text:
What is the mass of ozone present in 10.6 g of Na2CO3?
I got a question in my textbook asking for the mass of oxygen present in 10.6 grams of $\ce{Na2CO3}$.
Why are they asking about the mass of oxygen? Does this refer to $\ce{O2}$? Shouldn't it be the mass of ozone, $\ce{O3}$, present in 10.6 grams of $\ce{Na2CO3}$, since the formula of sodium carbonate contains three atoms of oxygen in its formula unit?
Would it make a difference if the question read "calculate the mass of ozone present in 10.6 grams of $\ce{Na2CO3}$"?

There is no ozone in Na2CO3, much like there is no arrow in a sparrow.

Similarly, there is no red oxygen O8 in potassium peroxodisulphste K2S2O8. https://en.wikipedia.org/wiki/Solid_oxygen#Red_oxygen

I think this is a valid question. It is somewhat surprising that we would ask about the mass of sodium present in sodium carbonate, if we think of sodium as the highly reactive element. However, if we think of it as a type of atom, we can split the compound into free atoms conceptually, and do the math. Adding to the confusion, we can think of oxygen as the atom or the element in its common form $\ce{O2}$. For the mass, it makes no difference, but for the amount of substance it would, so you have to be careful specifying what oxygen refers to (or have the experience to know).

Beautiful Ivan comment right there. Love them <3

@Jan Ivan just happens to be the best commenter in our community! One surf through all his comments will restore your faith in humor+science.

The question in the book is about elemental composition of the given substance, in terms of mass.
You are given the empirical formula, i.e. what could be obtained by elemental analysis, namely what elements composed the substance and in which relative ratio. This does not tell you about the spatial arrangement of the various elements and the way they bound each other. The empirical formula can be even "smaller" than the actual chemical entity, all that is important is that for a given amount all elements will be always present in that ratio.
The fact that you read O3, it does not mean that you have ozone inside a molecule or salt, and the three oxygens might be at different sites within the structure (by the way, ozone is a polar molecule itself not an element or elemental ion, why it should be found combined? ).
While this should answer your current question, I invite you to reconsider the previous chapters of the same book, which almost certainly explains what we call elements and why, the laws of chemical combination, and the different levels of chemical formulas.
Now let's move to the reason for why I've choose to answer instead commenting "there is not ozone" or simply ignore your post.
Once the above is clear, consider that in cases the content of a substance (not its empirical formula) can be given in terms of a formal equivalent of another one. This is pretty common discussing or even classifying substances for a specific purpose or effect, so it might be with drugs, fertilisers, etc.
In your case, it could have been the amount of ozone evolving from the sample if a process would convert in ozone all the oxygen contained in it. Note that the amount as mass won't change as a certain amount of oxygen remains the same in all allotropes :) which again makes your doubt quite unmotivated.
Realistic examples of this formal way to express the chemical content by formal equivalents are fertilisers. Their N-P-K numbers refer to elemental nitrogen, phosphoric anhydride and potassium oxide, respectively.  The latters are not the actual content of the fertiliser (as it wouldn't be ozone in your unfortunate example, too) but refer to the way as P and K were finally weighted in elemental analysis.
Another example is the carbon dioxide content of whatever fuel or combustible. There is not carbon dioxide itself in sugar, but its C content can be expressed by the carbon dioxide formed during its combustion.
Hope this helps.

To add to Alchimista's answer, suppose the question were about a sample containing 1 mole of ozone (and nothing else). You could ask two questions:

What is the mass of ozone in the sample?
What is the mass of oxygen in the sample?

Because all the oxygen atoms are in the form of ozone, the answer to questions 1. and 2. would be the same. On the other hand, if I asked about the amount of ozone molecules or the amount of oxygen atoms, the answers would be distinct (1 mole vs. 3 moles).

What is the mass of ozone present in 10.6 g of Na2CO3?

This question makes no sense because there is no ozone in sodium carbonate. However, you could specify a reaction (a fictional one):
$$\ce{Na2CO3 -> 2Na + C + O3}$$
For this reaction, you can ask how much ozone would be made from a certain amount of sodium carbonate. Of course, the mass of ozone produced would be equal to the mass of oxygen in sodium carbonate because sodium carbonate is the only source of oxygen in the reaction, and ozone is the only product containing oxygen in this reaction.

[Ivan] There is no ozone in $\ce{Na2CO3}$, much like there is no arrow in a sparrow.

This wisdom from our resident artiste of one-liners is a great analogy illustrating the misconception. If we go back to the specific question, we can see from the Lewis structures that there is no ozone in carbonate:
Carbonate:

Ozone:
Ozone has oxygen-oxygen bonds, which are lacking in carbonate. Carbonate does contain oxygen atoms bound to carbon, so asking about the oxygen content makes sense. Asking about the $\ce{O2}$ content or the $\ce{O3}$ content (or -ahem- the $\ce{O8}$ content) without giving a specific chemical reaction (that would specify the stoichiometry) makes no sense, on the other hand.


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Can sodium silicate be made insoluble?
In this question, the user asked about curing sodium silicate with calcium chloride. My question is, will the resultant product be insoluble in water? I know calcium silicate is only very slightly soluble, but I'm expecting it's going to be a bit more complicated than that.
I'm not a chemist, and wanted to know the answer to something I'm messing with at work to know if I need to hire a materials professional.

No, it is really as simple as that. You'll get calcium silicate, which is insoluble.

Where does the sodium go, in this case? Is it displaced into the water to become sodium chloride? And would the calcium propagate through a "thick" structure, or would this only be a surface coating?

Yes, Na remains as NaCl. And yes, propagation may be an issue.

That's alright, I only need a surface coating. If water can get to it, CaCl2 can get to it during the wash and then it's 'sealed'.

It really doesn't matter whether this is actual homework or not. I have decided to reopen the question, because I don't think it should have been closed as homework in the first place.

Thanks. Unfortunately, Ivan's comment answers my question perfectly, but as it's a comment rather than an answer, I can't mark it as such.

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Is oxygen above the critical point always supercritical fluid? Would it still appear to roughly follow the ideal gas law?
In this answer I've asserted (without a "Chemist's license") that as long as oxygen is above it's critical point in both temperature (154.5 K) and pressure (50.4 bar) it's going to be a supercritical fluid.
This comment by a user who I believe is an experienced diver says:

Scuba divers use the idea(l) gas law for air and oxygen bottles to calculate the amount of remaining gas for 200 down to 10 bar. This would not be possible if there is supercritical oxygen in the tank.

In other words, the ideal gas law works pretty well for oxygen between 10 and 200 bar; for a given temperature the amount of oxygen (number of atoms or mass in kg) is close to linearly proportional to pressure.
The comment proposes that this is evidence that the oxygen couldn't be a supercritical fluid.
Question(s):

When oxygen is above both 50.4 bar and 154.5 K is it always a supercritical fluid, end of story?
If it is, would it nonetheless still tend to follow fairly closely the ideal gas law?

Experienced divers do not have to be experienced physical chemists. Then again, there is no actual physical border between gaseous and supercritical states, so you may call it whatever you like. It is but a matter of convenience.

I've seen "amorphous" used for fluids above both critical pressure and critical temperature.  Basically they still approach ideal gases if the temperature is high enough and pressure low enough to give a low fluid density.

related https://chemistry.stackexchange.com/questions/37088/why-must-both-the-critical-temperature-and-pressure-be-exceeded-to-achieve-the-s

Yes.  Any fluid with a temperature is above critical temperature and the pressure above the critical pressure is by defintion a supercritical fluid.  Don't be mislead by all the claims that supercritical fluids are special and wonky with all sorts of amazing, bizarre properties.  This is true of some supercritical fluids near the critical point, but the farther you go away from the critical point, the more like a regular gas supercritical fluids become.  (The only exception here is that at gargantuan extremes of P and/or T, eventually fluids will become plasmas, metallic solids, black holes, etc.  But as long as it is a fluid of the same substance [no chemical reactions], then it will be supercritical.)
Yes, supercritical fluids can often be modeled by the ideal gas law.  A good look into why would take into the theory of corresponding states.  This theory says that most gases will most often have similar compressibilities $Z = \frac{PV}{nRT}$ when they are in "corresponding" states, which means that at similar temperature and pressure relative to each gas's critical point.  There are charts that you can use (approximately) for many, many real gases.  Things to realize:

The compressibility factor $Z$ is a measure of non-ideality.  Ideal gases always have a compressibility factor of 1.0.
You can use the chart to find approximate compressibility factors for many, many real gases if you know that gas's critical temperature $T_c$ and critical pressure $P_c$.  The "reduced temperature" is just $\frac{T}{T_c}$ and likewise for pressure.
If you check out the generalized compressibility factor chart, look at the dependence on reduced temperature!  At a reduced temperature of 2.0, the line is nearly flat at $Z = 1.0$.  This is an indication that the gas behaves nearly ideally.  It takes pretty close approaches of $\frac{T}{T_c}$ to 1.0 to get strong non-ideality.
Oxygen's critical temperature of 154 K means that a reduced temperature of 2.0 just requires 309 K, about ambient temperature.  At these temperatures, oxygen will be fairly ideal, regardless of temperature, up to reduced pressures of 5 or 8, which correspond to 250 to 400 bar or so.

This is really excellent! You've answered several additional questions that you figured I had, but didn't explicitly include. Nicely done.

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Why are we still using cis/trans system while the E/Z system is better than it?
I am currently learning about Alkenes. It is given that cis/trans isomerism is either ambiguous or doesn't even apply in cases where the alkene is trisubstituted or tetrasubstituted. To solve this problem, we make use of the E/Z system of classifying the diastereomers of alkenes. In some cases both the methods of classifying diastereomers works.
My doubt is, why are we still using cis/trans system while the E/Z system is better than it? Is the cis/trans system better than E/Z system in some way?

Not really. It is just that we got used to it.

cis/trans applies to concepts other than alkenes. Since you're describing similar relationships with cis/trans, the term is still valid with alkenes.

That is because the E/Z-system is an extension of the cis/trans-system. So both systems can be applied to a certain amount of structures and the E/Z-system is then able to treat more advanced systems.
From my perception, it seems that Chemistry is a very stubborn … or maybe just very conservative … field where especially new naming conventions get hardly accepted. At least not very fast. So we are still using Angstrom instead of nanometer or picometer and so on. This also means that we stick with the cis/trans-system as long as it fits and we only change in case we can’t apply it.

Ångstrom is 1e-10 so maybe you will get extra zeroes or have to prepend decimal places in certain disciplines

cis and trans are perfectly valid descriptors for linear disubstituted alkenes. The (E) and (Z) system is more comprehensive for alkenes, as it also covers trisubstituted and tetrasubstituted alkenes which cis/trans by definition cannot (although some might extend it anyway).
In most cases, it would be perfectly reasonable and possible to replace every alkenic cis/trans with a corresponding (E)/(Z). However, in practice cis and trans prevail for no other reason that ‘we did it before’ in a lot of contexts and systems. For example, Aldrich sells trans-cinnamic acid.
When you are not dealing with alkenes, but with coordination compounds, cis/trans cannot be replaced by (E)/(Z) as far as I am aware. Since it remains in use there, there is no harm in teaching it in other places, too.


