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Why is liquid oxygen more reactive than gaseous oxygen?
Liquid oxygen is somewhat hairy to work with, as, if it comes in contact with essentially anything combustible, it has an unfortunate tendency to detonate violently if looked at sideways.
Gaseous oxygen obviously does not do this, as can be seen from the fact that combustible objects can sit indefinitely in gaseous oxygen without exploding when touched.
This seems backwards, as, at a given pressure, liquid oxygen is colder than gaseous oxygen, and materials are generally more reactive at higher temperatures - thus, intuitively, one would expect liquid oxygen to be less reactive than gaseous oxygen, not more.  Why is this not so?

related: https://chemistry.stackexchange.com/q/2631/23561

@A.K. The related question lacks imho a good answer that addresses the present question. It would be nice to see a more complete answer.

I would guess that it would be partially related to the fact that liquid oxygen is much more dense than gaseous oxygen.

I would guess so as well. But, OTOH, it is opposed by the temperature dependence of reaction kinetics. Remember the empirical rule the reaction speed raise 2-4 times for each 10 Deg C.

Yes, liquid oxygen mixtures are dangerous and sometimes explosive, but most need some kick to start the violent process off. Starting the oxidation process isn't the issue: the issue is how fast it can go once started. Also don't forget that many things are very, very combustible in high gaseous concentrations of oxygen (same issue: fire proceeds much faster when there is more oxygen).

@matt_black: Very, very, combustible, yes.  Prone to detonating at the slightest provocation, no.

Since oxidation is generally strongly exothermic, a large part of the difference in reactivity can be explained by kinetics (rather than thermodynamics): the denser liquid brings oxidant (oxygen) into intimate contact with the oxidized substance throughout the reaction. Consumed oxygen is quickly replenished by surrounding liquid oxygen. In the gas the exothermic nature of combustion reactions accelerates reaction with oxygen gas by increasing the collision rate between oxygen molecules and the surface being oxidized. In the liquid this temperature effect is not as important because the oxidized surface is already bathed in oxygen molecules. Also, although the gas might be at a higher T, as assumed in the question, local heating of liquid oxygen around the oxidized surface due to the highly exothermic nature of most oxidation reactions can also add to the faster propagation rate of reactions in the liquid. The importance of thermal effects might also be reduced (if not trumped) by the acceleration of concerted events during the oxidation (such as sequential events at the same or neighboring sites or while in a short-lived activated state), since in the liquid collisions between surface and more than one oxygen molecule at neighboring sites will occur at rates that are orders of magnitude faster than in the gas.
This is admitedly an answer in need of some quantitative support.
Here's a "back-of-the-envelope" justification for the importance of density:
It turns out that if you use a model based on the Maxwell-Boltzmann kinetic theory then the collision frequency with a surface is proportional to $p/T^{1/2}$. Therefore, if you ignore the thermal activation energy for reaction, you can assume that the reaction rate is proportional to $p/T^{1/2} = \rho _m RT^{1/2} = RT^{1/2}/V_m$. This means we will have the same collision frequency for two different set of values of T/V when
$$\left(\frac{V_{m2}}{V_{m1}}\right)^2=\left(\frac{T_{2}}{T_{1}}\right)$$
Therefore, we can counter the effect of a 200 K drop in T by increasing the density (reducing the molar volume) by a factor of ~14. Now at STP the molar volume of a gas is ~20 L/mol, while for a compressed (liquified) gas the molar volume is 100 times smaller, say 0.2 L/mol. Therefore the increase in density more than makes up for the drop in T, at least within this simplified model.
In more general terms, we can explain away much of the loss of the kinetic advantage of higher temperature by the higher density in the liquid. So based on this simplified argument oxygen at a liquid-like density can be more reactive than a much warmer gas.
Note that the activation energy for reaction of $\ce{O2}$ is not negligible due to the relatively stable structure of the oxygen molecule. Therefore kinetic effects involving rapid heating and thermal activation of oxygen (due for instance to local accumulation of thermal energy) are probably also important to explain the reactivity of LOX.

What about temperature dependence of the reaction kinetics ? In sense of probability to gain the reaction activation energy ? (Arrhenius equation)

@Poutnik I try to explain that the high density of oxidant in the liquid overwhelms any temperature factors. There is the business of activation energy which I don't address. I assume other factors such as the acceleration of concerted processes (such as solvent cage effects) is far more important.

Well there is empirical rule reactions are faster 2-4 times eachv10 deg C. If we take just 2, and 200 Deg C different to ambient, it is $2^{20}$. About 1 million times slower. It does not seem to be matched by much more then  1 million times more frequent collisions. What you say may be true, but is strange, anyway.

@Poutnik I'll add an argument on this to my answer. Let's just say 3 orders of magnitude can be explained by the density alone, the remaining 3 orders require a different explanation.

It is possible the reaction gets activated, e.g by static electricity, as methane - air mix would.

@Poutnik well, I gave an attempt at a justification. Your counter-argument suggests other properties of a liquid (electrical) are also important. I'll leave it to someone else to write that into an answer.

Sure, I just wanted just to share different point of view.


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Why don't acyl chlorides give positive test with 2,4-DNPH?
In this post, Ron discusses the reasons why carboxylic acid derivatives such as amides, acids and esters do not give a positive result with 2,4-dinitrophenylhydrazine. He mentions that it is largely due to the fact that the carbonyl carbons in these compounds are not sufficiently electron-deficient to react with the hydrazine derivative due to strong conjugation within these compounds. What about acyl chlorides and possibly, acid anhydrides as well? These carboxylic acid derivatives are clearly rather electron-deficient, certainly much more so than aldehydes and ketones, as evident from their great ease of hydrolysis. Why don't they react with hydrazine to give hydrazone?
I think that the reason may be that $\ce {Cl^-}$ and $\ce {RCOO^-}$ may be good leaving groups that leave after the initial formation of the tetrahedral intermediate. In the case of acyl chloride, we may get a hydrazide instead of a hydrazone. This is likely the same for the case of acid anhydrides. Is my thinking correct?

You get the acyl hydrazide RCONHNHPh. These form easily with acyl chloride and anhydrides but the are not coloured llike hydrazones.

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Calculating the pH of a buffer made by a diprotic acid and its double salt
The concentration of the diprotic acid (tartaric acid) will be constant at $\pu{ 0.1 M}$. I want to find the concentration of its double salt (potassium sodium tartrate) needed to create a buffer of $\pu{ pH }$=x (for example,x= 4.0). How would I do this? Do I have to use the Henderson-Hasselbalch equation? Make certain assumptions?
I tried to set up two equilibrium expressions with $K_\mathrm{a1}$ and $K_\mathrm{a2}$ as follows:
$$K_\mathrm{a1} =\frac{[\ce{C4H5O6-}][\ce{H+}]}{[\ce{C4H6O6}]}$$
and
$$K_\mathrm{a2} =\frac{[\ce{C4H4O6^{2-}}][\ce{H+}]}{[\ce{C4H5O6-}]}$$
Can I multiply $K_\mathrm{a1}$ and $K_\mathrm{a1}$ to eliminate$[\ce{C4H5O6-}]$, and then get the concentration of $\ce{C4H4O6^{2-}}$ necessary by plugging in $\pu{0.1M}$ for$[\ce{C4H6O6}]$ and the target $\pu{pH}$ in the appropriate form in $[\ce{H+}$]? I feel like I'm making some big assumptions here, though.
(Also, please forgive me; I don't know how to use LaTeX or Tex.)
If I try to make the assumption that the second dissociation does not occur as it is too weak (which it isn't) I wouldn't have the concentration of $\ce{C4H4O6^{2-}}$ in my equation, which is what I'm trying to find.

Can I multiply Ka1 and Ka1 to eliminate [$\ce{C4H5O6−}$], and then get the concentration of C4H4O62− necessary by plugging in 0.1 M for [C4H6O6] and the target pH in the appropriate form in [H+]

No, because $\ce{C4H5O6−}$ is a one of the major species. In fact, if you add the tartaric acid and its double salt at equimolar ratios, $\ce{C4H5O6−}$ will be the only major species. As a first approximation, you can let this reaction go to completion. If you have an excess of the acid, there will be a mixture of $\ce{C4H5O6−}$ and acid as major species. If you have the double salt in excess, there will be a mixture of the double salt and $\ce{C4H5O6−}$ as major species.

Do I have to use the Henderson-Hasselbalch equation? Make certain assumptions?

If your ratio of reactants and products is such that you have a conjugate acid: base pair as the major species, you can use Henderson-Hasselbalch as the next step to estimate the pH. (If your major species are $\ce{C4H6O6}$ and $\ce{C4H5O6−}$, use $K_{a1}$, if they are $\ce{C4H5O6-}$ and $\ce{C4H4O6^2−}$ use $K_{a2}$.) In some cases, that's a good approximation, in others, not. You can test that by calculating the concentration of all species (based on the approximate pH) and then calculate Q for the two acid base equilibria to see how well they match K.

I would take different approach.

From given $pH$ and acidity constants, calculate ratios of the particular forms of the tartaric acid.
From given total molar amount, you get the molar amounts for the particular forms.
From molar amount of particular forms, you can get molar amount of a hydroxide to be added to the tartaric acid, or of a strong acid to be added to the tartrate.
Important thing is to know this is only approximation, as acidity constants are meant for activities and we assume activity coefficients are equal to 1. You may want to search for, how to calculate activity coefficients for ions.
If the final goal is practical, better is to take a tabelized compositions of verified buffer solutions, or to adjust the final $pH$ with help of calibrated $pH$ meter.

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Stability of resonance structures
Should the fact that all atoms involved having an octet, or the rule that more electronegative element should have negative charge and less electronegative element have positive charge, be given greater priority while determining the stability of resonance structures?

No formal charges if possible. The point is that when all atoms have octet and no charge separation, then that structure is a very good description for the real molecule. It will have an important weight and even more so if it happens for it to have degenerate companion, as in aromatic rings. Beside this concept, common (qualitative) rules are easily found.

Start by looking for structure with most complete octets then most number of bonds then look for if they have unlikely positive or and unlikely negative charge on the more electronegative and more electropositive atom respectively and then look for charge seperation

So what would be the answer of the above two questions?

I think I have answered.


