text:
Do catalysts shift equilibrium constant towards 1?
I want to be able to understand shifts in equilibrium from the maxwell boltzmann distribution. One thing I cannot get my head around is the effect of catalysts on the equilibrium position - supposedly it's none at all.
But consider these diagrams of the M.B distribution of both reactants and products of a reversible reaction.

The top diagram shows the distribution of the products of the reaction which equilibrium favours. It has a bigger concentration of particles. The line dividing the shaded region on these diagrams represents activation energy and the $dE$ represents the shift in activation energy due to the catalyst. We can see that the shaded regions, $A_1=A_2$ because forward and backward rates are the same at eqm. But clearly, the rate of the top reaction increases more as a greater fraction of particles have above activation energy. So adding a catalyst should shift equilibrium to the middle?
Basically, catalysts work better on reactants with higher concentration. So shouldn't the side with the lower concentration be favoured?

The energy difference between the reactants and products is unchanged by catalysis.

how is this related to the question?

re 'how is this related', everything! This is because the free energy is unchanged, since the reactants and products are the same, and as $\Delta G^0=-RT\ln(K_e)$  so $K_e$ is unchanged.

Oh ok I can understand that there's probably a link between free energy and changing equilibrium. I don't yet have much understanding of this area of statistical physics and chemistry (at all). Although my question was to specifically about approaching the problem using those graphs (albeit I am now aware that it's a terrible way to approach it in the first place).

It's not terrible, just a little bit flawed, and I wouldn't blame you for thinking that way. It is often taught that way in school, too, and my school was no exception: $A$ is the collision frequency and $\exp(-E_\mathrm{a}/kT)$ is the proportion of effective collisions, which can be represented by the M-B distribution. Qualitatively, the M-B diagram works, and personally I had always assumed that the maths would work out, until I tried it last night. So, good question, and it inspired (hopefully) a decent answer from me, although I can't say I'm anywhere near an expert on reaction dynamics.

Also, by the way, note also that the shaded area does not represent a concentration, in the conventional sense of the term (number of moles / volume). I get that perhaps you are referring to an "effective concentration" of particles with enough energy to react, though.

TL;DR Your Maxwell–Boltzmann diagram up there is not sufficient to describe the variation of rate with $E_\mathrm{a}$. Simply evaluating the  shaded area alone does not reproduce the exponential part of the rate constant correctly, and therefore the shaded area should not be taken as a quantitative measure of the rate (only a qualitative one).

There is a subtle issue with the way you've presented your drawing. However, we'll come to that slightly later. First, let's establish that the "proportion of molecules with sufficient energy to react" is given by
$$P(\varepsilon) = \exp \left(-\frac{\varepsilon}{kT}\right) \tag{1}$$
Therefore, for a reaction $\ce{X <=> Y}$ with uncatalysed forward activation energy $E_\mathrm{f}$ and uncatalysed backward activation energy $E_\mathrm{b}$, the rates are given by
$$k_\mathrm{f,uncat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f}}{kT}\right) \tag{2} $$
$$k_\mathrm{b,uncat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b}}{kT}\right) \tag{3} $$
The equilibrium constant of this reaction is given by
$$K_\mathrm{uncat} = \frac{k_\mathrm{f,uncat}}{k_\mathrm{b,uncat}} = \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{4}$$
As you have noted, the change in activation energy due to the catalyst is the same. I would be a bit careful with using "$\mathrm{d}E$" as the notation for this, since $\mathrm{d}$ implies an infinitesimal change, and if the change is infinitesimal, your catalyst isn't much of a catalyst. So, I'm going to use $\Delta E$. We then have
$$k_\mathrm{f,cat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f} - \Delta E}{kT}\right) \tag{5} $$
$$k_\mathrm{b,cat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b} - \Delta E}{kT}\right) \tag{6} $$
and the new equilibrium constant is
$$\begin{align}
K_\mathrm{cat} = \frac{k_\mathrm{f,cat}}{k_\mathrm{b,cat}} &= \frac{A_\mathrm{f}\exp[-(E_\mathrm{f} - \Delta E)/kT]}{A_\mathrm{b}\exp[-(E_\mathrm{b} - \Delta E)/kT]}  \tag{7} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \frac{\exp(\Delta E/kT)}{\exp(\Delta E/kT)} \tag{8} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{9}
\end{align}$$
Equations $(9)$ and $(4)$ are the same, so there is no change in the equilibrium constant.

The question then arises as to how eq. $(1)$ is obtained. The simplest way is to invoke a Boltzmann distribution, which almost by definition gives the desired form. However, since you have a Maxwell–Boltzmann curve, I guess I should talk about it a bit more. The fraction of molecules with energy $E_\mathrm{a}$ or greater is simply the shaded area under the curve, i.e. one can obtain it by integrating the curve over the desired range.
$$P(\varepsilon) = \int_{E_\mathrm{a}}^\infty f(\varepsilon)\,\mathrm{d}\varepsilon \tag{10}$$
where the Maxwell–Boltzmann distribution of energies is given by (see Wikipedia)
$$f(\varepsilon) = \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \tag{11}$$
At first glance, we would expect this to be directly proportional to the exponential part of the rate constant, i.e. $\exp(-E_\mathrm{a}/kT)$. Alas, it is not that simple. If you try to work out the integral
$$\int_{E_\mathrm{a}}^{\infty} \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \,\mathrm{d}\varepsilon \tag{12}$$
you don't get anything close to the form of $\exp(-E_\mathrm{a}/kT)$. Instead, you get some "error function" rubbish, and some nasty square roots and exponentials. (You can use WolframAlpha to verify this.)
Why is this so? Well, it turns out that there are other terms that also depend on $\varepsilon$ and therefore need to go inside that integral (they aren't constants and can't be taken out).
The simplest example is that faster molecules tend to collide more often, so even though the right-hand tail of the diagram seems to contribute very little to the "proportion of molecules with sufficient energy", it actually contributes more significantly to the overall rate because these molecules collide more often. In collision theory this is described using the "relative velocity" of the particles $v_\mathrm{rel}$.
There is also another complication, in that the Maxwell–Boltzmann distribution, the direction of the particles is not accounted for. (For more insight please refer to Levine Physical Chemistry 6th ed., p 467.) Therefore, there has to be yet another term that takes into account the direction of movement of the particles. The idea is that a head-on collision between two molecules is more likely to overcome the activation barrier than is a $90^\circ$ collision. The term that compensates for this is the "collision cross-section" $\sigma$.
If you go through the maths (and I don't really intend to type it out here, it's rather long, but I will give some references) then you will find that at the end you will recover the form $\exp(-\varepsilon/kT)$. Once you have arrived at this, it's very straightforward to see that the increases in rate of both the forward and backward reaction cancel each other out.
Now, as for the promised references, Pilling and Seakins's Reaction Kinetics pp 61-2 have a short outline of the proof. Atkins's Physical Chemistry 10th ed. has a slightly longer proof on pp 883-4.