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Preparation of acetate buffer from sodium acetate and hydrochloric acid

A $\pu{1.0 L}$ buffer solution of $\mathrm{pH}~4.43$ is made up of $\pu{0.30 M}$ sodium ethanoate and $\pu{0.20 M}$ $\ce{HCl}$ solutions. Calculate the volume of sodium ethanoate and $\pu{0.20 M}$ $\ce{HCl}$ solutions used to prepare this buffer solution.

The answer given for volume of $\ce{HCl}$ is $\pu{750 mL}$ and for the volume of volume $\ce{CH3COONa}$ is $\pu{250 mL}.$
My approach is to assume $\ce{HCl}$ as the limiting reactant and use ICE table to determine the equilibrium concentration of the ethanoic acid and sodium ethanoate at the end of the reaction. Then I use the Henderson–Hasselbalch equation.
The answer I got was $\pu{500 mL}$ for both the conjugate base and weak acid. Is the answer wrong?

This is sort of an odd problem. Typically one adds the chemicals dissolves the solids, then dilutes to 1.000 L in a volumetric flask. The problem is ambiguous if extra pure water  is needed.

To answer this question, we have to assume that the given buffer solution is an ideal solution of which the total volume of it always equals to the summed volumes of sodium ethanoate ($\ce{NaOAc}$) and $\ce{HCl}$ solutions when they are added together. It should also be assumed that $\mathrm{p}K_\mathrm{a}$ of ethanoic acid (acetic acid; $\ce{HOAc}$) is given ($\mathrm{p}K_\mathrm{a} = 4.76$).
When added together, $\ce{HCl}$ (a strong acid) and $\ce{NaOAc}$ (weekly basic salt) react completely to give $\ce{HOAc}$ and $\ce{NaCl}$, according to the following equation:
$$\ce{NaOAc + HCl -> HOAc + NaCl} \tag{1}$$
Suppose you added $x$ volume of $\ce{NaOAc}$ and $y$ volume of $\ce{HCl}$ to make the $\pu{1.0L}$ of buffer ($\mathrm{pH} = 4.43$). Thus if units of $x$ and $y$ are $\pu{L}$, then:
$$x + y = 1 \tag {2}$$
And, after acid-base reaction, according to the chemical equation given above and equation $(2)$, remaining amounts of $\ce{NaOAc}$ in $\pu{mol}$ is $(0.3x-0.2y)=\left(0.3x-0.2(1-x)\right)= (0.5x-0.2)$ and amounts of $\ce{HOAc}$ formed in $\pu{mol}$ is $0.2y= 0.2(1-x)=(0.2-0.2x)$. Thus, since total volume of buffer is $\pu{1.0 L}$, respective concentrations of $\ce{NaOAc}$ and $\ce{HOAc}$ in the solution are as follows:
$$\ce{[NaOAc]} = (0.5x-0.2)/1.0=(0.5x-0.2) \text{ and } \ce{[HOAc]} = (0.2-0.2x)/1.0 = 0.2-0.2x$$
Now you can substitute these values to Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[NaOAc]}}{\ce{[HOAc]}}$$
$$4.43 = 4.76 + \log \frac{(0.5x-0.2)}{(0.2-0.2x)}$$
Thus,
$$ \log \frac{(0.5x-0.2)}{(0.2-0.2x)}= 4.43 - 4.76= -0.33$$
$$ \frac{(0.5x-0.2)}{(0.2-0.2x)}= 10^{-0.33} = 0.468$$
Once you solved for $x$, you would get, $x=0.495$ and $y=1.0-0.495=0.505$ in $\pu{L}$, respectively (or $495$ and $\pu{505 mL}$). Thus, I'd say the given answers for your question is incorrect.


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Other metabolic reactions aside, what causes beta oxidation to stop?
To my understanding beta oxidation is when a fatty acid has its carbons plucked off in groups of 2 (acetyl groups) into acetly-coa which we can dump into the TCA.
At what point would this process stop ? lets say we have a short chain fatty acid like butanoate (4c) would it still transfer an acetyl group leaving an acetate molecule ?

Different enzymes specialize to work on different size ranges. If the fatty acid is too short, it might not bind to enzymes that work on large fatty acids. An enzyme capable of binding short fatty acids might have most binding determinants targeting the CoA moiety instead of the fatty acid itself.

At what point would this process stop ? lets say we have a short chain fatty acid like butanoate (4c) would it still transfer an acetyl group leaving an acetate molecule ?

It goes all the way as long as the CoA is attached.


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Why does an electron lose energy so that it falls back to a lower energy state than just stay at the higher energy state?
Suppose an electron in the 2nd shell gets excited and jumps to the 4th shell. Shouldn't the electron then exist in an unstable equilibrium in the 4th shell, than lose energy and drop down to the 2nd shell? What causes it to come out of the equilibrium position and lose energy? The teacher who gave me this problem to think about said that quantum mechanics plays a role in this. Can anyone please give an explanation?

Why does a stone fall to the ground, rather than just sit comfortably in the air?

In order for the excited state to decay it needs to be perturbed. An excited isolated system in vacuum can decay to a lower state by spontaneous emission which is caused by the fact that on very short times scales the vacuum is not empty. The energy-time uncertainty locally allows for large energies in a short time. In fact, so much energy that a electron-positron pair can be formed and annihilated during this short time. These vacuum fluctuations induce the decay.

@IvanNeretin I don't think that's the right picture.  After all, one could instead ask 'why does a stone fall off the mountain, rather than staying at the peak?'.  The answer is that in fact they don't spontaneously fall off the peak—some can rest there for centuries or more; to fall, they need to be perturbed.

https://en.wikipedia.org/wiki/Spontaneous_emission

@theorist - see Phosphorescence // Forbidden transitions shouldn't occur, but do so because of perturbations.

Actually, this question is not as bad as the comments may make you think. It's actually quite a sensible question, because when we solve the usual, Schrödinger's equation-based, QM problem of energy states of hydrogen atom (or any other atom), we get eigenstates of the Hamiltonian we inserted into the equation. These solutions should thus be stable for infinite time.
In real world, however, an atom isn't isolated. It interacts with the electromagnetic field, and this interaction results in spontaneous emission, which is the reason why an atom can't be in excited state indefinitely.

The energy here is a kind of potential energy. By very nature, it always tries to stay minimum in magnitude, so it is said to have a "potential" to change its energy into other form of energy.
Like stretch bow's potential energy change into (and transferred to) arrow's kinetic energy.
in case of electron jumps, the energy transformed is in the form light.

As Ivan Neretin has also hinted at the concept with another example of potential energy


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Zinc evaporation/toxicity at room temperature
I have a (slightly embarrassing) question regarding zinc powder.
I have recently broken an old mercury thermometer, and read that I can handle the mercury with powdered zinc - so I've sprinkled some on my floor. Too much probably. It turned out that unfortunately the powder is then quite hard to get out of the particular material of my floor (wood). I got out as much as possible, but I think there's still some left. So now I'm wondering if there is some danger if having all that zinc laying around.
Thanks anybody for the answer :)

If I were you, I'd be much more worried about mercury than about zinc.

@Peter Do you drive?  Zinc is all around you in your car's galvanized steel body panels.  I second Ivan's comment.

This isn't a personal medical question: it is a chemical safety question. And one that might be relevant to, for example, the best safety procedure in a lab. Which sounds like mainstream chemistry to me.

Zinc will amalgamate with mercury, but whether that reduces the vapor pressure of mercury very much is debatable. At best, the increased volume of zinc amalgam and the ability to amalgamate with fine droplets of mercury would make it easier to clean up the mercury as amalgam.
It might be better to spread some sulfur dust (flowers of sulfur) over the area. The sulfur should react with the mercury and tie it up as sulfide, with very low vapor pressure. Sulfur powder (and possibly the HgS) might also be easier to clean up than zinc dust, because sulfur is more like ordinary dust than a dense metal dust.
BTW, I love mercury thermometers. And I hate it when I break them!


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What is the physical state of vitamin E acetate at room temperature and normal pressure
Vitamin E acetate ((+)-alpha-tocopherol acetate) is sold as semi-solid/liquid by Sigma and their safety data sheet gives the melting point as 25 °C. On the other hand, internet databases such as Chemspider show melting points of either -28 °C or 28 °C. Supplements labeled either Vitamin E or Vitamin E oil are packaged as solid, gel capsules or liquid.
What is the normal melting point, and what is the physical state at room temperature (20 °C)?
Update
There are at least three products on the market, the single (+) stereoisomer (derived from biological sources, CAS 58-95-7), the racemic mixture, and the "all-racemic mixture" (8 stereoisomers, with some pairs being diastereomers and other pairs enantiomers). The latter is produced synthetically with non-stereospecific steps (CAS 7695-91-2). All have the formula $\ce{C31H52O3}$. To clear up any confusion, the physical state of the single isomer, of the (+/-) racemic mixture and the "all-racemic mixture" would be of interest.
Vitamin E, depending on the source, has different numbers of methyl groups on the aromatic ring; these forms are referred to as alpha, beta and gamma. The alpha form is more common in Europe while the gamma form is more common in the US (due to different diets).

The +28°C value on ChemSpider (your link) must be a typo. The source given there is a spreadsheet titled "BradleyMeltingPointDataset" that contains two entries for "vitamin e acetate," with melting points of -27.5 and -28°C, which is in agreement with all the other values listed on ChemSpider.

From https://www.alfa.com/en/prodspec/A14505 : "Appearance (Color): Clear, colourless to pale yellow; Form: viscous liquid; Assay (GC): > 96.0%."

Take a sample, put it on DSC, and check out. Maybe someone already made a paper, though... In such molecules (long non-polar chain, aromatic ring...) it's quite possible that melting and freezing temp. are different and whole business is complicated.

