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Are double spike isotopic standards suitable for quantification by isotope dilution mass spectrometry?
I would like to apply isotope dilution mass spectrometry (IDMS)* to quantify lead at trace levels. A good spike solution for the process would be the NIST SRM 991 which is an enriched material of almost only lead-206, but this material is now discontinued and the replacement is a radioactive material NIST SRM 983, which I think, will be a little difficult to import.
The NRC of Canada has produced the lead-204 and lead-207 double spike isotopic standard BLED-1 which contains 49% and 33% the two isotopes, respectively, and I was wondering if it could be suitable for IDMS considering that the natural relative abundance of those isotopes is low. I have only performed IDMS using the (single spike) NIST SRM 991, and although I think the same IDMS equations will apply to each pair of isotopes, would like to know please if there is something I am not taking into account.
*) An overview about this technique is provided by Vogel, J.; Prizkow, W. Isotope dilution mass spectrometry — A primary method of measurement and its role for RM certification. MAPAN 25, 2010, 135–164, doi 10.1007/s12647-010-0017-7. The author's copy on ResearchGate.

This is a good and well-asked question. Unfortunately I am not of the field and cannot answer it, but I hope some expert will. I'd like to take the opportunity though to point you to a proposal for a new site about mass spectrometry.

Yes, I think it would be suitable, although I am not an expert in IDMS either.  It seems like the new standard contains two stable isotopes.  This should give you a built-in check on the data you collect: if you treat Pb-204 as the spike isotope, and calculate the concentration of Pb in your sample from it, you should get the same (or very similar) answer as if you treat Pb-207 as the spike isotope.


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What molecules other than N$_2$ have wrong orbital ordering when calculated using Hartree-Fock?
N$_2$ is a canonical example of this aspect of the failure of HF method, for instance, see table 4.9 in "Modern Quantum Chemistry" by Szabo and Ostlund. Here it is shown that in HF, the HOMO of N$_2$ is of $\pi$ symmetry, whereas correlated method and experiments show that its HOMO should have $\sigma$ symmetry, which becomes HOMO-1 in HF.

Maybe this answer on a related question helps: https://chemistry.stackexchange.com/questions/154162/is-strength-of-pi-bonds-greater-than-sigma-bond-of-nitrogen-molecule/155418#155418

The linked question discusses N$_2$ only.

Yes, but the linked answer discusses the notion of "wrong" orbital ordering and the limits of the interpretability of orbitals in general.

The idea of the question is to address the symmetry of the highest occupied orbital in N$_2$ in HF method. The fact that there is a "experiment" column in the cited table above indicates that experiments can actually identify the symmetry of the lowest ionization potential of N$_2$, which is associated with the HOMO of the neutral. The next ionization potential is associated with ionization that brings the cation to its first excited state, and in HF this is approximated by the removal of HOMO-1 of the neutral.

Therefore, despite the idea of orbital being unphysical in the exact sense, it still bear certain degree of information in the approximate single determinant sense.

Yes, that is Koopman's theorem. So by "experiments show that its HOMO should have \sig symmetry" you mean, that the ion has \sig symmetry?

@libavius it is called Koopmans' Theorem as the s is part of the name.

@Libavius  I should have been more specific. The kind of ionization I was referring to is tunneling ionization, not single-photon ionization. The former requires no resonance between ionizing photon energy and neutral's energy levels but intensity must be high. In tunnel ionization, you are most likely ionizing the valence orbitals. Hence the $\sigma$ symmetry encoded in the experimental ion yield as a function of polarization angle of the laser comes from the symmetry of the HOMO of the neutral N$_2$.

The HOMO of N$_2$ obtained through HF has a $\pi$ symmetry, which is therefore inconsistent with experiments. However, the HOMO of this molecule obtained through  DFT (hence the concept of occupied orbital is still well-defined) using good functional has $\sigma$ symmetry. In this respect, tunneling ionization calculation on N$_2$ is better calculated using DFT rather than HF.

I haven't done correlated wavefunction methods such as CI for a while now. But as I recall even for these multideterminant methods where the number of occupied orbitals can be more than the number of electrons, you still typically have $N$ orbitals (where $N$ is the no. of electrons) whose occupation number is significantly higher than the rest of occupied orbitals, indicating that in these methods the HF characteristic is usually somewhat prominent.

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Can oxidation and reduction be described in terms of transfer of oxygen in the following way?
After reading and thinking about oxidation and reduction a bit, I tried to come up with a way to explain it to myself… Could someone verify if this is technically correct?
Consider 2 elements A and B where A is more electronegative than B.
(Also O represents Oxygen)
If in a reaction
$$\ce{AO + B -> BO + A}$$
So assuming that O is more electronegative than both A and B
In AO,

Oxidation Number of O = -2
Oxidation Number of A = +2

So we could say that O is “negatively charged” (or partially negative if it’s a covalent bond) while A is “positively charged” ( or partially positive if it’s covalent) and hence this is why they are attracted to each other (This part I’m not sure). Now given that A is more electronegative than B, when B comes near AO, A attracts electrons from B and oxidises B.

Oxidation Number of A gets reduced to 0 (A is reduced)
Oxidation Number of B increases to +2 (B is oxidised)

Now we could say that A is now “neutral” and B is “positively charged” while O remains “negatively charged”. Hence O is attracted to B forming BO.
Thus
$$\ce{AO + B -> BO + A}$$
From the above, we can thus say that a loss in oxygen causes reduction and a gain in oxygen causes oxidation.
Is this how redox reactions work or am I exaggerating everything?

You may overthink it. Why not to stay at oxidation/reduction as change of oxidation numbers, what is not limited to oxygen presence?

Your explanation reminds me of the historical definition  of the words "oxidation" and "reduction". In the early 19th century, oxidation was the reaction an element with oxygen. The only thing that the chemist could measure with precision was the weight, the mass, before and after the reaction. When iron or copper gets oxidized, it produces oxides which are heavier than the element. Now when one of these oxides reacts with hydrogen, it produces the original metal, but its weight is reduced. So the chemist said that this oxide has been reduced in metal by hydrogen. The oxide has lost some mass to be transformed into the metal. Its mass is reduced. So the element is also "reduced".

With the Latin root ducere (leading), one may understand reduction not only as action to diminish/to scale something down, yet equally to guide back/to return something into its earlier (elemental) state. Like education rooting in educere, to lead outwards.

I initially thought of reduction occurring when oxidation number got reduced to a lower value :)

@Unknown. You are right. But this is the present definition of reduction. In the time of Berzelius and Dalton, oxidation numbers were not known.

An easier approach would be to think about redox-reactions as an exchange of electrons, rather than of oxygen.  This is more general and equally allows oxygen to be oxidized (e.g., fluorine is even more electronegative than oxygen, as in oxygen fluorides).
Retain that an oxidation is the removal of electrons, and reduction is the addition of electrons. The «who gets the electrons» depends on the partners reacting with each other. In an allegory, think about an ox whose horns pierce and remove material from a tree (oxydation).  And think of reduction as bringing electrons back (based on the Latin root reducere) to heal the wound.

You can find examples where your thoughts fit what is actually happening. For example, a reaction of magnesium oxide with barium (I don't know if this reaction would actually happen):
$$\ce{MgO(s) + Ba(s) -> Mg(s) + BaO(s)}$$
Here, it is correct to speak of the negative and positive charges because we are looking at ionic compounds. You could show the ions explicitly:
$$\ce{Mg^2+ + O^2- + Ba -> Mg + O^2- + Ba^2+}$$
On the other hand, you could have a redox reaction involving molecular compounds:
$$\ce{H2O + C -> H2 + CO}$$
Here it does not make sense to talk about charges of atoms in the compounds (partial charges maybe, but they are also not necessarily correlated with oxidation states). The easiest way to figure out oxidation states in this case is to define the oxidation state of oxygen as -2 (except when oxygen is bound to itself or to more electronegative atoms), and use that to assign +1 to hydrogen in water and +2 to carbon in carbon monoxide.

From the above, we can thus say that a loss in oxygen causes reduction and a gain in oxygen causes oxidation?

No, unless this is the only change and the binding partner of the oxygen is less electronegative, as in your examples. Here is a counterexample:
$$\ce{F2 + O2 -> O2F2}$$
When you add water across a carbon-carbon double bond, this is usually not called oxidation either, even though one carbon gains a bond with oxygen and is oxidized. The reason it is not called oxidation is that at the same time, the other carbon gains a bond with hydrogen and is reduced.

Is this how redox reactions work or am I exaggerating everything?

This is how a subset of redox reactions work. In many redox reactions, there is no oxygen at all, e.g.:
$$\ce{Fe^2+ + Cu^2+ -> Fe^3+ + Cu+}$$
For a general modern definition of a redox reaction, we look at the transfer of electrons (as opposed to transfer of protons in acid/base reactions).

Yes I do know and agree that electron transfer is an easier way to understand redox. However for exam purposes I am supposed to learn redox in terms of gain/loss in electron, oxygen and hydrogen. So given that I am talking only when gain in oxygen does cause oxidation, then is what I said applicable? I was specifically concerned with whether when B is oxidised when it gains an oxygen, it’s not the oxygen which gains the electrons but rather it’s A? And for reactions where a gain in oxygen does cause oxidation, is what I said above applicable or are some details too frivolous ? Thanks

For covalent substances, especially organic compounds, counting bonds with more electronegative atoms (oxygen) vs less electronegative (hydrogen) works. In counting, you have to count double bonds twice, though. Artificially, you are breaking all the bonds, giving both electrons in a bonded pair to the more electronegative atom.

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Average distance of 1s electron from nucleus in He⁺ ion

What would be the average value of $r$, i.e.  $\langle r\rangle$, in the $\mathrm{1s}$ orbital of $\ce{He+}?$
$$
\text{a}.~\frac{3}{2}a_0 \qquad
\text{b}.~\frac{3}{4}a_0  \qquad
\text{c}.~3a_0  \qquad
\text{d}.~\frac{1}{2}a_0
$$

I have written the normalized wavefunction of $\mathrm{1s}$ orbital of $\ce{He+}:$
$$R_{(1,0)} = \frac{2 \sqrt 2}{\sqrt{\pi a_0^3}} \times a^{-2r/a_0},$$
but I could not proceed further.

