text:
What are the different types of van der Waals forces?
As per different websites on internet van der Waals force are of

Three types (Keesom force, Debye force and London dispersion force) @ Wikipedia
Two types (Dipole-dipole force/Keesome force and London dispersion force) @chemguide and chemwiki

While an expert website  also explained H-bonds to be a van der Waals force.
I'm confused. What is originally correct?

Believe Wikipedia before some other "expert" website that says "van der Waals' forces are forces that exist between MOLECULES of the same substance." Since ethanol and water aren't the same substance, I was really surprised to learn that there can't be any hydrogen bonding between them. ;-)

Wikipedia is right if you accept the IUPAC definitions:http://goldbook.iupac.org/V06597.html

When dealing with such forces and definitions, it is also important to remember that the distinctions are a bit artificial. In nature, you have the fundamental electromagnetic interaction and that's it.

The definition of van der Waals forces according to IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8., is as follows:

The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces. The term is sometimes used loosely for the totality of nonspecific attractive or repulsive intermolecular forces.

The phrase "other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules" would seem to imply that according to the IUPAC Gold Book, if a force is not an ionic or covalent chemical bond or electrostatic interaction, then it is a van der Waals force.

Slight correction on @airhuff's answer.
"if a force is not an ionic or covalent chemical bond or electrostatic interaction, then it is a van der Waals force." The IUPAC Gold Book's definition precludes "electrostatic interaction of ions or of ionic groups" from being VdW forces but all VdW forces are electrostatic in nature: forces between electrostatic dipoles, either permanent, induced or spontaneous.

https://doi.org/10.1351/goldbook.V06597 (direkt link)

There are too many definitions for something that seems to be incompletely understood and it is best to look at it operationally.  There are forces of attraction between molecules. We know this because it is observed. Different types of molecules have different forces. I will only talk about the weakest of these, the "pure" van der Waals forces between simple molecules that shouldn't attract each other, but do, such as the inert gases especially He and Ne and the nonpolar diatomics such as H2 and the Halogens and possibly to the alkanes, N2 and O2. London dispersion forces arise because of the temporary dipoles in molecules by "motions" of the electrons about the nuclei. I have a problem with this explanation probably because I just don't understand it and have trouble understanding why it is only attractive.  The explanation I was taught was that as molecules approached each other the electron clouds were displaced more or less tangentially exposing the nonvalence electrons to the nucleus of the other molecule The distortion of the orbitals is of slightly lower energy than the attraction hence a positive attraction until the molecules got too close. A simple conversion of kinetic to potential energy. This has a problem with H2 and He because no inner electrons but I can imagine a distance where there is an attractive force. With alkanes the slight bond polarity will give some attraction and perhaps more induced dipoles, with actual polar bonds the dipole dipole attraction is evident. Hydrogen bonding is a great leap upwards in attraction and I wonder why these are considered types of van der Waals forces when they can stand on their own.  So the various definitions seem, to me at least, a bit arbitrary.


text:
Is the structure of a crystalized protein the native one?
When one finds the 3D structure of a protein by crystalizing it and then making a X-ray experiment, how does one know that the geometrical configuration of the crystal is the same (or even close) to the structure the protein adopt in aqueous solution?

Well, it doesn't need to be, be it's still something. NMR is getting structures in solutions.

@Mithoron And they're doing fancy stuff with proteins NMR, but I don't see how they would get a whole configuration. That would be amazing.

@DGS Significant percent of protein structures was obtained with NMR not X-ray. http://www.rcsb.org/pdb/static.do?p=general_information/about_pdb/nature_of_3d_structural_data.html

@Mithoron And to think I have trouble assinging something that weighs around 500 g/mol.

I'll use quotes from B. Rupp, Biomolecular Crystallography (p. 7-8) to answer.
Generally the structure is similar...

Comparison of many nuclear magnetic resonance (NMR) solution structure
  ensembles with crystallographic structure has shown that the core
  structure of protein molecules remains unchanged compared with the
  solution state during crystallization. In addition, enzymes packed in
  crystals even maintain biological activity.

with some parts not visible in the experiment...

The maintenance of the core structure and of enzymatic function shows
  that crystal structures are a very good approximation of the native
  protein solution structure. Nonetheless, highly flexible or mobile
  regions, frequently the amino- or carboxyl-termini of the protein chain or
  flexible loops connecting secondary structure elements, can be
  poorly defined or even absent in the electron density and thus can be
  modeled only with limited confidence.

but beware...

In certain situation flexible and dynamic regions of a protein
  molecule can be rigidly fixed in a specific conformation as a result
  of crystal packing interactions. In most cases this represents just a
  snapshot of one possible conformation out of many and it must be
  understood that such a specific conformation may not locally represent
  the protein structure in solution. A simple safeguard against
  misinterpretations -- which is usually assignment of certain
  biological relevance to regions where that is de facto not warranted --
  is to display all neighboring, symmetry-related molecules in the crystal
  structure and examine if any intermolecular interactions are
  present that are a result of crystal packing. Such packing induced
  artifacts can also hamper for example drug discovery by altering or
  blocking binding sites and thus preventing an otherwise active
  substance from binding.

source: Stack Exchange

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

sort: votes


text:
Dipole moment of molecule (isolated vs in solution)
Would you expect a dipole moment determined in solution to be different from that in an isolated molecule? If so, why?

In short, yes. In more detail, that would depend greatly on the nature of the solvent (polar or non-polar, etc.)

To mention a prominent example: For water it is suggested that the increase in dipole moment upon solvation "originates from the decrease of the angle between the dipole vectors of the lone pairs on oxygen as the number of hydrogen bonds to that oxygen increases. Thus, the decreased angle, and the consequent increase in water dipole moment, is most likely to occur in environments with a larger number of hydrogen bonds, such as the center of a cluster of water molecules" (see: here).

I've actually been investigating this in my comp chem coursework. A large dielectric can support more charge separation and therefore stronger dipoles, but the effect diminishes. Going from an $\epsilon$ of 2 (cyclohexane) to 9 (DCM) gives a much bigger change in dipole than going from 9 to 80+ (water).

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


text:
How does concentration affect The Position Of Equilibrium?
I was taught that
when the concentration of the reactants is greater than the concentration of products then the position of equilibrium is shifted towards the left.
However when I used La'Chateliers Principle then we know that if you increase the concentration of reactants then the system wants to revert that change a hence produce a higher concentration of products (and subsequently a lower concentration of reactants) but wouldn't this then mean that the equilibrium would shift to the right?

Related: What does it mean to shift equilibrium?

You statement of Le Chatelier's Principle is a perversion that isn't true. The equilibrium does not depend on the sum of the concentrations of the reactants relative to the sum of the concentration of the products.
Your analysis is correct.

Words confuse sometimes. "Shifting" is usually used in the context of Le Chatelier's Principle, and is a verb. K is the equilibrium ratio of concentrations of species to their stoichiometic powers, and Q is the ratio of the concentrations of species raised to their stoichiometric powers at any point in time. When Q>K, there's more products than there should be. You can think of the reaction as "tilted" toward the right. The reaction will then "shift"/move to the left, trying to reduce its Q by forming reactants from products until Q=K.

This makes no sense. What are Q and K?!?


text:
Which pH approximation to choose when?
When I learned about pH approximations in school, I was told that each approximation fits to a certain acidity strength/pKa range. Those ranges are grouped in common categories (pKa borders are only approximately, as they change from book to book):

very strong acids: $\mathrm{pK}_a < 0$
strong acids: $0 \le \mathrm{pK}_a < 4$
medium strong acids: $4 \le \mathrm{pK}_a < 8$
weak acids: $8 \le \mathrm{pK}_a < 14$
very weak acids: $\mathrm{pK}_a \ge 14$

From this point on, the arguments were made and approximations for certain strengths of acids were evaluated. But is it possible to choose the "right" approximation only based on acidity strength? Do those equations really only apply to it's categories?
This question originates in my wish to implement a program on my calculator that computes the pH for a given concentration and pKa value. I thought it might be wrong to simply ask for the pKa value and choose one of the approximations solely based on that.

Preface
It is being assumed throughout this post that the concentration c equals the activity a. This is of course not true for concentrations above $\approx 10^{-3} \ldots 10^{-2}~\mathrm{mol\,L^{-1}}$ but it is common practice in schools and universities as long as the exercise is not very specifically about activities.
$$\mathrm{pH} = -\lg a(\ce{H+}) \xrightarrow{a=c} \mathrm{pH} = -\lg [\ce{H+}]$$
The other general assumption is that temperature is $22~^\circ \mathrm{C}$ and therefore $k_\text{W} = 10^{-14}~\mathrm{mol^2\,L^{-2}}$.
Given error contours are not "real measurements vs. calculations" but absolute differences between approximation n and the cubic equation that is not more simplified than using concentrations instead of activities.

General
If you have a calculator that can solve cubic equations, you don't need to simplify. In addition you only would have to remember one equation, no matter which pKa or concentration occurs (see above for real world restrictions).
Solve the following equation for x, with $x=[\ce{H+}]$:
$$x = \underbrace{\frac{k_\mathrm{W}}{x}}_{[\ce{OH-}]} + \underbrace{\frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}}_{[\ce{A-}]}$$
or as a single equation (that is harder to remember):
$$x^3+x^2 k_\mathrm{a}-x([\ce{HA}]_0 k_\mathrm{a}+k_\mathrm{W})-k_\mathrm{a} k_\mathrm{W} = 0$$
Results from this equation will be the reference pH values for the following comparison of pH approximations.
To calculate the pH for any monoprotic(?) base, simply replace [$\ce{H+}$] with [$\ce{OH-}$], $k_\mathrm{a}$ with $k_b$ and of course [$\ce{HA}$] with [$\ce{B}$].

Approximation 1 and 2
For very strong acids the approximation is that they are completely dissociated, which makes the resulting equation very easy:
\begin{align}
x &= [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg [\ce{HA}]_0
\end{align}
When am I able to use this approximation with an error that is smaller than $\Delta \mathrm{pH} \leq 0.1$? It's the blue area in the picture below ($\mathrm{pK_c} = -\lg [\ce{HA}]_0)$.

The simple rule is: $6.7 \gtrsim \mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5$.
Including the hydroxide ions from the autoprotolysis of water makes it a little bit more complicated and yields a quadratic equation in x:
\begin{align}
x &= [\ce{OH-}] + [\ce{HA}]_0 = \frac{k_\mathrm{W}}{x} + [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg \left(0.5 [\ce{HA}]_0 + \sqrt{0.25 [\ce{HA}]_0^2 + k_\mathrm{W}} \right)
\end{align}

This formula holds within: $(\mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5) \vee (\mathrm{pK_c} \gtrsim 7.4)$.
It's not surprising, that the range of usefulness increases over the whole are of diluted concentrations. And that the high pKa values are also covers is only due to the fact that the pH is 7 everywhere upon concentrations that are below $10^{-7}\ldots10^{-8}$.

