text:
Acid-catalyzed hydration of alkenes
So I'm basically doing an experiment where I measure the effect of pH on the yield of alcohol within the acid catalyzed hydration of ethene. Essentially, all other details aside, I get a mixture of ethanol and ethyl chloride solutions at the end of my experiment. I want to find the total amount of either. I want to use a simple method to determine the amount of each in the solution. I want to avoid using fractional distillation, if possible. I need help to do this.

If you are certain that there are no other halogenated byproducts or reagents remaining in your solution, and that any levels of bromine and iodine would be negligible compared to chlorine (fluorine would not be detected), then one of the total halogen analysis (TX) methods would be a good choice.
If this is an ongoing analysis that you need to set up, I suggest the Carius halogen method. Wikipedia gives a good overview here:

The Carius halogen method in analytical chemistry is a method for the
  quantitative determination of halogens in chemical substances. A
  known mass of an organic compound is heated with fuming nitric acid in
  the presence of silver nitrate contained in a hard glass tube known as
  carius tube, in a furnace. Carbon and hydrogen present in the compound
  are oxidised to carbon dioxide and water. The halogen present forms
  the corresponding silver halide (AgX). It is filtered, washed, dried
  and weighed.

Alternatively, depending on the equipment available to you, you could also do an oxygen bomb combustion of the sample with a bicarbonate absorbant, followed by either titration of the resulting chloride or ion chromatographic (IC) analysis. In the case of the IC finish, you can usually just use the IC eluent as the adsorbant.
If you are only going to do this analysis once, or only rarely, then I would strongly suggest a commercial TX anaysis using Standard Methods - For the Examination of Water and Wastewater, Method 9020. Method 9020B is for the analysis of total organic halogens (TOX) in water, which is not appropriate for your analysis as a whole because it includes a step for extracting trace amounts of organics from water onto activated carbon. But, then a TX analysis is performed on the activated carbon, which is what you would want to do for your compound rather than for the activated carbon. This is typically about an $80 analysis.
Regardless of the TX technique used, you would of course then assume that all of the halogen came from the ethyl chloride and calculate the ethyl chloride concentration from there.

Thank you for the answer. The carius method seems to require equipment that my school lab might not have. In particular, the carius tube and the furnace, though I am wondering if I can still do the same analysis without these apparatus. Suppose now that my solution also contained alcohol within it. In particular, I'm extremely sure that the alcohol present is only going to be ethyl alcohol. Is there any test that I can do to find the amount of ethanol in it? Thanks so much once again.

Do you have access to either gas or liquid chromatography equipment?

There really isn't a simple, non-instrumental quantitative analysis for alcohols. Going back to the commercial route, if you can assume the only O in your product is from the alcohol, then you could get an O analysis done for about $35-$45. But, if you were to go with that route, there is a combustion/ion selective electrode method for about the same price that would give more accurate results (for the ethyl chloride of course). I think the most likely method you'd be able to do yourself is the bomb combustion followed by gravimetric determination after precipitating Cl- with AgNO3.

I'm looking for methods which are extremely simple. Like, I was thinking that, since ethyl chloride's hydrogens are barely acidic and ethanol is way more acidic than ethyl chloride, would it not be possible for me to stick a pH probe into the final solution, measure the pH and calculate the yield of ethanol directly using acid-base equilibrium calculations?

OK, knowing what you do/don't have available is very helpful. Yes a fractional distillation should be a good way to go, with a couple caveats. If you are purely doing it for analytical purposes, and not trying to recover the ethyl chloride, that will be particularly easy. If you have a small distillation apparatus and enough material to sacrifice for the distillation, then you could just weigh the mixture, let the ethyl chloride boil off, then weigh the remaining ethanol. If you want to separate and keep the fractions, remember that ethyl chloride boils at 12C and won't stick around at RT.

About how many mL of your mixture do you have total? BTW, the pH idea is clever, but without data about the acidity of ethanol in that matrix (and there probably is none) it's just not going to work. Also, do you think that it is mostly ethanol or about what mixture would you guess; 50:50, 70:30, 95:5, etc. If you don't know that, that's fine, but knowing how much total you have would be helpful.

The acidity of ethanol in that matrix? If I know the pKa of ethanol and if i know how to account for the auto-protolysis of water in my equilibrium calculations, wouldn't that give me the amount of ethanol in the solution?
Also, I don't know what's the proportion of ethanol and ethyl chloride in the solution. However, I know that I will definitely have more than 50 ml of solution for any given experiment. Would that help?

But you only know the pKa of ethanol in water (~16) or selected other solvents. In DMSO, the pKa of ethanol is more like 30. You just can't predict the pKa of ethanol in an arbitrary solvent, as it's a function of the solvation of both the anion and cation (H+) in that solvent. I tried to think of something clever based on the fact that ethanol is water miscible but ethyl chloride is not, but ethanol would probably solublize the ethyl chloride (that's where I was headed with my how much of each question). I think your distillation idea is probably the best given what you have to work with.

Oh yea of course. Hmmm but suppose that I had it such that the amount of water is in excess as compared to the amount of ethanol produced in my experiment. Would that allow me to obtain results using the pH method?

Okay because my experiment basically involves me bubbling ethene into a solution of hydrochoric acid. The pH of each solution varies. I'm basically testing for the relative yield of ethyl chloride and ethanol. I have a few arguments for why one might be formed in a greater amount than the other.

Essentially, I could argue that because HCl is a strong acid, it will dissociate in water to form a lot of hydronium ions, which will then react with ethene to produce ethyl chloride. So, there'll be very little ethyl chloride formed. However, I could also argue that since HCl is a stronger acid than ethene, it is more likely to react with ethene and produce ethyl chloride. I don't exactly know how good these arguments are, exactly. I'm looking at the relevant mechanisms as well as the equilibrium constants for both reactions to see if I can make conclusions from there.

I'm more of an analytical type than an organic synthesis jock so I can't help much with that kind of stuff ;) Since your system already contains water from aqueous HCl and is in 1 phase, that shoots my solubility idea. Unless you've taken steps to dry the mixture and remove the HCl that kills my Cl / halogen analysis ideas. I really think pH measurements are a dead end. Distillation looks better all the time. The advantage is the huge difference in boiling points, the disadvantage is the 12C boiling point of EC and how to keep it from evaporating. Even at 0C it'll be gone in a hurry.

So in the end I should just stick to heating my solution till the ethyl chloride vaporizes and then measuring the mass of that before and after? How long should I heat it for and under what conditions?
Also, thank you so much. I'm not the analytical type, I'm more of the theoretical type in chem and physics so this kind of experimental stuff is beyond me.


text:
If NO3- were to get oxidised to some compound in the way SO4 2- gets oxidised to S2O8 2-, what would be that compound?
In electrochemistry, we learn of the sulphate ion getting oxidised to peroxydisulphate ion. If this was to occur in the case of the nitrate ion which has a larger discharge potential, what would be formed?

First, let’s take a look at both sulfate and nitrate. They both show similar characteristics when you compare the oxidation states: remember that nitrogen is in group 15 so it has five valence electrons while sulfur is in group 16 so it has six — therefore, the highest reasonable oxidation states for these elements should be $\mathrm{+V}$ and $\mathrm{+VI}$, respectively.
$$\begin{align}&\ce{\overset{$\mathrm{+V}$}{N}\overset{$\mathrm{-II}$}{O3}^-}&&\ce{\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O4}^2-}\end{align}$$
When oxidising sulfate $\ce{SO4^2-}$ to peroxidisulfate $\ce{S2O8^2-}$, sulfur cannot be oxidised any further; the only atom that can still be oxidised is oxygen. Hence, in an initial step, we generate $\ce{SO4^{.-}}$ with three oxygens of $\mathrm{-II}$ and a fourth of $\mathrm{-I}$. Two of these then dimerise forming a peroxo-bond.
$$\ce{^-\overset{$\mathrm{-II}$}{O3}\overset{$\mathrm{+VI}$}{S}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O3}^-}$$
Looking at nitrate, the only obvious theoretical would be to do the same, i.e. remove one oxygen electron to intermediately generate $\ce{NO3^.}$ and then have this dimerise to $\ce{O2N-O-O-NO2}$ or $\ce{N2O6}$. Interestingly, this would be a neutral compound unlike $\ce{S2O8^2-}$ which retains its charge. I don’t know whether it has been observed or not, but I would predict its formation from what I know.

$\ce{N2O5}$ is reasonably common in nitrogen oxide chemistry; I never came across $\ce{N2O6}$ in my grad work, though, so if it can occur it's a very minor player.

Scifinder has a few hits for the actual molecule. Didn't really investigate further

That was quite helpful and I am interested in knowing the presence of N2O6.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Location of double bond in 1,3-dimethylcyclohexene
Why is the double bond in 1,3-dimethylcyclohexene located between the alkyl subtituents? Does it have to be this way or is this just the most common form?

You are going through wrong way.. First, the structure of molecules comes then comes the name. And to your written molecule you have to consider one of the double-bonded carbon as the 1 carbon. Then, you have to count clockwise or anticlockwise for the number lpcant carbon. But in case of different counting available you have to ensure that on first differnce between the number of locant the name has the lowest number. and your given name is just the result of this procedure.

In unsubstituted cyclohexene, the double bond is always allocated the locant ‘1’. When alone, the locant ‘1’ is omitted in the name (see Subsection P-31.1.3.1 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)). Thus, the preferred IUPAC name is cyclohexene rather than cyclohex-1-ene.
However, if any locants are essential for defining the structure of the parent structure, then all locants shall be cited. Therefore, the name 1,3-dimethylcyclohexene is not in accordance with current IUPAC recommendations. The preferred IUPAC name for the compound that is given in the question is 1,3-dimethylcyclohex-1-ene.
Nevertheless, the practice of omitting locants when there is no ambiguity is widespread, and the incomplete name 1,3-dimethylcyclohexene is unambiguous provided that the other relevant nomenclature rules have been applied correctly, in particular:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

In substituted cyclohexenes, a low locant is given first to the double bond (the ‘ene’ ending) according to (e). After that, low locants are given to any substituents that are expressed as prefixes (e.g. ‘methyl’) according to (f). Therefore, it is clear that the double bond in 1,3-dimethylcyclohex-1-ene has the lowest locant ‘1’ even if it is not explicitly mentioned in the name 1,3-dimethylcyclohexene.

Thank you very much. My chemistry text book, dumbed down for physicians, does not mention anything about such intricacies of nomenclature. That there is a preferred and more precise way of naming is just not mentioned. I thought that there is some deeper reason that there could not likely be an Isomer with a double bond in another position.

@Max to be fair to your book: the recommendations are still quite recent (a mere four years), many books simply haven caught up yet. Nomenclature before that was a little bit less specific and the concept of a preferred IUPAC name is as recent as this. This doesn't mean that other names are incorrect, the basic rule should be that they are unambiguous.

