text:
Why do we say the position of equilibrium is shifted even when K is the same?
I realize that there are similar questions that have already been asked here, however none of them answer the question. The answers all say that (when K is the same) you "you haven't really shifted the equilibrium position in the way it sounds like" or something similar.
Many people also say that the position of equilibrium is roughly equivalent to Q. However, going by that definition, the position of equilibrium would be the same if you only change concentrations. Yet, many textbooks explicitly say things like "increasing the concentration of a reactant shifts the position of equilibrium to the products". I understand that more products are produced to cause Q to reach K again, but once you are at equilibrium Q is the same as before, Q = K.
Is there any specific definition of position of equilibrium? Or is the term only used when you are saying "the position of equilibrium is shifting right/left" to mean that the forward/reverse reaction is happening at a higher rate for the moment?
The second option seems to be what people answer with, but it just does not make sense to me that the position of equilibrium would only be a term used to refer to a behavior of a reaction that is disturbed from equilibrium, instead of referring a quality at equilibrium that can be compared with other equilibrium states.

Is there any specific definition of position of equilibrium?

No, it is a fuzzy concept and not very helpful. When a reaction is at equilibrium, there is a certain composition of the equilibrium mixture. However, this composition is not unique. There are many other compositions (sets of concentrations) that are also at equilibrium.
I will use a simple reaction to illustrate this:

The concentration of the reactant $\ce{N2O4}$ is plotted on the y-axis and the concentration of the product $\ce{NO2}$ is plotted on the x-axis. The blue curve shows reaction mixtures that are at equilibrium. If I start with pure $\ce{N2O4}$ at a partial pressure or concentration of 10 (arbitrary units, just for illustration), the reaction will proceed on the red straight line (slope -1/2 because of the stoichiometry). The reaction will proceed net forward until it reaches equilibrium, marked with the circle. However, if I initially have 8 (purple straight line) or 12 (green straight line), I will reach different equilibrium compositions. I can't reach the circled composition because I don't have the right number of atoms.

Or is the term only used when you are saying "the position of equilibrium is shifting right/left" to mean that the forward/reverse reaction is happening at a higher rate for the moment?

"The position of equilibrium is shifting right" should be replaced by "the reaction is going in a net forward direction" until it reaches equilibrium. I will use the same example to illustrate this:

In this example, we start out at equilibrium on the red line, and add some product. We are pretending the reaction is turned off as we add the product, so the reactant concentration stays the same. Then (like in a cartoon where the character notices they stepped off the cliff), we pretend the reaction is turned on again. It is not at equilibrium, and we see a net reverse reaction. You could talk about this net reverse reaction as a shift to the left, but it is describing the reaction, not the equilibrium. Equilibrium states are defined by the blue curve, which does not shift.
You could also add reactant (half as much), as the next panel shows:

This time, there is a net forward reaction, and you end up at the same equilibrium composition (because the reaction moves along on the same green line, constrained by stoichiometry).
Below are the other two cases involving removal of reactant or product. Again, I chose the example such that both cases have the reaction composition constrained to the same line, purple in this case.

Why do we say the position of equilibrium is shifted even when K is the same?

I don't know why, but we shouldn't. We should say that when we disturb an equilibrium by adding or removing a single reactant or product, the reaction will go in the direction of re-establishing an equilibrium (net forward or reverse, depending on the details). Notice I say establish "an equilibrium" because there are multiple compositions (or sets of concentrations) that satisfy the equilibrium condition. After adding or removing a single species, you can not return to the same equilibrium state (because the number of atoms changed).

Maybe you understand better with a simple numerical example.
Suppose we are dealing with an equilibrium reaction A + B <=> C + D, where the equilibrium constant $K$ is $K = \frac{[C][D]}{[A][B]}$ = $1$. Suppose the volume of the system (in liters), the equilibrium constant and all concentrations (in mole/L) are all equal to $1$.
Now add one supplementary mole A to the system. As a consequence,  Q = $2$. The system is out of equilibrium. To obtain a new equilibrium, a certain amount $x$ of A and B is destroyed producing $x$ new molecules of C and D. The new equilibrium state will be described by [A] = $2$ - $x$, [B] = $1$ - $x$, [C] = [D] = $1$ + $x$. This equilibrium has the same constant $K$ = $1$, and is described by :
$$K = \frac{[C][D]}{[A][B]} = \frac{(1 + x)^2}{(2-x)(1-x)} = 1 $$
With a little algebra, you find $x$ = $0.2$. The final system contains $1.8$ mole A, $0.8$ mole B, $1.2$ mole C and D.  This correspond exactly to the sentence the position of equilibrium is shifting right that you were discussing in the question. There is no change of kinetics.

Since concentrations at equilibrium are changing the kinetic rates must also change. the forward and reverse rates are equal not constant.

Is there any specific definition of position of equilibrium? Or is the term only used when you are saying "the position of equilibrium is shifting right/left" to mean that the forward/reverse reaction is happening at a higher rate for the moment?

Yes, there is a specific definition. It is the condition at which there is no net driving force for the (macroscopic) system to change. An imperfect mechanical analogue for this is a ball rolling in a frictionless basin. When it reaches the basin bottom the net force on the ball is null (the ball however has inertia which makes the analogy flawed; if friction occurred between basin and ball, however, the latter could exchange (lose) kinetic energy (as heat). The motion in the basin can be described as motion within a gravitational potential energy surface. The slope of the surface is proportional to the gravitational force driving motion.
In the molecular world the classical mechanical worldview must be supplemented by a statistical treatment. You can still describe reaction paths in terms of chemical potential (the analogue of gravitational potential) and even of chemical forces (although this is a weaker analogy since the vectorial property of a force, direction, has to be interpreted as the direction of a process such as a chemical reaction (what you call left or right).
In any case, at equilibrium the free energy of the system is at a minimum (and the net chemical force is null). Karsten's diagrams could be modified to show free energy (the sum of chemical potentials). The blue line would represent a potential energy minimum isoline.
Note that K is the equilibrium constant, not "the equilibrium". A condition of chemical reaction equilibrium is that Q=K. Equivalently that the chemical potentials of products and reactants are equal.

The technical term I would prefer is equilibrium state, not equilibrium position. This is a state that happens to be at equilibrium. If you take all of the equilibrium states in a multi-dimensional diagram showing concentration of species, temperature, and other relevant parameters, equilibrium is not at a certain position but rather on a hyperplane (n-1 dimensional, if your parameter space is n-dimensional).

@Karsten Position can refer to a specific coordinate in the state space of equilibrium states. Yes, there can be many such states, each one is defined by the conditions I outline, namely the equivalent of mechanical equilibrium, when chemical forces balance.

Think carefully about what is happening at equilibrium. First everything is constant T, P, numbers of moles of reactants and products etc. [Lets minimize such things as tidal forces from the time of day etc.] Everything is Rock Solid the universe is frozen in place! That seems a perfectly reasonable approach and some think that is the eventual end of the Universe, everything at equilibrium. Fortunately we have not arrived yet and There is continuous energy and matter flux that can perturb any [micro] system that attains equilibrium.
How is this possible? Why do reactions not proceed to completion and the entire universe fall into a minimum energy rock state? Because of the quantized properties of matter and energy and the constant interplay between them. Matter is constantly absorbing and emitting energy and this energy is contained in random molecular velocities and QM states of the molecules. This thermal energy or heat content is conveniently contained in temperature. This means that at equilibrium all the molecules of reactants and products have kinetic energy and their concentrations [chemical activities] satisfy the equilibrium constant.
All molecules are a relative energy minimum; an activation energy is necessary for a reaction to happen. Only an elementary reaction, a reaction where the products do not further react more quickly, can reach equilibrium. This means that the mechanism is identical in the forward and reverse directions. The direction of the reaction depends on the concentrations of the components, the two activation energies, and the number of molecules with sufficient energy to react; the latter is determined by the temperature. At equilibrium the reactions do not cease; the activities and activation energies are such that the forward and reverse reactions have the same rates. Changing a concentration will change a rate and the equilibrium will change to reestablish equal rates and equilibrium. The stoichiometry of the reaction determines whether the ratios change. A change in temperature changes the equilibrium position [the free energy changes] and results in a different equilibrium constant because the different activation energies respond differently to temperature.
This is why yields of reactions can be changed by manipulating reactant ratios or removing products as formed or by changing temperatures.


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Best ways to dissolve large amounts of candle wax?
I work for a company that occasionally rents out candles and (mostly glass) holders for events. Many times these holders come back covered in candle wax. I am looking for an efficient way to clean these. I know of many substances that can dissolve candle wax, but most of them require each piece to be cleaned by hand or the substance itself requires its own cleanup.
Can anyone think of a cost efficient way to clean hundreds of candle holders at a time without much physical labor?  Perhaps a solution that they can sit in overnight that could fully dissolve even thick wax buildup? It would be best if the solution was non flammable, and not too noxious.
Right now my best idea is heated vegetable oil to both melt and dissolve the wax, but that will have to be cleaned up as well.

Rather than a chemical method, can you use a physical one? How about placing the candle holders upside-down inside a low temperature oven and then just let the wax drip out? The chemical solutions are flammable and/or somewhat toxic, and you'd rather not have to deal with the solvent use and disposal.

Just like you need to rinse dirty dishes after soaping them, you also need to rinse your glasses after the initial solvent treatment. Otherwise a film of wax-containing solvent will remain, and even if the solvent evaporates, the wax won't.
If vegetable oil dissolves the wax successfully, you could then put the glassware in a dishwasher to remove the oil.

I wonder about dipping in a pot of boiling water. Not sure if that would be hot enough to melt whatever wax. The other problem with this would be the thermal shock. Don't know if that would break some of the holders. I also wonder about adding some dishwasher detergent to the water (dishwasher so it won't foam). This might work better if the candle holders were small votive candle holders and you could get fill the pot cold, then heat and left cool down overnight.

After giving this a bit more thought, it might be best to just use hot water. The wax will be hydrophobic so water may be better at removing it without leaving any residue. The wax should either float or sink. I’m not sure about candlewax density. I guess I’ll have to test it. This might remove the residue as well because wax can mix with water without dissolving if it is hot enough
If I did have to go with the vegetable oil, the dishwasher would probably work once most of the wax is removed. I would just have to make sure that there is very little wax left so that it didn’t clog the pipes.

The wax should float. If you allow the solution to cool, or throw in some ice, the wax should be hard and floating on the surface. Just skim the wax off. Definitely do not put melted wax down the drain.

