!

“And it is a nerve-wracking experience to put your ear against a propellant tank and hear it go "glub" -long pause- "glub" and so on. After such an experience many people, myself (particularly) included, tended to look dubiously at peroxide and to pass it by on the other side.”
― John D. Clark, Ignition!: An informal history of liquid rocket propellants

What "potency" do you mean? While probably nothing compares with catalase present in a humble potato, it's concentration in tissues is low. BTW if you somehow have access to 60 % H2O2, you should already know such things.

Well there is manganese dioxide, there is iron(III) nitrate, there is  iron(III)chloride.

Another thing is what's "liquids available to an amateur" supposed to be mean, when hardly any "amateur" has access to 60 % H2O2!


text:
is there any single non toxic compound/chemical which can give direct color change by react with pure ethanol?
I want single compound/reagent/chemical/paper/dye which gives colored reaction with pure ethanol and it should be non toxic as well.
By simply adding/diping it to pure ethanol it should give color change at room temperature.

Addition of Pottasium dichromate would do the trick.. Basically anything that reacts with the compouund that results in a change

@Safdar But the non-toxic thing is a problem! ;)

@Safdar Not sure what you mean. In any event, potassium dichromate and potassium permanganate are toxic. What the OP seems to want is essentially a test strip for ethanol. Or compound, etc. And it apparently needs to be specific, i.e., does not give false positives for methanol, etc.

@EdV well that scares me. We actually had to tests with Potassium permanganate to check whether a certain compound was an alkene or not. thanks for the clarification.

@Safdar No need to be scared of potassium permanganate: it has many uses, as discussed in the wikipedia article and elsewhere. But it does need to be respected as a strong oxidizer.

This PDF has a list of reactions that show different color changes for various reactions.. You might want to consider the Sucrose test.. However, i do not know the validity..

Does adding an acid to acidify the medium count as a reagent?

If you are looking for an edible reagent, I can offer egg whites (albumin) for some consumptions. The egg protein is denatured in the presence of ethanol, and a white precipitate is formed. This method is known as cooking egg without heat but scientifically used for purification of lysozyme and ovalbumin. I think this is the best easy-access and non-toxic reagent for detecting ethanol.

What would happen with pure methanol or pure isopropanol or with vodka (any proof)?

It seems all water-soluble alcohols denature albumin (https://doi.org/10.1002/bip.1978.360170907). It's better to know egg white is a non-toxic alternative, not a reagent mainly.

So then there is nothing special about ethanol. I think the OP’s question is in need of elaboration: what specificity or selectivity for ethanol do they require?

No, this is not a specific reagent to ethanol. Also, I'm not sure there is a quite non-toxic compound to detect ethanol.

That is the problem: the OP does not tell us enough about what they really want. For example, maybe the “test strip” would change color if and only if it was wetted with pure ethanol. Nothing else would cause a color change. Then it would be very useful in detecting adulterated potable alcohol. But it is hard to achieve total specificity/selectivity and also have adequate sensitivity. So I did not downvote, but I have a guess why others did so: what the OP wants probably does not exist. Of course, maybe they will edit their question to clarify exactly what they seek.


text:
Infra red stretching frequency of carbon monoxide ligand in borane carbonyl compared to that of free carbon monoxide
Why is the infrared stretching frequency of $\ce{CO}$ in the borane carbonyl $\ce{BH3CO}$ more than that of free $\ce{CO}$?
Is there any real explanation for this?
There will be a small amount of hyperconjugation of $\ce{H}$ atoms from boron to carbon which would increase the electron density of carbon closer to boron due to which the $\ce{CO}$ bond strength should weaken.
Am I wrong anywhere or there is any other theory that explains this?

Actually the stretching frequency of CO in the adduct increases, hyperconjugation disfavours the experimental results because due to hyperconjugation, the CO bond weakens.Only Sigma donation to p orbital of B atom will increase the CO frequency in the adduct.


text:
Do all reducing sugars show mutarotation?
Do all reducing sugars show mutarotation?
I studied that for a sugar to be reducing it must possess a free aldehydo or keto group. Also, for a mutarotation interconversion between α- and β-forms of compound are required.
If there is a free aldehydo group or keto group in the compound must it show mutarotation?
Note: I am a high school student so please try to explain in a simpler manner.

https://www.masterorganicchemistry.com/2017/08/17/mutarotation/

The compounds having hemiacetal in their structure give mutarotation.
Generally, all monosaccharides (glucose, fructose, ribose, arabinose, …) and disaccharides (lactose, maltose, …) give mutarotation, except sucrose.
Sucrose, despite being a disaccharide, fails to give mutarotation because it doesn't contain hemiacetal.


text:
Silver "polish" with aluminum foil left a hard yellow coat
I cleaned my silver teapot with the aluminum foil-baking soda-hot water method.
Usually this works well on my silver plated flatware.
However,  this time it left a hard yellowish coating.  I was only able to remove it by very strenuous manual polishing with a silver polish paste and when desperate, I even used toothpaste. The trade off was that it could scratch very finely,  but that was preferable to the yellow.
Now I'm afraid to use this method on unusual pieces.
--> How can I prevent this from happening again?
Are there kinds of silver and metal combinations that produce this reaction?
Is there an easier way to remove it?
Thanks!

Usually that happens when the metal is not 100% pure silver. Alloys don't clean very well with this Al foil method. Contact the manufacturer to find out if is 100% pure. If you also add a picture that would be useful.

It might help if you edit your question to state the exact steps in your silver cleaning method and also check your silverware: Sterling silver is 92.5% silver and 7.5% copper. But if the flatware is not that, then maybe it is silver filled or silver plated. If it is Sterling, there should be a “925” on it somewhere.

If the teapot has value to you, I suggest taking it to a professional jeweler and having them tell you exactly what it is and how much it will cost to restore its beauty. People here like to help, but nothing beats having an experienced professional examine the teapot and tell it like it is. All the best with this!

The source of the "hard yellowish coating" may be Silver carbonate (Ag2CO3), as per Wikipedia:

Silver carbonate is yellow but typical samples are grayish due to the presence of elemental silver.

I suspect the teapot in question is "electroplated Britannia metal" per Wikipedia with a nickel alloy as the base metal, as to quote:

After the development of electroplating with silver in 1846, Britannia metal was widely used as the base metal for silver-plated household goods and cutlery.[5] The abbreviation EPBM on such items denotes "electroplated Britannia metal". Britannia metal was generally used as a cheaper alternative to electroplated nickel silver (EPNS) which is more durable.

