text:
Making Sodium Citrate
I've been trying to make Sodium Citrate and I've seen online people saying just react sodium bicarbonate with citric acid in water (with correct ratios) and boil off/remove water once the reaction has ceased.
I've also seen online that some people suggest to use sodium carbonate instead.
I thought I would ask here as I assume at some point the bicarbonate would turn to carbonate when reacted/exposed to the heat anyway by releasing hydrogen and carbon dioxide but I'm not a chemistry guy so if anyone knows which is better or which would result in more fool proof way, please let me know!
Thanks!

The problem is that the 3rd acidic  H of citric acid has about the same acidity as CO2(aq), so there is more or less balanced equilibrium  $\ce{HCitr^2- + HCO3- <=> Citr^3- + CO2 + H2O}$. Boiling helps, if done properly, but Na2CO3 ensures the stoichiometry is achieved.

Thanks. I shall use sodium carbonate/Na2CO3 then. Cheers for your help.


text:
Why exactly does copper have a higher melting point than zinc?
It is said that copper has a higher melting point than zinc because of the d electrons in copper being involved in metallic bonding. However, copper has no unpaired electrons in its d subshell just like zinc. How can we explain this observation ?

why was this question closed? just curious

the question seems clear to me, why does copper show a higher melting point than zinc even though both have fully filled d subshells. (not sure about the validity of the premise)

@GauravSaiMaddipati This q. was completely rewritten and might be reopened, but copper does have an unpaired electron.

https://chemistry.stackexchange.com/questions/4766/melting-and-boiling-points-of-transition-elements/4815#4815

@Mithoron Copper does have an unpaired electron, but in its s subshell. The reasoning is given based on the electrons in its d subshell, which makes it confusing.

@wonderingwhy read the link sent by mithoron and your doubt may be clarified.

@GauravSaiMaddipati I did read the link and the other answers. It explains the anomalous behaviours of Cr,Mn and Tc and that is understandable. My doubt is strictly limited to the statement which says that 'the d electrons in copper are involved in metallic bonding due to which it has a higher melting point than zinc'.

Could you give a source for that statement?

There is a multiple choice question going around and that is chosen as the correct answer. For example : https://www.sarthaks.com/2987888/the-melting-point-of-copper-is-higher-than-that-of-zinc-because

source: Stack Exchange

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

sort: votes


text:
Is barium peroxide (BaO2) paramagnetic?
Is barium peroxide paramagnetic?
I couldn't find find a credible source that states if it's paramagnetic or diamagnetic.
I thought it must be cause it does have $\ce{O2^{-2}}$  in it but I'm unsure.
Whichever book I looked into just states its preparation method

It's a peroxide, not superoxide, so of course it's diamagnetic.

In both the $\ce{Ba^{2+}}$ and $\ce{O2^{2-}}$ ions all the electrons are paired. Can you tell from this what the magnetic response is likely to be?
Wikipedia gives a negative magnetic susceptibility, which goes along with what would be expected from the electron pairing.

I'm so sorry , I got it messed up.

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


text:
How feasible is it to separate acetone, butanol, and ethanol (and water) via simple distillation?
I have no experience distilling, though I understand the basics about it. When two liquids with different boiling points are mixed, you can sometimes separate them by boiling away one of the liquids, leaving the other behind. Sometimes multiple iterations of this process can enhance the process. If the important substance is the one that boils first, the gas form can be collected and condensed back to liquid. This is usually done with liquors and other alcohol products, but in principle should be possible with many substances.
I have been reading up on fermentation with Clostridium acetobutylicum bacteria. Some strains produce acetone, butanol, and ethanol in a 3-6-1 ratio. From my reading, it's not entirely certain what yields could be expected, but I would imagine these are low, and that what's left over in a fermentation vat is still mostly water. Plus, I suppose, some sediment and biological contaminants that won't filter out entirely.
Acetone's boiling point 58° according to Wikipedia, far below the boiling point of water, even significantly below the boiling point of ethanol. Can this be distilled out (first) to some reasonably high purity?
Would it then be simply a matter of raising the temperature to next remove the ethanol, and after that the water (both mostly waste products to be discarded responsibly)? Would the butanol be left in the vat afterward at some reasonable level of purity (contaminants aside)?
I am interested in both the acetone and butanol, and would like to be able to make modest quantities of both, but if I had to take one or the other the butanol is the more important of the two.

Here's a good review of some options: https://academic.oup.com/femsle/article/363/9/fnw073/2570161

@Andrew Thanks for the link. Putting this at the top of my reading list.

Though distillation can separate chemicals to some extent, water, alcohols and acetone form azeotropes, making complete separation by distillation impossible.
An azeotrope, or constant heating point mixture, has the same proportion of chemicals in the vapor as in the liquid. For example, when an ethanol/water mixture is ~95.6% alcohol, the ratio of the two in the distillate stays the same. To get purer ethanol, with less water, requires another method, such as use of a desiccant, e.g., anhydrous $\ce{CaO}$.
The Wikipedia article to which you're referring mentions the issue under Improvement attempts. Could you make some relatively impure acetone and butanol, as a demonstration? Certainly! Would it be economically feasible? "ABE fermentation became generally non-profitable, compared to the production... from petroleum." It would cost less to buy them than to ferment and distill. Acetone, for example, can be found for ~US$6/L.
BTW, it's interesting to read how Weizmann further developed that process to help the Allies make acetone during WWII.

The whole point of the exercise would be to not have to buy it. And I don't have any petroleum. Is there any way to know at what percentages this particular mixture maxes out? 95.6% ethanol is useful for several things for instance (though I think it can take several passes through the still before you get it that high?). 95% butanol might serve its purpose, if it could be separated as well as that.

The mix probably has several azeotropes look them up in a handbook. With a decent column acetone and 95 ethanol possibly could be distilled out.  Freeze out water from the butyl compounds.  possibly a sodium sulfate dehydrant and distillation could recover butanol . It sounds like a chemical engineering bad dream if not a nightmare, but I agree profit is not everything. Get a book on organic laboratory techniques and read it! Check out simple and fractional distillation.


text:
Inconsistency in between Nernst Equation and Gibbs Free Energy Equation
Part 1 - Derivation of the Gibbs Free Energy Equation: [copied from this]

Using the fundamental equations for the state function (and its natural variables):
\begin{align}
  \mathrm{d}G &= -S\mathrm{d}T + V\mathrm{d}P\\
  V &= \left(\frac{\partial G}{\partial P}\right)_T\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p
\end{align}
Here $\bar{x}$ represents molar $x$, i.e. $x$ per mole
\begin{align}
  \bar{V} &= \frac{RT}{P}\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1}
\end{align}
Defining standard state as $P = \pu{1 bar}$ and $\bar{G}=\mu$
$$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$
consider the general gaseous reaction $\ce{a A + b B -> c C + d D}$
$$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$
for "unit progress" in reaction. Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$
\begin{align}
\Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\
\Delta G &= \Delta G^\circ + RT\ln Q
\end{align}

Here, $Q=\ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$
Part 2 - Derivation of the Nernst Equation [Copied from this]

Realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$.
$$
$$The fact that $\Delta G = W_\text{non-exp}$ can be derived as under:
\begin{align}
\mathrm{d}S
  &= \frac{\delta q}{T}  && \text{(reversible case)}\\
\mathrm{d}U
  &= \delta q + \delta W_\text{non-exp} + \delta W_\text{exp}\\
\delta W_\text{non-exp}
  &= \mathrm{d}U - \delta W_\text{exp} - \delta q \\
  &= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S
  && \text{(const. $p$ and $T$)} \\
  &= \mathrm{d}H - T\,\mathrm{d}S \\
  &= \mathrm{d}G
\end{align}

Combining the above two, we get Nernst Equation:
$$E_{cell}=E^{\circ}_{cell}-\frac{RT}{nF}lnQ$$

The Q for this remains same i.e. $Q_p=\ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$
But we always use $Q_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$ in the Nernst Equation in terms of molarity instead of $Q_p$ (in terms of partial pressure). How is this justified?
Thnx :)

In the nernst equation by design all gases are described with partial pressures and aqueous ions by concentrations, it's mixed

@GauravSaiMaddipati any proofs for the aqueous ions using concentrations ?!

A substance is its chemical activity. Two factors are involved1. the internal energy of the substance This is contained in the number of molecules and the temperature. The second factor is the entropy or dispersion of the substance . This is best measured in mole fraction in gases, liquids and solids, no contradictions there. In liquids molarity is a reasonable substitute for mole fraction simply substituting a volume or a mass. The stand activity is defined as 1. Molar. For pure solids, liquids and gases the standard is mole fraction of one.  At high concentrations things get complicated.

Im sure there is a proof, but i have asked this question to my prof and this is what he replied, i didn't pursue it further. Im sure someone much more knowledgeable than I could give a better answer hence i just left a comment

Rather than being an inconsistency, it has to do with the freedom of choosing a mathematical expression for the chemical potential that is more convenient to you.
A general expression for the chemical potential is
$$ \mu_j = \mu_j^0 + RT \ln\left(\gamma_j \frac{\xi}{\xi_0}\right) \tag{1} $$
where $\gamma_j$ is the activity coefficient of species $j$, $\xi$ is some measure of the concentration, and $\xi_0$ is that measure of the concentration for a defined standard state.
Across chemistry you will see the next variations of Eq. (1)
\begin{align}
  \mu_j &= \mu_j^{\mathrm{(g)}} + RT\ln\left(\gamma_j^{\mathrm{(g)}}
  \frac{P_j}{P_0}\right) \tag{2} \\
  \mu_j &= \mu_j^{\mathrm{(l)}} + RT\ln\left(\gamma_j^{\mathrm{(l)}}
  x_j\right) \tag{3} \\
  \mu_j &= \mu_j^{\mathrm{(c)}} + RT\ln\left(\gamma_j^{\mathrm{(c)}}
  \frac{c_j}{c_0}\right) \tag{4} \\
  \mu_j &= \mu_j^{\mathrm{(m)}} + RT\ln\left(\gamma_j^{\mathrm{(m)}}
  \frac{m_j}{m_0}\right) \tag{5} \\
\end{align}
Below is a description:

Standard state
Measure of concentration
Activity coefficient

Pure substance in the ideal-gas state at a pressure of $\pu{1 bar}$
Partial pressure
Fugacity coefficient

Hypothetical ideal liquid solution at a pressure of $\pu{1 bar}$
Molar fraction
Mole-fraction activity coefficient

Hypothetical ideal 1-molar solution at a pressure of $\pu{1 bar}$
Molar concentration
Molarity activity coefficient

Hypothetical ideal 1-molal solution at a pressure of $\pu{1 bar}$
Molality
Molality activity coefficient

You can use any of the Eqs. (2)-(5) for any compound to obtain an expression of $\Delta G$. In all of these cases, the activity coefficient is going to be different depending on the expression you use. Of course, the measure of concentration is going to be different also for every species.
The only precaution you must take is that once you select the 'scale' for each compound, you have to be consistent with it. For example, if an equilibrium is attained you can calculate the equilibrium constant $K = \exp(-\Delta G/RT)$. When explaining in words what $\Delta G$ points too, you tell the reactants and products it refers to and the states you have considered for all of them. As you can see, depending on the different scales, you will have different $K$'s.
When someone reads your $K$ and the scale you used for every compound, that person will put correctly the measure of concentration for further calculations. If you just say $K = 50$ with no additional information, I won't know which $\xi$ to choose for every compound.