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Naive question on the SN1 mechanism
I was teaching the $S_N1$ mechanism to a friend. He asked a question a subtle question I can't shake off. Take a tertiary alkyl halide and aqueous sodium hydroxide. The first step is the formation of the carbocation. My friend asked how did the alkyl halide just do that all of a sudden. What about the presence of the $\ce{NaOH}$ made this compound do this if it doesn't even interact with it ( in the first step)? I first thought maybe it was just that the first step is something that is a feasible process, and that the alkyl halide would do (to an appreciable extent) that even without the $\ce{NaOH}$, but I am not sure.
Is my idea correct?

The solvation of the halide part led to energy being released which compensated for the bond being broken(which apparently seems "spontaneous")

So this would have happened even without the NaOH present?

Heard of the term solvolysis if not look it up


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Sulfuric acid symmetry point group
On the following page: https://cccbdb.nist.gov/pglist.asp It appears that the point symmetry group of sulphuric acid is $C_2$. I know that $C_2$ means that there is an axis of symmetry of order 2 (180 degree rotation).
But, I am not able to visualize it in the molecule.

In fact, you can see $C_2$ and other symmetry elements for yourself in 3D by opening the CIF [1, COD-2005680] with Mercury (free, available for Windows, Linux, MacOS):

Figure 1.
Unit cell of crystalline sulfuric acid $\ce{H2SO4}$.
Green lines parallel to crystallographic axis $b$ depict $C_2$ rotation axes.
This can be achieved by drawing the unit cell content with Calculate → Packing/Slicing…  and adding symmetry elements of interest via Display → Symmetry Elements….

Figure 2.
Single molecular unit $\ce{H2SO4}$ in two projections with the $C_2$ placed in drawing plane (left) and perpendicular to drawing plane (right).
References

Kemnitz, E.; Werner, C.; Trojanov, S. Reinvestigation of Crystalline Sulfuric Acid and Oxonium Hydrogensulfate. Acta Crystallographica Section C Crystal Structure Communications 1996, 52 (11), 2665–2668. https://doi.org/10.1107/S0108270196006749.

How I can a draw the molecule on Mercury? @andseliskç

@aprendiendo-a-programar I updated the answer with the link to CIF, which you can open with Mercury. This is an experimentally determined structure which you are not supposed to "draw" yourself.

I do everything you tell me and I don't get what you get.

@aprendiendo-a-programar What exactly you are experiencing problems with?

First of all, it is still not clear to me where I can find more molecules (such as water, benzene, ammonia,...). Also, once I open the sulphuric acid file in Mercury and pulse the "Packing/..." command I don't get the same thing that you hung in your answer. And, if I add the symmetry, numerous planes and lines appear, nothing to do with what you put.

@aprendiendo-a-programar 1. The question was solely about sulfuric acid. If you want other structures, search the COD yourself. 2. You are supposed to choose what area you want packed by ticking Pack box and selecting the coefficients for each axis $a$, $b$ and $c$ (I used defaults 1). 3. As I said, add symmetry elements of interest: read what each element stands for in that dialogue windows, and deselect those you don't want to see.

Okey, thanks. And, last but not least, how did you get the last images? That is, a single molecule with its elements of symmetry.

@aprendiendo-a-programar No prob. Figure 2 is a compilation of two screenshots of a single molecule depicted in Mercury along two separate crystallographic axes which I added a "$C_2$" label in Inkscape (image editor) to. I decide to do so for clarity for those users who won't bother opening the structure in Mercury:)

It's just that I'm currently learning the punctual symmetry of molecules and I'm new at this. And, a question, is Mercury able to predict the elements of symmetry in the molecule? Because I've seen this option and I don't know if it's the right one:
Display < More Information < Symmetry Operator List...
By the way, I have searched in the page that you told me the ammonia (drawing its structure in the option that there is for it) and it doesn't come out. What do you usually do to search in that database for molecules?

Mercury is just a viewer (and certain functions are not available in free version), and it just takes the symmetry info from the CIF. However, Mercury is probably the best crystallographic tool for the beginners I ever encountered (the rest is even more cumbersome or costs money). As for ammonia, here is one I found by adding "H3" and "N" to "1 to 8 elements" field and "2" and "2" to "number of distinct elements min and max" (search hints). Unfortunately, no COD structures have H assigned.

And apart from this program, would you recommend any other to appreciate the elements of symmetry? Sometimes I do them, and I have no way of knowing if it's okay. Because, for example, with the sulphuric acid in the tables on the internet I knew it was C2 and I didn't know where it was. So, what would you recommend me for? Thanks

I'd suggest to have a look at online tutorials such as Chemtube3d/Symmetry and Symmetry@Otterbein. Both offer similar 3D interactivity via JSmol, a JavaScript library for crystal structures, a desktop alternative to which is Jmol. You can also search Chemistry.SE (this site), I remember even I answered a couple of similar/related questions before.

In the structure as drawn here, the axis is perpenducular to the plane of the paper passing through the sulfur atom.

But, if I turn around that axis, every 180 degrees the molecule would not be as it was initially. @Oscar Lanzi

Which atoms would be out of place?  Look again... .


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Resonance in a non-planar molecule
I am well aware that the following compound is non-planar and hence, will not be anti-aromatic but non-aromatic.
But I wonder whether resonance will occur to some extent i.e. whether any three consecutive parallel p orbitals will lie in the same plane or not?

This can't show aromaticity as it doesn't obey huckel's rule you can think of non planarity as a measure so as to prevent anti aromaticity then why would it even planarize

Will it show resonance? Even to some extent?

Aree why would it if resonance would destabilize it

With this logic, there should exist no anti-aromatic compound.

If molecule is planar resonance will happen aromatic or anti-arometic , it doesn't matter. If non-planar than no

The p orbitals can overlap to some extent, if that's what you're asking. But it won't be termed as resonance


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Confusion about the SN1 and SN2 Mechanism
My book says, and I understand, that the $\mathrm{S_N1}$ mechanism is much faster for tertiary alkyl halides than primary. And the opposite for $\mathrm{S_N2}.$ This all makes sense. What I don't get is the conclusion it comes to next: hence, we predict that the primary alkyl halides will react primarily via $\mathrm{S_N2},$ and the tertiary via $\mathrm{S_N1}.$
The reason I get confused here is that nothing was said about the comparative rates for a primary alkyl halide via the $\mathrm{S_N1}$ or $\mathrm{S_N2}$ (they just compared primary and tertiary for $\mathrm{S_N1}$ and $\mathrm{S_N2}$). In my mind, which mechanism dominates for primary or tertiary should be dependent on the relative rates for that compound; not rates compared between primary and tertiary.
Is my argument sound? And if so, is there some literature you know of that explains this in such a manner?

Just measure the absolute rate. Then you know all relative rates.

If your reactant and product are chiral, you can tell from the stereochemistry of the product which mechanism was at play. SN2 shows inversion of configuration, while SN1 scrambles it (racemic mixture as product).

Your reasoning is correct. It's the relative rates for a given alkyl halide that matter, and it's difficult to draw conclusions about those, because the relative rates are not fixed, as the $\mathrm{S_N2}$ rate is dependent also on the nucleophile concentration.
We can write the rate of the $\mathrm{S_N1}$ as $k_1[\text{alkyl halide}]$ and the rate of the $\mathrm{S_N2}$ as $k_2[\text{alkyl halide}][\text{nucleophile}]$, so the relative rate at a fixed concentration of the alkyl halide is given by
$$\frac{\mathrm{S_N2}}{\mathrm{S_N1}} = \frac{k_2[\text{nucleophile}]}{k_1}.$$
The identity of the nucleophile will be a factor as well, since that will affect $k_2$.
So all in all, you can't make a general conclusion that one reaction type will always go faster than the other, although it is likely that there are alkyl halides for which one reaction type is faster in a large majority of cases.

the way I see it now is that for primary alkyl halides, the unstable carbocation slows down SN1 much, so generally SN2 wins out. And for tertiary, steric hindrance tends to slow down SN2 much, so generally, the SN1 wins out. That sound right?

Yes, you've got it.


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How could mass increase when sulfuric acid is added to calcium carbonate?
Context: I was doing an experiment in class that investigates the rate of mass loss over time when $\pu{20 mL}$ of sulfuric acid is added to $\pu{5g}$ of calcium carbonate. I did the experiment 3 times, each experiment involves mixing four, $\pu{20 mL}$ different concentration of sulfuric acid ($\pu{0.5 M}$, $\pu{1.0 M}$, $\pu{1.5 M}$, and $\pu{2.0 M}$) into $\pu{5g}$ of calcium carbonate.
Hypothesis: If concentration increase, the rate of mass loss should increase too
Result: Everything went out just fine, however, in each of the 3 experiments, when I add the $\pu{2.0 M}$ sulfuric acid to the calcium carbonate, the mass INCREASES (mass still loss overtime).
Could anybody please explain how could this be happening? As they react, carbon dioxide is produced and there should only be a mass loss. Also, this only occurs with the $\pu{2.0 M}$ sulfuric acid, the other concentrations work out just fine.
Could the electronic scale wasnt working properly? but this only happened with $\pu{2.0 M}$ acid.

I'm not sure how you were taking the weights. Did you account for the fact that the densities of the sulfuric acid solutions are different? In other words 20ml of 2 molar acid weighs more than 20 ml of 0.5 molar acid.

As MaxW pointed out in his comment, I also felt like this may be the case of densities of $\ce{H2SO4}$ solutions. Thus, I googled for densities and found few sites listing them. In this Steffen's Chemistry page, you'd find some trends. I just listed relevant densities and milarities of some 1-15% $\ce{H2SO4}$ solutions, and graphed them molarity vs density to find the relationship. I used data for  1-25% to graph for better reliability. Sure enough, there was a linear relationship of $y=0.0597x+1.0006$ with $R^2=0.9998$ reliability:

So, using the graph, I calculated the masses of your $0.5, 1.0, 1.5, \text{and }\pu{2.0 M}\; \ce{H2SO4}$ solutions. They were $20.609, 21.206, 21.803, \text{and }\pu{22.40 g}$, respectively. Thus, if you are using electronic balance with $\pu{0.10 g}$ accuracy, you may have readings of $20.6, 21.2, 21.8, \text{and }\pu{22.4 g}$ for one set. All except for last have closed to $\pu{21 g}$ weight. Last one is $\pu{22 g}$, a little bit more than others. Make sense? :-)

WOW Cheers Mathew! I am at school right now and i just saw this! You saved my life!

Does it matter if they are closer to 21 or 22?