Approximations 3 & 4
The following approximations are used for weak acids. It's the second easiest to remember, so it's used quite often. The assumption is, that $x \gg k_\mathrm{a}$.
\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a}} = 0.5 (\mathrm{pK_c} + \mathrm{pk_a})
\end{align}

It's application range is: $(\mathrm{pK_c} < 13.7 - \mathrm{pk_a}) \wedge (\mathrm{pK_c} < \mathrm{pk_a} - 0.6)$.
It is obvious, that the approximation is useful in a small area where the acid is not too diluted and it's pKa value is not too low. Seems to fit well for weak acids.
Again, what happens if $k_\mathrm{W}$ gets involved involved?
\begin{align}
x &= \frac{k_\mathrm{W}}{x} + \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a} + k_\mathrm{W}}
\end{align}

Again, the range expands over very weak acids and strong diluted solutions as expected: $(\mathrm{pK_c} < \mathrm{pk_a} - 0.6) \vee (\mathrm{pK_c} > 14.2 - \mathrm{pk_a})$.

Approximation 5
The last approximation is for strong acids that are not diluted. The amount of protons in solution is equivalent to the amount of dissociated acid but this time with taking $k_\mathrm{a}$ into account. As only the influence of the autoprotolysis of water is neglected, it is "the most accurate" approximation for high to medium concentrations. This can easily be seen as it includes the scope of both approximations, 1 and 2. It is again a quadratic equation in x.
\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}\\
\mathrm{pH} &= -\lg(-0.5 k_\mathrm{a} + \sqrt{0.25 k_\mathrm{a}^2 + [\ce{HA}]_0 k_\mathrm{a}})
\end{align}

It can be used if: $(\mathrm{pK_c} < 6.7) \wedge (\mathrm{pK_c} < 13.7 - \mathrm{pk_a})$.

Now there are all those approximations, with all their respective acceptable ranges but it's quite hard to compare five diagrams with another. I made two further pictures that show 1) which approximation is the best per (pkc,pks)-pair and 2) which approximation is the easiest to remember per (pkc,pks)-pair.
1) The best -- which function yields the lowest error in comparison with the exact equation?

2) The easiest -- which function yields the lowest error in comparison with the exact equation and if there is more than function with $\Delta \mathrm{pH} < 0.1$, which one is the easiest to remember?

So ... the pka ranges kind of resemble when to use which function but in the end it's a little bit more complicated. If $k_\mathrm{W}$ is neglected, the low-error-area (LEA) is quite small, only approx. 5 is able to yield acceptable values for a big range of concentrations and pka values. Approximations 1 and 3 should be used with care, as their LEAs are quite fancy. For approximations 1 and 3 it's often better to switch to their low-concentration counterparts 2 and 4 to be on the safe side.

-1 for thinking that you can calculate to an error of < 0.01 pH units up to a concentration of 1 M without considering activity.  pH = -log (H+ activity), Ka = a(H3O+)a(A-)/a(HA)a(H2O), Kw=a(H3O+)a(OH-)/(a(H2O))^2

@DavePhD Thank you for your comment. Is your problem about the very wide pKc scale or about the 0.01 error contour in the images? You might have noted, that this Q/A is not about the calculation of "real" pH values using acitivities but about which of the approximation yields the lowest errors for a certain (pkc,pka)-pair with respect to the "exact" equation. This is all within the very common treatment of schools (and also universities) to ignore the activity. I am well aware, that those assumptions are wrong for concentrated solutions and I have mentioned that a few times here on chem.se.

My criticism corresponds to the high concentration region of the graphs (0.01-1 M) where error would definitely exceed 0.01 units and even 0.1 units close to 1M.  Also the statement "no matter which pKa or concentration occurs" makes it sound like the cubic equation would give an exactly correct answer at all concentrations.  I don't seen how someone reading the question or answer would think they aren't about approximating real pH.

The new preface definitely helps.  There is still the issue of not including the concentration of water in Ka and Kw, in other words assuming that the water concentration is always the concentration of pure water, regardless of how much acid there is.  Obviously with a large enough molecule (an enzyme for example), the concentration of water would approach zero at even a low concentration of acid.  But for say benzene sulfonic acid at 1M concentration, how much error is introduced by the  assumption that water concentration is the concentration of pure water?

You are right, but that's way out of the scope of this Q/A. Have you ever seen any pH exercise, which does not want to explicitly emphasize on the limits of the "normal" calculations, that covers what you are questioning?

I don't see why it is out of the scope  of the question.   In a previous millennium, my research and publications involved determining the pKas of amino acid groups in proteins by NMR.  The protein solutions  would be concentrated in terms of mass fraction, but still much less than 1M, and I would consider the decreased concentration of water.   Have I ever seen a pH homework exercise that considers water concentration?  I'll try to find one.  My college physical chemistry text (Levine) included water activity in the Ka and Kw definitions and then explained approximations from there.

maybe an example of considering water concentration in an exercise is "As my student was taught to ignore water concentrations...Unfortunately for her (and for many other Singapore candidates in 2010), this was not what the Cambridge examiners were expecting. Since water was part of the reaction, examiners were expecting candidates to include the water concentration" http://www.wewwchemistry.com/2013/07/water-in-equilibrium-constant-expressions.html

I don't think that this is an example for what we are talking about here. It is an example for the explicit usage of water concentration to calculate equilibrium constants, not for explicit usage of water concentration during pH calculations.

I like the way the issue is discussed here: "As long as we are talking about diluted solutions we may safely assume water concentration is constant, and error introduced into our calculations will be rarely higher than the precision of dissociation constants used in calculations" http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

@pH13-YetanotherPhilipp thank you for your effort ,how can I prove this relation :
$$[A^- ]= \frac{[\ce{HA}]0 k\mathrm{a}}{x+k_\mathrm{a}}$$


text:
Why can't HS- participate in E1 reactions?
In my textbook (Organic Chemistry by Klein), it says that certain molecules can only act as nucleophiles and not as bases; thus they can only undergo substitution reactions, not elimination reactions. This group of molecules includes: $\ce{Cl- , Br- , I- , HS- , H2S , RS- , RSH}$. I know that while E2 reactions require strong bases, E1 reactions do not. Even a base like $\ce{H2O}$, which is pretty weak, can undergo E1 reactions. Then why can't something like $\ce{HS-}$ undergo a E1 reaction? For example, if we have a tertiary alkyl halide and $\ce{HS-}$, I would expect a mix of $S_N1$ and E1 products. However, my book claims there are only $S_N1$ products. Why is this the case?

I think "can not" is probably not the right description of its reactivity, and "does not" is more accurate.

$\ce{H2S}$ is a much better nucleophile than $\ce{H2O}$. $\ce{HS^-}$ is even more so. The diffuse nature of sulfur's electronic radius 1. makes it more acidic (less attracted to $\ce{H^+}$) and 2. gives it a higher HOMO. This allows it to react with the low LUMO of a C-X bond $\sigma ^*_{\text{CX}}$ rather than the small, dense s-orbital of $\ce{H^+}$.

@DGS But isn't $\ce{HS-}$ still a stronger base than $\ce{H2O}$? So shouldn't it undergo E1 better than $\ce{H2O}$ can? I understand that for $\ce{HS-}$, substitution is the predominant reaction, but does this mean that elimination doesn't happen at all?

Think about the competitive reactions. If it is a much better nucleophile, the nucleophilic on a carbocation attack would be a significantly faster reaction relative to deprotonation.

As has been pointed out in the comments, the use of the word only is a little misleading. What it should really say is that $E1$ is possible but $S_N1$ is so strongly favoured that to all intents and purposes the only product is the $S_N1$ product.
This can be explained in terms of the relative energies and sizes of the orbitals involved. $\ce{HS-}$ has a large, diffuse, high energy HOMO which can have much better overlap with the empty $2p$ orbital of the carbon than the small, low energy $\ce{C-H}~\sigma^*$. Therefore, the $S_N1$ reaction is strongly thermodynamically favoured and dominates.
If you compare this to something like $\ce{H2O}$, the HOMO is much more compact and lower in energy and so it has much better overlap with the $\ce{C-H}~\sigma^*$ and so $E1$ is comparatively much more favourable.


text:
Calculating the molarity of an acid solution

Concentrated hydrochloric acid is $37\% \ce{HCl}$ by mass and has a density of $1.2\rm~\frac{g}{ml}$. Calculate the molarity of a solution made by diluting $125\rm~ ml$ of concentrated $\ce{HCl}$ with water to a total volume of $2\rm~ L$.

I got $194.88\rm~M$ but that's not correct. First, I found the molarity of the $37\%$ solution to be $12.18\rm~M$; then I used the $M_1V_1=M_2V_2$ formula.
Finding Molarity
$$\rm\frac{100g}{1.20~\frac{g}{mL}}= 83.3~mL = 0.083~L$$
$$\rm\frac{37~g~HCl}{36.46~\frac{g}{mol}} = 1.015~mol~HCl$$
$$\rm\frac{1.015~mol}{0.083~L}= 12.18~\frac{mol}{L}$$ (This answer was marked correct on my hw)
Finding molarity of concentrated HCl
$$M_1V_1=M_2V_2$$
$$M_1\times(0.125)=(12.18)(2)$$
$$M_1=194.88~\rm\frac{mol}{L}$$

You had the correct formula but plugged in the numbers wrong. For dilutions the formula is: $$\mathrm{M}_1 \mathrm{V}_1 = \mathrm{M}_2 \mathrm{V}_2$$ where

$\mathrm{M}_1$ is the original molar strength
$\mathrm{V}_1$ is the original volume
$\mathrm{M}_2$ is the diluted molar strength
$\mathrm{V}_1$ is the diluted volume

The formula works because $$\mathrm{M} \mathrm{V} = \mathrm{(M)(L) = (moles/L)(L) = moles}$$ so regardless of how much dilution you have the same number of moles of reagent.
So for this particular problem the final part of the solution is: $$\mathrm{(12.18M)(0.125L)}=(x)\mathrm{(2L)}$$ or $$x = \dfrac{\mathrm{(12.18M)(0.125L)}}{\mathrm{(2L)}} = \mathrm{0.761 M}$$


text:
Naming of elements 113, 115, 117, and 118
Does anyone know of a blog, website, news server, or listserv to which one may subscribe to be notified when these elements get their official names?

According to this press release from 8 June 2016, IUPAC proposes names for the four new elements as follows:

nihonium and symbol Nh, for the element 113,
moscovium and symbol Mc, for the element 115,
tennessine and symbol Ts, for the element 117, and
oganesson and symbol Og, for the element 118.

The public review (comments by 8 November 2016) of the provisional recommendations can be found here.

Take that, tosylate users.

IUPAC will certainly have a press release on their news page, given that they are officially responsible for the names.
Currently, that's at this address: http://www.iupac.org/news/latest-news.html
Sadly, they don't appear to have an RSS feed. (Not even an Atom feed [rimshot noise].)

