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Why is concentration never zero for zero order reactions?
By formula of zero order reaction:-
$$C_t=C_o-kt$$
The concentration should be zero at $t=C_0/k$. But in some books and websites I saw that graph between concentration and time don't touch the X-axis. So does zero order reaction complete or not?

The reaction may attain equilibrium some point.

... and before that, you have subtract the rate of backward reaction from the forward reaction rate.

In your body, alcohol like many other drugs is ingested rapidly. It is then enzymatically catalysed in the liver, initially to acetaldehyde in a zero-order reaction with rate constant $\approx \mathrm{0.19 g/dm^3/hour}$. This is how you can work out if someone was above a legal limit at a given time before a test was taken. But at long times,  and thus low alcohol the mechanism changes to a more complex one. See Chapter 10 Q11 in https://chemistry-maths-book.com/

I think you're confusing zeroth with first order.

@Mithoron no it is a zero order reaction I have checked it

If the concentration never reaches zero, then it isn't following "zero-order" kinetics, at least not at all times.
If it were, as you say, the concentration would reach zero at time $t = C_0/k$.
A lot of surface-catalyzed reactions are successfully modeled as zero-order, at least some of the time.  When the availability of surface sites is limiting, and the concentration of bulk-phase (usually gaseous) reactants is high, reactions can seem zero-order. But as the concentration lowers, zero-order kinetics don't always make sense.  For example, when there is only one single molecule of reactant left, it seems unlikely that rate of reaction would be the same as when the concentration of reactant was high.  Maybe as the concentration of reactant lowers enough, availability of reactant, not of surface sites, becomes limiting.  That means that zero-order behavior has ended.
But if there were a mathematically perfect zero-order reaction that was zero-order at all times and concentrations, then you'd be right.  Concentration would go to zero at finite time.

There can be the forward reaction  $$\ce{A -> B + C}$$
with the rate of the zeroth order $$r_\mathrm{F}=k_\mathrm{F}$$.
There can be the backward reaction $$\ce{B + C -> A}$$
with the rate of the second order $$r_\mathrm{B}=k_\mathrm{B}[\ce{B}][\ce{C}]$$.
Then the net reaction rate is
$$r = r_\mathrm{F} - r_\mathrm{B} = k_\mathrm{F} - k_\mathrm{B}[\ce{B}][\ce{C}]$$
It can be seen the net reaction rate is of the zeroth only far from the equilibrium. When approaching the equilibrium, the rate is asymptotically approaching zero at nonzero, equilibrium concentration of $\ce{A}$.
Typical case are glasses with light absorption adapting to illumination, using reversible photolysis of silver chloride.

The combustion of a candle is zeroth order. During the combustion, the mass of wax is constantly decreasing at a constant rate. It goes to zero when the candle is totally burned.

Except you get some wax at the bottom - pretty much Curt's point.

If the candle is vertical, and burns in a calm atmosphere, the wax burns entirely.


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Why are the latent heat of vaporisation values unavailable for cooking oils?
Oil evaporates. We know this, it's called the "smoke point". So why is it then, that whenever I try to find reference values for the latent heat of vaporisation for cooking oil, such as canola oil or olive oil, that the value is undefined in these tables?
Is it unknown? Or is the purity of cooking oil so variable that no scientist dares publish even an average amount of kJ/kg required to change oil from liquid to gas? Why is this information unavailable?
Follow‑up question: Is the information actually available and I've been unable to find it?

The smoke point is not the sign of evaporation, but decomposition. Additionally, they are mixtures with variable composition even for the single kind of oil.

Interesting. So indeed the "purity" is a factor, but you're mainly saying that oil is decomposing rather than evaporating, and that is why this information (latent heat of vaporisation of cooking oil) is unavailable, right? That's interesting because I now presume that cooking oil does not exist as a vapor, only as a solid and liquid, if I'm not mistaken?

Exactly. Solid or liquid. If it had not been decomposing, its boiling point would have been MUCH higher then Teflon point of decomposition.

Thank you. I think you've answered my question… I'd be happy to accept your answer if you wanted to publish it as an answer!

The smoke point is not the sign of evaporation, but decomposition. Additionally, they are mixtures with variable composition even for the single kind of oil.
It means edible fats or oils are either solid or liquid. If they had not been decomposing, their boiling points would have been MUCH higher then Teflon point of decomposition.

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Medicinal chemistry: adding substituents to increase/decrease activity
In medicinal chemistry, it is possible that adding 2 different substituents does not increase the activity of the drug much when added separately, but when added together, the activity is increased a lot. An example of this was for the drug sorafenib.
But my question is can the other way around happen? Where 2 substituents increase the activity of a drug separately, but when added together, the activity of the drug is now slowed.
Regarding aromatic rings:
The classic examples for activators and deactivators are in regards to aromatic rings and in regards to increasing/decreasing pH and on whether they were put on ortho/para or meta positions. But my question isn't limited to just aromatic rings and for pH as a variable. But my question for that is - can 2 substituents that are both activators or both deactivators, and both on either ortho/para position or both on meta position, can 1 still increase activity or still decrease activity, when all else is the same?
Anyways, the classic example of sorafenib is as follows. It was shown to be slow activity with a phenoxy substituent, and with a isoxazole ring, but when both were bonded to "it" at the same time, showed an increase in activity. Where "it" is a urea that was bonded to a paramethylbenzene.  Thanks.

I would argue that, if starting with a single scaffold, there will be a "sweet spot" with regard to numerous ADME properties (eg Lipinski's rules) as you add/subtract groups. However the optimum will be potentially very different from the properties that optimize in vitro protein-drug interactions.

@BuckThorn my topic is about QSARs. Quantitative structure-activity relationships. Since your Google search link is for "drug reactions" that is less-specific than my topic, I fail to see how your phrase is better than mine. If you feel it is feel free to explain to me why that is.

I am simply asking for clarification but I admit my comment was a bit dense. You are evidently starting with one molecule and making modifications. Consider then your example, sorafenib. It is evidently an enzyme inhibitor, so you need to consider both ADME and how it binds to its target. Regarding the target binding, the effect of modifications depends on complementarity with the active site. Imo it would be difficult to generalize whether removals, additions or modifications help without consideration of what the target looks like, but someone else might have the detailed answer you seek.

A concrete example: https://pubmed.ncbi.nlm.nih.gov/31146197/

@BuckThorn my question has more to do with synergism and contraindications. For example, 2 acids can never go in the basic direction when added together or 2 bases when combined go in the acid direction, but it's not the case with this in medicinal chemistry on activities. This was technically a "yes/no" question and in other places I asked, I got the answer "sure why not" but I didn't explicitly ask for examples. If I did that this question may be shut down quicker.

In any event activity is measured on some uM the greek letter for mu. So 1.7uM is 10x faster than 17 uM. I realize I have a question regarding that. Now, regarding a false analogy, if by chance I saw 2 acids will act as a base if used together, I'm just asking if the "other way around" can happen. So so far in other chemistry forums, someone replied "sure why not."

I took up your example of sorafenib because it makes it easier to discuss the question. IMO there is nothing wrong with bringing up concrete examples as you did to make a question clearer. // Drug activity in micromolar units is not a measure of how fast a compound acts but rather of the dose (concentration) required for a certain result, for instance inhibiting a target enzyme.// Take sorafenib again. You can modify it this way or that and find that different modifications, taken one at a time, enhance binding, but some combinations taken two at a time decrease binding....

... whereas other paired combinations do the opposite. This is not surprising at all, because the effect depends on the size, shape, and chemical composition of the enzyme pocket where the molecule is binding.

BTW activity may be measured in "unit" units: https://www.biomol.com/resources/biomol-blog/guide-to-enzyme-unit-definitions-and-assay-design This is not the same as "u" used to mean "micro", as in micromolar. The article I linked to discusses the IC50 of the drug in micromolar concentration.

Drug chemistry is complicated: simple patterns are not the norm
For a drug to be effective it must meet many criteria, some simple and some very complex.
Bioavailability, for example, depends on the route of delivery and even this is sometimes complicated. Orally delivered drugs (eg analgesic pills) have to survive the stomach and also have to be absorbed into the body from some part of the digestive system. This may not be easy to achieve. Sometimes even small changes in a molecule will vastly alter its absorption, stability in the digestive tract or active concentration in the body. Sometimes small changes may make the molecule far easier for the body to metabolise, greatly lowering the bioavailability. And those things are the simple side of the problem.
Many drugs work by inhibiting specific enzymes. Here the problem gets even more complicated. Good inhibitors need to match and bind to an active site in a specific enzyme. Sometimes similar molecules will have some activity for a specific target and similar variants will have similar activity. For example, there are structural similarities between the commonest NSAIDs, paracetamol (acetaminophen), aspirin, ibuprofen and naproxen (they all inhibit the cyclooxygenase enzymes, though the details differ).
But enzyme active sites are complex 3D structures and the specific details of the active site, which will determine whether a specific molecule can bind to it, are far from simple. Sometimes a very small changes in the molecule makes a huge change to the binding ability potentially making the drug either less or more effective. Multiple changes can have synergistic or contradictory effects. And it isn't easy to work out why as we often don't know the actual structure of the key binding site or sometimes even which enzyme is the target.
So pharma firms, when they find something that works a bit, tend to get their chemists to make a wide range of slightly modified molecules and test them all in vivo in the hope some will be better.
It isn't simple and there are often no clear reasons why one molecule works better than another molecule. The variety of 3D structures (of drugs or binding sites) is very large and, especially when the target site is poorly understood, impossible to predict what modifications will work.
And these two considerations ignore another important constraint: whether a specific molecule will cause other side-effects by interaction with some other target.
Given all the constraints, it is not a surprise that it is hard to predict whether a molecular modification will have an effect that is good or bad. and you probably should not expect clear patters in most cases.

When you said "Many drugs work by inhibiting specific enzymes" why? Why would a drug need to prevent an enzyme from binding to a receptor? I thought majority of drugs is to bind to a receptor.

@NealConroy Perhaps I oversimplified. Some drugs activate, some inhibit enzymes (or proteins in general). The way drugs do that is to interact with receptor sites on enzymes (sometimes activating something, sometimes blocking something).

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Non-toxic alternative to borax for use in slime chemistry?
I understand that sodium tetraborate decahydrate ($\ce{Na2B4O7•10H2O}$) and water is the primary binder for PVA glue based slimes.
We have learned borax is not good for the ground water, or marine life, so I am trying to determine if there is a safer alternative substance that would produce the same cross-linking effect between the protein molecules of the glue and the borate ions $\ce{(B(OH)4)}$ of the borax solution. The hope is a non-newtonian fluid.
If not, is there something that could be added to the slime, to neutralize the borax, making it safely compostable?

@Shannon Strong. What is the nature of the glue used for making PVA Glue based slime ? Is it a protein ?