A discussion of this topic can be found at the start of chapter $9$ of:

J. M. Prausnitz, R. N. Lichtenhaler, E. G. de Azevedo, "Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd ed., Prentice Hall PTR, 1999.

I went through the book you referred and found the Fugacity Equation [Equation $(2)$ in your answer] is derived for Ideal Gases in ($2-37$) and generalised for all systems in ($2-40$) in the book. But I can't find the proofs for Equations $(3)$ to $(5)$ in your answer. Kindly tell me where to find!

@AsmitKarmakar You can't prove those equations, although the first one seems to be one. They are definitions. For example, look in Introduction to Chemical Engineering Thermodynamics (Smith, Van Ness, and Abbott, 7th edition) in Chapter 11.8. It says explicitly that $\mu_j$ for the ideal liquid solution model is defined (in bold). Then, he defines the Eq. (11.90) and arrives at Eq. (11.92) which is  the Eq. (4) in this post.

For Eq. (5) you may see the part named "Activities" in Physical Chemistry (Peter Atkins and Julio de Paula,  9th edition). It says "It proves convenient to write", and then pops out my equation for the ideal case (where $\gamma_j = 1$). Then he defines the activity in Eq. (5.61) (they use brackets [] in the book for definitions), and arrives finally at my Eq. (5).

Since $\mu_{j}$ is the molar Gibbs Free Energy, definitions of $\mu_{j}$ are equivalently definitions of Gibbs Free Energy. But Gibbs Free Energy should be the same everywhere! Or is it not? Is $G=H-TS$ valid only for Gaseous State? Nevertheless, how can there be three alternative definitions of $G$ for solutions? (Equations $(3), (4) & (5)$)

The definition of $G$ is what you have written. We have several ways to write the chemical potential of a species in a solution $\mu_j$. The one you read in books is derived for a solution of ideal gases. Then, you add the fugacity coefficient (arbitrarily) to extend them to real solutions. All the other forms are "copies" of this one, which are more suitable for other states of aggregation.


text:
How would you explain the general trend in melting point between Group 1 and Group 2 chlorides?
This question is based off of the attached chart. I notice how there's a general increase in melting point down group 2 chlorides and a general decrease (except for Lithium) down the group 1 chlorides. So, is the melting points of group 2 chlorides greater generally, or vice versa? Also, how would you explain these trends using electronegativity and intermolecular forces?

Does this answer your question? How do we predict the order of melting points of group 1 and group 2 halides? just check and let us know. If your question is different, please edit it.

Not exactly, I want to know specifically why group 2 halide trend is increasing where group 1 decreases, and if you could say generally which group has higher melting points

First off, we need to give magnesium chloride its due. Wikipedia reports a melting point of 714°C for the anhydrous salt (the lower figures are for hydrates losing water molecules); apparently the 415°C figure was wrongly recopied. Putting the magnedium chloride point at the correct position makes the trends much more consistent with a sharp change in both groups from  Period 1 to Period 2.
If the chlorides were 100% ionic, we should expect melting points to drop off with heavier metals as electrostatically driven lattice energies drop off with larger atoms. But only alkali metal chlorides from sodium onwards follow this model. For the rest, we should expect less than 100% ionic character to favor easier formation of small molecules/clusters in the liquid phase, leading to a lower melting point.
Fajans' Rules are relevant here. They favor more covalent character if the cation in an ionic model is small or multiply charged. "Small" implies that lithium chloride might have significant covalent character where the chlorides with larger alkali metal atoms would not, while "multiply charged" implies the Group 2 chlorides would have more covalent character (and thus would melt more easily with molecules/clusters) overall. Beryllium, with both small size and multiple charge in the ionic model, gives an especially low-melting chloride.
The covalency in the Group 2 case never appears to reach zero but does decrease with larger ions, so heavier Group 2 chlorides tend to melt at higher temperatures; whereas with the singly charged alkali metal ions only the lightest/smallest-atom one deviates from the trend expected from an ionic model.
With lithium chloride melting at a temperature closer to (corrected) magnesium chloride than to any other on the list aside from the much heavier caesium compound, we see yet another example of lithium resembling alkaline earth metals (apart from valency/oxidation state), especially magnesium, more than other alkali metals.
The reader should contrast the Group 2 behavior shown here with the corresponding results for oxides. Oxides, with a more electronegative nonmetal and a more compact anion, are more highly ionic than chlorides, and the decreasing trend we expect to see from the ionic model sets in for that case from magnesium onwards. See here.


text:
What is the enthalpy change for elimination reactions of alkyl halide - exothermic or endothermic?
There are a few questions in thermodynamics (with unverified data) that the addition reaction of ethene to give ethyl chloride has a +ve enthalpy change, thereby elimination to give alkene must be endothermic. The latter seems to be correct as in (ideal) E2 mechanism the TS is known to stabilized by alkene character. This, as per Hammond's postulate lead us to the idea that alkenes must be closer to TS in energy coordinate.
But, there are different sources which claim it otherwise include LG Wade - that it is exothermic

What was expected and given in some other sources like Prof. William Reusch chem.libretexts.org suggests otherwise - which corresponds with Hammond's postulate is the second image.
I understand, it may change with the nature of substrate and the base up to certain extent, but there could be some generalization, or at least the data about the reaction: ethyl chloride giving ethene under E2 conditions.
Could someone please verify the two opposite claims?
Thanks.

I have the reasoning of positive entropy favouring these reactions at high temperature, making dG more negative. Also, most of these energy profiles have been wrt dG and not dH. They do have here the energetics of electrophilic addition of alkenes, which is exothermic (actually they gave it as exoergic, dE may be dG, I guess):
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.11%3A_The_Hammond_Postulate


text:
Why is cyclooctene the smallest cycloalkene to have cis-trans isomerism?
What prevents cis and trans isomers from existing in smaller cycloalkenes, for example, cyclohexenes for more than a short period of time?

Try building a model of trans-cyclohexene, a good lesson to understand ring strain

You might also look at https://pubs.acs.org/doi/10.1021/jo00389a067 which also says trans-cycloheptene has been prepared, but half a half life of ~10 minutes at 274K

https://chemistry.stackexchange.com/questions/30940/relative-stability-of-cis-and-trans-cycloalkenes

https://chemistry.stackexchange.com/questions/78571/why-is-cyclooctateraene-shaped-like-a-tub-rather-than-a-chair cyclooctadiene with two trans bonds was made.

Whether cyclooctene is the lowest such alkene is not cut and dried. trans-cycloheptene is known as a transient species (Inoue et al 1 and Verbeek et al. 2). Calculations indicate that it converts to the more stable cis isomer through a diradical intermediate, not a double-bond rotation (Squillacote et al, 3).
Consider a planar $n$-sided polygon with $n-2$ vertex angles measuring $109.5°$ and the remaining two vertex angles measuring $120°$ and $240°$, the last of these a reflex angle at a concave vertex. Then the sum of angles is both $180°(n-2)$ (from the planar polygon formula) and $109.5°(n-2)+360°$ (from the specified angles), where these expressions closely agree at $n=7$. Thus trans-cycloheptene would be forced to have all ring atoms essentially is one plane, which is highly strained for a seven-atom ring with only one double bond (compare with the observed structure of cis-cycloalkenes). With an eighth atom in the ring, so that $109.5°(n-2)+360°$ is significantly less than the planar-polygon angle sum $180°(n-2)$, the forced planarity is relaxed and the structure is much less strained.
References

Yoshihisa Inoue, Toshihiko Ueoka,   Takayuki Kuroda  and  Tadao Hakushi. "trans-Cycloheptene. Photochemical generation and thermal trans-cis isomerization". J. Chem. Soc., Chem. Commun., 1981, 1031-1033. https://doi.org/10.1039/C39810001031.

J. Verbeek, J. H. Van Lenthe, P. J. J. A. Timmermans, A. Mackor, and P. H. M. Budzelaar. "On the existence of trans-cyclohexene". J. Org. Chem. 1987, 52, 13, 2955–2957.
https://doi.org/10.1021/jo00389a067.

Michael E. Squillacote, James DeFellipis, and Qingning Shu
"How Stable Is trans-Cycloheptene?" J. Am. Chem. Soc. 2005, 127, 45, 15983–15988.
https://doi.org/10.1021/ja055388i


text:
Does molten metal liquid chromatography exist?
Liquid chromatography is used to analyze and purify different chemical samples in a solvent. It is also used for industrial scale chemical purification. I was wondering if something similar exists for liquid (generally molten) metals.
I have not been able to find any examples online, but I may be missing some key vocabulary. It seems there would be some big advantages to purify metal in this way.
Lead for example has impurities of "arsenic, antimony, bismuth, zinc, copper, silver, and gold" per the lead wiki article. It also goes on to state that the impurities are typically removed via pyrometallurgical processes; which appear to be very energy intensive. I imagine liquid chromatography type purification process would work well for other metals with a lower melting point such as gallium, mercury, zinc, tin, and cadmium.
I imagine the solid material that would fulfill the role of the resin could be any material with a high amount of wettable surface area and can handle the elevated temperatures. Potentially could be zeolites, activated carbon, graphene, metal powders, fiberglass? I imagine selecting a material that is properly wetted by the liquid metal would be the trick.
Does this exist? If not, any thoughts on why it might or might not work?

I have to say this is a curious idea, but it appears to be fraught with very difficult problems, both from an implementation and a concept standpoint. For one, chromatography isn't all that useful for purifying substances deeply, but it is good at purifying them broadly. In a crude sense, in synthesis there are a combinatorially-exploded number of possible molecular substances which require differentiation (among small subsets of the whole, but very very many subsets) , but there aren't so many metals, especially if you're talking about just the elements or simple combinations of them.