I guess I could change my morning routine to include a visit to this page.


text:
Mechanism in formation of acetals

someone asked a question about this mechanism and there were 2 answers as shown below.

but these two answers seem contradict.
and my questions are
  1. which one is right?
2. does the protonation by H2SO4 and free proton both mean the protonation by H3O+?
3. I know sp2 hybridized C is more electronegative, but at the same time it's losing e- to oxygen through pi bond. is C=O still less polarized than C-O even considering pi bond?
4. in answer #2, it's saying like free proton(H3O+?) is less acidic than hydrogen attached to positively charged oxygen(ROH2+?). how is this right?
Hope somebody can help me understand it !!

The first answer is wrong, and the second answer is somewhat true (alcohols are more polar than carbonyls, but not for this reason), but unhelpful.

You use protonated ethanol because it is the most likely acid present.

Let me provide an analogy.
Sulfuric acid is a strong acid. When sulfuric acid is dissolved in water, it ionizes (reacts with water)to an extent that only the tiniest fraction of $\ce{H2SO4}$ is left:
$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$
In an aqueous solution, its fair to say that the concentration of hydronium cation outweighs the concentration of free sulfuric acid by quite a lot. If you were drawing an acid-catalyzed mechanism in water, would you use $\ce{H2SO4}$ or would you use $\ce{H3O+}$?
This reaction is done in ethanol as the solvent. The basicity of ethanol is pretty similar to water, and we have a lot of ethanol, so we can expect the sulfuric acid to ionize to a great extent. These acid-base equilibrium shenanigans reaction would occur regardless of whatever else we added to the mix.
$$\ce{CH3CH2OH + H2SO4 -> CH3CH2OH2+ + HSO4-}$$
Now, if we are doing an acid-catalyzed reaction in ethanol using a strong acid, which species should we use as the proton source when drawing a mechanism: $\ce{H2SO4}$ or $\ce{CH3CH2OH2+}$?
Note that this is an example of the "leveling effect". The strongest acid that can exist in your solution is the conjugate acid of your solvent, and the strongest base that can exist in your solvent is the conjugate base of your solvent. Anything stronger reacts with your solvent.

Thank you! Now that I saw your answer I realize I was asking so simple thing T_T.

@NKYu: You may consider it simple, but go easy on yourself. The fine details of this issue are overlooked by many organic chemistry students (and instructors). I find myself often taking the lazy way out and just using $\ce{H+}$. By the way, if you are new to the Stack Exchange experience, you can upvote good answers and accept answers that were most useful to you for your question.


text:
A way to turn KCl into KOH is needed
I figured out that KCl + NaOH may result in KOH but I think this reaction can not happen in an aqueous medium. So there I have a couple of questions.

Is there a way to make the reaction I stated above to happen?
Is there a way to turn KCl into KOH?

I think the reaction does not works very well, because you get an equilibrium of KOH and NaOH.
If you have KCl in an aqueous solution you can use AgOH to get KOH. AgCl precipitate and the solution contains only KOH. If you like to get KOH it is better to buy them.

On an industrial scale, the preferred method for the production of potassium hydroxide from potassium chloride is electrolysis (diaphragm, mercury, or membrane process).
$$\begin{align}
\ce{2Cl- &-> Cl2 + 2e-}\\
\ce{2H2O + 2e- &-> H2 + 2OH-}\\
\hline
\ce{2Cl- + 2H2O &-> Cl2 + H2 + 2OH-}
\end{align}$$
or
$$\ce{2KCl + 2H2O -> 2KOH + Cl2 + H2}$$


text:
Multiphasic liquid "flotation" experiment
I'm trying to create an experiment where two immiscible liquids are mixed and are then separated by density. Most of the container is occupied by the lighter liquid, and the small amount of heavier liquid sinks.
Then, a third liquid, immiscible with the other two is added, in amounts intermediate between the two. This liquid is lighter than both of the other liquids and floats.
However, this 3rd light liquid can wet the dense liquid, and after mixing it carries it upwards to the top of the container, basically creating a flotation process.
Here's a rough sketch of what I want to achieve:

I don't care if after the last step the dense liquid then sinks again through the light liquid. If the dense liquid is elevated, even temporarily, that's good enough.
This is slightly similar to industrial flotation processes that separate ore minerals from a slurry of crushed rock. Air bubbles "wet" surfaces of heavy minerals such as copper sulfides and lead sulfides, and float them to the top of a container. In my case I'm talking about three liquids and not a liquid, solid, and a gas.
The reason I'm asking is that I created something similar at high pressure and temperature (~1 GPa, 1000 C) in a sealed pressure vessel and I'm looking for a good way of demonstrating this phenomenon in a more accessible way.
Is there such a thing? I have access to common (and some uncommon) chemical reagents. The liquids have to be transparent, obviously, to see the entire thing. Preferably coloured, or possible to colour them.

That's a tough one. When two liquids "wet" each other, they usually also dissolve each other, so your 1st and 3rd would end up being just one liquid. You may be able to adjust its density so that it would float as a sphere in the 2nd liquid, but that's another story.

@IvanNeretin it also could be that instead of adjusting density, you could adjust the phase proportions..

Do you mean that a third liquid causes a phase inversion?

@Beerhunter what do you mean by phase inversion?

I read an older version of the post and missed that the 3rd liquid is also immiscible. I have experienced dissolving a (solid) product which has caused the normally less dense phase to actually be the lower layer in a water-solvent mixture. This is what I'm referring to as a phase inversion. Your query is different due to the most dense phase being dispersed in your least dense phase.

https://www.youtube.com/watch?v=GRPufqPgTeI
I played around a bit with this in the lab and this is what I came up with.
Basically, I made a three phase system with hexanes on top, water with fluorescein and green food colouring in the middle, and a perfluoroalkylether HT-110 on the bottom.
Just finding three mutually immiscible liquids was pretty challenging and unfortunately, like most highly fluorinated liquids, HT-110 is almost twice as dense as the other two liquids, but if you look at the video carefully, you can see that some bubbles of HT-110 are getting trapped in the hexanes as its surface energy isn't quite as high in hexanes as it is in water.
When I tilt the vial on its side, you can see these bubbles snapping back across in the interface.
It's not perfect, uses a goofy solvent, and I probably could have played with the liquid ratios/container shape to trap more HT-110 at the top, but it seems that something at least similar to what you're after can be made fairly simply.

So let's see if I understand what's happening. You first have the three liquids in layers. When you shake it, "as its surface energy isn't quite as high in hexanes as it is in water" - you mean that it "wets" it preferentially? Anyway, the bubbles of HT 110 are trapped in the hexane, and they can't sink down through the water because they "stick" to the hexane. When you tilt it, they sink again to the HT 110 again.

Would it be possible to add colouring to either the hexane or the HT 110? Does it matter if it's HT 110 or any other fluorinated liquid? Does it matter which hexane?

I actually tried adding a porphyrin dye to the hexane, but it wasn't soluble. There are many dyes that would be soluble in the hexane and not the water, I just don't have any handy. The HT-110 isn't likely to dissolve many normal dyes. I don't think it has to be HT-110. Most highly fluorinated liquids have these kind of omniphobic characteristics. (Sigma sells FC-40 and the like) It was a mixture of different hexane isomers—your normal lab solvent grade.

I don't know if I would call the interaction wetting exactly, as the term usually refers to liquid-solid interactions, but what I mean is that it's more energetically costly to have a HT-110/water interface than a HT-110/hexane interface, which means, for a certain size of droplet, gravity isn't enough to pull the droplet through the water.

That's why I was referring to "wetting" in quotes :)
Thank you! This is exactly what I was looking for.

Three phases is easy for the home chemist: 70% rubbing alcohol (isopropanol), table salt, and vegetable oil will do. The salt basically makes the water more polar so that the isopropanol is forced out.


text:
What does "parts by weight" or "parts by mass" specifically mean?

Equivalent mass may be defined as the number of parts by mass of a substance which combines with or displaces directly or indirectly 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine.

Why isn't it defined like "that mass of a substance which combines/displaces 1 gram of hydrogen"
Why does it use the term "parts by mass" ?

Parts by mass refers to the unit you are using..... actually equivalent weight is dimensionless...when you are using any specified unit, then though the equivalent weight quantities remains the same, the phrase  parts by mass gets replaced by that unit eg. gram equivalent weight.

Thank you for reply. makes sense. But then why relative mass or atomic mass dont use "parts by mass" in their defination ?

It seems that the unusual wording comes from the fact that we are used to having all of the atomic masses available to us simply because we have measured all of them. In the early days of chemistry, however, things were not so simple. Thus, the equivalent weight (technically mass since it has units of $\mathrm{kg}$ rather than $\mathrm{N}$) was used as a method for determining a weight relative to some predefined standard. In this case, hydrogen.
Some of the simplest reactions are acid/base reactions, and this explains why the scale is measured relative to equivalents (where one equivalent is one mole) of hydrogen. That is, in your quote from the wikipedia page, one mole of hydrogen, which has a mass of 1.008 grams, can react with 8 grams of oxygen or 35.5 grams of chlorine to form $\ce{H2O}$ and $\ce{HCl}$ respectively. Another way of saying this is that every hydrogen atom reacted with two oxygen atoms and every hydrogen atom reacted with one chlorine atom.$^1$ Thus, we see what is meant by the term "parts by mass". Notice this is not the same as the phrase "parts per mass". "Parts by mass" could be stated more clearly as the "parts of X measured by mass". That is, we are indirectly measuring how many parts of oxygen, chlorine, or whatever other X goes into a reaction by measuring the final mass of a substance which reacts with a known mass of 1.008 grams of hydrogen.
The wikipedia article notes that not everything reacts with hydrogen, so in order to find equivalent weights for some materials, you must measure its equivalent weight to a known mass of oxygen and then use the 8:1 conversion to hydrogen equivalent weights.
I hope this clarifies the idea. Don't get worried though, this does not contain any more information than simple atomic weights.

$^1$This is kind of cheating though because we know the weights of oxygen and chlorine, while chemists in the 1800s may not have known this.


text:
Trying to displace iron from ferrous sulphate heptahydrate using aluminium foil but nothing has happened
I started by dissolving the sulphate into water and filtering it as I noticed there was a dirty residue lying at the bottom. (I am using ferrous sulphate that is commonly used in gardening.) I then poured this solution into a jar along with torn pieces of tin foil (tin foil being a common name; the foil is made of aluminium). At the start the solution was fairly clear and greenish in colour. A few days later there is no sign of iron collecting at the bottom of the jar and there has been a slight change in colour. The solution is now slightly cloudier and looks more like apple juice. The orange colour makes me think there must be iron oxide being produced. I need to know whether my method for displacing the iron is correct and if elementary iron will be produced.

No, you can't reduce iron like that, for tin is less active than iron. Tin just sits there and does nothing. Meanwhile, iron is oxidized by air to $\ce{Fe^3+}$.

Take a look at an activity series for more info.