Just agreeing with @NicolauSakerNeto: easiest way to clean a candelabra is place it in an oven upside down through the top rack, with a cookie sheet underneath, keeping the temp just above the wax m.p. Of course, this could also be a serious smoke and fire hazard -- watch closely, and have an extinguisher handy (e.g., a box of baking soda.) This really belongs on https://lifehacks.stackexchange.com/ , not here, though.

source: Stack Exchange

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How do I find pH of an extremely low molarity strong acid?
Given
$$[HNO_{3}] = 4.53(10)^{-10},$$ find the pH of the acid. $K_{a}$ is not given, and it is at 25 degrees celsius.
Because the molarity is very low, as in less than 10^-6, the teacher tells us that the hydronium ions from water affect the pH, and so the pH is not exactly $-log(4.53(10)^{10})$. How do I calculate the pH then?
I remember the teacher creating a reaction based off of nitric acid and water, and that because nitric acid is a strong acid, that all the HNO3 would dissociate leading to [H3O+] = 4.53e-10, and then I tried putting this answer into the standard autoionization of water reaction (as 25 celsius), so
$$10^{-14} = (4.53e-10 + x)x$$ solve for x, then pH would be -log(4.53e-10+x), but this creates an imaginary answer

Let's not make this complicated. Because of the equilibrium shift: $$\ce{[H+] - [OH-] = [HNO3]}$$

The general equation is found in this answer https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566 and in particular look at the answer in the comments which is 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

There are two possible questions here, which have very different answers.
Question 1 is what you actually posed here: What is the pH of a solution of nitric acid where $\ce{[HNO3]} = 4.53 \times \pu{10^{-10} M}$? The problem with question 1 is that it really cannot be solved without a pKa for nitric acid. To answer it, you have to know how much nitric acid to add to a solution so that $4.53 \times \pu{10^{-10} M}$ of the acid remains undissociated, and the answer is very sensitive to the pKa.
Question 2 is: what happens if $4.53 \times \pu{10^{-10} mol}$ of $\ce{HNO3}$ is dissolved in water? This question can be solved (approximately) even if all you know is that nitric acid is strong. The rough argument goes like this: Pure water has $\ce{[H+]} = \pu{10^{-7} M}$. Even if the acid fully dissociates, it will add only $\pu{4.53 \times 10^{-10} M}$ more protons, which is negligible (only 0.45% of the concentration already present) so the pH is still roughly 7. Feel free to do the calculation more precisely than that.
The problem is, in the resulting solution, $[\ce{HNO3}]$ is not $\pu{4.53 \times 10^{-10}}$, but actually much smaller, because nearly all of the acid has dissociated. So question 2 is actually very different from the question you've stated. That said, it might be the question your teacher intended to ask.

I believe the teacher had in mind the total concentration, for strong and many weak acids equal at given dilution to concentration of the conjugate base.

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Why do most sulphate salts have water of crystallization in them?
I mean looking at most of the sulphate salts (there are so many but just to name some Epsom salt, blue vitirol, gypsum...) they have some amount of water crystallized within them (Epsom salt is MgSO4.7H2O, CuSO4.5H2O is blue vitirol and so on) so why is it like that? Is there any reason why water is crystallized in almost all metal sulphates
It isn't (maybe) related to hydration energy... nor (maybe) related to the size of the sulphate ion.

can anyone suggest me a better tag for this question please I am very confused

It is not fully clear what you are asking. Can you elaborate?

@Poutnik oh I meant looking at most of sulphate salts (there are so many but just to name some epsom salt, blue vitirol, gypsum etc..) they have some amount of water crystallized within them (epsom salt is MgSO4.7H2O , CuSO4.5H2O is blue vitirol and so on) so why is that? like is there any reason why water is crystallized in almost all such sulphates ? That is what I wanted to ask. Yes definitely looking at the length of my post surely told me it wasn't enough to justify

Add it rather to the question and you may ask for the question reopening via the respectively named link. Also, you may want to think longer over written text before posting it.

@Poutnik uh okay i mean this question is still dumb and maybe i shouldnt have asked it here
but yea i am editing it thanks

Now that you've changed the meaning of the question entirely, let's look into it. BaSO4 seems to crystallize without water; so does K2SO4. Still sure about "most"? Well, maybe the majority of all salts, not just sulphates, crystallize with water in them.

@IvanNeretin oh okay but I still think (not believe) that sulphates are the most to somehow grab water in them.....you would be right though and it just may not be sulphates (like the next best bet would be chlorides)........and yea i amn't sure so waiting for more responses :)

@IvanNeretin basically I am finding for some chemical bonding-ish solution if it even exists :')

related https://chemistry.stackexchange.com/questions/110728/hydrate-decomposition-equation-steps

Anion barely does anything, it's the cation that can coordinate many water molecules.

Water of crystallization is stabilized by electrostatic attractions, consequently hydrates are common for salts that contain +2 and +3 cations as well as −2 anions. Sulfate ion has -2 charge, so it is more prone to form hydrates.

@Mithoron yes that is true....

@Nilay oh is there any reason behind for which +2 and +3 attract more?

Sulfates commonly do have water of hydration, but so do many other salts such as chlorides. For instance, iron chlorides used in laboratory experiments at my job are commonly bought as $\ce{FeCl2•4H2O}$ and $\ce{FeCl3•6H2O}$.
One difference between sulfates and some other salts is the oxygen in the sulfate ion can form relatively strong hydrogen bonds with the protic hydrogen in water, enabling incorporation of additional water molecules into the crystal that aren't coordinated with the metal ion. Thus for instance magnesium ion commonly coordinates with six water molecules, but the sulfate salt commonly incorporates a seventh water molecules in its formula unit (Epsom salt) and can even have eleven at lower temperature.

"2-" is also important. After all, that's a great deal greater than 1-.

@IvanNeretin Curiosity strikes and turns out even our old NaCl also form hydrates: From cold (sub-freezing) solutions, salt crystallizes with water of hydration as hydrohalite (the dihydrate NaCl·2H2O NaCl·2H2O). In 2023, it was discovered that under pressure, sodium chloride can form the hydrates NaCl·8.5H2O and NaCl·13H2O.

@IvanNeretin yes true

@NilayGhosh oh thats cool

hey just a quick question don't you think flourides and nitrates would also have this phenomenon?

In the case of fluorides noncoordinated water of hydration is known in some double salts; see https://pubs.rsc.org/en/Content/ArticleLanding/1977/DT/DT9770002212.

@OscarLanzi ohkk thanks!


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Whatever happened to the use of perfluoroalkanes as breathable liquids or blood substitutes?
Some liquid perfluoroalkanes have very high solubilities and carrying capacities for gases like oxygen and carbon dioxide.
In the 1980s and 1990s there were experiments showing that animals could "breathe" when submerged in these liquids (see this on Wikipedia). A real example of this even appeared in the SciFi movie The Abyss (1989) where, in one scene, rats were shown breathing while submersed in one (for real, apparently).
There were related attempts at the time to manufacture artificial blood substitutes using the compounds.
Whatever happened to these ideas? Presumably the efforts to find effective uses failed, but why?

A 2020 article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7086064/

Nothing happened to ideas; it's just that there's not much point in using this. And I seriously doubt you'd like to drown and then vomit all this crap.

And a 2020 article on The Many Uses of Liquid Breathing: A modern day torture device or chemical savior? https://medium.com/stupid-learning/the-many-uses-of-liquid-breathing-123bff550e6c It's always a good concept in SF, where people have been modified to remove drowning reflex. Perhaps the rapid progress in teleoperation and robotics has made this less urgent.

@Mithoron There were other major uses not just breathable liquids. The big ones were to overcome serious medical problems where there is some justification even if there are also side-effects. And why ideas fail is interesting.

BTW, some fire-suppression systems use fluorine-carbon compounds. One was triggered at a data center where I worked, blowing staff through a plate-glass window. Though there were some injuries due to force, nobody suffocated. However, PFAS are now problematic: https://www.fireline.com/how-pfas-is-affecting-the-fire-suppression-industry/

The Discoveries article by Fahad Khan et al. (Discoveries (Craiova). 2020 Jan-Mar; 8(1): e104.
Published online 2020 Mar 18. doi: 10.15190/d.2020.1) discusses perfluorocarbon options (amongst others). From that article, Table 1 shows four different products.

Flusol-DA-20 (Green Cross Corporation, Japan) - FDA approved in 1989, discontinued due to side effects and limited success (patients still needed supplemental O2 is noted in the text).

Oxygen (Alliance Pharmaceutical, San Diego) - reached phase II trial, discontinued due to cost.

Oxycyte (Synthetic Blood International, Costa Mesa) - reached phase IIB trials, discontinued due to lack of enrollment in trials.

Perftoran (Russian Academy of Sciences) - rebranded as Vidaphor (FluorO2 Therapeutics, Boca Raton), "currently awaiting clinical trials" (which would seem unlikely in the US at least right now).

As noted in section "3.3 Limitations":

PFCs are ambitious products that in theory and chemically seem to be a viable option for clinical use. However, issues with manufacturing and development have led them to become mostly discontinued.

and

Due to initial clinical trials with these products showing lack of beneficial outcomes, no current trials are ongoing with PFCs.

So, the products have not lived up to their initial promise.


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How do you determine the path or process in which a gas attains it's maximum (or minimum) temperature in a cyclic process?
Say we have the graph of a reversible cyclic process, in which an isochoric, isobaric, and general process take place.

How would we determine in which path -  AB, BC or CA, the gas would attain it's maximum or minimum temperature?
It would have to be of the form y = mx + c ie. P as a function of V ie. it has to pass the Vertical Line Test to make sure it's a function. So, in this case, it cannot be the isochoric  process AB.
This is actually a graph from a very popular problem, so I'm aware the gas reaches it's maximum temperature during the BC process.
I'm aware of how to calculate the maximum or minimum temperature once we have determined the process or path at which it occurs thanks to this Stack exchange post. Writing the equation of the line, turning it into a quadratic equation, and determining the maxima or minima isn't too difficult.
But what if we had a Carnot Cycle as such:

We're given the pressure and volume at every point. In this case, how would we determine the path at which the gas attains it's maximum temperature? And how would we determine that temperature in this case (as each process is a curve)?

Assuming all the above processes are being performed reversibly over an ideal gas,  the gas follows the ideal gas state equation $$pV=nRT$$ in every moment.
As we are interested in temperature in the respective path segments, we can rewrite the equation as $$p = \frac{nRT}{V}.$$
Then we can, really or mentally, draw hyperboles as isoterms $p=\mathrm{f}(V)_T$.
For a given $T$, $pV = \text{const}$ and $pV$ is proportional to $T$.
For better visualization of the gas isotherms:

For every point of the process paths on the diagram, the absolute temperatute $T$ is proportional to the area of the rectangular with diagonal from the origin to the given point.
Therefore, either from the drawn hyperbole set, either from the rectangle area, we can see that:

The lowest temperature is at the point A.
The highest temperature is near the middle of the path segment BC.