Chromatography is a means of partitioning a mixture into various components. It can be applied to attempt to isolate one component, and once isolated it might be deemed sufficiently pure. Washing steps such as in organic synthesis (eg separatory funnel) might be closer to what you are thinking of.

Metals that are easy to liquefy (mercury, gallium, alkalis metals...) are easy to purify by distillation, so why bother with chromatography? Metals that are not are going to need a lot of heat, so the process will use just as much energy as pyrometallurgical processes so there is no clear energy benefit.

Another obstacle would be the very limited throughput of chromatography in general. For organics it is used mostly as analytical tool, and only in some cases to purify expensive pharmaceutical compunds after a synthesis. You cannot upscale it to bulk amounts and make a profit.

Chromatography is used to purify substances, IF there is no other way to do so, and the target material is extremely valueable. Not 30€ per gram, more like per milligram. Or scientifically valueable.

@matt_black, Looking at the example of Lead: "Specific heat of Lead is 0.13 J/g K. Latent Heat of Fusion of Lead is 4.799 kJ/mol. Latent Heat of Vaporization of Lead is 177.7 kJ/mol." There are clearly other difficulties with this idea as pointed out by everyone here, but there is enormous energy savings potential.

@And, I actually work in the beet sugar industry and we use chromatography to separate sucrose from molasses in the tons/hour.

@Karl, Definitely true for pharmaceuticals, but in sugar processing we use chromatography to extract sugar from molasses, as the process is faster than crystallization at lower purities. The current price of sugar is $0.23 per pound.

@ericnutsch Fascinating, my professors told me a different story. Thank you for enlightening me!

@ericnutsch Not a lot less than what I pay in a supermarket. Could you give a hint on how that chromatography might work?

@Karl, https://www.arifractal.net/technologies/chromatography/

@uhoh, https://www.arifractal.net/wp-content/uploads/2019/06/Molasses-Desugarization-in-the-US-Beet-Sugar-Industry-Recent-Update.pdf

Ah, good old moving bed. Forgot about that one, and I wasn't aware it had been upscaled.

@ericnutsch Thanks!

An obstacle of implementation is that the blend of metals has to be kept liquid all the time the chromatography runs, top to bottom of the column, while the equipment around has to sustain the mechanical strain and stress at elevated temperatures without reacting with the sample (which could cause new contamination, e.g., by formation of alloys).  With cryoscopy in mind, it is likely metals then purified have a higher melting point, than the submitted sample; potentially calling for additional heating to keep the process flowing.
Activated carbon and graphene you mention as candidates for the stationary phase could oxidize if the process is not anaerobic ($\ce{C + O2 -> CO2}$, and $\ce{CO2 + C <=> 2 CO}$), and already traces of carbon can substantially affect mechanical properties of metals (cf. $\ce{Fe}$ vs. $\ce{Fe3C}$ and its subsequent products).  Money aside, not every gas can be a suitable inert gas either (nitrides, e.g., $\ce{3 Mg + N2 -> Mg3N2}$).  Metal powders: potential creation of solutions, alloys, and new compounds.  There may be obstacles with zeolites and glass (what about the silicon in these silicates?), while fiberglass is a composite of glass fiber and some plastic polymer, the later possibly burning away, too.
A technique however used in e.g., semiconductor industry to further purify metals in the liquid state is zone melting (or zone refining).  This may work if thermodynamics are favorable (prompt: phase diagrams) and the starting material already is of somewhat high purity.  E.g., silicon which underwent a purification via distillation of intermediate $\ce{SiCl4}$.  Contrasting to chromatography, zone melting usually does not even attempt to split a mixture into multiple separate fractions A, B, C, D.  Yet, the directed crystallization of zone melting does not allow you to pass beyond an eutectic minimum (point C below), for instance to transform cast iron into steel, let alone pure iron.

(source)

Presuming you could address the engineering difficulties associated with heat, structure, oxidation, stationary phase selection, and alloying; would it be reasonable to expect to see a difference in migration rates of the different metals?

@ericnutsch I speculate this taps into recording diffusion coefficients (e.g., diffusion coefficient of liquid potassium into liquid sodium) e.g., by use of radioactive potassium as a marker, in contrast to e.g., diffusion of iron in liquid sodium. Perhaps there is an easier way to monitor the advance of a concentration profile in melts, larger atoms might advance slower than smaller ones as they experience more dynamic resistance and have a greater inertia.

Molten liquid chromatography does not exist currently. However the problems you mentioned for purifying metals is very conveniently done using ion-exchange chromatography. A lot of difficult to separate metals (rare earths) are separated or historically separated that way.
Given the high viscosity/ density of liquid metals diffusion of "analytes" would be slow, which means mass transfer kinetics between mobile phase(=liquid metal) and your solid refractory resin would be very slow. If we had a liquid metal as a mobile phase, the LC separation would be horribly bad and poor for these reasons.

Could you possibly put some ballpark numbers to it? or direct me to where to calculate the diffusion? As long as the transfer kinetic reduction was on the same order of magnitude as the diffusion reduction, I think there could still be some potential.

Not really, but you can identify metallurgical processes that rely on principles of phase separation and immiscibility of different components in molten salts, for instance during the production of sodium metal and chlorine gas from NaCl. In that process molten NaCl is subjected to electrolysis and the less dense molten sodium produced rises to the top and flows out of the chamber one exit while chlorine gas is evacuated separately. This is illustrated here:

Other molten salt applications: https://en.wikipedia.org/wiki/Molten_salt, https://moltensalt.org/whatIsMoltenSalt.html

Liquid chromatography is used to analyze and purify different chemical samples in a solvent... I was wondering if something similar exists for liquid (generally molten) metals.

The detail to remember is that, in molten metal, the metal atoms are separate individual molecules rather than being parts of larger complex compounds.  As such, a Lead atom's molecular size and shape is near-identical to that of a Bismuth atom, and so its travel-time through a Liquid Chromatography column would be imperceptibly time-differentiated from that of the impurity.
This idea would essentially require an isotope-differentiating level of discrimination from a organic-molecule-scale liquid-phase column technology - basically trying to do Mass Spectrometry separations in a benchtop Gel Electrophoresis type device.

If the metal atoms weren't actually "molten" but rather just complexed with organic compounds in such a way as to remain mobile in the liquid phase then you could remove the high-temperature engineering issues and would be working with larger molecules better suited to the technology.
Of course, if the same single metal-complexing molecule is used to hold both the pure metal and impurity atoms, then you're back to a situation of needing isotope-differentiating resolution on your column (to separate the MoleculeA-Lead complexes from the MoleculeA-Bismuth complexes).   On the other hand, if you use multiple complexing agents that will preferentially attach to pure metal and impurity separately (e.g. MoleculeA-Lead and Molecule-Bismuth) then you already have a chemical process that can discriminate the target molecule from impurities and could likely engineer a large-scale phase-separation process more efficiently than trying to do separation by a HPLC column time-of-flight differentiation process.

Chromatography doesn't just separate by size and shape, it can also separate by physical properties. (e.g. Organic HPLC typically separates mainly based on hydrophobicity rather than size or shape.) If the interaction of the lead atoms with the stationary phase were different than bismuth atoms, then they could potentially be separated. (As the one which interacts more tightly would be delayed more as it flowed through the column than the one which interacts less.)

@R.M. You are absolutely correct, but I believe that "size and shape" was perfectly reasonable for an ELI5-like first-order approximate explanation.

@AmateurDotCounter Beside chromatography mostly based on adsorption/desorption, and partition between mobile/stationary phase, there equally is the size exclusion chromatography. Here, larger polymer molecules typically elute ahead of the smaller ones.  But the sizes of the polymer molecules in question (and their differences) are larger, than individual metal atoms of a melt.


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Chemical Mechanism Behind Water-Soluble Ink
I came across water-soluble ink pens designed for creating temporary fabric marks, particularly in cross-stitch stamping. These ink dissolve in water, transforming water into a colorful liquid, while ensuring the fabric remains clean without any stains.
I'm interested in understanding the inner workings of these ink and their chemical composition. How does this type of ink function, and what specific chemical components enable it to dissolve in water without leaving stains on the fabric?
I appreciate any insights into the science behind these ink pens. Thank you

These are proprietary water soluble dyes. Only the company will know the exact composition. Search water soluble dyes. It must be a huge class of dyes which polar functional groups.

BTW, some dyes adhere to cotton and linen (cellulose-based fibers), others to wool and silk (protein-based), etc. Those inks might wash off one fabric, and not another.


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Volumetric flow rate ratio calculation
I'm using two tanks of nitrogen and oxygen and mixing them in different ratios to change the oxygen partial pressure in the system. System is opened to the atmosphere through the venting pipe which therefore sets the total pressure of the system to 1 bar. System is opened and gases flow through it. Schematic is provided to give better insight:
To create atmospheric composition of the mixture, mole fraction of oxygen needs to be 0.21 while nitrogen needs to be 0.79. Therefore, molar flow rate ratio, $A$ is: $$A = \frac {\dot n(O_2)}{\dot n(N_2)} = \frac {0.21}{0.79} = 0.266$$
Gas tanks are equipped with volumetric flowmeters. Relating volumetric and molar flow rates can easily be done if the assumption of ideal gas is valid. Since I'm using air at atmospheric pressure and high temperatures (500 – 800 $^\circ$C), the assumption should hold.
Volumetric flow ratio is: $$ \frac {\dot V(0_2)}{\dot V(N_2)} = \frac {\frac {\dot n(O_2)RT}{P(O_2)}} {\frac {\dot n(N_2)RT}{P(N_2)}} = A \frac {P(N_2)}{P(O_2)}$$
Since we're talking about the ideal gas, Dalton's law holds. Partial pressure ratio is equal to the mole fraction ratio of the components in the mixture. Therefore: $$\frac {P(N_2)}{P(O_2)} = \frac {1}{A}$$
Leading to the volumetric flow rate ratio being always equal to 1. This doesn't seem correct, what am I doing wrong?

I think you are wrongly relating the variables to the flow streams. Maybe the following image can help orienting yourself:

So, I will follow your procedure:

Nitrogen and oxygen amount balances
\begin{align}
  y_{\ce{N2},3}\dot{n}_3 = (1)\dot{n}_1 \rightarrow
  \dot{n}_1 = 0.79\dot{n}_3 \tag{1}  \\
  y_{\ce{O2},3}\dot{n}_3  = (1)\dot{n}_2 \rightarrow
  \dot{n}_2 = 0.21\dot{n}_3 \tag{2} \\
\end{align}
and taking the quotient we obtain $\dot{n}_2/\dot{n}_1 = 0.21/0.79$ which is what you obtained.