In a catalysed reaction the activation barrier is not lowered but the catalyst causes the reaction to proceed by a different route which occurs more rapidly.
Thus there is a third party a 'chaperone' if you like that controls the way the two reactive species interact. This can be a surface, or a metal-organic or an enzyme, the details do not matter, but what matters is that the catalyst causes the reactants to undergo different reactions to that which would occur in homogeneous solution to produce the same product.
As the final outcome is the same so is the equilibrium. The kinetic approach is outlined by @orthocresol based on the Arrhenius equation. From a thermodynamic view, which is the most general, at equilibrium $\Delta G^{\mathrm {o}} = -RT\ln(K_e)$ and as $\Delta G^{\mathrm {o}}$ is unchanged in the reaction (the reactants and products are the same)  so is the equilibrium constant $K_e$.
Now to your approach using the Maxwell-Boltzmann (MB) distribution. It is not at all clear that Maxwell-Boltzmann distribution applies to a liquid. It is based on ideas around the perfect or ideal gas, and point like particles colliding but have no other interaction. In a liquid molecules do interact with one another since a liquid has a  meniscus and is distinct from its vapour.
However, there still has to be a distribution of speeds and hence energy. On collision the energy is converted into increasing vibrational/rotational energy in the colliding partners until by random chance enough energy is reached to cause reaction. As collisions can occur at at least $10^{12} \pu{s^{-1}}$ only very occasionally, perhaps only one in a million collisions, is enough energy present to cause reaction unless the activation energy is very low. In these cases diffusion limits the bimolecular reaction rate constant to $\lt 10^{11} \pu{dm^3 mol^{-1}s^{-1}}$ and often far less but depending on solvent viscosity. Thus in these cases it is the rate of approach that limits reaction not energy, so clearly MB energy distribution does not apply.
In solution it is normal to describe the rate constant of a (non-diffusion controlled)  reaction using transition state theory. This involves calculating the properties of the transitions state and reactants using statistical thermodynamics. The weakness of this approach is that the transition state properties have to be 'guestimated' in some way.
Appendix
[ Using the MB distribution an Arrhenius type expression can be obtained for a bimolecular gas phase reaction. The expression for the rate constant at energy E is
$$k(E) \approx \int_0^{\infty} Q(E)\sqrt{E} ~f(E)dE$$
where $f(E)$ is the MB distribution in terms of energy (by substituting $E=mv^2/2$), and $Q(E)$ is the reaction cross-section. Under conditions of thermal equilibrium the rate constant is the product of reaction cross section $Q(E)$, the magnitude of the relative velocity $\sqrt{E}$ and the thermal distribution of relative speed $f(E)$. The simplest form of $Q(E)$ is to suppose that it is a hard sphere of diameter d (like a billiard ball) in which case it has the form
$$Q(E)= \pi d^2(1-E_a/E)$$
when $E \ge E_a$ and zero otherwise. Performing the integration and converting to temperature produces and Arrhenious type equation
$$ k(T) \approx \exp(-E_a/k_BT)$$
where $k_B$ is the Boltzmann constant.]


text:
Why do vinegar and other acids remove rust?
I have started a project on restoring a very rusty knife. I had used vinegar to remove the rust. A question dawned on me: why does vinegar remove rust? So being the thoughtful person I am, I googled it. Unfortunately, Dr. Google could only provide me with what it is doing (which I already know).
But what I really want to know is: what chemical reaction causes the rust to disappear?

Googling without result usually means you are asking the wrong question. There are millions of chemical reactions, nobody remembers them all. Instead, we remember that certain classes of compounds tend to react with each other in such-and-such way. You identified one of the classes correctly, it is acids. Now what is rust?

I don't know whether this applies for vinegar (I don't believe so). But some acids remove rust by making it soluble in water, and hence it comes off, whereas others may aggravate it and make it worse. Sometimes it etches into the rust and flakes it off. Hope this helps others

The acid can dissolve rust (e.g. $\ce{FeOOH}$). The chemical reaction is:
$\ce{3CH_3COOH + FeOOH -> Fe(CH_3COO)_3 + 2H_2O}$
EDIT: ($\ce{Fe(OAc)_3}$is NOT soluble in water) The acid forms a water-insoluble salt with the iron oxide, which then probably just crumbles from the rust layer and precipitates in the container.
Mind that rust also contains for example $\ce{FeO(OH)_3}$ and $\ce{Fe_2O_3·nH_2O}$. The mechanism is the same.
Also, the acid oxidizes the upper layer of the iron under the rust to form a rust-resisting oxide-layer.

Also, the acid oxidizes the upper layer of the iron under the rust to form a rust-resisting oxide-layer. Meaning it won't continue eating the base metal?  But as soon as you remove it from the acid, the surface rusts very quickly again, so that coating isn't stable?

Vinegar contains  $\ce{CH3COOH}$ (acetic acid) which reacts with rust $\ce{FeOOH}$:
$\ce{3CH3COOH + FeOOH -> Fe(CH3COO)3 + 2H2O}$
and $\ce{Fe(CH3COO)3}$ (iron(III) acetate) is water soluble.

The wikipedia page claims it is soluble in ethanol and the citation that the article uses clarifies that it is not soluble in water https://en.wikipedia.org/wiki/Iron(III)_acetate so do you have a source for the last sentence?

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text:
Hydrogen bonding in concave rings
Can hydrogen bonding occur if both O and H are in a ring having 6 members  however the ring is concave .
Example
3 hydroxycyclohexanone

It doesn't work.  The ring geometry holds the carbonyl oxygen and hydroxyl hydrogen too far apart.  If you have a molecular model kit, construct the molecule with it and see what happens to the oxygen and hydrogen atoms.
Hydrogen bonding between different molecules is not constrained by the ring geometry, and thus works perfectly well.