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Does emulsify mean to combine or break down?
In chemistry, emulsification is defined as the "an additive which helps two liquids mix" (Source: Wikipedia), "a mixture of two or more liquids that are normally immiscible owing to liquid-liquid phase separation"(Google's Knowledge Graph), and a substance that stabilizes an emulsion (see https://rockedu.rockefeller.edu/component/emulsifiers-and-surfactants/).  In biology, however, bile is defined as an emulsifying molecule that aids in the digestion, or breakdown, of fats (see https://www.hopkinsmedicine.org/health/conditions-and-diseases/biliary-system-anatomy-and-functions and other similar sources).  So, does an emulsifier stabilize, or keep together fats and hydrophilic molecules or does it break down the fat by surrounding it and dispersing it? It's almost as if there are two opposite definition of emulsifier in chemistry and biology.

Emulsification is a process, an emulsifier a substance that helps to form and stabilize an emulsion - a dispersion system liquid in liquid.

An emulsifier stabilizes an emulsion which is a process of phase change. Molecules are not being broken down, they are just being rearranged between phases.

Opposite? It's the same meaning, just perspective change.

It seems you have confused the related word forms.

Emulsification is a process of forming an emulsion.
An emulsifier is a substance that helps to form and stabilize an emulsion.
Emulsion is a heterogeneous dispersion system of two non-miscible liquids, with droplet size typically in range $\pu{10 nm} - \pu{100 \mu m}$.
To emulsify is performing emulsification to form an emulsion.

IUPAC quote from Wikipedia:

Note 3: The diameters of the droplets constituting the dispersed phase usually range from approximately 10 nm to 100 μm; i.e., the droplets may exceed the usual size limits for colloidal particles.

Emulsifiers have two major, related effects:

They act as surfactants, decreasing surface tension at phase boundaries. This cause decreasing the required work to divide droplets to smaller ones.
They act as stabilizers, adsorbing on phase boundaries, acting against merging two droplets at the contact.

These effects make the  formation of emulsions easier  and cause higher  thermodynamic and kinetic emulsion stability.
The mentioned eventual break of fats by bile has two aspects:

Forming fat emulsions from oils and liquefied solid fats, breaking them macroscopically, mechanically.
Aside of emulsifiers, the bile contains also lipase to enzymatically process  emulsified fats.

Wikipedia: Bile - function:

Bile or gall acts to some extent as a surfactant, helping to emulsify the lipids in food. Bile salt anions are hydrophilic on one side and hydrophobic on the other side; consequently, they tend to aggregate around droplets of lipids (triglycerides and phospholipids) to form micelles, with the hydrophobic sides towards the fat and hydrophilic sides facing outwards. The hydrophilic sides are negatively charged, and this charge prevents fat droplets coated with bile from re-aggregating into larger fat particles. Ordinarily, the micelles in the duodenum have a diameter around 1–50 μm in humans.
The dispersion of food fat into micelles provides a greatly increased surface area for the action of the enzyme pancreatic lipase, which digests the triglycerides, and is able to reach the fatty core through gaps between the bile salts. A triglyceride is broken down into two fatty acids and a monoglyceride, which are absorbed by the villi on the intestine walls. After being transferred across the intestinal membrane, the fatty acids reform into triglycerides (re-esterified), before being absorbed into the lymphatic system through lacteals. Without bile salts, most of the lipids in food would be excreted in feces, undigested.

An emulsion is just a colloidal system of liquid dispersed in a liquid phase just like smoke is a colloidal system of solids dispersed in air. In emulsions, one of the liquid phase is of typical colloidal dimensions (nano to micrometer size). Thus emulsify does not imply chemical breakdown, it simply means that one of the liquid phase has been "physically" broken down into nano or microdroplets in another immiscible liquid. This is why vigorous mechanical forces are needed to emulsify or make an emulsion. Look up the recipes for making mayonaisse at home.
Bile acids emulsify fats means that they help in making fat and oils smaller and smaller (no chemical decomposition) in size during the digestion process so that the body can absorb them easily.

Droplets of stable emulsions need not to have colloidal size.

IUPAC quote from Wikipedia:  Note 3: The diameters of the droplets constituting the dispersed phase
usually range from approximately 10 nm to 100 μm; i.e., the droplets
may exceed the usual size limits for colloidal particles.


text:
UV-Vis Absorption Spectra climbing at lower wavelengths
I have run a sample of Pyrene in Ethanol and got a UV Vis absorbance spectra that starts climbing as it reaches the lower wavelengths. I am using quartz cuvettes and was wondering what could be the cause of this rise at the shorter wavelengths.

Here is the literature spectra.

Here is my measured spectra.

Please add more instrumental details. Is it a single beam instrument or a double beam instrument. What was your blank?

the solvent and quartz of your cell (depending on type of quartz) will start absorbing at shorter wavelengths.  You can minimise this to some extent using a matched blank cell.

@AChem it is a double beam instrument, however on this test run I did not use a blank

You should not use air as a blank.

If you had given the reference for the literature spectra, I would be able to say for certain, but it's possible that they didn't use a quartz cuvette. According to this website, quartz starts absorbing below ~200 nm. Looks like it depends on the type of quartz, and brand of cuvette, you've used (from this link):

At wavelengths shorter than $200$ nm, everything absorbs the light, even glass, water or air. Absorption spectrum at these wavelengths must be determined in a vacuum.

@Maurice it didn’t occur to me how high that energy was (I’m used to working in wavenumbers)… but yeah, that makes sense


text:
How do I denote a specific rovibrational transition?
I am using a linelist from HITRAN, which gives me a whole lot of extra data and I want to denote the transitions accurately. But I don't know what to use for rovibrational transitions. The following parameters are given (separated by semicolons):
E'; u(E'); E"; u(E"); v1'; v2'; v3'; J'; Ka'; Kc'; s'; v1"; v2"; v3"; J"; Ka"; Kc"; s"; t'; t".
Now, using these, how do I denote a specific rovibrational transition?

Just to make it clearer, you should use the maths mode for the symbols (‘v’ is not a symbol that gets used, but ‘$\nu$’ and ‘$\upsilon$’ are).

Sorry for that, but this is how the .txt file shows all transitions. So I stuck to their method for this. I just need to know which notation I have to use for such a transition.

Oh... that's inconvenient... I'll take a look and see what I can do. What type of molecules are you looking at? (I think the symbols depend on the type of rotor it is, to a certain extent)

Very sorry for this late response, I didn't get notified. But the molecule in question is H2O! And if you can, do you have any links/references to further explanations on the notations?


text:
How does steam distillation work to extract a substance from a mixture, below its boiling point?
This is a small confusion I have regarding steam distillation. Let us consider a sample of water mixed with an organic compound that is insoluble in water. Let us also consider that the boiling point of this organic compound is at 70 °C at atmospheric pressure and the boiling point of water is 100 °C at atmospheric pressure.
According to what I have learned and what I have researched online, it says the two compounds in the mixture evaporate separately, independent from each other. So, the total pressure of the system at a given time would be the vapor pressure of water and the vapor pressure of organic compound combined.
My confusion is that the articles online says that this combined vapor pressure is higher than the vapor pressure when only one compound is available, so, the mixture reach atmospheric pressure faster so, it boils at lower temperature than 70 °C.
How can this be possible if these two compounds (water and the organic compound) are insoluble in one another, and if these two compounds evaporate independently from each other? I cannot understand it because I have also learned that when the environmental pressure increases it also increases the boiling point of that substance. Since these two compounds are not soluble in each other and evaporate independently, should not the vapor pressure of one compound increase pressure that affects the other compounds boiling point and evaporation rather than the vapor pressures of both these compounds affecting the boiling point and evaporation of whole system of these compounds combined (Because these compounds are separate and does not have intermolecular interactions like a binary mixture). What is the mistake I made in my thought process?
Edit: For example, take benzene and water mixture and their boiling points. Pure water is at 100 °C, benzene at 80 °C, benzene + water at 65 °C. The reason suggested is that, at 65 °C, the vapor pressures are 0.23 atm for water, and 0.77 atm for benzene. The total pressure over this mixture at 65 °C is 0.23 atm + 0.77 atm = 1.00 atm. So any mixture water + benzene boils at 65 °C.
Why does this combined vapor pressure reaching the atmospheric pressure lower the boiling point of both liquids, while these two liquids are in equilibrium with their vapor independent from each other? What kind of interaction does these two liquids has that causes this?

It is rather water distillation with organic vapor than steam distillation of the organic. As it is supposed the organic substance has (much) higher boiling point than water, typically essential oils from plants.

For 2 separate liquids, the partial pressure of each is the same regardless of presence of the other. And the total pressure is sum of partial pressures. In reality, there is deviation due limited mutual solubility.

Your problem looks like the mixture water + benzene, which are immiscible. At atmospheric pressure, water boils at $100$°C, and benzene boils at $80$°C. Whatever its proportions, a mixture  water + benzene boils at temperature lower than $80$°C, probably at $65$°C. Why ? The published tables mention that, at $65$°C, the vapor pressures are $0.23$ atm for water, and $0.77$ atm for benzene. The total pressure over this mixture at $65$°C is $0.23$ + $0.77$ = $1.00$ atm. So any mixture water + benzene boils at $65$°C.

@Maurice yes, What my confusion is why both of these liquids boil at a lower temperature? What interaction does these two liquids have that causes this.

@Mithoron No, it does not explain why the mixture boils at a lower boiling point.

I do not think it should be closed as a duplicate, if the OP does not recognize answers in the referred post as answering  his question. If there were provided no answers here, the confused user would be left stranded without the answer. Similarly for any future user confused in similar way.

@Poutnik I read that question too. It was quite different from mine, but after reading answers there and here. I think I properly understand what really happens in steam distillation.

With comparable boiling points like for water and benzene mixture at constant pressure, neither of liquids boils when sum of partial pressures of their vapors overcomes air pressure. They are just quickly evaporating, as this evaporation is pushing air and own vapors away, not being controlled by diffusion nor convection like ordinary evaporation.
In a way, it is "boiling without boiling", or "boiling from liquid surfaces" what is not real boiling as forming bubbles inside the liquid.
If the mixture is overheated above its pseudo-boiling point, the summary vapor pressure gets higher than environmental pressure and evaporation significantly speeds up, cooling the mixture back to the pseudo-boiling temperature. similarly as if a single phase liquid is overheated and starts sudden violent boiling, cooling itself down.
Such a distillation is usually arranged in the way the vapor of more volatile liquid (usually water, but not in this case) is brought to the heated mixture of liquids, stripping vapor of the other liquid and taking it away to the condenser. The vapor input and mixture heating is being balanced so the net volume change rate of the mixture is about zero. This arrangement covers evaporation loses of water (or more volatile liquid) during distillation, kinetically helps releasing vapors to gaseous phase and prevents mixture overheating, leading potentially to sudden true violent boiling.
One could have two separate liquid reservoirs in the common, heated open system with restricted outlet. Each of liquids would have vapour tension lower than sum of their vapor tensions, so neither of them would boil, but just evaporate. But there would be steady flow of vapors from the outlet, like if they were boiling.
With raising difference in boiling points, the reality converges to boiling of more volatile liquid. As the less volatile liquid contribution to the summary vapor pressure is minimal. The system temperature then frequently randomly change between the temperature for summary vapor pressure being equal to air pressure and the more volatile liquid boiling point. That depends of fine balancing of the evaporation rate and heat transfer rate.