Those volumetric flow rates from your second equation refer to streams $1$ and $2$
\begin{equation}
  \frac{\dot{V}_2}{\dot{V}_1} =
  \dfrac{\dfrac{\dot{n}_2RT_2}{P_2}}{\dfrac{\dot{n}_1RT_1}{P_1}} \rightarrow
  \frac{\dot{V}_2}{\dot{V}_1} = \left(\frac{\dot{n}_2}{\dot{n}_1}\right)
  \left(\frac{P_1}{P_2}\right) = \left(\frac{0.21}{0.79}\right)
  \color{blue}{\left(\frac{P_1}{P_2}\right)} \tag{3}
\end{equation}
since $T_1 = T_2$ as I see in your math. This is again what you obtained. However, the streams refer to $1$ and $2$. Thus, $P_1$ is the pressure of the nitrogen tank, and $P_2$ the pressure of the oxygen tank.

The mistake comes when you make the quotient in the last equation. We have that
\begin{equation}
  \frac{P_{\ce{N2},3}}{P_{\ce{O2},3}} =
  \frac{y_{\ce{N2},3}P_3}{y_{\ce{O2},3}P_3} \rightarrow
  \color{blue}{\frac{P_{\ce{N2},3}}{P_{\ce{O2},3}}}
  = \frac{0.79}{0.21} \tag{4}
\end{equation}
and the question is, are the blue parts in Eqs. (3) and (4) the same? The answer is no. One thing are the pressures of the pure compounds in their respective tanks, and another one the partial pressures of the constituents in stream $3$. A clear example would be an initial situation where the tanks are at the same pressure, hence $P_1/P_2 = 1$. However, the relation $P_{\ce{N2},3}/P_{\ce{O2},3}$ cannot be equal to one, and is in fact $0.79/0.21$. The pressure in stream $3$, upon mixing, is almost $P_3 = \pu{5 atm}$.

If you want to ensure, you have to do more calculations where the unknowns are $\dot{V}_3$ and $P_3$.
Firstly, we have the global balance of amount of substance
\begin{align}
  \dot{n}_3       &= \dot{n}_1 + \dot{n}_2 \\
  \rho_3\dot{V}_3 &= \rho_1\dot{V}_1 + \rho_2\dot{V}_2  \\
  \left(\frac{\color{red}{P_3}}{RT_3}\right)\color{red}{\dot{V}_3} &=
  \left(\frac{P_1}{RT_1}\right)\dot{V}_1 + \left(\frac{P_2}{RT_2}\right)
  \dot{V}_2 \tag{5} \\
\end{align}
Secondly, we have the global conservation of mechanical energy, under the assumptions that there is no work transfer and the friction is absent
\begin{align}
  \left(\frac{P_3}{\rho_3} +
  \frac{\color{red}{v_3}^2}{2} + gz_3\right)\dot{m}_3 =
  \left(\frac{P_1}{\rho_1} + \frac{v_1^2}{2} + gz_1\right)\dot{m}_1 +
  \left(\frac{P_1}{\rho_2} + \frac{v_2^2}{2} + gz_2\right)\dot{m}_2
  \tag{6}
\end{align}
where the $\dot{m}$'s are the mass flow rates, and $z$'s are the heights. This is Bernoulli's equation but with three streams. Note that in Eq. (6), the only unknown is $v_3$ since $P_3/\rho_3 = RT_3/M_3$ which is a value that you can calculate. After this, you can calculate $\dot{V}_3 = v_3S$, where $S$ is the area perpendicular to the flow. Afterwards, go back to Eq. (5) to know the final pressure $P_3$.

Yes, I thought that the pressure on which the gas is used, $P_3$ is what relates mass and volumetric flow rate. I'm not sure why this isn't the case on intuitive level, though. Why is it that pressure in the tank relates the two?

If that's the case, that means that the volumetric flowrate increases as the tank is emptied (pressure in the tanks drops) to keep the mass flow rate constant. Mass flow rate should remain constant in time due to mass conservation.

Oh well, then the situation is mathematically more involved. In here, all the balances are stationary (don't depend on time). I imagined that your tanks were an enormous source of nitrogen an oxygen, and they will not get depleted as to decrease its pressure (or $\dot{m}$ or $\dot{V}$). You should see that mass and volumetric flow rates upstream have constant values throughout the experiment. Is that the case?

I think $P_3$ is the total pressure of the mixture which is set to the atmospheric value as the system is opened to the influence of the atmosphere through the venting pipe.

Well, as far as I know, mass flow rate should be constant (mass conservation for the pipes, no accumulation condition anywhere in the pipes where the gases flow), but volumetric flow rate will change with pressure. In the real setup, pressure in the tanks will probably not change very much during the experiment and that will keep volumetric flow rate almost constant.

@DarioMirić Well, it is an interesting situation. If you set all the variables in tank 1 and tank 2, $P_3$ is derived from balances that I have shown you. These are conservation pricnciples, very well known, and it seems that you are not free to choose its value. I wouldn't worry too much anyways. I'd be interested in how things turn.

For the first seconds of the run, all the variables will change a little and then they will stabilise (they acquire a steady state value). For example, you go from 0 $\pu{m^3/h}$ in the volumetric flow rates to some constant value. These balances focus on what happens after that transient time.

There is no freedom, the fact that the system is opened to the atmosphere, sets the total pressure to the atmospheric.

Let us continue this discussion in chat.


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Developing aluminum mirrors, similar to tollens reagent but with Al instead of Ag
Is there a chemical process to deposit an aluminum mirror onto a surface similar to the tollens reagent process for silver? I should add that any liquid process would satisfy my curiosity, it does not have to use water. Just anything to reduce the aluminum ions into metallic aluminum nano particles
I have searched and searched but have found very little on the subject. Thank you!

To get a silver mirtor you reduce an aqueous solution containing silver ions. Aluminum ions might be a little harder to reduce.

Not possible in water environment. Aluminium is deposited from vacuum on clean glass and then protective layers are applied.

@Poutnik It is to aluminize a mirror. The technique scales well to work on small reflectors and larger ones.

@David Dawkins A mirror with silver exposed as first surface (similar to the ones which are aluminized) looses quickly reflectance e.g., by the formation of silver sulfide.  This is why silver coating typically is applied to the back and then covered by lacquer (an example).  The aluminized mirrors can afford to have the reflective surface first, because the $\ce{Al2O3}$ layer which forms is practically transparent (at least in the range of visible light).   But it contains pores (useful for embed dyes) which can trap dust.

While there are good chemical reasons why this is hard, economics also matters. It is relatively easy and cheap to sputter Al directly onto surfaces (though not, perhaps, in a home lab without a vacuum chamber).

@Buttonwood - I've worked with silvered diffraction gratings - they just have to be in a nitrogen purged environment. And one can put down very thin pore-free Al2O3 layers by atomic layer deposition these days.

@JonCuster Movies similar to «Downtown Abbey» sometimes depict how heavy silverware/cutlery made of silver was to be polished again, and again; not necessarily with a technique suitable for an optical device, or a mirror with ductile Ag on the primary surface...

@Buttonwood - i agree that one doesn't want to treat your optics like silverware, but when you want that extra percent or so of signal it is perfectly reasonable to use bare silver coated optics - it just makes the rest of the setup more complex. Silvered optics in a spectrometer are pretty easy since you just have to purge the box.

Aluminum cannot be produced as a metal from and aqueous solution, because $\ce{Al}$ reacts chemically with water, producing $\ce{Al(OH)3 + H2}$, due to the rather negative value of the $\ce{Al}$ redox potential: $-1.70$ V. This reaction is suppressed on usual aluminum foils or objects, because, in air, metallic aluminum is spontaneously covered by a protective layer of aluminum oxide, which is thin, continuous, transparent, and protective against oxidation. As soon as this protective layer is locally removed by a scratch, new $\ce{Al2O3}$ is immediately synthesized in this scratch by the atmospheric oxygen of air.
This layer can be dismantled from the metal by dipping an aluminum object or foil in a solution of mercury chloride $\ce{HgCl2}$ for a couple of seconds. In contact of $\ce{HgCl2}$, the metallic aluminum reacts and produces some metallic mercury that makes an outer layer on aluminum amalgam. This amalgam prevents the formation of the protective and continuous layer of $\ce{Al2O3}$. As a consequence, the more or less amalgamated aluminum pieces react quickly with oxygen or water and produce in a couple of minutes long white filaments of $\ce{Al2O3}$. This strange phenomena lasts as long as some mercury is still remaining on the outer aluminum surface. Later on, mercury is eliminated from the surface so that the protective layer of $\ce{Al2O3}$ is regenerated. The corrosion stops.

I should add, it does not necessarily have to use water as in tollens reagent. It can be any liquid. Just looking for a way aluminum ions into metallic aluminum nano particles for coating a surface


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Is boric acid aqueous solution corrosive to stainless steel?
I'm thinking of trying an aqueous boric acid solution in a nebulizer for a sinus infection, in a similar strength as for eyewash (anywhere from 1 tsp. in 4 oz. to 1 tsp. in 32 oz.). My nebulizer has a stainless steel screen that the vapor comes thru, and I need to know if the aqueous boric acid solution will have any corrosive effect on the stainless?  Thank you.

You think anything would sooner damage stainless steel than mucous membrane? Well, in the long run there would be many such things, but not in the short.

Boric acid is one of the weakest acids commonly used. Its $p$K$_a$ is equal to $9.24$. Its solution will have a negligible corrosive effect on the stainless.


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Mg2+ ion formation
Why does Mg not form a Mg+1 ion, even though its second ionization energy is much higher than the first ionization energy?
(I know that an ion should resemble the noble gas closest to the element from which it was formed (at least in the p and s blocks), but the question is why).
I saw that there is a similar question on the site, but I want to understand it atomically and not by enthalpy.
(I mean I want to understand it according to considerations such as whether because there is a decrease in the energy level from Mg+1 to Mg+2 (from n=3 to n=2) energy is released and the energy it emits is more than the energy needed to go from n=3 to n=2 or based on similar considerations)

If we are talking about monatomic ions, then we have to consider the balance between ionization energy and electrostatic attraction. When we go to $\ce{Mg^{2+}}$ instead of just $\ce{Mg^{1+}}$, the electrostatic attraction to anions such as chloride or hydroxide is stronger and the energy saved from that could be enough to allow the more costly additional ionization. Roughly speaking, at least under conditions we encounter everyday, an extra ion charge can balance out about $3000$ kJ/mol of ionization energy, and the second ionization energy of magnesium comes in well under that. However the third ionization energy, involving an electron from the neon core, is far too high.
Magnesium does form diatomic $\ce{Mg2^{2+}}$ ions in certain complexes; these are made more stable than a monatomic ion would be by a magnesium-magnesium bond (which may have some ion-electride character[1], i.e. a superposition of $\ce{Mg^+–Mg^+}$ with $\ce{Mg^{2+}[e_2^{2-}]Mg^{2+}}$). Such compounds are widespread enough to earn an article in Wikipedia.
Reference

Platts, J. A.; Overgaard, J.; Jones, C.; Iversen, B. B.; Stasch, A., "First Experimental Characterization of a Non-nuclear Attractor in a Dimeric Magnesium(I) Compound," J. Phys. Chem. A, 2011, 115, 194-200, DOI: 10.1021/jp109547w

Thank you very much for answering.
I just want to make it clear. If I understood you correctly, then it actually happens because the electrostatic attraction between a Mg2+ ion and a certain anion is stronger than the electrostatic attraction between Mg1+ and the same anion, and this additional energy is greater (usually) than the second ionization energy, did I understand you correctly?