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text:
Dissociation constant of molecular dinitrogen as a function of temperature?
I'm looking for the dissociation constant of molecular dinitrogen ($\ce{N2}$) as a function of temperature,
$$\ce{N2 <=>[$K$] 2N} \quad K=\frac{[\ce{N}]^2}{[\ce{N2}]}$$
Casas & Bermelo (1978) reported three data points for nitrogen (I couldn't access their references):
$$K_{298}=2\cdot 10^{-160},\ K_{2000}=10^{-3},\ K_{20000}=1.29\cdot 10^5$$

Are these the accepted values today? Is there a more complete database online?
Would it be correct to fit an exponential function using these data points to extract $K$ for other temperatures?

Update: If you plot the $\log(K_p)$ vs temperature it does not give a straight line over such a large temperature range so you are unlikely to be successful.
However, you can estimate $K_p$ as the molecular properties of dinitrogen are known so you can calculate the partition functions Z and so the equilibrium constant at any temperature. You need the vibrational frequency (and possibly anharmonicity), rotational constant and dissociation energy. Work out the partition functions for translation, rotation & vibration for dinitrogen and translation for N atoms and that is all you need.
The equations needed are given in most phys. chem. textbooks and specialist books on statistical mechanics. At a standard state of $1$ bar pressure
$$k_P= \frac{Z_{trans}(N)^2}{NZ_{trans}(N_2)Z_{rot}(N_2)Z_{vib}(N_2)}\exp(-E_0/(kT))$$
The experimental values at low temperature are likely to have the biggest error since dinitrogen is such a stable molecule, and it probably does not matter much how small the $K_p$ is its always tiny at low temperatures. Using the method above I calculated values as $2.4 10^{-160}, ~ 8.4 10^{-19}$ and $ 62400$ at the three temperatures you quote (using the dissociation energy as 9.756 eV).

did you get the constants from NIST?

There are also NIST records of ~e-33 at room temperature.

No I looked them up in Herzberg spectroscopy books. The constants are well known, however. I'm sure that the room temp values are difficulty to measure accurately as so little dissociates. It may not matter as the amount is so small, it will depend on what you need the number for. There seems to be uncertainty as to the dissociation constant.  It is not hard to write code to do the calculation, I knocked together a Python version quite easily.

The NIST data is for the rate constant not the equilibrium constant. When calculating values the atom Z has to be multiplied by its multiplicity, 4 in the case of  N atoms ($^4S_{3/2}$) and the rotational Z divided by 2 due to symmetry.


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In the nitration, why fluorobenzene has less products in the ortho position than Clorobenzene?
Is it because, due to the eletronegativity of Fluorine, his mesomeric effect will be less pronounced, which means the electronic density will be lower, than for the clorobenzes, in the ortho positions?

Mesomeric effect of fluorine is stronger. However, flourine has a stronger inductive effect too and its ortho position is more severly deactivated than chlorine's.

Halogens — though deactivating groups — are ortho-para directing. The inductive effect reduces the electron density at ortho position more than in the para position (inductive effect varies with distance). As fluorine is more electronegative than chlorine, you will get less ortho products with fluorine than with chlorine.

Halogens are deactivating groups even though they show $+M$ mesomeric effect. The inductive effect is stronger than mesomeric effect. As the halogens show $-I$ inductive effect, they are deactivating groups.
When it comes to selectivity, the mesomeric effect plays a role. As the halogens show $+M$ mesomeric effect, the resonance structure has higher electron density at ortho and para positions. Therefore, halogens are ortho-para directing.
When you take inductive effect into account for its directing properties, the para position is more favorable for electrophilic reagents because it has a higher electron density compared to the ortho position. The strength of the inductive effect decreases with distance. As the ortho position is closer to the halogen, the inductive is stronger at this position. Therefore, the major product for electrophilic substitution reactions for haloarenes is the para-substituted product.
Fluorine is more electronegative than chlorine and shows a higher tendency to withdraw electrons towards itself than chlorine. Therefore, fluorine shows a stronger inductive effect than chlorine. As fluorine reduces the electron density at the ortho position more than chlorine, you get a lesser amount of ortho substituted product with fluorine.


text:
Confusion on syn and anti
I came across the following compound syn acetaldoxime
However why is it syn because the lone pair having lowest priority is anti to H which is lowest priority on other side.
Please clarify

The lone pair and bulkier alkyl group is on same side so syn.

In an oxime, the syn or anti configuration is determined by the position of the lone pair on nitrogen with respect to the alkyl or aryl group. That is, syn configuration is one in which the lone pair on nitrogen and the alkyl or aryl group are in the same side, while, in anti configuration, they are on opposite sides.

By traditional convention, the drawing displays aldoxime in (syn)-configuration because both OH (of oxime) and remaining H (from the once aldehyde) are on the same side of the double bond.
Your right, on first sight (even on subsequent ones ...) it is puzzling.
An obvious alternative is to adhere again to the CIP-rules, where you check each side of the double bond for the substitutent of highest priority; which yields in the example provided by you the (E)-configuration.  Applying the CIP-rules comes with the advantage to be consistent with the naming around other C=C double bonds, and being applicable for ketoximes, too.
Side note:  Do not forget, often reactions affording the oxime yield a mixture of (E)- and (Z)-configurated product.  Albeit the (E)-aldoxime is thermodynamically favoured (sugar chemistry, for example), you may then recognise the patterns of the two isomers in the 1H NMR spectrum.


text:
What effect does diluting a buffer have on its buffer capacity?
What effect does diluting a buffer have on its buffer capacity? I am not worried about the effect on the pH, just the buffer capacity.

The buffer capacity β is a direct function of the concentrations of all protolytes in the solution. Thus, if the concentration of the protolytes decreases you should then expect a decrease in the buffer capacity.


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What is the structure of the hypothetical xenon octafluoride?
According to Frlec et al.,[1] theoretical studies indicated that there is a possibility (albeit slim) that neutral xenon octafluoride ($\ce{XeF8}$) might exist.
I am curious as to what structure this molecule might possibly have. In terms of hybridisation theory, I have only learnt cases up to seven-coordination ($\mathrm{d^3sp^3}$ or $\mathrm{sp^3d^3}$), where the molecular geometry is pentagonal bipyramidal. However, answers based on other frameworks (e.g. molecular orbital theory) would also be welcome.
My conjecture: could its structure possibly be hexagonal bipyramidal?

Frlec, B.; Holloway, J. H.; Slivnik, J.; Šmalc, A.; Volavšek, B.; Zemljič, A. An examination of the possibility of the existence of xenon octafluoride. J. Inorg. Nucl. Chem. 1970, 32 (8), 2521–2527. DOI: 10.1016/0022-1902(70)80296-2.