Consider the analogy: Imagine an open system with sensors for partial pressure of nitrogen and oxygen. Sources of these compressed gases would provide respective gas until partial pressures reach 0.21 atm for oxygen and 0.8 atm for nitrogen. If only one of gas sources were active nothing much would happen. But with both sources active, there would be steady flow of gas from the system.
The 2 non miscible liquid system effectively works like if there was a pump, pumping away vapor if total vapor pressure is higher than 1 atm, so the space above liquid is never fully saturated by vapors.
Imagine that as some temperature, the saturated vapor pressures of benzene and water are $\pu{0.8 atm}$, resp. $\pu{0.45 atm}$, in total $\pu{1.25 atm}$. But the pressure balancing with air keeps it at $\pu{1 atm}$, so the relative vapor saturation is just 80%. With quite high absolute vapor tensions, evaporation is fast.

Does this mean the vapor pressure of one component in the mixture is not trying to add extra pressure to other component, but rather reduces the net applied atmospheric pressure to the other component, so both of these liquids can boil (or fast evaporate) at a lower temperature? Am I correct?

No, they rather cooperate by summing their lower than ambient partial pressures, like any ordinary gases, so the summary vapor pressure equals or get higher than external pressure.

See also the answer update.

That is a great analogy and explanation. I understand it properly now. thank you.

At a given temperature isolated, pure substances have an intrinsic vapor pressure. When 2 or more pure substances are mixed that do not dissolve in each other [ie. each is "isolated"] each pure substance expresses its own vapor pressure. Therefore, in a closed container the total pressure is the sum of the vapor pressure of each substance. If the container is enclosed by a movable piston [in a distillation system the movable piston is the opposing atmosphere] and energy is supplied to the liquid the evaporating liquids will push back the atmosphere and there will be mass transfer [aka distillation]. The energy can be supplied by direct heating or by the introduction of steam. This technique was probably developed by natural product chemists why back when after they discovered that direct heat could burn their material but steam maintained a reasonably constant, safe, temperature high enough distill many volatile, but insoluble in water, substances.
This behavior is typical of gases and is expressed in the Ideal Gas Law that states the thermodynamic properties of gases depend on the moles present but are [reasonably] independent of their identities. Of course, there are real life complications because nothing is entirely insoluble in anything, and gases do interact. When it works steam distillation gives marvelous results. Using direct heating it is a fast way to remove water from a reaction by azeotropic distillation. It is also important in the kitchen! All those wonderful aromas that can emit from cooking when water is involved are from steam distillation of volatile chemicals. If you check under the lid after cooking meat with water there is a fine even layer of steam distilled more volatile lipids.
You are right in thinking everything can be more complicated. It just requires much deeper thought and research.

Does this mean the vapor pressure of the organic compound also acts as a vapor that was caused by the water.

No, at least not as a primary cause. A substance has a vapor pressure as a function of temperature and to a lesser extent the pressure of any inert gas present. That water vapor is an inert gas it does have an effect on the vapor pressure of other materials present. The main use of steam distillation is to effect a safe, flameless, heat source at a reasonable temperature and allowing the condensed water to entrain the small amount of volatile material allowing efficient collection.


text:
What is brown/red precipitate in battery acid
I am working on recovering some lead acid batteries that have lost some of their capacity. I drained all of the sulfuric acid out and there was an awful lot of brownish red precipitate in it. My understanding was the lead sulfate was white crystals, and so I'm curious what this contaminant was?

Lead oxides have various colors.

$\ce{PbO}$ is red or yellow.

$\ce{Pb3O4}$ is orange or red.

$\ce{PbO2}$ is brown or black.

Likely you have a mix of all of them.
BTW, this is not the safest project -- whole communities have been poisoned recycling lead-acid batteries.

You did something responsible with the acid (which is usually around 35% if memory serves) such as neutralize it to pH 7, right?
The best (and cheapest) way to neutralize sulfuric acid is with ash from a wood fire.  Wood fire ash is mostly potassium, calcium, magnesium, manganese, sodium, phosphorous, and carbon.  So combined with sulfuric acid you get sulfates of all those elements, such as magnesium sulfate and potassium sulfate.  All of which are essentially fertilizers.  But whatever you do, please make sure your H2SO4 is properly neutralized when you dispose of it.


text:
what will happen if we add lithium cobalt oxide into water?is it true that molecule will break and lithium will make LiOH and CoO2?
I have tried to dissolve powder coated on cathode of lithium ion batteries.

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

There is no molecule of lithium cobalt oxide.

There is no molecule of lithium cobalt oxide. There are negatively charged layers of $\ce{CoO2}$ intercalated with positive lithium ions.
$\ce{LiCoO2}$ and $\ce{CoO2}$ are extreme and unstable states of the general composition $\ce{Li_xCoO2}$. $\ce{LiCoO2}$ tends to form separate $\ce{Li2O}$ and $\ce{Co2O3}$. $\ce{CoO2}$ tends to release $\ce{O2}$, forming $\ce{Co2O3}$. These states are avoided by the cell under-/overcharge control.
$\ce{Li_xCoO2}$ can be written as $\ce{Li_xCo^{III}_xCo^{IV}_{(1-x)}O2}$.
Reaction with water gradually leads to respective hydroxid / hydrated oxide and oxygen.
$$\ce{4 Li_xCo^{III}_xCo^{IV}_{(1-x)}O2 + $(2+2x)$ H2O -> 4x LiOH + 4 CoO(OH) + $(1-x)$ O2}$$

Related discussion: https://chemistry.stackexchange.com/a/134739/17368


text:
Differences in UV/vis absorbance spectrum of tap water and distilled water
In my lecture, I was told that the absorbance of a substance can be measured by a colorimeter. By connecting the device to a computer, a graph showing the wavelengths and absorbance values can be shown. I suppose the graph is called a "spectrum".
My lecturer used deionized water in the experiment for calibration.
Then I wonder, if there are any differences (even subtle ones) in the spectrum of tap water and distilled/deionized water?
I understand that the result does not cause major effects to the measurement. What are those effects?

There is possible to not selectively determine nitrate context in water by water UV absorption near 220 nm.

It depends on the tap water. For example, if there are minerals that start to precipitate during water treatment, it might increase turbidity and scattering. A community near me has very high concentrations of Fe and Mn in the tap water, giving it a muddy brown-black appearance. (It's supposedly safe for drinking, but not appealing.)

This is a bit of a broad question, since it can be rephrased as "what compounds might be found in tap water that can be detected by direct measurement by UV spectroscopy".
Obviously one answer cannot address all sources of tap water, and the question does not specify a level of purity. At least one recent study (1) suggests UV measurement as a method of detecting some  contaminants of concern:

Contaminants in water were studied using ultraviolet absorption with light emitting diode and
deuterium lamp sources, and a thresholding detector. The absorption spectra of potassium hydrogen
pthalate, clothianidin, tryptophan, thiamethoxam, uric acid and metaldehyde were obtained in the
range 200–360 nm. Only metaldehyde was not suitable for detection in this range. For the other
contaminants, and mixtures of pairs of compounds, the transmitted signal could be approximately
described with a simple spectral model of the source–absorption–detector system. Combined
measurements at two wavelengths could allow relative concentrations in certain mixtures to be
determined, and real-time absorption measurements were demonstrated in a flume.

Regarding what effects these particular compounds had on the UV spectrum, obviously detection requires that there is a difference between pure (e.g. distilled) water and the contaminated one. From the results:

UV absorption spectra were measured for a range of contaminants in water (concentration 10 mg/L, optical path 10 mm in silica glass cuvettes). These are shown in Fig. 1 for six substances relevant to water quality. All are transparent in the soft UV and visible range. Clothianidin and thiamethoxam are neonicotinoid pesticides. Each has a pair of broad absorption features below 280 nm. Tryptophan is an amino acid whose UV fluorescence is similar to that of organic matter such as sewage and farm slurry 31 . This shows a broad peak around 280 nm (absorption here is also associated with fluorescence at 340 nm) and a sharper peak at 220 nm. Potassium hydrogen pthalate (KHP) is used as a calibration standard for TOC measurements, and shows a steady increase of UV absorption below 250 nm. Metaldehyde, a pesticide for slugs and snails, absorbs rather weakly across the UV range studied and we found that it was not possible to reliably detect this compound with UV LEDs of wavelength 250 nm and above. Uric acid is present in untreated sewage but is not normally found in unpolluted waters 32 . It shows broad peaks at around 235 and 290 nm, and a peak or edge at 200 nm.

Spangenberg, M., Bryant, J.I., Gibson, S.J. et al. Ultraviolet absorption of contaminants in water. Sci Rep 11, 3682 (2021). https://doi.org/10.1038/s41598-021-83322-w


text:
Distillation thermometer position
I am aware that in theory the position of the thermometer needs to be right at the transition point of the vapor into the condenser. In practice,I don´t reach the exact boiling temp. at this height but rather 2-3 °C less. When I put it down further it goes up to the correct one.
In some distillation setups ther is quite a huge distance between liquid and condenser (20 cm) and the vapor condenses already on the thermometer.
What does speak against a lower position? It is still going to read the temperature of the currently boiling fraction? But we were all taught that one must not do this.

The temperature drop is obviously due high heat loses on the way up. It is usually minimized by proper thermal isolation of the upward path. The favorite way in student labs is wrapping the glass by $\ce{Al}$ foil (IR reflection), cotton wool or PU foam as thermal isolation and another foil to wrap it, all tied by a thin rope or adhesive tape.
But, you are not really interested in the temperature near boiling liquid. You are interested in temperature of vapor before leaving to the condenser.
If e.g. water in the vessel is already boiling, but the thermometer shows just $\pu{90 ^\circ C}$, it means vapor is condensing on its well cooled way to the thermometer. At this point, the ascending path is acting like a trivial distillation column and you are measuring the temperature at the "column head", which has not been reached by the vapor fully yet, and part of the gas is still air (or more volatile liquid).

The liquid has temperature of its boiling point.
The gas at a given point has initially temperature gained by convection of moist air. At places with the local temperature below the local dew point, vapor condensation occurs and the local composition of the gas reflects the local liquid saturated vapor pressure at given temperature.
At surface colder than the boiling point, there is ongoing vapor condensation and surface warming until local vapor partial pressure meets the vapor pressure at surface temperature. Air or more volatile liquid vapor is present to top the total pressure to be equal to external pressure.
If there is too little of more volatile liquid, boiling does not keep pace with the heating and the temperature of liquid starts to raise to keep the needed partial pressure of vapor. Temperature of liquid will progressively raise, as it boiling point progressively raise, while temperature in the head stagnates, until liquid reaches the new boiling point. Soon after, the head temperature steeply raise toward the new boiling point.
Whatever you know about the column distillation, is applicable on the simple distillation as well, considering it is a very, very poor distillation column.