Yup, that's it.


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∆S for an isolated system
As far as I know, for isolated system ∆U=q=work=0. And ∆S=q reversible/T
So shouldn't ∆S also be equal to 0. But for an isolated system ∆S is positive. I know that entropy is the degree of randomness and disorderness of the system but here the formula doesn't seem to work.

Forget randomness and disorderliness. You have written quite correctly that $\Delta S = \frac{q_{rev}}{T}$ and also noted that in an isolated system $q_{rev}=0$. So when it is an isolated system AND the process is reversible indeed we have $\Delta S=0$ but otherwise in an isolated system when the process is irreversible, that is a spontaneous process, $\Delta S >0$.

But ΔS > 0, if some spontaneous reaction occurs in this isolated system.

Imagine an isolated system that is subdivided into subsystems, each of which is at thermal equilibrium.  There is a constraint between the two subsystems that prevents them from interacting.  Now you remove the constraint and allow them to re-equilibrate with one another spontaneously. Do you think that this is a reversible process?  If not, do you think that the entropy change is zero?

@Maurice as I have written above otherwise, and do not forget that $q_{rev}$ denotes the externally supplied thermal energy between two equilibrium states, which by definition is zero for an isolated system irrespective of the process being reversible or irreversible. So when the process is reversible and isolated $\Delta S=0$ but when it  is isolated and is not reversible, i.e., irreversible, $\Delta S \ne 0$. Experience and its formulation the 2nd Law says that it is then $\Delta S > 0$

And if you want to avoid any lingering confusion in your mind then always, always, and always think of thermal energy transfer as transport of entropy at some temperature. If you write $\Delta S \ge \Delta S_0$ where $\Delta S_0$ is the transported entropy into the system from the reservoir at temperature $T_0$ then the 2nd law states that the system's entropy change $\Delta S$ is never less than $\Delta S_0$ and $\Delta S- \Delta S_0\ge 0$ is the entropy generated by the process meanwhile the system's internal energy has increased by $T_0\Delta S_0$. Now there cannot be any confusion.


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Dot blotting immunoassay for visualizing the amyloid formation at different time points
We purchased the primary antibody approximately two months ago, and it has been stored in the freezer since then. Currently, we are attempting to assess its functionality through dot blotting. In our experiment, we utilized a protein concentration of 2 μM, with a 1:1000 dilution of the primary antibody and a 1:500 dilution of the secondary antibody. The membrane was allowed to incubate in the primary antibody for 48 hours using NFDM TBST buffer. Unfortunately, the results were unsatisfactory, showing no binding, including non-specific binding. We are seeking insights into potential issues and guidance on how to determine the effectiveness of the primary antibodies.


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Increasing energy density for flow batteries with water based electrolyte?
Given a flow battery with a water based electrolyte, would it be possible to increase the energy density of battery by precipitating the redox active species from the anolyte and catholyte solutions? I am assuming the main limiting factor in energy density of flow batteries is limits on how much of the redox active species can be dissolved in water.
To give an example, say both the catholyte and anolyte tanks have two chambers: the active chamber and the precipitation chamber. The active chamber is where the normal solution would exist. Above a certain concentration, the solution would be pumped to the precipitation chamber where the water and active species would be separated and the pure water would be pumped back to the active chamber decreasing its active species concentration. A naive way to separate the water and active species might be to evaporate the water and condense it back in the active chamber. Chemistry is not my strong suit, so I could be totally off base.

Please give us a real example of a stuff which could be precipitated from anolytic and catholytic chambers. Also what is an "active chamber" separated from a " precipitation chamber" ?

What would be precipitated is part of my question. If the water is evaporated, I would assume that due to the limits on solubility of the active species that something would precipitate out (maybe I'm wrong). The active chamber is the standard chamber for the anolyte/catholyte you see in a flow battery.


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Why does CO lack a Q-branch in its rovibrational spectrum as opposed to NO?
Since I assumed that if NO has a Q-branch, then so should CO. However, this explanation of rovibrational spectroscopy says otherwise: https://pages.vassar.edu/magnes/2018/12/15/fundamental-rovibrational-spectrum-of-co/
Is there a mistake or is there something I'm missing here?

"The Q-branch can be observed in polyatomic molecules and diatomic molecules with electronic angular momentum in the ground electronic state, e.g. nitric oxide" - https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Rovibrational_Spectroscopy CO has no unpaired electron.

NO has an unpaired electron, this generates electronic angular momentum about the NO bond axis. As a result the selection rules are $\Delta J=0,\pm 1,\Delta M=0,\pm 1$ and $\Delta K=0$ where $K$ is the non-zero angular momentum quantum number about the NO axis.  The caveat is that  the $\Delta J=0$ applies except for $J=0$. (spin orbit coupling also complicated the spectrum leading to a doubling of lines, and lambda type doubling (v.v small is als0 present))


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How long can pH solution calibration be used for?
I am using a pH meter regularly (6 to 12 time per week), used mostly for home-food processing (mostly measuring the acidity of whey).
The meter requires weekly calibration, which is quite time consuming.
I usually prepare the solutions and then discard them. However, I was wondering: how long can I actually keep the solutions for? What is their shelf-life? Do I need to prepare the solutions weekly or can I keep them longer?

If you are using ultrapure water and storing the solutions bottled in a dark cool place they should keep for months, I would think. One brand of unopened commercial buffer has an official shelf life of 2 years. Commercial solutions have been filtered to reduce microbial growth. Various websites suggest not keeping opened solutions longer than 6 months but hesitate to provide an exact time since this depends on the conditions in which they are kept.

If you want to deviate from the given procedures, you are on your own and you have to experiment and verify.
One way is, after the calibration, to save calibration solution(s) to be used as the control sample. You then periodically measure its pH whenever you measure true samples.
If the pH of this control sample deviates from its nominal value more than is acceptable for you, prepare new solutions and recalibrate. Note the needed time interval for this recalibration and then eventually decide to use this (or shorter) interval as your regular recalibration routine.
Note that it is not only about the pH drift of calibration solutions, but also about pH drift of the device itself (pH electrode). The mentioned measuring of control samples addresses both.

So you're saying it is not advisable to do it but not impossible and I should compare a fresh calibration solution against a preserved one over time and compare the delta?

Rather, doing so, it is expected you know what you are doing. I do not say comparing fresh vs preserved calibration. I mean continuing normal pH measurement and evaluating measured pH of preserved solution with its expected pH  value. E.g if the calibration solution has nominal pH 4.00 and if you meaure its pH after 4 weeks as 4.2, you know something is off and you need to recalibrate with fresh solutions.( at least 2)

Note that serious measurements in labs involve recallibrations before every measurement or every day, depending on measurement frequency.

pH 7 and 4 buffers are very stable if air contact is minimized. A fresh buffer should be used for every calibration. My procedure was to store the electrodes in pH4 buffer. Each morning measure the millivolt reading of the solution, rinse, replace with the previous pH7 buffer record the mv reading, calibrate with fresh pH7 buffer recording the mv reading in the log book. set the slope at pH4 with fresh buffer after recording the mv value. The difference should be the Nernst equation value ~175mv; the mv readings should be almost constant. I have found that electrodes slowly drift and eventually fail[unless broken first]. If a new bottle of buffer is used check the mv readings with the old and new they should agree. This takes less time to do than to read. After a while any problems with electrodes, meters or buffer solutions will be immediately apparent. I recommend Fisher Scientific color coded buffers.


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NaOH etching time and temperature for Stainless Steels (SS-304) for Surface Preparation for coating
To improve adhesion of TiO2 coatings on stainless steel sample I am treating SS-304 with 1M NaOH for functionalizing the surface with metal hydroxide groups as well as increasing roughness. What would be the general range of temperature of NaOH solution we required and duration of this process.


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Finding or designing a thermochromic that becomes transparent to NIR at room temperature
Are there known nanoparticles or organic materials that have the following properties:

absorb near-infrared light and convert them to heat at temperatures below a certain threshold
become mostly transparent to NIR at temperatures above the threshold.

The threshold should be somewhere between -10 and 30° C.
If such compounds do not exist, how would you design them? I'm thinking of somehow coupling a gold nanorod with a PCM that "breaks" the rod when changing phase, but unclear how that could actually be designed.


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how can I tell compton versus coherent scatter apart?
I am trying to get a better understanding of compton and coherent scatting, along with all the other interactions. It has seemed like a lot to memorize and I am trying to find ways to more easily understand them and tell them apart when being asked questions. I was reading today and I think I may have figured out a way to tell them apart but want to make sure my thought process is correct. Here is what I am thinking:
If I can remember that with coherent scatter, the ejected particle has to be in a ground state. which I think of something that needs to  be groudned so it is safe for use and will not cause electrical problems, meaning it has to be "neutral". So coherent ejects something that has to be neutral, or not charged. that means it cant be an electron or proton, therefore its a neutron? and it would make sence because the incoming xray in coherent colides with the nucleus where neutrons can be found. the scattered neutron would be an xray, which again is neutral and like a gamma ray except for its origin, being the nucleus. Now for compton, the incoming photon, ill say it is a xray to be specific, interacts with an electron that is loosly bound/valence electron, which has a negative charge. so really, in coherent, a neutral particle is released. in compton, it is a charged particle?
also with compton, is this only happening within a body?
edit: I read somewher else that in coherent, a photon witht the same energy can be scatterd.. but if it is ground state wouldnt it have no charge? therefor how would it have the same energy as the incoming photon? im confused all over again.

This is not a question of chemistry. The Compton scattering belongs to physics, The Compton effect is one of the three sorts of interaction happening when electromagnetic radiation interacts with matter, without producing any chemical reaction.