The double bond is there because it's there and the name the name says so.

With other words, there are other isomers which all have the six-membered ring with two methyl groups in the same substitution pattern, but have the double bond at different positions.
In order to distinguish between them, these would have different names.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How to apply Faraday's law to calculate the charge of metal ions?

Electrolysis of a molten ionic compound with a current of $\pu{0.50 A}$ for $\pu{30 min}$ yielded $\pu{0.700 g}$ of a metallic element at the cathode. If the element has a relative atomic mass of $150.0$, calculate the charge on the metal ions.

My attempt at solving:
$$n = \frac{m}{M} = \frac{0.7}{150} = \pu{4.67e-3}\\
Q = Ft = 0.5 \times (30 \times 60) = \pu{900 C}$$
The main problem I'm having here is that, how do I relate this unknown metallic ionic compound to the amount of electrons gained in this reaction?
After I obtain that can I just use $$Q = n\ce{e-} \times F?$$
I then solved for $$F = Q/n = \pu{1.92E5 F}$$. I was experimenting with some stuff and divided this by $\pu{96500 C}$ which got me $1.99$. The correct answer was $\pu{2 F}$.
Is this just coincidence because I don't see the connection between dividing by $\pu{1 F} = \pu{96500 C}$ when I already solved for $F$?

Use the formula given by @ Kai, but use SI units which means putting time into seconds to find $Q = It$ and $F = 96485.3  \pu{ C mol^{-1}}$. The charge z comes out as 2 to the nearest whole number.

When you're confused by a problem like this it is definitely time to use dimensional analysis.
For the amount of electrons $\text{mol}_e$
\begin{align}
\text{The ampere} &= \text{Amp} = 1.00 \frac{\text{coulomb}}{\text{second}} \approx 6.242\times10^{18}~\frac{\text{electrons}}{\text{second}}\\
\text{Farady} &= 26.801 \frac{\text{Amp}\times\text{hours}}{\text{mol}_e}\\
30~\text{minutes} &= \frac{30 \text{minutes}}{60~\text{minutes/hour}} = 0.500~\text{hours}\\
\text{mol}_e &= \frac{0.50~\text{Amp} \times 0.500~\text{hours}}{26.801~\text{Amp}\times\text{hours}\text{/mol}_e} = 9.328\times10^{-3}~\text{mol}_e
\end{align}
For the amount of substance of metal $\text{m}_m$
$$\text{mol}_m = \frac{0.700~\text{g}}{150~\text{g/mol}_m} = 4.667\times10^{-3}~\text{mol}_m$$
Charge per atom = $z$
$$z = \frac{\text{mol}_e}{\text{mol}_m}= \frac{9.328\times10^{-3}}{4.667\times10^{-3}} = 1.998 = 2.0 \frac~{\text{electrons}}{\text{atom}}$$

$$m=\left(\frac{Q}{F}\right)\left(\frac{M}{z}\right)$$
In the given question, we have to find $z$ which is the valency of the metallic ions.
Use  $Q=IT $ (in Ampere-Hours), $F=26.8$ Ampere hour per equivalent.


text:
What exactly is petuntse? (mineral)
What exactly is petuntse, the "china stone" used as one of two ingredients in porcelain along with kaolin?
The Wikipedia article (and books I have read) describe petuntse as a mineral containing feldspar. Well, fine, but there are a lot of minerals that are certainly not petuntse that contain feldspar. Feldspar is one of the most common minerals, so saying it is a material with feldspar in it, is not specific.
What is the specific composition of traditional petuntses used to make porcelain?

According to the Pottery Histories web site:

Petuntse
  Petuntse is a historical Chinese term for rocks with a high content of
  mica and feldspathic minerals that have been used as one of the
  ingredients in the manufacture of Chinese porcelain. It is generally
  similar in composition and use to China stone. Petuntse, with china
  clay, forms the base ingredients for hard-paste porcelains similar to
  those produced in China in historic times. See China stone.

and

China stone (Cornish stone, Cornwall stone)
  China stone is a
  feldspar-rich mineral derived from partially decomposed granite. China
  stone includes feldspar, mica, fluorospar, quartz and other
  granite-derived minerals such as kaolinite, but lacks iron-bearing
  minerals.  China Stone is used worldwide as a flux in the manufacture
  of ceramic bodies and glazes. In the United Kingdom, the material is
  mined in one small area of Cornwall.

I don't know if this meets the degree of specificity that you were looking for.


text:
Why is FeCl3 covalent?
There is an electronegativity difference of 1.2 between $\ce{Fe}$ and $\ce{Cl}$. Then why it is covalent?

Electronegativity alone can't tell you everything. For example, $\ce{MnO}$ is mostly ionic, while $\ce{Mn2O7}$ is markedly covalent.

Even from basic electronegativity difference you can tell it - 1,2 < 1,7 so why it  should be unusual according to you?

usually when metal is bonded to a non-metal, they are bonded together by ionic bond but this is not the only condition , the more important condition is the electronegativity that must be greater than 1.7 to be sure that the bond is ionic if not like in your example 1.2 then the bond will be polar covalent bond which means that the electronegativity difference is not enough for the electrons to be lost by iron and gained by chlorine but they are shared and shifted toward the chlorine atom as it has the greater electronegativity.


text:
Solubility of white phosphorus
Why is white phosphorus soluble in $\ce{CS2}$ but red phosphorus not? Is it something in relation with the structures of both? Please explain.

You can (and should!) use the \ce{...} environment to format chemical equations and formulae; see here for a basic intro and here for a complete documentation. P/S names of chemicals are not capitalised.

What do you know about white phosphorus, to begin with?

@IvanNeretin That white phosphorus is tetrahedral closed ${P_4}$ molecule and red phosphorus is kind of open structure formed from ${P_4}$ molecule.

That's it, then. To dissolve red phosphorus, you'd have to break covalent bonds, which is hard. Dissolving white phosphorus is much easier.

@IvanNeretin Then what kind of bonds are in $P_4$ molecule in white phosphorus among $P-P$ atoms aren't the covalent ?Do we need to break $P-P$ bonds Or  the interactions between two $P_4$ molecules while considering solubility?

Within $\rm P_4$ molecule, there are of course covalent bonds. No, we don't have to break those.

@IvanNeretin So are the interactions between $P_4$ molecules Van der Waals attaractions ?

That's right, and those are weak.

White phosphorus has P4 molecules packed into a crystal,these dissolve readily in Carbon Disulfide.

Whereas Red phosphorus is polymeric in nature. It is a derivative of white phosphorus where one P-P bond is broken and an addtional bond is formed with neighboring tetrahedron molecule resulting in a chain like structure .

Like all long chain polymers this will lower its ability to dissolve and its solubility in solvents


text:
Why is OH a good leaving group?
OH is a strong base which makes it a good nucleophile. Why then is it a good leaving group?
Also, why are bad nucleophiles good leaving groups? What are the criteria a good leaving group should fulfill?

I think $OH^-$ is not a good leaving group as far as I know. That is the reason why we protonate it and then water is released as a good leaving group.Strong bases such as OH−, OR−  tend be poor leaving groups, due their inability to stabilize a negative charge.For a good leaving group the stabilization of charge is one of the factors.

Iodide is a great nucleophile and an excellent leaving group.

What in the world made you think it is?

I can’t quite follow you because hydroxide is one of the lousiest leaving groups known to chemistry that are still considered potential leaving groups (mostly by students, that is).
Note that nucleophilicity and basicity are unrelated as are nucleophilicity and leaving group ability. Here are two examples:

hydroxide is a pretty good nucleophile, a very strong base and a lousy leaving group.
iodide is a pretty good nucleophile, a very weak base and a very good leaving group.

Good leaving groups are groups that are stabilized in the reaction medium. In general, the conjugate base for a strong acid is a good leaving group, and indeed, you can apply the same rationale for why something is the conjugate base of a strong acid to why it's a good leaving group.
This, of course, refutes your claim that hydroxide is a good leaving group since it is not the conjugate base of a strong acid.
Nucleophicility is a much harder concept to quantify in terms of what is important, and I think it's best if I not attempt a lengthy answer here.


text:
Electron cloud distribution
Can a please explain me what is electron cloud distribution in simple words?
Electron cloud is the area around which electrons are most probably found. What is meant by electron cloud distribution?

Related; http://chemistry.stackexchange.com/questions/39438

I just want to know what it is in simple words

Brief and short, please help me with that. Please!!!

@LokeshSangewar Read the paragraph bridging pages 45-46 in the following book: https://books.google.com/books?id=MxdxGPk-LqoC&pg=PA45&dq=%22term+electron+cloud+which+is+customarily+given%22&hl=en&sa=X&ved=0ahUKEwiA4I-v5KbSAhVBKCYKHX-AA34Q6AEIHDAA#v=onepage&q=%22term%20electron%20cloud%20which%20is%20customarily%20given%22&f=false That is the best simply answer.

Where the electrons are likely to be...

The electron cloud distribution describes the probability of finding an electron in some region of space around the nucleus at some instant of time.
Imagine you have a camera that can take close-up shots of a hydrogen atom showing the electron at an instant in time. Also suppose that camera can keep snapping pictures, say, a thousand times a second, each time on the same image so that you get a multiple exposure. As the camera runs, the multiple images of the electron will begin to form a cloud-like picture which is dense in some places and thin in others. For the single electron of hydrogen, the densest part of this cloud will form a sphere with the nucleus at the center. As you travel inward toward the nucleus, or outward away from the nucleus, the cloud will get progressively thinner.
The thickest regions of this cloud are thick because the electron was found there more often than in the thin regions. So, if we were to take just one snapshot, this is where the electron would be most likely be found. That is, the probability of finding the electron at a given time is greatest where the cloud is densest.
Other elecrton clouds in other element's atoms have different shapes. Do understand that this is a model that explains some chemical behavior but it is not the current "final" model of how electrons behave in atoms.


text:
What is the difference between volatility and evaporation?
I've been trying to find the difference between volatility and evaporation and I can see almost everywhere that the volatility is the tendency of a substance to vaporize. So Q1: what makes the volatility special? I understand that the fuels are generally volatile, and I understand that they vaporize faster than water, for example. But Q2: is this the only difference?
I was thinking that volatility also makes the vapor have a different chemical composition, or did I get it wrong?
For instance, the water (a non-volatile liquid) vapor is still $\ce{H2O}$, right? Q3: Will the vapor of benzene $\ce{C6H6}$ still be $\ce{C6H6}$ or it brakes into other compounds?

No it's still going to be benzene vapour.

Q2: Yeah vaporised water is still H2O but not H2O(aq), vaporised water is H2O(g). This is called change of state Q3: this is also true for the C6H6(g)

Thank you, both. So basically volatile substances are substances with low molecular forces and that's all, true? Nothing special with them when speaking about vaporization. Is this true?

