text:
How to measure purity of nitrogen gas in a "homemade" way?
Is there any indicator or a way of measuring purity of nitrogen? The lab-way of doing things is out of my reach and I was thinking about a DIY way. I want to decrease oxidising atmosphere in a container by introducing $\ce{N2}$ but I am not sure about the purity of it.

There is no easy homemade way. It would help if you say how you would produce it.

You probably mean $\ce{N_2}$ (typed as $\ce{N_2}$: on Chemistry.SE you can use mhchem to render chemical formulas with MathJax), not N3. Also, what is the source of your nitrogen?

Provide the target $\ce{N2}$ purity and procedure dependent minor components ( $\ce{O2}$, $\ce{CO2}$ etc.). BTW, is applicable replacing $\ce{N2}$ by $\ce{CO2}$, ? As it is easier to generate ( unless its acidobasic activity matters ).

Since most of the contaminants you might worry about (in particular oxygen) are more reactive than nitrogen, try measuring the amount of gas lost when such reactions occur. Measuring volume is easy and the lost volume will indicate the purity.

Thanks guys! I am burning the oxygen, creating CO2 and then filtering those gases via Ca(OH)2 aq.

First, you should have ~1% Ar, in addition to the N2. Second, what purity is needed? At home, you could use a diffraction grating, razor slit and camera to make a spectroscope,  but would the Ar confuse results?


text:
Why polysaccharides are not considered as lipids?
It is mentioned in my textbook that all lipids have one common trait which defines them :they are insoluble in water;they are hydropohic.
I want to ask ,since polysaccharides are also insoluble in water,what makes them not classified within lipids?

https://en.wikipedia.org/wiki/Lipid https://en.wikipedia.org/wiki/Polysaccharide

Like this your are going to classify all compounds insoluble in water in one big class whatever called. Say, hydrophobic water insoluble ones. Being polysaccharides insoluble in water might be their property but not the definition of them.

Polysaccharides are usually water-soluble.

Wood is not soluble in water and neither are stones. What is then the difference between wood and stones ?

Check your closest beginners’/general chem instruction lab. I bet they have starch solution.

Suggest:  Solubility of either lipids, or (poly)saccharides may be affected by side groups and chains, the length and chemical nature of the later.  Instead, see how sub-units are joined together.  Typical lipids are esters, e.g. of glycerol:

(credit)
In polysaccarides, sub-units bind together as polyethers like chitin or cellulose:

(enter link description here)
And while sharing this structure motif, table sugar is well soluble in water, chitin and cellulose are not.  Not only because of different size of the molecules, but because of their peripheral substituents.

source: Stack Exchange

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

sort: votes


text:
How does soap actually reduce surface tension of water
I was a bit confused on how soap reduce surface tension of water.I've read some answers like Why does soap reduce surface tension of water and Why do surfactants lower the surface tension
But the answers given don't clears my doubt that if the soap molecules come to the surface to separate the water molecules and weaken the hydrogen bond, wouldn't they form stronger ion dipole interactions and increase the surface tension.I mean they would attract the water molecules with greater force.
I do understand from the energy point of view but I wanted to understand it considering the forces.
I know that it is a repeated question but consider giving an explanation to my point.

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


text:
Is this stoichiometric question answerable? (unknown variable)
This question kind of borders on the fields of both semantics and math, as well as chemistry, but I still believe this is the right place to ask.
On a test I took, a question was posed as such:

Via titration we are adding a solution of 25 ml, with 0.5 M NaOH, to 25 ml of HCl. What will the concentration of the acid be?

Now, this question seemed a bit strange to be, and it was definitely new. I'm used to calculating concentrations after dilution and other stoichiometric calculations, but this question seemed like a misworded version of those other questions I've grown familiar to.
So, I asked my teacher if she could clarify. First I asked whether the 25 ml of HCl was pure HCl, or a water solution with HCl in it. She said it was pure HCl. Then I asked if she was looking for the concentration of the acid before ANY reaction between the NaOH and the HCl had occured. To this, she also said yes.
The options were:

0.25 M
0.5 M
0.75 M
1 M

I answered 1 M, my reasoning being that 25 ml of HCl is the same volume as 25 ml of water with NaOH. Obviously, this reasoning is quite flawed, but I didn't have anything better. The answer was wrong. However, when my teacher told me what was meant by the question afterwards, her explanation did not align with the question, nor her previous clarifications. She said that this was a simple case of:

$$C_1 \times V_1 = C_2 \times V_2$$

Where $C_2$ was the unknown, since the concentration of HCl was not specified. I don't see how this makes sense. All the question is saying, is that there is a 25 ml container, with HCl in it. It doesn't even say that this 25 ml consists of a water solution with HCl in it, but a more knowledgeable chemist would've known that HCl cannot exist outside of a water solution (that's what my teacher said at least).
Then the the question says that 25 ml of water, with a concentration of 0.5 M of NaOH is added. And from those two values, I'm supposed to find the concentration of HCl. That basically means:

$0.025$ l of H$_2$O$ \times 0.5$ M NaOH(aq) $= 0.025$ l of water $\times x$ M HCl (aq)

Which would then make the answer 0.5 M. Maybe she wanted me two add the two volumes, which would make sense:

$0.025$ l of H$_2$O$ \times 0.5$ M NaOH(aq) $= 0.05$ l of water $\times x$ M HCl (aq)

Which would then make the answer 0.025, which wasn't even an option.
But even if it was, it still wouldn't make sense to me. Just because you have a solution with a given volume and an unknown concentration, adding another solution with a known volume AND concentration won't just magically let you know what the unknown concentration in the original solution is, now in this new solution. Or am I wrong?

Use rather a meaningful title than a click bait.

@Poutnik Is this title better?

The answer is either trivial either there is no answer, depending on interpretation of the question.

Your teacher is giving you rather wishy-washy questions and answers. I lean towards the interpretation where 25 ml of 0.5 M NaOH is supposed to exactly neutralise 25 ml of x M HCl, and you are supposed to find x. After all, that's what a titration is about: you add HCl until you neutralise the NaOH exactly, which can be determined using a pH indicator. However, none of this is explicitly stated, and as per Poutnik's comment one could easily say that there is no reasonable answer.

The question as it is in the box cannot be answered. And I censor my own opinion about the teacher / teaching.

With a lot of assumptions, 0.5 M is a reasonable answer.
Here are the holes you have to fill in.

If it is a titration of NaOH and HCl, it is probably an acid/base titration where we stop the titration when the pH is neutral.

Titrations are often performed to figure out the concentration of a solution. You would titrate a smaller sample of the solution to figure out the concentration. Technically, the concentration of HCl after the titration is zero because all of it is supposed to react as you reach the endpoint of the titration. So it makes sense to ask about the concentration in the original solution.

The stoichiometry of the neutralization of HCl and NaOH is one-to-one:
$$\ce{NaOH(aq) + HCl(aq) -> H2O(l) + Na+(aq) + Cl-(aq)}$$

Of course, it is silly not to give any context when posing such a question on a quiz. I won't comment on the statements you attribute to your teacher other than to say that hydrochloric acid always contains water (pure HCl is a gas under ambient conditions).

Sorry, we simultaneously answered it. Hope the OP gets it now.

@MFarooq No worries. The teacher should read it...

Good point, it seems that the teacher is teaching the wrong stuff. Nothing annoys me more when I see the dilution formula being used for titrations.

I think I understand. Since the same amount of NaOH moles is required to neutralize the same amout of HCl molecules, then we know that the amount of moles must be equal in both of the solutions in order for them to neutralize each other. Since the volumes are equal, this means that the molarity of NaOH and HCl will be equal. However, for this to be clear, one must assume that neutralization is the goal. So, if all the values were the same, but HCl was switched out with H$_2$SO$_4$, then the answer would be 0.25, since double the amount of NaOH is needed to neutralize H$_2$SO$_4$?

@A.Kvåle, You are right, but always, look at the balanced equation.

The formula $$C_1 \times V_1 = C_2 \times V_2$$
should never ever be used for titrations calculations. It fails miserably in most of the cases (e.g. titration of sulfuric acid with NaOH). This formula only works when the concentrations are in normality, which is considered an obsolete unit in modern chemistry. The proper use of this relation is for calculating dilutions because it is simply a mass balance equation and it works perfectly.
The test question as you wrote is rather poorly worded but basically it is asking you to calculate the concentration of the 25 mL HCl sample if 25 mL of 0.5 M NaOH was required to complete the titration.
All you have to do is to write a balance equation.
From the balanced equation determine the stoichiometric coefficients, i.e., the mole ratio between NaOH and HCl.
Calculate the moles of NaOH = Molarity of NaOH x Volume of NaOH used (L).
Apply the right ratio, and if you have written the balanced equation correctly, you will see the mole ratio is 1:1.
So, m= the moles of HCl = moles of NaOH,
Molarity of HCl = m/Volume of the original HCl sample in L.
You do not care about mixing of volumes until and unless the titration is incomplete, but that would be a separate story.
The correct answer is listed in the choices.

But the box question cannot really be answered. I agree it probably meant this, but the "will" makes an already bad question totally non sense....

Either the OP wrote it from memory or he/she translated it into English from another language.


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$j$-$j$ coupling interaction in Magnetic Resonance Spectroscopy
I am studying Magnetic Resonance Spectroscopy and I have some difficulties in understanding the j-j coupling interaction. What does actually happen from the physical point of view to the electronic cloud of the neighbouring atom? When I try to understand in books like A Complete Introduction to MODERN NMR Spctroscopy of Roger S. Macomber (but also in websites) it seems to me that there are explainations more focused on quantum theory and its rules, but what I want to know is simply what happens physically at the electronic cloud and the consequent interactions ? Therefore why does this causes that in the spectrum, for example, of the Ethanol molecule (CH3CH2OH) there are three peaks related to -CH3 group? Or why in the Acetaldehyde molecule (CH3COH) there are four peaks related to -COH group and two peaks related to -CH3 group just for the j-j coupling interaction?
I really would like to see things clearer at the basis (so what happens physically to the electronic cloud and the consequent interactions) and thus being able to understand the number of peaks (while I think I have understood the chemical-shift interaction. I want also to specify that for the moment I have to consider just these two interactions because I am considering MRS applied to brain tissues which I can approximate as liquid so, for example, the dipolar interaction is considered average to 0).

I am not exactly sure if this is what you are asking, but the J-coupling happens primarily by the interaction between the spin of nuclei which are close to each other. Consider the simple case of two inequivalent nuclei with spin I=1/2. Then, those two nuclei can orient parallel to each other, or opposite to each other. These two states will have different energy. (You can use the analogy of a dipole to think about it)

@ShoubhikRMaiti, the energies of the states is one thing, but the splitting is manifested as a difference in the transition energy. Well, yes, those are just differences between energies of the states: but it's slightly more subtle than just saying that $E(\alpha\beta) \neq E(\alpha\alpha)$. Instead, the splitting arises because $E(\beta\beta) - E(\alpha\beta) \neq E(\beta\alpha) - E(\alpha\alpha)$. Or, more loosely speaking: the energy needed for the "spin flip" depends on the spin states of the surrounding electrons. This is a pretty simplified view, but it sort of works.