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What explains the relative order of melting and boiling of oxygen and nitrogen?
Why is the melting point of nitrogen ($\ce{N2}$) greater than that of oxygen ($\ce{O2}$)? After all, both are non-polar, and $\ce{O2}$ has more electrons than $\ce{N2}$.
In addition, why is the boiling point of both opposite? I mean, why is the boiling point of oxygen greater than that of nitrogen, even though the melting point of nitrogen is greater than that of oxygen?

Crystal structures are different, so the solid bonding is different. Details of this inter-molecular interactions are clearly different. If you want a simple answer it won’t happen…

This is a valid question which does not have a simple answer. Most questions involving crystal structures in one way or another are like that.

I understand, thanks.

For boiling point comparisons see here.

Thanks for the answer, but I didn't really understand its connection. If we talk about boiling point, then the boiling point of oxygen is higher than that of nitrogen because the London forces are stronger in oxygen, and there is no point in going to such a level of depth.

Please read my answer more carefully. Fluorine has more electrons but its bp drops off. We have to go more in depth to learn why.


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Some particles are observed inside the wall of glass vial
Recently, when I use glass vial, some particles are observed inside the wall. These particles are so small and transparent or maybe white-colored, so it is hard to observe during ordinary usage. But when I store some organic solvents, say acetone, the surface of particles are wet, then it can be observed if I carefully stare at it. It can be just dirt, but the problem is, although it is not dissolved in common organic solvents, it is dissolved in water. It is observed in unused vial, and also in ones that were stored for several years. Does anyone have this experience or have an idea to what it is?

Do you live near the ocean? Have you tested for salt? Another possibility is anything stored near HCl and NH4OH bottles (yes, they're usually labeled NH4OH, not aqua ammonia) may find NH4Cl depositing on surfaces. Again, test!


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Cation/Anion UV-Vis absorption
I want to understand how the ground state absorption of a molecule shifts if an electron is added or removed. I found these two publications 1 and 2 that observe shifts, but both times a redshift, be it anion or cation. I am specifically asking about the HOMO/LUMO gap (although then it should probably be called HOMO/SOMO or SOMO/LUMO for anion and cation, respectively) and not additional excitations.
Is there a general rule? If not how could the observation made by the publications be explained. (I am not aware that they explained it..)
I see this argument: "An extended π-system increases the absorption
wavelength to obtain colored materials with lower energetic
gaps" but I think it applies to modification of the molecule by adding atoms. I dont see how this works for anion/cation.
I'd also be happy if the answer was: There is no general rule, because... Or this question is unprecise, since your picture of electronic structure is too simple, you need to take into account that...


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Major product in chloro-sulfonation of toluene
According to Clayden's Organic Chemistry (2e) (p.485)

... and the acid chloride (tosyl chloride, $\ce{TsCl}$) needed to make tosylates can be made from the acid in the usual way (p. 215) with $\ce{PCl5}$ . It can also be made directly from toluene by sulfonation with chloro-sulfonic acid $\ce{ClSO2OH}$. This reaction favours the ortho sulfonyl chloride, which is isolated by distillation.

My question is, why is the ortho product favoured? If anything there should be steric hindrance due to the methyl group and the $\ce{sp}^{3}$ hybridized sulphur atom, destabilizing the ortho-product.
Edit 1: I've found the same question here which does provide an answer based on kinetic and thermodynamic control, however, I share the same doubt as the author at the end of the answer.
Edit 2: Some more data from the source quoted in the linked answer (see Edit 1)
So, I guess, it's just not detailed enough in Clayden? Please share your thoughts.

According to my teacher, steric hindrance is the reason for the reaction favouring para-toluenesulfonic acid during sulfonation of toluene.

What more than you found would you know? Steric isn't the issue. If it was t-butyl, or methyls on both sides, the situation would be different.

The fact that raising the temperature favors the para product is a clue. Higher temperatures, which allow a wider variety of choices kinetically and more reversibility of these choices, tend to favor thetmodynamically more stable products; lower temperatures favor products that form kinetically through a lower-energy intermediate.
Here, steric hindrance in the product certainly makes the para isomer more favorable thetmodynamically, so if we raise the temperature sufficiently we are sure to get that isomer as expected. But at lower temperatures where kinetics and activation energy are more important, the ortho intermediate has the advantage of being stabilized by hydrogen bonding.
Hydrogen bonding to hydrogen atoms attached to carbon? Well, the methyl group on the intermediate formed when the electrophile attacks toluene does not quite have your ordinary carbon-hydrogen bonds. With the intermediate becoming positively charged and electron-deficient, the methyl group will hyperconjugate with the ring: its bonding orbitals will overlap with the formally electron-deficient ring, producing an extension of the pi conjugation in exchange for weakening the carbon-hydrogen sigma bonding in the methyl group. We may represent this interaction as a contributing structure in which a carbon-hydrogen bond is exchanged for this extended pi interaction:

Imparting part of the electron deficiency and positive charge then primes the methyl hydrogen atoms for hydrogen bonding in several ways related to both the electrostatic and the molecular-orbital renderings of hydrogen bonds:

The methyl hydrogens gain positive charge and thus attract electrons from a potential acceptor atom in the electrophile.

When the bonding is shifted away from the methyl group towards the ring, the unoccupied antibonding orbitals shift in the reverse direction. Therefore there is more probability density for the electrons from the acceptor atom to overlap.

Withdrawing electron density from the bonding orbitals in the methyl group means there is less repulsion to oppose overlap with the antibonding orbitals which occurs in hydrogen bonding.

So when we specify the electrophilic function $E$ as the chlorosulfonyl group, the oxygen atoms can act as hydrogen-bond acceptors and they will, creating a third contributing structure in which the hydrogen bonding in effect replaces the lost sigma bonding from the hyperconjugated methyl group:

This extra delocalization can occur, however, only with the ortho intermediate as otherwise the methyl group and oxygen atoms are too far apart. The hydrogen bonding interaction will eventually disappear when the intermediate is deprotonated and becomes neutral again, but the intermediate has already done the deed of directing the reaction towards the ortho product. Unless, of course, the temperature is high enough to allow this choice to be reversed.


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Redox Reaction Half Equations
$$\ce{PbS(s) + 4 H2O2(aq) -> PbSO4(s) + 4 H2O(l)}$$
I been trying to find the half equations for a while now. How would you find the balanced half equations for this redox reaction? Do you just add water for the extra oxygens on the sulfate??? Assume acidic solution.

That's a possible option.

The two half-equations are :
$\ce{PbS + 4 H2O ->  PbSO4 + 8 H+ + 8e-}$
$\ce{H2O2 + 2 e^- + 2 H+ -> 2 H2O}$
Multiplying the second equation by $4$ and summing them gives :
$\ce{PbS + 4 H2O2 -> PbSO4 + 4 H2O}$


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Can Schiff's reagent respond to hemiacetal?
Glucose responds with both Tollens' reagent $(\ce{AgNO3})$ and Fehling's solution $(\ce{CuSO4}).$ But it doesn't respond when treated with Schiff's reagent (p-rosaniline hydrochloride), the reason being that it exists in its pyranose structure, which has a hemiacetal in place of an aldehyde group.
So, does this mean that the Schiff's reagent does not react with hemiacetal, while both Tollens' and Fehling's do? Also, what about Benedict's solution? Can it serve as a test for hemiacetals?


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What is the most accurate way to determine the amount of water in a material from XRF and LOD-105 results?
this might seem like a silly question but I want to be sure. I have the results from an external laboratory for the bulk chemical composition (by XRF) and moisture (by Loss on Drying to 105 degrees Celsius) (among other test parameters, PSD etc.) and I'm wondering what the most accurate way is to incorporate the amount of water into subsequent equations I'm using for a mathematical model.
For example, let's assume I have 1 g of the material with 50 wt.% Si and 50 wt.% Al (as an example) from its XRF data and 1.5% (or 1.5 g H2O / g Sample) moisture from the Loss on Drying method results. Which of the following methods below (or a different method) is better for quantifying the amount of water in the material better? Please provide a detailed and reasoned response.
$\textbf{Method 1:}\\
1 \text{ g Sample} \times 1.5 \text{ g H}_2\text{O} / \text{g sample} = 0.015 \text{ g H}_2\text{O}$
$\textbf{Method 2:}\\
1 \text{ g Sample} = X \text{ g Solids} + Y \text{ g Liquids} = X \text{ g Solids} + Y \text{ g H}_2\text{O} = 100\text{wt.% Solids} + 1.5 \text{ wt.% Liquids} = 101.5 \text{ wt.% Sample}\\
\text{Therefore: } Y \text{ g Liquids} = 1 \text{ g Sample} \times \left(\frac{1.5 \text{ wt% Liquids}}{101.5 \text{ wt% Sample}}\right) = 0.0148 \text{ g H}_2\text{O}$
Note that with this example the moisture % is small, and both methods produce a similar result, but with other samples the moisture % may be bigger and thus introduce more errors if I have the calculations wrong, so I want to be sure I'm doing things right.


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How do electrons travel through a fuel cell or flow battery with multiple cells?
I understand that if you have a fuel cell or flow battery with a single cell that electrons will go from the electrode and through an external circuit across the load which the fuel cell / flow battery is attached to.
But, if you have multiple cells in series, how are the electrons in the middle cells able to travel anywhere? Each cell is separated by a non-electrically conductive membrane. There are only external connections on the first and last cells. Where are the electrons traveling in the middle cells?

Internally, within any cell, electrons are not traveling. Ions are. Externally, between 2 electrodes of the same or different cells, electrons travel via wires as in any circuit, regardless of batteries.

Right, I understand electrons are not traveling internally. But the cells of a fuel cell, or at least of a flow battery, are not connected externally. The first and last plate are connected to an external circuit. The middle cells are not. Where's the connection to the electrode of each cell? Look at lead acid batteries or any battery that has multiple cells in series. The electrodes are welded together so there's a path for electrons. Where is this connection in flow batteries and fuel cells?

There is always a connection otherwise current could not flow. Just their geometry may create false impression there is none.

Linking two examples below. The only external connection to each of the electrodes is where voltage is monitored (as you can see with the connections on the side of each battery). So there has to be some internal connection from plate to plate where electrons pass. But the membrane covers the active area of each plate so it's totally unclear how this is done.
https://pubs.acs.org/cms/10.1021/acssuschemeng.2c03148/asset/images/medium/sc2c03148_0001.gif
https://www.researchgate.net/publication/307877246/figure/fig8/AS:403792488091649@1473283039586/Photo-of-5-kW-class-VRB-cell-stack.png

Wire-A-C-wire-A-C-wire-A-C-wire-A-C-wire / Between A and C, ions are passing according their charge sign.