Vitamin E acetate, or α-tocopheryl acetate, exists in two forms: D-form (solid/highly viscous at 20 °C) and a racemate (viscous liquid at 20 °C).
Interestingly enough, the difference in physical properties is well denoted in German Wikipedia article, whereas English Wikipedia omits this.
Also, I noticed many authoritative sources arbitrarily listing a melting point for one or another isomer without denoting the exact form, e.g. even CRC Handbook of Chemistry and Physics provides a single entry for vitamin E acetate with melting point −27.5 °C [1, p. 3-550], so no wonder online sources and sellers mix the things up a little.
More relevant details can be found in Handbook of Pharmaceutical Excipients [2, p. 32]:

d-Alpha tocopheryl acetate
Empirical formula $\ce{C31H52O3}$
Molecular weight 472.73
CAS number [58-95-7]
Synonyms (+)-(2R,4'R,8'R)-2,5,7,8-Tetramethyl-2-(4',8',12'-trimethyltridecyl)-6-chromanyl acetate; d-α-tocopheryl acetate; vitamin E.
Appearance A practically odorless, clear, yellow, or greenish-yellow colored viscous oil that may solidify in the cold.
Melting point 28 °C
Solubility Practically insoluble in water; soluble in ethanol (95%). Miscible with acetone, chloroform, ether, and vegetable oils.
Specific rotation $[α]_\mathrm{D}^{25} = +0.25^\circ$ (10% w/v solution in chloroform)
Comments Unstable to alkalis.
dl-Alpha tocopheryl acetate
Empirical formula $\ce{C31H52O3}$
Molecular weight 472.73
CAS number [7695-91-2]
Synonyms (±)-3,4-Dihydro-2,5,7,8-tetramethyl-2-(4,8,12-trimethyltridecyl)-2H-1-benzopyran-6-ol acetate; (±)-
  (2RS,4'RS,8'RS)-2,5,7,8-tetramethyl-2-(4',8',12'-trimethyltridecyl)-6-chromanyl acetate; (±)-a-tocopherol acetate; α-tocopheroli acetas; all-rac-α-tocopheryl acetate; dl-α-tocopheryl acetate; vitamin E.
Appearance A practically odorless, clear, yellow, or greenish-yellow viscous oil.
Density $\pu{0.953 g/cm3}$
Melting point –27.5 °C
Refractive index $n_\mathrm{D}^{20} = \pu{1.4950 – 1.4972}$
Solubility Practically insoluble in water; freely soluble in acetone, chloroform, ethanol, ether, and vegetable oils; soluble in ethanol (95%).
Comments Unstable to alkali. However, unlike alpha tocopherol, the acetate is much less susceptible to the effects of air, light, or ultraviolet light. Alpha tocopherol acetate concentrate, a powdered form of alpha tocopherol acetate, is described in the
  PhEur 6.0. The concentrate may be prepared by either dispersing alpha tocopherol acetate in a suitable carrier such as acacia or gelatin, or by adsorbing alpha tocopherol acetate on silicic acid.

D-form may appear as a highly viscous oil, however it will become solid over time [3, p. 232] (emphasis mine):

D-α-Tocopheryl acetate. CAS: 58-95-7. Obtained by vacuum steam distillation
  and acetylation of edible vegetable oil products. Occurs colorless to yellow, nearly
  odorless, clear, viscous oil. May solidify on standing, and melts at about 25 °C.
  Unstable in presence of alkalies.
DL-α-Tocopheryl acetate. CAS:7695-91-2. Colorless to yellow or greenish yellow, nearly odorless, clear, viscous oil. Unstable in presence of alkalies.
D-α-Tocopheryl acetate concentrate. Obtained by vacuum steam distillation and
  acetylation of edible vegetable oil products. May be adjusted by suitable physical
  or chemical means. Occurs as light brown to light yellow, nearly odorless, clear,
  viscous oil.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; CRC Press, 2016. ISBN 978-1-4987-5429-3.
Handbook of Pharmaceutical Excipients, 6th Ed.; Rowe, R. C., Ed.; APhA, (PhP) Pharmaceutical Press: London, 2009. ISBN 978-0-85369-792-3.
Food Additives, 2nd Ed., rev. expanded.; Branen, A. L., Ed.; Food science and technology; Marcel Dekker: New York, 2002. ISBN 978-0-8247-9343-2.


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Phenolphthalein Indicator mixed with base that first turns pink and then clear
What occurs in the following situation: After adding 2 drops of Phenolphthalein Indicator (an indicator that turns bases pink and stays clear for acids,) into a unknown solution, the solution turns pink. However, within a few minutes, the pink slowly fades and the solution becomes colorless. What is the unknown solution?

There probably is no one answer, but consider this:  commercial chlorine bleach can be expected to react with indicators which are organic dyes.

There is the 2nd acid/base transition at high pH, turning phenolphthalein colourless again, but I suppose it should be fast reaction, not taking minutes.

In strongly basic solutions, phenolphthalein is converted to its $\ce{In(OH)^3-}$ form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above $\mathrm{pH}= 13$.

wikipedia
So  the strong alkali is very probable reason, but bleachers or other compounds reacting with phph cannot be eliminated.

One limitation in this answer is any alkali can do this, all are sufficiently soluble.  The problem is actually the question, which is not well posed unless multiple choices are attached (and missing here).

@Oscar Lanzi No objection, but quality of answers should not be expected better than the quality of the question. The general alkali is as good answer as sodium hydroxide. Sure, bleachers or other compounds reactions with phph cannot be eliminated.

The indicator may slowly disappear after the titration because it is reacting with the air. As it reacts with chemicals in the air (CO 2 for example) which causes the indicator to be slowly disappear from the solution.


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Sulfur can violate the octet rule because it has an "empty D orbital" is there any other information the periodic table isnt telling us?
In sulfur or any atom with an expanded octet on the 3rd row, where does the idea that they have an empty D orbital come from ?

The periodic table hides the existence of quantum chemistry, including allowed electron quantum states.

but why? bc their energy is higher and therefore their uncertainty greater ?

@Poutnik The octet rule hides the existence of quantum chemistry. Atoms are not taught the octet rule, they go ahead and make compounds as they please. $\ce{SF6}$ exists because we observe it experimentally, and quantum chemistry is a good model to explain why it exists. Quantum chemistry also predicts  that we would not find any $\ce{OF6}$ lying around. If someone makes $\ce{OF6}$ anyway, we have to update the model.

@KarstenTheis We should really start teaching atoms what theories they are allowed to follow and which not so that they’ll be better behaved!

I'm voting to leave open because there is a high-quality answer.

Sulphur does not expand its octet. That was a hypothesis proposed long ago and vehemently supported by Linus Pauling. While Linus Pauling had a few very great ideas that moved chemistry forward, he also made some major blunders like the octet expansion hypothesis or seven grams of vitamin C a day. Possibly because of Pauling’s status and Nobel Prize, octet expansion remains to be present in high school textbooks to this day and even professors still believe in it.
Back in the day, a competing hypothesis to explain compounds such as $\ce{SO3}$ was proposed: charge separation. In a nutshell, the central sulphur atom does not take part in three double bonds (which would expand the octet) but rather forms two $\ce{\overset{+}{S}-\overset{-}{O}}$-type single bonds and one regular double bond for a total of eight electrons. On paper, these bonds can switch back and forth so that each oxygen gets one third of the double bond share (in reality, they do not switch; every bond has the bond order 1.333, all are equal length and so on. This is resonance, akin to what is found in benzene, and often called Y-aromaticity on this site). You can learn more about $\ce{SO3}$ in this wonderfully detailed answer.
In the case of $\ce{SF6}$, fluorine can only form single bonds to sulphur. So instead of a $\ce{\overset{+}{S}-\overset{-}{O}}$ bond we see a $\ce{\overset{+}{S}\bond{...}\overset{-}{F}}$ bond which is always opposite a ‘normal’ $\ce{S-F}$ bond, so together they form a $\ce{F-\overset{+}{S}\bond{...}\overset{-}{F}}$ 4-electron-3-centre bond with a similar resonance phenomenon; this type of bond probably won’t get introduced for a while in your curriculum.
Now what about this ‘empty d orbital’? Well, that actually ‘exists’ and can be derived from quantum mechanics. You may have heard of the $2n^2$ rule for predicting how many electrons can fit into a shell—but at the same time heard of the octet rule which means that after Ar the next electron gets added into the 4th shell although there is still space for ten more electrons in the 3rd shell. Well, quantum mechanics tells us that shells can be divided into subshells; we call these s, p, d, f, g, h, …. Quantum mechanics also tells us that the $n^\text{th}$ shell has exactly $n$ subshells in the order I have given them there (so 1st shell has only s, 2nd has s and p, 3rd has s, p, d; and so on). As sulphur’s position in the periodic table tells us that its valence shell is the 3rd shell, it follows that its 3d subshell of the third shell ‘exists’.
The term to exist is stretched a little here. Technically, these orbitals don’t exist at all, they are only ‘paths’ the electrons can ‘follow’ (also two wrong terms but they get the picture across). Only if an electron is excited to a higher energy level will a higher-level (sub)shell be observable. So without electrons, do the shells exist—does a falling tree make a sound if nobody hears it? On the other hand, the energy levels are there. They can be calculated, we can hurl energy at electrons and they can then get excited to these shells. Thus, in molecular calculations these unoccupied shells are usually called virtual: there, but not quite.
So in a lengthy discussion I have established that these empty d orbitals exist. Why can’t they be used in bonding? Well, the most obvious point is hidden in the aufbau principle. Basically, it is the more elaborate octet prediction and says that after putting all possible electrons into 3p the next electron must go into 4s because 4s has a lower energy level in the ground state than 3d. This shows the rather large energy gap between p and d orbitals of the same shell which is one reason why they do not participate in bonding to any meaningful extent.

So this means that sulphur atom has a net +2 charge on it in $SF_6$?

The discussion surrounding SF6 is often centered around two opposing hypotheses:
(1) Hypothesis # 1: SF6 actually obeys the octet rule, because the sulfur atom has a net positive charge and some of the S-F bonds are ionic in nature. For example, four single S-F bonds and two ionic S - F bonds. The six equal S-F bond lengths are then explained as a resonance among all possible placements of the ionic and covalent bonds.
(2) Hypothesis # 2: SF6 has twelve shared valence electrons which form six polar-covalent bonds. In other words, there are six single-order S-F bonds in SF6, and this sulfur atom does not obey the octet rule. In this case, some of the sulfur valence electron(s) are promoted to the 3d subshell.
Without quantitative calculations it would be impossible to determine whether hypothesis # 1 or # 2 is the correct one. The main requirements for determining whether hypothesis # 1 or # 2 is the correct one are: (i) a rigorously correct method to quantify spdfg populations of atoms in materials is required to correctly quantify the (non-tail) d-electron population of S in SF6 and (ii) a rigorously correct method to quantify bond orders is required to determine whether the sum of S-F bond orders is approximately four (as in hypothesis # 1) or approximately six (as in hypothesis # 2). Only recently have rigorous methods been developed to do this. The following journal article shows the sum of S-F bond orders in SF6 is approximately six, which supports hypothesis # 2:
T. A. Manz, “Introducing DDEC6 atomic population analysis: part 3. Comprehensive method to compute bond orders,” RSC Advances, 7 (2017) 45552-45581 (open access) http://doi.org/10.1039/c7ra07400j


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Hofmann and Zaitsev products
It is known that, with regard to the elimination reaction, the Zaitsev product is the thermodynamically more stable product, which is also often the more substituted alkene while the Hofmann product is the thermodynamically less stable product and is usually the less substituted alkene. I have always thought that the substituents refer to alkyl groups. This is because the basis of the thermodynamic stability lies in hyperconjugation with $\ce {C-H}$ and $\ce {C-C}$ bonds.
However, I recently encountered a test question in which we were supposed to identify the Zaitsev product, from the two given product alkenes, based on the groups attached to the alkene $\ce {C=C}$ bond. The alkene double bonds both had two alkyl groups attached to them. The difference between them is that one had an additional $\ce {-OH}$ substituent while for the other double bond, the two other substituents were just hydrogen atoms. On this basis, the teacher said that the alkene with the $\ce {-OH}$ substituent was more substituted and hence, it was the Zaitsev product while the other alkene was the Hofmann product. I would like to ask if this identification was made correctly. Can the $\ce {-OH}$ group be considered to have increased the degree of substitution of an alkene and hence, allow it to be identified as the "Zaitsev product"?
Of course, it is known that tautomerisation would occur for such an alkene product as it is an enol. However, let us ignore this fact in this case.