Many organic chemistry textbooks will give some handwavy explanations which you might find useful at this stage, but beyond that, I'd also strongly suggest learning the QM at some point. I suggest Keeler's book (there is an accompanying series of lectures on YouTube). From what I can tell, you're asking for a physical picture of what goes on with nuclei / electrons, and implicitly you want a classical description: so-and-so repels so-and-so. But this is bound to fail eventually. Electrons are not classical objects, and spin is decidedly not a classical phenomenon. We can do better :-)

@ShoubhikRMaiti I understood that the j-j coupling interaction is also called indirect polar interaction because there is not only the interaction between nuclei but also a nucleus that causes the polarization of the electronic cloud of the close atom and this causes like a sort of interaction with the nucleus of the same close atom ... but I do not know if I understood well. And why this causes different peaks on the MRS spectrum ? So I was asking for clarification.

Thank you @orthocresol♦ ! So you are saying that I can not interpret the number of peaks with considering the interaction between the nuclei and electronic cloud from a classical point of view ?

@Manuela It looks like you've already asked this question on Physics.SE. Please note that cross-posting is usually not permitted. If there is a solid reason for it, it would be nice to add a link to the duplicate question on the other network site, mention in both questions that it's a cross-post and briefly explain the motto for it.

@andselisk♦ I did it because nobody answered in physics so I thought it was a question more related to chemistry

I think there are some classical explanations that may sort of make sense, but many of them actually do not stand up to scrutiny, when judged against the quantum theory. On top of that, from your question it sounds like you are doing a reasonably serious project. So I think it would be a good thing for you to eventually read some of the QM. Re. crossposting, if nobody's said anything on Physics, then I'd suggest the easiest solution now is to just delete the question on Physics. (Btw, no need to include the diamonds when @-ing us.)

Because for example you can understand the peaks due to the chemical shift interaction considering the electronegativity of the different atoms of molecules so I thought I could do a similar reasoning for j-j coupling interaction. Ok @orthocresol thank you again this helps me .

About the question: can someone simply close the question on physics ? :) @andselisk

@Manuela I believe you can delete the question yourself, just as orthocresol suggested. Alternatively, you can ping moderators on Physics.SE and ask how to better handle this situation.

The thing is, you can draw empirical correlations between molecular properties and the values of the J-coupling: that would be very similar to what you say about electronegativity and chemical shift. Again, there are a number of textbooks that cover this, in particular those with a more organic chemistry focus (the specialised NMR books will not, because they simply take the J values for granted, as some experimentally determined parameter). So are you really asking about the factors that affect how large the coupling is, or are you asking about the underlying origin of the coupling?

In either case, I'm not going to attempt to give a full explanation (it is a bit too long for me today), but I would point you towards some of the books here https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry (Ctrl-F for NMR). Check them out, and consult a supervisor or tutor as necessary...


text:
Find minimum solubilty of AgCN

i know it has to be solved by writing an equation between CN- conc and solubilty but i am unable to write a non-linear equation,any help is appreciated

Option C is the right answer
Solubility of AgCN will be equal to sum of concentrations of  Ag+ and [Ag(CN)2]-

Now I get it, rereading the q.

@OscarLanzi: Could you tell me where I am going wrong with the this approach? Let the  initial concentration of $\ce{AgCN}$ be $x$, so we write $$\ce{\underbrace{AgCN}_{initially x} <=> Ag+ + CN-}$$ and so concentration of $\ce{Ag+}$ and $\ce{CN-}$ would become $y$, and that of $\ce{AgCN}$ be $x-y$. Now, we write $$\ce{Ag+ + 2 CN- <=> [Ag(CN)_2]-}$$ And if now $z$ amound reacts in this reaction, we get the final concentrations of $\ce{Ag+}$ as $y-z$, of $\ce{CN-}$ as $y-2z$ and that of $\ce{[Ag(CN)_2]-}$ as $z$.

@OscarLanzi: So that we have from the $\ce{K_\text{sp}}$ equation, $(y-z)(y-2z)=10^{-16}$ and from the $\ce{K_f}$ equation, $\dfrac{z}{(y-z)(y-2z)^2}=1.5 \times 10^{17}$, but this would yield 2 equations and 2 variables and hence a unique value for the concentration of $\ce{CN-}$, but we want the minimum value of $y$... so what is wrong with this approach?

@Light Yagami Check this
https://www.desmos.com/calculator/7r8tvfytgw

@Govind What do you want to convey by that? There are two solutions as I said, but one would yield negative values hence discarded but the one with positive values is legible, right? See this. What I wanted to ask is, what is wrong with this approach as this yields an exact solution, whereas I should have gotten an algebraic expression in which I would have to minimize $y$. But here I am getting a value of $y$ from the equations itself!

Actually the question conveys a different thing. The CN- concentration can be varied by us so as to get the maximum solubility of AgCN.


text:
Determine the rate law for the given reaction
I was studying for a PhD entrance exam and this question made me wonder about some topics in kinetics:
Consider the multi-step reaction below:
$\ce{A_2 → 2 A}$
$\ce{2A + X → Y + B}$
$\ce{B → A2 + Z}$
Which one can't be determined from the given equations:
a) The general balanced equation
False, just by adding the reactions, we have $\ce{X -> Y + Z}$
b) The identity of the catalyst
False, A2 is the catalyst: consumed in the first step and reformed in the last
c) If an intermediate is formed
False, B is an intermediate
d) The rate law for this reaction
e) None of the above
It is not stated which step is the rate-determining step.
However, since $\ce{A_2}$ is a catalyst, its decomposition can't be the slow step (it doesn't make sense to have a catalyst that slows the reaction).
The second step is termolecular and forms an intermediate, which makes it a strong candidate for rate-determining step.
B is the decomposition of an intermediate, which should be fast considering that intermediates in reactions are usually unstable species that are quickly consumed.
My main question is: can we determine the rate law for the reaction with the given information?
Secondary questions, which could be added as separate questions, would be:

Can the decomposition of a catalyst be the slow step?
Is the formation of an intermediate always slower than its consumption? In other words, are there examples that don't follow the Steady State Approximation (considering simple chemical reactions only, not enzyme-catalyzed)?

I think the criteria for something to be a catalyst is that it reduces the speed of the overall reaction. So, in theory, it should be possible to have a catalyst where the decomposition of the catalyst is the slow step, as long as it is faster than the uncatalysed reaction. However, I can't think of any example at the moment. (A case which might be similar to this is the catalysis of ammonia synthesis by Fe, where the binding of nitrogen with catalyst is the slow step)

As for your second question, I think there can be reactions where the formation of intermediate is faster than its consumption (particularly in the case of low energy intermediates).

$\ce{A_{2}}$ is a catalyst if and only if the rate of the reaction is increased by using it. In other words, without it, $\ce{X->Y + Z}$ is slower.

"it doesn't make sense to have a catalyst that slows the reaction" That's true for the overall reaction. The problem is that the overall reaction is $\ce{X->Y + Z}$, so when you add $\ce{A2}$, as long as the overall reaction is faster, then $\ce{A2 -> 2A}$ can be the slow step.


text:
Does the enthalpy of solution formula (LE + Hyd) change depending on the question?

Hey guys. I have been working on enthalpies and literally every video and website I’ve seen,
the formula for calculating enthalpy of solution = lattice enthalpy of dissociation + enthalpy of hydration, using which, I got a few questions correct. However, this particular question (picture added), I have used the same formula and gotten the answer -5273 $\pu{kJ mol-1}$.
The mark scene for this question states that the answer is -51 $\pu{kJ mol-1}$.
I know what they have done, I know that they have used        -(-2662) because the answer sheet has the working. I don’t understand why they gave done this Why have they done so?????
Thank you!!

What is the difference between lattice formation and lattice disruption? Which of the two happens when you dissolve calcium fluoride in water? Also, any ways you could start with -2611 and -2662 and get -51?

@KarstenTheis I know what they have done, I know that they have used -(-2662) because the answer sheet has the working. I don’t understand why they gave done this..

@KarstenTheis explained it in his first two sentences! Please read them carefully!

@EdV we have only done lattice formation in school.. I’m an a level student

If you form calcium fluoride crystals from the constituent ions, energy is released. Going the opposite direction requires an input of energy. So think about the signed energy balance (enthalpy balance) and forget about the formula stuff.

@EdV Oh!!!!! Thank you so much!!!! I appreciate the help

Apparently your problem is related to the improper use of negative signs. Let's solve it differently.
Suppose you start from the separated ions $\ce{Ca^{2+}}$ and $\ce{2 F^-}$ in the vacuum. This state has the energy zero by definition.
Now suppose you join the ions to produce $1$ mole of solid $\ce{CaF2}$.  The process is exothermic, and the final state $\ce{CaF2(s)}$ is at the energy level $-2611 kJ/mol.$
Now suppose you are doing another transformation with the same ions. You put them directly into water, so that the ions become hydrated in solution. You obtain a solution of $\ce{CaF2}$ in water, or a solution of $\ce{Ca^{2+}(aq)}$ and $\ce{2 F^-(aq)}$. This transformation is also exothermic, and at the end $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ are at the energy level $- 2662 kJ/mol$.
The final levels after these two transformations are $-2611 kJ/mol$ for the final solid, and $-2662 kJ/mol$ for the solution. These energy levels are not very different from one another.
So the passage from the solid state $\ce{CaF2}$ to the solution $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ is slightly exothermic, with $\Delta H = -51 kJ/mol$


text:
Is soap forming micelles in water an exothermic reaction?
I have seen it described as soap molecules releasing energy when forming micelles. Is that correct? Is it an exothermic reaction?
I mean a bunch of molecules uniting to form a micelle.

Reserve "reaction" to process involving a change in chemical nature of the species involved (change of the primary structure, to extend the terminology used in protein and bio- chemistry to material science and supramolecular chemistry. If we want, micelles formation is an example).

This overlaps to other questions written by you. From those thread you should already knew it, isn't?

Endothermic process but governed by entropic changes of the water, see the Wikipedia page https://en.wikipedia.org/wiki/Micelle


text:
Can soap make an liposome around virsuses?
I have read that soap does not kill bacteria and viruses, but it rather removes them, that is, it strips them off the skin sort of speak by forming micelles around them, which are then rinsed off.
But isn't it more logical that soap forms liposomes for viruses instead of micelles to remove them.

@z1273 This question seems to focus on the chemical side so it is much better kept here.