Evaporate is a phase change from liquid to gas without changing the molecule itself (or the atoms is the liquid is atomic).
Volatility is essentially the tendency of a liquid to evaporate. So a highly volatile liquid would show a great tendency to evaporate.
A measure of volatility is the vapor pressure. There is a very related term partial pressure. So if we think of air as being nitrogen and oxygen, then there is a partial pressure for nitrogen and a partial pressure for oxygen, and the sum of the two partial pressure is the actual pressure (the absolute pressure).
To make things a bit more confusing chemists think of absolute pressure. So because we are in earth's atmosphere there is always 1 atmosphere of gas pressure. So when you pump a car tire up to 30 lbs/sq inch, there is already 14.7 lbs /sq inch air pressure due to the atmosphere. So the absolute pressure inside the tire is 44.7 lbs/sq inch.


text:
How to calculate formal charge?
I am trying to understand formal charge and how it is being calculated. I have copied the image from organic chemistry as a second language below:

How do I calculate formal charge? In the book formal charge is calculated by adding up the electrons where each bond counts as 1 electron, so in the image for the oxygen with a formal charge of -1, we get a total of 7 electrons which is greater than the 6 that oxygen originally has. But for the octet rule I count the bond as 2 electrons (total of 8 electrons), so I am confused as to how the formal charge calculation calculation is being made?

Formal charge and octet rule follow a different formalism. This shouldn’t confuse you, you should simply keep the concepts separate. Note that oxidation state follows yet another formalism (all electrons given to the electronegative partner).

As Jan hinted in the comments, there are multiple electron bookkeeping methods which serve different uses. We want to keep track of electrons because it gives us an idea of what type of chemical behavior might be attributable to an atom or molecule.
Octet Rule
The Octet Rule (and the related 18 electron rule for some transition metal compounds) serves to give us some idea of whether the structure proposed meets quantum mechanical requirements for valid structures. It also gives us a quick estimate of reactivity.
As an example, second period elements do not exceed the Octet Rule, so you know your structure is invalid if you have 14 electrons on an oxygen atom.
Boron compounds often do not have complete octets on boron (for example $\ce{BH3}$), but that leads to boron compounds being powerful electrophiles and having some weird bonding situations (example: diborane $\ce{B2H6}$).
Both Lewis bond theory and valence bond theory attributes all shared bonding electrons to count toward the total number of electrons for an atom.
Formal Charge
Formal charge is attempting to identify regions of high and low electron density within a structure, which help predict how some compounds react.
For formal charge, we act like each atom in a structure is an ion that has assigned to it all non-bonding electrons and half of all bonding electrons. For your structure, we can learn where the negative charge on the anion is located and where it is not.
Oxidation Number
Oxidation number is one final way to approximate electron distribution, and therefor the propensity to undergo certain kinds of reactions. Oxidation number is calculated by assuming all atoms in the structure are ionic and assigning all bonding electrons to the more electronegative atom in the pair.
Comparison
So, for your compound, the anionic oxygen would meet the octet rule, have a formal charge of 1-, and an oxidation number of -2. The neutral oxygen would meet the octet rule, have a formal charge of 0, and an oxidation number of -2. If you needed to predict where a cation would be attracted to this compound, you could choose one of the two oxygen atoms appropriately.
The two carbon atoms would both have full octets and zero formal charge, but one has an oxidation number of -4 and the other has an oxidation number of +3. In this case, if you needed to predict where a reducing agent would react with this compound, you would be able to choose one carbon atom over the other.
For monoatomic species, the three counting methods will give you the same number of electrons and the formal charge will equal the oxidation number. As an exmaple: $\ce{O^2-}$ — 8 electrons, formal charge 2-, oxidation number -2.

So, octet rule gives us a quick estimate of reactivity (that means everything in period  2 reacts the same, O & C have same reactivity?
Formal charge says how a reaction might occur? In the case of the cation, it will attack the O with FC -1 1st? I need to read more about oxidation number (haven't read it yet), but that is telling you where the highest distribution of e- are, hence it makes sense Oxygen would be more reactive in my compound.
It seems that you are layering on different characterizations of a atom in order to model it's likely behavior in the system (aka the molecule)?

http://chemistry.stackexchange.com/questions/68875/why-are-electrons-shared-equally-when-calculating-formal-charge-but-unequally-w?rq=1
This one helps: basically the bonds are a combination of ionic and covalent bonds, pure covalent would share e- with both atoms while as pure ionic would make 1 atom hog all the electrons.  So would it be correct to think of them as a components of a vector and that vector describes the type of bond? Just like how AC current has 2 parts to it magnitude and phase?

The following is a quick check on for the formal charge:
formal charge = Valence Electrons - ½ (bonded electrons) - (non-bonded electrons)
Consider the following Lewis structure for sulfuric acid:

For the sulfur atom there are six valence electrons ($V=6$), eight bonded electrons ($B=8$) and zero non-bonded electrons ($N=0$). Therfore the formal charge of sulfur $q_{\ce{S}}^\mathrm{f}$ is
$$q_{\ce{S}}^\mathrm{f} = V - \frac{B}{2} - N = 6 - \frac{8}{2} - 0 = +2$$
Oxygen has 6 valence electrons, and when it is terminal, it has  6 non-bonded electrons and 2 bonded electrons:
$$q_{\ce{O}}^\mathrm{f} = V - \frac{B}{2} - N = 6 - \frac{2}{2} - 6 = -1$$
In the hydroxyl group it has 4 non-bonded electrons and 4 bonded electrons:
$$q_{\ce{O\color{gray}{(H)}}}^\mathrm{f} = V - \frac{B}{2} - N = 6 - \frac{4}{2} - 4 = 0$$

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


text:
Gibbs of vaporization of a substance at a non-standard temperature
I'm trying to solve for the gibbs energy of vaporization of ethanol at a temperature lower than its boiling point. I am given the heat capacities of liquid and gaseous ethanol, the standard enthalpy of vaporization, and the standard boiling point of ethanol.
I start with the form:
$\ce{dH=(C_{p,g}-C_{p,l})dt}$
Following integration, I get the form:
$\ce{\Delta H=\Delta H_{vap} +\Delta T(C_{p,g}-C_{p,l})}$
This form gives me the new enthalpy of vaporization at a non-standard temperature. However, When I try to enter this into the form:
$\ce{\Delta G=\Delta H - T\Delta S}$, with $\ce{\Delta S=\frac{\Delta H}{T}}$
This solution always results in an answer of zero (equilibrium). Intuitively, I understand that phase changes are typically reversible, however using another representation:
$\ce{\Delta G=-RTln(K)}$, with $\ce{K=\frac{P}{P_{0}}}$, the ratio of vapor pressure to standard pressure of the liquid (liquid activity = 1).
My question is, why do these formulas apparently conflate?

Alright, this is kind of fast but after thinking about it I think the issue with the problem is the definition of $\ce{\Delta S}$. My second try to compute this value involves starting with the form:
$\ce{dS = \frac{C_{p,g}-C_{p,l}}{T}*dT}$
Following integration, I arrive at the form:
$\ce{\Delta S = (C_{p,g}-C_{p,l})ln\frac{T}{T_b}+\frac{\Delta H}{T}}$ where $\ce{\Delta H}$ is determined in the way as in the question above.
Placing everything into the form $\ce{\Delta G = \Delta H-T\Delta S}$:
$\ce{\Delta G = \Delta H_{vap} +(T-T_b)(C_{p,g}-C_{p,l}) - T((C_{p,g}-C_{p,l})*ln\frac{T}{T_b}+\frac{\Delta H}{T})}$ where $\ce{\Delta H}$ is the non-standard enthalpy of vaporization.

Well, don't thank in posts or ask if they are ok. You can do it in comments, but as one can up/downvote answer or comment it then it could be said that all of them are waiting for review. As for your Q&A, for starters heat capacities don't change linearly.


text:
How do ionic, polar, hydrogen-bonding and non-polar compounds compare in TLC?
$R_\mathrm{f}$ decreases with increasing polarity. Does this mean a compound with an ionic bond would have the lowest $R_\mathrm{f}$? And nonpolar would have the highest $R_\mathrm{f}$?
Example: $\ce{RCOO- Na+}$ is ionic.  If the above is true it should have a lower $R_\mathrm{f}$ than cyclohexanol which has H-bonding?

No, your analysis is wrong.
The nice thing about TLC is that you can pick the "paper" and the solvent to suit your analysis. So if you use benzene as the solvent then it won't "drag" ionized compounds along behind the solvent front very well.  If you use water ionized compounds do get dragged along nicely.

The basis of each chromatography experiment — and that explicitly includes TLC — is that the retention factor is inherently dependent on the interactions between analyte, mobile phase and stationary phase. In silica gel–based chromatography, the staionary phase consists of $\ce{Si-OH}$ fragments and thus polar, hydrogen-bonding and ionic compounds interact well with the stationary phase while non-polar compounds don’t. Thus, as a general rule the more polar (i.e. more hydrogen-bonding, more ionic) compound will have a lower $R_\mathrm{f}$ value than the less polar one.
However note that $R_\mathrm{f}$ values are always dependent on stationary and mobile phase. By tuning your mobile phase, you can increase or decrease your compound’s $R_\mathrm{f}$ value to your desire — with the notable exception of very unpolar compounds that always run with the solvent front, practically all $R_\mathrm{f}$ values are possible. The most extreme case I actually used in the lab was $\text{ethyl acetate} : \text{methanol}\ 5:1$.


text:
Can an ethyl ester of L-proline be synthesised from the amino acid and ethyl alcohol in toluene with azeotropic distillation of water?
I was told that the ethyl ester of L-proline may not be synthesized from L-proline and ethyl alcohol in refluxing toluene, with an acid catalyst such as p-toluensulfonic acid for example, because toluene and ethanol and in this case water as well form an azeotrope. Is this true?
I do not have a supply of pure thionyl chloride, and that's why I am trying to use this route. I appreciate if anyone may answer this question.

Hello and welcome to Chemistry.CE. I always recommend taking the short [tour] to better familiarize yourself with the site. Regardless, good luck with you problem!

I would check the literature to be sure, but the rationale seems reasonable.

I checked in SciFinder the preparation of the ethyl ester of L-proline and to my surprise didn't find what I was looking for, but anyway, I will try to see if mixing pure toluene, ethanol and water in the proportions they are found in the azeotrope yields a two phase liquid in the side arm of a Dean-Stark trap.

I doubt (neither Scifinder, nor Reaxys at hand), an N-protection "simply by a Boc group" survives the PTSA-catalysed esterification until intentionally set to be cleaved off again.  But maybe Kociensky's  or Theodora Greene's compilations indeed know N-protections more suitable here.