Exactly what do you think is toxic about borates?

Boron's toxicity depends on how much you're making, and how you would dispose of it.
There's some naturally-occurring boron in the environment, and boron is a necessary for many green plants. Borax is commercially available for home laundry use. This Safety Data Sheet states it is, "Readily degradable in the environment."
Therefore, if you're using no more than the quantity used in home-laundry, and if the waste is sufficiently diluted, then disposal through community sewage is reasonable. Since the slime is so thick, sufficient dilution is almost inevitable to get it down the drain.


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Did my electrolysis experiment produce NaOH?
I recently conducted an electrolysis experiment on saltwater, hoping to obtain chlorine gas and sodium hydroxide. The experiment ran for approximately 12 hours, powered by a 9V battery. To my surprise, I ended up with a peculiar white substance. I suspect it might be sodium hydroxide, but I'm not entirely sure. The concentration of water is so high that further stirring doesn't seem to dissolve any more of this mysterious substance. Has anyone encountered a similar result in their experiments? Can you confirm if this white substance is indeed NaOH? Any insights or advice would be greatly appreciated! Thanks in advance for your help!

"a peculiar white substance." Where? What test for NaOH have you tried? 2. Choice of 9-volt battery is poor, since the current capacity is minimal.

Please do not electrolyze $\ce{NaCl}$ solutions, if you want to obtain chlorine gas. I explain why.
If salt water is electrolyzed, the first reactions on the electrodes are :
$$\ce{Cathode : 2 H2O + 2 e- -> 2 OH- + H2}$$
$$\ce{Anode : 2 Cl- -> Cl2 + 2 e-}$$
As a consequence, $\ce{H2}$ and $\ce{Cl2}$ are produced on the electrodes, and $\ce{OH-}$ is emitted in the solution at the cathode. But $\ce{Cl2}$ is partially soluble in water, and goes into solution at the anode. After a while the negative ions $\ce{OH-}$ migrate to the anode (like $\ce{Cl-}$) where they meet the dissolved $\ce{Cl2}$ molecules and react according to :
$$\ce{Cl2 +  2 OH- ->  Cl- + ClO- + H2O}$$ No free chlorine gas $\ce{Cl2}$ is produced any more at the anode, as it is consumed to create the hypochlorite ion $\ce{ClO-}$ The obtained solution is called "bleach". Now if this bleach is slightly heated, the hypochlorite ions disproportionate according to : $$\ce{3 ClO- -> ClO3^- + 2 Cl-}$$ So the solution contains now three negative ions containing chlorine atoms : $\ce{Cl-, ClO-, ClO3^-}$, instead of one !
And now $4$ possible reactions may occur at the anode, due to the negative ions  $\ce{Cl-, OH-, ClO-, and ClO3-}$. I will not give here the four different anodic reactions produced by these four ions.
So please do not electrolyze salt solutions to get gaseous chlorine without special equipment to prevent the contact between $\ce{Cl2}$ and $\ce{OH-}$ in solution.
I do not know the nature of your white substance. But it is not $\ce{NaOH}$, because $\ce{NaOH}$ is extremely soluble in water. It may be $\ce{NaCl, NaClO, NaClO3}$, or something else.


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What chemical(s) in peanut butter cause it to glow under UV light?
It is relatively well known that peanut butter glows in the dark under UV light (if you don't believe me, see this short YouTube demo by NileRed).
I've tried it myself and it is clearly a real thing.
It is less clear whether the effect is phosphorescence or fluorescence though the glow persists for seconds after exposure suggesting the former.
Which chemicals cause this and, for bonus points, is it fluorescence or phosphorescence?

Aflatoxins are famous for their intense fluorescense. Raw peanuts are especially vulnerable in supporting growth of yellow feared mold Aspergillus flavus, producing these toxins. Fluorescence under UV is frequent screening procedure to detect problematic lots. Notes that the test is not specific and many substances or materials show fluorescence. Nature: Fluorescence spectroscopy and  aflatoxin-B1

@Poutnik That would be a good suggestion were it not for the fact that food standards routinely test peanuts for aflatoxin levels and the glow effect seems to fairly universal in peanut butters. Unless the aflatoxin fluorescence is extremely bright at ppb levels.

I guess it is. It could be worthy to compare different lots.

Also for the HNQ readers: https://chemistry.stackexchange.com/questions/72595/what-makes-scorpions-glow-under-uv-light

There is an interesting comment on the subject in reddit with refs and all that attributes PB luminescence to phenolic compounds.

The reddit answers is I am afraid vague. By attributing luminescence to phenolics, he/she just suggested thousands of compounds. Black tea is rich in phenolics but it is non-fluorescent. In the middle of the answer trans-coumaric acid is proposed, which does not emit green light.

Many natural oils from vegetable sources show intense fluorescence under UV. Olive oil glows brilliant red under UV. In the peanut butter case, my first guess was turmeric, the yellow spice, because it glows green under UV. When we look at the ingredients of this peanut butter which shows soybean oil, among others and turmeric is absent.
https://www.kraftheinzfoodservice.ca/

Ingredients: Select Roasted Peanuts, Soybean Oil, Corn Maltodextrin,
Sugar, Hydrogenated Vegetable Oil (Cotton Seed And Rapeseed Oil),
Salt, Mono- And Diglycerides.

From this article on ResearchGate (Monitoring the Transesterification Reaction of Vegetable Oil to Biodiesel by Fluorescence Spectroscopy with UV Excitation: Correlation with Viscosity January 2018Orbital - The Electronic Journal of Chemistry 10(1), DOI:10.17807/orbital.v10i1.1026), we can see the emission spectrum of soybean oil.
So, it is very likely the oils in the product are fluorescing green.

Did you mean to put in links for the ingredients of the peanut butter and for the article with the graph?

Quick empirical test. My extra virgin olive oil has a red fluorescence under UV but my light olive oil seems to be blue. Neither shows any phosphorescence as seen in peanut butter (be interesting–if oils are the source–why it seems to achieve phosphorescence.)

Is there any reason why at 400nm wavelength all 3 of them had that dip ? Or is it just faulty/mini errors in the equipment

It is very likely delayed fluorescence rather phosphorescence. Delayed fluorescence is common in plants too. It is hard to distinguish between phosphorescence and delayed luminescence until and unless someone does magnetic studies during photoirradiation. Light olive oil might be adulterated with other oils or it could be completely non-fluorescent. The apparently "blue light" could be Rayleigh scattering. Are you viewing oils at 90- degrees to the light source?

@Or4ng3h4t, most likely a grating change. UV gratings might be different from Vis gratings.

Out of idle curiosity, do you have any pure, 100% (±salt) peanut butter to hand (it is commonly on sale here in Europe)? The ingredients list of Kraft peanut butter is quite complex and any of soybean / corn / rapeseed oil could contain quite the mix of naturally conjugated bond systems with varying length and varying proportions. Pure peanut would help a bit as it is, perhaps surprisingly, well characterised: c.f. https://doi.org/10.1002/ejlt.200900176

@AChem it could be delayed fluorescence hence why knowing the specific chemicals might be useful. As for olive oil, yes I'm viewing at 90°.

Total analysis of natural product without knowing the exact target class of analytes can be someone's Ph.D. thesis in food chemistry. What is surprising is the lack of work on peanut butter.


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Can I reuse LDPE bottles after trace element analysis?
I run some algae samples through mass spec a few times over the month to test the P content of algal samples. My (simplified) procedure was as follows:

Put a small portion of algae in a teflon vessel
Add nitric acid and hydrogen peroxide in the teflon vessel
Put that teflon vessel in a microwave digestion system and run it for an hour
Add deionized water to the solution
Put the sample on a hotplate and let it reduce over night.
Add more nitric acid to the solution
Transfer the solution to a 125 ml LDPE bottle
Add deionized water to the solution until it reaches 100 grams

I have decided to run this procedure a few more times so I bought 250 ml LDPE bottles for my experiments and to replace some bottles that I used from another lab. I am not experienced in chemistry at all and a tech has been helping me with these experiments, which I am grateful for. Today I asked the tech how I can clean the bottles to return them to the lab and he told me that those bottles were just thrown away after being used for trace elemental analysis.
I am honestly shocked because they cost like $300 for 30 of these bottles. I need to run over 200 samples and I can't afford to keep replacing these bottles. I have no doubt in the tech's experience and they have been teaching me, but I find it odd that these bottles can't be used again for trace element analysis especially since the teflon and PFA beakers can be reused multiple times and I never heated or reheated these bottles. I simply do not have enough knowledge to understand why I can't soak these bottles in HNO3 for 7 days and call it good? Have any of you tried to clean these bottles after trace elemental analysis?
Thanks!

He's got a point.

Well, if you were sure they won't affect the measurements... but I guess that would call for special cleaning protocol and testing for elements and adjusting the protocol... which would likely cost much more.

I clean the teflon and the PFA beakers using nitric acid so It seems odd that there is not a protocol for cleaning LDPE after trace element analysis. I mean, I am a beginner so there has to one somewhere right? I don't test if my cleaning of the PFA and the teflon has gone well because I assume there is a tried and true protocol that has already been done many times.

LDPE is quite different than PFA, perhaps there would be an issue with leftover traces of catalyst used in polymerisation? Wait for some specialist here or ask another IRL.

Apparently, the the lab buys economy LDPE bottles so they cost $2 which is way more cost-efficient than what I bought. I returned the bottles and will be using the economy bottles for the next run.


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Why Green Solution After Putting Bismuth Antimony in Nitrogen Acid
I am in a physics lab studying dislocation, and we submerged a bismuth antimony sample in nitrogen acid (about 50% concentration, if I remembered it right) as the first step to etch it. There were bubbles, and the solution turned green. We googled and found bismuth nitrite should be colorless, so the greenish solution became a mystery.
I suspect there might be residues of copper in the beaker, the only possibility I can think of because I am terrible at chemistry. But we did clean the beaker with soap and paper towel. Can I get some chemist insight to this situation?
Edit: I forgot to mention that the green color persisted even after we took the sample out.

I think I've heard of this happening in some cases of metal dissolution with nitric acid due to transient formation of lower nitrogen oxides which can be blue-green in colour (e.g. dinitrogen trioxide). You could also be getting some nitrosyl metal complexes.

@NicolauSakerNeto That was really helpful! I forwarded your insight to my lab members. Thank you so much.

What is "nitrogen acid"? Do you mean Nitric acid?

@NilayGhosh Oh yes sorry I kinda improvised on that name

Copper could have come from the sample. From Belmont Metals:

Bismuth as an alloying element can be used in the production of fusible alloys ( low melting point alloys ) and Bismuth based low temperature solders. Standard and special compositions are Bismuth alloyed with either Antimony, Cadmium, Copper, Indium, Lead or Tin.