Whether a mixture of phospholipids/lipids/detergents form micelles or liposome is not dependent on the hydrophobic "payload" but on the nature of the amphiphilic molecules.
Detergents form micelles (and bubbles and antibubbles, when there is air around). The micelles are fairly small (radius similar to the length of the detergent), and they don't have space for an entire virus (diameter on the order of 100 nm) or bacteria (diameter on the order of 1000 nm). Detergents that form micelles are able, however, to solubilize membrane proteins (diameter on the order of 10 nm) by forming a band of detergents around the hydrophobic parts of membrane proteins.
Phospholipids mixed with cholesterol form lipid bilayers and liposomes (the latter by extrusion or sonification). The distinct behavior has to do with differently charged head groups and the number of hydrophobic tails.
Some bacteria are killed by detergents because they break down the cell membrane, integrating the phospholipids and cholesteral into the micelle. Viruses can't be killed because they are not alive. If they are surrounded by a lipid bilayer, the detergent can also break this down, rendering the genetic material inside harmless.
No matter whether bacteria and viruses are destroyed by the soap or not, the soap allows the water to wash the germs away, and they end up in the canalization (e.g. coronavirus genetic material in Boston) instead of in people's bodies.

May I ask you how exactly does a soap inactivate a virus? The lipid bilayer that surrounds viruses is polar from the outside. How does the soap get into the intermediate zone to fix to the lipids. And if and when it does attatch to the lipids how does it pry the bilipid layer? Does the water outside cause a tractional pull on the soap and that results in the lipids detaching?

@Mhkhm That is an intriguing follow-up question, ready to be posted. I would argue it is more interesting and of higher quality than the original question, but of course that is for the community to decide.


text:
database for catalytic reaction mechanism
I am looking for a database containing detailed reaction mechanisms for (homogenous) catalytic reactions. Regular organic textbooks or papers generally draw some broad picture of how the mechanism may work, but generally, they skip many steps.
Is there any collection of detailed mechanisms?
I am primarily interested in catalytic reactions.
For clarification:

I am primarily looking for homogenous, metal-catalyzed, non-enzymatic reactions of small molecules.
Searching so far in places like Reaxys or SciFinder did not really result in much more than the bibliography.

Have you explored Reaxys or SciFinder?

To answer this question would require pages of text, because there are several types of catalysts. Some are homogeneous, some are heterogeneous. It may be a surface effect, or a cavity effect, or a geometric effect with enzymes, A whole lecture would be needed to answer your question. Sorry ! We cannot develop this chapter here.

@Maurice It is funny because I know extensive and legitimate databases for synthetic methods, crystallographic structures, proteins structures, small molecule biological activities, etc in spite of the fact that most of those topics "need a whole lecture".


text:
How do I tell the difference between an acid, base, and a salt solution?
I'm learning about acid-base equilibria. I know that a (Bronsted-Lowry) acid donates a hydrogen proton, and a base receives a hydrogen proton, but I've also heard about a Lewis acid base theory, where electrons are the key difference. What's the point of making it even more general?And I'm not sure exactly what an salt is. I know it contains a cation and anion, but don't acids and bases have that as well?

Further edit has made a chameleon transition to ( rollback applied ):
What characteristics make a salt acidic or basic?
Let's use the example $\ce{NaC_2H_3O_2}$ (sodium acetate).
It disassociates into $\ce{Na+}$ and $\ce{C_2H_3O_2-}$. Why is the sodium ion not react with water, but the acetate ion does?
How can you tell if a anion and cation of a salt will react in water (if placed in water)?

KHPO4 is all of them. An acid, a base and a salt.

H2O is an acid and a base, too.

@Poutnik so it's dependent on what it's reacting with? for ones that have more than one characteristics?

Sorry, should be KH2PO4 ( or K2HPO4 or KHSO4). The above applies to all of them.

What you just did is called a "chameleon question". After it was answered, it was edited to become very different, or follow up question, more or less destroying the original content and relation to the answers . This can be considered rude to those who put effort into answering and should be avoided. The better, recommended and expected way is to post a new question, referring eventually to the original one as the follow up question, if it applies .

@Poutnik sorry I didn't know.

Rather, all depends on what aspects or criteria you evaluate.

If the compound fits criteria for salts, it is a salt.
If the compound fits criteria for acids, it is an acid.
If the compound fits criteria for bases, it is a base.

And the main point is - these categories are not mutually exclusive.
Additionally, there are 2 meanings for an acid/base.

Fitting criteria of Broensted-Lawry theory, resp. Lewis theory.
A practical meaning: ability of forming ( strongly ) acidic or alkaline solutions.

Is it acidic ? It is an acid.
Is it alkaline ? It is a base.
Is it about neutral and conducting electricity well ? It is a salt.

A compound can be acid(1),base(1) or salt, while being base(2), as it is a stronger base(1) than an acid(1). And vice versa.

Addressing feedback:
E.g. classical soap is a salt AND a base. So is e.g. sodium  carbonate.
Like quantum objects manifest wave-like or particle-like properties, depending on scenario,  compounds manifest scenario dependent salt-like or base/acid-like properties.
Solid $\ce{Na2CO3}$ is more like a salt,  being ionic compound, while its solution is more like a base solution, with carbonate anion capturing hydronium ions and causing alkaline $\mathrm{pH}$.
$\ce{NaOH}$ as the typical strong base can be in its solid form formally considered as a sodium salt of the very weak acid $\ce{H2O}$ , what makes its anion $\ce{OH-}$ , when $\ce{NaOH}$ is dissolved, a very strong Broensted-Lawry conjugate base.

In the bases entry, why does Wikipedia show a picture of a soap stating that "Soaps are weak bases formed by the reaction of fatty acids with sodium hydroxide or potassium hydroxide." Soaps are weakly alkaline but they are salts not bases. What do you think?

See the A update.

@Wise sequoia well a salt is when an acid and a base react .For example when sodium (the base) and sulfuric acid (the acid) react it forms sodium sulfate(the salt) and hydrogen gas.


text:
Why do nitrogen-containing organic compounds give N2 as the combustion product?
My teacher gave the explanation stating that consider that first the nitrogen in lower state is oxidised to $\ce{N2}$ and then the further oxidation would stop as the $\ce{N-N}$ bond energy of $\ce{N2}$ is very high and can't be overcome under normal conditions.
But my doubt is can't the nascent nitrogen (formed during the formation of $\ce{N2})$ react further with $\ce{O2}$ and get oxidised to some higher oxides like $\ce{NO2}$ or $\ce{N2O}?$
The Latimer diagram shows that it is possible for oxygen to oxidise the nitrogen to higher states:

I don't think they do. If you search "combustion products of triethylamine" for example you will find phrases such as "Combustion products include: carbon dioxide (CO2), nitrogen oxides (NOx), other pyrolysis products typical of burning organic material."

Additionally, NOx is created during gas engine combustion from aerial N2, as equilibrium N2 + O2 <=> 2 NO is not so strongly shifted to the left at high temperatures.

Side note: "nascent nitrogen" is an obsolete term used to explain high reactivity in certain cases long time ago. These days, it's more appropriate to call it by what it really is: atomic nitrogen.

What else could it be? All nitrogen oxides are endothermic. I mean, yeah, they would form to some tiny extent, but not as the major product.

@govind Note that atomic nitrogen would help with kinetics. But it must fight the thermodynamics in the first place. Molecular nitrogen is thermodynamically highly preferred, forming nitrogen oxides just as minor compounds.

@Poutnik The electrode potential for conversion of NH4+ to N2O4 and NO to N2O4 according to the Latimer diagram are -0.89V and -1.033V. Also the conversion of NO to NO2 is spontaneous in atmospheric oxygen. so thermodynamics might not be the appropriate reason for molecular nitrogen being the major product.

is there something special for this rxn https://www.sciencedirect.com/science/article/abs/pii/0021961483901271

What is important is the standard electrode potential, or the equivalent standard reaction Gibbs energy for conversions of particular nitrogen species to $\ce{N2}$. Like jumping from a tree branch to another branch can produce energy, but molecular nitrogen is on the ground.

@andselisk Converted my (deleted) comment into a question: Why is the term “nascent nitrogen” still used in metallurgical industries?

@Govind not only do they give elemental nitrogen, they do give all sorts of nitrogen oxides. See these papers for more info.: https://pubs.rsc.org/En/content/articlelanding/1947/an/an9477200142/unauth... https://www.sciencedirect.com/science/article/abs/pii/S0926860X02000133 ... https://www.sciencedirect.com/science/article/abs/pii/0026265X66900531


text:
Chemical compound scaffolds name
in RDKit, there is this nice function to decompose a chemical compound into its scaffolds.
Is there a way to automatically name the scaffolds?
I usually search names in PubChem but this may not be practical for a large dataaset.

Look at your answer here https://chemistry.stackexchange.com/questions/98055/search-by-cas-registry-name-to-obtain-trackable-accession-ie-inchi/134993#134993 and use iupac_name as output.

@rapelpy thank you import cirpy   print(cirpy.resolve('O=C1c2ccccc2OCC1Cc1ccccc1', 'iupac_name'))   print(cirpy.resolve('O=C1c2ccccc2OCC1Cc1ccccc1', 'names'))  seem to work


text:
How much is 1 mμg?
So I stumbled on a research paper by Okuda and Takedatsu [1] that uses mμg as a symbol for the mass unit.
I couldn't find anything about this measure unit anywhere except a recommendation not to use it.
So how much is 1 mμg?
I suppose it means milli-micro-grams, so it should be 1 μg/1000, but it's just a guess.
Reference
Okuda, K.; Takedatsu, H. Absorption of Vitamin B12 in a Rectal Suppository. Experimental Biology and Medicine 1966, 123 (2), 504–506. DOI: 10.3181/00379727-123-31527.

It looks like the authors wanted to use a consistent unit, rather than say the doses were 10 nanograms and 2 milligrams, they went for micrograms for both.

Is there any chance this is subtle humour?  The unit could be read as "mug" when stating it as nanograms, or μmg (micro-milli-grams) removes that possible confusion.

This could happen in new publications as well when you use your graphing software incorrectly. You tell the software to graph some value, specify the unit as µg, and when the software gets an input value < 1.0, it might multiply by 1000 and prepend a m to the unit, thus converting 0.67 µg to 670 mµg.

Less than a mug.

Looks like in 1966 there were still residues of double prefix notation in use.
Here's an entry from Russ Rowlett's compilation of units on ibiblio.org (Rowlett):

millimicro- (mμ-)
an obsolete metric prefix denoting 10-9 or one billionth. This prefix has been replaced by nano- (n-).

Here's what is noted on wikipedia:

Double prefixes
Double prefixes have been used in the past, such as micromillimetres or millimicrons (now nanometres), micromicrofarads (μμF; now picofarads, pF), kilomegatons (now gigatons), hectokilometres (now 100 kilometres) and the derived adjective hectokilometric (typically used for qualifying the fuel consumption measures). These are not compatible with the SI.

Reference
Rowlett, Russ. "millimicro-". How Many? A Dictionary of Units of Measurement. University of North Carolina at Chapel Hill. http://www.ibiblio.org/units/dictM.html

I just wonder, why it was used mcg instead of self-offering mmg for $\mu$g ?

@Poutnik Not sure about mmg. I also noticed that unlike the Roman numeral system that make use of both addition and subtraction of values, the historical use of double prefixes appears to go in one direction only, either to greater mounts or to smaller amounts. These are my two megamicro-cents

@Poutnik : Once we're in the land of double prefixes, 1 mcg would be 1 milicentigram, equivalent to 10 micrograms.  There is a not-universal standard use of "mcg" in parts of the healthcare industry, putatively to minimize misdosing due to handwritten conflation of cursive "m" and mu.  Regarding "mmg", the micro- prefix was in widespread use prior to the SI, with published usage back to the 1870s.