The Baking Soda will interact with any soluble Ag ions to form the unstable Silver bicarbonate which breaks down to Ag2CO3 on warming.
The question is what is the source of Ag+?
My speculation is that the Al foil was completely coated given the surface area of the teapot, and the nickel-based alloy is notoriously cathodic (hence, the use of Nickel alloys in coins). So now, the Silver itself can be subject to an anodic attack in the presence of H+ (from amphoteric NaHCO3) and O2 (from air exposure), resulting in Silver ions.
This was a somewhat unique situation, hence the successful use of the Al/NaHCO3 process on the Silver-plated flatware.

(+1) Thought provoking answer: it will be interesting if EPBM appears on any of the items! I suppose the 6% Sb in the Britannia metal is no health issue. ;)

Thanks Ed V, yes thought provoking, but I also have an alternate path, perhaps less consumable for the general population, presented here, at a link you supplied:  https://chemistry.stackexchange.com/questions/35213/yellow-stains-after-cleaning-silver/136635#136635 which may actually be more accurate.

I deleted that link (though you have it in your comment) because I am now thinking that the OP should simply take the teapot to a jeweler and get a professional, hands on opinion. Ultimately, I think only that will be satisfying, even if it turns out the teapot has little intrinsic value. But it was a nice little puzzle and thanks for the Britannia metal information! By the way, there are already two close votes and I did not downvote or vote to close.


text:
Amine vs Amide Solubility
I had a question regarding the solubility of amines and amides.
I was looking into the solubility of butanamide and n-Butylamine, and it turns out that whilst butylamine is miscible in water, the solubility for butanamide is 163g/L
This would imply that amines have a greater solubility than amides. My understanding is that
the carbonyl group would add two lone pairs, which act as hydrogen bond acceptors and thus should mean amides are more soluble. (more hydrogen bonds)
It'd be much appreciate if someone could point me in the right direction or let me know how my reasoning is flawed.
(I have viewed this link but the answer given isn't too helpful: Solubility of Amides)
Thanks in advance!
Sources for Solubility Values
https://en.wikipedia.org/wiki/N-Butylamine
https://chem.libretexts.org/Under_Construction/Walker/Chemicals/Substance%3AB/Butanamide

amines are basic - that is very big factor in plus for dissolving 2) A solid simply cannot be miscible with liquid, there has to be some limit.

The mixing of two compounds is a process which requires consideration of three types of interactions: solute-solvent, solvent-solvent and solute-solute. You make a good argument for the solute-solvent interaction being stronger in mixtures of butanamide and water, compared to butylamine and water. However, as you say, this argument alone would predict a higher solubility for butanamide, which goes against the observed data. Also, in both cases the solvent-solvent interaction is the same (water with water), so this can't really be the source of an explanation.
By exclusion, we're guided towards considering the solute-solute interactions. At a first glance, both butylamine and butanamide display a combination of dipole-dipole and hydrogen bonding interactions, making it tricky to compare how their relative strengths change between compounds.
To make things easier, we can refer to additional experimental data. A nice way to gauge the strength of intermolecular interactions in a substance is to look at its melting and boiling points. According to Wikipedia, the melting and boiling points of butylamine are approximately -49 °C and 78 °C respectively. Meanwhile, for butanamide, the values are a whopping 115 °C and 216 °C! Clearly the mutual interactions between butanamide molecules are much stronger than those between butylamine molecules.
And there is our answer. You are correct that, compared to a butylamine molecule, a butanamide molecule should interact more strongly with water molecules. However, it turns out that butanamide molecules also interact very strongly with themselves, far more so than butylamine molecules. Taking everything into account, the effect of the solute-solute interactions is more prevalent; the water molecules have more difficulty keeping butanamide molecules apart, to the point that after a critical value of 163 grams of butanamide in a liter of aqueous solution (at 15 °C), the water molecules just can't stop the butanamide molecules from meeting and packing into a solid. Even if you add more butanamide, it doesn't visibly dissolve. Some butanamide molecules in the solid continuously manage to break loose from each other and get pulled into the aqueous solution, but they are replaced just as fast by molecules of butanamide already in solution which get too close and pack into the solid state.
You will find that this can be made into a very rough rule of thumb: when comparing the solubility of two roughly similar substances in a same solvent, if one of the substances has significantly higher melting and boiling points, they are likely to be less soluble.
Note that I have discussed only the influence of the enthalpic contributions to mixing. An entropic contribution also exists (favouring dissolution for both compounds), but in this case it can be safely neglected due to its comparatively small effect.


text:
Why does spontaneity of a reaction depend on temperature?
As we know, $$\Delta G=\Delta H - T\,\Delta S.$$
Both in my textbook and on the internet, it was given that temperature was a determining factor in the spontaneity of a reaction. For example, when $\Delta S$ is negative (unfavorable) even then the reaction can still be favorable at lower temperatures.
If $\Delta S = q/T,$ then wouldn't the entropy also decrease by the same factor on decreasing the temperature? In other words, why don't the temperature terms cancel leaving $\Delta G$ independent of temperature?

https://chemistry.stackexchange.com/questions/33953/trying-to-understand-entropy
Does this help?

Consider a melting point - the spontaneity of the reaction is clearly dependent on temperature.

If you decrease T and keep q constant, won't the entropy change increase (according to your equation)?

The issue here is in your definition of entropy. According to your definition:
$$\mathrm dS = \frac{\text{đ}q}{T}.\tag{1}$$
However, there is a small difference.
As you may know, the entropy of a system is a state function and so it only depends on the initial and final states. Therefore, if the entropy of the system were to be calculated for a specific path for which the parameters are simple to find, then the value would not change irrespective of the path.
The actual formula of change in entropy is:
$$\mathrm dS = \frac{\text{đ}q_\mathrm{rev}}{T}.\tag{2}$$
This value is calculated for a reversible isothermal process.
Now, let us assume that we have taken such an isothermal reversible process and discovered the value of $\Delta S$ and now used an isobaric process to discover the value of $\Delta H.$ Now, we solve for $\Delta G$ in our isothermal process (this is done so that we can assume constant temperature in our formula).
Therefore, for such a process
$$q_\mathrm{rev} = nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{3}$$
Thus, we get change in entropy as
$$\Delta S = nR\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{4}$$
Now, enthalpy would be a constant value that we discerned from the isobaric process.
Therefore, since $\Delta G = \Delta H - T\Delta S,$ we get
$$\Delta G = \Delta H - nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{5}$$
As you can see, there is still a temperature dependence.
This is because $\Delta S$  is temperature independent and so you would still have a temperature dependence for $\Delta G.$
Another thing that you may have forgotten to take into consideration is the fact that $q$ is dependent on the temperature.
The above example is that for an ideal gas expansion.
As Poutnik stated in the comments in case of a reaction, the calculation of $\Delta H$ becomes more complicated as we have to take into consideration the different molar capacities of the reagents and products which would $\Delta H$ would also be a function of temperature.
For standard value of entropy, we use the formula
$$\Delta S^\circ_\mathrm{rxn} = \sum nS^\circ_\mathrm{products} - \sum nS^\circ_\mathrm{reactants}.\tag{6}$$

Things get more complicated, as delta H =f(T), because of the Hess law and different molar heat capacities of reagents and products.