Copper might be only a minor element, but a little bit can go a long way when it comes to color perception.
One way to check for copper is to add thiourea to a sample of your solution. Thiourea will decolorize solutions containing copper(II) by reducing it to copper(I).


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Synthesis of 1-ethylcyclohexane-1,2-diol from cyclohexylethene
Here's my synthesis:

HBr
NaEtO
OsO4/NMO

But the problem is that for step 2, there are 3 possible alkene locations, with all of them having 3 substituents, so I would have all 3 forming (although 2 of them would result in the same thing). So I checked my professor's solution, but he used HBr twice. However, wouldn't the first HBr result in a hydride shift, making the 2nd HBr unnecessary?
Either way, this solution also allows the formation of alkene in 3 sites, just like my solution. Is that allowed?
This is my prof's solution:

A hydride shift in the first HBr reaction is certainly possible, but you will likely get a mixture of positional isomers of brominated products


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What green and orange solids did I get when I mixed copper acetate and sodium carbonate?
I tried to make copper carbonate by mixing copper acetate with a sodium carbonate solution. I was expecting to see a light blue solid form at the bottom of the beaker and instead I found a green powder distributed throughout the solution. I filtered it through a coffee filter and it started turning orange. The entire coffee filter is now covered in a bright orange powder.
I need to know whether the green powder and/or the orange powder is toxic or corrosive and also what it is.

When you mix copper(II) and carbonate, instead of getting pure $\ce{CuCO3}$ you get "basic copper carbonate," which has some amount of the carbonate replaced with hydroxide. This can range from blue to green. I've done this a few times with various salts and mixtures and I don't know what factors actually control the color. But green shouldn't be a surprise.
The most likely reason it turned orange is that some combination of your coffee filter and impurities in your chemicals ended up reducing some of it to $\ce{Cu2O}$, which is orange or red. But naturally, regardless of what this unknown substance actually turns out to be, you'll treat it as though it's toxic and handle it responsibly, right?

Does this answer change if I had used the copper acetate to electroplate a piece of nickel and it dropped some nickel into the solution or not?

If your solution is contaminated with nickel, some nickel carbonate is likely also present in the green solid. Nickel carbonate is green.


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How to break C-C bond in a cyclohexanone?
I cannot find a way to start this reaction. For the end, I found that a H that is attached to C with OTs group will be deprotonated forming double bond but before that I think I need to break C-C bond of ketone. Can someone help me on writing the mechanism of this reaction? Thank you.

The most acidic proton in the starting material is alpha to the sulfonyl group

There is a similar reaction on Chegg which doesn't lose the methyl group on the butyl chain: https://i.stack.imgur.com/ac8Mz.png. It also has some subtle visual hints about how the ring forms.

@Waylander So alpha in this context means the carbon directly bonded to sulfur?

"I found that a H that is attached to C with OTs group will be deprotonated forming double bond". The two carbons which will form the double bond have the "right" number of hydrogen atoms attached already (zero and two), so there is no need to deprotonate there.

@Karsten  yes, the protons next to the sulfonyl group are acidic enough to be removed by NaH

As for the second part of the mechanism, learn about Grob fragmentation. Then you will know what to due with the sulfonyl anion.

Something is wrong in the product. There is a methyl group present in the starting material that should be on the 6-member ring. That methyl group incidentally does not have stereochemistry expressed, but I suspect that it is quite important when it's so close to the norbonyl system...

Yes, the methyl group should be present in the product. It was mistakenly not added to the drawing.


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Position of H in formula
I couldn't come to a resolution with these 2 questions that use hydrogen in the middle of a formula that is equipped with a numerical prefix.
Why is is one inside the bracket and one outside? Am I mixing up one is a cation and the other is anion?
Aluminum monohydrogen phosphate: $\ce{Al2(HPO4)3}$
Copper(I)dihydrogen phosphate: $\ce{CuH2PO4
}$
The other question is, in hypochlorous acid (HClO), leaves out the 2 charges after the O?
ex) I wrote $\ce{HClO2}$.
Do we always not mention the charges in binary acid compounds?

Please don't put multiple unrelated questions into one post. Also, are you confusing oxyacids with binary acids? Finally, google and Wikipedia are your friends, for instance: https://en.wikipedia.org/wiki/Aluminium_dihydrogenphosphate The notation reflects the composition but can also reflect the structure.

You can technically put the H in bracket for copper dihydrogen phosphate and it mean the same thing. It is an ionic compound comprising of cation and anion. So, by taking that into consideration, we can write $\ce{Cu^{+}(H2PO4)^{-}}$. For simplicity, remove the brackets and charge to give $\ce{CuH2PO4}$. We provide brackets to identify that the entity inside the bracket is a single anion.

$\ce{Al2(HPO4)3}$ is a compound made of $2$ ions $\ce{Al^{3+}}$ and $3$ ions $\ce{HPO4^{2-}}$. In total, this produces $6$ positive charges, and $6$ negative charges. Both number of charges are equal in the formula. Total zero. This is needed to get a neutral substance. Now, in the formula, the positive ions is at left, and the negative charges are at right. So the formula is all right. It is not absolutely forbidden, but usually it is not necessary to include the charges in the formula.
Same thing happens in the substance $\ce{CuH2PO4}$, which is made of $2$ ions, one $\ce{Cu+}$ and one $\ce{H2PO4-}$. In total, this substance contains $1$ positive ion, and $1$ negative ion. The total is zero. This is needed to get a neutral substance. Now in the formula the positive ion is written first, then the negative ion. The final formula may contain the individual charges, but usually, these charges are not necessary. So they are not part of the formula.
For the last question, you should not compare $\ce{HClO}$ and $\ce{HClO2}$ which are different substances, without any connection between them. They cannot be exchanged or compared.

thank you i now well understand the use of numerical prefix in ion compounds. For the second question i forgot the difference between hydro_ acid and the existence of hypochlorite acid so the similarity between those 2 names had me confused. thank you


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cellophane bags
I am conducting an experiment to test the components of a cello bag. I believe that cello bags do not just contain cellulose, and as mentioned by many websites, only 92-98%, but not 100% of the bag is cellulose.
What do you think the bag is made up of?

Hydrophilic materials always contain small amount of moisture.  Cotton fibers are not 100% cellulose either. In cello bags, there may be residual chemicals from viscose process and speculatively some modifiers.
Wikipedia says there are used softeners like glycerine to prevent brittleness.


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Order of radical dimerization ability
In a question I came across recently, it was asked to arrange the following radicals in the ascending order of dimerization ability.

I have read about the triphenylmethyl radical undergoing Gomberg dimerization, but I couldn't relate it to this question. Please explain the order of dimerization.
Also, please suggest a valid source (books or journals) to read more about the same.

Hint: I which position the radicals bind?

They bind on the para position. So, does it mean that since the para position on B is hindered, it would be the least, in dimerization, and the order is A>C>B? Is there any other effect, or any other factor that has a role in dimerization, like radical stability?

Fluorine in p- has much bigger effect than just steric.

Could someone please explain the answer?


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Preparation of copper carbonate from copper acetate and sodium carbonate
I reacted copper acetate with sodium carbonate. It formed a mixture of sodium hydroxide, copper carbonate, copper acetate, and copper oxide. I wanted to know if there is a process to remove the sodium hydroxide and copper oxide to leave me with just copper carbonate. I was thinking that I could wash the solution through a filter with water since sodium hydroxide is soluble and copper carbonate is not but I’m not sure if this would work. That still leaves copper oxide. Any ideas on how to get copper oxide out?

Is there a way not to make copper(I)oxide in the first place? Do you know what the reductant in your system is? In Benedict's assay, it is the aldehyde groups of sugars.

Related (followup of): https://chemistry.stackexchange.com/q/177120/72973

I don’t know for sure

And also yes to not make copper oxide in the first place you can use copper sulfate but I did not have copper sulfate or access to sulfur or sulfuric acid. But I did have copper acetate on hand.


text:
Boiling point of pentanol isomers

Arrange the following alchohols and alkanes in increasing order of boiling point:
3-pentanol, n-pentane, 2,2-dimethylpropanol, n-pentanol

I understand that n-pentane would have the lowest boiling point because alkanes have a lesser BP than alcohols for the same number of carbons.
2,2-dimethylpropanol would be next because it has the most amount of branching and branching decreases BP as it makes the compound more compact (giving it the shape of a sphere).
But I am confused between n-pentanol and 3-pentanol, from intuition, 3-pentanol should have a lesser BP because it seems it will have more H-Bonding compared to n-pentanol and that is the right answer as well. But is that the actual reason? If not then what is the reason?

More or stronger hydrogen bonding usually increases BP. The purpose of the question is unclear; the differences among the many C-5 alcohols can be subtle.  To see the main differences from shape and location look up the pentanes and the butyl alcohols; then graduate to the pentyl alcohols.

As jimchmst said H bonding usually increases boiling point. All i could think of is, OH is like a "branch" in 3 pentanol which might be why its BP is less than n-pentanol


text:
Why is there a yellow solution formed upon reaction of silicon tetrachloride with water?
I've read somewhere that reacting the two will give a white solid and a yellow solution. I'm assuming the white solid is the silicon dioxide, but what is the chemical origins of the yellow colour of the solution?

How pure is the used SiCl4? There is chance there is dissolved FeCl3 impurity, forming yellow ferric chloride solution.

I don't have information on its purity, but I suspect you may be right.

The yellow color is indeed due to an impurity. According to Wikipedia:

Silicon tetrachloride is prepared by the chlorination of various silicon compounds such as ferrosilicon, silicon carbide, or mixtures of silicon dioxide and carbon. The ferrosilicon route is most common.[1]

Thus the most likely impurities are chlorine and iron, the latter converted to ferric chloride by the chlorination process. Either one may produce a yellow color in water solution.
Cited Reference

Simmler, W. "Silicon Compounds, Inorganic". Ullmann's Encyclopedia of Industrial Chemistry. Weinheim: Wiley-VCH. doi:10.1002/14356007.a24_001


text:
Reduction of di-N-substituted imidazole to imidazoline to remove counterion?
I would like to synthesize a di-N-substituted imidazole (compound 4) in its free base form or reduce it to generate an imidazoline (compound 5) to remove the bromide counterion to the imidazole ring.

I have tried the steps below resulting in the outcomes described below. Does anyone have any alternative suggestions on how to get compound 5 or somehow remove the bromide ion in compound 4? Thanks very much!