@Poutnik From Wikipedia: Microgram: "the recommended symbol in the United States and United Kingdom when communicating medical information is mcg." Perhaps "mc" is closer to "micro", which only has one letter "m".

This should be subtitled "Fun Ways To Abuse Units". I particularly like "hectokilometers". Maybe I'll use this to guilt my daughter when driving her back to college at the end of the week. "Do you have any idea of how far we have to take you? My heavens - it's almost FIVE HECTOKILOMETERS!!!". Fortunately, after 19 years she's used to me... :-)

@AndrewMorton Your point is well taken. The FDA noted as well that mcg is intended as microgram. I learned mcg to stand for millicentigram. Given the discussion here, I'm not so sure anymore.

@BobJarvis-ReinstateMonica: Wow. Forth and back, it's a megameter, a.k.a a femtozettameter, or just $\mathrm{fZm}$.

@Poutnik  I always thought the reason the abbreviation stuck is that "mcg" corresponds directly to the way it's pronounced, which strongly reinforces the connection.  Amusingly, I just checked my Mac's dictionary, and "mcg" is given prominently as the abbreviation, with $\mu g$ in much smaller type below.

authors: Bob Jarvis - Слава Україні


text:
Why is ephedrine optically active?
Acyclic amine in which three groups are different with a lone pair of electron is optically inactive and nitrogen is achiral because of the rapid pyramidal inversion.

Does similar inversion happen in ephedrine?
If yes, doesn't the inversion cancel the optical activity of the chiral carbon. So why then is ephedrine optically active?

Nitrogen is not the only chiral center here.

@Ivan Doesn't the inversion cancel the optical activity of the chiral carbon?

No, why? Inversion happens on nitrogen, not on carbon.

@Ivan One group attached to the nitrogen is bit bulky so does the inversion happen and is the nitrogen chiral?

I don't think inversion takes place here. But in any case, why inversion should affect or remove the chirality of near C?

@Alchimista Is the nitrogen chiral?

There are two chiral carbon centres in ephedrine, neither are affected by nitrogen inversion

@AsherAbraham I am not sure but you can easily search for this specific question. Perhaps no as for inversion is at work. But again there are two other stereo centers...

Why are you trying to cancel on optical rotation from things that are not completely identical?!

@Alchimista Nitrogen inversion most certainly happens here as there is nothing fixing the nitrogen atom into one defined configuration (including not in a planar structure). That’s obviously irrelevant for the question of the molecule’s overall chirality.

@Jan I know is irrelevant. I was just not sure about inversion, thinking of the bulkiness of one substituent. Nice to know.

@Alchimista I mean, nitrogen inversion is best thought of as the nitrogen swinging through the plane of the other atoms. They have minimal configurational changes, nitrogen moves far more ;)

Nitrogen will always perform nitrogen inversion if it is not configurationally fixed due to steric or electronic reasons. Fixing the configuration sterically could be achieved, for example, by a polycyclic compound as in 1 below. Fixing the configuration electronically is exemplified by the amide 2 (note that the amide is achiral; the nitrogen is forced into a planar configuration by amide resonance as shown).

Figure 1: examples of nitrogen-containing compounds where nitrogen inversion is not possible. 1 is chiral, 2 is not.
However, nitrogen inversion only concerns nitrogen atoms and prevents most nitrogens in organic compounds from being asymmetric even if they are attached to three different residues (including a lone pair). Ephedrine contains two additional asymmetric atoms both of which are carbon atoms. These carbons cannot perform an inversion similar to nitrogen inversion as they are covalently bound to four different residues. As there is no plane or centre of symmetry transforming one of these asymmetric carbon atoms onto the other, this makes the molecule as a whole chiral.
The effect of the asymmetric carbon atoms can be seen in the number of possible stereoisomers: both carbons can be either (R) or (S) configured giving 4 possible isomers ((+)-ephedrine, (–)-ephedrine, (+)-pseudoephedrine and (–)-pseudoephedrine). If nitrogen inversion were not a thing, we would expect 8 stereoisomers as then each of these 4 could occur with an (S) or an (R) configured nitrogen. However, only those four are known.
For an abundance of clarity: nitrogen inversion only affects nitrogen atoms and only turns an (R)-nitrogen centre into an (S)-nitrogen centre; non-nitrogen centres are unaffected.


text:
Is the pz orbital wave function three- or four-dimensional?
The wave function of the $2\mathrm{p}_z$ orbital is
$$Ψ = \frac{1}{4\sqrt{2π}}\left(\frac Z a\right)^{5/2} r \mathrm e^{-Zr/a}\cos θ.$$
I'm confused if this function will be a three-dimensional function or a four-dimensional function?
My professor told me that
$$z/r = \cos θ,$$
where $θ$ is the angle made by $r$ with the $z$ axis. So, I guess it will be four-dimensional since we would have to then plug in
$$r = \sqrt{x^2 + y^2 + z^2}$$
in the equation.
But when he plotted the function, he plotted a three-dimensional one! What is the correct answer and why?

The wave function is a function of $x, y$ and $z$. That is all. Why would you create a fourth dimension ?

The expression is in spherical coordinates, the coordinates as (r,theta,phi) - https://en.wikipedia.org/wiki/Spherical_coordinate_system .

If you plug in particular values of x,y,z and get a value of the function,which axis will you ploit it on?

You have to start with a simpler question: is $\sin x$ a one-dimensional or two-dimensional function?

two dimensional

I think we have found the misconception. Typically, when we refer to the dimension of a function, we are describing how many input variables it depends upon. In the case of $\sin(x)$, this is just 1. To plot the value of the function, we need one additional dimension, but that's not the same as saying $\sin(x)$ is two dimensional. $\sin(x)$ has a value for every point in one dimensional space. The pz orbital has a value for every point in 3 dimensional space.

A three dimensional function(assuming the domain is $R^{2}$ )will have a value for every point in 2 dimensional space. Similarly,a 4 dimensional function will have a value for every point in 3 dimensional space.So,shouldn't the orbital function be a 4 dimensional one?

That's not the way we use these words. A function has 3 variables, hence it is 3-dimensional. A graph of it is 4D, but that's another story.

Rather than debate semantics- the short answer you are looking for is that this function only needs two input values to be determined: r and theta. That is because the value of the function for each (r, theta) pair is constant with respect to the third variable phi in the spherical coordinate system

The wavefunction covers 3-dimensional space. Formally, it maps each point in space to a complex number. That is, it is a function that takes a point in 3-dimensional space as input and returns a complex number as output: $\Psi:\mathbb{R}^{3}\rightarrow \mathbb{C}$.
There are many ways to represent 3-dimensional space: cartesian coordinates $(x,y,z)$, spherical coordinates $(r,\theta,\phi)$, cylindrical coordinates $(z, r, \theta)$, to name the most common. We have these different ways because functions might look "nice" in one coordinate system and not nice in another.
As it turns out, the wave function for the $p_{z}$ orbital looks "nice" in spherical coordinates, that is, $\Psi(r, \theta, \phi)$. It looks less nice in cartesian coordinates, but you can do the replacement for $\theta$ and $\phi$ as noted in the question. However, people do not usually mix coordinates systems together because they are redundant. For example, no one writes $r\sqrt{x^{2}+y^{2}}$. They write either $r^{2}$ or $x^{2}+y^{2}$.
To conclude, the $p_{z}$ orbital you have written is a function of 3 variables: $r, \theta, \phi$. It just happens not to depend on $\phi$, so it looks like just a function of $r$ and $\theta$. Forget this concept of dimensionality because it's not doing you any favors.

you have written $r$ as the square root of $x^2+y^2$. Shouldn't it be the square root of $x^2+y^2+z^2$

Only in spherical coordinates. In cylindrical coordinates, it's just a function of x and y. Honestly, though, the point is don't mix coordinates. If you're hung up on what the definition of r is, then you've missed the point.


text:
How to draw "box"/square arrows in Chemdraw Prime 16.0?
I want to draw arrows like the one in the figure shown (with reagent a beside it) in ChemDraw 16.0
What is the best way to go about this?
I couldn't find anything like it in the arrow options in ChemDraw Prime 16.0

Go into the icon that has circles and lines. Make a horizontal and vertical line segment. Make a right arrow. Put the three together with the lasso. Finally, group them so that it is reusable.

One approach:

draw the upper horizontal, then the vertical line to form the angle, e.g. using a single bond each time in anticlockwise direction.
draw an reaction arrow «as usual» horizontally, pointing from the left to the right.  Depending on the style template, increasing the zoom may help to adjust its position relative to this angle when moving one of the two objects until the back of the arrow and the lower tip of the angle overlap.
join arrow with the vertical of the angle constructed earlier into one object
adjust size and position of this construct relative to the three text lines

While likely not designed for this purpose, it is possible to many molecular editors (not only ChemDraw) for a quick doodle / map about the buildings near the Chem department, or labels for the drawer because you can vertically / horizontally align shapes & objects quickly and export this in either vector, or bitmap format.
Be aware of one caveat: when structure cleaning, mark the true, meaningful structures, and not the whole page.  Or, instead of tearing a propane into said angle, use the line tool instead.  To some limited extent, this approach is working with the appetizer versions of ChemDraw, and ChemDoodle, too.


text:
Roasting of galena
In the metallurgy of $\ce{Pb}$, Galena is first concentrated by froth floatation process. It is then partially roasted to and self reduced to obtain $\ce{Pb}$.

Why is galena only partially roasted(that is in limited oxygen supply)?

My initial thought was that sulphur could further react with $\ce{O2}$ to form $\ce{SO2}$, but in the reaction:
$\ce{3PbS + 5O2(limited)-> 2PbO + PbSO4 + 2SO2\uparrow}$
we get $\ce{PbSO4}$, which I dont think reacts with $\ce{O2}$. Whats the problem here then?

Your equation is the sum of two different and independent equations, namely $$\ce{2 PbS + 3 O2 -> 2 PbO + 2 SO2}$$ $$\ce{PbS + 2 O2 -> PbSO4}$$ And these equations are no use for producing metallic lead. They are not a moderate oxydation, as the ratio $\ce{O2/PbS = 5/3}$. Producing metallic lead by moderate oxydation requires a reaction based on the equation $$\ce{PbS + O2 -> Pb + SO2}$$ And here the ratio $\ce{O2/Pb = 1 < 5/3}$

I'd take ratio of $\ce{2PbO :: 1PbSO4 }$ with a grain of salt.

@MaxW. You are right. The ratio $\ce{PbO:PbSO4}$ is arbitrary. I don't know why Newbie has chosen $2:1$. There is no reason.


text:
Volume of ideal gas molecule
Is there any reason as to why we assume the volume of ideal gas to be negligible in comparison to the container? What is the underlying reason to assume that the ideal gas molecules are point molecules?
Why can we not assume them to be non-point particles?