I think you are taking mass difference to calculate rate. I think you are okay on this.

There are several contributing factors to this outlier observation. First of all, when scientists did mass based experiments, even hundreds of years ago, they recorded mass up to four decimal places. The mass of CaCO3 should have been recorded to 0.1 mg for example 5.0612 g. When you say that mass increases, we do not know the details of mass change. In short, never record mass with one significant figure.
Secondly, how did you measure volumes of 20 mL acid? If you were using cylinders, they are never accurate. There is significant error in volume transfer. With a relatively viscous solution such as 2M H2SO4, if you transfer 20 mL, the transferred volume via cylinder is usually less than 20 mL.
This all about good laboratory practice. Coming to chemistry: when CaCO3 reacts with H2SO4, if forms an insoluble salt called calcium sulfate (CaSO4). With concentrated 2 M sulfuric acid, once it comes in contact with CaCO3, it may form a coating of CaSO4, which would slow down the rate of reaction. Did you see some unreacted solid mass?
Also conc. sulfuric acid is hygroscopic.

Thanks Mr.Farooq! I will make sure I apply these tips in the next experiment!


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Partial Aromaticity and Anti-aromaticity
The following resonance structures are possible for the given compound. (a) promotes both aromaticity and Anti-aromaticity whereas (b) promotes aromaticity. In the resonance hybrid, (b) contributes more as we can see. Will the resonance hybrid be called aromatic or anti-aromatic?

From a classification point of view,  there is no reason for not calling the molecule aromatic.
First, saying a molecule is aromatic doesn't imply that  aromaticity fully extends over it. A trivial example is, just to say, cyclohexylbenzene, or whatever aromatic molecule of your choice.
Now, let us assume the two rings being completely independent so to exclude any delocalisation. The phenyl moiety will be just aromatic while the Huckel antiaromatic cyclobutadiene ring will simply adopt a rectangular shape, so effectively avoiding the antiaromatic unstable configuration.
Again, there wouldn't be reason for not calling the molecule aromatic.
This is often a sort of misunderstanding when antiaromaticity is discussed. It is just a criterion to decide if full delocalisation within planar rings is attained. If the answer is no, because the system is in fact antiaromatic, then there is nothing actually specific to attain.
In short: while a ring that is Huckel-aromatic does actually tends to be aromatic, the Huckel-antiaromaticity is simply a relative high energy situation that a system will always try to avoid.*
Finally let us consider the actual case. The resonance structures you wrote indicate indeed a certain character of aromaticity in the cyclobutadiene moiety and a partial double character of the interring bond. This alone, let aside the above considerations, definitely tell us that the molecule is aromatic.
For more insights I've found the following papers, analyzing delocalisation in phenylcyclobutadiene experimentally (enthalpy of hydrogenation and EPR) and theoretically.
https://pubs.acs.org/doi/abs/10.1021/ja0525511
https://www.academia.edu/25340037/Aromatic_pathways_in_conjugated_rings_connected_by_single_bonds
As a very quick resume enthalpy of hydrogenation point to a destabilising effect of the Huckel-antiaromatic moiety, while electron spin resonance and calculation show overall aromatic rings current and an interring C-C bond length of about 1.44 Å, about midway from a single to a double C-C bond length.
*note that chemists are sometimes able to constrain structures in otherwise highly unfavourable systems. I won't be surprised if exotic beasts forcing an inner ring into antiaromaticity were synthesised. I might even have seen a related Q here in SE.


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Does higher Oxidation/ reduction potential translate to higher energy storage in battery?
Some articles refer to the fact that Lithium is preferred metal of choice for batteries because their half cell voltage is high (slightly above 3V)
If that is the case why can't we use other metals like Mn which can have higher oxidation potential (Mn can have +7 oxidation state which translates to higher oxidation potential)?

There is a difference in a high oxidation state and a large EMF. A number of factors are considered in making a battery, not just the EMF of the half cell. A battery can relatively easily have multiple cells to get a higher voltage.

Energy storage depends on the electromotive potential (i.e. difference between species in the electromotive series) and on the number of electrons available.
Li, for example, has an oxidation potential of ~3.04 V relative to hydrogen, but Al has one of 1.66 V, so Li has the greater potential. On the other hand, Li has only one freely available outer electron, but Al has three, so Al stores more charge per mol, giving three times the capacity in coulombs on that basis. On the gripping hand, however, Li has a mass of ~7 amu, and Al is ~27 amu, so an aluminum cell carries around the dead weight of unused neutrons and inner electron/protons.
So, depending on the needs, Li batteries weigh relatively little and provide comparatively high potential, aluminum batteries would hold more charge at a bit lower voltage, and even Zn is useful, though deficient compared to those others in volts and coulombs (it also has a bunch of inert neutrons, but there is no charge for them), because it is relatively inexpensive.


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Mechanism for formation of amides from ammonium carboxylate salts
One of the reactions that you can do with a carboxylic acid is to react it with ammonia or an amine, which produces the corresponding ammonium carboxylate salt. When this salt is heated, amide is obtained by elimination of water (although it seems that the yield is not that good).
$$\ce{RNH2 + RCOOH->RCOO- ^+NH3R->(heat)->RCOONHR + H2O}$$
So what sort of mechanism can we draw for the elimination of water? It seems to me a bit complicated because the amine is already protonated in the salt, so it does not have any lone pair remaining for nucleophilic attack. Does the acid base reaction go backward under high temperatures? Or is it something else?

Uhh..dehydration?

@NilayGhosh, but can we draw an electron pushing mechanism?

The question is about mechanism of amidation not from R-NH2/R'-COOH but from their salt R-NH3+/R'-COO-. The former is well known and the only mechanism of amidation. The salt R-NH3+/R'-COO- is always in equilibrium with R-NH2 + R'-COOH. Equilibrium constant increases with increasing base strength of R-NH2 and increasing acid strength of R'-COOH, and vice versa. Salts with higher constants have higher ionic characters, and their pyrolysis temperatures become higher for amidation, signifying that completely ionic salts cannot be amidated. If they are heated to very high temperatures, they will be not amidated but just degraded.

In this reaction the lone pair on the ammonia or amine reacts with the carbon attached to the oxygen atoms since it is partially electropositive due to the presence of extremely electronegative oxygen atoms. It does NOT react with the hydrogen on the alcohol group. This will give us an ammonium salt and the electrons involved in the pi-bond between the carbon and oxygen will be taken by the oxygen atom since carbon can't have 5 bonds .
On heating the compound produced above an intra-molecular acid-base reaction takes place protonating the alcohol group and making it a good leaving group .
The negative charge on the other oxygen atom forms a pi-bond again, forcing the protonated alcohol to leave as water ; leaving us with an Amide .
The following image shows the reaction mechanism :


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What are the reasons that make platinum a good electrocatalyst for oxygen reduction reaction?
Researchers are trying to find new electrocatalysts for energy related reactions in fuel cells in order to replace the expensive platinum-based materials.
I would like to know what are exactly the properties of platinum that make it have such a good performance (1 step reduction reaction of oxygen/4 electron process)?
Thank you in advance.

related? : https://chemistry.stackexchange.com/questions/34145/platinum-group-metals-why-are-platinum-and-palladium-great-catalysts-and-not-th?rq=1


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Number of stable molecules of artificial molecule XY4

Suppose element X has $5$ stable isotopes, and element Y has $6$ stable isotopes. Find the number of natural molecules, knowing that an X has a charge of $+4$ and Y has a charge of $-1$.

What I tried was this: for element X we have $5$ options, and for each element Y we have $6$ options. Therefore the total number of options would be $5 × 6 × 6 × 6 × 6$.
But this answer is clearly wrong, as it counts the same type of molecule, multiple items. (AAAB) and (AAABA) count twice which we don't want to happen. (A is an isotope of Y and B is some other isotope of B)
What can I do?

I think Mathematics SE they will have a formula at hands without tediously manually removing the multiple items

You can't answer this question. What are the valid geometries of the $\ce{XY4}$ molecule? Tetrahedral? Square planar? Distorted tetrahedral? See-saw? Technically, these are all different, and you would need to account for them or rule them out, which requires more information than what is given here. (I'm assuming that $\ce{X}$ is the central atom here. If that's not a valid assumption, then it's even more insane.)

There are 5 ways to choose the central atom.
Choosing the possible combinations of the other 4 items from 6 isotopes is complicated as this article explains. But it provides a formula for combinations with repetition that gives the answer of 126.
This implies that there are 630 distinct isotope combinations in the molecule (assuming that we can distinguish combinations of isotopes with the same atomic number which is possible with high resolution mass spectroscopy). If only integer atomic numbers count then the distinguisahble combinations will be lower. But the easiest way to work that out is to do a Monte-Carlo simulation not to look up probability formulae.


text:
Why do polyoxometalates have good electrochemical/ redox properties?
Many authors are using polyoxomethalates immobilised in carbon nanomaterial as electrocatalysts for energy related reactions (ORR, HER, OER). Other metal alloys have also shown to be successful as electrocatalysts for energy related reactions.
I would like to know what makes polyoxometalates have these good redox properties, is it just because the high valency of the metal/addenda atoms that makes the multiple-electron acceptance ?
If it is possible to give an explanation based on coordination chemistry (molecular orbitals), it would be very helpful.
Thank you in advance.


text:
How can I easily identify if a substance is an acid or base based on its chemical formula?
I know of the Arrhenius definition, Bronsted Lowry definition and the Lewis definition for acids and bases but I don't know how to apply them on the spot. Also, some chemical formulas (such as acetic acid) throw me off since it has an OH at the end which causes me to think that it's a base when it really isn't.
Any tips/ways to identify if something is an acid or base simply based on its chemical formula?

Sulphuric acid has even 2 hydroxy groups, so it is not the proper criteria. :-) The important thing is to understand properties of elements and  particular chemical bonds, so you can determine, if it is donor or acceptor of $\ce{H+}$ or an electron pair.

I was going to mark this as too broad, but it's not too broad, because the answer is quite simple: in general, you cannot determine the acid or base properties of a substance from its formula.
As a concrete example, consider the formula $\ce{C4H7NO3}$.
Here are two possible isomers for that formula:

The left compound is an acid. The right compound is a base.