Isn't tin foil made from aluminium though?

Aluminium, however,  is easily passivated in the presence of oxygen.

You are correct: Commercially available metal foil is typically made from aluminum, even if it's commonly called 'tinfoil' sometimes. As you say, technically, tin foil is made of tin; aluminum foil is made of aluminum.
Jannis Andreska is also correct: The passivating oxide on the aluminum is likely preventing the iron ions from reaching the $\ce{Al^0}$ and swiping their electrons.
So, you are correct again: The yellow/orange color you're observing is probably oxidation of the $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$, by $\ce{O2}$ in the air.

One possible approach to breaking the oxide would be to connect the foil to the negative terminal of, say, a AA battery, with the positive terminal connected to a junk stainless steel spoon, also submerged.  That might not be enough voltage to break down the oxide film, though; on the other hand, a higher voltage might lead to corrosion of the spoon and the release of iron and other metals into your solution.  May or may not be a problem for your experiment?
Regardless, I would recommend adding some table salt $\left(\ce{NaCl}\right)$ to your solution, both to increase its conductivity and also to introduce chloride, which may help to attack the oxide on the aluminum, in tandem with the applied potential.


text:
How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory?
The quantum theory of atoms in molecules is based on the topology of the electron density. This mathematical analysis allows to find critical points and hence has a unambiguous way of separating a molecule into atoms. Each atom can be assigned a basin, and integrating the electron density in it yields a partial charge of that atom (taking the charge of the nucleus into consideration).
In principle this analysis can be carried out from observed electron densities. It is also a more and more common tool to analyse calculated electron densities. Since it is only possible to solve the Schrödinger equation approximately, I also expect a dependency of the calculated charges on the level of theory. It is well known, that the Hartree-Fock method only accounts for Pauli correlation, while other methods also account for coulomb correlation. I would expect, that the higher the level of theory the more accurate the calculated charges are. Hence the following questions:

How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory?
Do calculated charges converge to a specific value with higher level of theory?
Are charges from calculated electron densities comparable to charges from measured electron densities?
(Optional) Are there more reliable methods to obtain partial charges?

This may be slightly related: http://dx.doi.org/10.1016/S0301-0104(03)00006-5

I gave this question a shot. Hopefully my answer is a good start. I certainly learned a lot from reading through the papers and chapters I reference. It's really shocking how we take for granted that atoms in molecules are really not well-defined outside of a context like AIM.

@jheindel . Nuclei are pretty unambiguous. Atoms are challenging.

I have been doing quite a lot of reading recently about the theory of atoms in molecules and think that I have semi-satisfactory answers to your questions, so I will share them here and perhaps others will add complementary answers as there actually seems to be quite an extensive literature on this subject, but there is still a lot of debate over these topics.
First, a quick note on how the charge density, $\rho$, can be used in the theory of atoms in molecules (AIM) to determine chemically intuitive properties. This is usually done by the definition of an index. For instance, the first index one might think to define is the electron number index, $N(\Omega)$
, where $\Omega$ is the space defined by the critical points surrounding a single nucleus. Essentially AIM claims that $\Omega$ and the very idea of an atom are interchangeable, so $N(\Omega)$ is conceptually identical to answering how many electrons are associated with a partiular atom. Hence, the AIM definition of the charge on an atom follows as,
$$
Q_{\Omega_i}=Z_{\Omega_i}-N(\Omega_i)
$$
where $Z_{\Omega_i}$ is the charge of the nucleus of atom $i$ in the basin $\Omega_i$.
Not surprisingly, $N(\Omega_i)$ is defined as[1],
$$
N(\Omega_i)=\int_{\Omega}\rho(\textbf{r})d\tau
$$
which is clearly just the charge density in a particular basin. This is interchangeable with a number density of electrons by division by the electric charge $e$.
These indices are always defined in terms of the charge density. The reason these are called indices, rather than something more exact, is because we are really computing a quantity which dimensionally satisfies the property we are interested in, and then interpreting it as the thing itself. The ambiguity comes from the idea of a basin, $\Omega$, which is mathematically well-defined but is not guaranteed to be physically exact for any reason.

How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory?

Ref. [2] is a review of the AIM theory with a specific emphasis on the electron localization functions (ELFs) to describe aromaticity. More on these ELFs in a moment. They present the calculation of the index $N(\Omega)$ for several diatomic molecules using DFT, HF, and CISD calculations. The total charge in each of the basins stays more or less the same for all of these methods. The value of $N(\Omega)$ generally only changes in the first or second decimal place for the calculations they present (so tenths to hundredths of an electron). The trend when Coulomb correlation is included is that $N(\Omega)$ decreases for what would traditionally be considered the more electronegative atom. That is to say that coulomb correlation makes the bonds more covalent and less ionic. For mostly ionic systems such as $\ce{LiF}$, the addition of coulomb correlation has a small effect because the electrons are already mostly localized in the basins [2]. All of this is to say that the actual charges computed based on $N(\Omega)$ seem to be relatively insensitive to coulomb correlation. This makes sense based on the fact that the primary features of the spatial distribution of $\rho$ will be dictated by the requirement of antisymmetrization of the wavefunction which is, of course, included in a HF description.
That covers the question of charges based on $N(\Omega)$, but this does not actually give us a picture of how electrons are shared between atoms. In other words, how localized and how delocalized are the electrons? In order to answer this question, ref. [3] provides definitons of both localization and delocalization indices, which are unique because the localization index describes localization of electrons to a single basin, while the delocalization index describes the sharing of electrons with either a specific basin, or with all other basin in a molecule.
While $N(\Omega)$ is not greatly affected by coulomb correlation, the localization index, $\lambda(A)$, and the delocalization index, $\delta(A,B)$ are. For instance, $\delta(A,B)$ which is a measure of the sharing of electrons between the atoms $A$ and $B$, changes by almost $1.0$ when going from the HF description of $\ce{N2}$ to the CISD description of $\ce{N2}$. In general, for homodiatomics, coulomb correlation has the effect of increasing the density in individual basins. And, consistent with the statement above about charge, the localization index for polar diatomics tends to increase the charge on the less electronegative element when a correlated wavefunction is used.
I would highly recommend reading ref. [3] as a place to gain some intuition about the behavior of these different indices.

Do calculated charges converge to a specific value with higher level of theory?

I have not been able to find any explicit discussion of this, but of course any method which approaches the exact wavefunction, and hence the exact density, must have the same limiting value for any index defined in terms of the density. This is really the power of the AIM model. It is completely quantum mechanical in the sense that the exact wavefunction will yield the exact basins. Whether or not these basins mean what AIM wants them to mean, however, is what people debate.
Another related point is how the various indices are affected by the particular basis set used. I have seen a couple papers which mention this, and note that the results really do not depend much on the particular basis set chosen[1]. The reason which is speculated for this behavior is that the partitioning used in AIM is in real space rather than in the Hilbert space to which the basis set belongs. Unsurprisingly, however, the indices are sensitive to the basis set when the density itself is very sensitive to the basis set. Open-shell molecules seem to be more complicated on this front [2].

Are charges from calculated electron densities comparable to charges from measured electron densities?

The primary purpose of ref. [1] is to compare the value of properties determined from AIM and from a density determined by x-ray diffraction of p-nitroaniline. This molecule was chosen, I believe, because it has a large dipole moment.
One complication in using the experimental density is that it is not as simple as just measuring the density and then having the whole scalar field in front of you. Rather, from what I understand, you measure the diffraction data and then use certain theoretical models to translate this diffraction data into a charge density. Thus, there is ambiguity in what charge density you actually measure. Ref. [1] uses the UMM and KRMM multipole formalisms which are discussed in this wikipedia article. The point of bringing this up is that the answer for the charges actually depends on which method you use to construct the experimental density.
Ref. [1] indicates that the KRMM multipole method which uses the AIM analysis of charges agrees much more closely with an AIM analysis of the charges from a theoretical density found from periodic-HF and periodic-DFT calculations (this is all solid phase). The p-HF and moreso the p-DFT charges agree very well with experimental charges which are calculated from KRMM using the AIM method. See Table 2 of [1].
The dipole moments also agree very well, but only after various refinements are made to both the partitioning of the experimental density as well as the use of the KRMM method. See Table 4 of Ref. [1].
One point which was mentioned in this book[4] (the relevant chapter is available on Google Books, ch. 3) is that the location of bond critical points, saddle points on the density between two atoms, can be very different for theoretical and experimental densities. Thus, properties derived from the basins can, in principle, be very different for the densities. It seems, however, that despite this problem of locating the bond critical point accurately, properties which are derived from the theoretical and experimental basins are not very different. This is because in the region of the saddle point, the density is usually quite flat, thus shifting the bond critical point does not have much effect on how much density is in each basin.

Are there more reliable methods to obtain partial charges?

Only briefly, I will point out that various population analyses are always possible to determine charges on atoms. The answer to this really depends on what you mean by reliable. In a sense, AIM is the only reliable method of determining charges on atoms in molecules because it is the only method I am aware of which bothers to define what an atom in a molecule actually is. Additionally, it is reliable in the sense that it exists in a completely self-contained mathematical framework. Population analyses, however, can be very dependent on the particular basis set used, and also are not invariant to unitary transformations of the orbitals. In this sense they are very unreliable.
Mulliken charges are a common form of finding charges from a population analysis, but these are known to very basis set dependent. A better scheme, I think, is the CHELPG scheme which fits the molecular electrostatic potential of a system and hence conserves the dipole moment when it assigns charges to atoms. This is a very nice property for a population analysis. Nonetheless, it will be sensitive to method and basis set depending on how much the electrostatic potential changes upon changing either of these.
I think that generally the AIM charges and charges from population analyses will not be too different, but one notable exception is given in ref. [4] on page 162 where it is noted that the AIM charge on boron in $\ce{(H3BNH3)2}$ is positive, $q(\Omega_{\ce{B}})=2.15$, while the Mulliken population analysis gives a negative charge of $-0.26$. So that's a very stark difference. I guess I would trust AIM, but I don't really know.

As a concluding point, everything I have discussed above calculates properties from indices based on $\rho$, but there is no reason to limit the definition of indices to the density. Rather, some authors have defined indices, for physically motivated reasons, based on the gradient of the density, $\nabla\rho$ and the Laplacian of the density, $\nabla^2\rho$.
An interesting application of one of these based on the gradient is for some alkanediols where there is no bond critical point between the two $\ce{O-H}$ groups despite the obvious presence of an internal hydrogen bond [5].
Ref. [5] is a really cool paper. Definitely give it a read.