I think I would need a sketch of the products …

To disregard tautomerization, perhaps the case of a halogen substitutent should be considered instead of the OH?

As to your first paragraph, to say that the Zaitsev product is the thermodynamic product does not imply that it was formed under thermodynamically controlled conditions. The ratio of Zaitsev/Hofmann (Z/H) products in the hydroxide elimination of 2-bromo-2-methylbutane is not necessarily the same as the Z/H ratio in the acid-catalyzed equilibration of the two isomers.

@electronpusher Yes, we could also consider a halogen substituent.


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Titration/ Separating Soluble Salts/ NaHSO4
When performing titration between NaOH and H2SO4, both Na2SO4 and NaHSO4 are formed (both are soluble in water), is there any method to obtain crystals of NaHSO4? (Maybe by separating two soluble salts or varying the number of moles of NaOH?)

It seems the question follows 2 different goals, the titration analytics and salt preparation. Can you clarify the question?

The question states that when titrating H2SO4 against NaOH, both Na2SO4 (H2SO4+ 2NaOH—> Na2SO4 + 2H2O) and NaHSO4 (H2SO4+ NaOH —> NaHSO4 + H2O) will form. Outline a method to obtain crystals of NaHSO4.

It seems the most used way is reaction of sulphuric acid with salts of volatile mineral acids :
$$\ce{H2SO4 + NaCl -> NaHSO4 + HCl ^}$$
Reaction with excess of the acid
$$\ce{H2SO4 + NaOH ->[H2SO4] NaHSO4 +H2O}$$
has several drawbacks:

It is much more exothermic than the former one
It releases extra water
It needs somewhat diluted solution not to be violent.
It needs evaporation of water from quite concentrated

The interesting option could be recrystallization of $\ce{Na2SO4}$ (anhydrate) from $\ce{H2SO4}$.

Thank you so much! Another question I want to ask would be is there a way to separate two soluble salts? Or in this case, Na2SO4 and NaHSO4?

$\mathrm{p}K_\mathrm{a2}$ of $\ce{H2SO4}$ is cca 2.0.  There will be the minor concentration of $\ce{SO4^2-}$ in $\ce{H2SO4}$ so solubility product of $\ce{NaHSO4}$ should be reached much sooner.


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Grammar of elimination reactions
I would like to double check the English grammar of a sentence of an elimination reaction, but I cannot find an answer.
I'd normally say "the elimination of A with X as the leaving group", where A is the initial compound.
So my question is saying "the elimination of X from A", correct? I've never been told not to, but I feel it is wrong chemically, so I would change it to "the leaving/loss of X from A", which unfortunately sounds casual.
EDIT.
So based on the answer below I was totally wrong —I am glad I asked!— and to recap with A as reactant and X as leaving group and optional parts in parenthesis:

✓ → ✗ — the elimination of A (with loss of X)
✗ — the elimination of X (from A)
✓ — the loss of X (from A)
✓ → iffy — the leaving of X (from A)
✓ → ✗  — A gets/is eliminated
✗ — X gets/is eliminated from A
✓ — A undergoes elimination (with loss of X)

How about "A undergoes elimination under [conditions] by loss of X"?

I'm voting to close this question as off-topic because there is no chemistry behind this question and it is more about grammatical correction. It is better suited in ELU.SE.

I don't think it's off-topic: chemical terminology is OK. However, I would also not feel comfortable offering a specific suggestion on how to phrase it, unless some more context is added. That said, phrases like "loss of chloride ion" are accurate and widely used.

I was not sure myself when posting. But I certain the English language stack exchange would say that "the elimination of X (leaving group)" is correct, while from the two comments and my own question the answer seems to be no: the leaving group is lost and the substrate is eliminated, while "the elimination of the leaving group" is wrong by a possibly unspoken convention (62k vs. 100k results in google for "elimination of leaving group" vs. "loss of leaving group").

"A gets eliminated" or "the elimination of A" is just plain wrong, typical sloppy lab slang by many (too often native?) speakers. Very common, but still terrible.
As you say, the small molecule X gets eliminated from substrate A. Chemical English is still just English, and that is the correct way to put the case in question into words in this language.
"A undergoes elimination" (short for "A undergoes an elimination reaction) is a widely used, and imo also correct, variant.

No, being native or not is not very important. It rather (il)logic. A literal translation of the elimination of A would be wrong in French Spanish etc as well. Also "the precipitate is filtered" is plain wrong but accepted. Terrible is also "vacuum evaporation" to mean evaporation in vacuum :) plus one anyway.

I (OP) may be a native speaker, but have a non-English name, so I thought that was a rather risky statement. :-D

So to clarify your answer @Karl, you are saying that the associated verb to eliminate should not be used in Chemistry? I thought it was okay, with the caveat that in English to eliminate is transitive ("I eliminate waste"), but in Chemistry I would have said it is intransitive or passive only ("The substrate is/gets eliminated via the loss of leaving group").
My question was specifically about the usage of "the elimination of leaving group" statements, which I believe is wrong (but another author did not).

@MatteoFerla Why should the correct way to say this in normal English suddenly  be wrong. There is no "Chemical" English, only British and American English (and a few more regional, orthograhic variants).

Sorry, I had not noticed that you had edited to include "elimination of A". Wikipedia uses it extensively though.

Nitpick: *that's or *that is


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How to deal with non-symmorphic space groups for orbital/vibrational analysis?
I am learning about how to apply group theory analysis of MO diagrams and vibrational in the solid state context (not isolated molecules).
I think the strategy for calculating representations and getting the the orbitals/vibrations with distinct symmetry is quite clear for normal point groups. Essentially you forget about the solid state part of the problem for the most part and just look at the unit cell as a large "molecule".
To incorporate crystal momentum, the unit cell becomes large enough to contain one wavelength of the corresponding wave. This is more complicated, but it seems like you just do a symmetry analysis of a larger unit cell with (typically) lower symmetry.
Now I am trying to understand how this process extends to nonsymmorphic space groups (not point groups). Here I have trouble finding character tables online, and the screw axes/glide symmetries throw off my intuition.
So my question is: how do we apply group theory for bonding and vibrations in the context of nonsymmorphic space groups? I guess it should be the exact same as usual, but it just doesn't seem intuitive how bonding should care about screw axes or glide symmetries that connect far away atoms.


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Can the rate of reaction of a Voltaic/Galvanic cell be determined theoretically?
I am conducting a chemistry practical in which I measure the voltage of a Zn-Cu Voltaic cell over time.
Assuming that the surface area of the electrodes are known, concentration of reactants are known, wire resistance is known (Could assume negligible for simplicity).
Is it possible to determine the rate of the reaction for this setup?
Could I then determine respective concentration of reactants? This would enable the use of the nernst equation to determine theoretical voltage.
Many thanks,

You deal with 2 independent things, thermodynamics, determining the Nernst equation, and kinetics, determining the electrode uquivslnce establishing and response to the load. The latter involves diffusion in free solution, diffusion in dielectric layer and kinetic rate of electrode reactions

Since battery scientists got Nobel this week, this question is worth pondering. It is not a trivial question. However consider this question:
A car battery has 12 V, whereas 9 V cells are also common. If we connect two 1.5 V cells and one 9 V cell on series, one can generate 1.5+1.5+9 =12 V, yet this arrangement cannot start a car despite producing the same voltage.
What is so magical about the lead-acid car battery and this series arrangement? All this shows is that voltage and current produced by battery (which is the rate, hence kinetics and hence the rate of reaction) are two separate things. The main problem in batteries, the rate of electron transfer, is always a limiting factor as to how much current can be drawn from them. So far the lead acid battery is still one of the best because it has a very fast electron transfer rate unparalleled by other batteries. I do remember there was another Nobel Prize winner (Marcus?) who did a lot of work on electron transfer kinetics on the electrodes.
The Nernst potential you see for batteries is the potential when a very very small amount of current is drawn (ideally none) from the battery. This way of measurement of the potential is called null point potentiometry. Under those conditions, the rate of reaction is nearly zero because no current is being drawn from the battery.
In short one cannot measure the rate of the electrochemical reaction just from the Nernst potential because it heavily depends on the amount of current being drawn from the battery. Indeed a non-trivial problem worth someone's PhD.

The rate of the chemical reaction perfectly correlates to the current drawn. Not sure if Farady got a doctorate for that, but they named the proportionality constant after him. ;)

Faraday never got a doctorate, masters or bachelors, but his experimental hand, ingenuity and original thinking was far better than hundreds of PhDs churned out by universities today. If you read Nature, time and again mention that enough PhDs please :-)  If someone can figure out how to control, actually speed up, electron transfer kinetics battery science will make great progress.

You can very easily determine the rate of the reaction, its the current flowing in your wire, divided by Faradays constant.
If you want to ask what the maximum current of your battery is:
That depends on which is the rate determining step in these:

electron transport in wire (sure not)
reaction at cathode
reaction at anode
transport of anion in electrolyte
transport of cation

Chances are that it is one of the latter, i.e. the maximum reaction rate depends (in first approximation) simply on the diffusivity of the slower of your two ions and its charge.
Then you have to wrap that, together with the voltage and dimensions (electrode size and distance), into an equation, and I would guess you wont come out too far from reality.
If your electrodes are large and closely spaced or your electrolyte more concentrated, then the last nanometers at the electrodes become important. Thats more complicated, large books have been written about it.

Ok, so we are supposed to estimate the limiting factor of the reaction and work from there?

Obviously. The steps can only work together, so the slowest one determines the speed.

For short, the limit on the battery is either external resistance in the circuit, in which case the battery will have its nominal voltage. Or interval resistance, in which case the voltage of the battery will drop accordingly.

@StianYttervik Each battery consist of two potential differences, connected by your external load (light bulb or whatever) and the electrolyte. If the resistance of the load is not much larger than that of the electrolyte, then you need to take that internal resistance into account.


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Work done in cyclic process
I have been told that when $\Delta V=0$, work done will be zero $(W=0)$. I know that, in cyclic process $\Delta V$ will be zero, since it's a path function and $\mid W \mid$ i.e. magnitude of work done will be the area enclosed by the curve. But my doubt is when $\Delta V=0$, shouldn't $W$ also be zero in such processes?

If the W had been zero, the fuel combustion engines would have stopped working.

@orthocresol it certainly wasn't homework problem...