Could try NaBH4 in DMF or NaHB(OAc)3 in AcOH as described in this paper  https://www.sciencedirect.com/science/article/abs/pii/S0040402017310906   but this does not look straightforward

I found someone's thesis showing that they were able to add a sulfur at position 2 of di-N-substituted imidazoles. Would I be able to replace the alkyl i have on my position 2 with a sulfur using the reaction scheme they describe in Figure 35 page 40? Just trying to see if I can somehow salvage all the excess compound 4 I made and turn it into something marginally useful for my experiments https://jyx.jyu.fi/bitstream/handle/123456789/86607/1/URN%3ANBN%3Afi%3Ajyu-202304262714.pdf

You could try but I would not be optimisitic about it.

Did you consider an anion exchange (the soft $\ce{Br^-}$ is substituted by say tetra fluoroborate $\ce{BF^-_4}$) subsequently followed by an electrochemical (i.e., regarding the potential adjustable) reduction?  Do you have the relevant data at hand, i.e. a CV? For the former, see e.g. 2012Molecules4007, for the later see e.g., 2018JChemEduc197; both sources are open access.  The scale could match well the Baran/Ika stirrer setup (see his lab page).

Continuation:  The company's advertising web site, a promotional video (small scale chemistry, this time without connecting Baran's Tesla to the reaction ...); Baran's talk Your Chemistry has Potential at ACS 2017 fall meeting, and a more recent tutorial like Simplifying synthesis with Electricity of the Kharkiv Chemical Seminar just in September, 2023.

As Ref.1 pointed out, aromatic nitrogen heterocycles in which the nitrogen contributes only one electron to the $\pi$-system are electrophilic as compared to benzene, and show a reduction by lithium aluminum hydride, an active reducing agent. Thus, nitrogen heterocycles are more electrophilic than benzene and are susceptible to attack by the hydride ion from a complex metal hydride anion (e.g., $\ce{NaBH4}$). However, in protic solvents, the intermediate cyclic enamines can undergo further reduction such as heterocyclic ring opening. For example, when $\ce{NaBH4}$ (1n $1:5$ excess) was used in refluxing 95% ethanol as the solvent to reduce heterocyclic quaternary salts such as OP's imidazolinium ion $(\bf{4})$, the intermediate cyclic enamine (in OP's case, it is $\bf{5}$) can further undergoes reduction to give ring opening product, a diamine within one hour (Ref.2). It is also possible to quickly generate various functionalized heterocycles including macrocyclic ones in addition to these highly substituted diamino derivatives.
Nonetheless, Shvydenko, et al. (2017) have published their result that described a way to reduce imidazolium salts without ring opening (Ref.3):

Keep in mind that this reference was first mentioned here by Waylander in his comment to the question (many thanks). However, the authors have reported that all products having structure $\bf{5}$ (similar to OP's expected product) is unstable, specifically in even slight acetic condition (e.g., in NMR samples in $\ce{CDCl3}$). Bigger the fused ring, more the instability. This means open chain compounds without a fused ring such as OP's might be less stable, yet we don't know that for sure. Thus, it is tricky to use these conditions, but it is worth trying it to see the outcome.
References:

Robert E. Lyle and Paul S. Anderson, "The Reduction of Nitrogen Heterocycles with Complex Metal Hydrides," Advances in Heterocyclic Chemistry 1966, 6, Pages 45-93 (DOI: https://doi.org/10.1016/S0065-2725(08)60575-9).
Erik F. Godefroi, "Reductive ring cleavage of 1,3-disubstituted imidazolium iodides by sodium borohydride," J. Org. Chem. 1968, 33(2), 860-862 (DOI: https://doi.org/10.1021/jo01266a089).
Tetiana Shvydenko, Kostiantyn Nazarenko, Kostiantyn Shvydenko, Sergey Boron, Oleksii Gutov, Andrey Tolmachev, Aleksandr Kostyuk, "Reduction of imidazolium salts – An approach to diazocines and diazocanes," Tetrahedron 2017, 73(49), 6942-6953 (DOI: https://doi.org/10.1016/j.tet.2017.10.053).


text:
What is the difference between combination reaction and addition reaction
In my textbook,two definitions were given separately

Combination: Two or more substances react to form one product.

Addition: Something is added to a chemical substance without elemination.

What is the difference between this two reactions?

Looks like they mean the same thing, but combination is for inorganic compounds while addition reactions are for organic comcounds.
https://en.wikipedia.org/wiki/Addition_reaction
But it looks like you can classify addition reactions, as a type of combination reaction.

CH SE site should not be your first choice to resolve textbook confusion, but rather the last one. If it is rather the last one, you should include compact summary of your prior looking for the answer.

"Combination" is an artificial term, a sort of baby talk, rarely used in that meaning outside a classroom. "Addition" is a real thing.

The difference in terminology is probably due to the fact that an organic compound may have an atom or radical added in different positions.
For example, propene reacts with $\ce{HI}$ to form both 1-iodopropane and 2-iodopropane. Though Markovnikov’s rule implies the majority product is 2-iodopropane, some iodine is added at the terminal carbon, i.e., forming some 1-iodopropane.
We say the iodine is added at a location, not combined at a location.


text:
Solubility of raspberry ketone in paraffin oil
I want to dissolve raspberry ketones (RK) in paraffin oil, but I'm not sure about its solubility in it. I want to use this mixture for the fruit fly's behavioral study. Is it possible to dissolve the RK pheromone in paraffin oil?

Based on nothing more than looking at the structure I think it would have some solubility. This paper reports the solubility in some organic solvents  https://www.sciencedirect.com/science/article/abs/pii/S016773221733218X

Do you mean raspberry ketone as a substance (3-(p-hydroxyphenyl)propanal), mix of similar substances or food dietary supplement formula? // For the substance itself, my opinion is it seems to be partially soluble, but better if little more polar solvent is used, what may not be applicable. OTOH, the concentration would not be probably high.

It is not soluble in paraffin oil, but soluble in glycerol, which is suitable for biological tests.

@Poutnik: Of course, raspberry ketone is a ketone. 4-(4-hydroxyphenyl)butan-2-one. But you knew that. ;)

@user55119 Hehe, you are right, I did not pay enough attention to the structure, for some strange reasons considering the ending methyl as hydrogen, not realizing it would not be a ketone. :-D


text:
Units of Activation Energy
Consider the reaction
$$\ce{aA + bB -> cC + dD}$$ whose reactants are given by A and B and the products are C and D.
Say the activation energy of this reaction is some $x$ joules/mole, But moles of what ( $A$ or $B$ or $C$ or $D$) ?
If I need $x$ joules of energy to react $A$ with $B$ with the correct orientation, To react $a$ moles of $A$, I would need $ax$ Joules of energy, if it was $x$ joules of energy to react $B$ with $A$, I would need $bx$ amount of energy to react $b$ moles of $B$. Since the reaction is the same the amount of energy needed should be constant to react a fixed concentration of reactants. I know I am missing some understanding here.

Think of it as "moles of reaction". If there are x moles of reaction, then xa moles of A react with xb moles of B.

Per mole of species in transition state.

To form 1 mole of TS?

@BuckThorn So to form c moles of C and d moles of D from a moles of A and b moles of B, I need x joules of energy?

If it is activation energy, you need that amount but will get some back. In general, molar quantities are per mole of reaction (enthalpy of reaction, Gibbs energy of reaction).

You won't need x Joules, you'll need x times the activation energy per mole (the activation energy will be in units of energy per moles of reaction).

It is the thermal energy (kinetic , vibrational /rotational). Very, very occasionally, by Boltzmann distribution, a pair of molecules have enough of this energy to react. We make this into a per mole basis for ease of comparison and usage with other molar quantities.


text:
Is mass of a gas taken into consideration while describing ideal gas law
According to Avogadro's law equal volume of all gases at constant temperature and pressure contain equal number of molecules.
But how is this possible ??
Imagine helium and radon gas.
In this case is $m_\text{Rn} \gt m_\text{He}$. As pressure is directly proportional to density of gas, if the volume is constant or equal for both the gases then mass should also be same in order to have same or constant pressure.
So, how the number of molecules will be same if $m_\text{Rn} \gt m_\text{He}$, there should be $N_\text{Rn} \lt N_\text{He}$.
So, how they will have same number of molecules at equal volume, constant temp and pressure ?

Think of a mixture of helium and radon, 50% of each atom type. Temperature is obviously constant. The partial pressure due to helium is the same as the partial pressure due to radon. The kinetic energy distribution of the two types of atoms is the same. Thus for a given kinetic energy the helium atoms are traveling faster than the radon atoms.

It is radon, not Raedon. Element names are not written capitalized. Proper text punctation is essential for text readability.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

The gas hydrostatic pressure is proportional to the mass of molecules. The kinetic pressure is not.
Evaluating an ideal gas at given temperature, volume and number of molecules, the mean translational kinetic energy of molecules is $E_\text{k}=\frac 32 k_\text{B} T$, where $k_\text{B} \approx \pu{1.38E-23 J K-1}$ is the Bolzmann constant
The mean quadratic speed of such molecules is then $v = \sqrt{\frac{2 E_\text{k}}{m}}=\sqrt{\frac{3k_\text{B}T}{m}}=\sqrt{\frac{3RT}{M}}$
and the mean impulse is
$$p=mv=\sqrt{3k_\text{B}mT}$$
Pressure due molecular kinetics is then:
$$p=\frac{\text{|d}\vec {F}|}{|\text{d}\vec {A}|}=\frac{\text{|d}\vec {p}|}{|\text{d}\vec {A}|\cdot \text{d}t}$$
As the pressure is proportional to both the speed and impulse:
$$p \propto |\vec {v}| \cdot |\vec {p}| = \sqrt{\frac{3k_\text{B}T}{m}} \cdot \sqrt{3k_\text{B}mT} = 3k_\text{B}T$$
Therefore, the pressure for given volume, temperature and number of molecules of an ideal gas does not depend on mass of a molecule.

In the ideal gas or perfect gas model, the gas atoms/molecules are considered as point masses. Further the law is only valid when the intermolecular distances are large and only elastic collisions occur. Thus, when the intermolecular distances are large, the size of individual atoms, and corresponding masses do not affect the system, through some form of interactions.
Van der Waals equation, amongst others, accounts for the sizes of particles and intermolecular interactions.
In short, masses and sizes do matter but not in the ideal gas or perfect gas model and Avogadro law is valid for such systems.


text:
Are there LCMS compatible buffer systems available for the high pH range 12-14?
I want to perform chemical stability studies for different pH Values. To achieve pH Values from 1 to 6 I can prepare solutions with trifluoric acid and formic acid and ammonium formate and acetate. I know that the very high and low values are outside the optimal buffer range, but it still works because the sample concentration is small compared to an high excess of testing buffer.