[1] Volkov, A., Gatti, C., Abramov, Y., & Coppens, P. (2000). Evaluation of net atomic charges and atomic and molecular electrostatic moments through topological analysis of the experimental charge density. Acta Crystallographica Section A: Foundations of Crystallography, 56(3), 252-258.
[2] Poater, J., Duran, M., Sola, M., & Silvi, B. (2005). Theoretical evaluation of electron delocalization in aromatic molecules by means of atoms in molecules (AIM) and electron localization function (ELF) topological approaches. Chemical reviews, 105(10), 3911-3947.
[3] Fradera, X., Austen, M. A., & Bader, R. F. (1999). The Lewis model and beyond. The Journal of Physical Chemistry A, 103(2), 304-314.
[4] Popelier, P. L. A., Aicken, F. M., & O’Brien, S. E. (2000). Atoms in molecules. Chemical Modelling: Applications and Theory, 1, 143À198.
[5] Lane, J. R., Contreras-García, J., Piquemal, J. P., Miller, B. J., & Kjaergaard, H. G. (2013). Are bond critical points really critical for hydrogen bonding?. Journal of chemical theory and computation, 9(8), 3263-3266.

(1) I think the basins of the ELF are also quite useful, at least in my own explorations. See, e.g., my answers as re charge-shift bonding (example). (2) The charges from non-AIM population analyses (e.g., Mulliken vs Loewdin) are often in poor agreement, regardless of the level of theory. Thus, AIM charges cannot agree well with both methods in such cases.

(3) I still owe Martin an answer to this question (and, well, still need to run the calculations needed to draft said answer), but I believe that cases do exist where including correlation leads to a sufficiently substantial shift in the density field to yield appreciably different $N(\Omega)$ values. (I tentatively name the sulfanilic acid zwitterion as one such system, but I need to actually figure out and perform a rigorous evaluation of this.)

All of these are very good points. The point you make about $N(\Omega)$ is very believable, and yes you make a good point about the fact that population analyses do not even agree with each other, so how could they agree with AIM? In general I sort of distrust population analyses, though something like CHELPG at least makes the charges less arbitrary. If you did some calculations that would be really fantastic. I'd like to see what some of these basins actually look like, as I shockingly didn't see any good depictions of individual basins.

@Martin No problem. I've been wanting to read about AIM extensively for a while now anyways. There's also a review by Bader and a book by Bader which I didn't mention in the answer that I would guess address many of these points. The review (I'll add link later) also shows more formal derivations which come from the action integral and provide an AIM version of the virial theorem. So that is probably of interest. The whole subject is very interesting. I'm not sure why AIM is supposed to be so controversial. It seems quite rigorous.

@jh Bader is really hard to read, unfortunately. I suppose he really knew his stuff as re the quantum mechanical formalisms, but his writing is just ... tough, sometimes.

@hBy2Py I actually get the feeling he intentionally complicates his writing. There was the whole thing with people not liking him and he basically pitted himself against everyone else so I sort of feel like he writes as technically as possible to make people think he's smart. Which he obviously is. I could be wrong but he was apparently very controversial so I'm just guessing. Maybe history of science SE could illuminate as to why he was disliked so much.

This is an excellent answer! I'm really sorry it took me almost two years to come back to it. It certainly warrants a bounty, too. Thanks again for writing this and being so thorough.

@jheindel This is quite the heroic effort into an answer, and the 200 point bounty is certainly well deserved. I read the conclusion section too, but is there a summary you can give as to how accurate Bader charges are compared to other methods? You mentioned that they are very basis set dependent, but is there a summary you can give as to the dependence on the level of theory? I'm sorry if I missed this while looking at your answer (which indeed has a lot of detail and probably does answer my questions somewhere).


text:
Comparison of rate of dehydration of primary, secondary & tertiary alcohols on cyclohexane
Rate of dehydration when given compounds are treated with conc. $\ce{H2SO4}$ is

According to me the answer should be $R > Q > S > P$ but my textbook solution gives the answer as R > Q > P > S (Sadly, there is no explanation).
Dehydration of secondary & tertiary alcohols in the presence of conc. $\ce{H2SO4}$ proceed through E1 mechanism whereas most primary alcohols proceed through E2 mechanism.
In the molecular R, the carbocation formed after the alcohol group leaves the molecule after protonation is a tertiary carbocation and not just that, it is resonance stabilized as well.
In the molecule Q, you get a secondary carbocation and this is resonance stabilized as well.
In molecule S, you get a secondary carbocation but there is no resonance stabilization.
In molecule P, the dehydration goes via E2 mechanism and there is a bulky cyclohexene group attached to the $\alpha$-carbon which should be contributing to steric hindrance.
If the question did not have molecule P, then its pretty much obvious that the answer is R > Q > S but with molecule P, I am not quite sure where to place it.
I thought that P will be least reactive since it proceeds via E2 mechanism and there is one bulky group attached to the carbon.
So how do I come up with a concrete answer to justify why molecule P should react faster than molecule S but slower than molecule Q.

Note that P can also form an allylic carbocation after protonation of OH group and subsequent loss of water.

P gets dehydrated via E2 mechanism? Carbocation isn't formed. Now its even more confusing, if it goes through E1 mechanism it will form an allylic carbocation. How do I now decide if it would undergo E1 or E2 elimination?

The rate of dehydration is related to the ease of formation of the carbocation and the energy of the carbocation intermediate.
The ease of formation of carbocation is tertiary>secondary>primary. That is the ease of abstraction of $\ce{OH2+}$ COULD BE be R>Q/S>P.
Let's look at a comparison of the energy of the carbocation intermediates in all these cases.

A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation $\ce{CH2=CH–CH2+}$ and benzyl cation $\ce{C6H5–CH2+}$ are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the $\pi$ bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge - and, therefore, stabilizes the carbocation.

Let's look at them individually,

P - allylic cation, primary alcohol
Q - allylic cation, secondary alcohol
R - allylic cation, tertiary alcohol
S - NOT an allylic cation, secondary alcohol

$\ce{R}$ is clearly the winner.
Now we may conclude that the charge dispersal is better in resonance than it is in hyperconjugation (explanation)
Here are the 3D diagrams of the P carbocation. Hydrogen cation doesn't look like it has to face significant steric hindrance to perform protonation of the alcohol.


text:
Fractional distillation of crude oil and resulting fuel ratios
I understand from the diagram below that different fuels condense at different temperatures. But I don't know anything about crude oil (and very little about chemistry in general).
Could I simply refine 100% of crude oil to gasoline by heating it up to 150 °C (value taken from diagram).
Put another way, are the ratios of end products from refining crude oil fixed?

Sorry, are you asking if: if you had a gallon of crude and let it sit and simmer at 150 degrees, would you wind up with a gallon of gasoline eventually? Because I'm not clear on what you're asking, but that is emphatically not how fractional distillation works.

Thank you, your comment was helpful. I was wondering whether I can only produce gasoline from crude oil. No Diesel, no kerosene, nothing else. Or is there a fixed relationship, e.g., 1 gallon of crude results in 0.x gallons of gasoline, 0.x gallons of Diesel, and so on.

You never get 100% pure anything from fractional distillation.

Crude oil is a complicated mixture consisting of a variety of different hydrocarbons and the particular composition is very variable depending on where it is found.
Distillation is merely a means of separating those components so it can do nothing to create more of one of them. And gasoline is, essentially, just a mixture of those components with similar properties and boiling points. There is little you can do by distillation to create more gasoline.
However refiners do have access to chemical means to alter the composition. Catalytic cracking, for example, converts constituents into different chemicals and this can allow a refinery to create more gasoline or diesel by breaking down the higher-boiling components into smaller parts. But this is much more complicated than distillation.

I see where you are coming from but I don't think you can simply get gasoline by heating crude oil to 150°C. There are few more important steps that should be taken before putting crude oil into the furnace.
Have a look. (Picture taken from Wikipedia)

Thank you for your answer. I was aiming at a different aspect, though. I was wondering whether distillation of crude oil always yields the same ratios of fuel or if that ratio can be influenced. E.g., you'll always end up with 20% gasoline, 20% Diesel, and so on. (I know very little about chemistry and just randomly picked numbers here.)


text:
Metal carbonate + carbonic acid?
As metal carbonates are basic would there be a reaction if a sample of, say potassium carbonate, was added to carbonic acid? If so what would be the products? A balanced equation would be useful.
Will this apply to other basic substances if they are added to an acid from which they already have an ion of?

There would be hydrocarbonates, and in case of other acids, their corresponding acidic salts.

In water  all carbonate will appear as the hydrocarbonate ion. The law of mass action applies as usual.

$\ce{Na2CO3 + H2CO3 -> 2NaHCO3}$, yielding sodium "bicarbonate", as an example, and in similar fashion, $\ce{Na2SO4 + H2SO4 -> 2NaHSO4}$, or sodium "bisulfate". The "bi" in this nomenclature meant two different anions, and is confusing, as it is the opposite of "bi" in mercury bichloride, two of the same cations. For that reason, sodium hydrogen sulfate or sodium acid sulfate might be preferred.

Would it be mercury dichloride?

Yes, but the "traditional" name was bichloride. See many detective stories for its use (https://marymiley.wordpress.com/2009/08/15/popular-poisons-part-ii-mercury-bichloride/).

The "bi" in bicarbonate question was also in a previous question here http://chemistry.stackexchange.com/questions/31083 I wouldn't agree with the statement "The 'bi' in this nomenclature meant two different anions" since it is an obtuse explanation.


text:
Can ionic and covalent bond be explained in terms of orbitals?
Covalent bonding is stated as overlap of atomic orbitals .In ionic bonding the composite ions are held together by their electrostatic charges that result from the transfer of electrons from one atom to the other. What is exactly meant by "transfer"
in terms of orbitals?

A plot of the expectation value for the electron's wave function would be heavily biased towards one of the two atoms, resulting in a net dipole moment.

For the ionic form of $\ce{AB}$ viz. $\ce{A^+B^-},$ the wavefunction can be written as $\psi_{\ce{A^+B^-}}(1,2)= \psi_{\ce{B}}(1)\psi_{\ce{B}}(2);.$

The "transfered" electron will be in an orbital that is mostly on the side of only one atom, i.e. the anion. That orbital will be slightly deformed by the presence of the cations around it, but no significant overlap with any of their outer orbitals (those which are as good as empty, because the electron has been lost to the other atom).
(Im writing "mostly" and "as good as" because there can always be a covalent contribution to the bond.)


text:
Compound formed from concentrated sulfuric acid and sulfur
In the reaction below, I know what A and B are, but I'm not sure what compound C is.
All I know about it is the following:

Deep yellow solid
Dication contains only sulfur
Raman and infra-red spectra of the dication show no common features

I think that the dication might be $\ce{[S4]^{2+}}$ since I think that this would be square planar and therefore have $D_{\mathrm{4h}}$ symmetry, which has no irreps that are both Raman and IR active, but I don't know what the anion would be.
The other likely options for the sulfur dication would be $\ce{[S8]^{2+}}$ or $\ce{[S19]^{2+}}$ but I think the fact that it's yellow would make it $\ce{[S4]^{2+}}$.