There does seem to be a good body of evidence, including your reference, for the existence of xenon octafluoride.
Regarding it's structure, according to VSEPR theory as tabulated here, the structure is square anti prism:

This is the xenon octafluoride dianion $\ce{[XeF8]^2-}$. Simple "xenon octafluoride" would be neutral $\ce{XeF8}$.

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text:
Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron?
We all know that $\ce{CuSO4.5H2O}$ is blue and it turns to $\ce{CuSO4}$ on heating which is white. I also learnt about coordination complexes and d–d transitions and how transition metal compounds are colored. Recently I came to know that $\ce{CuSO4.5H2O}$ is also a coordination compound and in its structure four water molecules are coordinated with $\ce{Cu^2+}$ and one is hydrogen-bonded to $\ce{SO4^2-}.$ So, I thought lets try to apply the theory here and see what causes the blue color of copper sulfate.
$\ce{CuSO4}$ has $\ce{Cu^2+}$ ions and I wrote down the electronic configuration as $\mathrm{(3d)^9(4s)^0}.$ Since no d–d transition is possible, it should be white.
$\ce{CuSO4.5H2O}$ is actually $\ce{[Cu(H2O)4]SO4.H2O}$ and the complex ion is $\ce{[Cu(H2O)4]^2+}$. $\ce{H2O}$ is a weak field ligand, therefore the electronic configuration is $\mathrm{(3d)^9(4s)^2(4p)^6}.$ But still no d–d transition is possible, so it should also be white!? What went wrong in my estimation?

Let's compare the two compounds, here I plotted the $\ce{Cu(II)}$ centers of $\ce{CuSO4.5H2O}$ and $\ce{CuSO4}$ from their crystal structure data.

As you can see, it changes from a $\ce{[Cu(H2O)4[SO4]2]^2-}$ to a let's call it $\ce{[Cu[SO4]6]^10-}$. So all effects that are typical for $\ce{Cu(II)}$ should remain, it is still an octahedron, there is still Jahn-Teller present, we only got the change from four $\ce{H2O}$ ligands to four $\ce{[SO4]^2-}$ ligands.
Those two are similar in their ligand field splitting, where the latter one is slightly weaker, so we expect a red-shift of the absorption energy. And that is also what Nagase et al. investigated on (Ref.1):

As shown partly in Fig. 1, the $\mathrm{d–d}$ band maxima of the hydrated and anhydrous copper sulfates were observed at $\pu{13200 cm-1}$ ($\ce{CuSO4.5H2O}$), [...] and $\pu{12500 cm-1}$ ($\ce{CuSO4}$), respectively, in their reflectance spectra. This is in agreement with the expectation from the spectrochemical series, and indicates that, in a series of complexes of comparable tetragonalities, the substitution of a coordinated water molecule with a sulfate oxygen weakens the apparent ligand field strenght about $\pu{150-200 cm-1}$. In addition to this shift of the absorption maxima, the following spectral changes were observed on dehydration: (i) the width of the $\mathrm{d–d}$ band becomes gradually narrower with the advance of dehydration; (ii) a broad and very weak absorption appears near $\ce{20000 cm^{-1}}$ in the spectrum of $\ce{CuSO4.H2O}$, and, with somewhat increased intensity, in that of $\ce{CuSO4}$; and (iii) a strong absorption (probably charge-transfer band) appears in the ultraviolet, and shifts remarkably to the lower frequency side with the advance of dehydration, indicating that the $\ce{[SO4]^2-}$ ions are getting more and more strongly bonded, as more and more sulfate oxygens are getting coordinated in the $x-y$ plane of the complex and shared among many $\ce{Cu^2+}$ ions.
[...]
Both the slight red-shift and the narrowing of the $\mathrm{d–d}$ band contribute to the disappearance of the blue color [...]

Or, as it is summarized in the book by K. Sone and Y. Fukuda (Ref.2):

Therefore, the disappearance of the blue color of $\ce{CuSO4.5H2O}$ caused by dehydration is only an apparent phenomenon, which is brought about by a relatively small shift of the $\mathrm{d–d}$ band to the red end of the visible spectrum (and partially into the near infrared) and perhaps also by the spontaneous pulverization of the dehydrated product; if we had eyes which could register radiation from near IR to near UV, we should certainly feel that $\ce{CuSO4}$ is much more darkly colored than $\ce{CuSO4.5H2O}$!
Since the ligand field strengths of $\ce{H2O}$ and $\ce{[SO4]^2-}$ are seemingly only slightly different from each other, these changes will only bring about a small shift in the $\mathrm{d–d}$ band towards the red. On the other hand, the $\ce{[SO4]^2-}$ ions, which are weakly held in the axial positions of $\ce{CuSO4.5H2O}$, are gradually brought into the equatorial coordination sphere, so that their CT interaction with the $\ce{Cu^2+}$ ion will be strongly enhanced. The observed spectral changes thus come about. Somewhat similar spectral changes are also observed when the double sulfates $\ce{M2Cu[SO4]2.nH2O}$ (where $\ce{M = Na^+, K^+, Rb^+, Cs^+}$ and $\ce{[NH4]^+}$) are heated.

References:

K. Nagase, H. Yokobayashi, K. Sone, “Spectrophotometric and thermal analytical studies on the dehydration of copper(II) sulfate and its double salts,” Thermochimica Acta 1978, 23(2), 283-291 (https://doi.org/10.1016/0040-6031(78)85070-9).
Kozo Sone, Yutaka Fukuda, In Inorganic Thermochromism (Inorganic Chemistry Concepts, Volume 10); Christian K. Jørgensen, Michael F. Lappert, Stephen J. Lippard, John L. Margrave, Kurt Niedenzu, Heinrich Noth, Robert W. Parry, Hideo Yamatera, Eds.; Springer-Verlag: Berlin, Germany, 1987 (ISBN 978-3-642-51019-9).

The $\mathrm{d^9}$ configuration is an unsymmetrical field, with the $\mathrm{t_{2g}}$ orbitals being completely filled while the $\mathrm{e_g}$ orbitals have only three electrons. So, according to the Jahn Teller theorem, these octahedral shapes will undergo slight distortions and will not be a perfect octahedron.
It is easy to see why it is so. Due to this unsymmetrical filling up of $\mathrm d$ electrons, some ligands will be repelled more as compared to other when they approach to form the complex. In this case, the $\mathrm{e_g}$ orbitals (which point directly at the approaching ligands) have two electrons in one while one electron in the second $\mathrm{e_g}$ orbital. So, the ligand along either $\mathrm d_{z^2}$ or $\mathrm d_{x^2-y^2}$ will be repelled more as compared to the other.
Hence it will experience some sort of elongation towards that side, and more importantly, to remove the asymmetry, further sub-splitting will occur in the $\mathrm{e_g}$ level, so even the restrictions of spin will be lifted. Clearly, the Laporte rule is no longer valid, and so, it can undergo $\mathrm{d–d}$ transitions. While in the anhydrous form, we can clearly see that a centre of symmetry will be present, and so transitions will be Laporte-forbidden.