The average value of $r$ can be found out, in case of $\text s$ orbitals by multiplying the volume of each thin spherical shell by the probability density at that $r$ and adding them up.
This can simply be accomplished by using integration as a limit of sum. We know that the probability density $\mathrm dP/\mathrm dV = R^2$. So,
$$
\begin{align}
⟨r⟩ &= \int_0^\infty r\,\mathrm dP \\
⟨r⟩ &= \int_0^\infty rR^2\,\mathrm {d}V\\
⟨r⟩ &= \int_0^\infty r\left(\frac{2 \sqrt 2}{\sqrt{\pi a_0^3}} \times e^{-2r/a_0}\right)^2\left(4\pi r^2\,\mathrm {d}r\right)\\
⟨r⟩ &= \frac{3a_0}{4}\\
\end{align}
$$


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Producing hydrogen sulfide—difficult to do much?
When I was young (nearly sixty years ago), my "chemistry set" taught me how to heat sulfur and candle wax to stink up the house.  It had no warning that hydrogen sulfide is toxic and explosive in large quantities.  Is that because people weren't as litigious back then, or because producing large quantities is hard to do?
The reason I ask is that near here, we had a major issue with hydrogen sulfide coming out of an SUV and affecting three first responders who tried to rescue the driver (deceased).  It was announced as a suicide, but I thought if a fellow wanted to end it all, that doesn’t seem like an easy way to do it.
And there is at least one known case of deaths in an SUV from hydrogen sulfide.  I’ve read a few other reports of problems from people leaving the vent hose off of a battery in the passenger area of a car.  Maybe this week's incident was not a suicide.  (Maybe putting a lead-acid battery inside a passenger compartment is a really stupid thing to do.)

Sadly, there are quite a few cases like you mention in the first sentence of your third paragraph and the less said about the whole thing, the better. Hydrogen sulfide has the unfortunate property of being extremely easy to detect by smell, until, shortly later, it cannot be smelled at all.

Toxic as it is, H2S is much more smelly than it is toxic. The idea was that the smell will compel you to stop heating before you have a dangerous concentration around. Also, a small amount of sulfur in a beaker won't produce much H2S in any case. Also, a house is a lot bigger than a car. So no, I don't think it presented a serious danger.

As @EdV states, the nose becomes accustomed to H2S -- see how H2S can cause anosmia: https://pubmed.ncbi.nlm.nih.gov/10805151/ . Though the first whiff can serve as a warning, our olfactory sense is not good at measuring the amount. However, the tiny bit of gas made from paraffin and a sprinkle of sulfur is harmless -- a bigger danger is igniting the furniture with a candle.

oh those old Chemistry sets were amazing and wondrous things, then came the personal injury lawyers (and probably some serious personal injuries)

The problem with H2S is that it is both highly toxic (albeit not quite at the level of HCN but surprisingly to me, too close) and  with uncommon properties such as at a threshold of exposure, a suppression of the ability to detect any smell.
The latter is particularly pernicious as one can, apparently, achieve a toxic dose while not even realizing that one is continued to be exposed to the extent of having actually received a fatal dose (reputedly, a 10 minute exposure at 50-100 ppm destroys the sense of smell and thus results in the loss of ability to sense its presence through smell).
Also, pathways of fatal ingestion also relate to a limited extent to skin contact, albeit sewer workers wearing a gas mask with extended exposure could, for example, actually fall victim to this deadly poisonous gas.
The victims with fatal doses have actually been referred to as "the walking dead" as they will, progressively sicken over a course of hours, suffering irreversible organ damage leading to  death.
I agree that general knowledge and labeling on H2S is deficient, and that there is an issue, as you noted: "It had no warning that hydrogen sulfide is toxic and explosive in large quantities".
I also view this a failing relating to the quality and availability of knowledge in the field of inorganic chemistry.  In the case of poisoning in a car with a battery, this knowledge base apparently extends to electrochemistry as well.
As such poisoning from H2S is, in certain locales (Japan being an exception), in my opinion, can be more likely be the result of absence of knowledge and proper labeling than intentional suicide.
Some confirming references include to quote:

Hydrogen sulfide is a highly toxic gas—second only to carbon monoxide as a cause of inhalational deaths. Its mechanism of toxicity is only partially known and no specific therapy exists for sulfide poisoning....No antidote is currently available for sulfide poisoning and treatment is largely supportive.

Per the CDC, to quote:

Hydrogen sulfide is well absorbed through the lungs; cutaneous absorption is minimal. Exposure by any route can cause systemic effects...However, although its strong odor is readily identified, olfactory fatigue occurs at high concentrations and at continuous low concentrations. For this reason, odor is not a reliable indicator of hydrogen sulfide's presence and may not provide adequate warning of hazardous concentrations...Inhalation of high concentrations of hydrogen sulfide can produce extremely rapid unconsciousness and death.

I appreciate all this info, and suspect the authorities were too hasty in calling the event a suicide.  However, that wasn't exactly my question ()in bold in the OP and in the title).  But I did learn a bit more here, so I appreciate it.

I doubt the article is accurate. Hydrogen sulfide is noticeable at less than 1 ppm and  up to 100 ppm causes only sinus irritation in a half hour exposure. It is very dangerous as 150 ppm can cause loss of smell sense in minutes. H2S is common in oil/ gas production so oil companies and governments have extensive rules concerning it. In particular it causes stress corrosion cracking of high strength steels so NACE MR 01-75, now ISO 15656 has detailed rules for H2S. Internal combustion engines do not produce it in quantity. I would like to see some technical analysis of how a lead acid battery can produce H2S in quantity. Also prior to 1940 , it was not unusual for an auto battery to be located under the floor of a car. I have seen a later model Cadillac with the battery under the back seat so apparently that is not a unique location. I have never seen another report blaming battery location for mysterious deaths. I got some information from SPE Sour Gas Design Monograph , 1993 ; I doubt it is readily available. Pardon my ramble but I like H2S it provided me a large portion of my career. PS ; From what I find on the net , in the US sulfur is limited to 15 ppm in gasoline, there are exceptions for off-road and other categories.

Sorry, some research and experience with H2S (and lack of knowledge in the general population) a critical aspect, which I have experienced (that is with a total suppression after some level of exposure to continue to smell the gas at all) is common. Even more problematic, this can result in achieving a fatal dosing and not immediately realizing that you have received a fatal dose. Actually referred to victims as the "walking dead" as a irreversible successive occurrence of organ failure resulting in death over time.


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How to Validate Choice of DFT Functionals and Basis Sets for atomic charge calculation of Fluorocarbons
I need to calculate the charges (electron densities) on each atom of a set of partially fluorinated hydrocarbon molecules by NPA. The question arises:
What is the currently accepted / best practice method for validating these calculations as a function of Functional and Basis Set for DFT based calculations? Since no "experimental" values for electron densities are available, I have tried evaluating the "correctness" of the chosen functional and basis set by correlating the predicted bond lengths and bond angles to experimental values. However, this seems to be far too simplistic to provide a meaningful estimation of the correctness of the choice.

As you said in the question, electron densities and atomic charges do not have any strictly defined experimental measurement. In such cases, I often see people comparing results to a high level method, such as CCSD(T). You can try that if your molecules aren't too large. I have also heard of experimental estimation of atomic charges using NMR, I don't know how reliable or useful those are, but it's worth looking into.


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Reaction kinetics of exothermic reaction
A colleague said we can’t dissolve a salt (whose solvation enthalpy is exothermic) faster if we increase the temperature (the solubility equilibrium product is not reached) because Le Chatelier‘s principle would favor the reactants.
For example, imagine dissolving NaOH(s) in distilled water, can’t this be accelerated by elevating the temperature?
I read a similar question on here, see: Factors that influence the kinetics of an irreversible exothermic reaction
How I see it is the following:
As Le Chatelier‘s principle only applies to systems in equilibrium, and we only dissolve a salt (whose enthalpy happens to be exothermic), no equilibrium is established. The same goes for exothermic reactions in general. As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless whether it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in the Arrhenius equation).
It is important to note that this is about kinetics, that is, how quickly the NaOH dissolves, in a non-saturated solution, that is about 1M, not about how much NaOH a saturated solution can contain at a given temperature.
Is my reasoning correct?

Check this discussion: https://web.archive.org/web/20150228170949/http://www.newton.dep.anl.gov/askasci/chem99/chem99198.htm

Does this answer your question? Why is heating increasing the solubility of sodium hydroxide?

Not really, as the most upvoted answer talks about maximum solubility of a concentrated solution of NaOH, not about if a temperature increase dissolves a certain amount of NaOH (solid) faster

A well-known and intuitive counter-example: burning (of everything). The reaction is exotermic and pretty much increases with temperature. This is how fires self-accelerate when not controlled.

In this case, whether it is endothermic or exothermic depends on the concentration because a saturated solution is not an ideal solution, chemistry.stackexchange.com/a/4464/72973

Typically, rates of uncatalyzed simple reactions increase with temperature. There are well-known examples where this is not the case (such as enzyme-catalyzed reactions where the enzyme denatures at high temperatures, or reactions with an intermediate that is at rapid equilibrium with the reactant in an exothermic step).
In this case, if the $\ce{NaOH}$ is soluble at both temperatures, it is likely that the increasing rate in the forward direction will make a bigger difference that the also increasing rate in the reverse direction.

As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless if it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in Arrhenius equation).

In this case, the reaction goes to completion rather than attaining equilibrium (if you stay below the solubility limit). Even for reactions that attain equilibrium, however, the rate with which they approach the equilibrium constant is proportional to the sum of the forward and reverse reaction (for a one-step reaction closes to equilibrium, it is possible to show this in a quick derivation).

Our current understanding of chemical kinetics is that all reactions with an activation energy are increased in rate by an increase in temperature. The question that is pertinent is "Which reaction is accelerated faster?". A temperature change at equilibrium influences the equilibrium and the overall composite rate.  Whether the overall direction of a given reaction is accelerated by an increase in T is determined by the actual value of delta G; if negative it is accelerated as written.
The choice of NaOH as an illustration is a bit unfortunate. As anyone who has tried it knows adding water to solid NaOH results in a severe exothermic reaction even to a boiling solution. Yet NaOH is more soluble at higher T. This is contradictory to the idea that an exothermic reaction is lessened at a higher T. What happens is that the dissolution reaction becomes endothermic as the concentration increases; don't ask me how or why.
The general answer: Displaced from equilibrium any reaction is accelerated by an increase in T provided there is an activation energy. At equilibrium the increase of both rates moves the equilibrium in the direction of the reaction with the higher activation energy. If the activation energy is vanishingly small the problem becomes removal of energy and the reaction doesn't exist.


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What is reversible change for ideal gas?
What are the reversible and irreversible changes for an ideal gas? Is it the change that keeps the ideal gas law during the change? Is any reversible change a sum of changes satisfying the ideal gas law $(pV = nRT)?$ If so, it looks to me like any change is reversible if it is done very slowly.

The concept of ideal gas has no bearing on this whatsoever. You are right, though, in that the changes which are done very slowly tend to be reversible.

See Reversible_process

thank you. Because the word "reversible" always comes with another word "isothermal" in my textbook, I was initially under impression that all the reversible changes were isothermal ones.


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Difference between threo/erythro and cis/trans nomenclature
According to what I have read, it seems that erythro isomers are defined in the same way as cis isomers and same holds true for threo and trans isomers. Is there any difference between these two set of terms or they mean one and the same thing?

You're mixing up geometrical and optical isomerism.