For the range of 7 to 12 I can achieve these values by mixing ammonium bicarbonate and ammonium hydroxide in specific quantities. Still, I am wondering if there is a way to also test the high pH range 12-14 with a system that is LCMS compatible. Normally these pH ranges are accessed by using sodium hydroxide. But this is not compatible with LCMS, because hard ions are harmfull for mass detectors. I know that normally silica columns for rp18 are used in the pH 2-8 range. The solutions are therefore obviously diluted with LCMS mobile phase that contains formic acid before they are injected in the systems to achieve an acceptable pH value and sample concentration. Still, sodium formate would not be volatile. I know, that it is possible to divert the first minutes of an hplc run into the waste before they enter the mass detector. Still, if your product is being deteriorated into very polar substances (which I suspect for my case), you would not observe these species. Is there a possibility to access these pH values with a volatile reagent that is not harmfull for LCMS detectors? Alternatively I am thinking of doing NMR testing, still here you would be limited to heteroatoms to avoid water contamination in the spectra and you can not time the measurements as flexible as LCMS runs.

The decomposition of a compound possibly could be monitored by UV-Vis (at pH this high perhaps better to substitute glass by an other window material), or ATR-FT-IR spectroscopy (Mettler Toledo for instance commercializes this as ReactIR), too.

Most of my compounds are UV inactive. That complicates things a lot. Especially purification appears to be very challenging in most cases because you can not rely on UV bus detectors. A specific compound is even UV inactive, very polar and inert for staining reagents. Here your only change is doing mass coupled purification, screen every fraction or a purification system that is either equiped with an ELSD, conductivity or refractive sensor. (The latter two I don't have seen in a main stream purification setup yet).

First off, as a comment implies you need the right column. Silica-based columns are not suitable for high $\pu{pH}$, even $\pu{pH }8$ can cause problems according to a now-deleted comment in the thread. AChem recommends a pure graphitic carbon alternative.
Assuming you have the right column, you may want to consider guanidine, an organic strong base made of carbon, hydrogen and nitrogen and forming a multiatomic cation upon protonation. With a $\pu{pK}_b$ of $0.4$ this will be $>90$% ionized below $\pu{pH } 12.6$, but becomes more of a true buffer above that level.

@AChem does this comment belong to the answer or the question?

The comment belonged to both the Q and A. Let me clarify. Injecting a highly basic solution will eventually create a void at inlet of the OP's HPLC column. Not a good idea to employ highly basic sample solutions or even basic mobile phase. Only Hypercarb column can handle basic pH which is made of pure graphitic carbon.

@AChem I prefaced my answer with your recommendation.

@OscarLanzi I am aware of the requirements of RP-Phase columns. I thinks I stated out, that the extreme pH - values of the testing buffer is reduced before injection. I am more interested in finding a solution for utilizing this extreme pH regime in a way that is not harmfull for the detector, even after neutralisation and will not overlap with the retention time frame of polar substances (I know, that is a lot all at once and this might not even possible).

@OscarLanzi I was not able to find a supplier for free guanidin. Only HCl salt. Have you used this base in this fashion already?

Guanidine is a good choice, +1 from me.  DBU and DBN might also be useful options.

@CurtF. Can you suggest a source for guanidine? The OP says he cannot find it  (and neither can I) because commercial auppliers sell only the hydrochloride salt.

https://cymitquimica.com/products/10-F062221/113-00-8/guanidine-free-base/

But also be aware that strong bases like this are going to suck up CO2 from the air relatively quickly and lose basicity.


text:
Sodium hypochlorite bleach and chromium mop handle
I have been using sodium hypochlorite bleach as a cleaning method in my house. During a two week period when I was at home I left the mop handle in the bucket with bleach in it. I did not know it was plated with chromium. I used the water once/twice on a small area of the house (3 square metres) before noticing that the mop handle had some "rust marks/browning". I tested the water and they found approximately 200 microrgams of chromium per litre in the water. It did not specify if it was chromium 3 or 6. My questions are:

Should i be concerned about the amount of chromium in the water? Assuming it was oxidised to Chromium 6, would the amount of 200micrograms per litre be dangerous to health? I did not use all the water in the bucket, just enough for those 2-3 square metres.

The mop handle on top also had some "rust" marks which I am guessing is from the hypochlorous acid/chlorine fumes released over the 2 week period from the bucket reacting with the handle. Any chromium 6 formed on the handle, would that have fallen off as well while i was using the handle onto my floor, adding to the chromium deposited on the floor? Or is it stuck onto the handle and does not come off?

Sorry for the specific question, any answer anyone can give will be very appreciated. x

Why in the name of Hades did you mop your floor with bleach? Instead of some cheap, simple, harmless, and highly effective household detergent?

In the toxicology books, Chromium(III) is said to be non-toxic. When ingested, Chromium(IV) is toxic : $1$ g Cr(VI) is lethal. $0.5$ g Cr(VI) produces stomach aches and inflammation. So $200$ micrograms is not dangerous to health.

Thank you very much for the answer. Could I also ask, from the research I have read, they all deal with millimolar concentrations. I guess I am dealing with micromolar concentrations here. What form will the chromium 6 be in sodium hypochlorite? Will it be sodium chromate, sodium dichromate? Or something else? So I can look for any further information. Thank you again.

Sodium hypochlorite solutions are basic, with a pH between $8$ and $14$. . So the  Chromium(VI) is present as $\ce{CrO4^{2-}}$ in such a solution.

Where did you got the info that 200µg of Cr(VI) are harmless? Soluble chromates are definitely carcinogenic, not to mention that soluble chromium species are known for inducing contact allergies, often on first contact. I wouldn't be in panic here, but definitely clean the contaminated areas with gloves and a regular household detergent.

The area has already been cleaned multiple times Karl. I did not specify this in my question: I mopped the area once with the water in the bucket (an area of two square metres). Once I saw that the handle had been eaten away I tested the BUCKET water (the water that remained) for the Chromium. So the 200micrograms per litre were in the bucket. It had around 3L so around 600 micrograms I know. That water was disposed of. I am hoping that the small amount of water I added to my floor contained much less Chromium. So if I used 100ml, then a much smaller amount than 200 micrograms.

From the CDC:

EPA has an enforceable maximum contaminant level of total chromium in drinking water of 100 µg/L (100 ppb) for public water systems [EPA 1999h].

Trivalent chromium is less stringent, but because air (and certainly hypochlorite bleach) is thermodynamically capable of ocidizing Cr(III) to Cr(VI).
To reduce the chromium, you may want to consult Prasad et al.[1].

Prasad, P.V.V.V., Das, C. and Golder, A.K. (2011), "Reduction of Cr(VI) to Cr(III) and removal of total chromium from wastewater using scrap iron in the form of zerovalent iron(ZVI): Batch and column studies". Can. J. Chem. Eng., 89: 1575-1582. https://doi.org/10.1002/cjce.20590


text:
Solubility of amlodipine besylate
In European Pharmacopoeia, the drug amlodipine besylate is characterised as slightly soluble in water and freely soluble in methanol (additionally, sparingly soluble in anhydrous ethanol and slightly soluble in 2-propanol). Why is this? Both water and methanol are polar protic solvents.

There is big difference between water and methanol. Also what is this "freely soluble" supposed to mean? Sounds like overstatement.

I understand water and methanol are different, however both are polar protic solvents, so I'm not sure why a substance would have such different solubilities between the two.
In regards to "freely soluble" that is a quote from monograph in the European Pharmacopoeia.

That's the problem what this difference is. I doubt it's as big as you seem to think. Finding more accurate info may be difficult, though.

Right that's what I'm having a hard time with. I'm a pharmacy student, and in an upcoming exam, will probably be asked to judge the solubility of a drug on the basis of it's structure. I picked this example as I can't figure out why it would be different. I've referenced a couple of organic chemistry texts and medicinal chemistry texts, and not come up with much.

According to OP, the monograph in the European Pharmacopoeia says:

the drug amlodipine besylate is characterized as slightly soluble in water and freely soluble in methanol (additionally, sparingly soluble in anhydrous ethanol and slightly soluble in 2-propanol).

Theory and application of solubility Parameters are at least beyond my scope. Yet, above excerpts doesn't make any sense. If amlodipine besylate (AMB) is slightly soluble in water and freely soluble in methanol, to my common knowledge in organic chemistry, it should be at least moderately (not sparingly) soluble in absolute ethanol based on increasing organic nature from methanol to ethanol (see my answer in this question to understand one of parameters in solubility theory). Thus, I'd say OP might have misunderstood the facts given in European Pharmacopoeia. I haven't seen this monograph, but I assume it might be talking about different salts of amlodipine (e.g., amlodipine acetate).
I have tried to find some reliable data in literature, but it wasn't as successful as I expected. Yet again, different publications give different values for solubility. For example, Ref.1 gives Experimental Mole Fraction Solubility $(X_{m,T}^{Exp})$ of AMB at $\pu{30 ^\circ C}$ as $2.56 \times 10^{-5}$ and $6.20 \times 10^{-7}$ in absolute ethanol and water, respectively. According to this results, AMB dissolves highly in absolute ethanol compared to that in water (definitely not slightly in water to sparingly in absolute ethanol). Note that, in Ref.2, even though it is from the same research group, $(X_{m,T}^{Exp})$ of AMB in water at $\pu{30 ^\circ C}$ as $6.30 \times 10^{-5}$ $(\pu{0.0051 M})$.
When the counter ion is changed from besylate to acetate, the solubility has changed dramatically (Ref.3):
$(X_{m,T}^{Exp})$ of amlodipine acetate at $\pu{30 ^\circ C}$: $4.370 \times 10^{-3}$ (in methanol); $6.951 \times 10^{-3}$ (in ethanol); and $7.737 \times 10^{-3}$ (in 2-propanol).
As I expected, the solubility has increased with the increasing organic nature of the protic solvent.

References:

Faroukh Sardari and Abolghasem Jouyban, "Solubility of 3-Ethyl-5-methyl-(4RS)-2-((2-aminoethoxy)methyl)-4-(2-chlorophenyl)-1,4-dihydro-6-methyl-3,5-pyridinedicarboxylate Monobenzenesulfonate (Amlodipine Besylate) in Ethanol + Water and Propane-1,2-diol + Water Mixtures at Various Temperatures," J. Chem. Eng. Data 2012, 57(10), 2848–2854 (DOI: https://doi.org/10.1021/je3005672).
Elmira Behboudi, Jafar Soleymani, Fleming Martinez, and Abolghasem Jouyban, "Solubility of amlodipine besylate in binary mixtures of polyethylene glycol 400 + water at various temperatures: Measurement and modelling," Journal of Molecular Liquids 2022, 347, 118394 (10 pages) (DOI: https://doi.org/10.1016/j.molliq.2021.118394).
Chao Cheng, Yang Cong, Cunbin Du, Jian wang, Ganbing Yao, and Hongkun Zhao, "Solubility determination and thermodynamic models for dehydroepiandrosterone acetate in mixed solvents of (ethyl acetate + methanol), (ethyl acetate + ethanol) and (ethyl acetate + isopropanol)," The Journal of Chemical Thermodynamics 2016, 101, 372-379 (DOI: https://doi.org/10.1016/j.jct.2016.06.014).