I think this is unlikely to work under the conditions described as the catalytic PTSA is tied up as its proline salt. If you reflux under D-S conditions you will just remove the EtOH due to the azeotrope with Toluene.


text:
Why do you need excess LAH to reduce esters?
I am little confused on this question. On one hand, I remember my teacher saying that $\ce{LiAlH4}$ is like $\ce{BH3}$ of hydroboration oxidation of alkene, where you one mole of $\ce{BH3}$ can oxidize 3 mole of alkene molecule. but on the other hand my textbook keeps on mentioning excess LAH when reducing ester? Can't one mole of LAH reduce four mole of ester molecules as well? Thanks.

A single reduction of a single gives a aldehyde, which is a more reactive carbonyl than an ester is.

I know that. I am asking why can't one mole of LAH reduce ester twice since it havs 4 hydrogens attacked to Aluminum.

LAH is a bit more fickle than borane. Some of the hydrides might not add. Depends on the binding of the reduced components to the central aluminum. They form an oxide, and the oxide can reduce the nucleophiliticity of the hydrides and increase steric crowding.

Lithium aluminium hydride or LAH is the more everyday form of $\ce{LiAlH4}$. When originally produced, most of it is indeed $\ce{LiAlH4}$ but that quickly partially hydrolyses giving the grey powder you may or may not know from the lab. As such, $\ce{LiAlH4}$ is practically never entered into reactions in pure form. Since it is impure, excess reagent is added to make sure the reduction job is performed as desired.
Also, remember to consider the reaction meachnism:
$$\ce{R-COOR' ->[LAH][- HOR'] R-CHO ->[LAH] R-CH2OH}\tag{1}$$
Note that the second step, the reduction of an aldehyde, is more rapid than the first as aldehydes are more reactive towards nucleophiles than esters. Thus, if you attempted to bargain, hoping to arrive at the aldehyde selectively, you would end up with under half of your reactant reduced to the alcohol and half of it still as an ester. Another reason to always add excess just to be sure.

For one, the action of $\ce{BH3}$ (or a borane in general) on an alkene in the course of a hydroboration is not an oxidation, rather than an reduction.  The second step, the action of hydrogenperoxide on the product of this addition (an alkane) is an oxidation.
For the reduction of an ester by LAH, there are two things to retain:

general rule: LAH reacts with a carbonyl compound, initially forming a lithium alcoholate and $\ce{AlH3}$, and subsequently they form the lithium alanate (decomposed during mildly acidic aqueous workup)
The first H of $\ce{LiAlH4}$ leads only to the aldehyde, and the second one to the alcohol you are able to isolate.


text:
How many particles are necessary to observe a well-defined state of matter?
The three ordinary states of matter $-$ solid, liquid, gas $-$ can usually be distinguished by a function of the strength of interparticular forces and distance. A shorter distance and a stronger force results in a solid whereas the opposite will yield a gaseous state. In other words, the states of matter are describing macrostates.
With clarity in mind, let us define a practical limit to observing different phases.

Definition. A practical limit of particles is such that a phase transition is concievably observable by experiment. This is the smallest number of individual particles that when probed give us the traditional phase diagrams (omitting different solid phases).

What is the practical limit for most purposes? I am probably looking for a 'number' of the form $10^a\ldots10^{a + \approx2.5}$ where $a$ is a positive integer.

So while an individual particle under standard conditions can be described as gaseous, it does not fit the definition.

An alternative wording (equivalent as far as I can tell) to this question is given by the Clapeyron equation. Apostrophes mark different phases.
$$\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{\Delta H}{T\left(V'' - V'\right)}$$

In which range of the number of particles do the entalphy of transition and difference in volumes of phases become measurable? This does not definitively mean that $\Delta H$ has to be a constant w.r.t. $T$ but rather a function of temperature $\Delta H = \Delta H\left(T\right)$ that has predictive capability.

Surprisingly few compared to a mole. The 'noise' in $10^6$ is $10^3$ so a million molecules would give reliable answers, ignoring any practical difficulties involved. There are also many 'single molecule' experiments, pulling apart individual proteins  using AFM for example, or fluorophores measured singly or in a single protein.

I learned it as the thermodynamic limit. In our lab, accurate phase transitions are simulated with 10^3 particles pretty easily via periodic boundary conditions

@porphyrin, @ khaverim So currently $PL = 10^3\ldots 10^6$ and the $a$ to beat is about $3$. If you convert your comments into answers, please provide a summary of the experimental techniques applied, including references if possible.

^ @khaverim (feel free to mark comment as obsolete once read)

This an interesting question, but it would seem that a different definition is needed for a solid, a liquid and a gas. For a gas you'd have to have enough so that the Maxwell-Boltzmann equation holds. For a liquid it would that you'd need  enough molecules to pull the "drop" into a sphere. For a solid I'm not sure. But solids have bulk, small particles, and nonoparticles as well as individual atoms.

This question is too broad without being more precise what properties and what specific materials are you talking about. Nanoparticles with few hundred atoms behave like well-defined states, and many properties can be well defined on smaller (less than 100) clusters if surface effects are considered in the description.

Computationally:
It depends on the material but in general $10^2 - 10^4$ atoms/molecules is a representative range. I learned this concept as the thermodynamic limit; i.e. the number of particles in a system at which increasing particle number does not change density significantly. For obvious reasons it's also called the macroscopic limit. In Statistical Mechanics we bridge the gap between small "countable" systems and macroscopic "mole-quantity" systems with the thermodynamic limit.
Our lab develops and applies models for small molecules like $\ce{H2, CO2}$, noble gases, hydrocarbons etc. for use in computer simulation. The usual method is to parametrize a model such that phase transitions (i.e. density changes as a function of temperature or pressure) are accurately mapped, and intermolecular potentials are comparable to highly accurate QM calculations (CCS(D/T/Q) methods).
One can reproduce accurate phase diagrams of noble gases using just 40-100 atoms by putting the system in a fluctuating periodic box and changing particle positions until an energy minimum is reached (the computational method is called Monte Carlo) under given $T$ and $p$ -- this is called the isothermal-isobaric ensemble, or $NPT$. You can achieve identical results using the $\mu VT$ (grand-canonical ensemble) wherein chemical potential, volume, and temperature are fixed, but $N$ is allowed to change. For professional model development it is more common to use $10^3 - 10^4$ particles for a good fit.
Below is a snapshot of output from a Monte Carlo program obtaining the density of $\ce{He}$ at STP using 100 atoms (the reported value is 0.1786g/L). You can see that $\ce{He}$ in this model is a near-ideal gas with compressibility factor $Z \approx 1$ :

I think the answer is more appropriately expressed via computational (Statistical-Mechanical) methods because the concept of the thermodynamic limit arose in and is used by SM. In general, the central-limit theorem is where the concept came from, which states that fluctuations to the mean of a quantity (density here, $N/V$) are expressed as $\frac{1}{\sqrt{N}}$, thus for $N=10,000$, the mean density can expectedly vary by $0.01$. For $N=N_A$, the variance would be certainly negligible.


text:
Melting Point of Amorphous Solids
My textbook says that amorphous solids do not melt at a fixed temperature,but at a range of temperature. It depends on the way the atoms/molecules are present in a given piece  of solid.

Since the melting point(range) depends on the whole solid piece , can it be said that melting point of amorphous solid is an extensive property?

No way! Extensive properties are proportional to the amount of material, much like price: you take twice as much, you pay twice as much. Melting point is not like that.

No matter how the structure is the heat given to the substance will be distributed over many molecule/ atoms. And the all molecules will more or less the average energy. So, on average temperature will be same across the substance. This will happen even if the substance does not conduct heat well but on a smaller region.
For breaking a bond, you need specific amount of energy. Boiling or melting needs a lot of bonds to be broken. Most molecules, on average will gain that energy on a specific temperature.
So, that being said, the m.p. and b.p are not extensive properties. Being amorphous or crystalline doesn't change the reasoning.

The question mixes two concepts.

As correctly stated by @Ivan Neretin and @Mockingbird, "melting point" is an intensive property, that does not change if you have 5 g or 200 g of a material. (See for an explanation for example here.)
There are many instances without sharp transition solid -> liquid, resulting what is called glass transition.  This is what your textbook likely refers to (first phrase of your question).

Polymers in general are one example, DNA and some proteins do not melt either and show reversible glass transitions.  (If you iron a shirt, you benefit from glass transitions, too.)
In contrast some small molecule crystalline materials undergo irreversible glass transitions , while heating up to the melt there is one large (across several K) range of phase transition; but after cooling the molten sample and re-heating, only a sharp melting point is observed.
You may monitor glass transitions indirectly by observing the intensity of fluorescence in function of the rising temperature; often used in DNA chemistry, for example.  Or you heat your sample gradually in reference with a known sample to compare with, by differential scanning calorimetry, the more general method.


text:
Does benzene vapor react with air in a closed barrel?
If we would have benzene in a closed barrel for quite a while at room temperature, some vapor would form fast due to its volatility, right? But let's say that when the barrel was closed, some air got trapped inside. So it's inside at any time. Is there any chance that the benzene would react with the components in the air at room temperature? Or would we simply have benzene vapor and air? Or would we have benzene, some products of reaction and air? I made a sketch for a better understanding of my question.

So which one is the real one? And if it is the second one, would the reaction be similar to combustion, or?Is temperature a major player in this? And if so, how strongly would it affect the reaction?

Note that the air (really it's the oxygen that may slowly react with the benzene) will also dissolve into the liquid benzene. In this case, for the reaction $\ce{C6H6 + O2 -> products}$, the concentration of one reactant (benzene) would be greatly increased. Depending on the solubility of $\ce{O2}$ in benzene, the reaction would probably mainly take place in the liquid phase (or maybe at the liquid/vapor interface). Still, regardless of where or how the reaction might take place, the key to your question is the kinetic issues discussed in the answer by Chester Miller.

Combustion is occurring, but the kinetics are very very very slow at room temperature.  So, on a time scale of days, weeks, and months, the system will be closer to the first picture.  On a time scale of decades, centuries, and millenia, the system will be closer to the 2nd picture.

to add a little to you nice answer, it is the activation energy of the reaction benzene/oxygen that is so large that the rate of reaction is so slow. The same is true of many organics molecules, those that make up you and me for example :)

Thank you both for your answers! So if one would add enough thermal energy to go over the activation energy of the reaction, then there will be a full combustion process, right?

Sure.  Just add a spark.


text:
Baby bottles accidentally steamed with soft plastic teether - melted - Bottles still safe?
I'm not sure who to ask so have been googling around -- My husband accidentally steamed our new teether for our infant son with the bottles. What he didn't remember/realize is that you're not supposed to (per the package notes) and found out quickly why when he opened the steamer and the plastic was molten and the gel on the bottom of our steamer. My concern is: I can wash and re-steam our bottles, nipples, bottle parts (some plastic), but should we be worried about some kind of chemical reaction that happened within the bubble of the steamer? I am not well enough versed in chemistry to know if this would be a problem. I'm just about out of bottles for the day so I'm hoping someone has the answer somewhere.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. I understand your predicament, but personal medical questions (and this one does fit the category) are off-topic on Chemistry. We can not safely answer questions for your specific situation as it is impossible to judge. For example, we would need to know the exact composition of everything in the bath to exclude any evil reaction...


text:
Does this freezing point depression problem make sense?