Imagine an aldol reaction--not condensation--between an enolate of 2,2-dimethylpentan-3-one and benzaldehyde. Two racemic aldols, displayed here as single enantiomers, may be formed,--namely, the anti and syn isomers when the carbon backbones (in red) are drawn in a zig-zag fashion. While these two diastereomers look like trans and cis isomers, respectively, these terms are not applicable because they are a function of how the structures are drawn. Heathcock[1] defined these isomers, respectively, as threo and erythro because the substituents were on the opposite side in the former and on the same side in the latter as in threose and erythrose (see footnote 5 in ref. 1). This choice of terminology was unfortunate in that conversion of the zig-zag conformation to the Fischer projection gives the opposite assignments of erythro and threo. The bottom line is ignore erythro/threo and use anti/syn.

The cis/trans terms are applicable to substituents on rings as is illustrated for the 1,2-dimethylcyclohexanes (red box). Trivial 1,2-disubstituted alkenes are often referred to as cis/trans isomers but Z/E is preferable.

C. H. Heathcock, C. T. Buse, W. A. Kleschick, M. C. Pirrung, J. E. Sohn, and J. Lampe, J. Org. Chem., 1980, 45, 1066.


text:
Pcb etching safety questions (HCl, H2O2, H3PO4)
I am about to embark on a pcb etching journey, however, I am not sure what safety precautions I should take when handling the products. I know that the reagents should be stored in their original, properly labeled containers, and that normal PPE rules apply. Lab coat, gloves, goggles etc.
What I am unsure about is, is weather or not the stabiliser used in the hydrogen peroxide, phosphoric acid, would be of any concern.
The reaction I am aiming to achieve is the conversion of copper to cupric chloride (Cu(II)Cl2), in a HCl and H2O2 solution. The end product will be stored, as CuCl2 is also a good etchant, and converts into Copper I Chloride when used as such. My main concern is the stabiliser H3PO4 (I assume, bottle just says phosphoric acid,in German phosphorsäure (AB.)) used in the peroxide.
Would this reaction or the following products create phosphorus fumes?
Would the phosphoric acid create gas in product solution as it is stored?
Is there any possibility of the material dissolving or damaging Polypropylene and Latex?
What extra precautions, other than PPE and workplace safety (I. E. Covering the work surface, doing the reaction next to an open window and common sense) do I need to consider?
I appreciate your help, input advice and improvement ideas!
Thank you!

Phosphoric acid is $\ce{H3PO4}$ and is one of the ingredients in cola soft drinks, among many other things. There will be no “phosphorus fumes”. Ps. Thanks for spelling phosphorus correctly!

Thanks for the answer. I'll let my autocorrect know, that you appreciate the spelling lol! I know that phosphoric acid is found in soft drinks. However, knowing that certain mixtures can react in an unexpected way, I thought I ask about possibilities. I am mainly thinking about how aqua Regia can dissolve gold, but the components can't.

Phosphoric acid is $\ce{H3PO4}$ and not $\ce{H3PO3}$ as given in the title of the question. It is a stabilizing agent of $\ce{H2O2}$, but it will not  produce any effect in the reaction of $\ce{HCl + H2O2}$ on copper. It will not react with polypropylene and latex. It will not produce any gas in your mixture. You don't need to consider extra precaution with your experiment. Simply don't put it on your skin or on your clothes. $\ce{HCl}$ and $\ce{H2O2}$ are corrosive.

Thanks a lot! I wasn't sure, if any further reactions of the products would be harmful. However, now this begs the question of recycling the chemicals....But that's still far off in the future and I will ask again if I don't find anything online. Thank you a lot for your answers!


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Does the phosphoric acid in drinks lead to leaching chromium from stainless steel 316?
I had considered using a small SS 316 pipe as a straw for sodas. Question being, would the phosphoric acid in drinks leach out chromium, and if so would it be carcinogenic? Hexavalent chromium is considerably carcinogenic.

It is pretty hard to oxidize chromium to Cr (VI) and phosphoric acid is not an oxidizing agent. If you check the chemical compatibility chart of SS 316, phosphoric acid is not recommended to be used with SS 316. Regardless of Cr, sodas are far more dangerous in terms of sugar content/artificalsweetners, than chromium or its compounds. Human body was not meant to consume free phosphoric acid, otherwise Nature would have provided it in fruits.

Many years in corrosion and never encountered leaching of chrome from stainless.  Some operations included moving hot super phosphoric acid around in 316 L lined rail cars.

The point of stainless steel is being highly resistant to acids...

Not all stainless steel are resistant to acids. SS 316 in question is an example.

I would say all are resistant, not immune  Concentrated nitric has no affect on 13 Cr  ( at least during a spot test).

@blacksmith37 Sometimes dilute acids will corrode, and stronger versions cause passivation

I don't think that whatever would or could leach from these pipes would be more hazardous than the drinks themselves. At least not in the concentrations expected from the leach process.

Technically, Phosphoric acid in drinks is not leaching Chromium from your Stainless Steel, but is contributing agent to its gradual galvanic corrosion. The process requires an anodic metal (found in steel alloy), source of H+ (acid presence), oxygen (air exposure) and a electrolyte (from the dissolved salts in the drink are the electrolyte.
Usually the corrosion takes some time (as in the case of my steak knives which were the victim of so-called pit corrosion from cutting a mango apparently rich in ascorbic acid (discussed here) that were left sitting in the sink for a few days), actually, more like a consequence of neglect.
Now the case with phosphoric acid drinks is more interesting because normally they have a dark coloration. The latter implies dyes which in sunlight generally possess some photocatalytic abilities. The latter can introduce solvated electrons and electron holes leading to radical formation with water (hydroxyl radical) and in this case with phosphoric acid, the phosphoric radical, which also could potentially be more active here than your drinks original dilute phosphoric acid.
Good news is that since this is electrochemical reaction, it is the iron metal (likely the most anodic metal present) which first corrodes (forming deeply colored iron salts) and not per se, the chromium metal (albeit, it may eventually be attacked when the Fe is depleted).
So basically you had all the required set of reagents (and may be even catalytic lighting) for a nice galvanic corrosion experiment, although completely unintended.

Ascorbic acid does not corrode stainles steel.

Thanks for reading my answer and your comment reflects the need for a source (apparently other my personally observed issue with my stainless steel steak knives) which I have now inserted with other sources as well. Note, the role of NaCl per the technical reference as well (likely in the drink).

Relating to Ascorbic acid/NaCl corroding stainless steel, may I suggest PICTURES here http://www.sciencemadness.org/talk/viewthread.php?tid=152955#pid620207 .

More technically correct, Ascorbic acid (A.A ) with NaCl in tap water with A.A at higher  concentrations may induce a chelating effect resulting in severe pit corrosion here.

With respect to my answer to the question, it remains correct, the reported corrosion is not the action of any one acid. Likely a more complex pathway leading to an assisted corrosion action.

Note, this is not a matter for a corroding acid as noted above, to quote: "Many years in corrosion and never encountered leaching of chrome from stainless. Some operations included moving hot super phosphoric acid around in 316 L lined rail cars. –
blacksmith37" which is clearly supports, instead, the literature noting a complex galvanic corrosion pathway for stainless steel with, for example, fruit acids and NaCl in transition rich tap water, and here via another acid.


text:
How can one determine if a chemical reaction is impossible to balance?
I asked this question on Math Stackexchange, but thought I'd get a more chemical view:
https://math.stackexchange.com/questions/4326904/determining-whether-or-not-a-system-of-equations-has-nonzero-solutions
I understand that for redox reactions in acidic conditions, water molecules and hydrogen ions can be added to a reaction in order to make balancing it easier. However, my teacher specifically told us that we should only add in water molecules and hydrogen ions if and only if at least one of these conditions are met (the accuracy of this may be debatable, but he specifically told my class that marks will be taken off on assessments if we do not follow this instruction):

There are zero hydrogen atoms/ions on one side of the reaction
There are zero oxygen atoms/ions on one side of the reaction
The reaction is impossible to balance by manipulating coefficients

The third point is what is giving me trouble, and why I asked for a mathematical explanation. How can we just look at a reaction and determine whether it is possible to balance or not?

Balancing just means that every atom on the left side appears also on the right side and net charge on the left side equals net charge on the right side. It does not mean that the balanced chemical equation is realistic or even plausible. The third condition may just mean that it is assumed that an acidic aqueous solution has water and hydrogen ions freely available for balancing purposes.

I stress that the teacher was just giving a set of "to do list". The meaning is in fact that all chemical reactions conserve atoms and charge and, thus, are inherently balanced. Further, all the linked threads have nothing to do with chemistry, in one case blatantly, in the other case is just simple mathematics that, perhaps disguised by the use of chemical symbols, was left open in Maths SE.

The reaction is impossible to balance by manipulating coefficients

I feel you over-interpreted the statement. Instead of a pure mathematical reasoning, think in terms of mass balance. New atoms cannot be created or destroyed during a chemical reaction.
Let us see the classical $\ce{Fe^{2+}}$ and $\ce{MnO4-}$ reaction.
The case of $\ce{Fe^{2+}->Fe^{3+} + e-}$ is trivial.
For $\ce{MnO4^{-}-> Mn^{2+}}$
Now what choices we can have? We have to rely on chemical knowledge and experimentally observed facts (i) the reaction is carried out in acidic medium and (ii) Mn(II) is formed.
Since oxygen atoms cannot be destroyed, and we know by experimentation that the reaction has to be carried out in acidic medium (in basic medium this reaction does not proceed), so $$\ce{MnO4^{-} + H+-> Mn^{2+} + hydrogen containing molecule}$$
What choices do you have expect to add water on the right hand side? Oxygen has to go somewhere. This is what your instructor meant.
I know a lot of mathematically inclined chemists have developed matrix algenra methods, but a matrix can only be set up when you know beforehand what products and reactants are there. Mathematics cannot tell us which molecule will be formed.
Go to Google Scholar and search balancing chemical equations + matrix.