So without insulation the correct temperature is at a lower point. My question now is how does the liquid differ from the vapor and why. Shouldn´t it be 100 °C in both? and if there was a higher boiling second phase, the temp. still would only reach 100 °C until the water is gone?


text:
Changes of colour of sunset yellow in water
I am conducting an experiment on “sunset yellow (FCF)”.
Based on my observations, the substance, when in powder form, is bright red in colour (observed with naked eyes). When I dissolve the powder in distilled water, the solution turns dark red. And the increase in colour intensity is significant.
I am not sure about the reason behind this phenomenon. It is said that FCF can act as a pH indicator through showing different colours. I shall point out that the “water” is distilled and thus should be neutral in nature. My guess is that pH is not a factor leading to this.

The solid is probably more concentrated than the solution, and then dilute solutions will often appear different in colour to concentrated ones. How we perceive colour is rather complicated which is why spectrophotometers (colorimeters) are normally used for this.

Pictures might have helped.
The sunset yellow FCF molecule is actually a disodium disulfonate with an additional phenolic hydroxyl group and a diazo function, with overall formula $\ce{C_{16}H_{10}N2Na2O7S2}$. In neutral water solution we would not expect the sulfonic acid groups to be protonated or the phenolic function to be deprotonated, but their oxygen atoms may hydrogen-bond to the protic hydrogens of the water. Such hydrogen bonding may affect the light absorption from the extended conjugated systems to which all the oxygen atoms belong.

Sorry that I don’t have a photo with me. But that’s a nice suggestion for me to propose better questions in the future. Thanks for your detailed explanation. :)

(+1) Especially for suggesting that “Pictures might have helped.” Practically everyone has a camera in their cellphone and there is no current legal requirement that said cameras can only be used to take “selfies”! ;-)

Additionally, in the solid phase, there may be intramolecular interactions that change the absorption. Packing effects may also twist the molecule, leading to differences in the $\pi$-system.

Powders are often paler than large crystals of the same substance, or of the substance in solution, because of diffraction, interference and quantum interactions of light with small particles.
Some other examples:

$\ce{CuSO4·5H2O}$, is pale blue as a powder but deep blue as larger crystals. Start with the powder, grow crystals, and crush them to see that change.

Colloidal gold makes glass red ("cranberry" glass).

That makes a lot of sense. Thanks!


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Why does Raman Activity Require Anisotropic Polarizability?
Normal mode of a molecule needs to involve the change in molecular polarizability (anisotropic polarizability) to be Raman active.
Explanation is provided in Physical Chemistry textbook by Atkins on the example of the rotational Raman spectra. Only the frequency of the electric field, $f_i$ is occurs in the induced dipole formula if the polarizability is isotropic. If it's anisotropic, two additional frequencies occur, $f_i - 2f_R$ and $f_i + 2f_R$ corresponding to Raman shift (Stokes and anti-Stokes lines) where $f_R$ is the rotational frequency of the molecule. In another words, induced dipole of the molecule has two additional frequencies of oscillation. This explanation is clear, but it's quite math based without much intuition.
Can you give more intuitive explanation of the relation between anisotropic polarizability and Raman activity?

The Raman effect involves scattering of a photon with a change in energy due to vibrations or rotations of the molecule. It is changes in the polarisability ellipsoid (either size, shape or orientation) that causes the effect. You can think of polarisability as the shape of the electron distribution in a molecule, we usually imagine this as an ellipsoid. In the case of a linear molecule such as $H_2$ the ellipsoid is prolate (cigar-shaped, rather than plate-shaped which is oblate) and so when it rotates end to end the polarisability changes in orientation, however, the same shape is presented to the electric field of the radiation twice in one rotation, this means that the $J$ quantum number changes by 2 in a rotational transition. In a classical model, the electric field of the radiation and the rotating polarisability can interact  and then energy can be taken from the radiation or given to it in the Raman transition.  Rotation about the bond axis shows no change in the shape of the polarisability and so there is Rayleigh scattering only and at the same frequency as the radiation. If the molecule is 'cold' and $J=0$ is the only level populated only one rotational frequency can be measured ($J=0\to 2$) even though two lines appear in the spectrum, Stokes and anti-Stokes. (In a real spectrum many rotational levels are populated as rotational quanta are usually small wrt thermal energy so many rotational levels may be populated and the spectrum has many lines.)
Added after comment:
Suppose that the radiation moves along x direction with polarisation in the z direction and polarises the molecule forming an ellipsoid pointing in some direction $x',y',z'$ and the ellipsoid has axis $\alpha_s$ and $\alpha_L$ for short and long axes. The long axis makes the angle $\theta$ to the radiation polarisation or $z$ axis. This angle changes as the molecule rotates. If the components of the polarisation are worked out on the z  and x axis they are
$$M_z=(\alpha_s+\alpha_L)E_z/2+(\alpha_s-\alpha_L)\cos( 2\pi 2v_{rot}tE_z )/2$$
$$M_x=(\alpha_s-\alpha_L)\sin( 2\pi 2v_{rot}tE_z )/2$$
what matters here is the term $(\alpha_s-\alpha_L)$ which is zero if the ellipsoid is spherical, i.e. the polarisation is not zero if the ellipsoid is prolate or oblate. The term $(\alpha_s+\alpha_L)$ causes Rayleigh scattering due to rotational motion.
(In passing vibrational Raman spectra can also be observed. In this case it is $d\alpha/dr$ that matters so a change in size of the polarisability is all that is needed and thus all diatomic molecules have a vibrational Raman spectrum.)

''In a classical model, the electric field of the radiation and the rotating polarisability can interact and then energy can be taken from the radiation or given to it in the Raman transition.'' That sentence is the point of my question. Why is it that such an non-elastic interaction only happens when the polarizability is anisotropic? I don't really see the reason for this, it might be that this can't be explained classically, and quantum mechanics can only provide the explanation.

I have added some details to my answer


text:
Cleaning Cobalt with Dodecylbenzene Sulphonic Acid
Here in Brazil we use a special detergent to clean up metals, based on Dodecylbenzene Sulphonic Acid, it is very effective at shining up all sorts of different metals.
I recently acquired 2 samples of 99.99% pure metallic cobalt for my element collection...
They were covered in some sort of weird blackish-grey-green-brown residue...
I then dumped the 2 pieces of Co in a small container with this special dish detergent... to my surprise, this blackish residue immediatelly started to come off. After some time, I was left with very shiny pieces of Cobalt.
Did some reaction actually take place here? What reaction? Thanks in advance.
As I understand, Cobalt is meant to be rather resistant to acids/corrosion in general.

The detergent is acid, probably dissolving cobalt oxides.

Funny. It seemed like some sort of residue or dirt coming off of the metal.

"based on"? Typical detergents are salts of alkylbenzenesulphonic acids. I think that's the kind of being based on. Acid itself would be highly unlikely and I wouldn't call it a detergent.

On the label of the detergent, the active ingredients listed are "dodecylbenzene sulfonic acid and glycolic acid".

These detergents are sold here as "Brilha Alumínio" - or somethink like "Aluminum Shine" in english.
I've been able to clean up several types of metals with it... Like Copper, in which the result is instantaneous, Bismuth, Aluminum and of course, Cobalt.


text:
Chemistry of Rutile, CdCl2 and CdI2
During lectures, we were told that 3d transition metal fluorides generally prefer a rutile structure. Chlorides prefer a $\ce{CdCl2}$ structure where $\ce{Cl-}$ adopt a CCP lattice and the metal cations occupy octahedral voids within alternate sheets. Bromides and iodides prefer a $\ce{CdI2}$ structure which is the same structure as $\ce{CdCl2}$ but iodide anions form a HCP lattice.
The difference between fluorides and the other halides is rationalised easily by the lack of covalency/directionality within the bonds. However, it is not immediately obvious why larger halides prefer to adopt a HCP lattice while chlorides prefer a CCP lattice.
My guess is that from a pure structure perspective, the metal-metal bond distance on average decreases from rutile > cadmium chloride > cadmium iodide. Comparing rutile and the cadmium salts, each octahedra in rutile is edge sharing with 2 other octahedra but corner sharing with 8 other octahedra while for the cadmium salts, each octahedra is edge sharing with 6 octahedra. The more edge sharing octahedra for the cadmium salts implies the metal-metal bond distances are short which can only be stabilised by covalency within ionic bonding. The ability to stabilise such structures and the directionality of covalent interactions makes is stable for 3d chlorides to iodides to adopt the structure of the cadmium salts. Between cadmium chloride and iodide, the metal ions are stacked on top of each other for the iodide salt resulting in shorter distances. The large size of an iodide salt allows for this structure to happen.
Above is my best guess at what is happening. The problem with the argument is that it does not explain why the cadmium iodide structure is more stable than the cadmium chloride structure for larger halides. Additionally, it does not appear that this small repulsion would have much effect on the relative stability between a CCP and HCP arrangement of anions. From an entropic point of view, there does not appear to be much of a difference in entropy between the two arrangements. I am very confused by this trend in nature and would like some help clearing up this confusion.

Often it is hard to figure out why some elements are FCC and some others are HCP (or, stranger, DHCP) - the stacking is a little different that is all. But that stacking makes the difference...

https://chemistry.stackexchange.com/questions/178137/chemistry-of-rutile-$\chem{CdCl_2}$-and-$\chem{CdI_2}$

https://chemistry.stackexchange.com/questions/178137/chemistry-of-rutile-cdcl2-and-cdi2

Dear @JonCuster, You have to edit my comment which I made on this website on Jan 15 at 7:05 AM from its current state itself to chemistry.stackexchange.com/questions/178137/chemistry_of_rutile_$\ce{CdCl_2}$−and−
$\ce{CdI_2}$ prior to deleting the recent comments which I am now posting you as that is urgent

@user142960 - I don’t have to do anything, much less make some unclear edit to a question I did not ask. Sorry, not my problem.

https://chemistry.stackexchange.com/questions/178137/chemistry-of-rutile-cdcl2-and-cdi2

@JonCuster-In such a case, you have to persuade OscarLanzi to make all those changes here, as the listed one is only his problem, prior to deleting the comment link which I made 9-10 mins ago in the area given here.