The solvent is water. Pure water has a freezing point of 0 degrees Celsius. How come, in this problem, the addition of solute results in an elevation of the freezing point from 0 to 0.46 degrees C? Is this possible? Or is this question just flawed?

If you'd see "Buffalo" written on an elephant's cage in a zoo, would you suspect a mistake, or suppose that you didn't know something?

I would assume that this is supposed to be $-0.46\ ^{\circ}\mathrm{C}$...

My thought as well.

Adding deuterium oxide would raise the freezing point.

I see no problem here. It's not a salt and it doesn't lower melting point but on contrary...

Truthfully it is very unlikely, but possible.

Ha, found it! http://chemistry.stackexchange.com/questions/57686/are-melting-points-for-mixtures-of-solids-always-below-the-melting-points-of-the

The solute does not need to be ionic but just be involatile relative to the solvent.


text:
How can you perform an acid-base titration with antacids if substances, like aluminium hydroxide, are insoluble?
I am trying to perform an experiment in order to determine the efficiency of different antacids. An online search later shows that you add an HCl solution in a solution created with a base/antacid. However, the substances do not dissolve in water and hence precipitate in the bottom. Is there a step I am missing?

Solids can react, so if you design the titration appropriately, this will still work.

In order to have the antacid neutralization reaction go fast add a known amount of excess acid and back titrate the extra acid with NaOH.

Grind the solid well and stir or swirl as you titrate.

So it doesn't matter if there is still solid in the solution? Just add acid until everything dissolves and then do the back-titration with NaOH?

That's what I would do, just like MaxW's comment. Good note by bpedit too; get that surface area up by grinding it until the consistency is as close to a "flour" as possible, then stir throughout dissolution and titration.


text:
Regioselectivity of an amination reaction of 2,6-dichloro-3-nitropyridine

In the formation of ethyl 4-(6-chloro-3-nitropyridin-2-yl)piperazine-1-carboxylate G, why is the substitution favoured at the 2-position? I would have said that the 6-position substitution is preferred as:

In pyridines, nucleophilic substitution occurs at the 2/6th or 4th position, and as chlorine is an effective leaving group compared to hydride this rules out position 4, leaving 2 and 6 as the most feasible...
The electron withdrawing nature of the nitro group on the ring would favour nucleophilic substitution at the positions ortho or para to itself.
The chlorine at the 2 -position is ortho, and the chlorine at the 6-position is para so both are feasible...
If I didn't know what the product was, I would have said that position 6 is therefore the site of substitution as the amine reagent is quite bulky and the 6-position is less sterically hindered due to the absence of an adjacent nitro group

However, this evidently isn't the case! I found another similar reaction where the same thing occurs:

Is it something to do with the nitrogen in the pyridine ring hindering the electron density transfer to and from the para nitro group? I'm really stuck!

I'd agree that nucleophilic substitution at C-6, leaving the nitro and piperazine substituents para to each other, would lead to the thermodynamic product.
The observed regioselectivity is therefore probably kinetically controlled. Although the nitro group activates both ortho (C-2) and para (C-6) positions towards nucleophilic substitution via resonance, it's worth bearing in mind that the nitro group is also extremely electron-withdrawing via the inductive effect in its own right.
This would make C-2 more electron-deficient and hence prone to nucleophilic attack.
Recall that in an SNAr reaction the nucleophile must attack the π* orbitals of the arene, which are above and below the plane of the ring. In this case, the nitro substituent is also (for the most part) in the plane of the ring, and doesn't lie along the trajectory of nucleophilic attack. So, the nitro substituent doesn't present all that much steric hindrance to the nucleophile.

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


text:
Which salt will best be able to buffer the addition of HCl?

From these three salts, which will be best able to buffer the addition of $\ce{HCl}$?

ammonium chloride

sodium carbonate

ammonium acetate.

I think sodium carbonate is the right choice because when it dissociates in water forms a basic solution, so when $\ce{HCl}$ mixes with sodium carbonate it will be the best buffer, but it's a weak base so I'm confused. So can someone explain which is best?

To pick the best salt in this case, we should pick the least acidic cation and the most basic anion.
While sodium cannot dissociate into $\ce{H^+}$ ions, ammonium can, so the ammonium salts will be more acidic than their sodium counterparts.
In addition to being the most basic ($\mathrm{p}K_\mathrm{b}(\ce{CO3^{2-}})=3.67$,  $\mathrm{p}K_\mathrm{b}(\ce{CH3COO^{-}})=9.24$, $\mathrm{p}K_\mathrm{b}(\ce{Cl-})=20.3$), carbonate is also a diacidic base whose conjugate acid is also a stronger base than the other two anions ($\mathrm{p}K_\mathrm{b}(\ce{HCO3-})=7.63$).
So, sodium carbonate is the best choice.

From the first line, can you tell me the reason for choosing most basic anion? I am struggling to understand this concept of buffer. Thanks alot.

Is it because the more basic anion will abstract $\ce{H+}$ from solution given by $\ce{HCl}$ ?

Of the top of my head...

Ammonium chloride

Definitely not this salt. There will be little free ammonia from the solvation of the $\ce{NH4^+}$ cation. The $\ce{Cl^-}$ is a spectator anion.

Sodium carbonate

The $\ce{Na^+}$ is a spectator cation. The $\ce{CO3^{2-}}$ will be mostly $\ce{CO3^{2-}}$ with a tiny bit of $\ce{HCO3^-}$. So the ratio of $\frac{\ce{HCO3^-}}{\ce{CO3^{2-}}}$ (and hence the pH) will change rapidly with the addition of $\ce{H^+}$.

Ammonium acetate

Some of ammonium will protonate the acetate anion. So you end up with a mixture of $\ce{NH4^+}$, $\ce{NH3}$, $\ce{CH3COO^-}$ and $\ce{CH3COOH}$. I think this solution would have the most buffer capacity for acid.
It seems you need to solve all of these for $\dfrac{d\text{pH}}{d\ce{H^+}}$ -- well the last two anyway...
SOLUTION
My calculus-foo is lost on this, so let's just brute force an answer. Let's assume we have 1.00 liters of 0.1 molar solutions of each of the salts. We'll calculate the pH for the salt, add $1.00\times10^{-3}$ moles of $\ce{H^+}$ and calculate the new pH. We can then calculate:
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}}$

Ammonium chloride

$\ce{NH4^+ <=> H^+ + NH3 }$
$\text{K}_\text{a} = 5.75\times10^{-10} = \dfrac{\ce{[H^+][NH3]}}{\ce{[NH4^+]}}$
Salt alone
assume $\ce{[H^+]=[NH3]}$ and $\ce{[NH4^+]=0.1}$ molar then simplify
$\ce{[H^+]} = \sqrt{\text{K}_\text{a}\times\ce{[NH4^+]}} = 7.56\times10^{-6}$
pH = 5.121
we can now solve for
$\dfrac{\ce{[NH3]}}{\ce{[NH4^+]}} = \dfrac{\text{K}_\text{a}}{\ce{[H^+]}} = \dfrac{5.75\times10^{-10}}{7.56\times10^{-6}} = 7.61\times 10^{-5} $
$\ce{[NH3] = 7.61\times10^{-6}}$
Salt plus 0.001 moles acid
So there isn't any significant amount of $\ce{NH3}$ to protonate and the acid added can be added to the initial concentration of  $\ce{H^+}$ from the salt.
final $\ce{[H^+]} = 7.56\times10^{-6} + 1.00\times10^{-3} = 1.008\times10^{-3}$
pH = 2.997
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{2.997 - 5.121 }{1.00\times10^{-3}} = -2124$

Sodium carbonate

$\ce{CO3^{2-} + H2O <=> OH^- + HCO3^-}$
$\text{K}_\text{b} = 2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$
Salt alone
assume $x = \ce{[HCO3^-] = [OH^-]}$ and $\ce{[CO3^{2-}]} = 0.100 -x$ molar then simplify
$0 = x^2 + 2.14\times10^{-4}x - 2.14\times10^{-5}$
$x = 4.52\times10^{-3}$
$\ce{[OH^-]} = 4.52\times10^{-3}$
$\ce{[H^+]} = \dfrac{K_\rm{w}}{\ce{[OH^-]}} =2.21\times10^{-12}$
pH = 11.655
Salt plus 0.001 moles acid
Now if we add 0.001 moles of $\ce{H^+}$.
We'll essentially neutralize some $\ce{OH^-}$ and make some $\ce{HCO3^-}$, but we know that
$\ce{[HCO3^-] - [OH^-]} = 0.001$ or $\ce{[HCO3^-]} = 0.001 + \ce{[OH^-]}$
Let $x = \ce{[OH-]}$
$2.14\times10^{-4} = \dfrac{\ce{[OH^-][HCO3^-]}}{\ce{[CO3^{2-}]}}$
$2.14\times10^{-4} = \dfrac{(x)(0.001+x)}{0.1 - (0.001+x)} = \dfrac{x^2 + 0.001x}{0.099-x}$
$0 = x^2 + 1.214\times10^{-3}x - 2.1186\times10^{-4}$
$x = 0.004036$
$\ce{[OH^-]} = 0.004036$
$\ce{[H^+]} = \dfrac{1.00\times10^{-14}}{\ce{[OH^-]}} = 2.48\times10^{-12}$
pH = 11.606
$\dfrac{d\text{pH}}{d\ce{H^+}} = \dfrac{\text{pH}_2 - \text{pH}_1}{1.00\times10^{-3}} = \dfrac{11.655 - 11.606}{1.00\times10^{-3}} = 49$

Ammonium Acetate

Salt alone
Salt plus 0.001 moles acid

You have not added your analysis for Ammonium acetate, but nevertheless a good and detailed answer.

$\ce{CO3^- <=> OH^- + HCO3^{-}}$ that does not appear right, the charges and atoms are not balanced. I can't make sense of it.

@Martin-マーチン - No new Nobel prize chemistry just plain sloppiness. I fixed it.

And user ringo no doubt deduced the right answer using chemical logic not math.


text:
How to create a 3 M solution of ammonia from 25% ammonia

We have ammonia $25\%$. We also know $\rho_{\text{ammonia}} = \pu{0.86 \frac{g}{cm^3}}$, $\pu{25 ^\circ C}$. We want to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.

This means that there are some calculations we need to do in order to produce $\pu{3 M}$, $\pu{0.5dm^3}$ ammonia.
I try to calculate the required mass by the following method: $\text{m}=\text{C}\cdot \text{V}\cdot M \cdot 0.25$. Is this method correct?