The reversible work is $W=\int{PdV}$ over the path of the process.  This integral is not zero for the cyclic process (involving V changing) even through $\Delta V$ from beginning to end of the process is zero.  So, whoever told you that $W = 0$ when $\Delta V=0$ is wrong.

I'm not sure about my answer but I think this is how it goes.
$\Delta G=-W$ (useful work);
$= \Delta H-T \Delta S$
$= P + \Delta V -T \Delta S$
Therefore even though $\Delta V$ is zero it does not imply that all the other terms are zero too. Therefore $W$ is not zero.


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Why CO2 is not reducing the pH of NaOH as the way it is expected
I am using $\ce{CO2}$ for the reduction of pH of $\ce{NaOH}$ solution ($\pu{2.5g}\; \ce{NaOH} + \pu{250ml}\; \ce{H2O}$, with $\mathrm{pH}\; 13+$).
I am introducing $\ce{CO2}$ with different dosages ($\ce{50cc, 100cc, 150cc, 200cc, 250cc, 300cc, 600cc, 900cc, 1200cc}$, and $\pu{1500cc}$) using syringe into $\pu{35 ml}$ solution of $\ce{NaOH}$ prepared as mentioned above. The $\mathrm{pH}$ first increases and then starts decreasing but stop at the same level of initial $\mathrm{pH}$ value of the solution (The amount of $\ce{CO2}$: $\pu{50cc}=\pu{0.03g}$).
Trials run for $\pu{60 min}$ after that, the samples are preserved at $\pu{4 ^\circ C}$ and $\mathrm{pH}$ measured again after $\pu{24 h}$, but the value does not reduce.
Ideally, if everything works properly, the $\mathrm{pH}$ of the solution should decrease gradually. However, instead of $\mathrm{pH}$ reduction first it increases and then decrease at the same level, when we mix $\ce{CO2}$.
I will appreciate if someone share his or her experience or comments as to what is the possible cause of this unexpected trend and how to fix it.

You're trying to neutralize with gaseous CO2? Well, that's like least effective way that could be still used in practice. Also I don't see why you'd think it'll work like you added strong acid...

@Mithoron Well, I would at least expect pH does not increase...

@Poutnik And I would expect OP to point out clearly that pH rises instead of get lower. In that case he's both neutralizing it in wrong way and using wrong tool to measure pH. May be a matter of too low pH for glass electrode.

He  did point it clearly, you may have missed it. But it is a well known issue most of glass electrodes do not behave properly at pH 12+.

You add max. $0.9g$ or 20 mmol of a weak acid to a solution containing 9 mmol of a strong base. Assuming that you loose no gas. Are solution and CO2 in two connected syringes?

@Mithoron, Maybe you are right there is something going wrong, however I used three different kinds of pH meter,  present in the lab, but the phenomenon is the same in all. eg., the pH at the initial stage is 13 when I inject the CO2 into the solution, pH raised up upto 13.3 and then start decreasing. and when it reaches back to point pH 13 or 12.8 it stops. It is might be possible that the amount of CO2 using for the process is insufficient, but the behavior (rais up in pH) is dramatical.
the use of CO2 is necessary because it's less harm and available in abundant quantity..

@Karl, I am using a syringe of 300cc. might be there are losses, to prevent the losses I'm using a stopper to block the gas inside of syringe. But the gas is directly injected into the test tube (size 50 ml, sample=35ml) through a pipe which opens inside the NaOH solution.  during the phenomenon its handle with care that there is no gas losses.

The pH increase by 0.3 units is just an instrumental artifact of the glass electrode.

There are a so-called acid error and an alkaline error with glass electrodes. Which means that the pH electrode shows a systematic bias when your solution is too acidic or too basic and if it contains lot of sodium ions. The value which you quote for +13 are unreliable. So if you are bubbling carbon dioxide into NaOH solution, you are indeed consuming NaOH and converting it into sodium carbonate. The Wikipedia article on glass electrode is poorly written however if you search the term "alkaline error of glass electrodes" you will hundreds of better results.
May be you should follow the pH changes spectrophotometrically using a proper indicator (dyes) in NaOH.
Also keep in mind that you shouldn't store NaOH in glass containers because glass starts to dissolve in NaOH (very slow process though)

NaOH especially destroys the glass electrode.

@M. Farooq, your suggestion is appreciatable.


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Do alpha hydroxy ketones give iodoform test
So, we know that alpha hydroxy ketones give tollens test as the tautomerise into their enol form and the then have an equilibrium between the 2 keto forms.
So going ahead with the same logic. Do alpha hydroxy ketones which can tautomerise to give a methyl ketone group, can they give positive iodoform test?

https://chemistry.stackexchange.com/questions/73313/why-do-%ce%b1-hydroxy-ketones-give-tollens-test https://chemistry.stackexchange.com/questions/94620/in-what-way-could-benzoin-give-tollens-test

Huh, I misunderstood your question. Your parallel isn't good. It's rather a duplicate of https://chemistry.stackexchange.com/questions/104242/why-does-ch3chohch2ch3-give-iodoform-test

Depends on the structure of the alpha hydroxy ketone. Oxidation of an alpha hydroxy ketone to an alpha diketone is not the same as an iodoform test.


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Ambiguous oxidation state of phosphorus in phosphorous acid
I was doing a question in which we were required to find the oxidation state of $\ce{H3PO3}$. I knew the structure that is:

Now, phosphorus being less electronegative than oxygen gets $+2,+1,+1$ charge from respective oxygens. Now we see that in case of hydrogen, it is taught that phosphorus being more electronegative than hydrogen thus it gets $-1$ charge from it, and finally adding all the charges we get Oxidation state of Phosphorus $=+3$.
Now my doubt is that I found on the internet that hydrogen is more electronegative than phosphorus that is, electronegativity of hydrogen$=2.2$ whereas electronegativity of phosphorus$= 2.19$.
Therefore, phosphorus must get $+1$ from hydrogen thus oxidation state of phosphorus $=+5$ but I know this is wrong and the probable reason I could figure out is that hydrogen is more stable in $+1$ form. Therefore, oxidation state of phosphorus is $-1$.
Please tell me if this reason is correct or please add to it.

People usually tend to assume that hydrogen is less electronegative than phosphorus (at least for the purposes of this question), which makes the latter's oxidation state +3. Don't overthink it. They are all arbitrary anyway.

Generally we consider their electronegativity equal
If you go in such depths then you will have a very hard time in thinking about oxidation states of phosphorus sulphur nitrogen in their oxy acids In competitive exams you need to remember all these structures as there are not any generalizations there


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Why is my washing powder not to be used for pre-wash?
Apologies but I can't find a more suitable stack exchange website to ask this question.  I feel the technical answer is chemical in nature so I landed here.
For at least the past two years (and on-and-off longer than that) I've been putting a little bit of my washing powder into the prewash compartment, as well as the main compartment in my washing machine, and then doing a pre-wash before the main cycle.
A few days ago I noticed on the washing powder I had been using there is a red cross over "Wool/silk/prewash" - Implying that I should not prewash (at least not with this powder)
My clothes have seemed fine all this time (except there has been the odd occasion where I've been slightly paranoid that they are not being washed as well as they could be) but I wonder if I am damaging the clothes and/or the washing machine by doing this.

That "prewash" cycle is only meant for clothing which is really dirty. Otherwise its a waste of time, energy, water and detergent. Anyway, what type of washing powder are you using? Wool and silk (protein based textiles) are sensitive to some detergents.

I may stop using it then.  I am a little bit clean/tidy obsessive so I've used the prewash because it makes me feel like I am getting cleaner clothes out of it.  I use the 'Bold Lavender & Camomile' biological powder (UK).  I only mention wool and silk because the red cross is over the three "wool/silk/prewash" I don't wash anything wool or silk.

That is just an ordinary washing powder. No idea why its "bio". The enzymes wont work during a short prewash cycle, and you have to wash out the perfume again in the second step. Otherwise, it think that "no prewash" sign is a nobrainer.


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Can the human body withstand the heat from setting plaster?
There was an incident a couple of years ago where a teenage girl during an art lessen got her hands stuck in a bucket with plaster. As the plaster set, the heat from the chemical process caused her fingers to melt (it says in the article that the temperature reached 60 °C; I am not sure how this can melt your fingers, but this seems to be the case). As a result, she had to have all but two fingers amputated. The lesson we seem to learn is that sticking your hands into a bucket with plaster is a bad idea.
At the same time, plaster casting the human body is something that people do all the time, both in art and for medical purposes. Of course, the layers are thinner in this case. So, what are these people doing differently than in that school incident? Could they be using a different kind of plaster that generates less heat?

Your cast has a far larger relative surface than that bucket. Also the incident was in Boston, if you read that link carefully. Plaster of Paris, not in.

Also you can mix the plaster with a large amount of inert filler if you absolutely want to put your hand in a full bucket, instead of just wrapping it in bandage with some plaster added. And think about how to get your hand out again without heavy machinery. Seems the stuff only becomes hot after it has already set ...

I once fixed a boat hull outside with some two component epoxy. It was under direct sunlight and I did the error of wanting to do some final smoothing of an uneven blob with a hand. That glue was smoking hot. Using it under direct sunlight was probably an oversight. UV light accelerates many things epoxy-like. Getting my hand out of the rapidly heat-shrinking/deteriorating glove was fortunate.

The thermal effect depends strongly on the amount of plaster and the surface area of the  object filled with plaster. A bucket filled with plaster will reach a much higher temperature than a thin layer of plaster (as used in fixing broken bones).

What are these people doing differently than in that school incident?

The plaster casting of the human body is a totally different process compared to the incident at school. At school, it was a liquid plaster of Paris paste ready for molding. Nowadays, plaster casting is done in medical purposes differently.

First they put thin stockinette (e.g., Delta Net®) covering the injured joint or around it covering all of the area that needs to be cast.
Then, put padding (e.g., Soffban®) over the stockinette. Wrap padding at least twice over the area.
Then apply wetted plaster of Paris bandage (Trade names: Gypsona®, Specialist®, Platrix®, or Biplatrix®) over the padding.
To secure the Plaster of Paris bandage, wrap around elastic or cohesive bandage (e.g., Elastomull® or Tensoplus®) over.

As you see, no liquid plaster of Paris paste has been used. The process just uses wetted plaster of Paris bandages, the extra water of which is also removed before wrapping around the injured surface. The videos of this, this, and this would help you understand the process clearly.

The thermal effect depends on the amount of water used. If you mix plaster of Paris with 0.6 times its amount of water (in weight) at $20$°c, the temperature will not go higher than $37$°C.

You can't make such a strong blanket statement.  For instance, it doesn't account for the temperature of the water being used.


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Does the distillation of alcohol cause the fractioning of isotopes and carbon-12 enrichment?
In water, distillation causes fractioning of the $\ce{^{18}O}/\ce{^{16}O}$ oxygen isotopes (Wikipedia).
Does something similar happen to the $\ce{^{13}C}/\ce{^{12}C}$ and possibly $\ce{^{14}C}$ isotopes in alcohol when it is distilled out of a watery brew?