Along with the downvoting,how about somebody actually points out the mistake in my answer?

As far as I'm aware, the Laporte rule remains valid in this case. It is, in general, valid for any centrosymmetric complex (i.e. has an inversion centre). The Jahn–Teller distortion from $O_\mathrm h$ to $D_\mathrm{4h}$ doesn't break this inversion symmetry. It is precisely because of the Laporte rule that the colour of Cu(II) complexes – which arises from forbidden d–d transitions – is very weak (contrast this with the intense colour of $\ce{MnO4-}$ which is LMCT, not d–d transitions).

And by "forbidden" I mean "very weakly allowed". In real molecules, the Laporte rule doesn't entirely forbid d-d transitions, because there are mechanisms by which it can be bypassed, such as spin-orbit coupling. Of course, in Cu(I), there are no d-d transitions to begin with, so it doesn't even matter whether they're allowed or not. Cu(I) can absorb EM radiation, but unlike Cu(II), it just isn't in the visible range, because the excitation energy is substantially larger than for the d-d transitions in Cu(II).

The blue color which arises when we dissolve $\ce{CuSO4}$ is indeed due to d–d transition. The reason behind this is the splitting effect.
The $\ce{Cu^2+}$ ion is hydrated by four $\ce{H2O}$ molecules, i.e it forms a $\ce{[Cu(H2O)4]^2+}$ complex. Due to the water molecules, the d-orbitals which were degenerate (of same energy) prior to the hydration, are split up into two discrete groups having different energy levels (they are called $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ groups). The energy difference (i.e the energy of photon which will be released due to the transition) between the groups is corresponding to the blue color seen.
Refer to crystal field theory (CFT) if you are willing to know more.

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text:
Error hunting: Guidelines on detecting error cancellation and locating their sources as much as possible?
It is known in computational chemistry, when computing e.g. electronic ground states energy and geometric paramters (bond lengths, angles, torsions etc.) of a given molecule using an ab initio or DFT methods with some basis sets, there are situations where a certain combination of method and basis will give good agreement with the experiment due to error contributions in that method/basis combinations having opposite sign of similar magnitude and thus cancel out.
This phenomenon is common in e.g. MP2 calculations of small molecules, where the electronic energy is usually in good agreement with experiment due to favourable error cancellations
However, not all calculations can benefit from error cancellations, and the presence of them will often mean they are quite sensitive to the nature of the molecule and the method/basis used to model them, making said method have less say on its reliability on similar molecules (e.g. with one ligand being replaced or rearranged in configuration)
Therefore, I wonder whether there are general guidelines in terms of what sequence of calculations to be carried out to test whether a good result is due to favourable error cancellation, or that the method faithfully reproduce the experimental result.
In particular, ideally if the error cancellation can be traced all the way down to the details and implementation of the theoretical method and basis set, such as the error cancellation in some molecule is due to the dipole interaction of a oxygen functional group with one of the aromatics, or that e.g. a polarisation function on oxygen atoms in that particular basis set contribute a term that is of opposite sign to another term in gaussian orbitals that model the delocalisation of electrons in an aromatic system.

Any good references that explores error hunting in computational chemistry and attempt at localising their sources?
(preferably in inorganic systems)

This lengthy paper by a researcher I personally find very knowledgeable explores in detail why mp2 is such an exceptionally good approach in quantum chemistry. It discusses many of your questions.  The literature cited there might provide you with further reading on other methods https://dx.doi.org/10.1063/1.4966689


text:
Why is water taken in calculating K, equilibrium constant
For the given question,

I was trying to solve it but I was not considering the concentration of water in the calculation (because water is a pure liquid and its concentration is supposed to be taken as 1), but then I needed the volume to be able to calculate the answer. I saw the solution,

Is it taken here because the reactants are organic and thus water would probably not be the solvent?

The question they gave you is somewhat flawed, in that there are four compounds and you don't know the range of their mutual solubility.
To really answer the question, you would need information such as from Solubility, liquid–liquid equilibrium and critical states for the quaternary system acetic acid–ethanol–ethyl acetate–water at 293.15 K Fluid Phase Equilibria vol. 313, pages 46–51.
The reactants, ethanol and acetic acid, are completely miscible; however, ethyl acetate only has limited solubility in water.
When you evaluate the equilibrium constant, all the reacting compounds, including water, which are in a single phase, must be included.

All reactants and products are in the liquid phase and mixed together in one vessel, so the total volume at equilibrium, $V$, is the same for each concentration we need to find $K$.  Also, all reactants and products are in one to one molar ratio in the balanced chemical equation.
Therefore, if we start with $\pu{3 mol}$ of each reactant, and reach equilibrium hen there are $\pu{2 mol}$ of the ester formed, we can be sure that there were also $\pu{2 mol}$  of water produced, and $\pu{2 mol}$  of each reactant onsumed, leaving only $\pu{1 mole}$  of each reactant in the vessel.
So, at equilibrium, the concentration of each reactant is $\frac{\pu{1 mol}}{V}$, and the concentration of each product is $\frac{\pu{1 mol}}{V}$, where $V$ = the magnitude of the total volume in completely arbitrary units, as it cancels out of the equation along with the units:
$$K=\frac{[(\frac{2}{V})\cdot{(\frac{2}{V})]}}{[(\frac{1}{V})\cdot{(\frac{1}{V})]}}= 4 $$

Welcome on the Chemistry SE! We have quite good Latex support, type in $\frac{2}{V}$ and you will get $\frac{2}{V}$.


text:
Galvanic cells/redox - confused regarding diagram

I'm not sure if I got this wrong because I interpreted the cell incorrectly. When they have an arrow pointing in to the tube for $\ce{O2}$, does this imply that $\ce{O2}$ is a reactant? If so then B is definitely wrong, which was my answer.
Is there any other reason why B is wrong? I wasn't sure if it was right or wrong when I did this question.