Before any chemical equation can be balanced the reactants and products must be defined; otherwise, it is just a math problem and might not even be possible.  A simple reaction:  stannous ion is used to reduce ferric ion to ferrous ion in some iron analyses:  Fe+3 + Sn+2 = Fe+2 + Sn+4. This equation is easy to balance and does not require any water even though Sn+4 hydrates and water is most likely involved in the reaction. 2Fe+3 + Sn+2 = 2Fe+2 + Sn+4.  Lets write a wrong equation without oxidation and reduction: Fe+3 + Sn+4 = Fe+2 + Sn+2! Now try to balance it.
Redox reactions with poly atomic ions such as MnO4- or CrO4= in water solution need acid to react with the oxygens. This acid comes from H+ in acidic solution or H2O in neutral or basic solutions. The acids take part in the reaction and are consumed; that is the main reason that they are used to balance the equation, not just to make it easy.  Fe+2 + MnO4- = Fe+3 + Mn+2 to remove 4O= you need 8H+. 8H+ + Fe+2 + MnO4- = Fe+3 + Mn+2 + 4H2O. The manganese is taken care of but we have 4 extra + charges on the left each Fe+2 adds a +charge to the right so add 4Fe+2 and 4 Fe+3 to give:  8H+ + 5Fe+2 + MnO4- = 5Fe+3 + Mn+2 + 4H2O.  Balanced with no need for oxidation numbers or half reactions [although they can and do help].
A solid-state reaction, here there is no water or acid and I making up a possible reaction:  FeO + KMnO4 = MnO2 + Fe2O3 + K2O.  This balances easily using changes in oxidation numbers but lets do it chemically. First thing there are odd-even problems so fix them. 2FeO + 2KMnO4 = 2MnO2 + Fe2O3 + K2O; 10Os on the left 8 on the right. Each 2FeO adds one O so we need 4 more FeO!  6FeO +2KMnO4 = 2MnO2 + 3Fe2O3 +K2O.
The point of all this is that balancing equations really is about learning chemistry, identifying reaction conditions and determining all the products of a reaction; it is not learning tricks and rules. Do learn the various methods, oxidation number change, redox half reaction balancing, the chemical method I illustrated that is particularly useful in changes in functional groups in organic compounds.

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@jimchmst. $\ce{FeO}$ and $\ce{K2O}$ are not common substances at all. They are difficult to obtain and will not be available in the usual laboratory or by mixing aqueous solutions.

FeO is far from uncommon. It's seen all the time as part of the scale formed on hot-rolled steel (the rest is mostly a different iron oxide, Fe3O4).

I stated that I made up the reaction to demonstrate that an equation could be balanced easily if all the products and reactants were listed. That is a "possible" solid state [possibly in a melt] reaction and invoking acid or water is not appropriate in an attempt to balance.

The key trick to balance an equation that may involve, for example, a Mn compound undergoing a change in oxidation states from the left to the right side of the equation, is to focus on the fact that the same # of Mn atoms must appear, in total, on both side of the equation. The adjustment factor to the Mn compound, for this to occur, should be somewhat apparent (for example, going from 2 atoms to 3, requires 3 and 2 coefficients to equate each side at 6). Then use this information to assist in computing the required # of oxygen and say hydrogen atoms appearing outside of the Mn compound to arrive at a total balanced equation. This can be may easier, at times, by also noting that each side of the equation's total charges must also be equal (if the sum +/- charges is +1 on one side, it must match the other). Note, that an electron has a charge of -1 and two electrons -2, etc.  Also, atom count and charge balance are both required (albeit, the latter often not emphasized as often naturally occurring) for a valid equation, in accord with the preservation of matter and energy.
Do your best to follow the teacher's rule, however, the last rule "The reaction is impossible to balance by manipulating coefficients" requires adjusting in practice.
If the actual resulting mathematics indicates answers for each coefficient that are unexpectedly large (irrational), present the results as a raw unbalanced equation. Clearly, some inputs are missing or not applicable for the reaction you are trying to balance (that is, to correctly explain).
Not surprising, there is a problem with universal rules for balancing chemical reactions, as all reactions are not transparent as to how precisely the observed products were actually formed.
This is evident, for example, when the experimentally estimated coefficients are decimal amounts. Evidently, the suggested reaction equation may a summary of several random occurring more elemental reactions (usually, these can be radical reactions, which are reversible and have temperature sensitive forward and backward reaction rates).
A larger problem arises when the reaction occurs in an open system with say the amount of oxygen from air placed into the system is not known. In case again of say fractional weights and some inputs unknown, how does one present results?
My recommendation is honest reporting, as an unbalanced equation, or otherwise, the ability to speculate on true reaction pathways could be lose under a perception of false certainty.
Per examination of the literature, this problem is also resolved by stating "+ other products". The latter is acceptable, in my opinion, as an honest statement of lack of knowledge as to underlying mechanics and/or product estimate accuracy issues.
Hope this helps!


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Why do diluted sulfuric acid and hydrogen peroxide etch copper only when chloride ions are present?
I've mixed 30 ml 10% sulfuric acid with 20 ml 3% hydrogen peroxide and put a drop of this solution onto copper foil. The solution quickly removed the oxide layer from the copper but did not dissolve it any further within the hour that I left the drop on the copper. When I wiped the solution off with a white paper towel, I couldn't see any blue or green tint, indicating that there was (almost) no copper dissolved in the solution.
After that, I added (roughly) 100 mg of sodium chloride to the solution and again put a drop of it onto the copper foil. (To clarify, I put 100 mg sodium chloride into the entire remaining ~50 ml of solution.) It began to oxidize and dissolve the copper, turning its surface brown. When I later wiped the solution off, the paper towel was stained greenish-blue.
A dilute mixture of hydrochloric acid and hydrogen peroxide also readily dissolves copper.
I've been unable to find any explanation for this even after an hour of googling. I've found this Chemistry Stack Exchange answer that deals with a similar reaction; however, it doesn't explain the role of the chloride ions.
Why does this solution only dissolve copper when chloride ions are present? (That is, how does the chloride help oxidizing the copper and why isn't hydrogen peroxide enough?)
I've got three theories, but I have no idea which of them (if any) is correct. (I'm sadly not a chemist.)

The hydrogen peroxide oxidizes chloride to $\text{Cl}_2$, which then attacks the copper. The problem with this theory is that hydrogen peroxide has $E^0=1.776V$ in acidic solutions while chlorine only has $E^0=1.396V$, meaning that the peroxide should attack the copper more readily than the chlorine, yet it evidently doesn't.

The $\text{Cl}_2$ produced as in theory #1 is further oxidized to form $\text{HClO}$ (or even further to $\text{HClO}_n$), which then oxidizes the copper. This sounds like the most likely reaction to me.

The chloride ions somehow catalyze the decomposition of hydrogen peroxide into radicals, which then attack the copper.

In you first paragraph it seems like the limiting reactor is the solution, drop is not enough try doing the opposite by means putting small piece of copper foil to a beaker of the solution be cautious this reaction is very exothermic! (it should work even with a weak acid like vinegar).

@Avon97 That's exactly the thing that stumped me - while vinegar works, dilute sulfuric acid doesn't! Not even in a large excess as you suggested. The answer that I linked explains that vinegar reacts with hydrogen peroxide to form an organic peroxy acid which attacks the copper; sulfuric acid on the other hand doesn't do that and hydrogen peroxide alone seems to oxidize copper only very slowly, if at all, as Poutnik explained in the answer to this question.

For the record, only hot concentrated sulfate acid directly reacts with copper metal, otherwise, the chemistry around copper can be quite complex including the role of chloride to otherwise facilitate the reaction.
Here is a sourced exposition which I would recommend for graduate students/researchers to assist in answering the question on the role of chloride and more insights in copper related reactions.
First, in the presence of $\ce{H+}$ and $\ce{O2}$ (or $\ce{H2O2}$) there is an electrochemical aspect (think of $\ce{Fe}$ in place of $\ce{Cu}$ here in air with an acid source in common rust formation) with the reported reaction for copper:

$\ce{2 Cu + 1/2 O2 + H+ -> 2 Cu+ + OH- }$

So, the chloride serves certainly in the role of an electrolyte.
An associated copper electrolysis based reference which notes on Page 15, for example, the half cell reaction (which is implied per the above) for copper (and noted other metals), namely:

$\ce{1/2 O2 + 2 H+ + 2 e- = H2O}$.

The above simple copper metal reaction with oxygen and an acid source also strongly suggests a basic salt formation as can be actually found in Equation 7 per a Wikipedia discussion on Dicopper chloride trihydroxide with respect to the commercial preparation of the basic copper salt. This source interestingly cites the use of $\ce{NaCl}$ as a complexing tool to form a soluble mixed cuprous compound. So, the second answer on the role of chloride is as a complexing agent.
Actually, this associated electrochemical oxidation reaction occurs correspondingly not only with iron but several other transition metals as well, as noted below:

$\ce{4 Cu/ 2 Fe/ 2 Co/ 2 Cr + O2 + 2 H+ -> 4 Cu+/ 2 Fe++/ 2 Co++/ 2 Cr++ + 2 OH- }$

Interesting and supportive, the above noted electrolysis reaction with copper can also be derived from a series of radical chemistry reactions. Namely, per a 2013 radical reaction supplement, "Impacts  of  aerosols  on  the  chemistry  of atmospheric  trace  gases: a  case  study  of peroxides radicals", I cite the following reactions:

R24: $\ce{O2 + Cu+ → Cu++ + •O2− (k = 4.6*10^5) }$

R27: $\ce{ •O2− + Cu+ + 2 H+ → Cu++ + H2O2 (k = 9.4*10^9) }$

R25 $\ce{H2O2 + Cu+ → Cu++ + •OH + OH− (k= 7.0*10^3 ) }$

R23 $\ce{•OH + Cu+ → Cu++ + OH- (  k = 3.0×E09 ) }$

Where the implied net reaction, in the case of copper, confirms the claim reaction above, namely, a net reaction of:

$\ce{O2 + 4 Cu+ + 2 H+ → 4 Cu++  + 2 OH- }$

where the reactions above clearly suggest again the need for a soluble cuprous presence.
The mechanics of the etching of copper with $\ce{H2O2}$, $\ce{NaCl}$ and a acid source would not be likely complete without noting the role of radicals created via a Fenton-type reaction with copper:

$\ce{Cu+ + H2O2 +  H+ →  Cu++  + .OH + H2O }$

$\ce{H2O2 + .OH -> .HO2 + H2O }$

which notes, of import here, the formation of the hydroperoxyl radical (•HO2, pKa 4.88), which is interestingly, a very acidic radical as noted in this article: "Radical-Enhanced Acidity: Why Bicarbonate, Carboxyl, Hydroperoxyl, and Related Radicals Are So Acidic". The acidic hydroperoxyl radical could be an explanation of the reaction proceeding even with select weak (as in chelating) acids, which again requires a soluble cuprous as provide by chloride.
Note, however, more recent 2013 work "Fenton-like copper redox chemistry revisited: Hydrogen peroxide and superoxide mediation of copper-catalyzed oxidant production" noting mainly just a reactive Cu(III) species creation and little hydroxyl radical per se. The absence of $\ce{.OH}$ argues for no chlorine radical from:

$\ce{.OH + Cl- = OH- + .Cl}$

As well as no free chlorine presence per the reaction:

$\ce{.Cl + .Cl = Cl2}$

However, the superoxide radical anion and at pH < 5, the hydroperoxyl radical, are still found in solution per this work.
I hope this helps to explain the role of chloride in the reaction's progression, which is still a matter of ongoing research.

Excellent answer +1. Take a look at the Pourbaix diagrams I cite which indicate that chloride ion promotes the "cuprous presence".