Thank you for your answer. I didn't misquote the monograph, below is a direct copy paste from Ph. Eur. 11.2.
CHARACTERS
Appearance: white or almost white powder.
Solubility: slightly soluble in water, freely soluble in methanol, sparingly soluble in anhydrous ethanol, slightly soluble in 2-propanol.

I didn't mean to insult you. But people make mistakes (including me). In that case, may be European Pharmacopoeia did it here. :-)

No insult taken! As a student, I understand the mistakes lies much more often with me. In a few weeks I have an exam in Arzneistoffanalytik (drug analysis) and looking at the past exams, these sorts of solubility questions have come up often. Unfortunately, the instructor is not exactly the most approachable, so I'm trying to figure it out.


text:
Storage of air and temperature sensitive reagents
What's your method for storing air/moisture and temperature sensitive reagents? For example:

A solid. Should I just blow some argon, screw the lid, put parafilm and put it in the fridge? Can a dessicator with silica be put in the fridge?

A liquid. Do you trust the septa seal that come with some bottles? I guess that after a few punctures, air can get inside the bottle. Moreover, sometimes I get bottles of air sensitive reagents with no septa. What is your procedure with this? Use a rubber septa or schlenk flask, maybe?

In general, do you trust your methods or purify your reagent before using it? For example, for something like $\ce{BF3 * OEt2}$.

Thank you in advance.

I am not much of a synthetic chemist, but here are my 2 cents. Schlenk + argon line + septa is good for one day's work, not for storage. Other than that, methods vary. Some things may sit in a fridge, some in a glovebox, and some have to be purified or prepared each time before use.

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.

Your question can not be easily answered. The only valid reply must be: It really depends on the individual case and the reactivity as well as physical properties.

Is your compound easy to evaporate? If not, it is good practice to
flame dry a Schlenk flask and to flush the flask multiple times with argon.
In the best case, you have dried the argon prior with driderite for
example. When the substance is transfered to the flask either through
the sept or in argon counterflow, it can be put under high vacuum for elongated periods of time to remove residual water and oxygen and than again flushed with
argon. Best to to this multiple times. For some hygroscopic compounds, also additional heating might be necessary in combination with high vacuum to get it really dry.

If your compound can be easily evaporated, and has a boiling point significantly different from water, destillation in a previously dried apparatus might be an option. I would add activated molecular sieve (see below) to the sample and  collection vessel to keep the water isolated.

As it was pointed out already, septs are prone to puncture and more
often you use them, more leaky they will get.  For longer term
storage, I put para film around the sept. Personally I found, that
silicon septs are better seals, but not as chemically resistant in
comparison with rubber. I perform some reactions which destroy the sept after one single use, so keep this in mind that you have to replace them more or less often.

When you are using a chemical that is sealed by the manufacturer, use always a balloon with inert gas, ideally argon because it is heavier than air to repressurize. Always use a balloon that was flushed several times beforehand with inert gas, otherwise it will contain residual water and oxygen.

Melting and sealing glas valves gives you a very good seal, but you need special equipment (oxygen flame, inert atmosphere). In combination with overpressure this might even be dangerous. Also, you can not easyly remove your compound from such a container.

If your substance of choice is a liquid, you may want to put
activated, that means very dry molecular sieve to the flask, to keep
it dry for a longer period of time. The proper way of activating
molecular sieve is to heat it to 400 degrees Celsius under high
vacuum until the pressure stabilizes. But be aware that there are
substances that will deteriorate the molecular sieve. You can not use
it for hydrofluoric acid or a reaction which contains oxalylchloride because they will attack
the silicon. I have however used molecular sieve for drying TBAF. If
a substance is extremely wet, molecular sieve is also not the perfect
choice to dry it, a better way would to start with $\ce{Na2SO4}$ first and
than apply molecular sieve. I had to do this once for wet
acetonitrile that was delivered in a fresh, sealed bottle. Molecular sieve can be obviously also used in the fridge or the freezer but continued freeze cycles may limit the lifetime.

For storing water sensitive solids I would use an excicator. The best way probably would the use proper high vacuum grease and use Schlenk flasks that are put under vacuum and than placed in the excicator. Subsequently the excicator itself would be put under vacuum. For the drying agent I use anhydrous $\ce{CaCl2} that was heated to 250 °C under vacuum for a couple of hours.

Regarding temperature dependence of compound stability, one has to keep in mind, that if the flask and its internal atmosphere are cooled down in the freezer, this creates a negative pressure and might suck humidity into the flask. You could counteract this be cooling a flask to freezer temperature under protective atmosphere first. If I have a compound that is very prone to hydrolysis and is kept in a freezer I put it immediately to the Schlenk line before it warms up, so the moisture is evaporated before it can react.

For substances that need to be cooled and still kept dry, I would use a Schlenk flask under argon and put these in a plastic container either with silica gel or $\ce{CaCl2}$ to avoid condensation water upon warming up to room temperature.

There are substances that are so reactive, that they are delivered to
you in sealed glas valves. Example are TMSI or
N-(trimethylsilyl)dimethylamine. In these cases it is best to order
just quantities that can be readly consumed in a small timeframe,
because they will often attack most sealing materials (TMSI
especially). On the other hand, due to their reactivity some substances like this might be used to dry your reaction mixture very efficiently.

Other ways to dry liquids, especially solvents, are the usage of elemental sodium or destillation over benzophenon. Here it is always necessary to check the compatibility. For example one can not use sodium with DCM but with THF.

Useful info on drying solvents:
https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Drying_Solvents
http://books.google.com/books?id=SYzm1tx2z3QC

When you are not sure if a sample contains water, it is always good
practice to do a quick proton NMR to check for a residual water peak.
This will obviously just work when the water is not reacting itself
with your compound or only very slowly. In this case you can test the
fresh compound and also deliberately add a little bit of water to see
how the spectrum changes to have a reference  chemical shift for hydrolysis
deterioation product. For many compounds, this works especially well
with heteroatoms like 31p. (Here you can give very easily give an estimate in percent for deterioation just by comparing the integrals.

A special case might be the drying of gases: As mentioned above one can use an indicator drying agent like driderite that changes it's color when it absorbed specific quantities of water. Alternatives would be the stream though concentrated sulfuric acid or through granulate of $\ce{NaOh}$, $\ce{CaCl2}$...

In some cases it is not sufficient to keep a compound or reaction dry but it is necessary to remove oxygen beforehand. For liquids and solvents I flushed solvents with argon. Helium might also work in some cases. Otherwise sodium benzophenone ketyl not only reacts with water but with oxygen and creates a blue color, when no oxygen, water or peroxides are present.

Thanks for the comprehensive answer. However, I have so quick doubts that I hope you could help me with.

If the reagent has to be stored under inert atmosphere, as is sensitive to oxygen and moisture, would you store it anyways in a vacuum desiccator or simply in a Schlenk flask under argon/nitrogen?

Does your institution pose a problem if you store reagents in recipients without a proper sticker with pictograms and information? Or you place the sticker (or a new one) on the new recipient?

Three thoughts:

As soon as you pierce the septum, it is perforated. Depending on what you aim to store in the bottle, the material of the septum and its age (not age, but what happened to the septum [light, temperature, time of exposure to chemicals so far including the lab dishwasher, etc]), you can't trust the thing as much as you did prior to this. It is noteworthy that lids and septa are available in different materials. The gradual move away from fluorinated materials (PFTE/Teflon) with new seals of different chemical resistance and mechanical properties (more prominently seen on seals on GCMS vials) is a story on its own.
You can degrade a chemical in a bottle closed by a septum, but the cause isn't the perforated septum in first place.  If the chemical is stored in the fridge to compete decomposition, and your first puncture (esp. across an already pierced septum) is the needle to draw the reagent solution instead of the one with gentle overpressure by the Schlenk line to add dry nitrogen (meh), or argon (better for its density, but more expensive), chances are increased that for pressure release you draw (humid) air back into the bottle.  Or you open the still cold container, and humidity condenses on the walls, this water starts to decompose your chemical.  The cross on Aldrich's SureSeal septa helps you (like a grid) to minimize the number of punctures, though.
What do you understand as storage?  If it is long term, infrequent draw, and your chemical is compatible to glass (there a multiple sorts of glass), than sealing it in a glass vial still is the most reliable option in different volumes, white glass/brown glass, under own atmosphere/under inert gas; just a matter of equipment and some training for dexterity.

As for the purification of chemicals, the typical goto in organic chemistry (though not only covering $\ce{C_xH_yO_z}$, but a fair amount of inorganics), is Perrin/Amareggo' Purification of Laboratory Chemicals.
The idea of putting a dessicator into a fridge is foreign to me: typically too heavy to support it on the grid, it occupies too much volume per volume of the sample of interest (don't forget the valve to evacuate the inner), it is too expensive if this falls to the floor and shatters (not only for the chemicals to store + the drying agent).

Fresh parafilm over septa each time your use it will help preserve the contenrs

To store air-sensitive and temperature-sensitive reagents:
Air-Sensitive Reagents:

Use air-tight containers.
Inert gas can help displace air.
Store in a cool, dry, dark place.

Temperature-Sensitive Reagents:

Follow recommended storage temperatures.

Use appropriate storage equipment.

Prevent temperature fluctuations.
Label containers clearly, monitor conditions, and handle with care, following safety guidelines.

Regarding the third bullet, mentioning BF3·OEt2, two answers.
1, As has been noted by at least one other, in the safe practice of preparative chemistry, especially involving anhydrous/air-sensitive reagents, there are no good, sufficient answers to overly general questions. Purification, handling, and storage of each reagent, product, solvent/solution, is an individual problem to be addressed.
2, With regard to the specific reagent mentioned as an example, BF3·OEt2 [CAS No. 109-63-7]—taking this not as a general example, but as a specific question of how that reagent should be handled, purified, stored:
Veronica Cornel addresses the BF3.Et2O reagent in her entry in the Encyclopedia of Reagents for Organic Synthesis (EROS), here: https://onlinelibrary.wiley.com/doi/10.1002/0471264180.or013.01
Cornel addresses the observation that advantitious air oxidizes (and so darkens) commercial preparations, making redistillation prior to use the standard course for careful work (because the if reagent titre varies with time since distillation, a portion used in chemistry that is not redistilled is being used without knowing actual concentration). She cites Zweifel and Brown's 1963 Organic Reactions chapter with regard to use of Ca(OH)2 for capturing impurities and addressing problematic bumping. [https://doi.org/10.1002/0471264180.or013.01]
She also generally addresses this reagent's precautions, and handling and storage, but in this regard we refer to our longer post here, that such reading is not sufficient to teach these things.