@user142981 (and other unregistered variants) All I see is you linking back to the question. No idea what you want changed or why, or what for. I’ll bow out now.

https://chemistry.stackexchange.com/questions/178137/chemistry-of-rutile-cdcl2-and-cdi2

https://chemistry.stackexchange.com/questions/178137/chemistry-of-rutile-cdcl2-and-cdi2#autocomment381885

@JonCuster-You might see me to link back to the question, as I want you to get https://chemistry.stackexchange.com/questions/178137/chemistry-of-rutile-$\chem{CdCl_2}$-and-$\chem{CdI_2}$, changed into chemistry.stackexchange.com/questions/178137/$\ce{CdCl_2}$-and-$\ce{CdI_2}$ and Oscar Lanzi will surely be ready to pursue it.

On completion, you shall delete these further comments


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On the geometry of quadra-coordinated coordination complexes
How do we tell whether a complex with a four coordinated central metal ion takes on a square planar geometry or a tetrahedral geometry?

Rough rule of thumb: if it's not d8 it's tetrahedral. If it's d8 and not nickel it's square planar. If it is nickel you either look it up, do the experiment or run the calculation. All assuming no steric constraints.

@Ian Bush. What do you mean with : If it is nickel, run the calculation !

@Maurice stick it in your favourite quantum chemistry package and run a geometry optimisation in both symmetries at the best level of theory you can afford. Select the one with the lowest energy.

I asked my teacher about it today. He said that for the first transition series elements, d⁹ and d¹⁰ configurations always form tetrahedral complexes, any other configuration is square planar under the influence of a strong field and tetrahedral under the influence of a weak field. For second and third transition series elements, all quadra coordinated complexes will be square planar


text:
Role of MnSO4 in permanganometric titration
This might sound like a stupid question, but I have been unable to wrap my head around it.
Recently in a lab I had to titrate H2C2O4 with KMnO4 in acidic medium (H2SO4). However, in the work, it requested for a few drops of a diluted MnSO4 solution. What role does it play in the reaction? It appears as one of the products of the reaction, but doesn't seem to play an active role in the reaction itself.

Both the oxidating ion $\ce{MnO4-}$ and reducing ion $\ce{C2O4^2-}$ are anions and therefore electrostatically repulsing each other, what makes their reaction slow.
$\ce{Mn^2+}$ ions serve as catalyst, being cycled between $\ce{Mn(II)}$ and $\ce{Mn(III)}$ oxidation states. As manganese ions have the opposite charge that both reagents, it is easy to understand it must speed up the reaction a lot.
Note that the formal standard reductions potential of $\ce{Mn^3+}$ is comparable with the one for $\ce{MnO4-}$.
\begin{align}
\ce{4 Mn^2+ + MnO4- + 8 H+  &-> 5 Mn^3+ + 4 H2O}\\
\ce{2 Mn^3+ + C2O4^2-  &-> 2 Mn^2+ + 2 CO2}
\end{align}
As the parallel mechanism is also possible:
\begin{align}
\ce{3 Mn^2+ + 2 MnO4- + 2 H2O &-> 5 MnO2(s) + 4 H+}\\
\ce{MnO2(s) + C2O4^2- + 4 H+  &-> Mn^2+ + 2 CO2 + 2 H2O}
\end{align}
See also the Pourbaix diagram for manganese from wikipedia:

Good explanation but in acidic solution the parallel mechanism is doubtful

I would not be so sure.


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Electroless plating
One of the simple methods to plate metal layers onto non-conductive surfaces is to attach graphite onto the surface. And this is what we call electroless plating.
Here, I take the electroless plating of plastics as an example.
The bonding between graphite and the plastics is predominately van der Waals’ forces. I have no ideas of how graphite and plastics can be struck together right before electroless plating, and how metal can be plated onto the surface of “graphite-plastic surface”. I was never taught that a bond can be formed between graphite and metal, or between plastic and metal.
Can anyone give me some ideas?

There is no significant bond to speak of. The structural integrity is provided by the metal shell itself, once it is thick enough.

Electroless plating is not limited to graphite usage and does not require surface conductivity. Electroless plating is chemical deposition of metals. There is requirement for metal adhesivity, but not on surface conductivity.

Do you mind providing me with more information about metal adhesivity? (Although I am more interested in the method in which graphite is used, frankly)

I have no handy resources about metal adhesivity but the famous low adhesivity of silver om nonmetalic, resp. plastic surfaces and the book "Surely you are joking, Mr. Feynman". About the graphite, I think it is not related to electroless plating, but to electrolytic plating of non-conducting surfaces.

You're correct, there is very little adhesion between graphite and most other surfaces -- after all, graphite can be removed from paper by rubbing with an eraser.
As mentioned in comments, a continuous metallic shell enclosing an article helps prevent peeling. Once that integrity is violated, the plated shell readily peels off.
Some alternatives provide more durable metalization:

Evaporating metal onto glass or plastic, such as aluminum
Etching with chromic acid-sulfuric acid, then dipping in tin(II) chloride ($\ce{SnCl2}$) and palladium chloride ($\ce{PdCl2}$) solutions to make an adherent conductive layer


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R/S designation
Why is my answer wrong ? (Tutors answer in blue bubble)

OH is highest priority, the carbon groups are second highest priority, H is lowest priority

So you need to distinguish between the carbon groups.
Let direction A be through the central bond and direction B be through the terminal phenyl or methyl group for either chiral center. If you look two positions away from the carbon you're evaluating, you see the second hydroxyl oxygen (which I suspect you missed) along direction A but only carbon and hydrogen atoms along direction B, so direction A is #2 behind the direct hydroxyl connection.


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The colour of sunset yellow powder (red, orange, and yellow)
I asked about the colour of sunset yellow in water in the previous post (Changes of colour of sunset yellow in water).
And here, I would like to raise another question based on sunset yellow.
Sunset yellow powder comes in several different colours, including, red, orange and yellow. And as mentioned, mine is red. I am intrigued by the reason behind the colour variations.
I come up with two possibilities (and feel free to point out my mistakes):

Sunset yellow is in crystalline form. And the colour variation is caused by the amount of water of crystallisation in the crystals. This idea is attributed to the fact that anhydrous copper(II) sulphate and hydrated copper(II) sulphate show different colours. And I wonder if this phenomenon can be applied to sunset yellow.

The manufacturers treat the sunset yellow powder with acidic chemicals, since sunset yellow, as a pH indicator, shows red-orange
colour in acidic solution.

Dissolve a sample and measure its pH, maybe?

Another possibility is the fineness of the crystals or powder. See the answer I've added for your previous question, https://chemistry.stackexchange.com/questions/178125/changes-of-colour-of-sunset-yellow-in-water


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Why is benzene more soluble than chlorobenzene in water?
Wikipedia lists chlorobenzene's solubility in water at 0.5 g/L at 20C, and benzene's at 1.79 g/L in 15C and 1.84g/L at 30C.
Chlorobenzene is very slightly polar, shouldn't it be more soluble? What other effects are taking place here?

For that you need to know how interactions between aromatics and water work. For starters why benzene is much better soluble, than for ex. hexane.

Solubility is one of the parameters that have never been perfectly explained and understood. A lot of different theories have been proposed to explain this or that solubility. These theories explain correctly plenty of substances. But not all. Never all. From time to time a new theory is published in the Journal of Chemical Education to explain solubility in term of electronegativity, dipole moments, entropies and other parameters. But there are always plenty of exceptions. If you discover the final solution of this problem, tell it quickly : it will be the next Nobel prize in chemistry.

Determining solubility is more complex than just evaluating interactions of a solute with the solvent at infinite dilution. If that were all that is required - if solubility depended only on enthalpy of solvation, then chlorobenezene would indeed be more soluble. The free energies of solvation are -0.87 kcal/mol for $\ce{C6H6 }$ and -1.12 kcal/mol for $\ce{C6H5Cl}$, from the Minnesota solvation database (Ref 1).
The free energy of solvation describes only the first step in the process of solubilization up to the solubility limit. As more solute is added, second-order effects that include solute-solute interactions and changes to the solvent structure must be considered. See eg Ref 3 on the range of intermolecular interactions possible between chlorobenzene molecules.
Predicting solvation free energy values accurately is not simple either, see for instance Ref 2.
References

Marenich, A. V.; Kelly, C. P.; Thompson, J. D.; Hawkins, G. D.; Chambers, C. C.; Giesen, D. J.; Winget, P.; Cramer, C. J.; Truhlar, D. G. Minnesota Solvation Database – version 2012, University of Minnesota, Minneapolis, 2012.

Sergei F. Vyboishchikov and Alexander A. Voityuk. Solvation Free Energies for Aqueous and Nonaqueous Solutions Computed Using PM7 Atomic Charges. Journal of Chemical Information and Modeling 2021 61 (9), 4544-4553. DOI: 10.1021/acs.jcim.1c00885

Lloyd Muzangwa, Silver Nyambo, Brandon Uhler, Scott A. Reid; On π-stacking, C-H/π, and halogen bonding interactions in halobenzene clusters: Resonant two-photon ionization studies of chlorobenzene. J. Chem. Phys. 14 November 2012; 137 (18): 184307. https://doi.org/10.1063/1.4765102


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Does distance exist in thermodynamic coordinates?
Consider a $P$-$V$ space. The question is a simple one: can we measure distance between two states $(P_1,V_1)$ and $(P_2,V_2)$?

Using simple Euclidian distance $d^2 = (P_2-P_1)^2 + (V_2-V_1^2)$ does not seem to work because the units of pressure are different than those of volume and these, or the squares of these, cannot be added owing to the difference in dimensions/units.
But we could divide the pressure and volume by some standard value and obtain dimensionless quantities.

Now, the pressure and volume, which are to be interpreted without any dimension or unit, could be used to measure distance between two thermodynamic states.
Does this imply anything relevant and is this approach valid? Are there other approaches to measuring distance between two thermodynamic states?

Perhaps see 'Wasserstein metric' and related ideas, ex: https://arxiv.org/abs/1912.08405

I guess you might start by asking yourself what significance you would like to attribute to the distance in question.  If you don't have anything particular in mind then any metric at all will serve, including the ones you propose to construct by choosing pairs of reference pressure and volume.  But then, why bother?  If you do have something in mind, then that should help in searching for something appropriate.

This is an old problem to find a natural metric for a thermodynamic process connecting two states. What you are suggesting does not really work because it is not natural in the sense that any other metric of the form $k_p(p_1-p_2)^2+k_v(V_1-V_2)^2$ would equally be justified.
The answer is not simple. One famous result that is originated in trying to improve on the works of Weinhold [4] is the so-called horse-carrot theorem of Salamon and Berry, for details see [1][2][3]; here I just quote it from the very easily readable [1]:
Theorem 1 (Discrete Horse-Carrot Inequality) In a process of $K$ near equilibrations of a simple system to states along a given quasi-static locus, $K \gg 1$, the entropy production is bounded below by
$$ \Delta S_u \ge \frac{L^2}{2K} \tag{1}$$
Here $L$ is the thermodynamic length of the path
$$L = \int \sqrt{-\sum_{i,j} \frac{\partial^2 S}{\partial X_i\partial X_j}dX_idX_j }\tag{2}$$
and the $X_i$ are a complete set of extensive variables of the thermodynamic system.
A closely related, continuous time version is given by
Theorem 2 (Horse-Carrot Inequality) In a sufficiently slow process
in which a system traverses a given quasi-static locus, the entropy production
is bounded below by
$$\Delta S_u \ge L^2 \frac {\bar\epsilon}{\tau} \tag{3}$$
where $L$ is the thermodynamic length of the quasi-static locus in Eq. (2), $\tau$ is the duration of the process and $\bar \epsilon$ is a mean relaxation time.