I think OP assumed no phase change single component systems.

My note does not consider phase changes. With a single component, there is no reaction.

@Poutnik, you misunderstand me. I meant that the OP was considering the expansion of ideal gases.. Adding this conception would be helpful to the question.. However I do not know this well enough and so cannot elaborate.

"Spontaneity of reaction"

Alright thanks. Should I add it to my answer or would you craft an answer for that.. It was my misunderstanding then. This was the simplest way to explain.

Put it into your answer. :-)

I. You've left out the differential from your equations (1) and (2).   Equation (1) (the incorrect one) should read:   $dS= \frac{\text{đ}q}{T}$. And equation (2) (the correct one) should read:   $dS= \frac{\text{đ}q_{rev}}{T}$  For more on this, see my answer here: https://chemistry.stackexchange.com/questions/136237/difference-between-Δq-and-dq/136254#136254
II.  $\Delta S$ and $\Delta H$ for chemical reactions are not independent of T.

@theorist, i haven't said the $\Delta S$ is independant of T. The example given is for an gas expansion. The end of the answer states so/

You did say "$\Delta S$ is temperature independent".  More broadly, his question was about chemical reactions ("spontaneity of a reaction"), so I'm not sure why you were using an ideal gas expansion here.  $\Delta S$ for chemical reactions is temperature-dependent.


text:
In which nucleophilic aromatic substitution rate is faster m-fluoronitrobenzene or p-fluoronitrobenzene?
In which nucleophilic aromatic substitution rate is faster: m-fluoronitrobenzene or p-fluoronitrobenzene?

SNAr on m-nitrofluoro species is by several orders slower than p-nitro

The rate of reaction depends on stability of intermediate. The given reaction follows $\ce{S_N2Ar}$ mechanism through "Meisenheimer intermediate". So, the stability of "Meisenheimer" primarily drives the reaction.

ortho and para isomer are almost same except their position, which is the main reason why is the former reacts much slowly than the latter.

In the ortho isomer, the negative charge of the ring isn't delocalised with $\ce{NO2}$, whereas in the para isomer, it is delocalised, as seen in the following image,

(Source: Master Organic Chemistry)


text:
What will be the major product when conjugated alkene (like 1,3-butadiene) is reacted with HBr in presence of R2O2?

What will be the major product when conjugated alkene (like 1,3-butadiene) is reacted with $\ce{HBr}$ in the presence of $\ce{R2O2}$?

As we know, two products are formed when no $\ce{R2O2}$ is used because of conjugation in double bond and carbocation. But when $\ce{R2O2}$ is used, free radical is formed.
So, will there be the same case in the following question as well? Also, if answer is yes, then what will be the major product?

Please convert the image into a better form.. You can use this to do so.

The 6-member ring is highly suggestive...


text:
Comparing bond angles in carbonyl dichloride and carbonyl dibromide
Background
I was studying VSEPR and hybridization theory and one of the popular questions in many textbooks was comparison of bond angle of $\ce{NH3}$, $\ce{NF3}$ and $\ce{NCl3}$.
$\ce{NF3}$ had smaller bond angle as electrons were nearer to the more electronegative fluorine atom, thereby reducing bond pair - bond pair repulsion (of N-F bonds) and thus allowing contraction of bond angle.
Further, in $\ce{NCl3}$ bond angle was larger than both $\ce{NH3}$ and $\ce{NF3}$ due to steric repulsion between large Cl atoms. So the order was $\ce{NCl3}$>$\ce{NH3}$>$\ce{NF3}$.
Now, a question that I later saw in one of my textbooks was the analysis of bond angles of $\ce{COCl2}$ and $\ce{$\ce{COBr2}$}$, which have trigonal planar geometry. According to my textbook, $\ce{COCl2}$ has smaller $\ce{O-C-Cl}$ angle, less than $\mathrm{120^{\circ}}$, (and thereby greater $\ce{Cl-C-Cl}$ angle). The angles for $\ce{COBr2}$ were not given.
My Approach
Now if we apply the reasoning given for $\ce{NCl3}$ and $\ce{NH3}$, $\ce{NF3}$ in this question comparing $\ce{COCl2}$ and $\ce{COBr2}$, we would find that the $\ce{C-X}$ bond pair electrons would be more polarized towards X, if X is Cl, and thus analogously to $\ce{NF3}$, it should contract leading to a smaller bond angle than in $\ce{COBr2}$.
Also, due to steric repulsion, $\ce{C-X}$ bond angle in $\ce{COBr2}$ should be greater than $\ce{COCl2}$.
Solution given in the book
The author has stated that there are four electrons in the $\ce{C-O}$ bond
Thus, bond pair repulsion between electrons of $\ce{C-O}$ bond and $\ce{C-X}$ bond would be much greater than any repulsion between the electrons of the $\ce{C-X}$ bonds.
As electrons in the $\ce{C-X}$ bond would be closer to carbon atom (and thus $\ce{C-O}$ bond pair electrons) in $\ce{COBr2}$, therefore a greater amount of repulsion would be present between them. This repulsion would lead to contraction of Br-C-Br bond angle leading to an angle less than that of $\ce{COCl2}$.
However, no data was provided in support or against it. Also no reasoning was given why they would be greater or less than $\mathrm{120^{\circ}}$, only the above reasoning for comparing the bond angle was provided.
I have not been able to find any bond angle data. My teachers are also not sure. Also, the textbook I referred to is infamous for having lots of errors.
Which reasoning is correct, or is it something else entirely?
What is the actual data?