Feel free to flag questions or answers that seem to be a safety issue. The community can close questions they feel are off topic. The moderators can do so as well.


text:
What happens to the ions after dissociation in an aqueous solution?
I'm in high school studying about Buffer Solutions. We had an example of a solution made up of Ammonium Hydroxide and Ammonium Chloride for a Basic Buffer.
NH4OH ⇌ NH4 (+)  +  OH(-)
NH4Cl ⇌ NH4 (+)  +  Cl(-)
What is up with the reversible sign? Do the compounds, once dissolved in water, break into ions, and then form the reactants again? This is a petty doubt perhaps, but I need conceptual clarity.
Now, if I add a strong acid, the H(+) would be accepted by the OH(-) from Ammonium Hydroxide, right? But then where would the Cl(-) go?
And if I add a strong base, the OH(-) will be accepted by an Ammonium ion. What about the rest of the ions? Do they just stay suspended? Or do they react or something, and form the reactants once again?
Say, I want to mix something else in there, shouldn't it technically react and form newer compounds?

There is no NH4OH in water solutions. Ammonium hydroxide is in context of ambient conditions a myth. There is equilibrium NH3(aq) + H+(aq) <=> NH4+(aq)

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

Atoms and radicals move all the time. In ionic solutions, they're particularly mobile, and do continually form new relationships.
Some groups are quite stable; e.g., in $\ce{NH4Cl}$ solution in water, $\ce{NH4+}$ mostly "sticks together". In solution, it is free to wander around the beaker, surrounded by a bevy of covetous negative ions. Lose the water through evaporation, however, and the $\ce{NH4+}$ and $\ce{Cl-}$ neatly form up again into a neat lattice.
Water itself is an example: $\ce{H2O <--> H+ + OH-}$ and yet other species. On the average, at room temperature, only about 2 molecules in 109 are ionized, yet that still gives it a somewhat ionic character.
Viewing an animation of salt dissolving in water might help you visualize the process.


text:
Why 40J energy was wasted in case of a non spontaneous process by the system?
If there is a system which while undergoing a process produced 30J energy and wasted 40J energy therefore the useful energy was -10J and this explains the process was non spontaneous because ∆G is coming out to be positive where ∆G is negative of useful work done. According to me the system which already had 30J of energy which was not sufficient for the process therefore to complete the process it gained 10J energy more and got its process completed and from this I understood that the process was non spontaneous because to complete it extra 10J was given to the system so the whole process took 40J to complete in which 30J was used by the system itself but 10J was given to the system. So according to me 40J energy is used. But my teacher said that if  a system produces less energy and wasted more then the useful work done is negative and he gave this example that a system produced 30J and wasted 40J so what I wanted to ask was I didn't got that how 40J was wasted if the system produced only 30J? According to me 30J was contributed by the system to the process and extra 10J was given to the system for the process....so according to me 40J total was used and not wasted!

It is always good to summarize and/or complement the verbal description  by describing the problem/task with symbolic algebraic expressions. And, keeping it that way until all is ready to just plug-in literal numbers with units. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger  permanent value.  //
You may find useful  formatting mathematical/chemical expressions/formulas.

What is the meaning of this post ? If a system produces $30$ J and wastes $40$ J, it can be understood that the system produces $70$ J, but that only $30$ J is a useful work, the rest ($40$ J) is lost as heat given to the environment. Is it right ?

@maurice Can you give me an example where the useful work done is negative. So that ∆G comes out to be positive.

The first sentence about 55 words, the second one about 80 words. Such sentences are very hard to comprehend. One of guidelines for writing technical texts is that average sentence length should be 20-25 words, with recommendation to avoid sentences longer than 35 words.

Any reaction like $\ce{Zn + 2 H+ -> Zn^{2+} + H2}$  can produce useful work.  Why should Delta G be positive ?

Even James Joyce used paragraphs in Finnegans Wake. Riverrunonsentences.

@Maurice if in any reaction ∆G comes out to be positive then in that case the reaction is non spontaneous. If the reaction is non spontaneous then the useful work done should be negative such that ∆G comes out to be positive because ∆G is negative of useful workdone

Consider a system consisting of two heat reservoirs of temperatures $T_2 = \frac 43 T_1$.
The first process would be the external thermal engine, taking heat $Q_2=\pu{40 J}$ of heat from the warmer reservoir $T_2$, producing work $W=x \lt \pu{10 J}$ and passing the heat $Q_2 = 40 - x\ \pu{J}$ to the colder reservoir $T_1$. In this case is $\Delta G \lt 0$.
The second process would be the external thermal pump, taking heat $Q_1=40 - x \ \pu{J}$ of heat from the colder reservoir $T_1$, work being externally provided $W=x \gt \pu{10 J}$ and passing the heat $Q_2 = \pu{40 J}$ to the warmer reservoir $T_2$. In this case is $\Delta G \gt 0$. The work provided by the system is negative, what means it is work provided to the system.

The 40 Joules of energy wasted in a non-spontaneous process by the system is due to the fact that the system must overcome a certain energy barrier to initiate the process. In a non-spontaneous process, the system doesn't naturally proceed from its initial state to the final state, so energy input is required to drive the process against the natural tendencies.
This energy input could be in the form of work or heat supplied to the system to make it change its state. Essentially, it's the energy needed to make the process happen even though it is not favored by the system's inherent thermodynamics. In contrast, spontaneous processes occur without the need for external energy input because they are energetically favorable based on the system's conditions.
So, the 40 Joules of energy represents the additional energy required to force the system through a non-spontaneous change or to push it over an energy barrier preventing the process from occurring on its own.

can you pls explain me a process where a system itself produces less energy but wastes more energy than produced which will lead to negative useful work done and hence positive ∆G


text:
Predicted H NMR spectrum shows two signals for NH2 group

There are two singlet signals for NH2 group around 2.4 and 3.05 ppm. Spectra is predicted (from MNova), so there was no water present. Why the signal of NH2 group isn't a broad singlet?

What is your question?

@PAEP Plausible option: why the signal of the amino group isn't a broad singlet (as in e.g., aniline, example in SDBS 905), but split into two.

@Lucia Boldišová There are multiple tools (as in programs) and methods (as in what parameters, level of theory, etc) to predict NMR spectra.  Can you please extend your question by a note to disclose at least the source, name (and because it evolves, the version / year of publication) of the software used to compute the data displayed?  There is a hunch (MNova), but to know for sure would provide future readers an avenue in lines of «... and I have already seen this kind of prediction provided by so-and-so».

Sorry I forgot to mention it, we used MNova for prediction.

The amino group can be conformationally locked by an intramolecular hydrogen bond:

by formation of a five-membred cycle.  (And in organic chemistry, the  formation of five, and to some lesser degree of six membered cycles anyway is favourable.)
Contrasting to a terminal amino group in e.g. n-butylamine, the two hydrogens of said amino group then no longer experience the same intensity of the magnetic field / averaged over time no longer are magnetically equivalent. Though it depends on the solvent used, temperature and concentration of the sample, and the spectrometer parameters how much of this effect eventually is observed in the actual experiment, I consider this as a good feature in the algorithm to predict the H NMR spectrum.

This sounds plausible; but I wonder whether the algorithm is too smart for its own good. I am not sure if the C(O)–N bond is configurationally stable, and if so, whether the isomer shown (R1 and R2 cis) is favoured. Even in N-methylacetamide (R1 = R2 = Me) the trans isomer is significantly more stable (e.g. https://doi.org/10.1021/jp8106556), and here R1 and R2 are even bigger...

@orthocresol In the structure posted by the OP, O (of carbonyl) and H (of amino group) present a path of 5 atoms; O-C-N-N-H.  In N-methylacetamide your paper describes, it however only equates to 4; O-C-N-H (with a ratio of the trans conformation over the cis conformation greater than 95:5 for H2O, CHCl3, C6H12). How does the energy surface look like for a rotation round the torsion angle O=C-N-N? So I speculate «the click» by the intramolecular hydrogen bond plus a little pi-pi interaction of the aryls yields a conformation not too costly (energy wise) vs the trans conformers.

@orthocresol Re algorithm: The OP doesn't disclose the origin of the prediction, Mestre/MNova however were a plausible reference (the optional plugin does cost some extra, eventually can help / automatically assigns heteronuclear correlation spectra, too). Though the companies tend to keep their algorithms closed source, there is e.g. «NMR signal processing, prediction, and structure verification with machine learning techniques» as a review by Carlos Cobas on behalf of Mestrelab in Magn. Reson. Chem. (https://doi.org/10.1002/mrc.4989, note the public review on publons straddles ACD).

Of course, that is very reasonable speculation. :) This is probably Mnova, yeah.


text:
Acidity of metronidazole
Metronidazole has two $\mathrm{p}K_\mathrm{a}$ values: $2.57$ and $15.42.$ The basic group is the imidazole moiety. Does metronidazole have acidic groups?

There are no acidic groups in metronidazole. The $\mathrm{p}K_\mathrm{a}$ you are quoting relates to the protonated form of the molecule (usually hydrochloride or benzoate). The most easily removed proton will be the $\ce{-OH}$ which is no more acidic than a standard alkyl alcohol.

Ok thanks! I just don't get it why the pKa of metronidazole is so low (2.57) when the pKa of imidazole is 7?

The nitro group is strongly electron withdrawing which reduces the basicity of the imidazole ring

@user139229  If you are happy with the reply please accept it

As Waylander correctly predict that presence of no acidic groups in metronidazole framework. The reported $\mathrm{p}K_\mathrm{a}$ value is related to the protonated hydrophilic form of the molecule (e.g., usually hydrochloride salt).
Metronidazole is a imidazole-based compound, namely 2-(2-methyl-5-nitroimidazole-1-yl)ethanol. All reported $\mathrm{p}K_\mathrm{a}$ values of the imidazole-based compounds are attributed to the protonated nitrogen atom at position 3 $(\ce{N-3})$ of the imidazole ring (Ref.1). Again, as Waylander pointed out, excerpts from Ref.2 suggest following:

Owing to the −NO2 group attached to the $\ce{C-5}$ position, metronidazole shows a weak basic character compared to the $\mathrm{p}K_\mathrm{a}$ value of the imidazole ring with a $\ce{N-3}$ atom ($\mathrm{p}K_\mathrm{a}$ 7). The introduction of an electron-donating group at the $\ce{N-1}$ and $\ce{C-2}$ positions of the imidazole ring of metronidazole increases the $\mathrm{p}K_\mathrm{a}$ value, but this effect is obstructed by the electron-withdrawing $\ce{−NO2}$ group.