I think the arrow for $\ce{O2}$ just means there's oxygen gas in that part of the apparatus.
The reason B is wrong is that conversion of hydrogen peroxide to oxygen gas and water is the spontaneous reaction at these conditions. Assuming SLC implies 1 atm of pure oxygen, then everything is under standard conditions so we don't need to bring Nernst into it. The half reactions are as follows with the corresponding potentials.
$$ \ce{H2O2 + 2H+ + 2e- -> 2H2O} (\Delta E^{\circ} = +1.78) $$
$$ \ce{H2O2 -> O2 + 2H+ + 2e-} (\Delta E^{\circ} = -0.7) $$
so the net reaction is
$$ \ce{2H2O2 -> O2 + 2H2O} (\Delta E^{\circ} = +1.08) $$
Since $\Delta E^{\circ} $ is positive, this reaction is spontaneous.
The fact that this reaction doesn't happen means that there's a high kinetic barrier. I think that's what A is getting at, though I have a little bit of an issue with how it's worded. Regardless it's still better than B.


text:
How to calculate the gas-liquid interaction for dissolved HNO3 in water with air?
Thanks for reading this and (maybe) helping me in advance!
I have to face the following problem:
I have got $\ce{HNO3}$ dissolved in water (roughly $\pu{50\%}$) and I want to know how much $\ce{HNO3}$ will be in the air directly above it. I got classes about liquid-gas phases, but it was a long time ago and therefore I am not capable anymore to engage such a question.
As far as I can recall the pressure and temperature are important, which are $\pu{1 atm}$ and $\pu{20-25^oC}$ (but I want to calculate it for a range between $\pu{20-100^oC}$). I can imagine that it might be hard due to the fact that a part of the $\ce{HNO3}$ will split into ions $\ce{H+}$ and $\ce{NO3-}$ (till it reach its pKa), so that will give an equilibrium towards the liquid phase. But in the end there must be some $\ce{HNO3}$ in the air due to gas-liquid equilibrium.
Is there anyone that can tell me how to calculate this or how to engage this question? Any advice is appreciated!
Edit: During further investigation I found out that the problem is even more complicated since $\ce{HNO3}$ will split into $\ce{NO}$, $\ce{NO2}$ and $\ce{N2O4}$ (nitrous gases). If possible I want all the concentrations of these gases at those temperatures.

I think you could treat the effect of decomposition as negligible for such temp. and conc. but it still wouldn't do you much good if you don't have software like aspen or sth.

Thank you for your reply! (sorry for the late response) I might do it with Aspen then, since there is nearly no tutorial how to deal with this without software.
But as far as I know HNO3 decomposes allready at room temperature. The NO2 will desolve almost immediately and the O2 and H2O will be gone of course. But NO2 will not go further into NO and N2O4 according to you?
And about Aspen; Aspen doesn't have chemical reactions in its database right? So how does Aspen encouter the decomposition of the nitric acid?


text:
DSC Metal hysteresis (melt/cryst)
So we were analyzing Polymers using DSC and calibrated the system using Indium. I expected it to be a single transition only as Indium only has 1 known crystal structure and as I never worked with DSC nor seen a real thermogram before I was quite surprised to a relatively large hysteresis between metling and recrystallizing. I expected it to be a big gap for polymers but for something as uniform as a metal with only one crystal structure it forms I thought the point of melting and crystallizing should be quite close together.
I remember the lecture on solid state chemistry about DSC's of crystals where Ehrenfest and displacive/reconstructive changes in for example quartz-phases had hysteresis and I think displacive ones didn't have any but it was never really explained what caused this gap. I always imagined that it takes more time to form a symmetry like going against entropy and this is why more energy has to be taken out of the system.
But is there a better explanation for pure metals? Does this gap tell me anything about the structure in the end? I imagine that structures which tend to me rather amorphous might have a smaller gap because it's not so much symmetry involved in the process which would then also imply why it is big (relatively, I haven't compared it to any other compound yet) for Indium.

Disclaimer: the following is based on using DSC data for organic materials.
Is the melting temperature of Indium (156 C) just at the temperature where your sample exhibits a phase transition / decomposition?  In general, I suggest to use the flat indium pellets and to record their melting for calibration of the DSC, rather than their recrystallisation. This is likely described in the manual of your DSC.  (I does not hurt to have a look at DOI: 10.1007/s10973-015-5186-8, or similar, either.)
Because of its transient character, it is worth to keep an eye on the heating rate / cooling rate while recording DSC data.  Provided enough sample material, sometimes a quick screening with 10 K/min is followed by a second recording at a rate of 4  K/min (or slower).  While the endothermic peaks related to melting may be sharp, the ones for the recrystallisation are often shifted towards lower values because without proper seeds of crystallisation, supercooling may occur, especially organic samples and polymers.
I recommend to consider cyclic DSC, instead of a simple single-heating experiment, as this offers you additional insight about the sample in question. It is applied to characterize pure metals (like DOI: 10.1098/rspa.1996.0113), and may shed a light on the degree of crystallinity and processes of reorganisation of your polymer sample, too.  In the following, a DSC of polylactic acid:

(source)
Sometimes these differences about recording the melting point (pristine sample, after 1st / 2nd heating cycle) are spot on small molecule materials, too.  They may appear even if the sample did not incorporate solvent molecules.


text:
E and Z configuration

In the above we need to find the configuration of the two double bonds whether (E) or (Z) according to me the one with the cyanides should be (E) and the other (E) however my book say that the correct answers are (E,Z) (in the same order as above) . Please explain how ?

Please show your precedence order; that way it would be easier to spot an error, if any.

Are you familiar with the Cahn-ingold-Prolog sequences?

I get (E) for both alkenes as well.

Chemdraw gives (E) for both alkenes also.

You are correct; the answer in your book is wrong. I get (E) for both alkenes.
For the alkene on the upper right:

$\ce{COOH}$ has precedence over $\ce{CH2OH}$ because the first carbon atom is bonded to $\mathrm{ \{O,O,O\}}$ and the second is bonded to $\mathrm{\{O,H,H\}}$.
Clockwise around the ring has precedence over counterclockwise. While both initially are $\mathrm{\{C,H,H\}}$, the next carbon atom clockwise is $\mathrm{\{C,C,H\}}$ while the next carbon atom counterclockwise is $\mathrm{\{C,H,H\}}$.

For the alkene on the lower left, the nitrile groups win in both cases:

On the left: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{N,H,H\}}$
On the right: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{C,C,H\}}$

I was thinking along those line, but wanted to confirm isn't it priotities decrease clockwise?