I believe your textbook is incorrect with regard to COCl2 bond angles. It is the Cl-C-Cl bond angle that is less than 120; it is approx. 111 degrees (making the O-C-Cl angle 124), at least in the solid phase. The crystal structure was reported in 1952 (Acta Cryst 5:833). The Schupf computational lab at Colby College gives the same bond angles for COBr2 as for COCl2, but I don't know if that is experimentally verified or not.

This data is available on the Computational Chemistry Comparison and Benchmark DataBase.These would be the values of $\ce{COCl2}$, $\ce{COBr2}$ and $\ce{COF2}$

The reasoning you utilized for comparing the bond angles of $\ce{NH3}$, $\ce{NF3}$, and $\ce{NCl3}$, while sensible, is simplistic and works best only for a few cases where weighing the factors of steric repulsion against bond pair-bond pair repulsions is feasible. It seems that you have encountered this very problem in the case of comparing the bond angles of $\ce{COCl2}$ and $\ce{COBr2}$.
To circumvent these issues, a better method(less prone to failures) would be using Bent's rule. Now, there has been a lot of discussion on this website over what this rule actually is (see here) and why it is physically useful (see this). The Wikipedia article on the same has a wonderful intuitive discussion under the "Justification" tab for why this rule exists. However, I will not be delving into the origins of this rule, and just state outright that the basic premise of this rule is that not all hybrid orbitals are equivalent under a given hybridization scheme. The percentage s-character and p-character (that is, the amount of s-orbital or p-orbital present in a given hybrid orbital as a percentage of the whole) varies as per the electronegativities of the substituents.
A statement of this rule was provided by Henry A. Bent as follows:

Atomic s character concentrates in orbitals directed toward electropositive substituents.

The idea is that the amount of p character will increase (alternatively, the amount of s character will decrease) in those bonds which have more electronegative substituents (that is, the atom other than carbon in the bond), while simultaneously the amount of p character will decrease(alternatively, the amount of s character will increase) in bonds towards less electronegative substituents in such a manner that the net sum of percentage s and p characters over all the bonds in the molecule comes out to be $100$%. This is essentially a redistribution of the orbital energies across bonds to minimize the energy of the system. Further details on this rule can be found in the links mentioned above.
Once you are aware of this rule, applying it to your case becomes a fairly simple task. I will try to compare the $\ce{Cl-C-Cl}$ and $\ce{Br-C-Br}$ angles here, the orders for the $\ce{O-C-Cl}$ and $\ce{O-C-Br}$ angles can be easily inferred from there. From a classical hybridization perspective, the carbon in both these molecules uses $\ce{sp^2}$ hybrid orbitals for bonding, and all the three hybrid orbitals thus generated should possess equivalent atomic orbital's character ($33.33$% s character and $66.66$% p character approximately). But as per Bent's rule, since $\ce{Cl}$ and $\ce{Br}$ are both more electronegative than carbon, they will increase p character towards their direction (that is, in the $\ce{C-Cl}$ and $\ce{C-Br}$ bonds respectively). Since $\ce{Cl}$ is even more electronegative than $\ce{Br}$, the p character increase in the $\ce{C-Cl}$ bonds will be more than that what will happen in the $\ce{C-Br}$ bonds.
Hence, we can say that more p character will be directed in the two $\ce{C-Cl}$ bonds as compared to the two $\ce{C-Br}$ bonds. This directly bears a relation to the bond angle: more is the p character in the constituent bonds, lesser will be the bond angle(you can see this from a fairly easy observation: $\mathrm{sp}$ hybrid orbitals are generally at an angle of $180°$ with $50$% p character, the $\mathrm{sp^2}$ hybrid orbitals are generally at an angle of $120°$ with $66.66$% p character and the $\mathrm{sp^3}$ hybrid orbitals are generally at an angle of $109.5°$ with $75$% p character).
Hence, Bent's rule predicts that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ should be lesser than the $\ce{Br-C-Br}$ of $\ce{COBr2}$. Experimental data says that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ is $111.83 ± 0.11°$, while the $\ce{Br-C-Br}$ of $\ce{COBr2}$ is $112.3 ± 0.4°$. Hence, the prediction made by Bent's rule is confirmed.
Notice that the difference between the given bond angles is quite small (around $1°$). This is also somewhat expected, as the electronegativity difference between chlorine and bromine is also quite small (approximately $0.2$).
The advantage of employing this method is that you don't have to consider conflicting factors like those you stated in your approach. The rule explains these angle variations based on Coulson's theorem, from which the following relation pops out:

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$
for $\theta \in (90^\circ,180^\circ)$

($\theta$ stands for the bond angle while ${P}$ and ${S}$ stand for p character and s character respectively)
As you can clearly see, when the p character goes up, the value of $\cos \theta$ also goes up if you closely observe the simplified form $\cos \theta = 1 - \frac{1}{P}$. This means that the value of $\theta$ will decrease for the given range of $\theta$ (as $\cos \theta$ is a decreasing function for $\theta \in (90^\circ,180^\circ)$)
Hence, you can generally predict things quite accurately with Bent's rule without getting confused with contrasting factors, as they are usually taken care of while determining the atomic orbital characters.

Ok great answer. One thing I want to clarify is whether the double bond with oxygen has any role or not? Does it affect bent's rule? Also if we take formic acid, would you expect any significant difference in p character of both oxygen atoms?

The double bond doesn't pay any special role as such,but it does help to determine the initial bonding scheme. What I mean to say is, the doubly bonded oxygen along with the two $\ce{C-X}$ bonds help to determine an initial hybridization scheme of $\ce{sp^2}$ for both the molecules,after which a more realistic picture can be determined by reassignment of s and p characters in accordance with Bent's rule

As for formic acid, it may initially seem like both the oxygens have different p characters directed towards them by looking at the separate equivalent Lewis structures.However,it is well known that both these oxygens in formic acid are involved in resonance,ans therefore neither of the two canonical structures gives the true picture;in reality a resonance hybrid exists which has equal double bond character between both the oxygens.Hence,the p character between both the oxygens would ultimately be equal(in a rough sense, the p character would be somewhere between $\ce{sp^2}$ and $\ce{sp^3}$)

@YusufHasan, the two oxygens in formic acid cannot be equivalent, unless you allow the proton to hop between the two (which is not resonance). Perhaps you meant to talk about formate ion.

@orthocresol Yes, I meant to talk about the "proton hopping" but couldn't find a better term for it. How would you frame this equivalence under which the proton is equally likely to protonate either of the two oxygens?