Note that if some oxidant is stronger thermodynamically ( a higher redox potential), it does not necessarily mean it oxidizes faster, as reaction kinetics aspects play their role as well.
Chlorine reaction with copper in acidic solution -- $\ce{Cl2 + HCl}$ dissolves gold as well -- is faster than with hydrogen peroxide. The formation of copper(II) chlorocomplexes helps too.
$$\ce{Cu(s) + H2O2(aq) + 2 H+(aq) -> Cu^2+(aq) + 2 H2O}$$
$$\ce{2 Cl-(aq) + H2O2(aq) + 2 H+(aq) -> Cl2(aq) + 2 H2O}$$
$$\ce{Cu(s) + Cl2(aq) -> Cu^2+(aq) + 2 Cl-(aq)}$$
$$\ce{Cu^2+(aq) + n Cl- <=> [CuCl_{n}]^{+2-n}}$$
Forming oxoacids of chlorine in acidic solution is not preferred, if there are chlorides yet.
$$\ce{HClO(aq) + H+(aq) + Cl-(aq) <=>> H2O + Cl2(aq)}$$
$$\ce{HClO2(aq) + 3 H+(aq) + 3 Cl-(aq) <=>> 2 H2O + 2 Cl2(aq)}$$

Thanks a lot for your explanation! Is there a way to check whether the solution contains oxoacids? (Not because I need to know for any particular purpose, I'm just very curious about the reaction mechanism of this seemingly simple etching process.) I thought about repeating the experiment with iodide instead of chloride since molecular iodine won't attack copper as far as I know. If it still dissolves the copper, that'd mean oxoacids are present; if it doesn't, then it's just halogen molecules. Do you think that experiment would work?

Note the reaction  $\ce{2 Cu^2+(aq) + 5 I-(aq) -> 2 CuI(s) + I3-(aq)}$

You're right, that reaction would also occur. So, if there's any $\text{HIO}_n$ in solution, I'd first get $\text{Cu}^{2+}$, which then gets reduced to $\text{CuI}$ that'll precipitate out. If there's just $\text{I}_2$, nothing should happen. I'll give it a try unless there's any other flaw in my train of thought. Edit: Nevermind, $\text{I}_2$ apparently also attacks copper.

Equilibrium between chlorine and chloride/oxoacid/anion can be derived from Redox potentials.

potentially related Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves?

This should have been the accepted answer. Already upvoted.

It is entirely possible that the reaction mechanism is affected by oxidation of the copper, not the chlorine.
Copper is often reported to have only +2 as a stable oxidation state, the +1 state tending to disproportionate. This is true in the hard-base environment of water solvation, but with softer bases a rather different outcome emerges. Iodide ion is well known to reduce aqueous copper(II) to a precipitate of copper(I) iodide, rather than to the metal. Thiourea forms a soluble copper(I) complex in a similar way; we see this complex (or rather, we don't see it) when a solution of copper(II) sulfate is de-color-ized by adding thiourea.
Chloride ion is less soft and nonreducing towards copper (and towards the acidic peroxide) but still has a stabilizing effect. Compare the Pourbaix diagrams[1] for copper solvated by water alone:

with that obtained when chloride ion is added at unit activity:

With only water solvating the copper, copper(I) is a metastable species in the presence of the metal, which is stably oxidized directly to the +2 oxidation state. But with the chloride ion present the copper is oxidized directly to sparingly soluble copper(I) chloride instead, with a soluble complex formed at low concentrations just below that field. With the stabilized copper(I) species thus available, copper becomes appreciably dissolved at a lower potential and therefore with lower activation energy with the chloride ion addition than without.
Reference
1.
King, Fraser. (2010). Critical review of the literature on the corrosion of copper by water. https://www.researchgate.net/publication/242469529_Critical_review_of_the_literature_on_the_corrosion_of_copper_by_water


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Can alkylboron amines undergo Suzuki reaction with bromobenzene?
$$\ce{NH2-CH2-CH2-CH2-B(OR)2}$$
I don't know what the name of this compound is, but can the Suzuki reaction proceed under basic conditions with an aryl halide and a general palladium catalyst in the case of an alkylborane with a free amine attached to the end?

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Your structure will undoubtedly exist as a cyclic structure and possibly lose ROH.


text:
Why do active methylene compounds give Sn2 reactions?

I dont understand why they are strong nucleophiles when the negative charge is resonance stabilised. Also it is bulky, is there any chance of elimination?

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What exactly is a "strong" nucleophile? IMHO, for the purposes of organic chemistry, it suffices to know that it is a nucleophile. Nucleophilicity scales do exist, but aren't really needed if you just want to draw a curly arrow mechanism. Also, if the charge weren't resonance stabilised, then lots of bad things happen. Alkyllithiums will react in all sorts of conceivable ways (as a base, as a nucleophile, lithium–halogen exchange). Tempering the negative charge helps to keep the reactivity under control.

I'm in high school and I've been taught while comparing nucleophilic strengths(I'm really sorry) , nucleophiles with delocalized charges are poorer nucleophiles because they cannot donate electrons as effectively as the ones with localized charges. I haven't really dealt with Sn1,Sn2 questions involving resonance stabilized nucleophiles before this.(Pardon me for any mistakes, English is not my first language)

Hi Jen, welcome to ChemSE. Your question is quite insightful for a high school student. Whether or not the nucleophile/base is resonance stabilized is not the main issue. What counts is the relative rate of the irreversible SN2 vs. E2 reactions. I did a quick Chemical Abstracts search of this reaction and found yields of 16%, 21% and 40% for the SN2 product. These yields are modest at best. They may indicate substantial E2 elimination is competing with displacement. I will dismiss poor technique in isolating the product. ;)


text:
How to estimate the pH of multi-component buffer (phosphate buffer in equilibrium with CO2 rich atmosphere)?
I wonder how the pH of buffer solution can be estimated if buffer consists of several systems in equilibrium to the gas phase. For example, (based on growth medium for one of the bacteria) buffer contains:

$\pu{3 mM}$ $\ce{KH_2PO_4}$
$\pu{3.75 mM}$ $\ce{Na_2HPO_4}$
$\pu{2.5 mM}$ $\ce{NaHCO_3}$
solution is sealed in a glass container with a gas phase containing 80% $\ce{N_2}$ and 20% $\ce{CO_2}$ under $\pu{1 bar}$ overpressure ($\pu{2 bar}$ total pressure)
liquid takes approx. half of the full volume of the glass container

To my understanding, there are simultaneously 2 buffer systems taking place (phosphate buffer and bicarbonate), but I don't get how to combine that in calculations.

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Consider:  0/ [H]+ iand [OH-] is negligible wrt other ions 1/ Initial concentrations H2PO4-/HPO4^2-  and HCO3-/CO2(aq,tot). 2/ Shift in concentrations for the 1st pair equals respective shift for the 2nd one. 3/ Concentrations for both pairs must lead to the same [H+]

You can always use the generally applicable approach of constructing the set of inventory and equilibrium equations, with simplifying approximations  by application of justified strong inequalities.


text:
What happens with electric charges in a salt water battery?
When building a simple salt water battery, consisting of a zinc-anode, copper-cathode and an NaCl-solution, I noticed the following:
When closing the circuit, the current decreases over time. (No surprise, since either charge adds up around the electrodes or $\ce{Na^+}$- and $\ce{Cl^-}$-ions get depleted.) When I gently slide my finger along the zinc-anode (under the water), the measured current increases very little (and continues to decrease over time). When I do the same at the copper-anode, the current increases much more than at the anode (and continues to decrease over time).
I understand that at the zinc-anode $\ce{Zn^2+}$ dissolves into the water. And I understand that at the copper-cathode the electrons will interact with the water, not the $\ce{Na^+}$.
Now, I have 3 questions, where no. 3 is my main question:

Close to the anode: Will $\ce{Cl^-}$-ions just float next to the $\ce{Zn^2+}$-ions independently or will they form $\ce{ZnCl_2}$?

Close to the cathode: What exactly is happening there? Since the voltage I measured is ~0.7V, it can't be electrolysis of water, since 1.23V are required for this to happen. I would assume due to autoprotolysis of water, $\ce{H_3^+O + e^- -> 1/2H_2 + H_2O}$. The autoprotolysis leaves behind an $\ce{^-OH}$. I assume it bonds with the Na, giving $\ce{NaOH}$.

Why do I measure a much greater temporary increase in electric current when I gently stirr the water close to the copper-cathode, compared to stirring close to the zinc-anode? Shouldn't both water-volumes around the two electrodes be equally neutral? On one side, $\ce{Zn^2+}$ gets cancelled out by $\ce{2Cl^-}$, on the other side $\ce{2OH^-}$ get cancelled out by $\ce{2Na^+}$?

Here is the battery:

And here is where I got my ideas for the reactions happening. (I'm not a chemist.)

Be aware it is not really Zn/Cu galvanic cell, but Zn/H2-on-Cu cell. By fiddling with cathode, you remove hydrogen and improve access of electrolyte to copper.

Ah, that's interesting. So the hydrogen stays close to the cathode, forming kind of a shield around the cathode? I thought it would rise and leave the water eventually. I should mention that I was not able to see any bubbles around the cathode, even after waiting for a long time... But I thought the bubbles might be too small.

Related: https://chemistry.stackexchange.com/q/159681/79678.

I know, that was me. I am still trying to figure out some details I am not totally sure about (see my 3 questions). Thanks for helping earlier btw.

copper-anode $->$ copper-cathode

There are some mistakes in your questions.

At the anode, $\ce{Zn}$ will form $\ce{Zn^{2+}}$ and not $\ce{ZnCl2}$. Simultaneously the anode attracts anions $\ce{Cl-}$ to compensate for the new positive charges created at the zinc plate. The $\ce{ZnCl2}$ compound does not exist in solution. but it can be obtained after evaporation of the aqueous solution.

At the cathode, the only reaction is : $\ce{2 H2O + 2e^- -> H2 + 2 OH^-}$. Simultaneously, the copper plate will attract $\ce{Na^+}$ to compensate for the new $\ce{OH-}$ ions created on the copper plate. So the solution near the cathode is made of $\ce{Na+ + OH-}$ ions. But the solution does not contain $\ce{NaOH}$. This substance may be obtained after evaporation of the solution.

When stirring the cathodic plate, you remove the hydrogen bubbles adsorbed on the cathode surface. Without stirring, the surface where $\ce{H2O}$ can produce $\ce{OH-}$ decreases slowly with the time, because the hydrogen bubble prevents further production of $\ce{OH-}$ ions. The current is proportional to the surface of the electrode. As the surface of the plate where $\ce{OH-}$ ions can be created decreases, the current decreases

Dear Maurice, awesome! Thank you so much. This is exactly what I was looking for. Now I understand it.

Let me just add one more question: Shouldn't the electrolysis of water require at least 1.23V? I was only able to measure ~0.7V.

@Rainer_Zoufal  Electrolysis is the opposite process as a galvanic cell providing power. Furthermore, water electrolysis means creation of hydrogen AND oxygen, what is not the case.