The $\mathrm{p}K_\mathrm{a}$ value of metronidazole was reported as 3.039 in Ref.2. In a different study using the capillary electrophoresis method, the $\mathrm{p}K_\mathrm{a}$ value of metronidazole was determined in the aqueous $30−60\% (v/v)$ methanol media (Ref.3). The average $\mathrm{p}K_\mathrm{a}$ value was calculated to be 2.49. A closely related value of $2.43 \pm 0.03$ has also been reported, which had been determined by the spectrophotometric method in a solution containing methanol−dioxane, acetonitrile, and water (Ref.4).
References:

Sheng-Feng Cui, Li-Ping Peng, Hui-Zhen Zhang, Syed Rasheed, Kannekanti Vijaya Kumar, and Cheng-He Zhou, "Novel hybrids of metronidazole and quinolones: Synthesis, bioactive evaluation, cytotoxicity, preliminary antimicrobial mechanism and effect of metal ions on their transportation by human serum albumin," European Journal of Medicinal Chemistry 2014, 86(30), 318-334 (DOI: https://doi.org/10.1016/j.ejmech.2014.08.063).
Kader Poturcu and Ebru Çubuk Demiralay, "Determination of $\mathrm{p}K_\mathrm{a}$ Values for Some Benzimidazole and Imidazole Group Drugs Using the Reversed-Phase Liquid Chromatography Method," Journal of Chemical & Engineering Data 2020, 65(11), 5617-5626 (DOI: https://doi.org/10.1021/acs.jced.0c00764).
Marina Shalaeva, Jeremy Kenseth, Franco Lombardo, and Andrea Bastin, "Measurement of Dissociation Constants ($\mathrm{p}K_\mathrm{a}$ Values) of Organic Compounds by Multiplexed Capillary Electrophoresis Using Aqueous and Cosolvent Buffers," Journal of Pharmaceutical Sciences 2008, 97(7), 2581–2606 (DOI: https://doi.org/10.1002/jps.21287).
Gergely Völgyi, Rebeca Ruiz, Karl Box, John Comer, Elisabeth Bosch, and Krisztina Takács-Novák, "Potentiometric and spectrophotometric $\mathrm{p}K_\mathrm{a}$ determination of water-insoluble compounds: Validation study in a new cosolvent system," Analytica Chimica Acta 2007, 583(2), 418-428 (DOI: https://doi.org/10.1016/j.aca.2006.10.015).


text:
What causes ionisation in redox reactions?
I'm trying to understand what's behind the ionisation of atoms in a redox reaction. I apologise if the answer to this question is obvious. I've searched around, but, surprisingly, I haven't found a clear answer on this.
Many many years ago, in secondary school, I remember watching a demonstration of a redox reaction of magnesium and oxygen (Mg2(s) + O2(g) --> MgO).
This involved a piece of magnesium being placed over a flame, after which a rather intense light would be emitted as the magnesium reacted with the oxygen, leaving behind MgO.
My understanding of this reaction is as follows: two valence electrons break free from the Mg atom and become part of the O atom, resulting in two ions, both of which have a stable octet configuration (8 electrons in their valence energy level). The electrostatic forces of the two ions result in an ionic bond, MgO.
Assuming the above is correct:

How did the two electrons break free from the Mg atom? What caused that ionisation? Was it the heat from the flame? I've read somewhere that the heat from the flame only speeds up the reaction, and that the reaction would take place even in the absence of the flame, but at a considerably slower rate. If this is true, then again, what's causing the ionisation? Is it the heat from the ambient environment? Is that heat enough to ionise an Mg atom?

What forces are behind the two "escaped" electrons joining the O atom? The O atom is neutral, so there should be no electrostatic forces pulling in those two electrons. So what's behind the ionisation of the O atom?

They simply bind to each other, that's it.

Who says both electrons from one Mg end up on the same oxygen?

The step-wise mechanism of the reaction might well depend on whether a flame is applied (what is the source and composition of this flame?). This question would largely be answered with an explanation of the location and mechanism of the reaction.

@Mithoron The ions bind to each other due to the electrostatic forces, but what I don't understand is how they become ions. Some amount of energy is required for the two electrons to break free from the Mg atom, right? Where does that energy come from? Is it the heat from the flame? And what forces attract the two electrons to the O atom? The O atom is electrically neutral, so what forces pull the electrons towards the O atom?

I'm saying your "understanding" is thoroughly wrong - premise is false, and you should have looked for earlier questions with this annoying misconception.

@Mithoron OK, I apologise if I've misunderstood something obvious. I've made a number of attempts to look for similar questions and have not found anything that answers my question. Could you please point me to an example of a question with this misconception?

For example https://chemistry.stackexchange.com/questions/164456/spontaneous-reaction-between-chlorine-and-a-hypothetical-atom

Unfortunately, this doesn't appear to be obvious to people in general, while it should be. $$\ce{Mg(g) + O(g) -> MgO(g)}$$ - nature of bonding in MgO molecule is barely relevant.

The fundamental mistake you're making here is treating each atom in the reaction as though it's isolated. If you do that, the reaction will make no sense. You have to look at them together.
The energy to remove the electrons does come from heat. Yes, magnesium does oxidize at room temperature, but no, there is not enough heat at room temperature to ionize the magnesium atom. That is, room temperature does not provide enough energy to rip the electrons completely away from the magnesium atom and fling them out into space. We know this because magnesium does not spontaneously emit electrons at room temperature. (But if you heat metals to high enough temperatures, there will be enough thermal energy to start emitting electrons; that's how vacuum tubes work.)
When magnesium reacts with oxygen, the electrons do not get flung out into space. In fact, the electrons barely go anywhere. They are still right next door to the magnesium atom, attached to the oxygen. That requires a lot less energy than completely ripping them away; that's why the reaction can proceed at room temperature. In other words, it is the formation of the ionic bond that makes all of this possible.
As for oxygen, it's easy to rationalize how the first electron attaches. This is similar to how a statically charged object, like a balloon, can cling to a neutral object. Atoms are neutral, but they are polarizable. As an electron approaches a neutral oxygen atom, it repels the electron cloud slightly, exposing a little bit of juicy positive charge that is enough to make it attach. This is a relatively weak effect. Attaching an additional electron to oxygen gets you 141 kJ/mol. By contrast, removing an electron from neutral oxygen costs 1300 kJ/mol. But 141 kJ/mol is still substantial on a chemistry scale.
The second electron is harder to understand, because now there is an excess negative charge on the atom, and it's hard to see how you could attach another electron. In fact, it does not happen. The oxygen dianion is not stable in a vacuum; it will emit an electron. But once again, the reaction is not occurring in a vacuum. There is a positively charged magnesium atom right next door, which provides an attractive force that prevents the second electron from flying away. Once again the ionic bond is critical; the reaction doesn't work without it.

Thanks for this! So just to be clear, the heat at room temperature is what allows for the two valence electrons to break free from the Mg, but only when they're in close proximity to the O atom? So if we attempted the reaction in an environment with a much lower temperature, nothing would happen?

@JohnO'brien Electrons don't ever "break free" in chemistry, unless it's plasma chemistry. Your atoms just bump into each other creating highly polarised bond that is called "ionic". That does not mean there are separate ions in a molecule, just that ionic component of bonding is greater than covalent.

Yes, reducing temperature slows reactions. The atoms can only bond if they collide with enough energy to overcome the activation energy. This becomes less likely at low temperatures. Eventually it will essentially stop. I don't know how cold you have to make magnesium before oxidation becomes negligible... probably pretty cold.


text:
Why does a Sealed baggie of dry ice lose weight before popping?
Why does a sealed baggie of dry ice lose weight over time at room temp, even before it pops and the gas escapes?  I thought it might be due to buoyancy, perhaps? As the bag expands, it displaces air.  But if that is the case, the why would it matter since CO2 is actually MORE dense than air, right?  Please help me understand.

Simply because the bag is not 100% air-tight. CO2 keeps escaping.

Aside of gas diffusion through the walls, bag expansion by 1 L causes measured mass on scales is decreased by approx. 1.2 g. as that is equivalent to the buoyancy change.

The density relative to air is (mostly) irrelevant. As the volume expands, the buoyancy increases relative to the volume/mass of displaced air. Ships float because of the displaced mass of water (this is purely dependent on the mass they carry, not the density of the substance making up the mass).


text:
Why this Lewis structure for HNO3 nitric acid is wrong?

I know that this Lewis structure of HNO3 is wrong, but I don't know the reason why, except that here the side atoms (oxygen) are connected via a bond. This structure fulfills the octet rule and also all atoms have zero formal charge. So what rules does this Lewis structure of HNO3 violate?

Indeed, this structure does not violate any of the said rules. It is wrong simply for the reason that it does not exist.

Lewis structures are a highly simplistic way of understanding bonding and bear almost no relation to the actual structures of real molecules. So the fact that this structure obeys the rules of lewis structures tell you nothing at all about whether this is the right structure of the compound. it isn't.

This is all missing the point - formal charges are not a problem - that's the point. Otherwise why would the concept be needed at all? Your stupid "minimization of charge" rule is so you wouldn't come up with crazy dianion-dication with charges on neighbouring atoms.

The structure isn't incorrect. It just has a high formation energy.

Structure
Heat of Formation ($\pu{kJ mol^{-1}}$)
Reference

236.69
MolCalc

-159.02
MolCalc

Because of the low heat of formation of the correct structure, the incorrect structure spontaneously converts to the correct structure. The structure that you have proposed may be an intermediate in some reactions.
Notice that the vibration of the correct structure at $662.09 \pu{cm^{-1}}$ shows some resemblance to the incorrect structure.

How did MolCalc obtain these enthalpies of formation? What method (PM3?), conditions, phase? The number $\pu{-159.02 kJ mol^-1}$ isn't something I was able find for ether gaseous or aqueous phase under standard conditions in the more or less reputable sources (e.g. NIST, Argonne). What uncertainty is expected for the thermodynamic properties calculated with the level of electronic structure computations used by MolCalc?

The exact structure that was inputted would be useful as well. Has a geometry optimization been done? I assume so because you quote vibration frequencies, but would be good to be clear.

@IanBush According to https://pubs.acs.org/doi/10.1021/ed400164n, the errors in standard enthalpies of formation and vibrational frequencies for PM3 are approx. $\pu{32 kJ mol^-1}$ and $\pu{160 cm^-1},$ respectively. These errors and rigorous information about used QC method should be clearly stated alongside the reported data. Also, would be interesting to compare the results from DFT calculations with the semi-empirical method.

Structures with nitrogen atom bound to several oxygen atoms are generally energy rich, which would be boosted by the stress of three-member ring. Such rings are generally unstable and reactive. Also, a peroxy group is very reactive even not being strained by the tight ring.
There are many molecular entities with structure not violating any from the listed rules. But in spite of that, they do not exist, as such a list is never complete. Sometime, the missing rules are not relevant, sometimes they are essential.