The wires shown in both images are not current carrying leads. They're for sensing voltage.

I have not referred to any particular wires in the pictures.  Circuit-wise, batteries of fuel or flow cells are not different than batteries of ordinary galvanic or electrolytic cells.

Suppose three cells are coupled in series. First a cell Zn/MnO2, then a cell Cu/Ag, and then a Daniell cell Zn/Cu. The middle cell Cu/Ag has a negative pole made of Cu, and a positive pole made of Ag. In the middle cell the electrons are coming out of the negative Cu electrode (anode). They are immediately sent through a copper wire to the positive pole (MnO2) of the first cell. In this first cell, the electrons are getting out of the anode (zinc) electrode. But these electrons are not sent to its own positive pole (MnO2). No. They are sent to the positive pole (cathode) of the 3rd cell, which is a copper plate from the Daniell cell. The cycle is closed when the electrons produced in the zinc anode (negative pole) of the Daniell cell (3rd cell) are sent through a copper wire to the silver cathode (positive pole) of the middle cell

Can you point to a diagram, paper, video, etc. that shows the electrode of each cell in a flow battery or fuel cell being connected to the next cell with an external wire? I have never seen this hence my confusion. I have seen these connections in conventional batteries (e.g. lead acid), but not in flow batteries or fuel cells.
I understand in principle what should be happening.

@DavePiae867, Wires are not magic. If the anode of one cell touches the cathode of the next, electrons flow from one to the other. The anode and cathode are simple conductors, just as is the wire.

@DavePiae867 Direct contact of electrodes of adjacent cells (like stacking "button" cells or columns of AA or AAA cells or the original Volta's Cell column) is fully equivalent to connecting them by wires (if we neglect wire resistance).

@DRMoishePippik I have not seen any fuel cell or flow battery design in which the electrode of one cell directly touches the electrode of the next cell. There's an ion selective membrane that covers the entirety of the active area on each cell. Each cell/plate is non-conductive outside of this active area. Which is why I'm asking if you can provide any sort of literature, diagram, picture, etc. that would clarify this.

@Poutnik I would appreciate seeing anything that describes how/where this internal contact is made when there's a membrane between each cell that covers the conductive active area.

If a membrane is put between each cell that covers the conductive area, the cells are insulated and deliver no currant.

@DavePiae867  There is no reason to put any membrane between cells. You must misinterpret the scheme or the physical look. If there are membranes, they are within cells.

@Poutnik Sure, I should've said plate rather than cell. There is a non-electrically conductive membrane between each plate which separates the electrically conductive active area of each plate. I still have no explanation for where electrical contact is made.

@DavePiae867. Which electrical contact are you mentioning ? Between what and what ?

Nothing much can be deduced from those photos. There is needed a schema of the battery of such cells and cells themselves to have something to talk about.

Every electrochemical cell is circuit-wise like a black box with two galvanic contacts, containing particular cell design geometry and technology, that is irrelevant with respect to the circuit schema.
A battery of such cells is serialisation of such 2-contact black boxes, galvanically connecting respective contacts of adjacent cell and connecting the end contacts of the whole cell battery to external circuit.
Being a fuel or flow cell is irrelevant – the schema is the same.
Below you can see the vanadium flow cell schema from ResearchGate.
A battery of such cells is assembled exactly as for any other cells, galvanically connecting external cell contacts with the opposite sign contacts of adjacent cells. The porous arrangement of electrodes is intended to maximize the contaxt area between the active electrolyte and the conductor, to have minimal internal resistance and maximal specific power.

Note that the schema
$\mathrm{...(E_1|El_1||El_2|E_2)-(E_1|El_1||El_2|E_2)-(E_1|El_1||El_2|E_2)}...$
where $\mathrm{E_1}$, $\mathrm{E_2}$ are electrodes, $\mathrm{El_1}$ and $\mathrm{El_2}$ respective electrolytes and $\text{||}$ is a selective ion permeable membrane.
the schema can be simplified as
$\mathrm{...(E|El_1||El_2|)(E|El_1||El_2|)(E|El_1||El_2|)}$
what can be confused with wrong one:
$\mathrm{...||(El_2|E|El_1)||(El_2|E|El_1)||(El_2|E|El_1)....}$
(...) means formal interpretation of a single cell in a battery.

Please describe the EXACT path of electrons in a fuel cell or flow battery. I can do this for batteries. If I place several Li ion batteries in series, I am forming a contact between the electrodes in each battery. In each battery, the cap is welded to the positive electrode. The negative electrode is in contact with the casing. When you connect the cap (+) of one battery to the casing (-) of another battery, you form an electrically conductive circuit. Or take a battery which is composed of several cells, such as a lead acid battery, I can point to the welded joint between each electrode.

@DavePiae867 but it is exactly the same for fuel cell batteries and flow cell batteries as for classical cell batteries. Just technology and geometric arrangement is different.


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Why is the ketone not attacked by the Grignard reagent/cuprate in this case but a double bond is?

According to the book the answer is (a)
Here I assume transmetallation occurs and then the nucleophilic carbon of methyl attacks the ring but what I don’t understand is why the ketone remains unaffected?

The book is correct. This is a classic example of a 1,4-addition (also called a conjugate addition) of an alkyl nucleophile to an α,β-unsaturated ketone. Normally, a Grignard reagent such as MeMgBr would indeed attack the carbonyl carbon in what is referred to as a 1,2-addition (also called a direct addition). However, treatment of the Grignard with a cuprous halide (CuI, in this case) changes the reaction entirely.
This can be explained well by hard-soft acid-base theory, or HSAB theory.
Methylmagnesium bromide is a relatively "hard" nucleophile and therefore preferentially reacts with the hard electrophile (the carbonyl carbon) in a 1,2-addition. After treatment with a proton source, product D is afforded.
But, in the presence of CuI, MeMgBr reacts to form an organocuprate with the stoichiometry of Me2CuMgBr. The details of the cuprate formation — and of the true structure of the cuprate itself — are complicated and still somewhat unclear. Nevertheless, this resulting organocuprate species is a much softer nucleophile than is the Grignard. It therefore preferentially attacks the softer carbon electrophilic site. A lone pair on the oxygen of the resulting enolate anion then reforms the carbonyl double-bond as the π-electrons of the enolate C=C double-bond attack nPrI to form a new C-C single bond, affording product A.


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Is a hydrogen atom 'locked' to a given oxygen atom in ice?
Ice, Hydrogen bond, and Ice rules Wikipedia articles seem to imply that the covalently bonded hydrogen atoms are fixed to their given oxygen atom and do not change positions as long as the ice remains solid. This would be in contrast to liquid water that self ionizes.

Is this actually true in reality? For example if you were to put a pure H2O ice crystal into a cup of water containing an increased concentration of D2O (or HDO), would the deuterium atoms diffuse into the ice crystal structure after some period of time? (presuming the temperatures were perfect such that the ice crystal did not grow or shrink) Or would the makeup of the ice crystal remain unchanged if you removed it and tested it say a month later?

Self-ionization actually still occurs in solid water. In the "standard" phase, $\mathrm{I_h}$, the autodissociation constant is $k_w = 2 \times 10^{-20}$ at -4 °C (ref). I'm guessing a Grotthuss-type mechanism to shift the hydrogens around still takes place.

https://chemistry.stackexchange.com/questions/50704/what-is-the-ph-of-ice?rq=1

https://chemistry.stackexchange.com/questions/64719/are-there-any-solids-that-undergo-diffusion

This example isn't good - crystal would melt and reform in a dynamic equilibrium. Over time you'd get pretty much a new one, except it wouldn't be via diffusion, well, not only diffusion.

@Mithoron Your comment made me consider the inverse experiment: a cube of solid $\ce{D2O}$ in liquid $\ce{H2O}$ between 0 °C and 3.8 °C (under ambient pressure). Naively, one might imagine that if H-D exchange throughout the crystal did not happen, the $\ce{D2O}$ would never melt, and if it melted, it would be evidence of such exchange. However, of course this completely disregards the surface of the $\ce{D2O}$, treating it only as a bulk solid - in all likelihood it would melt regardless. But perhaps very careful study of the melting rate could tease out the effects of exchange in the bulk.

Just as electrons exist in a "cloud", whole atoms and molecules do as well. The classic double-slit interference experiment has been performed with deuterium atoms and even molecules massing more than 25 kDa, such as tetraphenylporphyrins with fluoroalkylsulfanyl chains.
Then given that position and velocity of a particular hydrogen cannot be determined precisely, they could be interchanging continually -- though calculating the probability of such interchange is decidedly (no pun intended) difficult.
As a similar phenomenon, I find amazing the simple, inexpensive, Esaki or tunnel diode, in which electrons cross a barrier (well, disappear on on side and appear on the other) by quantum tunneling. Is that not happening to the hydrogens in different positions in ice? See this question on proton tunneling.

It is https://chemistry.stackexchange.com/questions/16592/why-is-proton-transfer-so-fast https://chemistry.stackexchange.com/questions/20377/significance-of-proton-tunneling

But even without it there would be proton transfers along h-bonds.

Transfer along H-bonds is a peculiar mode of transportation, in that it can only take your proton to the nearest oxygen and back.


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Why is the ionic product of water also the equilibrium constant of dissociation of water?
This answer presents a derivation of the value of ionic product of water at $25^{\circ}\text{C}$.
The relation $K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G^{\circ}}{RT}}$ is used for the derivation.
But why is the equilibrium constant $\mathrm{K_w=[H^+][OH^-]}$ and not $\mathrm{K_d=\frac{[H^+][OH^-]}{[H_2O]}}$ or $\mathrm{K_i=\frac{[H_3O^+][OH^-]}{[H_2O]^2}}$?
After all, the reaction $\ce{H2O <=> H+ + OH-}$ is used for calculating $\Delta_\text{r}G^{\circ}$, so intuitively it appears that $\mathrm{K_d}$ should be used.
 How can using any of the equations be equivalent when we have specified which reaction we are using while calculating $\Delta_\text{r}G^{\circ}$? What am I missing?

There is this habit to implicitly  involve [H2O] in the constant, assuming diluted solutions. It obviously fails for concentrated solutions, but then concentrations fail as well, not following activities any more.

Yes, but how is this assumption accounted for in $\mathrm{\Delta_r G ^{\circ}}$? I am not aware a similar analogue of this assumption for free energy.

These constants are not true thermodynamic constants. But ones using explicit concentrations are not either.

@Poutnik Could you please elaborate by answering this question? I don't understand.