I would add, that acetic acid is an acid compared to water but is a base compared to hydrochloric acid. So the question is only relevant if we define the substance supposed to be neutral.


text:
What is the best way to calculate the mass percentage of Calcium Carbonate from water?
Where I live it's believed that local tap water is really bad for drinking because of the amount of limestone. As a science experiment for school I wanted to approximately calculate the percentage of calcium carbonate in tap water so I can make the conclusion: Is that amount of drinking $\ce{CaCO3}$ actually the problem or something else?
This question might be similar to DK4's question but I'm trying just to calculate not extract, only by using items you can find in a typical house (or in a local shopping center).
If calcium carbonate isn't the issue what is?
Edit: As pointed out in comments, percentage of $\ce{Mg}$ is also applied.

You can't do a "decent" analysis without some basic chemistry lab equipment. // In general you'd need to consider both Mg and Ca which is known as the water hardness.

You can buy / borrow / order online some water hardness analysis kit for hobby aquarists. The kits are using a solution drop counting to determine amounts of the total and carbonate hardness. For your intentions they are precise enough.

See: https://en.wikipedia.org/wiki/Hard_water

Define "bad for drinking."

"bad for drinking " = Water that may cause health issues

Well, eating limestone, while not necessarily palatable, is unlikely to cause health problems, unless you ingest so much that it causes indigestion from over neutralization of stomach acid. On the other hand, because limestone is not actually soluble in water, you'd be hard pressed to drink enough. Technically speaking, water itself may cause health problems. Over consumption of water leads to over dilution of electrolytes. In the extreme case, too much water leads to drowning, which is definitely bad for health.

Water contains negligible amount of $\ce{CaCO3}$ or $\ce{MgCO3}$, as these are almost insoluble in water.
What you are interested in are bicarbonates $\ce{Ca(HCO3)2 + Mg(HCO3)2}$. Their content is known as the temporary or carbonate water hardness that can be expressed as $\ce{CaCO3}$ equivalent.
The complement to the carbonate hardness is the permanent hardness, formed mostly by $\ce{CaSO4, MgSO4, CaCl2, MgCl2}$. The sum of the temporary and permanent hardness is called the total hardness.
Bicarbonates ($\ce{HCO3-}$) are not threat to health themselves, as they convert to $\ce{CO2}$ in the stomach and are present in blood as the component of the blood $\ce{pH}$ buffer.
What can be a subject of health concern is the total content of $\ce{Ca^2+}$ (and in lesser extent $\ce{Mg^2+}$). It is again expressible as $\ce{CaCO3}$ equivalent.
Calcium/magnesium content is not measurable by resources usually found in households or local shop center.
The best way is to ask the water provider for the data.
The 2nd best way, close to home-like determination, is to obtain the set for aquarists from a pet shop for determination of carbonate and total water hardness. I am a former aquarist, and this was available even in 1980s in the former communist Czechoslovakia.
Determination is done by counting  the drops of the colored solutions, mixed with the tested volume of water, until the color changes.
Carbonate (in fact bicarbonate) hardness determination is based of reaction of an acid with bicarbonates: $$\ce{HCO3- + H+ -> H2O + CO2 ^}$$
Drop counting ends when yellow color (methyl orange dye indicator) changes to red. The dye turns red by the first excess of the added acid, as has red color in acidic solutions.
Total hardness is based on reaction of $\ce{Ca^2+ + Mg^2+}$ with ethylendiamintetraacetic acid (EDTA) in alkaline solution, enriched by the addition of the dye indicator Eriochrome black T,
$$\ce{Ca^2+ + H2EDTA^2- -> 2 H+ + CaEDTA^2-}$$ The EDTA first reacts with free $\ce{Ca^2+ + Mg^2+}$. The drop count ends when red turns to blue, as EDTA captures $\ce{Ca^2+ + Mg^2+}$  bonded to the Eriochrome black T.
As result, you obtain the carbonate (temporary) and total (temporary + permanent) hardness in 2 possible units: $\pu{mmol/l}$ or $\pu{dGH}$ (the degree of General Hardness).
$\pu{1 mmol/l = 5.6 dGH}$.
$\pu{1 mmol/l}=\pu{100 mg/l}= 0.01\%$, as molar mass of $\ce{CaCO3}$ is cca $\pu{100 g/mol}$.
The typical total hardness of drinking water is $\pu{1\!-\! 3 mmol/l}$,  i.e. the equivalent of $\pu{100\!-\! 300 mg/l}$, resp. $0.01\!-\! 0.03\%$ of $\ce{CaCO3}$.

@Mathew Mahindaratne : Thanks, Mathew, for thorough editing.  One objection: Unless American English is mandatory here,  colour and centre  is proper spelling in British English.

Thanks for the valuable point. I know they are correct but I also confused on what format been preferred.  I guess both should be okay, but changed for the consistency. :-)

As you have already understood the detailed chemistry of Hardness in water by above-mentioned answers in great detail, I am going to answer How you can calculate %mass of Hardness in Water in a cheap but crude way by which you can help your school students to find out.
So Basically, Hardness also causes basicity of water and just by simple Acid-Base or EDTA titration of small amount of sample tap water with Acids such as H2SO4 or HCl or EDTA (Ethylenediaminetetraacetic acid) by which you will get the volume of Base present in water which can be converted to Mass units.
Though it is a very crude and inaccurate method but you can explain school kids by letting them find the amount the Alkalinity in water.
Here I am attaching a link where you can find the stepwise procedure and calculation to calculate Total hardness in water by using EDTA which you can find in your school chemistry Lab.
http://www.uobabylon.edu.iq/eprints/publication_2_2630_250.pdf (Please read the very first procedure in this document)
Please ask for any further clarification or queries.
Keep Learning :)


text:
The distance that CO2 molecules travel in the air before they collide with each other
Well the title covers most.
But further at 410 ppm, 1 atm and 23 °C.
I believe they call it the mean free path of CO2-CO2 collisions. So not just the mean free path of CO2 in air colliding with every other molecule substances.
I have zero workexperience with gases so I'm having troubles with this one.
I'm Dutch (sadly) so excuse me in advance if my English isn't correct.

This is complicated in that the CO2 molecules would collide multiple times with N2 and O2 molecules before colliding with another CO2 molecule. So is this what you really want to calculate?

I think you should calculate the average interval then assume that about x molecules are CO2 in y total.

You might be able to approximate with the main free path for any molecule in air and hen adjusting for the concentration of CO2.

You need to calculate the probability of one (CO2) molecule being the nearest neighbour of another.
Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$ (on a spherical surface). This must be the probability than no molecules exist closer to the victim molecule,  which occurs at a distance $r$, and that this molecule exists in the shell $r$ to $r+dr$. Thus,
$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$
where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).
Solving this gives
$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$
This function is zero at zero goes through a maximum and is  small at large. distances, see figure.
If you want a single number then the average distance between molecules is
$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$
where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.
( $\Gamma(4/3)$ is the gamma function, sort of equivalent to a factorial but for a non-integer number)
The figure below shows $w(r)$ the probability of a near neighbour at distance $r$ for some concentrations.

If what you want is the total path taken accounting for all collisions then you would have to start with one the of the $w(r)$ distributions corresponding to your concentration and do a Monte Carlo computer calculation to work out the total path, not easy and v long calculation probably. There will be a statistical mechanics way of calculating this I'm sure but I'm not aware of it at present.

Ok, thanks very much!! (To bad I don't have software for Monte Carlo simulation). But this was very helpful. Thumbs up.


text:
The effect of a voltmeter on the electron flow of a voltaic cell
Due to a voltmeter’s strong resistance nature, how could electrons flow from anode to cathod? It is more reasonable to assume that a voltmeter would stop the redox reaction because the electrons could never reach the positive electrode. When I measure the potential energy between two electrodes with a voltmeter, is the reaction happening? If the reaction is not taking place, how come the potential difference exists?

Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.

Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember
$$V = iR$$
thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work.
The problem with a voltmeter is that you can imagine that a battery has an internal resistance. So the more current flowing through the battery the less voltage will be measured by the voltmeter. Thus the cell voltage measured by a voltmeter would be low, at least theoretically.
$$V_\text{measured} = iR_\text{meter} - iR_\text{battery} = i(R_\text{meter} - R_\text{battery})$$
Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_\text{meter} \gg R_\text{battery}$ so:
$$V_\text{measured} = i(R_\text{meter} - R_\text{battery})  \approx iR_\text{meter}$$
In the not to distance past voltmeters drew too much current to give an accurate reading for a cell, so a Wheatstone bridge was used. But even though the galvanometer of the Wheatstone bridge is used to detect "no" current, it still has some finite sensitivity, so the ideal of absolutely no current flow is certainly not reached. The other weakness of a Wheatstone bridge is the need for high precision resisters. So to measure the galvanic cell potential a Wheatstone bridge no longer has any practical usefulness.

I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?

Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.

@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.

@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.

An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.
Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu.  The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.


text:
Contribution of Tetrahedral and Octahedral voids in HCP
I cannot find anywhere what the contribution of atoms situated at octahedral or tetrahedral voids in a HCP unit cell would be. I need to know this to be able to calculate the formula of a compound with HCP unit cell, when the positions of the atoms in the lattice are given.
What I mean by 'contribution' is how much of the atom in that lattice point belongs to one unit cell. Eg:- in CCP packing the contribution of atoms at the corners is 1/8 th.
I know that in a CCP unit cell the octahedral voids are there at the body center and the edge centers- so the contribution of an atom from both locations is different- 1 and 1/4 th. So I'm not sure whether the positions of OTs or TVs in a hcp unit cell are different (for the atoms of the same type) if so could someone tell how the contribution would be?

I cannot find anywhere..

Couldn't find? You can figure it out on your own, with me, right now.
Disclaimer: By "hexagonal unit cell" I assume you mean hexagonal prism, which comprises of 3 primitive hexagonal Bravais lattice.

Octahedral Voids (OhV)
The one on the left displays octahedral voids(visibly surrounded by 6 atoms).
The OhV is at the center of the highlighted octahedron, and lies on the plane containing four of the six lattice points in the octahedron.
Note that these lattice points belong to the unit cell, and the OhV must hence lie inside the unit cell.
This is the same for the other 5, amounting to a total of 6 OhV per HCP unit cell.
Being completely inside, its contribution is taken as 1.

BONUS: If you remember, No. of atoms per unit cell is 6, which
tallies with the no. of OhV given by the formula $N_{OhV}=N$ where
$N$ represents no. of atoms per unit cell.