Let us continue this discussion in chat.

My point with dimensional analysis is to carry the units through the calculations so that you don't get twisted around... so not $$0.82\times 100=82;\pu{g}$$ but rather $$0.82\times 100=82;\text{g solution per 100 ml of solution }$$ not $$(0.25\times82);\pu{g}=20.5;\pu{g}$$ but rather $$(0.25\times82)\pu{g}=20.5;\text{g }\ce{NH3}\text{ per 100 ml of solution}$$ Other than that the calculations are fine. Good work!!

Since you have opened a chatroom, I have cleared the comments.

$\text{Produce a solution of }3\;\text{M}\;0.5\;\pu{dm^3}\;\ce{NH3}$

$\\\text{Ammonia}\;25\%\;\text{(mass-percent)}\\\rho_{ammonia}=0.82\pu{\frac{g}{cm^3}}$
Suppose 100ml
$$0.82\times 100\times 0.25=20.5\;\pu{g}\;\ce{NH3}\text{ per 100 ml of solution}$$
$$n_{ammonia}=\frac mM=\left(\frac{20.5}{17.031}\right)\pu{moles}$$
$$C_{ammonia}=\frac{n_{ammonia}}{100\times10^{-3}}\approx 12.037 \;\pu{M}.$$
To decide $V_1$, which is needed in order to prepare the solution of $0.5 \;\pu{dm^3}$ $3\;\pu{M}$ ammonia,
$$V_1=\frac{C_2V_2}{C_1}\approx0.1246 \pu{dm^3}=124.6\; \pu{ml}.$$

You mostly just reiterated the calculations that the OP did. :-( // Nitpicking ... Some of this isn't really right either. For instance the density should be grams per ml or kilograms per liter , not grams per liter (OP's mistake...). I'd rather that you specified that the ammonia was 25% by weight. for the equation calculating $n$ note that it is $n_{ammonia}$.

@MaxW I'm the OP. I will add that it is 25 mass-percent. And I added my own calculations as an answer, to show that they were correct to any future reader reading this post. Should it not be g/dm_3 or must it not be g/dm_3 in order for a correct caluculation, just so I understand? I may change it, anyway. Just want to know.

 Sorry. I didn't notice that you were in fact the OP. You were absolutely right to add the calcs...


text:
In the graph provided, which of the substances is the most efficient in acting as an antacid?

If every dot represents the addition of 0.1 g of the respective base. I suppose that the best would be the one that reaches the highest pH, in the shortest time, for example Calcium Carbonate, but if that is true, why are other antacids used that reach lower pH?

You shouldn't just consider the efficiency of antacid. Most efficient antacids can also be very dangerous to human health!

I read that there are two dangers with using antacids. First is alkalosis, when the pH goes higher than 7, and acid rebound when the stomach will produce even more acid to neutralize the extra base. So if that is not it what would be measured as efficiency?

Your experiment is an interesting approach to this question. Your result basically show that carbonate based antacids behave one way and the aluminum hydroxide / magnesium hydroxide ones another.  The latter two (the hydroxides) are so insoluble in water (though obviously soluble in acid) that they seem to take much longer to dissolve than the calcium carbonate and Rennie (calcium carbonate). I think that you will find that the carbonates act the fastest, though maybe won't last the longest. This just a measure of relative speed of dissolution though, not efficiency or quality.

@airhuff So what do you suggest for judging efficiency?

Speed isn't everything.  Some manufacturers will sacrifice speed (which they all know how to get) in the interest of more controlled pH, cutting down the risk of alkalosis or acid rebound described in other comments (@Theodore).

?!? The x-axis is labeled in minutes, yet you stated "every dot represents the addition of 0.1 g of the respective base. " So what is going on?

Normally, the efficacy of a drug is determined by the area under the curve of the entire time course of the drug in the body.  The graphs here only show the initial rate of the change in pH but not the entire time course after the peak pH is reached.  I assumed that these results were generated by an acid-base titration experiment in the laboratory and not generated in vivo.  Therefore I cannot answer which antacid is more effective and/or clinically safe.
Although calcium carbonate shows a rapid initial rate of pH neutralization, it may not be suitable for all patients.  Too rapid a pH rise in the digestive tract may cause nausea and vomiting and may trigger an “acid rebound”, meaning that the body tries to compensate by secreting more acid to maintain the normal gastric pH for food digestion, making the heartburn even more unbearable.
Some patients take calcium carbonate as a source of calcium, but one also needs vitamin D in order for the calcium to be absorbed.  Besides, calcium carbonate yields carbon dioxide gas with acid which is still acidic, and bloating.  For patients with compromised parathyroid function, too much calcium in the blood (hypercalcemia) after the bones are already saturated with calcium may start to deposit in the heart and in the kidneys to form calcium oxalate kidney stones (different from uric acid stone).  Therefore calcium antacid may not be the best antacid for all patients.
If calcium antacid is not tolerable, one can use aluminum hydroxide (Aludrox®) which is a good alternative.  However aluminum hydroxide can cause constipation.  Magnesium hydroxide (milk of magnesia) is another good alternative, but magnesium hydroxide can cause diarrhea because magnesium is a smooth muscle relaxant.  Therefore the product Maalox® or Mylanta® which combines aluminum hydroxide with magnesium hydroxide will be easier on the stomach.  Also, there are products that contain simethicone (which is an anti-flatulent) as an extra ingredient to help with bloating.
The choices for an antacid are out there, make a good, sound one.

Acid-base chemistry has different pharmacokinetics from regular small molecules. The drug will react in-situ on the order of minutes in the stomach, and opposed to being processed by the liver or just secreted via kidneys. What you see via the graphs is what you get in terms of efficacy, though clearly not in terms of side-effects as your answer so nicely points out.

Thank you for the comment.  I was answering the part of the question that asked why other antacids are used.

Don't get me wrong. This is not a complaint about your answer. I am simply extending discussion about the first line of your answer.

You are welcome.


text:
What is the reason of the color difference for the cis/trans isomers of tetraamminedichloridocobalt(III) chloride?
The cis and trans isomers of $\ce{[CoCl2(NH3)4]Cl}$ are quite famous as one is purple and the other one is green which is a huge difference in color. Now I can't find any information to why this is the case?
I remember they told me in the lecture on organometallic chemistry when we were discussing substituted carbonylmetallates that a trans-effect occurs where a π-donor trans to the carbonyl (which binds to the metal's d-Orbitals via its antibonding π*-orbitals) increases the electron density in the metal and therefore also increases the the bond's strength ($\ce{M-CO}$) which then changes the bond length. A π-acceptor ligand on the other hand decreases the e-density.
In case of an octahedron the trans isomer would only possess same ligands trans positioned to each other while the cis isomer has mixed ones, too.
But is this the explanation I am looking for or is there another effect I just can't think of right now?

Hello and welcome to Chemistry.SE. If you have any questions about the site and how it works I suggest starting off with the short [tour]. Best of luck!

If you search a library for a copy of Inorganic Chemistry by Miessler and Tarr, they have an excellent diagram showing which explains the phenomenon.
$\ce{Cl-}$ ligands have lone electrons that can donate into the metal, as you said yourself. The trick is to recognize what effect this has on the $t_\mathrm{2g}^*$ and $t_\mathrm{2g}$ orbitals. Remember that the five frontier orbitals in an octahedral configuration are the $t_\mathrm{2g}$ and $e_\mathrm{g}^*$ orbitals.
When a $\ce{Cl-}$ ligand pi-donates into the metal center, it can only interact with the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ $(t_\mathrm{2g})$ orbitals because the $\mathrm{d}_{x^2-y^2}$, $\mathrm{d}_{z^2}$ $(e_{g}^*)$ orbitals have only a sigma-type interaction with the ligand (for them, there is no node, draw this to ensure yourself). When these ligands like $\ce{Cl-}$ pi-donate, they fill the $t_\mathrm{2g}$ bonding orbitals, widening the gap between them and the open the $T_\mathrm{2g}^*$ antibonding orbitals. These antibonding orbitals become much closer in energy to the $e_{g}^*$ orbitals, so you reduce the octahedral field splitting energy $\Delta_\mathrm{O}$ which redshifts the absorbance of the compound.
The reason the trans isomer is more redshifted (green as opposed to purple) is due to the fact that in the trans isomer, the ligand-metal pi overlapping does not clash. On the other hand, if you draw the ligand-metal pi overlapping generated by two cis $\ce{Cl-}$ ligands, you will find that there is always one orbital that contains a lobe where the ligand overlap does clash. When this clash does exist, it becomes less favorable for the $\ce{Cl-}$ ligands to pi-donate into the metal via that clashing orbital -- so, the $t_\mathrm{2g}^*$ antibonding energy level is closer in energy to the $t_\mathrm{2g}$ bonding energy level, and further in energy from the $e_\mathrm{g}^*$ energy level. In the cis compound, field splitting $\Delta_\mathrm{O}$ is higher and the color should be purple instead of green.
If you are having trouble seeing the clash, draw a $\mathrm{d}_{xy}$ orbital on a metal $\ce{M}$, and then draw two $\ce{Cl-}$ ligands such that $\ce{Cl-M-Cl}$ is a right angle (cis ligands). When you try to show pi-donation from both ligands, you will find that there is one lobe of the $\mathrm{d}_{xy}$ orbital that is used by both; that is the clash that is disfavored.

An easy rationalisation starts off with symmetry considerations — and indeed any rationalisation should start off with symmetry considerations. The cis and trans complexes have a notably different symmetry.
For $\ce{\textit{trans-}[CoCl2(NH3)4]Cl}$, the point group is $D_\mathrm{4h}$. This is a point group with a rather high symmetry, meaning that there will be few and generally larger electronic transitions between orbitals.
$\ce{\textit{cis-}[CoCl2(NH3)4]Cl}$ features the $C_\mathrm{2v}$ point group. This has a very low symmetry meaning that few orbitals will be truly degenerate. In turn, this means that there will be many more different possible transitions and in general that these transitions will have smaller energy differences.
Note that what I have said in the two paragraphs above are broad generalisations. I did not attempt to draw up generic molecular orbital schemes — which you should always do for a proper discussion. Depending on the results of an MO discussion, it could well be that the larger transitions are found in the cis-complex, contrary to what the above paragraphs may imply.

I know you only wrote about broad generalisations, so this is just an addendum one might want to keep in mind: The point group of $\ce{$trans$-[CoCl2(NH3)4]Cl}$ is only $D_\mathrm{4h}$ for the complex ion, if you ignore the hydrogens. I think this is fine, however, it should be mentioned. And I would expect that rotation of the $\ce{NH3}$ ligand is very fast at room temperature, so that they hardly make a difference. Otherwise the point group would also be $C_\mathrm{2v}$ for the complex ion.