If you assume point particles and no inter-particle forces, as simplifying approximations, then you have an easy path to this useful approximation: pV = nRT, i.e., the ideal gas law. Of course, you can assume non-negligible particle sizes and assume that inter-particle forces exist, so then the equation of state will necessarily be more complicated.

If it weren't negligible, we would expect different behavior of molecules of distinct size. However, the ideal gas law is not dependent on particle size. You might try to derive a gas law for uniform volume particles rather than zero-volume particles but then you would expect that chosen uniform volume to somehow appear in your alternative gas law (for example, the volume of the gas could not be smaller than the total volume of particles). In the ideal gas law, the volume of the gas approaches zero as the temperature approaches zero (with fixed n and p).

The experiment shows that boiling water at 100°C produces a vapor which is about $1000$ times more cumbersome that the liquid water from  which it is produced. As in this process the molecules have not been chemically modified, it means that in the vapor, the volume occupied by the molecules is negligible, as if they were point particles.

I'd use voluminous  instead of cumbersome, but Maurice makes a good point. Chemistry typically uses 3 or 4 significant figures at best. The equation PV=nRT isn't going to yield 15 digits of precision. // Van der Waals equation makes some rudimentary corrections to the ideal gas equation.

Saving the best for last [ ;-) ], see equation of state in wikipedia! Loads of increasingly complicated equations of state! Imagine being a beginning chemistry student and having all that dropped on you! Hence the practical reality that beginning chemistry courses are heavily simplified (and often seriously over-simplified) so as to be manageable and so that there are survivors!

Not really what you ask but note that this, while being an approximation, points to the a very basic fact. A P "component" unrelated to force and occupancy of space but intrinsic to the existence of particles.

@EdV - Better than the equation of state link is the link to the Wikipedia article for real gas.

@MaxW Thanks, I stand corrected! ;-) Glad I never went beyond van der Waals when I taught general chemistry!

Because it gives simpler-to-derive laws which are often very good approximations
Clearly real gases do not always follow the ideal gas laws. They mostly liquefy under some conditions, for example, and, under those conditions they are clearly not ideal.
But in practice gas laws are used for things far away from those non-ideal regions. When we are applying the laws we are usually applying them to gases where the deviations from ideality are small. At normal lab conditions (1 atm pressure, room temperature) most of the gases you will ever deal with will be as close to ideal as matters within measurement error. Given that you want the simplest law possible and that law is easily derived by making the assumptions that particles have no size and don't stick together.
More complicated gas laws can and have been developed (van der Waals equation was a 19th century development of a simple extended law that captures some features about liquefaction and non-ideality under cold or high pressure conditions). But the deviations of the results from the simplest ideal gas equation are small under normal conditions (van der Waals calculations for carbon dioxide differ from ideal results by 0.5% under normal lab conditions). This is such a small amount that more complex equations are not worth bothering with unless you know you are using very low temperatures or high pressures.
And, when you are in one of those regions, you will find that the gas equations get very complicated very quickly. It takes a lot of extra empirical data and mathematical complexity to handle those cases and under normal conditions this just isn't worth the effort. Why waste effort when you don't gain any practical benefit?


text:
What is "silanol activity"?
In a previous post about how pyridine failed to move much from the baseline of a TLC, I was told to look into "silanol activity". However, I was unable to find a textbook-style description. All the articles I've seen where silanol activity was discussed assumed it was a known prerequisite.
Could anyone explain silanol activity, or direct me to an online explanation please?
Furthermore, if it is referring to an acid–base reaction, would that imply that what I'm seeing on the TLC is no longer pyridine, but rather $\ce{C5H5NH+}$?

It is the reactivity H+ and SiO- in Si-O-H surface.

Silanol activity is perhaps an informal term which is quite frequently used in the chromatographic literature to explain the poor band shape of primary amines. Other amines show this too but to a lesser extent. That is why you will not find a formal textbook definition of silanol activity. One might also see the term silanophilic interactions used occasionally. If you look at the surface of silica gel, you will see that a silanol group, is like an alcohol analogue, except that silanol is slightly a stronger acid.
If we disperse silica particles in water, and let it stay there for sometime, the pH of water is < 7 (around 3–4, if I remember correctly). So the point to keep in mind is that silica is acidic due to its ionizable silanols.
There are different types of silanols as drawn in this paper: Sunseri, J.; Cooper, W. T.; Dorsey, J. G. Reducing residual silanol interactions in reversed-phase liquid chromatography: thermal treatment of silica before derivatization. J. Chromatogr. A 2003, 1011 (1-2), 23–29. DOI: 10.1016/S0021-9673(03)01070-7.

Now if you are doing chromatography of relatively strong organic bases, the silanol may be deprotonated by the amine (acid–base). The locally ionized silanol Si–O− groups can now act as a cation exchanger, which is a pretty strong interaction. The reason for seeing the streaking or tailing, as it is called in the formal literature, arises from those dual modes of retention. One mode could be regular partitioning or adsorption and the other
one is this additional cation-exchange behavior.
There are dozens of chromatographic tests for assessing the silanol activity of a given silica gel. Not all silica gels were created equal!
As your question

would that imply that what I'm seeing on the TLC is no longer
pyridine, but rather $\ce{C5H5NH+}$

Remember that ionization is an equilibrium process and hence dynamic. You are seeing the combined effect of both, pyridine and protonated pyridine.


text:
Do side chain carboxylic acids need to be protected in peptide synthesis?
Consider the synthesis of the dipeptide Glu–Ala (image). The amino acid Glu (glutamic acid)  has two carboxyl groups, where one of them should reacts with the amine group from the other amino acid. Does the side chain carboxyl group in Glu have to be protected also, or do we just assume that it does not react?

You should protect it.

If it is in aqueous solution, no reaction would occur (neither the one desired one nor the two undesired ones).

Murphy's Law applies. If it can possibly react it will

Well our profesor said that we need to draw the mescanism and just assume it reacts. But thank you for your answers:)

related: https://chemistry.stackexchange.com/q/125224/102629

Yes, the side chain carboxylic acid requires protection.
The two most commonly used peptide coupling strategies use Boc or Fmoc on the amide.
The side-chain carboxylate is typically protected as a tBu ester in case of Fmoc synthesis (the compound is commercially available under CAS 71989-18-9) or as a benzyl ester in case of Boc synthesis (as before, CAS 13574-13-5).
Not protecting these side chains would give coupling mixtures as both carboxylic acids can and will react with the coupling reagents.


text:
Electrochemistry -gold recovery
I was wondering how to answer  a question I got from my students about the possible recovery of gold from solution. We just finished the reaction between silver ions and copper metal and they wondered why copper wouldn't displace gold from solution in an electrochemical reaction. The redox potentials suggest that it can be done but there is no documentation of this reaction anywhere. I can't try the reaction because we have no gold compounds. If it could be done you would think that gold recovery but companies would use the technique since copper is a lot cheaper.
Please advise.

This all depends on what you have as the gold-containing solution. If you just have an aqueous gold chloride solution, adding zinc powder will work, and zinc is cheap compared to copper.

Read about overpotential.

Short Answer: Yes, based on the electrochemical/reactivity series, it is possible to percipitate gold from solution using metals higher on the series like copper or zinc. However, it isn't done often because gold is highly unreactive, and when gold salts are formed, they are formed deliberately under atypical conditions like very high temperatures or acidities.
It would be possible to displace gold from it's ionic salt (gold (iii) chloride) using zinc or copper. This is because, as you mentioned copper is higher on the activity series than gold.
The likely reason to why companies don't use single-displacement reactions to retreive gold from solution is be cause gold salts rarely exist. Gold is extremely unreactive (see here) Gold (III) chloride is normally synthesized by reacting gold metal with chlorine gas at 180 degrees Celsius. This temperature is higher than anywhere you'd find on Earth's surface, and chlorine gas is too reactive to exist in that state in the atmosphere. These conditions are typically found only in the lab and not out in nature. Another example of a man-made gold salt is aqua regia (a 3:1 ratio of hydrochloric acid and nitric acid) which can dissolve gold. To recover the gold, ions are displaced using ferrous sulfphate reducing agent (source).
If you are interested in gold recovery, here is an excellent article on a variety of methods.


text:
Aspirin Hydrolysis at pH 2
Conducting an experiment, I found that very little salicylic acid was evolved performing hydrolysis at $\mathrm{pH \ 2}$ and at $\pu{45 ^\circ C}$. In fact, far more salicylic acid was evolved at $\mathrm{pH \ 7}$. This comes as a surprise to me because I expected an acid-catalyzed ester hydrolysis to actually quicken the rate of reaction and thereby evolve MORE salicylic acid.
Is aspirin generally more stable at lower $\mathrm{pH}$s then?
UPDATE 1: This graph shows that aspirin hydrolysis doesn’t lend itself well in low $\mathrm{pH}$ conditions, specifically at $\mathrm{pH \ 2}$.
UPDATE 2: No answer has been found — a reword to this question can perhaps be “Why is aspirin less susceptible to hydrolysis at $\mathrm{pH \ 2}$ when theoretically, lower $\mathrm{pH}$s are advantageous for ester hydrolysis?”

Source: https://compoundingtoday.com/Newsletter/Science_and_Tech_1112.cfm
What seems to be the explanation to this?

Let us continue this discussion in chat.

Found something that might help you. See this. No paywalls. And this too.

To understand the pH rate profile of the Aspirin hydrolysis, we have to look at all the possible mechanisms through which the hydrolysis may take place.
A good rule of thumb for interpreting a pH rate profile is: going from left to right, when the line curves upward (i.e. the rate increases suddenly), a new mechanism is dominating the reaction. When the line curves downward (i.e. the rate drops suddenly), that generally indicates that there is a change in ionisation/dissociation or that the rate-limiting step has changed for the same mechanism.
If you look at the graph for Aspirin, you will see that the graph bends up two times, around pH 2.5 and pH 8.5. So, there must be 3 different mechanisms throughout the whole pH range. (There is also a downward bend at pH ~5, that's not important at the moment).
So you can divide the graph into 3 segments (separated by blue line):

We need 3 mechanisms—(A), (B) and (C). Let's deal with them one at a time.

(A)

At pH < ~2.5 i.e. strongly acidic enviornment, the main components in the solution are the protonated form of aspirin (pKa of aspirin ~ 3.5), and $\ce{H3O+}$. A mechanism can be drawn:

This is the usual acidic ester hydrolysis mechanism. I haven't drawn the last few steps for brevity.
The rate of the reaction is: $$\ce{r=k_1[aspirin][H3O+]}$$ $$\mathrm{usually\,written\, as,}\;\; \ce{r=k_{obs}[aspirin]}$$
Which means, $\ce{k_{obs}=k_1[H3O+]} \Rightarrow \ce{log(k_{obs})=log(k_1) +log([H3O+])} \Rightarrow \ce{log(k_{obs})=log(k_1) - pH}$. You can immediately see why the graph decreases linearly as pH increases.