[1] P. Salamon: Thermodynamic Geometry, Udine notes, https://salamon.sdsu.edu/
[2] J. Nulton, P. Salamon, B. Andresen, and Qi Anmin: “Quasistatic Processes as Step Equilibrations”, Journal of Chemical Physics, 83, 334-338, 1985.
[3] P. Salamon and R.S. Berry, : “Thermodynamic Length and Dissipated Availability”, Physical Review Letters, 51, 1127-1130, 1983.
[4] https://www.researchgate.net/profile/Frank-Weinhold-2/publication/258879552_Thermodynamics_and_geometry/links/0f31752eff48885858000000/Thermodynamics-and-geometry.pdf


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How to calculate the pH of a mixture of an acid with a buffered solution?
Being a physicist by training, having worked in software development for the last 10 years. Having heard my last chemistry lecture 15 years a go I am a bit lost in regards to how one can calculate the pH of a mixture of an acid with a buffer.
I have acetic acid solution at pH 1.8, to be mixed with phosphate buffer (ph 7) at ratio 1:10. What is the pH of the mixture?
First attempt
I tried to work out the mixture based on concentrations I calculated from the pH.
$$10\cdot 10^{-pH(Mixture)} = 10^{-1.8} + 9\cdot 10^{-7}$$
But I don't think this accounts for the buffer trying to neutralize some acidity of the mixture. So after some thought I remembered that there was this equation with $pK_S$.
Attempt with Henderson Hasselbalch
Remembering the Henderson Hasselbalch equation, I am stuck at how to apply it in this situation.
$$\mathrm{pH}=\mathrm pK_S +\log{\frac{[\ce{A-}]}{[\ce{HA}]}}$$

The system is not sufficiently defined, as there must be known the total phosphate concentration or equivalent data. Otherwise, the final pH is undefined. // Also, pH 1.8 gives by extrapolation mass percentage of acetic acid more then 80%. At such conditions, pH cannot be easily defined, predicted nor measured. Therefore, all the task is rather made up without touch with reality.

Neither of your solutions are well enough defined to be used in calculating the pH of a mixture. You need to define total acetate and total phosphate species. You also need to define how the pH of each solution was adjusted/made.

Hmm, then I have to gather more information. A colleague is receiving a shipment and was trying to anticipate at what ratio they would have to mix with the buffer so that the buffer would stay within its pH range

There is the concept of buffer capacity which indicates how much acid or base can be added so that the pH change is 1 unit. So you'll also need to know how much of a pH change your colleague is willing to tolerate.


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Why does graphite have a higher standard molar entropy than diamond?
It is known that graphite (5.7 J/K mol) has a higher standard molar entropy than diamond (2.4 J/K mol) (data from Physical Chemistry by Atkins). I understand that this must be due to the structural differences between the two allotropes, however I have not been able to find any textbooks or scholarly sources addressing them. A few qualitative possibilities are as follows:

Diamond is a well-ordered 3D lattice, whereas graphite is a series of graphene sheets stacked on top of each other held together by van der Waals forces. Thus, each individual graphene sheet can slide, creating an additional degree of freedom (or, more possibilities) that diamond does not possess.

Graphite's structure includes delocalized electrons, while diamond's bonds just have localized electrons. Thus, the increased possibilities of locations for the delocalized electrons cause graphite to have a higher entropy than diamond.

The bonds in diamond are weaker than the bonds in graphite (in terms of bond order), so diamond's vibrational force constants are less than graphite's vibrational ones. Therefore, diamond's vibrational energy levels that are occupied are few and spaced apart, so there are fewer ways to distribute the atoms compared to graphite.

I would like clarity on which of these contributes to thermodynamic stability the most, ideally quantitatively.

Your 1 is the main factor.

Hmmm, #3 sounds like it's the wrong way around. Weaker bond implies smaller force constant implies smaller frequency and spacing and greater entropy.

Don’t forget the electron degrees of freedom (or bond degrees if you wish). The entropy is defined for high quality crystals, so #1 is out.

The measured heat capacity $C_V$ of graphite at a low temperature, say 100 K is greater than that of diamond. A characteristic measure of the heat capacity is the Debye temperature defined as $\Theta=h\nu/k_B$ where $\nu$ is the maximum phonon frequency and $k_B$ the Boltzmann constant. The value for diamond is $\sim 1900$ K and for Graphite $\sim 410$ K. This means that on average diamond has higher frequency vibrational quanta than graphite and this means that the entropy is greater for graphite than diamond because more energy levels are populated in graphite.  As the entropy is $\int_0^T C_V/T dT$ the area under the curve of heat capacity divided by temperature is greater for graphite than diamond and thus so is the entropy.
The graphs below show the calculated heat capacity and $C_V/T$ for diamond and graphite. You can see that the area under the curve, say, to $300$ K is greater for graphite than diamond and so the entropy is greater.

It's not just greater, it's a lot greater. I'm a little surprised. You'd think the stronger in-plane bonds of graphite would constrain degrees of freedom in that direction exceptionally well.

@Buck Thorn The picture is a bit misleading, when integrated up to $5000$ K the graphite is $\approx 1.7$ times that of diamond, not so different to the ratio in the question. I got $3.4$ J/mol/K for diamond and $5.7$ for the graphite. I guess the vibration frequency in graphite is smaller so more levels populated.


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Is this mechanism of reduction using LiAlH4 followed by hydrolysis of an amide correct?

Hello everyone. I had been given a question on reducing an amide group into an amine group using any reagent I saw fit as homework. I had used LiAlH4, which provides H- and does an acid base reaction with nh-h+, and then uses its pi electrons to kick out carbonyl pi bond back to the oxygen, followed by addition of more H- to form R-ch2-nh followed by hydrolysis which gives r-ch2-nh2. so how correct is this?

related https://chemistry.stackexchange.com/questions/50534/why-does-lialh4-reduce-an-amide-to-an-amine-but-only-reduce-a-ketone-aldehyde-t

How correct? Well, moderately ;) It's very sketchy, but goes in right direction.


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Electrolysing two salts with the same anion
Assume that I prepare solutions of sodium chloride (NaCl) and potassium chloride (KCl) in equal molarity, then combine them together to form a single homogeneous mixture. Then I electrolyse this mixture.
While the reaction at anode would be simple enough:
$$\rm Cl^- \rightarrow Cl + e^-$$
What would be the reaction at the cathode? Would both potassium and sodium ions be converted into their neutral elements (and thence into their hydroxides) in equal amounts or would one of the cations be converted disproportionately more than the other? That is to say, would all potassium be first converted into its hydroxide before sodium and then sodium ions will start electrolysing or will there be an equal (by molarity) concentration of both the hydroxides from the very start?
Many thanks.

None of potassium nor sodium ions are reduced to the element. H3O+  and H2O are reduced to H2 and H2O or OH-. Potassium and sodium ions are just bystanders there.

@Poutnik that's a shortcut version of what really happens. The alkali metal cation is reduced to its elemental form, but being extremely reactive, they react with water to generate OH ions and hydrogen. But it's just splitting hairs, really. The question is, what would happen once the solute is evaporated? If half of the chloride ions have been oxidised to elemental chlorine, would those half stick predominantly with potassium or be distributed equally between potassium and sodium?

No, alkali metal ions are not reduced. such a reduction is the shortcut, simple but misleading version. When evaporated, there would be mixture of NaCl, KCl, NaOH, KOH, NaClO3, KClO3.

@Poutnik ok. The primary question remains.

There is no difference, compared to involving just one of the salts.  The result is mix of results for singular salts.

The more interesting question, with fewer distractions due to what happens to water, would be what happens if you electrolyse a mixture of just sodium and potassium chloride as a molten salt.

In case of molten salts and some solid-state reference electrode, the current as the function of the cathode potential would manifest two steps, probably partially overlapped, resembling common polarographic or voltammetric curves. The first one very probably for sodium ions.

When electrolysing solutions of mixed alkali metal chlorides  $\ce{MCl}$, where $\ce{M=Na or K}$, the result is more or less the linear combination of results with pure salts, with deviation due different salt solubilities.
These processes occur:
On the cathode:
$$\ce{2 H3O+(aq) + 2 e- -> H2(g) + 2 H2O(l)}$$
$$\ce{2 H2O(l) + 2 e- -> H2(g) + 2 OH-(aq)}$$
$\ce{OH-}$ anion then electromigrate to the anode.
The former reaction is dominant in acidic pH, the latter one at alkalic pH.
On the (inert, like glassy carbon) anode:
$$\ce{2 Cl-(aq) -> Cl2(aq,g) + 2 e-}$$
$$\ce{Cl2(aq) + 2 OH-(aq) -> H2O(l) + Cl-(aq) + ClO-(aq)}$$
$$\ce{3 ClO-(aq) -> 2 Cl-(aq) + ClO3-(aq)}$$
The net result:
There are present ions $\ce{M+(aq)}$, $\ce{H3O+(aq)}$,$\ce{Cl-(aq)}$, $\ce{ClO-(aq)}$, $\ce{ClO3-(aq)}$, $\ce{OH-(aq)}$.
During evaporation, $\ce{H3O+(aq)}$ recombines with excess of  $\ce{OH-(aq)}$ and hypochlorite  decomposes to chloride and chlorate.
The solid rest after evaporation consists of $\ce{MCl}$, $\ce{MOH}$ and $\ce{MClO3}$.
Due relatively low solubility of $\ce{KClO3}$, chlorate salts will be enriched by potassium in expense of its context in hydroxide and chloride fraction.


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How can I isolate products that cannot be cleanly distilled or visualised by TLC for column chromatography?
I am a first year PhD student in the UK, and without giving too many details of my work as it is industrially funded and my sponsors are a bit touchy over IP, I am struggling to isolate the products of a series of reactions I have done. The general reaction is as follows:

I have found that yields (by NMR) are best with a significant excess of alcohol compared to the methylene source, so typical conditions are 8.3 mmol of methylene source, 83 mmol of alcohol in 5 mL of 1,4-dioxane. The rub is that my supervisors want me to get isolated yields, which I am struggling to do. Here is a few examples of the kind of products I am talking about.