Around the ring from the alkene in the upper right, clockwise > counterclockwise since CCH > CHH at the second carbon atom out.


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The effect of NaCl on the ability of an acid to protonate benzoate
I've been doing several experiments using soda to quantify the amounts of caffeine and benzoate. Because of our methods of analysis, we had to decarbonate our soda. The experiment offered two ways for decarbonating the soda: shaking it or heating it. Then, to make sure all the benzoate in the soda is converted to benzoic acid, we add a small amount of $\ce{HCl}$ (to protonate it.)
Several years ago, I did a similar experiment to quantify the amount of caffeine. To decarbonate the soda, I added salt ($\ce{NaCl}$.)
I'm wondering if adding salt would be inappropriate in the more recent experiment. None of the lab procedures mention adding salt and, to me, this is the easiest way to decarbonate soda. Therefore, I'm thinking there may be a specific reason the salt method was left out of the procedure.
My thought is there could be some interaction between $\ce{Na+}$ and $\ce{Cl-}$ that could "neutralize" the $\ce{HCl}$ so to speak, making the protonation of benzoate less effective.
Perhaps it would make sodium benzoate instead of benzoic acid? Perhaps they just didn't think to add salt. Any thoughts on this?

Some salts added obviously will alter the pH value of the soda ($\ce{NaHCO3}$, for example), and others not (like the example provided by you, $\ce{NaCl}$). Changing the ionic strength by addition of salts may render your mathematical model to describe the composition of the analyte more complex.
Hence, let me I suggest an alternative to outgassing the soda: immerse the open container into an ultrasound bath.  This works rapidly, however ensure the container is widely open and may accommodate the foam briefly generated.  Otherwise, it may surprise you as a sticky geyser.  Second hint: gently approach the active regions in the ultrasound bath to moderate the gas evolution.
(Degassing solvents with ultrasound is a common preparation for HPLC, too.)

When the salt concentration is increased, some of the water molecules are attracted STRONGLY by the salt ions, which decreases the number of water molecules available to interact with the polar CO2 of the soda.
As a result of the less availability for solvent molecules, CO2 molecule find it difficult to remain in the aqueous medium, hence escape.
This would then be possible with a number of other salts as well, if these are having good extent of hydration.
If this is so, please let us know.

Na+ + BzCOO- ◄► BzOONa is an equilibrium. Part of the answer is that increasing [Na] will shift the reaction towards the right. I guess heating the liquid makes sense, I've no idea how consistently you can remove the excess CO2 by shaking. I've also no idea how NaCl changes the solubility of CO2. Have you considered doing simple weight change experiments to compare the three techniques? You should. You should also consider that, in general, adding anything to an analyte increases the chances that there will be interference. Since the analytical techniques (as well as the work-up processes) weren't specified (to be honest, I wouldn't be able to help any more if they were, I'm almost certainly not familiar with the chemistry behind them), there's no way for me to say whether NaCl would interfere or not. I'm sure you understand there are other chemicals present in the "soda", so that your idea of adding NaCl would require pretty stringent verification of its "non effect" on the results. I sure wouldn't go there. I've been involved in the creation of analytical techniques and the process can be quite tedious as well as lengthy. (Co-incidentially, CO2 absorbed from the atmosphere turned out to influence not only the NaOH reagent solution (virtually no storage stability), but also the result of the titration (both time and mixing speed) itself. I sure wouldn't go there. But that's what literature searches are for.


text:
sn2 mechanism help please?
My mechanism for this conversion is an intramolecular attack sn2 to displace Cl- anion. OH- can then attack this aziridinium intermediate, but will the nucleophile not attack preferentially at the most substituted site (where greatest positive charge resides i.e. tertiary center)? This gives the incorrect product?

Steric factors are more important here. The positive charge resides on the N and the nucleophile attacks the least hindered site hence the product is as drawn.


text:
Why did phenol respond to Brady's reagent?
I was teaching a friend a few basic organic analysis experiments, during which I was showing him, how to get an alcohol, and took a sample of phenol.
The neutral $\ce{FeCl3}$ test went fine(which i mentioned, was only for phenols), with an ultra dark violet colouration, but just to show him a negative test, i added some of Brady's reagent, and surprisingly, it gave a yellowish colouration (something close, a mix of yellow and orange, couldn't trust my eyes, since i was kind of in disbelief). The samples in the lab are mostly pure, and not mixtures of different organic compounds. To satisfy him, i gave a possible explanation, saying that the phenol might've got oxidised to hydroquinone, and hence, might've given the test.
What probably would've happened there? Any chances my explanation was right?

Reagent has color by itself there needs to be precipitate.

Im not sure,  just saw the solution turn yellowish, didnt check for a precipitate.  Probably thats why it happened @Mithoron


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Disrupting the chemistry of oobleck so that it behaves like a newtonian fluid
Oobleck (a mixture of cornstarch and water) is a non-newtonian fluid, so it acts like a solid when under pressure, and like a liquid otherwise. My question is this, is there anything you could mix in with oobleck (a friend suggested vinegar, I don't know why) that would cause it to stop being non-newtonian (more specifically, to act like an ordinary liquid). The motivation for this is that I'm going to have a large amount of the stuff, and will need a practical way of disposing of it, like dumping it down a drain.
Thanks in advance for any answers!

Cornstarch and water is a great way of thickening sauces ;) P/S please don't really do this

Disposal isn't a problem - add more water as you dump it down the drain.
There are two extremes to ooblek:

Almost all cornstarch and almost no water: this is a dry solid or slightly wet starch. No big deal.
Almost all water and very little starch: this is some water with a little bit of starch mixed in. Sometimes it clumps or falls out of suspension, but it's also no big deal.

The non-Newtonian mixture occurs at intermediate ranges.
Since cornstarch will degrade in the sewer system (e.g., feed bacteria), it's safe down the drain. Alternatively, I compost it.
But if you're worried about non-Newtonian behavior, you just add more water.

As the old  saying goes, "dilution is the solution to pollution".


text:
How do hydrogen bonds break the ionic bonds of a crystal lattice?
In the case of water dissolving sodium chloride. I was under the impression that hydrogen bonds were much weaker than the ionic bonds in sodium chloride, so how do they overcome the ionic bond strength? Do the ions simply get swarmed with enough water molecules so that the combined dipole interaction forces break the ionic bond? Or is it possible that because the sodium chloride dissolution reaction is endothermic the hydrogen bonds on their own are not enough to break the ionic bonds so the reaction must draw energy from the surroundings to do so (the reaction must be entropically favorable to do so, right?)?