The reaction is given here.
$$\ce{S + 2H2SO4 → 3SO2 + 2H2O}$$
Sulfur react with sulfuric acid to produce sulfur dioxide and water. Sulfuric acid should be concentrated solution. The reaction takes place in a boiling solution.
$\ce{[S4]^2+}$, $\ce{[S8]^2+}$ or $\ce{[S19]^2+}$ is formed when sulfur is dissolved in oleum. The following is a paraphrase of an inorganic textbook.

The yellow species is indeed the $\ce{S4^{2+}}$ cation, as reportrd by Nilay Ghosh's reference. Its relative stability and symmetric structure, amenable to delocalization of electrons, goes along with an inorganic version of aromaticity.
The counterion in the sulfuric acid medium would surely be leveled to bisulfate ion, $\ce{HSO4^-}$. The number of bisulfate ions is actually more than twice the number of $\ce{S4^{2+}}$ ions because the reduction process of concentrated sulfuric acid inherently generates more $\ce{HSO4^-}$ ions, and $\ce{H3O^+}$ ions, along with the acid reduction product. If we make the likely simplifying assumption that the acid which is reduced forms sulfur dioxide, we may render the reaction
$\ce{4S + 5H2SO4 -> S4^{2+} + SO2(g) + 4HSO4^- + 2H3O^+}$

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text:
DL vs dl notation

As the L-isomer of glucose, it is the enantiomer of the more common D-glucose.

Source: Wikipedia
As far as I know enantiomers or Optical isomers are non superimposable mirror image structures with chiral centres and they are represented by d and l isomers, not D and L forms. Is there a difference between the capital and lower case designations?

@DGS As D and L denote the configuration wrt the glyceraldehyde and not optical activity then why should they be considered ways for denoting enantiomers?

Because glyceraldehyde is chiral at the alpha-carbon, so if you branch out from there to form the amino acid or sugar you want, you'll get one or the other enantiomer. Diasteromers of glucose aren't called glucose. There is (+)- or (-)- glucose, and then there is (+)- and (-)-galactose, which is a diastereomer of glucose but not either form of glucose.

Related: What is the difference between D and L configuration, and + and -?

The notation d, l, and dl (for dextrorotatory and laevorotatory, respectively) is used to designate the sign of optical activity. According to Basic terminology of stereochemistry (IUPAC Recommendations 1996), this notation is obsolete and its usage is strongly discouraged. The recommended notation uses the prefixes (+), (−), and (±), respectively.
Since optical rotation is a distinguishing characteristic of enantiomers, the notation using d, l, and dl as well as the notation using (+), (−), and (±) can be used to distinguish the enantiomers of a chiral molecule. However, without further information, the absolute configuration (R or S) of the chiral molecule is indeterminable. Thus, it remains unknown whether the dextrorotatory enantiomer (d and (+), respectively) is the R enantiomer or the S enantiomer.
The notation D, L, and DL is based on the arbitrary convention according to which d-glyceraldehyde was named D-glyceraldehyde, l-glyceraldehyde was named L-glyceraldehyde, and the racemate dl-glyceraldehyde was named DL-glyceraldehyde. (Note that the stereodescriptors D, L, and DL shall actually be written in small capitals, which cannot be correctly typeset here.) The enantiomers were taken to have the absolute configuration represented by the Fischer projection.
Today, we know that D-(+)-glyceraldehyde actually is (2R)-2,3-dihydroxypropanal.

D- and L- isomers vs d- and l- isomers: DL is newer, dl is old school. Both are ways of denoting enantiomers.
Sometimes they don't match up, though. Wikipedia says nine of the nineteen L-amino acids are dextrorotary, so they are "d", but their configuration is " L."
This is because lowercase deals with optical rotation (d- = (+); l- = (-) ), while D- and L- deal with configuration: "the version synthesized from naturally occurring (+)-glyceraldehyde being considered the D-form."
More here

D and L sugars are not the same as d and l sugars. These two abbreviations denote entirely different configurations. D and L denotes the position of hydroxyl group at the asymmetric carbon of a monosaccharide, whereas d and l denotes the rotation of plane polarized light i.e.
d: Dextrorotatory
l: levorotatory

While not a bad answer, I don't see what it adds that isn't covered by Loong already.

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text:
Solubility of Alum

What is solubility product of $\ce{K2SO4. Al2(SO4)3.24H2O}$?

Can we consider $K_\mathrm{sp}$ of  salts  $\ce{K2SO4}$ and  $\ce{Al2(SO4)3}$ or something else to find solubility?
I know that Sulphate ions are coming from two different salts. That's were my brain stop working.  Is it possible to solve this question something like $K_\mathrm{sp}=\ce{[Mg^2+][OH−]}=s \times (2s)2$

The very idea of using $K_{sp}$ for well-soluble salts does not sit well with me, but whatever. Now, there are no two different salts here; $\ce{K2SO4. Al2(SO4)3.24H2O}$ is one compound and should be treated as such. The data for $\ce{K2SO4}$ and $\ce{Al2(SO4)3}$ have nothing to do with it.

Here is a chart with some data for potassium alum https://en.wikipedia.org/wiki/Alum#Solubility

No you should not because $\ce{KAl(SO4)2}$ has a different crystal structure and thus different enthalphy and entropy of formation values than the pure independent salts.
The $K_\mathrm{sp}$ of a substance is given by:
$$RT~\ln\left(K_\mathrm{sp}\right) = \Delta G_\mathrm{sol}^\circ = \Delta H_\mathrm{sol}^\circ - T\Delta S_\mathrm{sol}^\circ$$
Producing potassium alum is a reaction of its own:
$$\ce{K2SO4 + Al2(SO4)3 -> 2 KAl(SO4)2}$$
This independent reaction has its own $\Delta G^\circ$ of formation and thus the reactants to product solution have a different energy and thus different solubility.
Further this double salt affects your cation and anionic species of solution.  For example the potassium may now dissociate to form a $\ce{K+}$ and a more stable $\ce{[Al(SO4)2]-}$ counter ion which will affect solubility as well.


text:
Why is KI oxidized by H2SO4, whereas NaBr isn't oxidised?
My textbook says that alcohols react with sodium bromide and sulphuric acid to form the corresponding alkyl bromides. At the same time they also mention that sulphuric acid mustn't be used with potassium iodide ($\ce{KI}$), because $\ce{HI}$ formed would be oxidized to $\ce{I2}$. Won't $\ce{HBr}$ formed from $\ce{NaBr}$/$\ce{H2SO4}$ oxidise to $\ce{Br2}$?

The answer might be lying in the bond enthalpy of H-I and H-Br and the less stability of HI as compared to HBr.

The original question has already been answered, but one way to work around the problem is to use phosphoric acid to make the hydrogen iodide. See https://en.wikipedia.org/wiki/Hydrogen_iodide.

What book is this? Would you share its title?

Iodide ions are much more powerful reducing agents than bromide ions:
$$\ce{2I- ->I2 + 2e-} \hspace{1cm}E^{o}=-0.54\ \mathrm{V}$$
$$\ce{2Br- ->Br2 + 2e-} \hspace{1cm}E^{o}=-1.0873\ \mathrm{V}$$
Iodide ions are so powerful, in fact, that they are capable of reducing sulfur from a +6 oxidation state in $\ce{H2SO4}$ all the way to -2 in $\ce{H2S}$ via the following half reaction:
$$\ce{HSO4- +9H+ +8e- -> H2S + 4H2O} \hspace{1cm}E^{o}=+0.80\ \mathrm{V}$$
The full reaction being being:
$$\ce{HSO4- +9H+ +8I- -> H2S + 4I2 +4H2O}\hspace{1cm}E^{o}=+0.26\ \mathrm{V}$$
A quick glance at the reduction potentials tells you that the analogous bromide reaction has a negative reduction potential and is therefore non-spontaneous.
[Source]

What about this video?
Solid Halides with Concentrated Sulfuric Acid
https://youtu.be/0vT0pkKMTDI
Isn't bromine produced, even without heating? Perhaps the OP should have mentioned cooling the reactants to prevent oxidation of bromide from happening?


text:
Is NaH2PO4 and Na2HPO4 buffer a neutral buffer?
My teacher mentioned in class that a mixture of two salts derived from polybasic acid functions as a buffer solution. But is it a neutral buffer or acidic buffer?

You ask whats the pH of this buffer?

No. But what type of buffer it is?....yeah in one way pH may be.

What type of buffer? There's no such classification I'm afraid.

It is true that dissolving equal proportions of the dibasic and monobasic sodium phosphate salts in water solution yields a buffer that is very close to pH 7 - for all practical purposes, neutral pH.


text:
Why does bromine react with iron(II) but not iodine?
My question comes from the answer (C) to this question:
Firstly, (correct me if I'm wrong please) I thought that the answer could either be C or D since the other answers are negative ions which can't react with iodine or bromine since they (as diatomic molecules) have a full shell of outer electrons. Now, from looking at the question, I knew it had something to do with oxidation and reduction as well as displacement reactions, however, as iodine and bromine aren't on the reactivity series, we can't determine which would react with which metal. From there, I just had no more ideas. Could anyone provide me with a hint or a guide as to how to do this ?

All four of the ions are reducing agents. But usually halogen gases' tendency to undergo reduction to become halide ions decreases down the group (along with their reactivity). Out of the $4$ ions, only $\ce{Fe^{2+}}$ has a mild enough reducing power.
$$\ce{2Fe^{2+} +Br2 -> 2Br^- +2Fe^{3+}}$$
But $\ce{Fe^{2+}}$ is not strong enough to reduce $\ce{I2}$ as it is relatively stable.
In fact, $\ce{SO3^{2-}}$ has the strongest reducing power among the four options, as it can reduce both $\ce{I2}$ and $\ce{Br2}$.
$\ce{OH^{-}}$ is a mildly strong reducing agent, it can react with both $\ce{I2}$ and $\ce{Br2}$, and its typical reaction is
$$\ce{4OH^- -> 2H2O +O2 + 4e^-}$$
$\ce{Mn^{2+}}$ is the least reducing, because it is the product of a strong oxidizing agent that has been reduced:
$$\ce{MnO4^{-} + 8H+ +5e^- -> Mn^{2+} +4H2O}$$
Here, permanganate ion $\left(\ce{MnO4^-}\right)$ is a very strong oxidizing agent; the product thus would be weakly reducing.
Actually, a similar idea can be seen with $\ce{SO3^{2-}}$ (sulphite) ions. Since it is a strong reducing agent, we can expect the product formed after it is oxidized is a weak oxidizing agent.
$$\ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e^-}$$
Indeed, sulphate ions alone are weak oxidizers, and stable.
I suggest you refer to E.C.S. (electrochemical series) to know whether a reaction would occur. It is easy to find online; e.g., at Wikipedia.

Nice answer! Thanks for wandering over here from Math.SE. :-)

@Brian Thanks for the warm welcome! I had some free time so I decided to come over see what I can learn here :)


text:
Gas solubility in water and alcohol
I know that a certain gas is soluble in water but not in alcohol. How can I find the solubility of the gas in mixture of water and alcohol? Do I use a simple linear relation? The gas in question is not a noble gas, but does not react with either liquid.