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Why is 1413 µS/cm specifically the "standard" conductivity standard?
I'm doing some work on conductivity sensor calibration, and noticed that all the conductivity standard suppliers offer a 1413 µS/cm standard solution. Why is this oddly specific value the "standard"?

It may be conductivity of some round concentration of KCl, that is usually used.

There are several values for different calibration levels. 1413 µS/cm is the value for 0.01 mol/l KCl(aq), and 12880 μS/cm is for 0.1 mol/l KCl(aq), both at 25 °C. Don't have time to find a decent reference and post a proper answer, maybe someone else could catch up.

In short, conductivity is the reciprocal of the resistance so in essence you are just measuring the resistance of a solution. Making a 0.01 M solution of KCl (aq) is straightforward and the eleectrical resistance (aka conductivity) can be measured very accurately. This is why a 0.01 M KCl conductivity standard has an odd value of 1413 µS/cm. This NIST document fully explains the theory behind conductivity measurements and how to make primary standards.
https://www.nist.gov/system/files/documents/srm/260-142-2ndVersion.pdf


text:
How can one change the color of Iron Oxide by changing the composition or hydration?
How does the color of rust change depending on its makeup and state, or more to the point, how would one put the rust into that state?
For example, Wikipedia shows Iron(III) Oxide as being red in the α hydrate state and yellow in the β hydrate state, while a brown pigment is shown here as being iron oxide with a +3 oxidation state (which I understand to be Iron(III) Oxide, whose color above depends on its hydration state). Finally, a black pigment can be made using Iron(II,III) Oxide.
So my question is: I have crude brown Fe2O3(aq) (which I made from simple NaCl hydrolysis of iron); If I let it dry I presume I will get brown pigment. What do I have to do to it in order to make it enter the α or β hydration states? Finally, how would one convert Iron(III) Oxide into Iron(II,III) Oxide?
I understand most of this information is explained on the wiki pages that I linked, but I'm having a hard time sorting through it and would love if someone could help me cut through it and let me know what's more or less possible in a very small home lab, as opposed to some sort of industrial process that I likely don't have access to. Thank you for your patience as I begin what I hope to be a wonderful new hobby.

Rust ( formed at ambient temperature in moist conditions) , is not Fe2 O3 or any other simple oxide. It is various hydrates and hydroxides.  I have seen about a thousand XRF patterns of various rust samples done for corrosion evaluation. At ambient temperature Fe2O3 only forms under very specific conditions of fretting corrosion. It will form at modest temperature near 700F at long times . FeO ( black )will form at high temp like 1800 F. Fe3O4 forms at intermediate temperatures.  We never heated rusts to try to alter them.

@blacksmith37 Okay, so I got my crude rust from electrolysis and it's rather brown. Is there anything I can try to do to refine it, i.e. remove some of those extra hydrates and hydroxides?

I never tried to modify the rusts.  I expect heating will will alter it but have no experience.

Per a source involved in the manufacturing of pigments, to quote:

iron from the Fool’s Gold is dissolved in the sulfuric acid. If the acid is properly neutralized, it will precipitate out as a hydrated iron oxide yellow, which... can be heat-treated or “burned” to produce a red color (Figure 1).

So yes, there is, at least one path to creating a select iron pigment and thermally altering its color.


text:
Energy levels of carbon and selection rules
Here is an atomic physics problem I am trying to solve:
The carbon ground state is $[He](2s)^2(2p)^2$. Suppose to excite an electron to the configuration $[He](2s)^2(2p)(3s)$. The spectroscopic terms associated to this configurations have energies $-3.7eV$ and $-3.9eV$. The ground state has energy $-11.3eV$.
Which emission lines will be measured in a decay to the ground state?
What I have done so far:
Considering the $P$ orbital associated to the ground state there are two particle with spin $s = \frac{1}{2}$, therefore $S = 1$ and $L = 1$, therefore we have a triplet $J = 0, 1, 2$.
Considering the excited state, we have for the $2P$ orbital that $S = \frac{1}{2}$ and again $L = 1$, therefore we have two levels associated to $J = \frac{1}{2}, \frac{3}{2}$. For the $3S$ level we have instead $L=0$ and $S=\frac{1}{2}$, therefore $J = \frac{1}{2}$.
Now I consider the selection rule according to which $\Delta J = 0, \pm 1$. If this is true, then none of the excited levels (which have non-integer $J$) can decay to the ground state (which has integer $J$), so the answer would be: we observe no lines at all.
My question:
Is my reasoning correct? Am I doing something wrong?

By chance have you over-interpreted the question? Did they just want to calculate the energy differences and convert that into wavelegnths?

Seems you're overestimating how "forbidden" things are. https://en.wikipedia.org/wiki/Atomic_carbon#Electromagnetic_properties

So, yes, you're overthinking.

Thank you all for your replies, however I am still pretty confused.
As I understand selection rules are approximately right, namely they tell us if some decay is less likely than others, but it can still happen. Then the solution would be "I observe 3 lines, 2 from the doublet $L^2_{1/2, 3/2}$ to the ground state ($S^3_0$) and one line from the singlet $S^2_{1/2}$ to the ground state"?
Plus, I could tell the wavelength of the decay from the doublet from the question, but not from the singlet. Is this right? Thanks!

Afaik, those selection rules are for a transition to be radiative and observable spectroscopically. Transitions 'forbidden' by selection rules can still happen, although non-radiatively, just that they won't be observable spectroscopically.

Have a look at the NIST Atomic Spectra Database and search for "C I". This gives you the configurations, term symbols and level energies and might help you to get on track.

@TheLearner Radiates just fine, only slowly - phosphorescence instead of fluorescence.


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What happens when $V=Nb$ in the Van der Waals equation (divergent)?
The VdW equation:
$$\left(P+a\left(\frac{N^2}{V^2}\right) \right)\left(V-Nb\right)=Nk_BT$$
when the intermolecular forces are zero $a=0$, so $P=\frac{Nk_BT}{V-Nb}$ which diverges at $V=Nb$ for fixed temperature.
I'm simulating some hard sphere collisions, with $r_{sphere}=0.5, R_{contianer}=10 $. From some curvefit I got $b \approx 2.2$, so when $N>143$ (or less than 50% of the volume/area),the VdW equation breaks down.
Does that imply a different law holds when the number of particles becomes large? But isn't $b \propto r^3$ anyway, so it should account  for the case when the number of particles are large?

If $V=Nb$, then container volume is 4 times total volume of  molecules, according to the geometry used in the reasoning of van der Waals real gas model.
Molecules of a gas ( or rather a fluid ) at such a state are not able to freely move anymore, forming for each other not passable obstacles. Therefore the statistical pressure model breaks down. Signs of this progressively occur when $V \gg Nb$ is not true any more and the former is approaching the latter.
Additionally, at such a density, intermolecular interactions are not  negligible.
Note that this and similar states are far, unjustified extrapolations of applicability of the van der Waals model. This model is not intended for high pressure/density nor states near the critical state, as the model breaks physically much sooner than mathematically.


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Is the value for the Boltzmann Constant different in 2D?
So I'm simulating some hard sphere collisions and want to determine the Boltzmann constant in two dimension, using the equation
$$PA=Nk_BT$$
where $A=\pi R^2_{container}$, $P=(averageImpulse/time)/(2\pi R_{container})$. $E_k=k_BT$(deg of freedom=2).
I'm plotting several graphs of $P$ against $T$ so that $$k_{B,2D}=\frac{A\cdot Gradient}{N}$$
but cannot get a consistent (constant) value for $k_{B,2D}$? In the sense that $k_{B,2D}$ keeps decreasing as I keep decreasing the radius of the molecules? Does this imply that no such a constant exists?
The spheres are too small to obey the VdW equation (at least 1000 times smaller than container)

Isn't everithing dimensionally incorrect at start? At the moment I don't see how this comment relates to your question but it seems an issue.

Why would you expect the Boltzmann constant to be dependent on dimension in the first place? Secondly, Boltzmann's constant is one of the seven constants of nature on which the SI units are based, it seems unlikely you can derive it with a simulation...

https://arxiv.org/pdf/1106.1273.pdf may be of interest

@Paul assuming the first equation holds (the arxiv link I give suggests it does under appropriate conditions) and provided everything is measured in a consistent set of units I don't see why a value for $k$ can't be obtained. Who knows what units, but it should be a constant. However there is a lot of devil in the detail - is the velocity distribution properly equilibrated? Are there finite size effects that need to be accounted for? (Not sure) Are the stats for averageImpulse/time any good? And probably a whole load of other things.

@Ian_Bush, the LHS of the equation by the OP is in Newton and the RHS in Joules, so something is wrong. Obviously, the equation is a modified form of the ideal gas law and indeed a simulation of pressure times volume should give you a constant (i.e. the energy of the system), however this is not the same as determining the Boltzmann constant. In fact, you can only define temperature when using the Boltzmann constant.

It's not determining the Boltzmann constant. It's the equivalent of the Boltzmann constant in 2d which is being evaluated, and this will clearly have a different value and different units from the one we know and love

This is a weird statement, what do you mean by 'equivalent of the Boltzmann constant in 2d'? In essence, the Boltzmann constant is a conversion factor from energy to temperature and vice versa. From the equipartition theorem you can basically see that there is no dimensional dependence of the Boltzmann constant (each degree of freedom contributes $kT/2$ to the energy).

@Paul that is the point. k should hould be impose along each degree of freedom. At best one can look for a kind of Maxwell Boltzmann  distribution (I thin is what OP is dealing with) for a 2 D case.

I think this pdf can be useful to OP. K is taken independent of the number of dimensions, as in some of the comments above. http://fab.cba.mit.edu/classes/864.11/people/dhaval_adjodah/final_project/write_up_boltzmann.pdf


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what happens to the electrons in low-voltage conductivity of salt water
Suppose you take a 100 millivolt voltage source, and connect it using a pair of copper wires to a bath of salt water.
A current will flow through the water. If I'm not getting things completely wrong, it is possible to leave the experiment running for enough time that say 10 Coloumbs of charge will pass through.
If I'm not getting things completely wrong, there will be a voltage gradient of about 100mV across the bath, which will pull positive ions towards the negative electrode and negative ions towards the positive electrode. Electrons will move from the negative terminal of the voltage source towards the negative electrode and from the positive electrode to the positive terminal of the voltage source.
Now the question - what happens to the electrons that reach the negative electrode, and from where do the electrons that get to the positive electrode come from?
One possible answer would be electrolysis of stuff in the water, but with the voltage so low, I can't see where the energy to electrolyse things would come from.
Another answer, that I've never read of but I think might be true, is that copper might dissolve from the positive electrode, giving off an electron, and plate on the negative electrode. This will be hard to see because 10 Coloumbs of Copper ions is like 3 milligrams of copper, so not a lot. This doesn't consume energy, because in equilibrium, concentration and electric potential will balance the inherent energy of the $Cu^{2+}_{aq} + 2 e^{-}_{copper} \leftrightarrow Cu_{s}$ reaction. Is this correct?