An attempt to simplify (to make it a bit more comprehensible to a curious highschooler) and explain some of the terminologies used in Xahoc's answer. This is not intended to be a standalone answer and should not be considered as one; please read through the rest of the answers for a thorough explaination.

All coordination compounds have a net stabilizing energy called crystal field stabilization energy (CFSE), due to a crystal field effect with respect to a spherical symmetrical (think of ligands approaching the central metal atom (CMA) equally from all the directions, like a sphere) ligand field; CFSE always has a negative or a positive value and one of its fundamental uses is the prediction of stability of a complex (greater the magnitude of CFSE, greater is the stability). I use a simple formula to calculate CFSE:
$$ \pu{CFSE= -0.4\Delta_\circ (n_{t_2g}) + 0.6\Delta_\circ (n_{eg}) + n'P}$$
The formula doesn't hold much importance here, although the term $\Delta_\circ$ does. What's that? Glad you asked. Before we dive in, something about crystal field theory:

The CMA is assumed to be at origin and the six ligands are approaching the CMA from $+x, -x, +y, -y, +z, -z$ directions. Now the ligands are approaching along the axis, and $d_{x^2-y^2}$ and $d_{z^2}$ orbitals have their lobes along the axis, so they:

Face more repulsion from the incoming ligands, which increases their energy content. And did I tell you that they're codenamed $e_g$?
Form a sigma bond, which explains:

When a $\ce{Cl-}$ ligand pi-donates into the metal center, it can only interact with the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ $(t_\mathrm{2g})$ orbitals because the $\mathrm{d}_{x^2-y^2}$, $\mathrm{d}_{z^2}$ $(e_{g}^*)$ orbitals have only a sigma-type interaction with the ligand $\dots$

The remaining three orbitals (namely $d_{xy}, d_{yz}, d_{zx}$) are code-named $t_2g$ and are at a lower energy level.

This splitting up of the orbitals into two groups is called crystal field splitting. The energy difference between those two groups of orbitals is precisely what our $\Delta_\circ$ measures. Well, I've been talking about octahedral complexes all along, $\Delta_\circ$ has an evil brother $\Delta_t$ which goes for tetrahedral complexes. So now we know: $\Delta_\circ$ is crystal field splitting energy. Did you say energy? I think I've studied this formula somewhere:
Energy = $\displaystyle \frac{hc}{\lambda}$
Seems good, let's apply it:
$\Delta_\circ = \displaystyle \frac{hc}{\lambda_{absorbed}}$
Wait! What's up with that 'absorbed' thing? Nothing much, but it appears that complexes seem to know some unicorn magic. So what they do is: they absorb (a wavelength corresponding to) a particular color and emit a wavelength that corresponds to another color. Don't worry, some noble souls have deciphered that, they follow this:

Absorb orange, emit indigo; absorb blue, emit red, etc... Green, however is a mysterious case. Also, please keep in mind that:
If $\Delta_\circ$ is less, $\lambda_{absorbed}$ is more and $\lambda_{observed}$ and vice-versa. Good to go. When Xahoc says,

you reduce the octahedral field splitting energy $\Delta_\circ$ which redshifts the absorbance of the compound.

They mean that a higher wavelength of light (on a blue-colored tone) must've been absorbed (reddish colors have a higher wavelength).

Why does $\Delta_\circ$ decrease?
Excerpts and images that follow are directly from Miessler et al. Content is copyrighted to their respective owners.
Understanding this can be quite tricky unless you're acquainted with angular overlap method. Nevertheless, let's give it a shot.
What's angular overlap method?

This is a method of estimating the relative magnitudes of molecular orbital energies within coordination complexes. It explicitly takes into account the orbitals responsible for ligand binding as well as the relative orientation of the frontier orbitals.
Modern computational chemistry allows calculations to predict geometries, orbital shapes and energies, and other properties of coordination complexes. Molecular orbital calculations are typically based on the Born–Oppenheimer approximation, which considers nuclei to be in fixed positions in comparison with rapidly moving electrons.

I'm skipping the description of $s$, $p$ and $d$ orbitals for the sake of simplicity. Further, have a look at this orbital diagram. (click on the picture for a larger image)

The beautiful colors of many transition-metal complexes are due in part to the energy difference between the $t_2g$ and $\textit{e*g}$ orbitals in these complexes, which is often equal to the energy of photons of visible light.

As an example, if the energy levels of $\textit{e*g}$ decrease, $\Delta_\circ$ decreases. A similar thing is at play in our case, albeit this time the molecular diagrams are different and $\Delta_\circ$ corresponds to a different set of orbitals, if you're still curious I strongly recommend to read about this theory, further simplifications as such are beyond the scope of this answer.


text:
Observability of Orbitals and Orbital Energies
This question comes from some thoughts I had after reading this question.
First of all, is an orbital an observable? I know the answer to this question is no because there is no "single-orbital operator" or whatever you'd like to call this. I brig this up because exactly what this means should be a part of a thorough answer. Plus an orbital is a state and we don't observe the eigenvectors but the eigenvalues.
Second, is the energy of an orbital an observable? It's more tempting to think that the answer to this is yes, but if it's true that there is no "single-orbital operator" then there must be no corresponding eigenvalue which represents the energy of this orbital. As I understand it, if one runs a HF calculation and sums the energies calculated for the individual orbitals, this is the total electronic energy. In the answer linked above, however, it is said that there is an infinite number of unitary transformations for a given set of orbitals to another set. Is there any correspondence between the energies of these transformed orbitals and the original set of orbitals? By that I mean, the orbitals themselves change, but would the optimized energies of each orbital change and it is the only the wavefunction and total energy which stay constant?
Finally, how do the answers to the above questions connect with Koopmans' theorem and photoelectron spectroscopy? That is to say, Koopmans' theorem says that the energy of the HOMO as calculated from Hartree-Fock corresponds to the first ionization energy of the system. Additionally, in photoelectron spectroscopy, it sure seems like people are observing the energies of individual orbitals, but I don't really know anything about this technique in detail so that may just be a misunderstanding.
Additionally, is Koopmans' theorem still true once we perform a unitary transformation on the Hartree-Fock wavefunction?

The reason I lump all these questions together is because I think in order to sufficiently answer the question of whether or not an orbital energy is observable, all of these points should be addressed, so I just asked them all at once. I hope it's not overkill.

This won't answer rather than comment only the the first question "is there a mean to observe an orbital".  If you equate "orbital" with "local high electron density", than I remember single diffraction crystallographic characterization of a Pt complex, nicely crystallized.  After structure solution and structure refinement, the difference Fourier map still revealed significant electron density close to the late transition metal cation that was tentatively attributed to d-orbitals not involved in bonding with the ligands around.

Well I mean observable in the quantum mechanical sense. i.e. the eigenvalues of a Hermitian operator.

@Buttowood The Q-peaks around have atoms in diffraction experiment are caused by method. We're using a spherical model for atoms (ellipsoid is still sphere) and this can't deal with ED on bonds. If you go for the mulitpol refinement this rest electron density is well described if non-spheric function (dipoles, quadrupoles etc) are used. Even if they are mathematically very similar to orbital functions they are not orbitals.

No, they are not observables, though there are many observables that can be approximately calculated from them.

I started writing up an answer, but then stumbled upon this answer in the physics.SE, that both mathematically and argumentativly explains why orbitals aren't observables. Maybe this can be of help for you.

You may be interested in http://physicsworld.com/cws/article/news/2013/may/23/quantum-microscope-peers-into-the-hydrogen-atom

Just something I didn't notice at when I initially viewed this question, but the sum of the orbital energies from an HF calculation is not equal to the total energy calculated from HF because if you simply sum the orbital energies, you wind up double counting the exchange and coulomb interactions.

Since the last part of this question got answered (https://chemistry.stackexchange.com/questions/69503/after-a-unitary-transformation-is-koopmans-theorem-still-valid), you might want to edit that in (in whatever way you wish, of course).

Related: https://chemistry.stackexchange.com/questions/16207/evidence-of-orbitals

@orthocresol so to clarify, are you looking for references that answer the first two bolded questions?

@Tyberius, I didn't really intend to be specific about which portions should be backed up with references - I just wanted a thorough answer, really.

Excluding the trivial case of a one electron system, where the wavefunction for the system is an orbital (1 electron wavefunction) and so could be measured via the electron density, orbitals are not observables. There is no known Hermitian Operator that returns orbitals (or their energies) from its expectation value. So is that the end of the discussion? Not necessarily.
In a 2006 paper, Shabazian and Zahedi$^1$ argue that what constitutes an observable depends on the conventions we establish for describing a system. They give as an example the Quantum Theory of Atoms in Molecules (QTAIM) developed by Richard Bader$^2$. Strictly speaking, the normal formulation of Quantum Mechanics doesn't allow us to describe subsystems within a molecule. We form molecules from atoms, but once the molecule is formed, there are no longer atoms; rather there is just a single wavefunction which describes the whole system. This is horribly inconvenient from a classical chemists perspective, as much of our understanding of chemistry comes from subdividing molecules into important functional groups that give the major properties of the molecule. To resolve this, Bader proposed a definition of an atom in a molecule by defining boundary conditions for a atom based on flux of the electron density. This gives an unambiguous, experimentally measurable way of determining observables like atomic charge, which without this set of external conditions is not considered an observable at all. While this has as of yet not been done, Shabazian and Zahedi argue that orbitals could be made observables by establishing a similar set of external conditions.
So while orbitals are not currently observables, what is and isn't an observable can change as new theoretical models develop.

Shahbazian, S. & Zahedi, M. Found Chem (2006) 8: 37. https://doi.org/10.1007/s10698-005-8247-4
Bader, R. Chem. Rev., 1991, 91 (5), pp 893–928
DOI: 10.1021/cr00005a013

I disagree with your last paragraph. Whether or not something is an observable is an experimental question. Also, I see what you're saying about how QTAIM allow experimental electron densities to be mapped to atomic charges, but QTAIM is not unique so one cannot really say you are observing the "true" atomic charges. Also, there is no unique way to go from an election density to a set of orbitals. This, I think, is why orbitals cannot be observable. The hard part of this question I think is explaining what is really measured in photoelectron spectroscopy.

@jheindel I'm not an expert in QTAIM, but I'm curious in what sense it isn't unique. In regards to the orbitals, I want to clarify that I'm not (and the paper isn't) saying QTAIM leads to observable orbitals, but that a similar formulation could lead to an experimentally measurable quantity corresponding to something like an orbital. In terms of photoelectron spectroscopy, what we observe is the energy of moving from a ground state to an excited state or vice versa. It can be thought of as the energy to move between different PES.

Right but the question is more whether or not these transitions correspond to orbital energies in a meaningful way. I don't think it can be said that they do. I just mean not unique in that there are many ways to partition a system into atoms and write down molecular properties in terms of these atoms. QTAIM just chooses the mathematically simplest one which leads to nice equations.