If this were an ideal situation in which there were really negligible solubility in ethanol and significant solubility in water, then you could just calculate solubilities in mixtures as being linear with the fraction of water.  But, your first sentence seems unlikely for a real situation.

So long as we assume that this is an ideal situation in which there is really insignificant solubility of the gas in ethanol but significant solubility in water, then yes, you can assume that the concentration in the mixture is linear with the water fraction of the liquid. In other words:
$${H_m = H_w * x_{w}}$$
Where:
${H_m}$ = Henry's law solubility of the gas in the water-ethanol mixture
${H_w}$ = Henry's law solubility of the gas in water
${x_{w}}$ = mol fraction of water in the liquid phase


text:
How do you find the heat capacity of a calorimeter?
My instructor gave me the formula:
$$C_\mathrm{cal} = \frac{C_\mathrm{bomb}}{\text{steel}+C_\mathrm{water}}$$
I know that the heat capacity for water is $4.184~\mathrm{J\,g^{-1}\,^\circ C^{-1}}$ and I calculated the value for $C_\mathrm{bomb}$, which is $356.7555~\mathrm{J\,g^{-1}\,^\circ C^{-1}}$. The part I don't understand is where it says steel. Is “steel” the same as “$C_\mathrm{bomb}$”?

You have this so discombobulated I can't tell what you are trying to do. Basically if $\mathrm{C}{\mathrm{t}}$ is the total heat capacity for the system , $\mathrm{C}{\mathrm{w}}$ is the heat capacity of the water, and $\mathrm{C}{\mathrm{cal}}$ is then heat capacity of the calorimeter, then $$\mathrm{C}{\mathrm{t}} = \mathrm{C}{\mathrm{w}} + \mathrm{C}{\mathrm{cal}}$$ so in order to raise the whole system by 1 degree C you have to supply enough heat to raise the water 1 degree and the calorimeter 1 degree.

Thus $$\mathrm{C}{\mathrm{t}} \Delta \mathrm{T}= (\mathrm{C}{\mathrm{w}} + \mathrm{C}_{\mathrm{cal}}) \Delta \mathrm{T}$$

If $\mathrm{c}\mathrm{w}$  is the specific heat of water, 4.184 J/g⋅°C, and $\mathrm{g}\mathrm{w}$ is the mass of the wtaer, then $$\mathrm{C}\mathrm{w} = \mathrm{c}\mathrm{w}\mathrm{g}_\mathrm{w}$$

What are $C_{cal}$, $C_{bomb}$ and $C_{water}$ supposed to be? If they are heat capacities, the formula is wrong.


text:
Is calcium hydroxide or potassium hydroxide more efficient at removing carbon dioxide?
In biology, we commonly use $\ce{KOH}$ in respirometers to remove carbon dioxide. A common test in chemistry to detect the presence of carbon dioxide is to bubble it through limewater, turning it milky. This has led me to wonder which is more efficient in removing carbon dioxide—$\ce{KOH}$ or $\ce{Ca(OH)_2}$—when both solutions are sitting in the atmosphere (i.e. they’re just passively absorbing $\ce{CO_2}$ and $\ce{CO_2}$ is not bubbled through either solution). An explanation would be much appreciated.

KOH would be better since Ca(OH)2 has limited solubility. This of course infers a much more concentrated solution of KOH.

$\ce{KOH}$ (and $\ce{NaOH}$) absorb $\ce{CO2}$ more quickly and completely than $\ce{Ca(OH)2}$, but they are much more dangerous to work with: caustic to skin, deliquescent (forming a highly concentrated solution that can pour out of the apparatus), etc. For that reason, "lime water" is preferred for classroom use to demonstrate the presence of $\ce{CO2}$.

Is KOH deliquescent? I know NaI is and KI isn't, so I thought maybe the $\ce{Na^+}$ may be the culprit with NaOH, also.

The OP asked about solutions, not using the hydroxides as  solid reactants. I still think capturing $\ce{CO2}$ with a concentrated $\ce{KOH}$ solution would be more efficient than a $\ce{Ca(OH)2}$ solution. $\ce{KOH}$ isn't very good for a demo since the solution would stay clear until a massive amount of $\ce{K2CO3}$ was created. Just a little $\ce{ CO2}$ would give a ppt with $\ce{Ca(OH)2}$.

Agreed, Ca(OH)2 is a better CO2 indicator, but the alkali hydroxides are more effective absorbents. BTW, I've found it hard to get a sufficiently concentrated "lime water" so that a precipitate is immediately noticeable (perhaps the water was a bit acidic?).

Limewater contains about a gram per liter of calcium hydroxide.  When you figure in the stoichiometry of the reaction and the density of $\ce{CaCO3}$ you are likely to end up with something like one volume of precipitate per thousand volumes of solution.  So you are not going to get copious white clouds of precipitate.  Such is life in the world of alkaline earths.


text:
Endothermic Reaction with Specific Details
I need some information for a book I'm writing.
Let's say you're in a confined space, about 8 feet by 8 feet. The area is being heated externally slowly (a few degrees a minute) to about 3,100 degrees F (1700 C). You have ammonium thiocyanate, $\ce{NH4SCN}$, and barium hydroxide octahydrate, $\ce{Ba(OH)2.8H2O}$, 20 g T. You combine them.
Since the room is getting so hot, would the room get equally cold and equally as fast? If you put the chemicals inside of a crate, closed it up, would it take longer to cool off the room? How long would it take for it to do this? Would the cold kill the person inside the room? (Aside from the ammonia burning the lungs and toxic barium because that's easy to fix in a story)
If there's just no way, is there a different chemical reaction that I could have the character use that won't kill her?

For the crate part, that would depend a lot on the properties of the crate. Also, the important question is the heat flux into the room; although I suppose that could be calculated with some degree of accuracy (neglecting walls, floors, etc.). Another factor that would influence this is if there is air flow in the room.


text:
Degree of dissociation of acetic acid at infinite dilution

Afew drops of acetic acid are added to a large excess of water at $\pu{25 ^\circ C}$. In the limit of infinite dilution, what percentage of acid remains unionised? The $K_\mathrm{a}$ of acetic acid is $1.85\times10^{-5}$.

The answer is supposed to be 0.54%, but I'm not sure how to reach this value. I know that given the pH we can use the definition of $K_\mathrm{a}$ to calculate the ratio of $\ce{CH3COOH}$ to $\ce{CH3COO-}$, but how do I find the pH?

As the dilution becomes infinite for the amount of acetic acid, then the acetic acid's contribution to the solution's acidity falls, and the pH of the water is controlled by the acid dissociation of water. So the pH falls to 7.00. Thus,
$$\begin{align}
\frac{(10^{-7})[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 1.85 \times 10^{-5} \\
[\ce{CH3COO-}] &= 185 [\ce{CH3COOH}]
\end{align}$$
The fraction of unionised acid, $x$, is therefore given by
$$\begin{align}
x &= \frac{[\ce{CH3COOH}]}{[\ce{CH3COOH}] + [\ce{CH3COO-}]} \\
&= \frac{1}{1+185} \\
&= 0.54\%.
\end{align}$$


text:
Does standard electrode potential of a reaction depend on stoichiometric coefficients?
Take a reaction like this
$$\ce{Fe^3+ + 3e- -> Fe}$$
now will the standard electrode potential change if we take 2 moles of $\ce{Fe^3+}$ and 6 moles of electrons?
$$\ce{2Fe^3+ + 6e- -> 2Fe}$$

If you think of the Nernst equation, you'll see that: $$\Delta G° = -nFE°$$
Since $n$ corresponds to the number of electrons transferred, the ratio $\frac{\Delta G°}{n}$ remains constant, because any change in the stoichiometric coefficients will multiply both terms by the same value.
Therefore, we see that the potential is an intensive property, not depending on the dimensions of the system under study.


text:
Ionic Compounds and finding Anions
K2HPO4 is an ionic compound with an -2 Anion and a +1 Cation. The thing I'm finding hard to understand is why the four Oxygen atoms don't add to the Anion charge.
E.g. H+1 P-3 O-2 O-2 O-2 O-2 = charge of -8.
I'm told I'm making this harder for myself but it's really frustrating not knowing why the 4 Oxygen atoms don't affect the charge.
Edit: I think I might have confused myself or confused my tutor because I'm mucking up the charges etc. Sorry if it's hard to understand

Why, they do add. But P is neither minus nor 3.

The Hydrogen Phosphate anion is made up of Covalent bonds. The overall charge of $-2$ that it has can be seen by its tetrahedral structure:

The four Oxygen atoms do not each contribute a $-2$ charge to the molecule, but rather two of them contribute a $-1$ charge each.
I think you are confusing Oxidation state, or 'charge' and valence state.
Oxidation state is the formal number of electrons added/removed from an atom when it forms a compound. Here, we have a covalent anion so the electrons are only formally removed, but it is still a useful formalism and can be used here to calculate the $-2$ charge :
Oxygen's valence state, the number of valence bonds used in bonding, is $-2$ and for Hydrogen it is $+1$ as you correctly said. Here, these are the same as their Oxidation States. But for Phosphorus, we must assign a $+5$ Oxidation state to it in order to obtain the overall $-2$ charge. Note that Phosphorus is not actually losing 5 electrons to form a $+5$ cation, this is a formalism, as aforementioned.
Because the anion is covalent, you cannot just treat each atom as an ion in its 'normal' oxidation state and try to add them up to obtain the charge.
You CAN do this for $K^+$ and $HPO_4$$^-$$^2$ because these form an ionic compound, so the charges simply add: Two $K^+$ and one $HPO_4$$^-$$^2$ will combine to produce the overall $0$ charge ionic molecule (As shown in the diagram, but for $Na^+$ rather than $K^+$.


text:
How many electrons are around the platinum in an ethenetriplatinate complex?

In $\ce{[Pt3(C2H4)]-}$, how many electrons are there around platinum?

While solving this problem I thought three $\ce{Pt}$ formed a triangle with ethylene above. But I was confused about how the $\ce{C2H4}$ gave electron pairs to $\ce{Pt}$. In addition, the answer gives $16$, but I only see the initial $10$ electrons around a $\ce{Pt}$ and $2$ given by another two $\ce{Pt}$.
What is the exact structure of this coordination compound, and how can it be deduced?