Looking at the percentage differences of the different isomers may be instructive.

Yes, It does. But the effect would be weaker, because of smaller relative mass difference of isotopes, and of the ethanol mokecule. Additionally, different mass of C atoms has much lower impact on ethanol intermolecular bonding, than has O mass on water bonding.

@Poutnik Selection among ethanol molecules might be limited, since most of ethanol evaporates out of the brew and water (with a bit of ethanol) stays behind. When evaporating water partially, as with Oxygen, all molecules are subject to selection. I wonder if fractioning still takes place for carbon.

The first ethanol fractions are enriched by 12C. As most of ethanol is evaporated, the last fractions are in contrary  impoverished, with total balance about neutral.

@TAR86 Can you expand on that? How would the enrichements of isomers relate to that of isotopes?

@HannesH I have not found a clean definition of molecules differing only in isotopes (same structure otherwise). I used isotopes for that. Basically, you are looking at 47 vs 46 g/mol for ethanol and 20 vs 18 for water in the examples given. The percentage differences are rather dissimilar.

@Poutnik, I would have guessed the same as you, but the literature (see my answer) surprisingly says that during ethanol distillation, the distillate is enriched in 13C relative to the starting material.

Does the distillation of alcohol cause the fractioning of isotopes[...]?

The short answer is it depends.

Distillation of ethanol from water can be done nearly to completion.  There are published methods for determining isotopic composition of sugars in fruit juices which involve fermentation of the juice to yield ethanol, followed by nearly quantitative distillation of the ethanol from the resulting brew.  Then, the 13C content of the ethanol is measured.  If the ethanol distillation step (or fermentation step) substantially altered the isotopic composition of the sugars, this method would not be suitable.  The main thing which determines the isotopic composition of the ethanol in this case is the type of photosynthetic metabolism used by the plants from which the juices or syrups are derived.  Quoting from the linked paper:

The fruit juices or maple syrups are completely fermented with yeast, and the alcohol is distilled with a quantitative yield (>96%). The carbon-13 deviation (delta13C) of ethanol is then determined by IRMS. This parameter becomes less negative when exogenous sugar derived from plants exhibiting a C4 metabolism (e.g., corn or cane) is added to a juice obtained from plants exhibiting a C3 metabolism (most common fruits except pineapple) or to maple syrup.

If distillation is not done to completion, there can be a fractionation.  In fact this has been measured fairly accurately in 1990 for distillation of pure ethanol.  Here is a figure from that paper.

It plots the ratio $R_D$ of 13C to 12C of the distillate ($D$) normalized to the same ratio $R_Q$ for the starting material $Q$, as a function of the fractional yield or progress $\rho$ of the distillation (so $\rho = 0$ corresponds to no distillation and $\rho = 1$ corresponds to complete distillation).  It also plots the same curve for the remaining liquid $L$.

As you can see, at $\rho = 0$, when only a few molecules of ethanol have distilled, the liquid (unsurprisingly) has an unchanged isotopic composition relative to the starting material, and the tiny amount of vapor that does come over has an isotopic composition that is enriched in $\ce{^{13}C}$.  That's right, for pure ethanol at least, distillation is a process with an inverse isotope effect.
As $\rho \rightarrow 1$, the isotopic content of the distillate tends to 1.0,  consistent with the idea that for quantitative distillation, isotopic composition doesn't change.
At 50% completion, the distillate has about 0.2% more $\ce{^{13}C}$ than the starting material, and the remaining liquid has about 0.2% less $\ce{^{13}C}$ than the starting material.  So even though the composition does vary, the effect isn't very big.

Does the distillation of alcohol cause [...] carbon-12 enrichment?

No, contrary to what I would have guessed before I read the literature, distillation of alcohol can cause carbon-13 enrichment.

Great answer, and surprising indeed. Any thought on how this inverse isotope effect could come about?

User @Paul supplied a nice answer and comment pointing to  this 1970 paper discussing the inverse isotope effect for 13C-methanol, where is also seen.


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Why does the displacement reaction of zinc in copper(II) sulfate solution result in dark metallic copper?
It is commonly known that when zinc metal is placed in a solution of copper(II) sulfate, a displacement reaction occurs, and elemental copper is deposited onto the decomposing zinc metal. A practical demonstration of this shows the zinc turning black from the deposited copper.
My question is, why is the deposited copper black? I thought metallic copper was red. Moreover, the dark color can be seen instantaneously, meaning the color cannot result from copper oxidizing.
I recently read a paper that claimed the copper is deposited in the form of plasmonic nanoparticles, and said nanoparticles display excellent solar absorption (and low thermal emittance, in fact). Any ideas?

All metals look black or dark when presented as a sufficiently thin powder. The metals display their color only when the dimension of the grain is greater than the wavelength of the light, namely 400 to 700 nm. This is about 1000 times bigger than the atoms. When prepared from a precipitation reaction, the metal atoms are never arranged neatly in a regular fashion. So they do not reflect the light like a metallic plate.


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Why does ammonia have a pKa of 32.5 at −33 °C?
According to Wikipedia, ammonia has a $\mathrm{p}K_\mathrm{a}$ of $32.5$ at $\pu{-33 °C}.$  Why does it say that? Is it a typo? Isn't anything above $12$ is a very strong base? And both bases and acids are slightly stronger at higher temperatures, not lower, correct?

That is the boiling point of ammonia. Pure liquid ammonia will have different properties than ammonia in solution. Also, when you switch solvents, it becomes difficult to compare pKa values.

The reaction described by the pKa of ammonia is $$\ce{NH3 <=> NH2- + H+}$$ In water, ammonia acts as base. A pKa of 32.5 means that the amide anion $\ce{NH2-}$ is a strong base. The temperature dependence is determined by whether the reaction is endothermic or exothermic.


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Calcium ion and sulfate did not form precipitate
I was in the chemistry lab earlier this week. We were doing a qualitative experiment (no concentration given for any solution).
I added an unknown solution that contains $\ce{Ba^2+},$ $\ce{Ca^2+},$ and $\ce{Mg^2+}$ with $\ce{(NH4)2SO4},$ I got barium precipitate and added $\ce{(NH4)2C2O4}$ and another precipitate appeared. I was told the second precipitate is $\ce{Ca^2+}.$
But I recall from solubility mnemonic rule, I should be able to see precipitate from both $\ce{Ca^2+}$ and $\ce{Ba^2+}$ after adding ammonium sulfate. Why didn't I? What are the mechanics behind this? How did calcium just sit there as a spectator ion?

Can you please tell, where did you get this handout from, Can you please share other helpful ones if possible. Please !!

Do you mean the CASH N' Gia Solubility Rule sheet link? I saw the mnemonic at reddit a while back and googled it. This is what I found. I think there's another sheet for precipitation rule sheet called "CCOPS", which is developed by UTexas at El Paso.

Here it is: CCOPS are insoluble salts: C-Carbonates and Chromates, O-OH (hydroxides), P-Phosphates, and S-Sulfides. In using the full mnemonic, PLs introduce sets of exceptions as HAPpy: HAP for Hg, Ag, Pb as exceptions for CASH N’ Gia solubility for halides; and as HAPpy (Hg, Ag, Pb) and CBS (Ca, Ba, Sr) for sulfates. see here: https://www.researchgate.net/publication/267322054_CCOPS_Chase_CASH_N'_Gia_A_Solubility_Rule_Mnemonic

While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at numbers.
\begin{array}{lcr}
\hline
\text{salt} & \text{mass solubility} & \text{molar solubility} \\ \hline
\ce{BaSO4} & \pu{2.45e-3 g/l} & \pu{1.05e-5 mol/l}\\
\ce{CaSO4} & \pu{2.1 g/l} & \pu{1.54e-2 mol/l}\\
\ce{CaC2O4} & \pu{6.7e-4 g/l} & \pu{5.23e-6 mol/l}\\
\hline
\end{array}
To calculate your actual experiment, you would have to transform these values into $K_\text{sp}$ values and then calculate how much would precipitate at each given step. Without wanting to do that tedious calculation, you can still arrive at the general conclusion:

addition of sulphate precipitates practically all barium ions

addition of sulphate precipitates a large part of calcium ions
but: a non-neglegible amount of $\ce{CaSO4}$ remains dissolved

addition of oxalate precipitates practically all the remaining calcium.

In fact, the solubilities of $\ce{BaSO4, SrSO4}$ and $\ce{CaSO4}$ are such that a saturated solution of $\ce{SrSO4}$ will always give precipitate when barium is added but never with calcium or strontium; while a saturated solution of $\ce{CaSO4}$ will give a precipitate if either strontium or barium is added but never with calcium. This can be used to identify the heaviest alkaline earth metal in an unknown sample.

Where can I find molar solubility chart? Thank you.

@Molly_K I found solubilities on Wikipedia. You can also Google, check your favourite textbook or ask if your library knows something.

Thank you for your answer. With the Ksp and solubility table, all make sense now. I was quite thrown off when there's no concentration was given for any of the solutions. Ha.


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Visualization of all hcp voids
There are 12 tetrahedral voids and 6 octahedral voids in hcp unit cell but I can't find any image that can show all tetrahedral voids and all octahedral voids in hcp unit cell (the common picture in Google shows 2 octahedral voids and 3 tetrahedral voids). Could someone help me to draw or find visualization image of all voids in hcp?
Thank you.

A unit cell is not what you think it is.

On the left: http://ruby.chemie.uni-freiburg.de/Vorlesung/Gif_bilder/Strukturchemie/packungen_luecken_c.png (German: Lücke = void)


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Can atomicity be infinity?
My chemistry teacher told us that graphite or diamond had an atomicity of infinity. However, the definition of molecule in my testbook is

A molecule is the smallest part of an element or compound which can exist on its own under ordinary condition.

It doesn't make sense to me because if a molecule has an atomicity of infinity, it it is not a discrete compound anymore. It contradicts the definition of molecule.

That's why graphite and diamond are not  made of molecules.

but my notes from my teacher actually classify them as molecules

Then I disagree with your teacher quite strongly.

I don't see what one gains, and in which context, by speaking about infinite atomicity in non molecular solids. Atomicity presumes molecularity. And atomicity is taken as 1 for carbon allotrops except fullerenes. Which also point to the relative usefulness of the term. Speaking about biatomic, triatomic, ... seems more immediate. Also the definition of molecule in the book is wrong. It doesn't always work. You might find cases in which it fails. "Part of a substance" might be already better. ..

@IvanNeretin would you still disagree with the teacher if they classify them as single macromolecules?

@Jan No. Then again, what is the point of talking about molecules in this case?

Introducing the idea of "infinite atomicity" can be useful to highlight the importance of particle size on the properties of metallic or network covalent solids. As the surface to volume ratio of discrete particles becomes sufficiently large, properties of crystalline solids such as melting point, absorption wavelength and reactivity begin to diverge significantly from the limiting values observed for larger crystals. A pronounced change often occurs as the nanoparticle regime is entered, at ~$\pu{ 0.1 \mu m}$, which corresponds to ~$\pu{10^8 atoms/particle}$ or so, well short of infinity, but converging to that limit (presumably heading closer to infinite atomicity will bring you into the regime where gravitational forces become very important, but I am guessing your teacher did not mean this).
But to answer your question, I found the wikipedia useful. On network solids, it states that

In a network solid there are no individual molecules, and the entire crystal or amorphous solid may be considered a macromolecule.