Here is what I have tried so far:

Simple aqueous workups. This generally gives me a pure fraction of product at the end to get a nice NMR spectrum of, but an awful lot of product is lost in the process. I think the -O-CH2-O- bridge gives the products a certain level of solubility in water, so while I am washing out the alcohol, I am taking a lot of product with it.
Distillations. I've worked out the boiling point gaps between the alcohols and those of the products that are well characterised. Methanol and dimethoxymethane have an azeotrope so can't get it pure, ethanol and diethoxymethane have only a 10 C difference in BP and I can't get pure product with my equipment (have tried a few times, best was like 80% product). IPA and butanol should work, just haven't done it yet. For the larger alcohols, such as the cyclohexanemethanol and some of the ones I haven't included, this probably isn't feasible, however. The presence of 1,4-dioxane (BP 101 C) complicates this, but a lot of these reactions will proceed (if significantly more slowly) neat.
Column chromatography. This is the obvious answer, but the problem I am having is that the products don't seem to be very visible, or at all visible, by TLC. There is just no functionality in a lot of cases for a stain to react with and visualise. Dimethoxymethane (the methanol product) boils at 42 C, so I don't know how I would concentrate the fractions without losing product, assuming the column worked in the first place.

These reactions have been shown in literature. I have found dozens of references for the synthesis of dialkoxymethanes, but the vast majority of examples are of various substituted benzyl alcohols which obviously have a UV chromophore. The few papers that use small alcohols generally only report NMR yield, and those that do column simple alcohols (never methanol, ethanol or IPA, generally just butanol or longer) do not report the visualisation method used aside from 'visualised by UV' which can't be possible as there is no UV chromophore. No clue how they did the column for these, its infuriating.
Paper that shows large scope but gives no visualisation method aside from 'UV'- https://pubs.acs.org/doi/full/10.1021/acs.joc.2c02637
Paper that shows simpler alcohols but no isolated yield - https://chemistry-europe.onlinelibrary.wiley.com/doi/10.1002/cctc.201902332
Can give more examples if you like.
Does anyone have any advice? I do not have a wealth of experience with columns or TLC so please let me know if I am missing something obvious. My supervisors don't have any immediate suggestions and no one in my group can think of anything either.

What TLC stains have you tried?

@Waylander Have only tried KMnO4, although I have looked through a few different sites that describe the best stains to use for specific functional groups (https://www.reachdevices.com/TLC_stains.html for example). It was my understanding that KMnO4 was the best choice, and I can get a faint spot from my crude product mixture that is more than likely product. When it comes to TLCing the fractions from the column, the product is now so dilute that I can't see it in any of the fractions.

2,4-DNP may show it as the dioxymethylene products are masked aldehydes. Also CAM might be worth trying see https://pcwww.liv.ac.uk/~jxiao/links/TLC%20Stains.pdf

I think it is just too difficult to get the dimethoxy methylene product clean. Quote the HMR yield and move onto to reacting other alcohols whose products will be easier to purify. If your industry supervisors want that result they can find a purification method. And yes, I have been an industrial PhD supervisor in the UK

If it's not necessary to isolate a product to go on with successive steps of a synthesis route, then you don't do it. Do you need the isolate? That being said: have you tried to crystallise the stuff in a -20 or -40 °C fridge? That doesn't give you a very high yield (and you need a few tries to work out the best solvent/temp. combination), but likely a good amount of pure substance, reducing the amount that needs workup by destillation/chromatography. Btw. while you're standing close to the fridge: have you tried the aqueous workup with ice cold water?

Remove the lower-boiling compound as an aceotrope with water? (No idea if that's a stupid idea if both form an aceotrope. ?)

Countercurrent liquid–liquid extraction?

@Karl  I think this is a synthetic methods project rather than a target synthesis so the isolated yield of a particular product made by the method is an important parameter.

@Waylander I'll give the 2,4-DNP stain a go, thanks. Dimethoxymethane is produced industrially as a precursor to a series of fuel additives for diesel so purification methods exist, they just aren't feasible on lab scale (big fancy chemical engineering style distillation things) so I don't think its a big deal for me to not isolate it. Thanks for your advice, will update when I get a chance.

@Karl I like the ice-cold aqueous extraction idea, will definitely try that next week. Waylander is right, however, in that this work is more about the scope of products this reaction could possibly make rather than being a step in a longer synthetic route. Crystallisation is a potential answer for some of the bigger products, I will also try this. Thanks for your advice!


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What is the product: Ethyne + (conc)KMnO4?
In cold and alkaline medium it gives ethane-1,2-dial. But what happens in conc KMnO4? Does it give CO2? Also, what will happen to other triple bond containing hydrocarbons?

Further oxidation of oxalic acid will give 2 x CO2. Other hydrocarbon alkynes will give the pair of carboxylic acids corresponding to cleavage across the triple bond


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Mathematics behind the graph between Gibbs energy and extent of reaction
When the graph between Gibbs energy and extent of reaction is plotted, such a curve is obtained:

I am unable to understand how we got to the concavity of this graph (which can also be seen as the monontonic nature of the graph during various segments of the reaction). And that is what I am interested in knowing. I want to see the math worked out.

https://chemistry.stackexchange.com/questions/6827/derivation-of-relationship-between-equilibrium-constant-and-gibbs-free-energy-ch https://chemistry.stackexchange.com/questions/162450/why-can-one-set-the-gibbs-free-energy-change-to-0-when-calculating-the-standard

The math is fairly straightforward if we can start from $$\Delta_r G = \Delta G ^\circ + RT\ln Q$$
Let's use the example of a very simple reaction $\ce{A <=> B}$ with an equilibrium constant of $K=1$ (so $\Delta G^\circ=0$) and assume that the activities of both A and B are equal to their concentrations and our starting concentration of A is 1 M.
In this simple case, $Q=\frac{[\ce{B}]}{[\ce{A}]}=\frac{\xi}{1-\xi}$, which, combined with $\Delta G^\circ =0$, gives us that $$\Delta_r G = RT\ln \frac{\xi}{1-\xi}$$
As noted on the plot you showed, $\Delta_r G$ represents the slope of the plot, so to reproduce the plot, we integrate the right-hand expression and get $$G(\xi)=RT\left(\ln(1-\xi) + \xi\ln\frac{\xi}{1-\xi}\right)$$
$$=RT\left((1-\xi)\ln(1-\xi) + \xi\ln\xi\right)$$
A plot of this function in the range $0 <\xi <1$ gives something resembling your original plot, except symmetric due to the choice of $K=1$.
ADDED:
If part of your question is why $\Delta_r G = \frac{dG}{d\xi}$, we can derive that as well. The total free energy of the system is $G = G_A + G_B$. For the separate components, the molar free energies (ignoring the reference concentration term (1M) for simplicity) are $$\overline{G}_A=\mu_A^\circ + RT\ln[\ce{A}]$$ and $$\overline{G}_B=\mu_B^\circ + RT\ln[\ce{B}]$$.
Since these are molar free energies, we need to multiply each by the relevant amount. Let's assume our system is 1 L, so the concentration and the amount of A are both $1-\xi$ and for B are both $\xi$.
Substituting in, we have $$G(\xi) = G_A + G_B$$
$$= (1-\xi)\overline{G}_A + \xi\overline{G}_B$$
$$= (1-\xi)(\mu_A^\circ + RT\ln(1-\xi)) + \xi(\mu_B^\circ + RT\ln\xi)$$
Recalling that we have set $K=1$ so that $\mu_A^\circ = \mu_B^\circ$, we can do a lot of rearranging and cancelling and get the integration result from above: $$G(\xi)=RT\left((1-\xi)\ln(1-\xi) + \xi\ln\xi\right)$$

It should be $G=\bar{G}_An_A+\bar{G}_Bn_B$

@ChetMiller - as described in the text, the $\xi$ and $1-\xi$ terms represent $n_A$ and $n_B$ respectively. $G_A$ and $G_B$ are total energies rather than molar and so are not multiplied by $n$. That is, $G_A = \overline{G}_An_A$.


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Will increasing the initial concentration of a reactant result in more time for the reaction to go to completion?
I understand that increasing the concentration of a reactant increases its instantaneous reaction rate, but shouldn’t the amount of time for the overall reaction to go to completion increase?
For example, in the 2015 AP Chemistry frq 5, the answer to part (b) is that increasing the concentration of the food colouring increases the time for the reaction mixture to reach an absorbance near zero.
https://www.khanacademy.org/science/chemistry/studying-for-ap-chemistry-exam/2015-ap-chemistry-free-response-questions/v/2015-ap-chemistry-free-response-5

Yes, it will, so what exactly do you want to know?

Reactions never go to 100% completion. A reaction is considered as achieved when the  amounts of reactants and products tend to respect the equilibrium constant of the reaction.

It is somewhat complicated by the fact that you need to be very careful when making general statements in chemistry. There are frequently exceptions. For instance is (1) Zero-order reaction? (2) The other reactant  the limiting reagent? (3) Is the exothermic such that the increase in temperature greatly accelerates the rate of reaction? (3)


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How do you perform an ion exchange of a monovalent cation with Br- counterion to a divalent or trivalent anion?
I tried to react an organic compound I have synthesized containing a di-N substituted imidazole bromide salt with Ag3PO4 with water at room temp to try to replace the Bromide with PO4(3-). I dont think the reaction worked based on elemental analysis.
What is the best way to perform this ion exchange? Im thinking of trying to charge Amberlite IRA400 with sulfuric acid or phosphoric acid and using that resin to exchange the Br- ions. Would that work?
**Edited to add picture of sulfate salt form of the compound:

I think the issue is with your cation - not terribly tolerant of things like silver, perhaps.

If the initial monovalent cation is called $\ce{M^+}$, coupled with an anion $\ce{Br-}$, it may react with $\ce{Ag3PO4}$ in a reaction like $$\ce{3 M+ + 3 Br- + Ag3PO4 -> 3 AgBr + 3 M+ + PO4^{3-}}$$ Depending on the nature of the ion $\ce{M+}$, the final group of ions $\ce{M+ + PO4^{3-}}$ may form an insoluble precipitate $\ce{M3PO4}$. It is not necessary to use an ion exchange resin like Amberlite.


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NMR splitting of equivalent, but separated protons
I’ve been trying to understand why equivalent sets of protons that are coupled behave as equivalent to a single coupled pair. For example in the spectrum for Diethyl ether:
Both groups of Hb are split by both groups of Ha resulting in a quartet.
However shouldn’t the Ha protons be independent of each other resulting in 7 possible unique spin configuration? Why is this not the case?

The two ethyl groups are identical and the methylene protons are enantiotopic, i.e., identical. Thus, n + 1 = 3 where n = 2, a triplet. The spins are up-up, up-down, down-up, down-down. The second and third spins are identical. The intensities of the methyl triplet are 1:2:1.