It is very probably, that they missed three ligands, possibly chloride

Looking up your proposed sum formula, I arrived at a quantum mechanical paper that used said complex (triangular $\ce{Pt3}$ moiety, whereonto an ethylene molecule coordinates onto the longer edge of the triangle) as a modelling study for the interaction of ethylene with platinum metal surfaces.[1]
As far as I can tell,[2] the authors in now way account for a surface of platinum; their three platinum atoms in vacuum are supposed to model an entire platinum surface. I cannot say whether that type of approach is typically used or whether it is considered good by quantum chemists.
Searching further, I also found a 1981 paper describing $\ce{[Pt3(C2H4)(cod)2(cot)2]}$, a neutral complex that includes two additional cycloocta-1,5-diene ($\ce{cod}$) and two additional cycloocta-1,3,5,7-tetrene ($\ce{cot}$) ligands.[3] In this complex, the bonding situation and electron count can be well answered, but that wasn’t your question.[4]
Furthermore, I find a complex $\ce{[Pt3(C2H4)]-}$ to be highly questionable: Platinum is in the 10th group, ethylene is a singlet molecule so adding a single additional electron will create a radical anion. Another reason why your question and answer cannot be serious, since in no way can a radical end up at $16$ electrons. I would even go a step further and say that it is likely an unstable complex that would decompose depending on which species are close to it. Note that the authors of the first paper calculate an uncharged complex $\ce{[Pt3(C2H4)]}$.

Notes and references

A. Cruz, V. Bertin, M. Castro, Int. J. Quantum Chem. 2000, 80, 298. DOI: 10.1002/1097-461X(2000)80:3<298::AID-QUA3>3.0.CO;2-2.
I am not a quantum chemist, and my last calculation efforts date back to a research project during my master’s studies. Those acquainted with methods will have to comment on how good the paper is.
N. M. Boag, J. A. K. Howard, J. L. Spencer, F. G. A. Stone, J. Chem. Soc., Dalton Trans. 1981, 1051. DOI: 10.1039/DT9810001051.
Two platinum atoms with 18 electrons in a square-planar environment, one with 16 electrons in a trigonal environment. The former coordinated by $\ce{cod}$ and $\ce{cot}$ (each contributing two double bonds), the latter coordinated by two $\ce{cot}$ molecules (each contributing one double bond) and an ethylene. The three platinum centres are too far apart to be considered bonded, according to the authors.
The bonding of the two 18-electron platinums to their respective $\ce{cot}$ unit is described as $\ce{cot^2-}$ and $\ce{Pt^2+}$, resulting in oxidation states of $\mathrm{+II}$ and $\mathrm{\pm 0}$ for platinum.

There should simply be PtCl3(C2H4) as permeakra said.

@Mithoron That may be possible; however, I prefer not to engange ni such speculation ;)


text:
How are SN2 transition states stabilised by adjacent double bonds and carbonyl groups?
It is my understanding that transition states in an SN2 reaction have some filled and some empty orbital character, as we have the backside of the C-X σ* beginning to fill with electrons from the HOMO of the nucleophile, forming a C-Nu σ bond in the process.
If we have an adjacent carbonyl group to the α carbon, I have learnt that the TS is then only stabilised by the  C-O π*, i.e. by electron withdrawal and not by the C-O π.
However, with adjacent double bonds, I have learnt that the TS is stabilised by both the C-C π and the C-C π*.
I was wondering why this would be the case, as looking at the shape of the orbitals, there seems no reason that the C-O π could not align itself so that the partly-empty p orbital in the TS is parallel to the lobes of this MO, therefore stabilising the TS as new lower energy MOs form.
Also, with regard to the best representation of these structures, I always see partial negative charges on the Leaving group and the Nucleophile in the transition state, but never the positive charge located on the α carbon. Surely if with adjacent carbonyl groups we only have stabilisation by electron withdrawal, it is valid to put the partial positive charge there?

I think both the homo and the lumo will stabilise the ts in both cases but with a carbonyl the polarity means that the lumo has a large coefficient on carbon but the homo has a small coefficient. So it's dominated by the interaction with the pi star in the case of a carbonyl.

@loolipop I didn't realise until I'd written an answer, but could you cite a source for the proposed different modes of stabilisation that you mention....

Unsaturated groups conjugated to the site of attack (aromatic rings, alkenes, carbonyl groups) all tend to have the effect of increasing the rate of an SN2 reaction at that centre.
As an example, displacement by nucleophilic iodide in acetone as a solvent is ca. 30x faster for allyl chloride than ethyl chloride, with benzyl chloride being faster still.[1]
The effect was initially thought to be steric, given that unsaturated groups are sp2 hybridised (and therefore planar) and allow the nucleophile a less hindered approach, but many empirical and computational studies now dispute that, with a stereoelectronic effect[*] widely accepted (though sterics may play a small role).
Empirical Evidence for a Stereoelectronic Effect
One of the earliest indications that the effect was stereoelectronic was via a series of experiments conducted by Payne (shown below) [2].The open (acyclic) compound reacts ca. 8000x faster than the closed (cyclic) compound. To understand why, we have to consider the orbitals involved...

The Orbital Explanation
If we accept a stereoelectronic effect (coming back to Payne’s findings later),  we then have to decide whether the effect is a ground state effect (weakening of the C-X bond, raising the energy) or a transition state effect (stabilisation of the TS, lowering the energy). To quote Fleming (emphasis my own):[3]

[This conjugation] may partly operate by raising the energy of the starting material, but most probably [the conjugating groups] are primarily operating to lower the energy of the transition structures.

In practice, computational studies [4] prove this to be the case, with the pi bond interacting with the principal axis of the SN2 transition state (sigma-pi conjugation):

Calculation of bond lengths also agrees with this mode of conjugation, with the C-C bond length decreasing upon stabilisation of the TS:

Hopefully, by this point, Payne's experimental results are obvious. In the cyclic structure, there is no way of allowing the pi-orbitals on benzene to correctly align with the SN2 transition state (given that we know the TS is linear, not bent).
One final thing to notice is that in both the alkene and carbonyl case (as in your question) the dominant effect is thought to be as above (sigma-pi conjugation).
—
[1]: March, J.; Smith, M. B. Advanced Organic Chemistry Wiley: Hoboken, New Jersey, 2007
[2]: King, J. F.; Tsang, G. T. Y.; Abdel-Malik, M. M.; Payne, N.C. J. Am. Chem. Soc. 1985, 107, 3224
[3]: Fleming, I. Molecular Orbitals & Organic Chemical Reactions:Reference Edition Wiley: Hoboken, New Jersey, 2010
[4]: Bach, R. D.; Soddens, B. A.; Wolber, G. J. J. Org. Chem. 1986, 51, 1030
[*]: Sorry Jan, couldn’t help myself with the bold, I’m not shouting, I promise…

In case of an adjacent carbonyl group, the carbons attached to the leaving group and oxygen are electrophillic due to the presence of a σ * and a π* low-energy orbital respectively. These combine to form a  LUMO which is even lower in energy. So each carbon becomes more electrophillic due to the other. Now, the nucleophile may attack the C=O but it is reversible. In such a case, Nucleophillic substitution is highly favourable as the displacement of leaving group is irreversible.
As for the allylic carbon, the carbon in the transition state is electron deficient as it forms "partial bonds" involving 2 electrons, which is remedied by the electron donating ability of the  π bond, stabilising the T.S., thereby increasing the rate. The stearic hindrance is small enough to be neglected(compared to a more expansive sp3 carbon)

Without an image showing the alignment, I can make no sense out of this.

In case of Beta Carbonyl groups, Carbon  attached to Oxygen through a double bond has a partial positive charge due to inductive effect and the nucleophile in most cases have negative charge. Therefore due to the electrostatic force between this partial positive charge and negative charge of nucleophile (electrostatic forces are very strong forces in chemistry), the nucleophile is attracted towards reaction site i.e. near the carbon bonded with leaving group. This increases the rate of reaction.
The double bond has a high electron density. Therefore, it helps the nucleophile pull the small carbon loop in the backside attack.
Same applies for the ajdacent lone pair.


text:
Chair conformation chirality

I have two questions regarding this solution:

I can't see how the structure drawn is the enantiomer.
How do I rotate the ring inversion chair form to obtain the enantiomer?

There's something wrong here because the two conformations are not mirror images.

1.) The mirror plane is orthogonal to the green line drawn between the respective center of the two enantiomers.
2.) The rotation is around the red axis. The red marked carbon atom shows the rotation angle.


text:
How to explain the structure of HNO3 with a Molecular Orbital diagram?
Take an example, $\ce{HNO3}$

The total valence of $\ce{HNO3}$ must be 24. By using formula $\pi =6\mathrm{n}+2-\mathrm{V}$ where $\pi$ is number of $\pi$ electrons, $\mathrm{n}$ is number of atoms excluding all hydrogen atom, and $\mathrm{V}$ is total valence, the $\pi$ is $2$. It means that there would be one double bond in the structure of $\ce{HNO3}$ and the remaining bonds other than $\pi$ bond are $\sigma$ bond.

I understand the double bond has 1 $\pi$ electron and 1 $\sigma$ electron. It means 10 electrons are used for the bonds and 14 electrons are remaining.
I know it would be impossible to violate octet rule because the calculation of $\pi$ already stated.
I have trouble how to spread the 7 lone pairs. Why does the $\ce{O}$ in the single bond $\ce{N-O}$ have 3 lone pairs, while the $\ce{O}$ in the double bond has 2 lone pairs? I don't expect the answer like "Just spread the lone pairs to somewhere".
I'm worried about the odd number of unshared electrons.
I know the $-$ means rich of electron and $+$ means lack of electron. To check the formal charge in each atom is easy, but how to calculate the formal charge, if I don't know where to put the $-$ and $+$.
I think this question is not a duplicate of The Lewis structure of HNO3. I already saw that question , but I'd like to ask here again, because I need the answer for my misunderstanding, even though the answer would be general structure of $\ce{HNO3}$.
To be precise, I just don't know where I need to put unshared electrons. It's like I'm missing the bridge. I know resonance, as arrow-pushing, and mechanisms are required in organic chemistry.

So how http://chemistry.stackexchange.com/questions/7174/the-lewis-structure-of-hno3 doesn't answer this? Then you seem to have many problems making this too broad instead.

http://chemistry.stackexchange.com/questions/4828/formal-charges-how-to-work-them-out

@Mithoron Those questions are not helpful. I've edited question because those are not exactly what I want.

You have 24 valence electrons. Counting the electrons: 6 of these will go into the $σ$ framework and there are 14 electrons in non-bonding, either $sp^3$ or $sp^2$ hybrids, on the oxygens. This leaves 4 electrons for the $π$ system.
We can think of the Nitrogen and Oxygen in the double bond being as both being $sp^2$ hybridised, with vacant $p$ orbitals orthogonal to the plane of the molecule. These 4 remaining electrons will then go into the $π$ system formed by the overlap of these 2 orbitals, which overlap to form 2 molecular orbitals, the $1π$ and the $2π$ MOs.
We expect the $1π$ to have greater contribution from the Nitrogen due to the lower orbital energy of the $p$ in N compared to O, and the $2π$ to have greater contribution from the O instead of the N.
A full MO calculation of the molecule would also take into account the non bonding $sp^3$ electrons on the oxygen, of which one lobe will be able to interact slightly with the $ p$ orbitals, giving a delocalised system over the Nitrogen double and single bond, exactly as we see drawing resonance structure for the molecule.

This explanation with MO makes sense. Thank you.