On atomicity it provides the following comment:

All metals and some other elements, such as carbon, do not have a simple structure but consist of a very large and indefinite number of atoms bonded together. Their atomicity cannot be determined and is usually considered as 1.

Is it true that there are $\pu{10^8 atoms}$ in $0.1\mu m$? I think the diameter of a single atom is about $1\mu m$. Or am I missing something?

You can calculate the number of atoms in a sphere 0.1 microns in diameter, starting from the density of a solid, say gold.

@LearningMathematics You are missing a factor of at least 1,000 (or even 10,000) in your measurement of atomic size. Atomic radii are usually fractions of 1nm.


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compare and contrast a 2p6 orbital to a 4s1 using quantum numbers to help
Would you be able to help me understand how to compare and contrast a 2p6 orbital to a 4s1 using quantum numbers to help?

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.   As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.  Please [edit] in your full reasoning or thoughts on this. See Homework.  Otherwise, the question may get closed as "low effort homework-like question."

$2\mathrm{p}^6$, $4\mathrm{p}^1$ are not orbitals.
It is a partial description of an atom electron configuration, saying there are 6 electrons in 3 orbitals $2\mathrm{p}$, respectively 1 electron in the orbital $4\mathrm{s}$.
$2$ in $2\mathrm{p}$ means the (main, energy ) quantum number $n=2$. $n$ can have values 1, 2, .....
$p$ in $2\mathrm{p}$ means the ( orbital, orbital angular momentum) quantum number $\ell=1$. $\ell$ can have values 0, 1, ...  , $n$ - 1.
The ( magnetic ) quantum number $m$ ( $-\ell \le m \le +\ell$ ) is for the axis projection of the vector of the electron orbital momentum ), determining e.g. there is

1 orbital $n\mathrm{s}$
3 orbitals $n\mathrm{p}$, called $\mathrm{p_x, p_y,p_z}$
5 orbitals $n\mathrm{d}$, called $\mathrm{d_{{xy}^2}, d_{{xz}^2}, d_{{yz}^2}, d_{x^2-y^2}, d_{z^2}}$
7 orbitals $n\mathrm{f}$, being seldom explicitly named.

See also Wikipedia.

In a hydrogen atom, or ions with just 1 electrons, orbitals with the same $n$ have the same energy(*).
In multielectron atoms, electrons in different orbitals have different interaction with other electrons and their energy differ. Their energy increases in order $\mathrm{s < p < d < f}$
Generally, but with multi exceptions, the energy ( and filling order) increases in this order:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 4d 7p

The naming has historical origin in spectroscopy. s as sharp, p as principal, d as diffuse, f as fundamental.

s means $\ell = 0$
p means $\ell = 1$
d means $\ell = 2$
f means $\ell = 3$

Orbitals have 3 meanings:

A wave function being  a particular solution of the Schroedinger wave equation, with integer constants $n, l, m$ called quantum numbers.
An electron quantum state with the particular $n, l, m$ values
A 3D shape, describing the probable electron occurence, with borders defined by occurence probability threshold, corresponding to 1 and 2.

(*) Relativistic corrections for Schroedinger  equation reveals and explain observations it is not fully true.


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Appropriate container for storing/keeping chromic acid?
I am planning to make chromic acid, however I the information for keeping chromic acid are conflicting. Can it be stored in a glass bottle even though it's used as a glass etchant? Also I saw some information that it's dangerous to keep it in plastic containers. What's the common practice?

The common practice would be to prepare chromic acid only if strictly necessary, and in an adequate environment (lab, special workshop, ...) and your question doesn't suggest it is the case.

Plastic is surely no good.

It's not a glass etchant.

it used to be stored in glass back in the day when I was doing my PhD and was used as a glass cleaner not an etchant

Chromic acid can well be stored in glass bottles with stopper/lid but should be kept away from any organic compounds. Chromic acid is a dangerous oxidising agent and incompatible with various organic compounds. Also, it should be well marked with proper label and hazard symbol to distinguish it from other mineral acids. It has a decent shelf life and can be kept for long period without decomposition if kept well.
Also make sure it remains separated from mineral acid. Chromic acid is known to form solution with sulfuric acid forming chromosulfuric acid which is ironically used to clean laboratory glasswares.
Chromic acid is not a glass etchant.


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Is it really possible for a liquid to stay afloat a gas (vapor)?
I have a book at home called Chemical Thermodynamics for Process Simulation (I have the first edition) and at the very beginning of chapter 2 of the book one may read the following:

No reference is given. I have never found anything about it so far and I am curious to know whether this statement is true.

Yes, supercritical liquid may have higher density then liquid.

The crucial point is that CO2 at high pressure, and moderately above room temperature, becomes a super critical fluid and not a gas.

The difference between a liquid and a gas is not as clearcut as the question implies.

There is no fundamental reason why a not yet supercritical gas cannot also have a higer density than some liquid at the same conditions, is there?

@Karl There isn't, but even in different conditions that would be rather extreme example.

The critical temperature of CO2 is 35C.  Above this temperature, no matter high the pressure, CO2 can't be liquified.  So it is certainly possible to find a pressure where its density is higher than that of water.

@ChetMiller Water in normal state - sure. Water under equally high pressure - not necessarily; it's density is also getting higher and later it turns into "hot ice" phases.

Density of liquid water just below the critical point ( cca 374 Deg C )  is near 0.2 g/mL. I suppose it is possible to compress gaseous CO2 at this temperature above this density. OTOH, hard to a priory tell, how CO2 dissolution in water affects the density and phase thermodynamics.

Yes! Here's a video of Cody floating NaK (liquid) or ethanol/water (liquid) on compressed xenon (gas):
https://youtu.be/AsP4yMY-a6U


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Aluminum Oxide + Epsom salt ⮕ aluminum sulphate + Magnesium oxide
I was wondering that if you wrapped some Epsom salt in aluminum foil and set it on fire would you get aluminum sulphate and magnesium oxide, I couldn't find anything about this on the internet

OK, I'll bite.  How do you expect alumina to take $\ce{SO3}$ away from magnesia when the latter is a much stronger (Lux-Flood) base?

I thought that because Oxygen is much more reactive than Sulphur the oxygen atoms will go to the magnesium one and the Sulphur atom will go to the aluminum atom

You will get magnesium and aluminium oxides.

Aluminum oxide does not react with sulphates, even at high temperature. Anyway, aluminum sulphate does not support high temperatures : it is easily decomposed at around 700°C.


text:
Finding pH of acid in water
How would you find the pH of an acid dissolved in water? Would you need to take the fact that the pH of water is already 7 into account and go on from there?
Say, for example, that you add 0.0500 mL of 0.00100 M HCl into 50 L of pure water, what will be the pH of the new solution?
I made the mistake of just finding the moles of the HCl placed into the pure water, and finding the new concentration by dividing by the litres of water present. This gave me a pH of 9, which does not seem to make any sense as acids are meant to lower pH!
Some help would be wonderful!

It will be same as earlier.

Related:  https://chemistry.stackexchange.com/questions/107852/ph-of-very-dilute-acids?r=SearchResults.

In this solution, there are two sorts of $\ce{H+}$ ions, those coming from added $\ce{HCl}$, whose concentration is $10^{-9}\ \mathrm M$, and those coming from the water which are unknown, and so described by $x$. But the concentration of $\ce{OH-}$ is also $x$. So $K_\mathrm w$ may be written:
$$K_\mathrm w = (x + 10^{-9})x =  10^{-14}$$
If you solve this equation, you find: $$x = 0.999\times10^{-7}$$
The total concentration of $\ce{H+}$ ion is $1.0001\times10^{-7}$; $\mathrm{pH} = 6.999$.

Most of $\mathrm{pH}$ calculations make simplifying assumptions and one must determine if they are justified for the particular case.
Simplification for strong acids $$\mathrm{pH}=-\log{[\mathrm{acid}]}$$
implies $\pu{[\mathrm{acid}] >> 10^{-7} mol/L}$.
Because water dissociation constant $K_\mathrm{w}=\ce{[H+][OH-]}=\pu{10^{-14} mol^2/L^2}$,
neutral water contains $\pu{10^{-7} mol/L}$ of $\ce{H+}$.
That means the very most $\ce{H+}$ ion must come from acid dissociation and not from water dissociation.

If OTOH very most $\ce{H+}$ are from water dissociation as in your case, another simplification says $\mathrm{pH}=7$, no matter how much of acid you added.

For general case:
$$c_{\mathrm{acid}}=\ce{[A-]}=\ce{[H+]} - \ce{[OH-]} = \ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}}$$
$${\ce{[H+]}}^2 - \ce{[H+]}\cdot c_{\mathrm{acid}} - K_\mathrm{w}=0$$
$$\ce{[H+]} = \frac{c_\mathrm{acid} + \sqrt{{ c_\mathrm{acid}}^2 + 4\cdot K_\mathrm{w}}}{2}$$
The limit case for $c_\mathrm{acid} << \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = \sqrt{K_\mathrm{w}}$$
$$\mathrm{pH}=\frac 12 \mathrm{p}K_\mathrm{w}$$
The limit case for $c_\mathrm{acid} >> \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = c_\mathrm{acid}$$
$$\mathrm{pH} = -\log{(c_\mathrm{acid})}$$

Simplification for weak acids $$\mathrm{pH}=0.5 \left(\mathrm{p}K_\mathrm{a}-\log {\ce{[acid]}}\right)$$
implies negligible water dissociation and also negligible relative acid dissociation $\ce{[HA]} >> \ce{[A-]}$.
So one can assume:

The acid concentration $\ce{[HA]}$ is not affected by acid dissociation.
$\ce{[A-]} = \ce{[H+]}$


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If gases assume the volume of the container that they are in, then wouldn't a gas that isn't enclosed in a container have an infinite volume?
But if the volume was infinite for gases not enclosed in a container, then how would the ideal gas law even function?
Thanks to anyone who can point out the piece of this puzzle I am not considering!

An absence of container does not mean an absence of external pressure. If gravity is involved, it ends with limited volume of unconstrained gas in the Earth atmosphere. As very minor fraction of molecules crosses the escape velocity.

Yes a gas that isn't enclosed will have infinite volume. What's the problem with that?

@Ivan Neretin Finally the gas own gravity become equal the repulsion pressure, as the gravity decreases with r^2, but the pressure with r^3. So no infinite volume. It is seen in astronomical scales.

@Poutnik In effect, you are talking about the Jeans instability. Yes, that's a thing; I ignored it.

@Ivan Neretin Yes, it is closely related by the critical density for given volume.