@YusufHasan Equivalence depends on the timescale that you are talking about, see my answer at https://chemistry.stackexchange.com/a/134072/16683. Unless specifically in a context where the proton transfer is considered "fast", I would not say that the oxygens are equivalent. That's certainly not the case here, because when you talk about s- and p-character, you're essentially referring to an optimised ground-state geometry (any sort of vibrational motion that changes the bond lengths or angles will affect the s- and p-character)...

... indeed, these optimised geometries are exactly the kind of "instantaneous" or "static" picture that I mention in that answer. Even though proton transfer is fast compared to some other chemical processes, it still takes a finite time for it to happen, and so cannot be incorporated into the "static" picture. Reusing the photography analogy: a person can walk between two places many times, but it's not ever possible to take one photo where they're at both places simultaneously.

The thing is, you don't need to go to these lengths to talk about formic acid, because the entirety of your comment applies to the formate ion. (I notice you describe it as a "resonance hybrid", which is true for formate, but certainly not formic acid.) And your answer itself is quite fine; I was only commenting about your comment.

@orthocresol Oh,I see.. So when we are discussing about s and p character in general, are we using a measuring method "slower" or "faster" than the proton transfer(as you talked about in your answer)?

I'll refer you to my second comment; I guess you hadn't read it yet. Sorry for spamming. :-)

@orthocresol Ok,I guess I understood after reading your comments and answer, that while resonance in the formate ion is something that creates equivalence even in the instantaneous or "static" picture for the carboxylate anion, proton transfers are subjected to a finite transfer time,and so they don't reflect equivalence in a static frame for the carboxylic acid.So, different p characters should be directed towards the two oxygens in formic acid,while my original comment above would apply to the "formate anion" more suitably than "formic acid" itself. I hope I got this correctly,thank you :)

That's a good summary!

@YusufHasan which oxygen would you expect to have more p character in formic acid or, any ester where there is no "proton hopping" and why? Also would the difference be significant?

@ManitAgarwal-Elpsycongroo In that case, I would say that more p character would be directed towards the doubly-bonded oxygen from carbon's hybrid orbitals.This is because as per the classical hybridization theory, the doubly bonded oxygen would be roughly $\ce{sp^2}$ hybridized, while the -OH oxygen would be $\ce{sp^3}$ hybridized. This means that, the C=O oxygen would be slightly more electronegative than the hydroxyl oxygen(more s character means more electronegativity)....

...Hence, the carbon will direct it's p character towards the more electronegative of oxygens(the double bond). The difference shouldn't be very significant though

The reasoning provided in the book conforms with the computational data and hence can be presumed correct. As the $\ce{C-X}$ bond lengths of increases, the position of electrons moves away from the $\ce{CO}$ (carbonyl electrons), and hence they are repelled less. Another proven method for the claim can be using the bent's rule provided by Yusuf Hasan. To support the above claims:
Here I present the $\ce{X-C-X}$ bond angles as calculated by Geometrical optimization by method Hartree-Fock on the basis set: Basic 3-21G using GPU accelerated engine  provided as TeraChem. (The computations were iterated for lowest possible error by computational analysis)
The computation finished in {1.92sec,5.20sec,12.63sec,38.80sec} for all halogens {F,Cl,Br,I}

The data below each image corresponds to the Natural Bond Orbital Analysis as suggested by Orthocresol. The currently essential p-character data for Carbon-Halogen and Halogen-Carbon bonds have been depicted as below. The computations were made by Engine: Gaussian, using method Hartree-Fock on the basis set: Basic 3-21G. (Here BD stands for a 2-centered bond with the labeled atoms[i])

$\ce{COF2}$

$\ce{COCl2}$

$\ce{COBr2}$

$\ce{COI2}$

Here we can observe that the p-character for $\ce{X-C}$ increases quite significantly (as per considerations of Electronegativity) as we go from Fluorine to Iodine and for $\ce{C-X}$ bond, the p-character decreases distinctly enough as we go from Fluorine to Iodine.
This computational evidence is in support with the arguments stated in the book, hence they are supposedly accurate.
Note:The data above has been calculated and the experimental error (considering the experiments as true value falls below 1.5% for Phosgene$^{[1]}$ and 0.33% for carbonyl fluoride$^{[2]}$ which seems a reasonable and acceptable precision.
1.Nakata, M.; Kohata, K.; Fukuyama, T.; Kuchitsu, K. (1980). "Molecular Structure of Phosgene as Studied by Gas Electron Diffraction and Microwave Spectroscopy. The rz Structure and Isotope Effect". Journal of Molecular Spectroscopy. 83: 105–117. doi:10.1016/0022-2852(80)90314-8.
2.https://pubchem.ncbi.nlm.nih.gov/compound/Carbonyl-fluoride

@Safdar Also, the links you have provided except for the wikipedia one are goverened by computations (that's the thing we werent trusting)... For a counter I refined my computational observations and provided a pubchem cite.

@Safdar My computations conforms to the methods of computation adopted by Tera chem. As a known fact, the bond angle is always subject to change due to uncontrollable environmental factors, that does mean that the computational data is correct for idealized environment. COI2 doesnt exist, that does not mean an optimization cannot be predicted.

@Safdar The very first line of the answer says computational data, which directly applies the fact that many things are idealized, so an experiment can never be 100% conforming with the computed data. You have no evidence of your claim that the ans is wrong. Considering the experimental data obtained (as cited wikipedia), the error margin falls below 1.5% and the data for COF2 as provided on #cccbdb site conforms with error margin of 0.33%. I think this much precision would be enough to lay computed claims.

It's one thing to optimise the geometries, find the bond angles, and say that it matches the trend. It's another thing to claim that the rationalisation given (about electron repulsion etc.) is valid. Your results only show that the bond angles increase going from F -> Cl -> Br -> I. They don't show why this is so. Showing why something happens can be a rather involved process that often necessitates further analysis of the computational results, e.g. orbital analysis (NBO), etc.

Also why would optimize with 3-21g? If you have TeraChem and GPU it seems a waste

@orthocresol Agreed, I will consider NBOA as well.

@CodyAldaz I did an accurate as well, but there were no bundle definitions for Iodine...so had to resort to basic 3-21g.