(+1) Have you actually measured the current flow with 100 mV applied potential and saltwater electrolyte? This is very easy to do and would be informative. I do not see how copper could be oxidized at the anode and copper ions reduced at the cathode: there should not be any redox reactions happening at the applied potential.

This voltage is too low to cause chemical decomposition. Just remember that with direct current, whenever current flows in the external circuit, electrolysis (=chemical decomposition by redox process) must occur in the cell.

It is not directly the voltage, but any created .H radicals. The latter, embedded on Al have been implemented to convert PbS to Pb and H2S. The embedded H was the result of an acid/metal process, chemical not electro in nature but no difference. Here Cu2O + 2 .H to Cu and H2O, same reaction! Reference is in Hydrometallury,

Reference Link: " PbS + 2 •H = Pb + H2S (5) "  (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&lpg=PA818&dq=what+metals+react+with+hcl+to+form+nascent+hydrogen&source=bl&ots=yEBZaIFL_T&sig=9383ZW7NCXAoSNmmny2sdrxXIF0&hl=en&sa=X&ved=0ahUKEwjEjpnWjejaAhVP4VMKHe1gBZE4FBDoAQg6MAU#v=onepage&q=what%20metals%20react%20with%20hcl%20to%20form%20nascent%20hydrogen&f=false ). Copy link, reaction is bottom of page.

Actually, you do not have pure copper wires, as likely air exposure has added a less bright presence of $\ce{Cu2O}$. Even better, use copper tubing, which if not very tarnished, heat them in air. Also, in this experiment, I recommend pre-boiling the solution formed from adding salt to distilled water to remove any dissolved oxygen content.
So, in your very low voltage experiment, what I expect (based on my galvanic copper battery cell experimentations) is that the cuprous layer on the electrode receiving the electrons is reduced to copper metal:

$\ce{ H+ (aq) +  e- (aq) ->  .H (aq) }$

$\ce{ Cu2O (s) +  2 .H (aq) ->  2 Cu (s)  + H2O}$

So, expect your electrodes to no longer look exactly the same (based on my related experience, it looks like someone cleaned one), an electrode could be visibly more bright from a small amount of fresh copper formation.
Now, what could occur if all the cuprous oxide is consumed? I cited a source, Page 29, Table 2.1 with a correspondingly slow reaction, the dissipation of a solvated electron in water:

$\ce{ H2O (l) +  e- (aq) -> .H  + OH- (aq) }$

where the hydrogen radical could either self react liberating hydrogen gas or, per the nature of the metal present, become surface embedded.
Caution: If you wish to run this experiment at high voltage, please note that the resulting chlorine formation interacting with water may create some hypochlorous acid (HOCl). The  latter HOCl does react directly with copper metal. To quote a source:

The reaction between copper and HOCl is known to generate strong oxidizing intermediates, most likely highly reactive Cu(III) (59–63). HOCl is also capable of oxidizing other transition metals, including iron (64, 65) and manganese (66, 67).

The presence of copper ions can present disposal issues as such ions are quite toxic to select organism (like fish) and even to plants,.etc.
See if you could actually verify my expectations on the electrodes.

Check the quoted reaction, it is not balanced.

The actual mechanics is via the .H radical cited as the major reducing species in solutions.

A good source for radical chemistry is Buxton, and the hydrogen atom is discussed. To quote:
"The hydrogen atom is the conjugate acid of e-(aq), and it is the major reducing species in acidic solution, Eq. (3).
e-(aq) + H3O+ ---> H (aq) (3) "

A galvanic version of this reaction is simply place Cu tube (or old copper penny) and a piece of carbon in NaCl with a bit of lemon juice.  Microwave and in minutes the Cu anode is as bright as a new copper penny,

Related reaction: Reference Link: " PbS + 2 •H = Pb + H2S (5) "  (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&lpg=PA818&dq=what+metals+react+with+hcl+to+form+nascent+hydrogen&source=bl&ots=yEBZaIFL_T&sig=9383ZW7NCXAoSNmmny2sdrxXIF0&hl=en&sa=X&ved=0ahUKEwjEjpnWjejaAhVP4VMKHe1gBZE4FBDoAQg6MAU#v=onepage&q=what%20metals%20react%20with%20hcl%20to%20form%20nascent%20hydrogen&f=false ). Copy link, reaction is bottom of page.

So the anode copper will be oxidized to produce circuit electrons, and that will produce the energy needed to reduce random stuff at the cathode?


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Briggs-Rauscher reaction doesn't oscillate and turns blue with Vitamin C
I tried to recreate the Briggs-Rauscher reaction  by using a solution of $\ce{H2O2}$, $\ce{KIO3}$, $\ce{H2SO4}$, $\ce{CH2(COOH)2}$, $\ce{MnSO4}$ and a soluble starch. I used this recipe.

In container 1, I added 295 ml of distilled water, and 205 ml of $\ce{H2O2}$ at 30% concentration.

In container 2:  500 ml of distilled water 21.5 g of $\ce{KIO3}$ and 2.15 ml of $\ce{H2SO4}$.

In container 3: 250 ml of distilled water, and 8 g  of $\ce{CH2(COOH)2}$ and 1.7g of $\ce{MnSO4}$ and 0.15 g of starch.

For the experiment, I added 10 ml of each container in the order 1, 2, 3 in a container with a magnetic stir plate. This was a clear mixture, whereas according to the recipe it should have been yellow and blue oscillating.
After waiting about 5 minutes I added 0.5 ml of Vitamin C. The things I used for vitamin C were: a 500mg tablet dissolved in water (50g or 100g), orange juice, apple juice pomegranite juice etc. What happened was: instantly when I added the vitamin C the solution turned blue. It then became transparent again within 2 to 40 seconds. (2 seconds was apple and orange juice, the 40 seconds in the high concentration of vitamin C). The solutions stayed clear for  the remainder of my time in the lab, which was up to 2 hours, depending on the trial.

The questions are:

Why was the solution clear when I added the first three containers? I was expecting yellow and blue oscillating.

Why did the solution turn blue when I added vitamin C, and then go back to clear? I was expecting the vitamin C to stop the oscillation to blue, and keep it at yellow for a time proportional to the vitamin C content. Instead adding the vitamin C made it turn blue for a length of time proportional to the concentration of Vitamin C. What happened?


text:
How does this coin-and-strike-paper trick work (and is it safe)?
When I was a kid I remember a thing an adult showed me where you would take a piece of strike paper from a matchbox, rest it on a "copper" coin (I have no idea what these coins are made of), and set it alight.
This left a residue which you could rub on your fingers, which would glow when you rubbed your fingers together.
What is that?  I remember it only worked with copper coins, so what does the coin contribute, here?
I was going to show this to my own kids, but I realised I don't know how safe it is.  We did a lot of dumb things back then, so I shouldn't assume it's safe just because I'm still alive.

Strike paper contained red phosphorus, which upon heating could convert to white phosphorus, which in turn could be collected on the cold metal surface. Yes, it will glow. No, by today's standards it is not considered safe.

The type of coin seemed to be relevant to whether or not it worked, and it only glowed when you rubbed it.  Is it possible that it produced something else besides (or alongside) white phosphorus?  As the experiment seems overcomplicated if that's all that is needed.  Maybe the coin just provided thermal regulation so it wasn't all burned up and lost, and the rubbing was just a catalyst for oxidation.  I don't know.  In any case it sounds like white phosphorus must be present and this isn't for kids.

@sh1 White phosphorus is not meant for anyone. It is best left untouched.

I’m voting to close this question because not only is it not clear in terms of a "chemical question", it is about "playing with fire" which is something we should not encourage or condone here

An opening comment from Wikipedia on phosphorus:

The glow of phosphorus is caused by oxidation of the white (but not red) phosphorus — a process now called chemiluminescence.

Now, my take, first from a source on the history of matches to quote:

The modern "safety-match"
The match head now consists of a paste containing an oxidant such as KClO3 and phosphorus sesquisulfide (P4S3), instead of a mixture or red phosphorus and sulfur. (P4S3 was possibly first suggested in 1872 by Frederick Zaiss of Philadelphia in US Patent 125,874 on coloured and scented parlour matches). These "safety-matches" were consider non-toxic and unaffected by the atmosphere and did not react readily with water.

Further, not surprising in the presence of some water (from touching, also adding NaCl), a source of H+ by the reaction:

$\ce{P4S3 + H2O = H2S + H3PO3 + H3PO2 }$

Claim, a metal/air battery cell is formed (from placing the extracted P4S3 mix on a copper penny in air) liberating electrons, which can further react creating the hydrogen radical:

$\ce{H+ + e- <-> .H}$

Source: See Science Direct cited Reaction (1), which per Buxton, is claimed to be the major reducing species in acidic solutions.
As such, upon placing P4S3 (after being touched with the human hand adding NaCl and moisture) on copper metal in air could create a galvanic corrosion cell with a possible further attack of the moist, now acidic P4S3, by generated hydrogen atom radicals, liberating P4 as follows:

$\ce{6 .H + P4S3 = P4 + 3 H2S}$

where this reactions mirrors the following cited reaction: $\ce{  PbS + 2 •H = Pb + H2S}$ (5) from the Hydrometallury literature.
Finally, as already noted, the liberated P4 undergoes oxidation in air resulting in chemiluminescence.
The reaction may not be safe as the indicated liberated substance, per my suggested chemistry, is white phosphorus, albeit, likely in small amounts.


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How do I separate an Indium-Gallium alloy into Indium and Gallium?
I've recently learned more about Indium and Gallium, and I find them quite interesting, especially how an alloy can be manually made simply by rubbing them together. I'm curious though, is there a convenient way to get them back apart?
Also, say for example I had them in a test tube after mixing them together (or even after separating them), how could the metal be removed from the container cleanly? I know they both dissolve in acids, but how would I then precipitate them back out of the acid?

The Gallium-Indium binary system is assessed by Bridget C. Rugg and Tim G. Chart in Calphad 14(2) 115-123 (1990). It is a very simple system, exhibiting essentially no solubility of In in Ga, and limited (< 5at.%) solubility of Ga in In. There are no intermediate phases, and a shallow eutectic about 15K below the Ga melting point at about 14 at.% In.
The diagram (comparing their Calphad calculations with experimental data) is:

So, how to separate (at least roughly) the constituents? One answer is using zone melt refining, although there will be limits to what you can do. I will presume starting with a ~50/50 atomic ratio. You will need something to melt the material in (repeatedly), while maintaining a temperature gradient.
So, at the 50/50 point, it should be molten at 350K. as you lower that temperature, you hit the liquidus line and the solid that forms is indium with a few atomic percent of Ga. Say you arrange the temperature gradient so that forms on the bottom of you test tube. Keep lowering the temperature, causing more indium to solidify, and the remaining melt to get progressively more gallium rich. Eventually you will hit the eutectic point, and the last bit to solidify will be at 289K and 14 at.% In in Ga.
Cut that last bit off - that is as pure of Ga you can get with this technique (sadly).You can now reprocess the In-rich portion to refine more of the Ga out, but given the fairly low solubility of Ga in the In to begin with, there might not be much point to that.
Not perfect, but it pretty much works.