From a strictly mathematical sense, all those equations are equivalent for pure water. Now, which one you decide to label as the autodissociation constant (as in, preferentially over the others) is another matter (and one which is largely less interesting). Don't get too worked up over labels, nature doesn't care what we call things.

The thermodynamic constant of water dissociation can be written as$$K_\mathrm{D}=\dfrac{a(\ce{H3O+(aq)}) \cdot a(\ce{OH-(aq)})}{\ce{(a(H2O(l)}))^2} \approx \\
\approx \dfrac{a(\ce{H3O+(aq)}) \cdot a(\ce{OH-(aq)})}{1^2} \approx \dfrac{[\ce{H3O+(aq)}][\ce{OH-(aq)}]}{1^2}=[\ce{H+}][\ce{OH-}]= K_\text{w} $$
$a$ as the thermodynamic activity is defined as
$$\mu = \mu^{\circ} + RT \ln a$$
where $\mu$ resp.  $\mu^{\circ}$ is chemical potential defined as
$$\mu = \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_j, j \ne i}$$
The standard chemical potential is defined for pure water, therefore activity of the pure water $a=1$.
OTOH the standard chemical potential of ions is defined as extrapolation from very diluted solutions to concentration $\pu{1 mol L-1}$, putting
$$\lim_{c \to 0}{(a)} =  c$$

minor correction - denominator should not be squared since you used H+ in numerator rather than H3O+,

@Andrew Thank you.  Well, $\ce{H+(aq)}$ implies $\ce{H3O+(aq)}$. Just because we are formally used to H+ or H+(aq), it does not mean we do not need 2 H2O for it.


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Ratio of amounts of different mono substituted alkanes (by halogenation)

My problem here is the I'm unable to use the information for reactivity of C-H bonds to calculate the amount of mono chlorides. If anyone can help me in the process, that would be great.

The relative probability of making a primary monochloroalkane related to a tertiary monochloroalkane is exactly $1:1$, out of this decane isomer.
To show how this result is obtained, the number of primary carbon atoms is counted and found to be $5$. The number of tertiary carbon atoms is counted and found to be $3$.  So the number of CH bonds is $5·3 = 15$ for primary carbons, and is $3·1 = 3$ for tertiary carbons.
As far as we can know it, the probability of removing a H atom from a primary C atom related to a tertiary atom is $1:5$.
So the relative probability of removing a H atom from a primary C atom in this molecule is $15$ x $1 = 15$. The relative probability of removing a H atom form a tertiary C atom in this molecule is $3$  x $5 = 15$.
As a consequence, the relative probabilities of forming a primary chloroalkane with respect to a tertiary chloroalkane are the same : $15:15$, or $1:1$.

Thanks man, I wasn't able to figure out what was going in the question


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Mesomeric form and organic reaction
I'm looking to find all the mesomeric form of the compound A. I don't know how to take it because there is triple bond. I think one bond of the double bond of ceton can go on the oxygen, but after that i'm quite lost.
Also, i want to find what is the product(B) after the reaction with NaH and also what this compound B can product (D) after the reaction with compound C.

Thank you for your time and your help

$\ce{A}$ is methyl propynoate. It is not a ketone.

Yes, sorry, it's an ester.

And you don't really need to think about mesomeric structures in this q.

the most acidic proton is on the end of the triple bond

If the bromide has the apparent relative stereochemistry (3S*,4R*)-3-bromo-4-methylhexane and given the branching and the presence of a secondary bromide, elimination to (Z)-3-methylhex-3-ene is a/the product D.

First stage
At the first stage of the reaction scheme, sodium acetylide B of a terminal alkyne A is obtained:

Second stage
The second stage of the reaction scheme is alkynylation of C via the SN2 mechanism with the inversion of configuration from (S) to (R) (see the image):


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How universe will end if everything is at same temperature?
I was studying the second law of thermodynamics and there was written "It implies that the universe will end in a 'heat death' in which everything is at the same temperature. This is the ultimate level of disorder; if everything is at same temperature no work can be done, and all the energy will end up as the random motion of atoms and molecules". Can anyone explain this more clearly?

What is the use of such a question that will never happen ? You could also ask what will happen in the Universe if the Sun and the stars do not exist ?

To run a heat engine you need a temperature difference. A hot and a cold side. No temperature difference means no work can be done.

@Maurice https://en.wikipedia.org/wiki/Einstein%27s_thought_experiments are of no use, for example? I would argue the application of scientific methodology to hypothetical situations is fundamental to growing one understanding of the subject.

@JonCuster That is a very good observation, power generation from thermodynamic cycles would be impossible. However, I don't think that work will disappear (as the OP's book suggests). Water can fall by nature like in dams, and we can use it to generate electric energy. Other ideas are possible. Maybe not everything is lost in this future...

Additionally, if matter flows naturally across the earth, friction will always play a part and there will be temperature differences, even though they are very small. E.g., some potential energy at the top of the dam will be converted as heat once it hits the bottom, and molecules will suffer a temperature increase once they absorb the heat.

Plainly, "Every valley shall be raised up, every mountain and hill made low; the rough ground shall become level, the rugged places a plain." If there is no potential energy difference, since all is at the same level,  no work gets done. Very much like a political entity.

@Ian Bush. You are right. Einstein's gedankenexperiment exist. But they have a use : to check the validity of the relativity theory.

Implications get everyone in trouble. The Second Law states that a system at equilibrium has a constant Free Energy and can do no work. The Laws state what happens under certain constraints or conditions, but they are oblivious to the mechanism of maintaining those conditions. The Third Law might be an exception it purports to be absolute starting at absolute zero, while sidestepping zero point energy. The Earth resides in an almost steady state of gravity and radiation. Thankfully it is nowhere close to equilibrium altho we are making a rash attempt with Global Warming. There is difficulty in predicting what happens in a simple situation here on Earth [What happens if a dead battery ostensibly at thermodynamic equilibrium, E = 0  is left too long in a device] so what hope is there to predict the fate of the Universe. It is nice to study the Universe and ponder the fate of the Universe; it is wise to consider our own fight against equilibrium and to use our own steady state wisely.

No work can be done by a system that is in a thermodynamic equilibrium.
It consists from several partial equilibrii:

mechanical equilibrium
chemical equilibrium
radiative equilibrium
thermal equilibrium.

The above textbook quote limits itself just to thermal equilibrium, that is required but not sufficient condition for thermodynamic equilibrium. At the thermal equilibrium, work cannot be done by heat engines using thermal balancing of reservoirs of different temperature.  The typical example is the closed medium Stirling engine, doing exactly that - providing work using temperature difference.
But, if some of other three conditions are not satisfied, there is still possible of a work done. Note that universe thermal equilibrium is reachable in mathematically finite, but practically infinite time.


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How to remove hexane from nBuLi
As described in a paper, the authors removed the hexane from the nBuLi solution "under reduced pressure".
I was wondering if anyone had any experience on how to proceed safely with this procedure. I was also wondering if anyone could confirm to me that nBuLi not in hexane and under vacuum is indeed stable.
Flower, K. R.; McGown, A. T.; Miles, P. J.; Pritchard, R. G.; Warren, J. E. Isolation of $\ce{1,4-Li2-C6H4}$ and Its Reaction with $\ce{[(Ph3P)AuCl]}$. Dalton Trans. 2010, 39 , 3509. https://doi.org/10.1039/b925355f.

Once upon a time solid butyllithium was commercially available, so yes it is stable if air and moisture are rigourously excluded

You state only a part of the story.  But once hexane is removed under reduced pressure (Schlenk line?), what do you want to do?  Maybe the swap of solvents (just a guess) is not necessary as nBuLi equally is commercially available as a solution in diethyl ether, THF, etc. And of course, the scale of the intended operation (both volume and initial concentration of the solution to «process» -- 10 mL 2M in hexanes is easier to manage than the 11 M stock).

Thanks! @Buttonwood This question was more out of curiosity on how the authors of the paper managed to remove the hexane from the nBuLi/Hexane solution without it catching on fire. From the other answer it seems possible they dried it on the schlenk line.


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Why isn't the dielectric constant for molecules like methanol higher than water using inductive effect?

The dielectric constant of a more polar molecule should be higher than a molecule which is less polar(polarity can be measured by dipole moment), so methanol should have higher dipole moment than water as $CH_3$ increase electron density around $O$ more than $H$ (Inductive effect), so methanol should also have higher dielectric constant than water.

Similarly dipole moment or dielectric constant of ethanol > methanol, and dipole moment or dielectric constant of dimethyl ether > ethanol. In reality the opposite of what I said happens. Why?

I read this Comparing the dipole moment of water and ethanol but, the answer given still doesn't answer the question why? He says it's because $H$ is more electropositive than $C$ but that contradicts inductive effect.

Why do you mention the dielectric constant ($\epsilon$) at all? As soon as you bring up $\epsilon$ the discussion must also address intermolecular interactions. For water this makes a large difference. Limit your discussion to gas phase molecular dipole moments. Then you compare an intramolecular property.

@BuckThorn I want to compare dielectric constants of these molecules. I don't understand how intermolecular interactions can influence value of dielectric constant, as my understanding of a dielectric is a substance with net charge 0, and the molecules have a positively charged part and a negatively charged part. The molecules can be jumbled up in many orientations. But when I apply an electric field onto the substance, all the molecules will experience the field's electric force and experiences torque until the molecule moves inline with the field. So now every molecule is in 1 orientation.

So now the dielectric substance is polarized. The resultant electric field will now be different as the molecules's electric field will interfere against the external applied electric field.

dielectric constant = (Applied electric field) / (Net electric field)
So if dipole of molecule is more, the electric field created by molecules which opposes the external applied electric field will increase and net electric field decreases so dielectric constant is more (by using the formula). So dielectric constant is only proportional to dipole moment?

Even if the molecule doesn't have a positively charged part and a negatively charged part, that is dipole moment = 0. It can still act like a dielectric, as the electric field can pull on the electrons instead and polarize the molecule.

What I meant with my comment is that you might want to limit the comparison to molecular properties, ideally obtained in the gas phase. As soon as you bring up a macroscopic property such as the dielectric constant for water and alcohols you have to take into account the structure of the liquid and intermolecular interactions such as hydrogen bonding. That doesn't mean you throw away everything you've learned. It means that things get more complicated. Perhaps someone will answer and provide an explanation.

yes i want to know how it gets more complicated. Intermolecular interactions or liquid structure shouldn't play a part I think because all that is happening is that the molecules are becoming in line with the electric field, that is the molecules stop rotating at a point where they experience no force from the field. But maybe the liquid structure or intermolecular forces can stop it from rotating? but then the substance will have dieelectric constant = 1, as there is no opposing electric field from the molecules which doesn't happen here with these molecules.