@orthocresol check it out now.


text:
What acids, solvents is lithium-6 fluoride soluble in (other than HF)?
$\ce{LiF}$ is quite insoluble in water, and I'm trying to use some water-soluble acid or solvent to dissolve $\ce{^6LiF}$ (95% enriched) in. I'm not a chemistry person, I'm a radiation detection person, and I have very little idea of chemistry. Could anyone please help?

It might help to say more about your goal: do you want to prepare a different lithium salt or clean the LiF off of a sample cell or whatever.

@EdV thank you for replying! I'm trying to prepare a water-soluble 6Li compound. It can be any salt, hydroxide, or anything, as long as it has 6Li and is water soluble. My goal is to detect neutrons using the 6Li, but my detector is set up such that i need it to be in a water-soluble form.

The obvious path is also definitely not something a non-chemist should ever attempt: only an experienced fluorine chemist, with proper lab facilities, safety gear and calcium gluconate ointment (just in case) would maybe do that and I will not elaborate. With apologies to bush pilots, there are old fluorine chemists and bold fluorine chemists, but no old, bold fluorine chemists. If I think of something safe, I will post a comment or answer. And maybe someone else here will do so: lots of really good chemists here. Stay tuned.

@EdV I enjoyed that answer! Are you talking about HF, by any chance? We definitely want to steer clear of any HF in our detection system. And if you are talking about something else, I apologize in advance. In fact, I can use a different compound of Li other than LiF if possible, but again, my knowledge in the matter is very limited. I will surely stay tuned for an answer. THank you!

Yeah, HF. Lithium chloride is highly soluble in water, so if you could just get that, then problem solved. As for fluorine chemists, it is kind of a joke that they might be missing digits, etc. I was giving a tour (to parents of prospective undergrads) of our new chemistry lab building one time, and mentioned that joke. Sure enough, one of the people on the tour was a chemist and he verified that the joke was no joke. I stopped talking about that on subsequent tours.

@EdV LiCl will work I think. Thank you! Perhaps I shall research a little more into this joke and use it sometime if I'm around chemists! :)

Glad to be of help and have an upvote!

$$\ce{2 LiF(aq) + CaCl2(aq) -> CaF2(s) + 2 LiCl(aq)}$$


text:
Sodium metal from electrolysis of molten sodium bisulfate?
My question is relatively straightforward. Can sodium metal be extracted from molten sodium bisulfate via electrolysis?
I have done it before with molten sodium hydroxide, but there is such a small temperature range to keep the salt molten and prevent the extracted sodium from dissolving back into the melt.
I then find out that sodium bisulfate melts at a much lower temperature than sodium hydroxide. This gives me much more wiggle room for error.

Metallic sodium reacts with molten sodium bisulfate to produce hydrogen gas, according to : $\ce{2 Na + 2 NaHSO_4 -> H_2 + 2 Na_2SO_4}$. So you will get $\ce{H_2}$  instead of $\ce{Na}$.

The melting point of anhydrous $\ce{NaHSO4}$ is $\pu{315 ^\circ C}$. But, it decomposes to $\ce{Na2S2O7 + H2O}$ at  $\pu{315 ^\circ C}$ (Wikipedia).

You may use $\ce{NaNO2}$, which melts at $\pu{271 ^\circ C}$. Keep the bath below  $\pu{300 ^\circ C}$.

Another problem: The sulfur(VI) is also reducible. Assuming you actually use the dehydration product $\ce{Na2S2O7}$ (liquid range $401-460°\mathrm{C}$) which would avoid the possibility of hydrogen reduction, your cathodic reaction could produce sulfur dioxide or sulfite ion ($\ce{S(VI)->S(IV)}$) instead of sodium metal. Halides, which are not reducible, make for a more predictable cathodic reduction reaction.


text:
Dissolution of iron(III) hydroxide
Problem
From the theoretical tour of the Mendeleev Olympiad in Chemistry:

a) Calculate the minimal amount of sulfuric acid that should be diluted with water to yield $\pu{100 mL}$ of solution that could completely dissolve $\pu{0.4 g}$ of rust.

Answer

 a) After the dissolution of a $\pu{0.4 g}$ portion of rust, the total concentration of iron(III) in a $\pu{100 mL}$ portion of solution is $\displaystyle\frac{\pu{0.4 g}}{\pu{107 g mol-1}\times\pu{0.1 L}} = \pu{0.037 M}.$ As there are no competing reactions (sulfate ion does not form stable complexes with iron), it is easy to find the minimal solution acidity after the dissolution using the solubility product $K_\mathrm{s} = [\ce{Fe^3+}][\ce{OH-}]^3$:

 $$[\ce{H+}] = K_\mathrm{w}\left(\frac{[\ce{Fe^3+}]}{K_\mathrm{s}}\right)^{1/3}  = \pu{0.0097 M},$$

 that is $\pu{0.49 mmol}$ of sulfuric acid in a $\pu{100 mL}$ solution. In addition, $\pu{0.037 M}\times 3\times\pu{0.1 L}\times 1000/2 = \pu{5.6 mmol}$ of sulfuric acid has reacted with iron hydroxide, therefore, the total amount of sulfuric acid required for the dissolution is $\pu{6.1 mol}$ (2 points).

My thoughts
This task seemed incredibly easy, but the solution overcomplicated some aspects, from my point of view.
The concept of "rust" was simplified in this problem since it was stated from the beginning that here "rust" means, in fact, $\ce{Fe(OH)3}.$ So, I wrote the chemical equation (which, because it is a neutralization reaction, should have an astonishingly large constant and is an equilibrium practically shifted fully towards the products):
$$\ce{2 Fe(OH)3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O}$$
$\ce{Fe(OH)3}$ has a molar mass of $\pu{106.87 g/mol}$. Finally, the amount of sulfuric acid is
$$n(\ce{H2SO4}) = \frac{\pu{0.4 g}}{\pu{106.87 g mol-1}}\times\frac{3}{2} ≈ \pu{5.61 mmol}.$$
After checking the solutions, not only I realized I were wrong (even though my final answer was part of theirs), but I also were confused by the explanations given. For some reason, the solubility product was included, even though its constant is so insignificantly small in comparison to the constant associated with the reaction I wrote above.
What I thought was happening in this system was that $\ce{H2SO4}$ would neutralize the rust and that was basically it. While I am familiar with the concept of equilibrium and I understand that $\ce{H2SO4}$ doesn't nullify the concentration of $\ce{Fe(OH)3},$ in the light of the huge constant the neutralization is expected to have, the solubility product, $K_\mathrm{sp},$ should be negligible.
However, it seems that not only it is not negligible, but it also seems to be able to influence the final result significantly. But why is that? Why do we bother calculating the additional acidity from the hydroxide ions resulted in the solubility reaction? Why can't we solve the problem stoichiometrically, the way an inorganic chemist would approach it? Please correct my reasoning wherever you consider it wrong.