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How to Improve Phosphoric Acid 85% leaching of Natural Titanium Rutile without increasing impurity?
Using magnetically separated natural titanium rutile from a mine in South Africa to leach titanium for oxide production using H3PO4 85%. (The reason for using H3PO4, is that it separates out any Fe from the leachate making the Fe residue soluble and allowing it to be washed out with water after gravity separation.) After leaching the rutile sand under reflux at 270 C° twice, the rest of the natural rutile sand left over does not produce leachate any more with 100% H3PO4(85%).
This is a very high grade natural rutile with 93% Titanium.
The Natural Rutile on a white plate.
After some research and experimentation, I found out that the Natural Rutile that wont leach has very strong bonds that the strength of the H3PO4 simply can't break and oxidise. It was suggested that I try adding 5 too 10% of Nitric Acid to the H3PO4 which would help break/crack the rutile and allow leaching to commence. The problem with this was that the Nitric Acid lowered the boiling point of the H3PO4, and also increased the fume/evaporation rate of the acid sol. I don't have a pressurised reaction vessel or chamber for the process and under simple reflux this now became a problem.
It was suggested to use 5% Nitric Acid(65%) and 10% Sulfuric Acid(98%) with the Phosphoric Acid(85%), the Sulfuric Acid brought the boiling point up nicely and almost stopped the Nitric Acid from fuming out (Only slight fuming that did not escape the condenser easily), this worked very nicely, but the new problem is that prior to adding the Sulfuric Acid, all my amorphous leachate after washing was a brilliant white. After adding the Sulfuric Acid the washed leachate is a white with a very slight yellow tinge to it, which I can't seem to remove and get it back to the brilliant white that is required for the Ti oxide.
H3PO4 only leachate after separation, which is a brilliant bright white after washing and drying.
The leachate color after adding sulfuric acid
Can any of you suggest an effective way forward that does not require specialised acid or chlorine reaction chambers? Or a simple chemical route to clean the yellowish leachate and make it a brilliant white? I am open to any suggestions.
Thank you for your time and consideration.

Here is something that I read in a hydrometallury text (see Hydrometallurgy in Extraction Processes, Vol I, by C. K. Gupta and T. K. Mukherjee, Page 15) as an aid in dissolving ores based on the science surrounding so-called activity coefficients.
It centers on simply increasing the effectiveness of an acid dissolution by actually using the same (or more dilute) parent acid combined with a related salt at high concentration. So, for example, in place of strong $\ce{HCl}$ employ strong or weaker $\ce{HCl}$ with say $\ce{FeCl3}$ (where the higher valence chloride ion is apparently more effective than say $\ce{FeCl2)}$.
In the current context, try working with $\ce{H3PO4}$ at varying concentrations and a select phosphate salts in high concentrations.
As such, you may be able, as per your question, actually "Improve Phosphoric Acid 85% leaching of Natural Titanium Rutile without increasing impurity".

I will give that a try, thanks for the link to an awesome read and intro to using chloride salts with bases.


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Validity of the adiabatic equation in irreversible processes
I read here that the ideal gas equation and all state equations are valid only at thermodynamic equilibrium. In irreversible processes, the system is not at thermodynamic equilibrium. Also, for reversible processes, I know that the process has to be incredibly slow. However, I was taught that the adiabatic equation $P_1V_1^\gamma = P_2V_2^\gamma$ is valid for all adiabatic processes, and that adiabatic processes can be carried out quickly (irreversibly) or slowly (reversibly). Is this true, or am I misunderstanding any of it?
Related to my question, this question states as fact that "adiabatic expansions and compressions involving ideal gases are reversible and follow $PV^\gamma$ constant, provided the process is quick enough." However, it was my understanding, as I mentioned, that quick processes are irreversible. So some clarification about speed and reversibility of processes would be helpful.

Slow enough to be considered reversible, fast enough to be considered adiabatic. E.g. pressure change in free air leads to adiabatic compresion/expansion of air. But if too slow, there would be heat transfer with surrounding.

And if there were perfectly insulating walls, wouldn't a faster process be unnecessary? As we're considering ideal gases, wouldn't this ideal scenario mean that adiabatic processes could happen slowly?

Sure, an isolated system can perform an adiabatic process in infinite time. For real cases one must evaluate if idealizing assumptions are true. As no process can take infinite time, truly reversible processes do not exist. but many processes can be considered reversible with negligible error.

Your "adiabatic equation" is valid only for adiabatic reversible processes of ideal gases.  Adiabatic expansions and compressions involving ideal gases are reversible and follow this equation only if the process is slow enough.  Rapid adiabatic expansions and compressions are irreversible and do not obey that equation.


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IUPAC naming of purines and pyrimidines
From Wikipedia, the IUPAC name for guanine is 2-amino-1,9-dihydro-6H-purin-6-one besides 2-amino-6-hydroxypurine and 2-aminohypoxanthine.
I have some difficulty understanding the 6H part in its name since there's a keto group at the 6th position unlike in cytosine (4-aminopyrimidin-2(1H)-one) where the 1H clearly refers to the hydrogen at the 1st position
What does 6H exactly mean?

So-called Indicated H https://goldbook.iupac.org/terms/view/I03004

@Alchimista Thank you! But a thing still bugs me; the indicated hydrogen in cytosine gets added at position-1 which is also evident in its name. While in the case of guanine, it(hydrogen) apparently gets added at position-1 and not 6 as shown in its IUPAC name. Am I getting it right?

Related: https://chemistry.stackexchange.com/questions/110664/bibenzimidazole-nomenclature-indicated-h-and-locants

@Nex the numbering fixes the double bonds with respect to purine numbering, guanine is named as a 1,9-dihydropurinone. The H in 1 isn't indicated H. See also Nilay Ghosh comment.

The original parent structure purine ($\ce{C5H4N4}$) has the maximum number of noncumulative double bonds for this kind of structure. Compared to an ideal unsaturated compound, however, purine has one extra hydrogen somewhere. The position of this extra hydrogen must be indicated. A few different isomers are possible; for example, the original parent structure could be 9H-purine.

In the given compound, a double bond of the original purine is missing. Thus, the parent structure is a bit more saturated with the equivalent of two additional hydrogen atoms ($\ce{C5H6N4}$). Such saturation is described using ‘hydro’ prefixes. The locants of the original indicated hydrogen and the two additional hydrogens are 1, 6, and 9. The indicated hydrogen gets the lowest locant, so the name of the unsubstituted parent structure is 6,9-dihydro-1H-purine.

In the substituted purin-6-one, however, the numbering is different because a double bond was removed to make room for the ketone. According to Rule P-58.2.3, the indicated hydrogen is placed at the position that will accommodate this principal characteristic group. Therefore, the name becomes 1,9-dihydro-6H-purin-6-one.

Am I right in thinking that there is no hydrogen at or near position six? If so, could someone add to the answer an acknowledgement of this and an explanation of what exactly is meant by "indicated hydrogen"? Is it in fact not a hydrogen atom at all, but a configuration of bonds?

@MatthewChristopherBartsh IUPAC Gold Book: indicated hydrogen

Thanks for the link. I already read it after seeing it in Alchemista's comment. There is a slight problem with it, however. In all the examples given in the link there is an actual hydrogen atom at the position referred to. There is no example analogous to the "6H" of 2-amino-1,9-dihydro-6H-purin-6-one where, if I am correct, there is no actual hydrogen atom at position 6. The nearest hydrogen being attached to the nitrogen at position 5. Could someone please confirm that the nearest H to position 6 is attached to N5?

@MatthewChristopherBartsh In the final product 2-amino-1,9-dihydro-6H-purin-6-one, three hydrogen atoms of 6,9-dihydro-1H-purine are substituted with other groups. So, yes, there is no hydrogen atom left at position 6. However, no, there is no hydrogen attached to position 5 and there is no nitrogen at position 5.

I meant to ask, "Could someone please confirm that the nearest H to position 6 is attached to N1?" You are right that there is no N5. I got confused about the numbering; my diagram has no numbering and left and right are reversed and I confused left and right I think. Thanks for confirming that there is no hydrogen at position 6. It sounds like "6H" refers to a hydrogen atom in a molecule that could be a precursor in a chemical process that synthesizes 2-amino-1,9-dihydro-6H-purin-6-one or perhaps in a "parent" molecule in a taxonomy of molecules. I find it very hard to understand.


text:
Problem with partition function when calculating equlibrium constant?
For example a reaction is given as ($\pu{T = 1000 K}$),
$$\ce{2Na(g) <--> Na2 (g)}$$
The equilibrium constant,
$$\ce{K_{c} = \frac{\frac{q_{Na_{2}}}{V}}{(\frac{q_{Na}}{V})^2}}$$
But when considering the partition function for $\ce{Na2(g)}$. Teacher multiplied by another term which I don't understand.
The partition function for $\ce{2Na(g)}$,
$$\ce{q_{Na} = q^{t}.q^{e}}$$
The partition function for $\ce{Na2(g)}$,
$$\ce{q_{Na_{2}} = q^{t}.q^{r}.q^{v}.q^{e}.e^{β(D_{o} - \frac{hν}{2})}}$$

Where dose this $\ce{e^{β(D_{o} - \frac{hν}{2})}}$ term come from why it has been used in the partition function from my understanding its coming from Morse Oscillator. Looks like it has some thing to do with bond dissociation for a diatomic molecule but I don't understand how did it get resemble to the partition function.

Teacher took $\ce{q^{e}}$ for $\ce{Na(g)}$ as $\pu{2}$ and for $\ce{Na_{2}(g)}$ $\pu{1}$ how did that work according to my understanding,

The electronic partition function is given as,
$$\pu{q^{e} = g_{o}e^{-βE_{o}} + g_{1}e^{-βE_{1}} + g_{2}e^{-βE_{2}} ....}$$
Teacher only considered the ground state energies how did the other terms got neglected if ground state energy is $\pu{0}$ then only degeneracy matter.Further more I did not find any doubts about the other partition functions they are ok for me.

The Morse potential term arises from the nuclear degree of freedom term, thus for your diatomic $q = q_{\rm trans}q_{\mathrm rot}q_{\mathrm vib}q_{\mathrm elec}q_{\mathrm nucl}$. Your expression for $q_{\mathrm elec}$ is missing the first term - $\omega_{e1}$ - which is what the term converges to in a common approximation. The factor of 2 is simply because there are 2 $\ce{Na}$. Finally, your teacher copied this example (exactly) from p. 144 (section 9-2 A) in McQuarrie ${\it Statistical, Mechanics}$ (1976), so you can look there to clear up the rest of your concerns.