As Ivan Neretin said, where is the problem? In particular for an ideal gas there won't even be gravitational attraction. See also the answer, which basically says the hypothetical ideal gas in that space will asymptotically tend to equilibrium.

@Alchimista The gas laws ignore gravity, but that doesn't mean it won't have an effect. You can, and atmospheric scientists would, add gravity and get results consistent with the ideal gas law but accounting for gravity as well. Th gas laws ignore this because, at any scale smaller than planetary, gravity is irrelevant.

Of course! @matt_black. The question refers/is dictated by the ideal gas: "how would the ideal gas law even function?" quotes OP

I give you a case. Suppose you opened an airtight jar containing an ideal gas. Suppose by some mysterious power, you are able to observe the molecules of the gas. What do you think you'll see on opening the jar?
Obviously you'll see the molecules dissipating in space(the correct word would be diffusing). Initially they suppose on opening occupied 100ml, then they occupied 200, then 300... Thus at every point of time the molecules have a finite volume, the value may be very large, but still its finite. Thus we say a gas "will occupy infinite volume" but only after "infinite time".

When radioactive gases are released, you can trace them around the world. During the Fukushima Daiichi nuclear disaster, iodine-131 was released between 12 and 15 March 2011. Elevated levels of the isotope were recorded on the east coast on March 27, 2011 (i.e. about two weeks later). With such large volumes of gas as in the atmosphere, transport mechanisms like wind will play a larger role. See https://pubs.usgs.gov/of/2011/1277/report/OF11-1277.pdf

Yes you're right. Iodine-131 was also detected in the air at the time of Chernobyl.

Yes. But with several complications you don't normally use the gas laws to deal with
The ideal gas law is useful in laboratory situations where you can control the various components in the law (volume, pressure, amount of gas, temperature). And it usually describes the equilibrium reached when those factors are held constant. In the situation you describe, while the laws still work, you may never be able to observe the equilibrium state of the gas.
Consider practical experiments such as releasing a container of hydrogen held while you were spacewalking in the vacuum of space. The lack of a container means the hydrogen molecules will spread out eventually occupying an infinite volume. But you can never see this state as the individual molecules travel at a finite (and not very fast) speed. Even at the end of the lifetime of the finite universe, they may not have spread far enough to reach the equilibrium predicted by the gas law when the volume is infinite.
The more constrained case of opening a container of hydrogen to the atmosphere in a laboratory has a similar constraint. The gas sees a nearly infinite volume (the entirety of the earth's atmosphere is pretty large compared to one cylinder of hydrogen). But there are more constraints than in space. The hydrogen will diffuse into the atmosphere (more slowly than it would expand into space) and perhaps constrained even more by the walls of the lab. The expansion is also limited slightly by gravity which might contain the ultimate volume occupied to be the same as the volume of the atmosphere (but we normally ignore the complication in the ideal gas law). But, again, you would never really see the final equilibrium predicted by the gas law as it would take a long time to reach and other factors would prevent it being reached (some molecules in the upper atmosphere would be swept into space rather than being constrained by the atmosphere).
In both these cases the final equilibrium state predicted by the gas law is still the same. But you can never observe it as it takes too long to happen and other complicating factors will become more important and will dominate the outcome.
The gas law is not wrong but it is sometimes irrelevant.


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Best ways to compress and store tons of CO2?
We are looking at doing planet changing work, and removing CO2 from the atmosphere. The issue is storing the CO2. Please let us know the best ways to compress millions of tons of CO2, as we have limited storage space. For example, how much could we compress a ton of CO2 gas? What size container would a ton of CO2 fit in, when fully compressed? Also, would turning tons of CO2 into liquid be more space effective? How much space would a ton of CO2 gas that has been turned into liquid take up?

Compression of CO2 mens also CO2 generation unless everything is green powered.  People looks for sequestration, as in the answer. Or to some kind of fuel production, that again must be green to be effective.

To remove 100ppm CO2 from the atmosphere of Earth would require removing about 1800 billion tons of CO2.  If you solidified this CO2 into dry ice you would be producing 1200 billion cubic meters of CO2, or about 1200 km^3.  This would be enough to fill half of Lake Victoria, or cover the entire land area of Hong Kong in a meter thick layer of dry ice.  So... good luck with that.

The global average $CO_2$ in the atmosphere is around 400 ppm.

@M.Farooq And we certainly don't want to remove all of it.  That would be catastrophic.

7z a CO2Storage.7z "Tons of CO2"

Poor carbon dioxide has been unnecessarily defamed as a greenhouse effect. This is not only gas in the atmosphere which is an infrared absorber. Guess what, water vapor is a major culprit as well along with methane (recall cows digestive system). Nature has a very delicate balance, carbon dioxide is plant's food as well.
I recall some people doing PhDs on carbon dioxide remediation in chemical engineering. One of options is to pump carbon dioxide into deep rocks with a hope that it will convert into carbonates, mainly calcium carbonate in deep layers. One can then study all types of phase diagram of carbon dioxide etc.
One green solution is a massive drive for the plantation of trees. This option is far better than more expensive options.

The plantation approach is a great way to restore Earth, but is there any more land left that can sustain vegetation... Nearly every bit appears to have been claimed for construction or housing purposes.

At least the forests which have been destroyed by humans should be re-grown. There is a trend of roof top gardening in some developed areas. And every tree/plant one person grows counts even in a medium sized pot in a balcony.

Just for precision and independent of any debate. Removal of water from the atmosphere happens continuously and fast by rain cycle. CO2 is also fixated but on more slow rate. A quick fixation is done by diatoms rather than plants. But it seems that more CO2 means less diatoms. That is why CO2 is at focus. That CO2 is plant food is not a mechanism of fixation. CO2 → plant→ digestion → CO2 & CH4. It only works if we let forests grow up.

Water vapour naturally precipitates and methane can be burned into CO2. CO2 is not unnecessarily defamed because it persists the longest for long term effects. Also as alchimista mentioned the more plants there are (which also respire) the increased animals that feed on them.

This does not answer the question whatsoever.

Feel to write a separate answer. There is no restriction :-)

The fact that $H_2O$ and $CH_4$ are both also infrared absorbers does not mean $CO_2$ is being "unnecessarily defamed".  E.g., $CO_2$ is a forcing agent, $H_2O$ is not.  (see: https://grist.org/article/water-vapor-accounts-for-almost-all-of-the-greenhouse-effect/ )

It is true that water vapour is the most dominant greenhouse gas. The problem is that water vapour is also the dominant positive feedback in our climate system and amplifies any warming caused by changes in atmospheric CO2. The important point is that the amount of water vapour in the atmosphere is controlled by temperature, therefore any increase of temperature due to CO2 will be amplified by water vapour. Sources: https://doi.org/10.1175/JCLI3799.1 and https://skepticalscience.com/water-vapor-greenhouse-gas-intermediate.htm

In case you're actually trying to design a serious scheme for extracting and storing carbon dioxide, you might want to start by reading the Wikipedia articles on carbon capture and storage and carbon sequestration.
For the impatient, the summary is that it's simply not practical to store that much $\ce{CO2}$ in containers of any kind. Instead, you either:

react it with minerals to form carbonates (which are stable, but require you to first mine lots of suitable minerals and then somehow deal with the fact that the reactions are very slow under typical conditions);
try to do something useful with it, such as:

making plants grow faster (which works up to a point, but would require huge greenhouses to do on a large scale, and isn't that much more efficient than simply releasing the $\ce{CO2}$ into the atmosphere and growing the same amount of plants outdoors would be) or
using it as a chemical feedstock (which typically requires you to spend considerable energy to drive off most of the oxygen from the $\ce{CO2}$, as there's not much you can do with $\ce{CO2}$ without de-oxidizing it first — more energy, in fact, than you obtained by burning the carbon with oxygen in the first place); or

pump it underground and hope and pray really hard that it stays there (which isn't that much of a stretch if you e.g. pump it into a seam that previously held oil or natural gas, since clearly that hadn't leaked in the past; of course, your seam now has a borehole in it that it didn't have before, which introduces an extra leak risk — also, when done with oil or gas fields, this isn't so much a way of removing $\ce{CO2}$ from circulation as it is a way of offsetting the $\ce{CO2}$ emissions from burning the extracted oil or gas).

Pumping the $\ce{CO2}$ into the ocean and letting it dissolve there as carbonate ions is also an option that some have considered, but comes with some serious risks like acidifying the ocean (since carbonic acid is, surprise surprise, an acid), disrupting deep-sea ecosystems and potentially creating dead zones on the ocean floor, and of course the risk of the $\ce{CO2}$ coming out of solution and bubbling back up into the atmosphere over time (or, if you're really unlucky, all at once).
Another option that the Wikipedia articles (currently) only mention in passing is storing the carbon not as $\ce{CO2}$ but as elemental carbon, a.k.a. charcoal.  This is most easily done before you burn the fuel, by pyrolyzing it to split it into carbon-rich charcoal and hydrogen-rich syngas.  You can then bury the charcoal underground (where it's pretty stable and even makes a decent soil additive) and burn the syngas for energy (which releases $\ce{CO2}$, but nearly not as much as if you'd burnt the original fuel directly; if the original fuel was from a non-fossil source, you can even end up with negative net carbon emissions).

As in how much you could compress $\ce{CO2}$ would be the point when it gets converted to dry ice. Surely a solid would take even lesser space than liquid. As for liquid, you need to maintain a pressure of approx 5 atm at 31.1 °C to liquify $\ce{CO2}.$
If you want, you can have a look at the article World can ‘safely’ store billions of tonnes of CO2 underground on carbonbrief.org (accessed 2019-10-13). They here speak about how we can store large amounts of $\ce{CO2}$ in depleted oil fields, natural gas reserves or other such underground storage reserves. Even MIT has a PDF on carbon storage and sequestration. Have a look!


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Is there any reasonably inert magnetic oxide?
I spent yesterday evening looking for data on magnetic oxides. What is easily available isn't well organized, so I ask here in hope someone can remember.
I'm looking for a magnetic oxide (that is, an oxide that can be held in place by neodymium magnet in presence of liquid flow) that has magnetic properties at 350 Celsium and do not react with $\ce{H2S}+\ce{H2}+\ce{CO}$ mixture at this temperature at 50 bar. Common nonmagnetic oxides for this conditions are many $\ce{Al2O3, SiO2, TiO2,}$ to name a few. But with magnetic oxides I hit a wall.
Note: No, ferrites are not an option as they are easily reduced. Same with cobalt and nickel oxides. At first I looked at chromium and manganese componds, but compounds I checked lose magnetic ordering well below room temperature or are easily reduced.
Any suggestions?
Note: At the moment I'm researching compounds of chromium, manganese and middle rare earths, with little luck.

What is the form of the oxides?  Is cladding a possibility?

@OscarLanzi I'm looking for a lab-grade catalyst support. I.e. water- and acid-resistant porous sub-millimeter polycrystalline grains or nanoparticles. I'm not limited to individual compounds, solid solutions and ceramics might fit the bill. I'm doubtful about suitability of coated magnetite.