Even this query has a down vote. Someone must be really sick and disturbed to read this question.

The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations.
Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized:
$$\ce{Fe(OH)3(s) <=>[H2O] Fe^3+(aq) + 3OH-(aq)}$$
$$K_\mathrm{sp} = s \times (3s)^3 = 27s^4 \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{27}\right)^{\frac 14} = 1.008 \times 10^{-10}$$
This is $[\ce{Fe^3+}]$ in water. Thus, $[\ce{OH-}]$ would be $3s = 3.024 \times 10^{-10}$.  This value is much smaller than autoionization values of water. Therefore we have to consider $[\ce{OH-}] = 1.00 \times 10^{-7}$ for this calculation. Hence, if revise the calculation accordingly:
$$K_\mathrm{sp} = s \times [\ce{OH-}]^3 = (1.00 \times 10^{-7})^3s \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{1.00 \times 10^{-21}}\right) = 2.79 \times 10^{-18}$$
Thus, when autoionization of water is considered, $[\ce{Fe^3+}]$ in water at $\mathrm{pH} \ 7$ is much smaller (the actual value). I did these calculations to show the effect of autoionization of water. Similarly, when real $[\ce{Fe^3+}]$ is high in the solution like the case (a) here, $K_\mathrm{sp}$ plays a role.
Let's re do the calculation again considering $[\ce{Fe^3+}] = 0.037 M$. If all $\ce{Fe(OH)3}$ stay dissolved, them we can find the $[\ce{OH-}]$ in the solution using $K_\mathrm{sp}$ calculations:
$$K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3 =  0.037 \times [\ce{OH-}]^3 \ \Rightarrow \  \therefore \ [\ce{OH-}] = \left(\frac{2.79 \times 10^{-39}}{0.037}\right)^{\frac 13} \\ = 4.22 \times 10^{-13}$$
Thus, you need to keep $[\ce{OH-}] = 4.22 \times 10^{-13}$ to keep the solution away from precipitating back to $\ce{Fe(OH)3}$. Thus, $[\ce{H+}]$ should be at:
$$ [\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{1.00 \times 10^{-14}}{4.22 \times 10^{-13}} = 0.024$$
Thus, after completely dissolving rust, you need to add additional $\ce{H2SO4}$ to keep the $[\ce{H+}]$ of the $\pu{100 mL}$ solution at $\pu{0.024 M}$. I think you can calculate that amount in $\pu{mmol}$ easily now (Keep in mind that you already added $\pu{5.6 mmol}$ of $\ce{H2SO4}$ to dissolve $\pu{0.4 g}$ of rust).
Note: Any deviation from the given answer should be due to the $K_\mathrm{sp}$ value used here (The given answer didn't give that numerical value).

This is a tricky question. You have answered it well. I am also somewhat confused. I am wondering that $K_{sp}$ is an equilibrium value, calculated from the equilibrium concentrations of iron (III), and hydroxide ion. I am imagining that the aqueous solution is saturated with rust. How "valid" it is put the total dissolved iron concentration, after acid dissolution there in this expression: $K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3$. There is no "rust" anymore after dissolution.

@M. Farooq: After dissolving all rust, the $[\ce{Fe^3+}]$ is large enough so that $Q_\mathrm{sp} = [\ce{Fe^3+}] \times (1.00 \times 10^{-7})^3 \gt K_\mathrm{sp}$, the condition to precipitate. So you add extra acid to get $Q_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH}]^3 = K_\mathrm{sp}$,

Excellent explanation! I now understand the point of further addition! So we only have to take into consideration the necessary condition for precipitation-dissolution, imposed by the constant. Now when I think about it, it has become obvious. Thank you very much!

The solubility product of $\ce{Fe(OH)3}$ is about $\ce{10^{-39}}$. So its influence on the real solubility of $\ce{Fe(OH)3}$ is not significant. The following equilibrium constants are much more important : $$\ce{[Fe(H_2O)_6]^{3+} <=> [Fe(H_2O)5(OH)]^{2+} + H+ ; \ ...\ K = 10^{-3.05}}$$ $$\ce{[Fe(H_2O)_5(OH)]^{2+} <=> [Fe(H_2O)4(OH)_2]^{+} + H+ ; \ ...\ K = 10^{-3.26}}$$  Another difficulty is the fact that these ions has a tendency to dimerize, according to : $$\ce{2[Fe(H_2O)_6]^{3+}<=> [Fe(H2O)_4(OH))]_2^{4+} + 2 H+ + 2 H_2O \ ...\ K = 10^{-2.91}}$$ So it is not easy to calculate the pH of hydrolyzed Fe(III) solutions.
Ref. R.N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon Press, Oxford $1984$, § $25.3.4$, page $1265$.

The Olympiad Tour provided only the values for Ksp and Kw, so I suggest we stick to these constants only. Yes, I appreciate correctness and rigour, but let's refrain ourselves from overcomplicating the solution. I would like to know why does the official solution (which I am by no means refuting) take into consideration the solubility product  and why isn't the fully stoichiometric approach (which is widely used in inorganic chemistry) correct?


text:
Naming a given organic compound according to IUPAC naming rules

Would the given organic compound be called "3,4-diethyl-3,7,7-trimethylnonane" with IUPAC naming rules?

I don't really know if this should be called 3,4-diethyl-3,7,7-trimethylnonane, because it could also be called 6,7-diethyl-3,3,7-trimethyl something.
I am trying to do this by IUPAC rules.

Which locant set do you think is lower, 3,3,4,7,7 or 3,3,6,7,7?