(B)

When pH > ~2.5, the solution contains both the protonated and the deprotonated form of aspirin, along with a lower concentration of $\ce{H3O+}$ as the solution is not strongly acidic anymore. As the rate of the reaction has increases suddenly, that must mean there is another mechanism in effect now, which overshadows the previous acid catalysed mechanism.
This part is difficult, because multiple mechanisms can be drawn, and it is difficult to distinguish which one is actually happening. Here the graph bends down once, and that happens around pH 4, close to the pKa of aspirin(3.5) which should clue you in to the fact that the mechanism here is somehow affected by the deprotonation of aspirin.
The rate rises at first, until pH 4 and then it stays almost constant until pH 8.5 (where (C) starts). This indicates that the mechanism is not catalysed by acid or base (because then the graph can't stay flat against changing pH), and that the mechanism involves the deprotonated form of aspirin (because once all the aspirin is deprotonated, the graph becomes flat).
One possible mechanism:

Note that we can't draw a mechanism where water attacks the ester directly, because water is not a good enough nucleophile, and also because the anion has to participate in the reaction somehow, as I said earlier.
Another possible(*) mechanism would be direct attack of the carboxyl anion on the ester:

Here an anhydride intermediate is formed, which is then hydrolysed as usual, and I have not drawn that part of the mechanism.
(*) I remember hearing in a lecture, that this mechanism via anhydride has been proven to be wrong, based on isotope substitution studies. I can't find a reference right now, so I will leave the mechanism in the answer, but I am pretty sure the first mechanism is the correct one.
Here, $\ce{[aspirin-]}$ and $\ce{[aspirin-H]}$ represents the deprotonated and protonated form of aspirin, respectively. The rate of the reaction is: $$\ce{r=k2[aspirin-]}$$ But the way the experimental measurements are taken means that the rate is expressed in terms of the total aspirin concentration:
$$\ce{r=k_{obs}[aspirin]}$$
From the acid dissociation equilibrium, we can write:$$\ce{[aspirin]=[aspirin-H] +[aspirin-]=\frac{[aspirin-][H3O+]}{K_a} +[aspirin-]=[aspirin-]\left( 1 +\frac{[H3O+]}{K_a}\right)}$$
We can substitute into the first equation: $$\ce{r=k2[aspirin-]= \frac{k2}{\left( 1 +\frac{[H3O+]}{K_a}\right)}[aspirin]}$$
Which gives:
$$\ce{k_{obs}=\frac{k2}{\left( 1 +\frac{[H3O+]}{K_a}\right)}}$$
In the region pH ~2.5 to 4, $\mathrm{k_{obs}}$ increases as $\ce{[H3O+]}$ decreases. After pH ~4, $\ce{[H3O+]<<K_a}$ so, $\mathrm{k_{obs}\approx k_2}$ and the graph becomes almost flat.

(C)

After pH 8.5, the main components of the solution are the deprotonated form of aspirin and $\ce{OH-}$. So, the base-catalysed mechanism becomes dominant.

As all the aspirin is in deprotonated form, we can write
$$\ce{r=k_3[aspirin][OH-]}$$
Which means, $\ce{k_{obs}=k_3[OH-]} \Rightarrow \ce{log(k_{obs})=log(k_3) - 14 + pH}$
So the graph rises linearly with increasing pH.
[P.S. The solvent does not appear in the rate equations.]

Beautiful description of the mechanisms! Thank you.


text:
Why are these two methods of converting from ppm to g/m3 equivalent?
I've found two methods to convert from $ppm$ to $g/m^3$. They both appear to give equivalent results, but I don't understand why. Can anyone join the dots for me and explain why they are equivalent methods, or show a derivation for Method 1? I've found a derivation for Method 2.
Method 1 - Definition
$$C[g/m^3] = \frac{C[ppm] \cdot \rho}{1000}$$
Where $C[ppm]$ and $C[g/m^3]$ represent gas concentration in units of $ppm$ and $g/m^3$ respectively and $\rho$ represents gas density.
Method 2 - Definition (As derived here)
$$C[g/m^3] = \frac{C[ppm] \cdot M \cdot P}{R \cdot T \cdot 10^6}$$
Where $M$, $P$, $R$ and $T$ represent molar mass, pressure, ideal gas constant and temperature respectively.
Method 1 - Example - Carbon Monoxide (CO)
Assuming $C[ppm]$ as $100ppm$ and CO density as $1.15 kg/m^3$ at $20^{\circ}C$ and $100000Pa$
$$C[g/m^3] = \frac{C[ppm] \cdot \rho}{1000}$$
$$C[g/m^3] = \frac{100 \cdot 1.15}{1000}$$
$$C[g/m^3] = 0.115$$
Method 2 - Example - Carbon Monoxide (CO)
Assuming $C[ppm]$ as $100ppm$ at $20^{\circ}C$ and $100000Pa$.
$$C[g/m^3] = \frac{C[ppm] \cdot M \cdot P}{R \cdot T \cdot 10^6}$$
$$C[g/m^3] = \frac{100 \cdot 28 \cdot 100000}{8.3145 \cdot (20+273.15) \cdot 10^6}$$
$$C[g/m^3] = 0.115$$

different but related in Earth Science SE: ppmV versus μmol/mol for atmospheric CO2 - how to convert?

Using the ideal gas law
$$
pV = nRT = \frac{m}{M}RT
$$
you can express the density $\rho = \frac m V$ in terms of molar mass, pressure, ideal gas constant, and temperature:
$$
\rho = \frac{pM}{RT}
$$
With the units you use in your second example, this yields a density in $\mathrm{g\,m^{-3}}$, whereas you insert a density in $\mathrm{kg\,m^{-3}}$ into your first equation. Thus, I will write your first equation as
$$
C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6}
$$
where the density is in $\mathrm{g\,m^{-3}}$.
Inserting the expression for $\rho$ derived above into this equation, we get
$$
C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6} = \frac{C[\mathrm{ppm}]}{10^6}\frac{pM}{RT}
$$
which is identical to your second equation.


text:
pKw, pKb, pKa and NEN6411
I am trying to understand how to account for pH-temperature relations during industrial pH measurements.
Do I have it right?:

The impact temperature has on the pKa of strong acids are in practical terms non existent as the equilibrium at any temperature lays very far to the right.

The impact of temperature on pKa are more significant for weaker acids.

The impact of temperature on pKw is only significant around the point of neutrality. At lower pH's the few hydronium ions produced by H2O + H2O are so few that it bares no meaning on the logarithmic pH scale. In other words, the pH-temperature relationship is important when dealing with neutralization processes, but not so much when having pH setpoints set to anything in the milder acidic region and below.

Compensating for the pH-temperature relationship caused by the temperature dependent pKa is only important when doing measurements on processes where the temperature of the process differs from the temperature of the final product and where the targeted pH level during process control has to match the targeted pH level of the end product (at another temperature). For example: say I want to produce an end  product with a pH at 5 at temperature X. In order to make this product the temperature of my fabrication process has to be something other than temperature X. In this case, temperature compensation to account for changes is pKa becomes important - otherwise the end product won't have the desired pH.

in a different scenario the pH setpoint of a process becomes a means to an end. It's the pH of the process that determines the chemical reactions that takes place in that particular process (perhaps the hydronium ions merely acts as catalysts). In this case, temperature compensation in no way should be applied. This because the pH changes casued by changes in pkA are real and if the process demands say pH 3 for an optimal chemical reaction then it doesn't matter what the pH would have been at another temperature. The important thing is what the temperature is in the process right here, right now.

while pH changes that comes from changes to pKa actually do mean that a solution becomes more or less acidic, changes to pKw doesn't result in a solution becoming more or less acidic even though pH changes. This is because an equal amount of hydroxide ions a produced. So only the point of neutrality is changed.

Do I understand things correctly so far?
If so, then my question now is: What happens when we move into the alkaline region? I mean, pKb is, according to Le Châteliers principle, temperature-dependent as well, right? When looking at this graph https://ibb.co/LRs7G5y (algorithm based on NEN6411 (now
DS/EN ISO 10523:2012)) it seems as if alkaline solutions ALSO becomes more acidic with increases in temperature. Why is this? Does it mean that the equilibrium constant pKb runs in reverse of pKa when temperature increases (less OH-)? Or does it mean that pKw plays are much larger importance once we wander into the alkaline region? I don't, however, see how pKw play any importance on the temperature-pH relationship of alkaline solutions. I mean, if we have an alkaline solution of say pH 12, then any extra/or less hydronium ions produced by changes in pKw would be dwarfed by the proportional much larger amount of hydroxide ions.
Or do I get this wrong?
Last question: Why does the graph https://ibb.co/LRs7G5y indicates that no significant pH-temperature relationship exists below pH 6? Is it because the acid gets so strong at this point that the changes to pKa becomes miniscule relatively speaking?
Help would be much appreciated.
/Lars

The followup of  temperature-effects-on-pka-and-pkw.

Why do you need to know the pH? Do you need to know the hydrogen ion activity, the hydroxide ion activity, or some other parameter? For strongly basic solutions, the pH is often a proxy for the hydroxide ion concentration or activity (a major species) rather than the hydrogen ion concentration (a species with very low concentration in that case).

This site is designed to answer a single focused question rather than to check "my understanding of the topic". People often sneak in multiple questions on the side, but it is important to know the focal question (the problem you want to solve) for receiving meaningful answers.

The impact temperature has on the pKa of strong acids are in practical terms non existent as the equilibrium at any temperature lays very far to the right.

Practically, yes.

The impact of temperature on pKa are more significant for weaker acids.

Yes.

The impact of temperature on pKw is only significant around the point of neutrality. At lower pH's the few hydronium ions produced by H2O + H2O are so few that it bares no meaning on the logarithmic pH scale. In other words, the pH-temperature relationship is important when dealing with neutralization processes, but not so much when having pH setpoints set to anything in the milder acidic region and below.

$\mathrm{p}K_\mathrm{w}$ changes apply to all $\ce{H3O+}$ and $\ce{OH-}$, not just to those from water autodisscociation. If $\ce{[H3O+]}$ ( [] is the convention for molar conceentration ) is determined by $\ce{[OH-]}$, or if some process depends directly on $\ce{[OH-]}$ and not on $\ce{[H3O+]}$, it is significant.

Compensating for the pH-temperature relationship caused by the temperature dependent pKa is only important when doing measurements on processes where the temperature of the process differs from the temperature of the final product and where the targeted pH level during process control has to match the targeted pH level of the end product (at another temperature).
For example: say I want to produce an end product with a pH at 5 at temperature X. In order to make this product the temperature of my fabrication process has to be something other than temperature X. In this case, temperature compensation to account for changes is pKa becomes important - otherwise the end product won't have the desired pH.

Yes. E.g. for acetic acid, $\mathrm{p}K_\mathrm{a, \pu{25^{\circ}C}}=4.75$. That is the $\mathrm{pH}$ of solution acetic acid / sodium acetate in molar ratio 1:1.  Let imagine for illustrative purposes $\mathrm{p}K_\mathrm{a, \pu{50^{\circ}C}}=4.5$. If you used at $t=\pu{50^{\circ}C}$ for the final process $\mathrm{pH}=4.75$, there would not be reached 1:1 ratio.

in a different scenario the pH setpoint of a process becomes a means to an end. It's the pH of the process that determines the chemical reactions that takes place in that particular process (perhaps the hydronium ions merely acts as catalysts). In this case, temperature compensation in no way should be applied. This because the pH changes casued by changes in pkA are real and if the process demands say pH 3 for an optimal chemical reaction then it doesn't matter what the pH would have been at another temperature. The important thing is what the temperature is in the process right here, right now.