Tetrahedral Voids (ThV)
I'm going to be frank: it IS not as easy as locating OhV.
The figure above absolutely freaked me out when I tried to visualize ThV's. I'm not keen on recommending that.
Instead, this 3D rendition should help immensely.
Type 1 ThV

The Type 1 Thv's are the ones inside in unit cell completely(so again, contribution=1).
Notice that I have marked 2 tetrahedrons. Also note that each tetrahedron we form will tell us about the ThV its existence creates.

RED: Note that there will be another 2 like this one in the lower half. That makes a total of 6 Red-type ThV in the entire HCP unit cell.
GREEN: Note that there will be a total of 2 of the green-type ThV, the other one belonging to the inverted tetrahedron just below it.

That gives us a total of $6+2=8$ Type 1 ThV.
Type 2
Let's do a little recap to see which ThVs have we counted so far(displayed in red)

I constructed a 3D model of 3 HCPs, which is required to visualize the remaining.

Now I will mark the ThV forming lattice points in red.

As you can see, lattice points from 3 unit cells are involved.
A quick analysis tells us that there will be 2 such Tetrahedrons for every vertical edge.
A quicker analysis tells us that since each ThV so formed is present in 3 HCP unit cells, or is "shared" by them, Its contribution must be $\frac 13$.
We count the number of such tetrahedrons to be 12 per unit cell; with contributions, ThV Type 2 will be $12*\frac 13$ $=4$.
Ultimately, we get:
ThV type 1= 8
ThV type 2= 4
Total ThV= 12.
Which brilliantly tallies with the formula for number of ThV($N_{ThV}=2N$).
I hope managed to enlighten you.
TL;DR

OhV contribution is 1, no issues there.
ThV does not have a single "contribution", but rather could be 1 or
$\frac 13$ depending on which ThV you're talking about.

The above answer is quite comprehensive, and i would like to complement that with this site, where the JSmol diagram of the HCP lattice helped me visualize it well.


text:
how does mass of carbon 12 is like this?
1u is defined as per carbon 12
so 1 atom of c have 6 neutrons and 6 protons both having mass more than 1u (Google)so total mass of carbon 12 is more then 12u (also there would be some mass of electrons) but it should be exactly 12u as u is defined such.So what should i interpret.I am confused how this is possible
I am just a school student just entered chemistry

The relevant term is "mass defect". Very simply, some mass/energy is used to hold the charged particles together.

@TAR86 Or, to put it another way, the mass of a nucleus does not, in general, equal the sum of the masses of its components because of nuclear binding energy.


text:
Any suggestions for a highly exothermic reaction in liquid phase but the reactants and products must be pH neutral?
Need a heat producing reaction for my lab experiment.

Use an electric kettle.

Has to be a chemical reaction, no physical heating

Why chemical reactions and why pH neutral ? What liquid is supposed ? The community likes more extended description of scenario and the own effort of the requester, before helping to provide answers.


text:
Determination of the most acidic hydrogen in an organic compound

Identify the most acidic hydrogen in the compound below:

Can someone please explain why is $\ce{H^\mathrm{a}}$ in the compound below more acidic than $\ce{H^\mathrm{b}}$?
This is what is written in the answer key although the conjugate base obtained from losing $\ce{H^\mathrm{b}}$ is stabilized by both resonance, and the inductive electron withdrawing nitro group.

Anion without Hb is not stabilized by resonance. NO2 is in the wrong position with respect to that OH.

It is stabilized by resonance due to the phenol ring. The stabilizing role of the nitro group is inductive.

Well, actually the stabilizing role of NO2 is not just inductive.

Well, another way of looking at it is that the Nitro group is ortho to the ring at the right, and thus decreases electron density in that ring.


text:
Determination of major product in crossed aldol condensation
The title speaks for itself really.
I know that if we have a conventional ketone (with 3+ α-hydrogens) and an aldehyde with less than or equal to 3 α-hydrogens, the ketone acts as the nucleophile and attacks the aldehyde, which consequently forms the major product.
Let us take an example of a case where we have acetophenone and acetaldehyde. Both have 3 α-hydrogens. Also, the acetophenone is sterically hindered due to the phenyl group. Which is the major product here?

Depends on the reaction conditions. Are you setting up to get the kinetic product or the thermodynamic product?

@Waylander it wasn't mentioned in the question. Actually this was a question asked in an exam. What will be the product in each case? And why so?

There are two main steps in any aldol condensation:-
1. Formation of enolate ion
2. Using the enolate ion as a nucleophile.
The 1st step is a reversible step. If you think that the kinetically favoured enolate ion will be formed fast then you are wrong as it also disappears faster. So, in any case the one which is more thermally stable will form in more amount except in the cases where the formation of stable enolate is too slow or the backward reaction of kinetic favoured enolate is blocked or not enough time has been given to let the reaction reach equilibrium.

Now, I will tell you about above strategies to block reverse reaction or make formation of stable enolate too slow:-
1. Use a strong electron withdrawing group at alpha position to the newly formed CH2(-)COR so that the electron can spend most of time away from the original carbon.
-This blocks reverse reaction
2. Use a bulky base or somehow hinder the alpha carbon to make the extraction of proton hard which will in turn make the reaction slow.
3. Using Non-Protic Solvent stops the protonation of the enolate ion which in turn blocks the reverse reaction.
Now, the answer for your question, the conjugate base of acetaldehyde is more stable when compared to the former but the attack of conjugate base of acetophenone is more easier as compared to latter because of steric reasons and the more electrophilic character of aldehyde and when you make cannonical structures, the number of structures which has negative charge on oxygen is greater than being away which makes it a good nucleophile when you take weighted average of all. So, the major product should be 1-phenyl-but2en1one.

I think that a mixture should get formed as the conditions are not so much sharp. Though i provided what i think

the product you've mentioned is the one we get after heating. In the answer, they've kept the alcohol intact (the one formed after the enolate attacks the aldehyde). By the way, which exam are you preparing for? And I'll think about it and get back to you when possible.

I'm preparing for JEE as well. My exam is on the 8th (second half). All the best to you!


text:
How can I reduce sodium chloride levels in water, without removing other minerals?
How can I reduce sodium chloride levels in water, without removing other minerals?
Good day,
I make a sole from Himalayan Salt which I add to my Reverse Osmosis treated water to ensure my body isn't rob of essential minerals, but the levels of sodium chloride are very high. How would I remove some of the NaCl without affecting the other minerals?

What you are asking is not possible using any conventional system that you would be able to obtain. Your only real solution is to take a different approach. Buy the individual salts (e.g., KCl, NaCl, CaCO3) and make a mixture yourself.

If you remove the NaCl from Himalayan salt there isn't much left.

Other than fluoride, drinking water is not a useful source of minerals. You get all you need from a normal diet. As for fluoride, most toothpastes and some mouthwashes contain supplements.

Removing a major component does not make sense. Better is to balance it with other ions, as $\ce{K+, Ca^2+, Mg^2+, HCO3-, H2PO4-, SO4^2- }$.
But the gain and safety to provide nutrients by the form of dissolved salts is questionable. I think, providing minor mineral nutrients by $\ce{NaCl}$ of whatever origin is a bad idea in the first place.


text:
How is energy released when a chemical bond is formed?
Chemical reactions are classed as endothermic or exothermic.
Defn: exothermic

(of a reaction or process) accompanied by the release of heat.

The heat is released when the chemical bonds in the product are formed.
Defn: heat

heat is energy in transfer to or from a thermodynamic system, ... The
  mechanisms include conduction, ...; or radiation between separated
  bodies; or ....

So, I have three questions:
1 - if the heat is released via conduction, is this equivalent to an increase in the kinetic energy of the product(s) compared to the kinetic energy of the reactant(s)?  Also, how does the formation of a chemical bond produce kinetic energy?
2 - if the heat is released via radiation, how is radiation created in the formation of a chemical bond?
3 - are conduction and radiation the only forms of heat transfer when a bond is created?
Edit - to clarify the question - the issue is addresed in the OPENSTAX AP chemistry text - which covers one example of 2 H atoms coming together to form H2 in detail as follows:

When the atoms are far apart ..... by convention the sum of their
  energies is 0. As the atoms move together their orbitals begin to
  overlap.  Each electron begins to feel the attraction of the nucleus in
  the other atom.  In addition, the electrons begin to repel each other,
  as do the nuclei.  While the atoms are still separated, the
  attractions are slightly stronger than the repulsions, and the energy
  of the system decreases. (A bond begins to form.) ... At some specific
  distance between the atoms, which varies with the atoms involved, the
  energy reaches its lowest  (most stable) value.  This optimum distance
  between the two bonded nuclei is the bond distance between the two
  atoms.

The text doesn't specify what energy is decreasing, and why.  But I can answer these questions I think.  The energy that is decreasing is the potential energies of the attractive fields of the atoms, and that energy is being transformed to kinetic energy as the atoms accelerate toward each other.
So we have the nuclei accelerating towards each other.  What happens next?
According to the text ... continuing where we left off

The bond is stable because at this point the attractive and repulsive
  forces combine to create the lowest possible energy configuration.

So there is no mention of the release of energy, and the atoms just arrive at the stable combined state.
Here is what I would expect, given that the nuclei are accelerating towards each other - at some point the repulsive forces become stronger than the attractive forces, and the atoms begin to decelerate.  If there is no damping force on the motion the atoms will oscillate about the point where the attractive and repulsive forces are equal.  So, there must be some damping force on the atoms, that causes them to come rest at the bond distance of H2.
Now comes the mystery, which is, how does the formation of a chemical bond release energy?
What happens to the kinetic energy of the two nuclei accelerating towards each other?  If the two hydrogen atoms are coming together head on there will be a collision that would appear to cancel out the motion, i.e. kinetic energy, of both atoms.  If there is a damping mechanism that generates the heat, what is it?
Since the energy is released as heat, I think it must be in the form of kinetic or electromagnetic energy, but what is the specific mechanism?
Or, am I trying to get too much out of the kinetic model?

Further, the subject is covered in my high school kid's chemistry text, OPENSTAX Chemistry, in the chapter 'Chemical Bonding and Molecular Geometry' as follows: 'plus the energy released when all bonds are formed', and 'the excess energy is released as heat', and that's it.

I think the issue is that you ask a couple different subquestions and any one of these has the potential to turn into a long answer depending on how much you know and how much detail you want. I think the overall question is interesting and it got a good, so I'll vote to reopen, but I think a full answer to the question could  cover excited states, various types of radiative and non radiative transitions, how collisions exchange energy between a system and its environment, and so on.