@FadedGiant i think it would be the first one, but is that the reason why it is correct? I thought it was because it also needed to be numerically lowest -> highest

Yes, your naming would be correct. For saturated hydrocarbons, the first point of difference rule takes precedence after the longest carbon chain. To apply this rule, the locants of each set are ordered numerically in a list from lowest to highest, and the series that has the lower number at the first occasion of difference is determined as the lower locant set, regardless of what comes after this first point of difference. In this case, "3,3,4,7,7" would be the lower locant set than "3,3,6,7,7", because 4 is lower than 6.


text:
Alkoxide metathesis
I am attempting a synthesis for acetylacetone. The preparation I wish to carry out is described on orgsynth by condensation of acetone and ethyl acetate with a base catalyst.
The problem for me is that I only have $\ce{t-BuOK}$, no THF and $\ce{t-BuOH}$ in sight, only $\ce{EtOH}$. First I searched for solubility of $\ce{t-BuOK}$ in ethanol, only to find nothing. It gave me some ideas why there is no data for the solubility so I assumed already that they react (Why didn't I think of this in the first place). I instantly looked up for this procedure but I could not find anything satisfactory for this specific reaction, only patents about metathesis reactions of alkoxides with higher carbon chain alcohols. On the Wikipedia site there is a line of information that was somewhat useful, which states that "many alkoxides are prepared by salt metathesis from sodium ethoxide", but still not satisfactory since its the reverse what I am looking for.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{EtOH}$ is 15.9 while $\mathrm{p}K_\mathrm{a}$ of $\ce{t-BuOH}$ is 16.5 in water. Looking at the values in DMSO the value of $\ce{EtOH}$ is still lower, so perhaps it can be generally stated the ethanol molecule has more tendency to deprotonate. Assuming and without cited literature by using excess $\ce{EtOH}$ (and as a solvent of course) I am expecting them to produce $\ce{EtOK}$ and $\ce{t-BuOH}$ by equilibration the mixture. I think that which base I use does not really matter in the reaction, but I'm short of solvents for solving $\ce{t-BuOK}$ efficiently and a little tertiary alcohol wont mess up my experiment. I am very confident about this simple metathesis reaction, but I would like to hear your ideas too before doing anything.
So my question, in short is the following: Will $\ce{t-BuOK}$ react with $\ce{EtOH}$ to produce $\ce{EtOK}$ and $\ce{t-BuOH}$?
I am of course aware of the dangers of these compounds, and have a basic experience in doing simple preparations like these.

The key question here is whether you can do the condensation using t-butoxide in EtOH and in my opinion you can. Whether the functional base is t-butoxide or ethoxide is immaterial, it will do the reaction.

It is my intuition too, but that is another topic my question is not about the condensation.

The answer is yes, it will be in following equilibrium:
$$\ce{EtOH + t-BuO- <=>[$K$] EtO- + t-BuOH}$$
$$K = \frac{[\ce{EtO-}][\ce{t-BuOH}]}{[\ce{EtOH}][\ce{t-BuO-}]} = \frac{[\ce{EtO-}][\ce{H3O+}]}{[\ce{EtOH}]} \times \frac{[\ce{t-BuOH}]}{[\ce{t-BuO-}][\ce{H3O+}]} =  \frac{K_\mathrm{a}(\ce{EtOH})}{K_\mathrm{a}(\ce{t-BuOH})} \\ = \frac{1.26 \times 10^{-16}}{3.16 \times 10^{-17}} \approx 4$$

Edit:
Although the individual $\mathrm{p}K_\mathrm{a}$ values used to determine the $K$ in this equilibrium are calculated for solutions in water (some of those are only available in DMSO other than water), I assumed the ratio might be closer to the real value. However, an empirical conversion method that transforms $\mathrm{p}K_\mathrm{a}$ values of arbitrary organic compounds from one solvent to the other is introduced in this reference (Ref.1). Unfortunately, the full paper is behind paywall, but its abstract states that:

An empirical conversion method (ECM) that transforms $\mathrm{p}K_\mathrm{a}$ values of arbitrary organic compounds from one solvent to the other is introduced. We demonstrate the method’s usefulness and performance on $\mathrm{p}K_\mathrm{a}$ conversions involving water and organic solvents acetonitrile ($\ce{MeCN}$), dimethyl sulfoxide ($\ce{Me2SO}$), and methanol ($\ce{MeOH}$). We focus on the $\mathrm{p}K_\mathrm{a}$ conversion from the known reference value in water to the other three organic solvents, although such a conversion can also be performed between any pair of the considered solvents. The ECM works with an additive parameter that is specific to a solvent and a molecular family (essentially characterized by a functional group that is titrated). We formally show that the method can be formulated with a single additive parameter, and that the extra multiplicative parameter used in other works is not required. The values of the additive parameter are determined from known $\mathrm{p}K_\mathrm{a}$ data, and their interpretation is provided on the basis of physicochemical concepts. The data set of known $\mathrm{p}K_\mathrm{a}$ values is augmented with $\mathrm{p}K_\mathrm{a}$ values computed with the recently introduced electrostatic transform method, whose validity is demonstrated. For a validation of our method, we consider $\mathrm{p}K_\mathrm{a}$ conversions for two data sets of titratable compounds. The first data set involves 81 relatively small molecules belonging to 19 different molecular families, with the $\mathrm{p}K_\mathrm{a}$ data available in all four considered solvents. The second data set involves 76 titratable molecules from 5 additional molecular families. These molecules are typically larger, and their experimental $\mathrm{p}K_\mathrm{a}$ values are available only in $\ce{Me2SO}$ and water. The validation tests show that the agreement between the experimental $\mathrm{p}K_\mathrm{a}$ data and the ECM predictions is generally good, with absolute errors often on the order of 0.5 $\mathrm{pH}$ units. The presence of a few outliers is rationalized, and observed trends with respect to molecular families are discussed.

For example, $\mathrm{p}K_\mathrm{a}$ values of phenol in water, DMSO, methanol, and acetnitrile are illustrated in following diagram:

Reference:

Emanuele Rossini, Art D. Bochevarov, Ernst Walter Knapp, "Empirical Conversion of $\mathrm{p}K_\mathrm{a}$ Values between Different Solvents and Interpretation of the Parameters: Application to Water, Acetonitrile, Dimethyl Sulfoxide, and Methanol," ACS Omega 2018, 3(2), 1653–1662 (https://doi.org/10.1021/acsomega.7b01895).

Seems these are the values for water - not quite what they are in organic solvents, also seems OP needs more practical answer...

Thank you for your time and answer. I will not mark this answer yet, will wait for one day if someone wants to answer too. I too have calculated this value, since this gives some guidance to what is happening in this situation. The only thing comes to my mind is that these constants are in water. I remember some literature claiming that in glacial acetic acid HCl is a very weak acid, so possibly this calculation result is far from reality, since i would be using EtOH as a solvent, an overwhelming number of EtOH molecules would further hinder the dissociation of t-BuOH.

Very nice addition to your answer, this is certainly more than i bargained for.