Yes, if I understant the paragraph well.

while pH changes that comes from changes to pKa actually do mean that a solution becomes more or less acidic, changes to pKw doesn't result in a solution becoming more or less acidic even though pH changes. This is because an equal amount of hydroxide ions a produced. So only the point of neutrality is changed.

This depends on what you mean by acidity. If you mean by acidity the activity or concentration of $\ce{H3O+}$, it is affected by $\mathrm{p}K_\mathrm{w}$ changes.

If so, then my question now is: What happens when we move into the alkaline region? I mean, pKb is, according to Le Châteliers principle, temperature-dependent as well, right? When looking at this graph https://ibb.co/LRs7G5y (algorithm based on NEN6411 (now DS/EN ISO 10523:2012)) it seems as if alkaline solutions ALSO becomes more acidic with increases in temperature. Why is this? Does it mean that the equilibrium constant pKb runs in reverse of pKa when temperature increases (less OH-)? Or does it mean that pKw plays are much larger importance once we wander into the alkaline region? I don't, however, see how pKw play any importance on the temperature-pH relationship of alkaline solutions. I mean, if we have an alkaline solution of say pH 12, then any extra/or less hydronium ions produced by changes in pKw would be dwarfed by the proportional much larger amount of hydroxide ions.

For weak acids, the primary parameters is $\mathrm{p}K_\mathrm{a}$. Then for their conjugate bases, $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a}$.
For weak bases, the primary parameters is $\mathrm{p}K_\mathrm{b}$. Then for their conjugate acids, $\mathrm{p}K_\mathrm{a}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b}$.
For solutions of strong acids, $\mathrm{pH} \approx -\log{[\ce{H3O+}]}$,  $\mathrm{pOH} =  \mathrm{p}K_\mathrm{w} - \mathrm{pH}$.
For solutions of strong bases, $\mathrm{pOH} \approx -\log{[\ce{OH-}]}$,
$\mathrm{pH} =  \mathrm{p}K_\mathrm{w} - \mathrm{pOH}$.

The graph you mention is for strong electrolytes, so $\mathrm{p}K_\mathrm{a}$ is not applicable.

Ok, so as I understand it the auto dissociation constant of H2O governs ALL solutions and not just solutions that hovers around the point of neutrality. Kw determines the total possible content of hydronium and hydroxide combined. More hydronium results in less hydroxide (and vice versa) according to Le Chatelier. This means that if I dissolve 1 HCl molecule in a solution, not only will a hydronium ion come into existence but a hydroxide ion will disappear (in order to satisfy the constant pKw). But if this is true, where does the OH- get its H+ from (so that it can return to being H2O)?

I am still no sure, however, how all of this ties into basic/alkaline solutions being more temperature dependent than highly acidic solutions (and neutral solutions as well for that matter). If I increase the temperature of a basic solution, why exactly is it that this solutions has a larger shift in pH than an acidic solution? What exactly happens chemically speaking?
I am also no quite sure what you mean by this:
"If [H3O+] is determined by [OH−], or if some process depends directly on [OH−] and not on [H3O+], it is significant"

OH- does not have "it's H3O+". It does not matter if its origin if from water auto-dissociation or acid dissociation. H3O+ and OH- appear and disappear all the time. At equilibrium both rates are equal. Out of equilibrium, one direction is faster.

We are again at pH + pOH=pKw(T) . For strong acid/base solution, pH/pOH are given, respectively. pOH/pH are then respectively the function of pKw(T) and pH/pOH. //  Some reactions involve OH-vand not H3O+.They depend directly on pOH=pKw(T)) - pH.

1 L of water contains 10^-7 mol/L OH-. if 1 mol of HCl(g) is added, about 10^-7 mol of OH- react with any H3O+( mostly from HCl), so just 10^-14 mol OH- would remain.


text:
Dielectric constant for gases possessing van der Waals forces
I've learnt that water can dissolve ionic bonds because it possesses a high dielectric constant.
In the cases of covalent compounds (especially the non-polar covalent compounds), where the molecules are held together by mainly van der Waals forces of attraction. They are said to possess a higher dielectric constant.
In the case of inoic compounds where they are held together by electrostatic force of attracrion, water is a solvent that can dissolve almost all the ionic compounds ,the reason stating  it to have "high dielectric constant".
Then why is that covalent compounds cannot be dissolved by an ionic solvent such as water? And is there any connection to this topic as to  how most of the non-polar covalent compounds are immiscible in ionic solvents?


text:
Would 10.0 mL of water at 10 °C have the same mass as 10.0 mL of water at 25 °C?
I am reading a book and it gives the following answer:

No, the density of water changes with
temperature. 10.0 mL of water at 10 °C would
have a higher mass than 10.0 mL of water at
25 °C because the density is higher at 10 °C.

For me it doesn't make sense. In other part of the book it states "Mass is defined as the amount of matter in an object". How can a change in temperature change the amount of matter of an object? So mass of an object is dependent on temperature?

Heating up 10 g of water does not change its mass, as you say. But that's not what this question is about. This is about two different amounts of water: 10 mL of water at 10 °C (let's say in a measuring cylinder), and 10 mL of water at 25 *C (in, let's say, a different measuring cylinder). The amount of water in the two measuring cylinders is not the same.

In fact, heating up 10 mL of water does not change its mass, either. However, it will change its volume, such that at 25 °C it is no longer 10 mL. Hence, 10 mL of water at 10 °C is not the same thing as 10 mL of water at 25 °C. Hence my suggestion to think of them as completely different samples. They are not the same object.

$m_1 = \rho_1V;$ $m_2 = \rho_2V.$ When $\rho_1 > \rho_2,$ $m_1 > m_2.$ What is the problem? And, yes, comparing two identical bodies at different temperatures will result in minuscular difference in mass (hotter one will be a slight bit heavier), but this is not what your textbook (which, by the way, should be properly referenced) is trying to say.

Thanks for your comments, I understand now!

Please find a better book. See if you can get hold of Linus Pauling's College Chemistry: An Introductory Textbook of General Chemistry

@M.Farooq why? Is the sentence too obvious?

@Alchimista, Apparently the book didn't deliver the concept of density very well.

Changing temperature doesn't change the mass of a fixed amount of a substance. But it does change the density. So a 10mL sample of water at 25 celcius will still contain the same amount of substance at 10 celcius. But it will not longer occupy exactly 10mL, but will be a slightly smaller volume.
Since you specified the conditions as having fixed volume, you would need to add a (small) extra amount of water to get 10mL at the lower temperature. So, at fixed volume, there would be more water.
Think carefully about all the conditions you specify in comparisons and be consistent.


text:
Why constant volume combustion gives higher energy than constant pressure?
I have noticed when I have done some combustion equilibrium that enthalpy is higher when doing a constant volume combustion than pressure constant combustion.
Why is it so ? What are the reasons explaining that ?
Is it because at constant pressure combustion the variation of enthalpy is equal to heat transfert $\Delta H = Q $ whereas at constant volume combustion the variation of enthalpy is $\Delta H = Q + V\Delta p$ ?
If so that suppose that heat transfer $Q$ is the same for both constant pressure and constant volume combustions. Is that so ?
Thank you

Combustions of organic stuffs produce a greater number of gaseous moles than in the reactants. When done at constant pressures, a part of the energy produced by the combustion must be used for repelling the outer atmosphere. This is a work and it costs some enegy. This is why a combustion at constant pressure produces less energy than at constant volume.


text:
How to prepare a 10% alcoholic KOH solution?
I am confused actually. A 10% alcoholic KOH solution means that there is 10g of KOH in 100 ml of ethanol, or 10g KOH in 100 g of ethanol or something entirely different?
Also, I understand that the solution is to be clear. I tried the 10g of KOH in 10 ml of ethanol, and somehow it is a turbid solution. Any thoughts on why it is so? And what could be done to make it clear?
Thank you so much!

Since you've got a solid and a liquid, the 10% must be by weight. So 10 grams of KOH to 90 grams of solvent.

KOH pellets are ~15% water. Adjust your weights accordingly.

If there is no further clarification in the specification of the 10% (like 10vol% or 10mol%), then this should indicate a mass fraction. Thus, 10% of the mass of the solution is supposed to be $\ce{KOH}$, so you would dissolve 10 g in 90 g ethanol to get the desired solution.
I don't know the precise solubility of $\ce{KOH}$ in ethanol but I assume that, if you are following a protocol, the 10% solution should be clear. Obviously, you have tried to get a 50-something% solution, which is not what you actually want. You might want to dilute your solution with whatever amount of ethanol you need to fill 100 g in total.

$\pu{38.7 g}$ $\ce{KOH}$ per $\pu{100 g}$ ethanol at $\pu{28 °C}$ according to http://chemister.ru/Database/properties-en.php?dbid=1&id=325 itself citing Рабинович В.А., Хавин З.Я. Краткий химический справочник. - Л.: Химия, 1977.


text:
Specific Heats for Non-Constant Processes
Why can the specific heat at constant Pressure ($C_p$) be used to calculate the change in enthalpy as $\text{d}H = C_p\text{d}T\newcommand{\d}{\text{d}}$ for processes that are not even at constant pressure?
I have $\d H = C_p\d T + V\d p$. Why do we neglect the $V\d p$ term? For gases, $V$ the specific volume, can be very large, so an increase in pressure could have a non-negligible effect on the enthalpy. Is it so? Or not?

The enthalpy of an ideal gas is a function only of temperature, but not pressure.  For a real gas (or other substance), the effect of pressure on enthalpy would also have to be included.

But isn't Cp defined for processes at constant pressure? @ChetMiller

See my answer below.

@ChetMiller I updated my question to express my concerns further. Does this explain my question better?

Your equation for dh is incorrect.  It should read $$dh=C_pdT+\left[v-T\left(\frac{\partial v}{\partial T}\right)_P\right]dP$$If you substitute the ideal gas law into the 2nd term, what do you get?

Still a little confused... How did you derive that dh equation? And how would you sub in the IGL? @ChetMiller

Ah, I get it. v= RT/p, so dv/dT = R/p. RT/p - T(R/p) = 0. I'm still confused how you derived that dh formula. How did you obtain it?

The derivation is in every thermo book.  But, it starts with dh=Tds+vdP.  I'm not going to provide the rest of the derivation here, since it is so widely available.

Alfred: Are you asking why, when the starting and ending pressures are the same, you can use CpdT to calculate dH of any substance, even if the process was not at constant pressure? [If yes, then see Light Yagami's answer.] Or are you asking why, for an ideal gas, you can use CpdT to calculate dH, even when the starting and ending pressures are different? [If yes, then see Chet Miller's answer.] –

Cp of an ideal gas doesn't depend on pressure.  Cp of any substance is defined as the partial derivative of enthalpy respect to temperature at constant pressure.  But, if the enthalpy of the substance is independent of pressure, then it doesn't matter if the pressure is constant.  However, determining Cp by measuring the amount of heat required to change the temperature by a certain amount of the substance at constant pressure is still a standard method of measuring Cp.