This is very ill-posed question. How heat can be transferred macroscopically has little to do with molecular process of bond formation. Heat can be also transferred via convection, but what it has to do with single molecule being formed?

@Tyberius - I'm not looking for a full treatment, just one simple example would do, suitable for a high school class.  To that end I've edited the question with an example from a high school text that describes the formation of a bond, but leaves out the part about the release of energy.

Also, the college board has an article, 'Ending Misconceptions About the Energy of Chemical Bonds', link: https://apcentral.collegeboard.org/courses/ap-chemistry/classroom-resources/ending-misconceptions-about-energy-chemical-bonds  which does not answer the question.

Rather than deleting the question (which we will not do because it is answered) and asking it anew, it is preferable to [edit] the question for clarity. Now this is a bit of a problem here, as the question is already answered. In such cases it would be best to ask a new question as a follow-up to this question. When in doubt, please consult [meta] to ask for advice.

https://chemistry.stackexchange.com/questions/55914/how-is-heat-physically-released-in-an-exothermic-process

It is easier first to think of what happens when a bond is broken in an exothermic reaction. The energy is initially released into vibrations,  rotations and translational motion of the two fragments. This energy is then lost as these fragments collide with other gas molecules or with solvent molecules and so eventually ends up as heat.
In exothermic bond formation the extra energy vibrationally and rotationally excites the new molecule and this is then lost via collisions with surrounding gas or solvent molecules and again ends up as heat.
(If the new molecule were isolated in space (so no collisions) it would remain 'hot' until the energy is radiated away from vibrational levels and until it (radiatively) comes into equilibrium with the surroundings. This is a  slow process, by many orders of magnitude,  compared to collisions in solution or in a gas at normal atmospheric pressure )
In a very few types of reaction an electronically excited state is produced in the product and then photons radiate (as visible light) some of the energy away. This is called chemiluminescence.

Energy is not released when a bond is broken.  It requires energy to break a bond.

The question is about an exothermic reaction in which case energy is released

Still, energy is not released when a bond is broken, even in an exothermic reaction.
You need to separate two atoms which have an attractive force between them. That has to require energy.
In an exothermic reaction, generally, more bonds form than break, so there is an energy surplus.

You are not quite correct. You are effectively talking about the activation energy, or energy barrier, between reactant and product,  when this is surmounted more energy is released in an exothermic reaction than needed to overcome the barrier.

Energy is never released when a bond is broken in isolation. An exothermic reaction involves stronger bonds forming than are broken, it is the formation that releases the energy. Activation energy doesn’t affect this as it only involves the energy that must be overcome during the reaction pathway, not the potential energies of the bonding states before and after.

No, you are wrong.

Not to reopen an old argument, but @Withnail was right (i.e. you spend energy breaking bonds and get energy back when bonds form, thus an exothermic reaction means you spent more energy breaking reactives than got back from forming products). You have evidence for it in this FAQ: https://www.studysmarter.us/explanations/chemistry/physical-chemistry/bond-enthalpy/ and in these questions/answers: https://chemistry.stackexchange.com/questions/136989/is-there-an-example-of-a-bond-with-negative-bond-enthalpy?newreg=c0e74d4a0caf4cfd94f1f1f5a058651d and https://chemistry.stackexchange.com/a/13583

@Rusca8 If you just break a bond say $\ce{H_2\to H\cdot  + H\cdot}$  of course this needs energy but in a reaction say $\ce{F + H2 \to HF + H}$ then this is exothermic and the HF product is rotationally, vibrationally and translationally excited to account for  the $130$ kJ/mole energy released. In fact so much vibrational energy is available in the HF product  that this can be made into an infra red laser.

But the point was (for all of them) that the reaction is exothermic cause you're getting more energy from creating HF than you spend breaking H2. Specifically, your 130kJ/mol are there because you're getting 565 kJ/mol from forming HF and spending 432 kJ/mol on breaking H2. So the breaking of a bond is indeed always endothermic. It's just that the product you're forming releases more energy than the energy you spent.

yes of course, the question was ' how is energy released...' which I think I have answered.


text:
Why does a weak base, which is also a salt, dissociate completely in water?
I'm answering a problem where $\ce{KOH}$ and $\ce{KOAc}$ dissociate nearly as much as each other in separate aqueous solutions at equal concentrations. The problem essentially assumed both completely dissociate.
It's my understanding that weak bases are not meant to dissociate completely, yet $\ce{KOAc}$ is a weak base. However, $\ce{KOAc}$, and $\ce{KOH}$, are both salts.
I understand that a salt is the combination of a charged anion and cation.
The rules that determine solubility state that Group I elements and acetates are always soluble. Thus, it does make sense that $\ce{KOAc}$ would completely dissolve.
Is it always true that if a weak base is also a highly soluble salt, it will still dissociate completely into its component ions, like a strong base would? Would this trend be generalizable to a weak acid that is also a highly soluble salt?
edit: I was trying to simplify the problem, but I made an error in 'transcribing' it. The problem stated that separate solutions of $\ce{KOAc}$ and $\ce{KOH}$ essentially produced the same amount of ions in solution (ie., they conducted electricity to an equal extent). After reading on solubility and dissociation, both compounds dissolved equally as much, producing ionic compounds. But only the strong base $\ce{OH-}$ dissociates completely. Although relative ratios of ions are different in each solution, the amount of ions remains the same.
My apologies for a badly worded question.

What do you mean by "dissociation of the weak base"? There are two separate but coupled equilibria here, and I think you are confused because you're mixing the two processes together.

I'm aware that weak bases do not fully dissociate. But, if the weak base is a highly soluble salt, then it does dissociate completely. I'm assuming you mean the 2 separate but coupled equilibria are acid-base and salt dissociation?
What I'm wondering is why the fact that $\ce{KOAc}$ is a soluble salt takes precedence over the fact that it is a weak base in determining whether dissociates.

You seem to be confusing solubility with dissociation. In solution both the KOAc and K$^+$ and OAc$^-$  exist in equilibrium, but for KOH only K$^+$ and OH$^-$ exist. See https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566 and https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355 for more details.

"I'm aware that weak bases do not fully dissociate." This statement doesn't quite make sense, and I think is at the root of your confusion.

@Zhe Why does that statement not make sense?

I give you acetate. It's an anion. It doesn't dissociate into anything.

@porphyrin Sir as the answer given here says weak bases do not dissociate , is it really true ? As such what i think is that most bases whether weak/strong fully/partially dissociate isnt ?

And Sir may you once check my question asked( recently posted )?

Weak bases, that are not salts, do not dissociate, but partial react with water as
$$ \ce{R-NH2 + H2O <<=> R-NH3^+ + OH-}$$
$\ce{KOAc}$ is not weak base, but a salt.
The weak base is $\ce{OAc-}$, that is created by dissociation of the salt $\ce{KOAc}$ in water, which undergoes protonization:
$$\begin{align}
\ce{KOAc &-> K+ + OAc-}  \\
\ce{OAc- + H2O &<=> HOAc + OH-}  \\
\end{align}$$
$\ce{KOH}$ is not a salt, but a base. If it were a salt, I would be curious what are the related base and an acid that create this salt.

$\ce{KOH}$ is a potassium salt. It dissociates into ions $\ce{K+}$ and $\ce{OH-}$ the same way salt $\ce{KOAc}$ does.

Salts dissociate, but not everything, what dissociates, is a salt. Would you consider sulfuric acid a salt ? Rather, $\ce{KOH}$ is a ionic compound.

@chompion If you consider $\ce{KOH}$ as a salt, what is the acid and the base that mutually react to create the "salt" $\ce{KOH}$  ?

The strong bases we usually work with are confusing in terms of the Brønsted-Lowry definition. It would be better to start with something like $\ce{CH3CH2O-}$ or butyl lithium as the first strong base (clearly proton acceptors) and then discuss NaOH (makes hydroxide, does not accept protons unless you count self-exchange) later.


text:
Percent Dissociated from Titration Curve

Question 818 references the titration curve. Answer is A because $\ce{H+}$ conc $= 10^{-4}$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $0.1 - 0.0001$, which is about $0.1$. So then its $\frac{0.0001}{0.1}\times 100 = 0.1\%$.
Where in the world are they getting the original concentration? how did they get $0.1$ as the concentration of undissociated acid?

I'm not sure I follow your logic.
For a monobasic acid S $(\ce{HA})$ dissociation degree $α$ is
$$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$
where $c_\mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_\mathrm{b}$:
$$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies c_\mathrm{a} = \frac{c_\mathrm{b}V_\mathrm{b}}{V_\mathrm{a}}$$
Finally, taking $V_\mathrm{a} = V_\mathrm{b} = \pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[\ce{H+}] \approx \pu{1e-4 mol L-1}$:
$$α = \frac{[\ce{H+}]V_\mathrm{a}}{c_\mathrm{b}V_\mathrm{b}} = \frac{\pu{1e-4 mol L-1}\cdot\pu{50 mL}}{\pu{0.1 mol L-1}\cdot\pu{50 mL}} = \pu{1e-3}~\text{or}~0.1\%$$

Thank you! I appreciate your help. I seem to have missed the entire purpose of the titration... so the idea is that you are setting the molar amount of base to the molar amount of acid specifically at the equivalence point, then solving for the original concentration of acid. then you divide the hydrogen ion concentration by this value


text:
Why do electrocatalysts based on polyoxometalates and carbon materials have a weaker activity/performance in acidic media?
I am testing now some doped N and S doped carbon nanomaterials alone and immobilized in POM for the purpose of developing a nanocomposite that can serve as an electrocatalyst for energy-related reactions in order to replace Pt.
The purpose in this work is for the reaction to proceed in 1 step, which means to have a 4 electron transfer.
In KOH 0.1 M they are giving better results (3.0 - 3.6 e), while in H2SO4 0.1 M the number of transferred electrons goes down to 1.5-2.
I was wondering what might be the specific reason of the acidic media (hydrogen cations) to lead to this ?
Thank you!


text:
Possibility of non-hexagonal graphene
I understand that monolayer carbon naturally forms in a regular hexagonal tiling.
Is it, however, possible (if difficult) to form carbon monolayers of a different structure?
A 3-12-12 tiling, for example…

(with carbon atoms at the vertices and bonds along the edges) would allow two single bonds and one double bond per atom, just like in normal graphene.