@jheindel I would say you are right. If the answer to the first and second question are no (i.e. if orbitals and their energies aren't observables) then the photoelectron spectrum can't depend meaningfully on the orbital energies or else that would constitute the orbital energies being observables.  My answer really boils down to the first paragraph, but I was hoping to give somewhat of a different perspective just to avoid giving an answer that basically just said "no".

@jheindel I think in terms of uniqueness, you could make the same argument about velocity or energy, which can vary based on what you choose as a reference frame. This is somewhat philosophical, but I think what we consider observables is more of a construction rather than something inherent about the universe.

Well I mean uniqueness in the mathematical sense. That is, QTAIM uses critical points of the electron density to define a surface which contains each nucleus only one. There are certainly other surfaces which could be made to contain each nucleus and hence would also be a valid definition of an atom. All I'm saying really is that QTAIM is not exact. If it were then it would be the correct way of partitioning a molecule into atoms.


text:
Does exotic potassium exist?
I was reading a book called 'Hundred and Seven Stories of Chemistry'.
There is a chapter about how the attributes of different elements can change in presence of enormous pressure and temperature. The book says in presence of enormous pressure and temperature potassium's 19th electron can enter the 3d orbital. Then, it would no longer be a metal.
Is it possible? If yes, is there any theoretical calculation possible for the pressure and temperature needed for this exotic behaviour?

Related: metallic hydrogen and solid oxygen.

"No longer a metal"?  Still a metal I would guess, but it would differ in some ways from an alkali metal.

@KlausWarzecha Metallic hydrogen turns me on! The core of the planet Jupiter contains this, probably due to high pressure and temperature.

related https://chemistry.stackexchange.com/questions/27920/why-is-nacl3-possible? https://chemistry.stackexchange.com/questions/19913/since-the-formation-of-diamond-from-graphite-is-endothermic-and-the-entropy-decr https://chemistry.stackexchange.com/questions/115429/what-would-superionic-water-ice-look-like

Another interesting paper (arXiv:1310.4718 [cond-mat.mtrl-sci]) on this topic. I think you should be able to access this one, but I found three passages worth noting in the article.

Peculiarities in structural behaviour of K under pressure are based on specific features of its electron configuration: K opens in the Periodic table the first long period with 3d transition metals. The empty 3d electron band in K is just above the valence electron 4s level and under pressure there is s – d transition [16]. In heavy alkalis Rb and Cs that have d-levels in their cores electron transitions start at lower pressures than in K. In lighter element Na the 3d level is well above the 3s valence electron level therefore fcc in Na is stable up to a very high pressure of ~100 GPa.

It seems that this is saying that the structural change away from FCC marks where the electron transition occurs. In the paper, they say that the FCC structure stops being the most stable for $\ce{K}$ after $\pu{20 GPa}$ (giga or billion pascal) at $\pu{300K}$.

We suggest that in potassium at densities higher than V/Vo ≈ 0.3 there is not only s – d electron transfer but also core electrons start to participate in the valence band and K becomes a polyvalent metal with the number of valence electrons counting to 2 in oP8 and even to 4 in oC16.

This is saying that for some of the even higher pressures, that even some of the core electrons are moving to the valence band.
This passage from the introduction of the article above does seem to support your view that, for some cases, compression can make a metal less metallic.

Na, one the best metal in terms of conductivity and reflectivity, becomes insulating and transparent at around 2 Mbar [4]. Lithium becomes superconducting [5] below 20K at pressures around 0.5 Mbar and at higher pressures shows a series of complex structures even with a semiconducting behaviour [6,7].

We wouldn't expect metal to exhibit insulating behavior.

Indeed an interesting paper. Now, at V/V0 of 30% it is not so surprising that more electrons come in to play - it would be more surprising if they didn't!. Further, in both the papers that we have referenced I see no mention of the high-pressure phases not being metals.

I would agree that there is nothing about them not being metals. If anything, the trend for most materials seems to be that they become more metallic under greater pressure.

Uh ... alkali metals under normal conditions are fcc?  Not bcc?

@Oscar Lanzi I left out that they all start bcc and are that way up until around 5-10 GPa depending on the element. The authors note the transition out of fcc is when they start to see this strange electronic character.

Why heavy alkalis Rb and Cs that have d-levels in their cores electron transitions start at lower pressures than in K?
I found the change in reflectibility very interesting. Why would s-d transition cause this?

First, a point that is more like a comment: I'm not sure that just because electrons get put in to d-orbitals (and thinking of crystal band structures in terms of atomic orbitals is of questionable value anyway) that it would not be a metal. Many elements (one could even say most elements) with d-electrons are quite happy as metals.
You might want to start with L.F. Lundegaard et al., Phys Rev B80 020101(R) (2009), where they do DFT calculations on potassium at different pressures. I will quote from their conclusions section:

The observation of the oP8 structure in K reveals it to
  have a pressure-induced structural behavior like that of Na,
  as well as having behavior like Rb and Cs as indicated by the
  observation of the tI4 and oC16 phases. The similarity to Na
  is unexpected in relation to the recent electronic structure
  calculations of Ma et al.$^{19}$ As noted, they found K, Rb, and Cs to have the same high-pressure structural sequence, and
  they attributed this to these heavier alkalis’ being dominated
  by d electrons arising from s-d transfer, whereas the ultrahigh
  pressure behavior of Na arises, they suggest, from p-d
  hybridization. Thus our results question this picture of a
  simple dependence of structural stability on d-electron dominance
  or p-d hybridization.


text:
Why does pyridine make such a big change from SNi to SN2?
When an alcohol is treated with SOCl2 alone, retention through SNi takes place, whereas adding pyridine makes the reaction follow SN2 mechanism and inversion takes place. What concept lies behind this occurrence?

Both cases share, that the reaction of the alcohol and the thionyl chloride yields an alkyl chlorosulfite under liberation of HCl. Then

Either the C-O bond of the alkyl chlorosulfite brakes, a (tight) ion pair of carbo-cation and chlorosulfit anion is formed.  The later decomposes into $\ce{SO2}$ and $\ce{Cl-}$, the latter reacting with the carbo-cation.  This yields the rentention of configuration observed.

.OR.

In presence of pyridine, HCl generated does not leave the reaction mixture (as above), but reacts with pyridine to yield pyridinium chloride.  Pyridinium chloride is able to react along SN2 on the alkyl chlorosulfite.  Seeing this variant from the beginning till the SN2 of the alkyl chlorosulfite yields the inversion of configuration observed.


text:
Why green leaf turns brown in Brine solution?
I immersed a semi dry green leaf in glass of concentrated NaCl solution . I left it in mild sunlight for around 3 hours . Solution before was clear but afterwards it became reddish brown and the color of leaf turned brown. I want to know why this happened? Please help.

It is most likely the $\ce{Na+}$ is formed a complex salt with chlorophyll (bulky group) subsequently displacing $\ce{Mg+}$ cation, even though $\ce{Na+}$ doesn't have extra electronic shells for complex ion stabilisation, instead the chlorophyll (through the $\ce{N}$ atoms with lone pairs) provided extra stabilisation of the positively charged $\ce{Na+}$ cation.
Chlorine dissociated from $\ce{NaCl}$ to form aqueous chloride ions $\ce{Cl-(aq)}$ forming a typically greenish-brown solution you noticed during your experiment.

Are you saying that chloride ion is greenish brown? If so, why is a sodium chloride solution not brown?

Not necessarily, because when isolated it exist as pale greenish-brown gas, however due to the presence of different molecules and aqueous environment several solvents are present together with the chloride ions, and may inhibit the Chloride ions to explicitly show it's typical greenish-brown colour, remember other minor products are also present in solution with the Cl- ions.  Also to note in the leaves there is also Fe³+ ions , subsequently can be attributed to the overall color of solution, if we to extract all the "impurities" to leave only chloride ions we would get a color close to green

NaCl would not be green because of the lack of "outer shells with vast electrons" so would not get excited to 3d states where it absorbs light, emit some to produce color like what transition elements do. For our complex solution above, the chlorophyll group ( with N lone pairs of electrons) allowed the complex salt to be activated hence emit some light energy to produce color.


text:
Toluenesulfonate vs. methylsulfonate
Are there any advantages/disadvantages of using toluenesulfonate instead of methylsulfonate as a leaving group in E2 and SN2 reactions? I know they are both excellent leaving groups but how do they compare to each other? Is there a green chemistry aspect to it?
Thank you!

From the point of atom efficiency, methylsulfonate is advantageous compared to the p-toluenesulfonate.
On the reagent itself, the methanesulfonate is more polar than the p-TsOH; so the later dissolves better in less polar organic solvents than the former.  Methanesulfonate even may be used to dissolve some metal salts.
If used to generate a better leaving group, on workup, a methanesulfonate enters the aqueous phase easier than the tosylate.
p-TsOH is a cheap, widely available solid (sometimes sold as hydrate) you may prepare by your own in large quantities; methanesulfonate however is a liquid, hence potentially easier to add continuously, and more expensive (even more so, per mL, the triflates).
Methyl sulfonates are able to alkylate DNA, hence potentially carcinogen.  For tosylates, currently there is not such evidence (but obviously the dust of this strong acid is considered harmful).

As a personal note, among other reasons I found consultation of journals like Organic Process Research & Development just under this aspect (including green chemistry) as useful in two ways: i) publications describing how running a particular reaction within a chain of others works well / better than an other -- especially on larger scales of application--, and ii) the sections in line of "Some Items of Interest to Process R&D Chemists and Engineers".
Addendum:  There are dedicated compilations to figure out scope and limitations of chemical reagents.  To mention only two, Fieser & Fieser and EROS edited by Leo Paquette (there is an e-EROS as electronic ressource, too) are multivolume references.

Do you have any citations at hand for literature on MSA alkylation of DNA?

@hBy2Py Thanks to NIH's policy, one I find in PubMed's data bases is Lundin et al. in Nucleic Acids Research, 2005, Vol. 33, No. 12
3799–3811, doi:10.1093/nar/gki681, found at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1174933/pdf/gki681.pdf

In my experience mesylate forms more quickly using std conditions of MsCl/Et3N/DCM.


text:
Changing Zinc acetate dihydrate to Zinc acetate
For some Chemistry work in the laboratory, I needed Zinc acetate but laboratory only has Zinc acetate dihydrate. Is there any way to convert hydrate form to anhydrous form? Maybe by heating or any other method?
If by heating, how much time and temperature I should give?

Heating would work.

Wikipedia is your friend...Zinc Acetate

Decomposes at 237 °C (459 °F; 510 K)
    (dihydrate loses water at 100 °C)

So heating to 120-130 °C for several hours would remove the water, but not be hot enough to decompose the salt. The dried salt will be hygroscopic of course so store it in a desiccator.