Specific heat capacity $\ce{C_p}$ can be used to calculate the enthalpy change even for the processes that are not at constant pressure as enthalpy is a state function, that is, it doesn't matter by which process you reach the end state, the value of enthalpy change will remain the same. Same is the case when we measure the change in internal energy, it is given by $\Delta U= n \ce{C_v}\Delta T$ where $\ce{C_v}$ is the specific heat capacity at constant volume, but this equation is used for all processes for the same reason as stated above, that is, internal energy is a state function and is independent of the path taken to reach there.

But why can Cp still be used to calculate enthalpy? I understand that, but isn't Cp specifically defined at constant pressures?

Yes, it is defined at constant pressure. Alright, take it this way, suppose the process occurs at constant pressure, then you don't have any problems calculating the enthalpy change, right? Now we know that enthalpy is a state function which means if we go by any other process also enthalpy will come out to be same. But for those processes we cannot evaluate the enthalpy change in a different way, that is why we always use Cp for any process because it doesn't matter by which process we go answer will come out to be same.

It follows from the familiar result:
$$\Delta H=q_p$$

 $$\Delta H=\Delta U+\Delta (pV)$$
 Pressure is constant:
 $$\Delta H =\Delta U+p\Delta V$$
 By First Law, $\Delta U=q-p\Delta V$, so,
 $$\Delta H=q_p$$

By using the fact that
$$C_p=\dfrac{q_p}{\Delta T}$$
We get
$$\Delta H=C_p\Delta T$$
And take my note: $\Delta H$ is a state function so it doesn't matter how the process is carried out.
Note, that the above answer is for an ideal gas, which the OP used in his question.
Hope this helps. Ask anything if not clear :)

This is correct only for an ideal gas.  For a real gas, H depends on pressure.

@ChetMiller: The OP had an ideal gas in his illustration.

It’s important to make it clear the this only applies to an ideal gas.

@ChetMiller: done :)


text:
Application of Le Chatelier's principle
I have the following problem:

The following reversible system is at equilibrium:
$$\ce{P_{4(s)} + 6Cl_{2(g)} <=> 4PCl_{3(l)}} $$
What happens with the equilibrium if $\ce{P_4}$ is added.

Normally, by applying Le Chatelier's principle the equilibrium should be shifted to the right, however, for some reason this does not apply to the above equation.
Why does increasing or decreasing the quantities does not change the equilibrium in the above system?

It would help yourself, if you wrote down the equation for the equilibrium constant.

The solution twists on the fact that $\ce{P4}$ is a solid and in a different phase than the $\ce{Cl2}$. By definition in this situation the activity of the $\ce{P4}$ is 1 in the equilibrium equation no matter if a microgram or a metric ton is left.

@Poutnik Thank you for the comment! From what I understand we don't include liquids and solids when the equilibrium constant is derived. However, I was hoping for more details on that, i.e. why?

@MaxW So, it is primarily relater to how the constant is defined, right? Could a slightly more detailed explanation be provided? In terms of the phase state or something like that? Is it experimentally verified that changes in those two phases do not change the equilibrium or something like that?

@Ziezi - We typically write equilibrium expression with concentrations, but really activities should be used. Think of it this way. You dissolve sugar in water until the solution is saturated. Once saturated it doesn't matter if there is a single grain of sugar  or a kilogram of sugar  undissolved. No more sugar will dissolve.

Things would get more complicated, if P4(s) is soluble in PCl3(l).

This question might help. The answer explains why pure solids and liquids have constant concentration.

The maximum concentration (or the pressure) of $\ce{P4}$ in the vapor phase does not depend on the amount of $\ce{P4}$ in the solid phase. It depends only on the temperature. It is like water in the atmosphere. The vapor pressure of water vapor does not depend on the amount of liquid vapor in contact with the humid air.

Well it is beacause $P_4$ is in solid state so it is not included. $$K = [PCl_3]/[Cl_2]$$

Firstly, as an approximation, we can assume that only gases and species in aqueous solution are input into the equilibrium constant expression.
Since $\ce{Cl_2}$ is the only species in the reaction that qualifies (as it's in gaseous state), $K_p$ will be:
$$K_p=\frac{1}{P_{\ce{Cl_2}}^6}$$
Because no change in temperature is taking place, $K_p$ is a constant, which means $P_{\ce{Cl_2}}$ is also a constant since:
$$P_{\ce{Cl_2}}=\left(\frac{1}{K_p}\right)^{1/6}$$
Now, since $\ce{Cl_2}$ is the only gas, the total equilibrium pressure $P$ will only depend on $P_{\ce{Cl_2}}$:
$$P=P_{\ce{Cl_2}}$$
This means that no matter how much phosphorus we add, the total equilibrium pressure will not change, since $P$ does not depend on phosphorus.
Let $P_1=$ total equilibrium pressure before adding more phosphorus.
Let $P_2=$ total equilibrium pressure after adding more phosphorus.
$$\Delta P=P_2-P_1=0$$
When we say "equilibrium shifts", (at constant temperature) what we mean is that $K_p$ remains constant, but $K_x$ does not. $K_x$ is the equilibrium expression in terms of molar fractions.
$K_x$ is a "false" or "fake" equilibrium constant in the sense that, while it has an analogous expression with respect to $K_p$, it is not actually always a constant (even when temperature is not changed).
This is due to the fact that $K_x$ changes at the expense of $K_p$ in order for the latter to remain constant, and the relationship between them is:
$$K_p=K_x\times P^{\Delta n}$$
In this case, if we can show that $K_x$ does not change after adding phosphorus, then there is no shift in equilibrium taking place.
Let:
$K_{x1}$ represent $K_x$ before phosphorus is added.
$K_{x2}$ represent $K_x$ after phosphorus is added.
$$K_p=K_{x1}\times P_1^{\Delta n}=K_{x2}\times P_2^{\Delta n}$$
Rearranging:
$$\frac{K_{x2}}{K_{x1}}=\left(\frac{P_1}{P_2}\right)^{\Delta n}$$
$$K_{x2}=K_{x1}\left(\frac{P_1}{P_2}\right)^{\Delta n}$$
Here we can observe that $K_{x2}$ = $K_{x1}$ when at least one of two things happens:
(1) $\Delta n=0$
(2) $P_2 = P_1$
In our case:
(1) is false since $\Delta n=-6$
but...
(2) is true because earlier we said $\Delta P=0$, so $P_2=P_1$
In conclusion, $\Delta K_x=0$, which means there is no shift in equilibrium for this reaction after adding more phosphorus.


text:
How is lapping slurry made?
I've recently gained an interest in the whole topic of lapping and the creation of flat surfaces.
Lapping is done in different ways, but the one I'm interested in is essentially just rubbing 2 surfaces together with an abrasive in the middle. The slurry is a liquid with the very tiny abrasive particles suspended in it (usually things like diamond, aluminium oxide, silicon carbide, etc..).
These particles can be different sizes of course, for example 1/4 micron or 9 microns. And they can make this slurry to these "tolerances".
A layman such as me looking at their cheap prices is lead to believe that maybe this isn't such a complicated manufacturing process. I don't know much about chemistry, but I am mighty curious about this.
How are these big batches of really tiny abrasive particles made to within such a small tolerance for cheap? Thank you!

I guess they get particles of all sizes, then fraction them by sedimentation speed.

The abrasive substance is synthesized or mined (e.g., garnet), then pulverized or ground up, then size sorted via sieves, etc.

Pretty much exactly as they make different grades of sandpaper.

Adding to the question: why are various lanthanide oxides used for lapping glass, in preference to other abrasives? See http://sfsidewalkastronomers.org/index.php?page=notes-from-john-dobson-s-telescope-making-class

@EdV are sub-micron sieves really that easy to come by? And how do you pulverize or grind diamond?

@TrisT Sieves are for the bigger stuff. As Ivan noted, sedimentation (like field flow fractionation) can be used for smaller sizes. Small size synthetic diamonds are prepared as such: no need to crush bigger ones!

@DrMoishePippik Well, I have the ceria, and a fair number of other abrasives, e.g., zirconium diboride, but I would never try making my own Dobsonian telescope mirror! Those folks are wild! But thanks for the link!

@EdV so you synthesize for example diamonds, and that process makes little diamonds of all shapes and sizes, so then you just sort them. I just looked up this FFF process, it it's pretty genius, never would have guessed sedimentation could take that form so thank you for clarifying that too. This is a lot of stuff that I can search further and actually get results on, thank you!


text:
Ellingham Diagram for carbon
Consider the following reactions:
$$\begin{align}\ce{C + O2&->CO2}\tag{1}\\[1em]
\ce{2C + O2&->2CO}\tag{2}\\[1em]
\ce{2CO + O2&->2CO2}\tag{3}\\[1.5em]
[(2)+(3)]/2 &= (1)\end{align}$$

Since $G$ is a state function, why aren't the straight lines corresponding to these reactions in the Ellingham diagram concurrent?
I have checked the graph from other sites but found none to be concurrent.
Is there some kind of error in data or extrapolation of the graphs or any other specific reason for this?

$\Delta G$ is not a state function, $G$ is a state function.

I was asking for the reason why the lines aren't concurrent

If we write ∆G for 2nd and 3rd rxn and add the two, it should have a value twice as for 1st.

So all the points on graph of 1st should be the midpoint of the corresponding points on graph of 2nd and 3rd

Hmm, I need to think about it then, seems correct...

Why has this question received downvotes? It's a reasonable question ,right?

@MilanPaul See my answer.

related: https://chemistry.stackexchange.com/q/126977/102629

@Govind Are you preparing for JEE?

@ Light Yagami Yes

You are indeed right, I confirmed that it is the fault in the diagram itself. See the diagram by wikipedia:

Whether the lines are exactly concurrent depends on how the data were obtained.  In the case of carbon oxidation, if we draw on measured thermodynamic data for oxidizing $\ce{CO}$ to $\ce{CO2}$ and also for oxidizing $\ce{C}$ to $\ce{CO2}$, then we could calculate the "data" for oxidizing $\ce{C}$ to $\ce{CO}$ by doubling the first reaction and subtracting the second (assuming all reactions are based on one mole of oxygen, which is standard for Ellingham diagrams), in which case the three lines are forced to be concurrent.
But if we draw on independent measurements for all three reactions and they are all subject to experimental errors, the errors may not combine mathematically in the same way as the theoretical reaction stoichiometries.  The error (per mole of oxygen) that occurred when $\ce{C}$ was oxidized to $\ce{CO}$ may not be exactly twice the $\ce{C/CO2}$ error minus the $\ce{CO/CO2}$, in which case the three independently determined free energy lines would also not match up.  If all three experiments had a similar amount of experimental error, then the deviation from concurrency in the diagram would provide a rough measure of this error.  So having the lines not meet exactly could actually provide an additional bit of information behind the data.