Welcome to Chemistry.SE! Could you use an image that shows the arrangement of atoms? It is not clear what  your graphic represents.

I remember answering a similar question an year or so ago: Is this structure stable? — the same pattern, but with nitrogen atoms. Probably might be interesting for you (also check out the comments to that question).

@andselisk USPEX sounds interesting. I should try and look into that.

related https://en.wikipedia.org/wiki/Penta-graphene https://en.wikipedia.org/wiki/Phagraphene


text:
Miscibility of Hexane and Ethanol (Anhydrous vs 96%)
In my lab, I mixed 8:2 (mass:mass) of Hexane:Ethanol
(Hexane: 0 ppm of water content, analytical grade, & ethanol: 4% water content, pharmacy grade)
At room temperature, and at -20ºC, there was a clear phase separation.
I repeated the experiment, but using anhydrous ethanol instead (1300 ppm of water content).
Now, at room temperature, and at -20ºC, there was no phase separation.
I added enough water to the mixture to make it equal to the first mixture, and the phase separation came clearly into view.
My question is, is this kind of very specific data of miscibilities available somewhere? Ethanol 96% is what is commonly used, yet pure ethanol behaves differently in this case.
Any link or insight into the importance of such little water is appreciated.

With anhydrous ethanol and hexane you have a binary system (hexane:ethanol), with 96% Ethanol and hexane you have a ternary system (hexane:ethanol:water). Yes they have different phase diagrams. Why would you expect anything different?

I'm just surprised that a small amount of water makes such a difference. Plus, any polar solvent is going to have a decent amount of water in it. So next time I see one used in a text, I'll be doubting just how anhidrous it is.

Welcome to StackExchange Chemistry. Water definitely does not mix with hexane, so there will be two phases for sure. Now, ethanol has a choice between the more polar phase (water) and the less polar (hexane). Most of it will go to the polar phase.

This is kind of an extreme case. Anhydrous ethanol and hexane are "just barely miscible"; the next step up in polarity (anhydrous methanol) is not miscible with hexane in all proportions. Furthermore, 4% water represents a significant amount of "contamination" with the least miscible common solvent. Even non-anhydrous solvents are often purchased/used with <1000 ppm water, and 1% is 10000 ppm. For most purposes solvent miscibility tables work fine even with slightly impure solvents, but for these aqueous mixtures, you may just have to test things yourself on a small scale.

Theory and Application of solubility Parameters are beyond our scope. However, an interesting viewer should have at least read through collection of research by Charles M. Hansen, published during 60s (Ref.1). One of the parameters discussed in this book is Hildebrand solubility parameter ($\delta$), which is a good indication of solubility, according to Wikipedia. The principal utility of Hildebrand solubility parameter is that it provides simple predictions of phase equilibrium, based on a single parameter. These predictions are often useful, particularly for nonpolar and slightly polar systems without hydrogen bonding. According to the tabulated $\delta$ values, ethanol is polar, and has relatively higher value ($\pu {12.92 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$) than that of hexane ($\pu {7.24 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$) (Ref.1). Yet, it is interesting to note that hexane is soluble in absolute ethanol, but is insoluble in methanol ($\delta_{\ce{MeOH}}=\pu {14.28 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$), albeit the difference of relevant Hildebrand solubility parameters (${\Delta\delta}$) is just $\pu {1.36 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$ (from Ref.1 and 2).
It is an interesting aspect of the Hildebrand solvent parameters that the $\delta$ value of a solvent mixture can be determined by averaging the $\delta$ values of the individual solvents by volume. For example, if you mix $V_\mathrm{A}$ of solvent A with $V_\mathrm{B}$ of solvent B, the mixture will have Hildebrand value of $\delta_\mathrm{Mix}$, which would be given by:
$$\delta_\mathrm{Mix} = \delta_\mathrm{A}\left( \frac{V_\mathrm{A}}{ V_\mathrm{A}+ V_\mathrm{B}}\right) + \delta_\mathrm{B}\left( \frac{V_\mathrm{B}}{ V_\mathrm{A}+ V_\mathrm{B}}\right)$$
Let's see what is the $\delta$ of 96% ethanol ($\delta_{\ce{H2O}}=\pu {23.5 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$):
$$ 23.5\left(\frac{4}{100}\right) + 12.92\left(\frac{96}{100}\right) = 13.34 = \delta_{96\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$$
Let's see now, what is the value of $\delta$ of 90% ethanol:
$$ 23.5\left(\frac{10}{100}\right) + 12.92\left(\frac{90}{100}\right) = 14.0 = \delta_{90\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$$
Even though still $\delta_{90\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$, the concept is clear. The value is more than $\delta_{\ce{EtOH}}$ and getting closer by each addition of water droplet and to been not miscible with hexane. Therefore, it is safe to say that these data would gives you some insight to what you are looking for.
Note: For a review of solution systems: See ref.3

References:

C. M. Hansen, In The Three Dimensional Solubility Parameter and Solvent Diffusion Coefficient: Their Importance in Surface Coating Formulation; Danish Technical Press: Copenhagen, Denmark, 1967.
C. M. Hansen, "The Three Dimensional Solubility Parameter Key to Paint Component Affinities: 1. Solvents Plasticizers, Polymers, and Resins," Journal of Paint Technology 1967, 39(505), 104-117.
R. Wiśniewski, E. Śmieszek, E. Kamińska, " Three-dimensional solubility parameters: simple and effective determination of compatibility regions," Progress in Organic Coatings 1995, 26(2-4), 265-274 (https://doi.org/10.1016/0300-9440(96)81583-9).

Takashi Moriyoshi, Yasuhiro Uosaki, Hiroyasu Matsuura, Wataru Nishimoto: (Liquid + liquid) equilibria of (water + ethanol + n-hexane) from 0.1 to 200 MPa at 298.15 K. J. Chem. Thermodyn. 20 (1988) (5) 551-557
IUPAC-NIST Solubility Data Series 69. Ternary Alcohol–Hydrocarbon–Water Systems. 1999
Chih-yung Huang, Pey-yu Chung, I-Min Tseng, Liang-sun Lee: Measurements and Correlations of Liquid-liquid-Equilibria of the Mixtures Consisting of Ethanol, Water, Pentane, Hexane, and Cyclohexane. The Open Thermodynamics Journal 4 (2010) 102-118


text:
Which acid/base does a strong base/acid react when added to a buffer solution?
I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $\ce{CH3COOH}$ and $\ce{NaCH3COO}$ with $K_a = 1.8 \times 10^{-5}$, we might want to find what the pH will become if $\ce{0.010 M KOH}$ is added. Given the dissociation of acetic acid:
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$
My question is, which acid ($\ce{CH3COOH}$ or the conjugate $\ce{H3O+}$) will the base $\ce{KOH}$ react with? My teacher said that the base will react with $\ce{H3O+}$ to shift equilibrium in the forward direction, but why wouldn't it react with $\ce{CH3COOH}$ since it also is an acid?
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.

"My teacher said that the base will react with $\ce{H3O+}$ to shift equilibrium in the $\color{red}{reverse}$ direction,".I think to shift the equilibrium in the $\color{green}{forward}$ direction.

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?

You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$\ce{H3O+ + OH- <=> 2 H2O}$$
If you add up the two reaction, you get a third one:
$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$
You can go either way (use the second or the third reaction) to explain that $\ce{OH-}$ is part of a neutralization reaction. As a consequence, the concentration of $\ce{CH3COO-}$ increases and that of $\ce{CH3COOH}$ and hydronium decreases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $\pu{e-5}$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.

Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.

"that the concentration of $~\ce{ OH−}$ $\color{red}{\text{drops}}$ and the concentration of $~\ce{ CH3COOH}$ $\color{red}{\text{increases}}$."I think:that the concentration of $\ce{ OH−}$ $\color{green}{\text{increases}}$ and the concentration of$\ce{ CH3COOH}$ $\color{green}{\text{decreases}}$.

Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?

@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.

@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.

The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.

Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.
However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.


text:
Will hydrogen chloride dissolved in ethanol react with magnesium?
Magnesium will react with aqueous HCl to form magnesium chloride and hydrogen gas. Simple. But I've just read in my textbook that HCl in a solution of ethanol no reaction will take place. There is no explanation given.
I know HCl is soluble in both solutions, but I cannot figure out what accounts for the difference in reactivity with Mg. I suppose it must be due to waters ability to act as a Bronsted-Lowry base and form a hydronium ion, but I also can't see how this would be an effective intermediate in the reaction mechanism.
I am lost, and would appreciate any expertise in this area.

I find the textbook's claim to be highly suspicious.

I agree. Which textbook is this?

The textbook is Hodder IB Higher Level Chemistry


text:
Determining Kc for the equilibrium involving iron thiocyanate using spectrophotometric data

$$\ce{Fe^3+(aq) + SCN-(aq) <=> [Fe(SCN)]^2+(aq)}$$
$\pu{12.0 mL}$ of $\pu{0.00110 mol L-1}$ $\ce{Fe^3+}$ was added to $\pu{6.00 mL}$ of $\pu{0.00140 mol L-1}$ $\ce{SCN-}$. The absorbance of the resulting solution was $0.625$. If the constant $(\varepsilon b) = \pu{3800 (mol L-1)-1}$, determine the value of $K_c$ that should be reported for the equilibrium.
a. $\pu{1.64e-4}$
b. $\pu{3.03e-4}$
c. $\pu{4.67e-4}$
d. $\pu{5.69e-4}$
e. $957$

Answer given: e.
Given what I know about $K_c$ it should be:
$$K_c = \frac{\left[\ce{[Fe(SCN)]^2+}\right]}{[\ce{Fe^3+}][\ce{SCN-}]}$$
which gives me the answer a. I tried calculating $C$ by manipulating Beer's law, but I still don't end up with $957$ no matter what I try.
I would appreciate it if someone could walk me through the steps of solving this problem.

answer choice e is just 957? Without even going through the question, it seems like it would be noticeable that this option is several orders of magnitude larger than any of the other options and is also the only option greater than 1.

I noticed that and on the test I would pick e, I just don't understand how to actually arrive at that answer through calculations. I would like to understand it because I feel like the question won't be so straight forward on the real exam.

To start with, its often helpful to make an ICE table when dealing with problems of chemical equilibrium.
$$\begin{array}{cccc}
\begin{array}{c|ccc}
\hline
 & \ce{Fe^3+} & \ce{SCN-} & \ce{[Fe(SCN)]^2+} \\ \hline
\mathrm{initial} & 1.32\times10^{-5} & 8.40\times10^{-6} & 0\\
\mathrm{change} & -x & -x &+x \\
 \mathrm{equil} & 1.32\times10^{-5}-x & 8.40\times10^{-6}-x & x\\ \hline\\
\mathrm{conc.} & \frac{1.32\times10^{-5}-x}{V}& \frac{8.40\times10^{-6}-x}{V} & \frac{x}{V}\end{array} \end{array}$$
Then we can use Beer's Law to find x.
$$\frac{x}{V}=\frac{A}{\epsilon b}=\pu{1.645\times10^{-3}mol/L} \to x= (1.645\times10^{-3})(.018)=\pu{2.961\times10^{-6}mol}$$
With x, we can find all the concentrations:
$[\ce{Fe^{3+}}]=5.69\times10^{-3}$
$[\ce{SCN-}]=3.02\times10^{-4}$
$[\ce{[Fe(SCN)]^2+}]=1.64\times10^{-4}$
Finally, we can then compute $K_c$:
$$K_c=\frac{[\ce{[Fe(SCN)]^2+}]}{[\ce{Fe^3+}][\ce{SCN-}]}=956.77$$

Thank you very much. I wasn't finding the concentrations properly and instead I was using the initial concentrations instead of the equilibrium ones. Very informative answer.

@Codeblockz definitely look at MaxW's answer as well. It can help with understanding a problem to see slightly different ways of getting to the solution.

Wonder when ICE tables became the rage. I don't ever remember seeing one before I came onto this site. -- When I was in chemistry classes there were just four elements, earth, air, water and fire. ;-)

@MaxW Thats an interesting point. It was what I was taught back in high school, so its at least been around for the last decade or so. However, I couldn't find the original source of the idea. Might be a worthwhile question here or History of Science and Math depending on if they allow pedagogical history.

I left high school in 1969.

@MaxW: It has been around for a while. But then, it wasn't called ICE. It's a cool name though! :-)

@MathewMahindaratne Do you have a sense of when it was made or previous terminology used to describe it? I have opened a question looking for when the method was made.

@Tyberius: Wasn't sure when is it introduced. But, I came to USA as a graduate student in 1989 (to Marquette University, Milwaukee, WI) and they were using the method then in Gen Chem classes (30 years ago! :-)).

The problem is a bit strange since the details of the absorbance measurement aren't detailed very well. I'll assume the book defines the relationship as:
$$A = \epsilon bc$$
where:

A=absorbance
$\epsilon$ = molar attenuation coefficient
$b$ = path length
$c$ = concentration

So the concentration, c, of the $\ce{[Fe(SCN)]^2+}$ is:
$$ c = \dfrac{A}{\epsilon b} = \dfrac{0.625}{3800} = 1.64\times10^{-4}$$
Considering the dilutions...
$\ce{[Fe^{+3}] + [Fe(SCN)]^2+} = 0.00110 \times \dfrac{12}{18} = 7.33\times10^{-4}$
$\ce{[SCN^-] + [Fe(SCN)]^2+} = 0.00140 \times \dfrac{6}{18} = 4.67\times10^{-4}$
But there is $1.64\times10^{-4}$ of $\ce{[Fe(SCN)]^2+}$ so:
$\ce{[Fe^{+3}]} = 7.33\times10^{-4} - 1.64\times10^{-4} = 5.69\times10^{-4}$
$\ce{[SCN^-]} = 4.67\times10^{-4} - 1.64\times10^{-4} = 3.03\times10^{-4}$
Therefore:
$K_c = \dfrac{1.64\times10^{-4}}{(5.69\times10^{-4})(3.03\times10^{-4})} = 951$
I'd assume the small difference is just due to a difference in rounding somewhere...

@MathewMahindaratne - Made the correction. Thanks for pointing out the error.


text:
Understanding the setup of a chemistry lab about redox
In chemistry lab we were looking at oxidation and reduction reactions. To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals. We mixed metals like $\ce{Fe}$ with solution like $\ce{Cu(NO3)2}$ to see if a reaction occurred.
I'm wondering why is it not necessary to mix a metal like $\ce{Fe}$ into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)? We mixed $\ce{Fe}$ with $\ce{Zn(NO3)2}$, $\ce{Cu(NO3)2}$, and $\ce{Ni(NO3)2}$.

my guess is there would be no reaction and nothing would happen. if that is the case is there something going on chemically that wont allow a reaction?

That's right, but what reaction would you expect if the chemical reasons would  allow it?

that's the part I cant figure out, haha

OK, what is the reaction in one of the cases where there surely is a reaction - say, with Cu(NO3)2?

IMHO, the reaction. $$\ce{Fe^2+ + Fe <=> Fe + Fe^2+ }$$ is not as crazy as it may look, as there would be ongoing dynamic exchange of ions between the metal and the solution. :-)

I'm wondering why is it not necessary to mix a metal like Fe into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)?

First remember what the purpose of your experiment was:

To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals.

In an electrochemical cell setup, $\ce{Fe}$ would be in two-way equilibrium with $\ce{Fe^2+}$:
$$\ce{Fe + Fe^2+ <=> Fe + Fe^2+}$$
When power was drawn from the cell you then get:
$$\ce{Fe -> 2 e- + Fe^2+}$$
Where as with another salt such as copper which is less reactive, the copper will oxidize the iron in one direction with no reverse reaction.
$$\ce{Fe + Cu^2+ -> Cu + Fe^2+}$$
or with zinc which is more reactive no reaction will occur at all:
$$\require{cancel} \ce{Fe + Zn^2+ \cancel{\ce{<=>}} Zn + Fe^2+ ->Fe + Zn^2+}$$

Thanks for the explanation. I knew there wasn't a reaction but didn't fully understand why.


text:
Enolate ion opening an epoxide?
Does anyone know if it is possible to open an epoxide and do a 2-C extension using an enolate ion, say, of acetone? Here's an example of what I mean:

Based on what I know of enolate chemistry, this shouldn't be problematic, but I've never seen it.
I have done some preliminary literature searching and checked out masterorganicchemistry.com, etc. but have not been able to find anything about this.

Never say never! :-) Here is the answer for your question (Ref.1):

Abstract: We report here that (1) enolate anions of five- to seven-membered cycloalkanones nucleophilically open cyclopentene and cyclohexene oxides in 57−76% yields and with 4−8:1 diastereoselectivity; (2) enolate anions formed regiospecifically via kinetic deprotonation of 2-cyclohexenone and 2-cycloheptenone open cyclohexene oxide in 60−62% yields and with 32−95:1 diastereoselectivity; and (3) an aryl methyl ketone enolate anion opens a monosubstituted epoxide as the key step in a short synthesis of the $\gamma$-hydroxyketone (GHK) aglycon of the natural product curculigine.

Graphical Abstract:

The product(s) and selectivity are depicted in following scheme:

Reference:

G. H. Posner, J. P. Maxwell†, M. Kahraman, "Mild, Fast, and Stereoselective Epoxide Opening by Ketone Enolate Anions. Application to Synthesis of the Norlignan Curculigine," J. Org. Chem. 2003, 68(8), 3049–3054 (DOI: 10.1021/jo020744q).

That's somewhat interesting that the O-alkylation decreases when the counterion is changed from Li -> Na -> K; I would have expected it the other way around (Li should associate tightly with the negatively charged oxygen).


text:
Who developed the concept of ICE tables?
Many general chemistry courses (at least in the US) or online tutorials solve problems of chemical equilibrium by using ICE (sometimes RICE) tables, which track how the moles of reactants and products change going from an initial state to equilibrium.
I'm curious when this technique became so widespread and who developed it? My understanding is that ICE tables are a somewhat modern pedagogical tool, but I have yet to find a source for the idea and when it was popularized.

Looked in "Books"  with Google for "ICE table" & Chemistry. First I found was Chemistry, Principles & Practice by Daniel L. Reger, Scott R. Goode, Edward E. Mercer, Saunders College Pub., 1997

Are you sure this is not just marketing? I mean, is this not just a book-keeping device? You solve these problems the same way no matter what you call them, right?

@NightWriter it is marketing, but someone had to be doing the marketing. Somebody had to convince every general chemistry textbook writer that this was the best way to represent these problems. An analogus situation happened in math where a lot of elementary school students aren't taught long multiplication, but instead use tables. It's all the same math, but it supposedly makes a difference in how well students can understand what they are doing.

ICE tables are mentioned in an 2003 J Chem Ed article by S. Watkins:

A systematic approach to chemical reaction calculations, which is an extension  and  elaboration  of  the  equilibrium  table  method first  observed  by  this  author  in  the  text  by  Masterton  and Slowinski (4) nearly thirty years ago, is presented. Most texts utilize an “ICE” (initial, change, equilibrium) table when discussing equilibrium (5), but this author has found none that apply the technique systematically to stoichiometry problems.

The textbook cited is Masterton, W. L.; Slowinski, E. J. Chemical Principles, 3rd ed.;Saunders: Philadelphia, 1973; p 355.
So according to this, they have been around at least since 1973.

This article discusses the history and application of the concept of "extent of reaction". ICE tables are a hidden way of introducing the extent of reaction, which is a concept that can be used to solve multi-reaction problems distinct from the usual system of equations from mass and charge balances.

I think this is the answer or close to it. I will look around for that original textbook and accept this answer in a day or so.

As expected the concept of equilibrium calculations is more than a century old idea. The algebra has not changed. Check internet archives for classical general chemistry textbooks to see how equilibrium calculations were done in 60s-70s. If you are specifically looking for the representation of "initial, change, equilibrium" tables, this is also four decades old. Books from the early 80s used this approach. See for example "Chemistry with inorganic qualitative analysis" by Thenard (1980). These days everyone likes to coin fancy acronyms and words. Gold colloids became gold nanoparticles, and "initial, change, equilibrium" became ICE!

My main interest is in the representation. I know the math of it likely hasn't changed in over a century, but the structure of how it is taught has. I'm interested in what convinced people to make this the de facto standard for this type of problem and when. Based on your answer and MaxW's comments on the linked post, it likely started being used in the 70s.

In problem solving, teachers tend to develop protocols (algorithms) so that everyone can systematically follow the approach. ICE is perhaps one of them.

Thats what I expect. However for it to be widely popularized, I think some had to be the first to include it in a textbook or write a chemical education article about the idea.