Ion-dipole interactions, not hydrogen bonds. Hydrogen bonds are between water molecules.

Your presumption that it is hydrogen bonding that breaks the ionic bonds is wrong. Forget H-bonds. It is true that they exist in both pure water and in a sodium chloride solution. Water molecules surround both the Na+ and Cl- ions. The interaction there is dipole-ion. Na-OH2 and Cl-HOH. As is obvious, only one of those involve an H atom. (note that the "-" I used is not intended to suggest either a covalent or an ionic bond, rather it is intended to show the electronic interaction between the two oppositely charged species (the Cl & Na ions have full integer charge, the H & O atoms have only partial opposite charge, mostly induced.) This question can be answered just by hand-waving about the thermodynamics: the energy of the solvated Na and Cl ions is less than that of those two ions in the crystal. So, on average, the ions will leave the crystal and enter the liquid (until their concentration is so high that equilibrium is established between the two phases). So, while it is true that HOH...Cl bond can be called a H-bond, thinking about it this way has no obvious advantage as far as I can see. The dipole-ion interactions that are occurring are indiscriminate. Each ion is surrounded by lots of (oriented) water molecules. To learn more see https://en.wikipedia.org/wiki/Solvation_shell and especially https://en.wikipedia.org/wiki/Solvation

You should be more concerned with formatting, IMO.


text:
Typical values of activation energies?
Where can I find typical values of the Arrhenius activation energy for arbitrary reactions? I only want to have orders of magnitude estimates, but so far I have not been able to find any good resource that lists activation energies for different types of reactions.
There are tons of papers in which people determine precise values for very specific reactions, but I am looking for a kind of database or reference that has a large collection of values. In particular, those for biochemical reactions would be useful to have.

You can roughly estimate based on the expected half-life and the Eyring equation. See: http://chemistry.stackexchange.com/questions/61602/how-to-relate-a-reaction-barrier-to-the-time-the-reaction-needs-to-proceed

That is because there are no typical values, and this is why only individual values are quoted.  Fast reactions have small activation energies, slow ones large ones, but what is fast, slow and small or large is subjective.  What is true, however, is that because of the exponential dependence  of rate constants (1/rate const = lifetime) relatively small differences in activation energy cause a huge difference  in reaction lifetime as pointed out by @Zhe.

NIST has an kinetics database which might be helpful to you.
In there own words:

The NIST Chemical Kinetics Database includes essentially all reported
kinetics results for thermal gas-phase chemical reactions. The
database is designed to be searched for kinetics data based on the
specific reactants involved, for reactions resulting in specified
products, for all the reactions of a particular species, or for
various combinations of these. In addition, the bibliography can be
searched by author name or combination of names. The database contains
in excess of 38,000 separate reaction records for over 11,700 distinct
reactant pairs. These data have been abstracted from over 12,000
papers with literature coverage through early 2000.
Rate constant records for a specified reaction are found by searching
the Reaction Database. All rate constant records for that reaction are
returned, with a link to "Details" on that record.
Each rate constant record contains the following information (as
available):
a. Reactants and, if defined, reaction products;
b. Rate parameters: $A$, $n$,
$\frac{E_a}{R}$, where $k = A(\frac{T}{298})^ne^{[-(\frac{E_a}{R})(\frac{1}{T})]}$, where $T$ is the
temperature in Kelvins;
c. Uncertainty in $A$, $n$, and $\frac{E_a}{R}$, if reported;
d. Temperature range of experiment or temperature range of validity of a
review or theoretical paper;
e. Pressure range and bulk gas of the
experiment;
f. Data type of the record (i.e., experimental, relative rate
measurement, theoretical calculation, modeling result, etc.). If the
result is a relative rate measurement, then the reaction to which the
rate is relative is also given;
g. Experimental procedure, including
separate fields for the description of the apparatus, the time
resolution of the experiment, and the excitation technique. A majority
of contemporary chemical kinetics methods are represented.

It has a simple reaction search feature as well as an advanced database search feature which is quite intuitive to use.


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What volume of water will dilute an acidic solution back to neutral?
Now we are talking about simple dilution, not acid-base neutralization, so we expect asymptotic behavior that never quite gets us back to pH=7; that is, an infinite amount of water would be required. However, most pH probes read to two decimal places, so diluting to reach pH=6.995 is a sufficient end condition. So, given an initial pH of $pH_0$ and an initial solution volume $V_0$, what additional volume $V$ of pure water would be required to dilute the pH to 7.00 (6.995)?
I've wondered this for a while, and today I attempted to solve it. I may have found the solution, but I am skeptical of its implications. If I have 1 L of a solution at pH=2, my derived formula indicates it would require additional volume of about 8.6 million liters of water for the solution to read neutral. Is this reasonable?
My derived formula:
$$V=\Big(\frac{1}{1-10^{-0.005}}\Big)\Big(10^{6.995-pH_0}-1\Big)V_0 \\
\approx \Big(86.36\cdot10^{7-pH_0}-87.36\Big)V_0$$
Derivation:
Moles of $H^+$ from original soln: $n_0 = [H^+]_0V_0$
from added water: $n = [10^{-7}]V$ (via autodissociation)
Total moles: $n'=n_0+n$
Total volume: $V'=V_0+V$
New molarity after addition of $V H_2O$:
$$[H^+]'=\frac{n'}{V'}=\frac{[H^+]_0V_0+[10^{-7}]V}{V_0+V}$$
$$\Rightarrow V=\Big(\frac{[H^+]_0-[H^+]'}{[H^+]'-[10^{-7}]}\Big)V_0$$
(Note that $V \to \infty$ as $[H^+]' \to 10^{-7}M$, an infinite volume is require to exactly reach neutral pH).
Letting $[H^+]'=10^{-6.995}M$ (paragraph 1), $[H^+]_0=10^{-pH_0}$, and multiplying top and bottom by $10^{6.995}$, obtain:
$$V=\Big(\frac{10^{6.995-pH_0}-1}{1-10^{-0.005}}\Big)V_0$$
Question: Is this derivation valid? Are there assumptions I should or should not have made? Would it truly take 8.6 million times the original volume to dilute a pH=2 solution back to neutral pH=7.00?

Since water establishes an equilibrium with $\ce{OH^-}$ as well as with $\ce{H^+}$, I don't think you can leave that out when considering what happens in the system as you add water.

The very idea of diluting a solution back to (almost) neutral with pure water is unreasonable, precisely because it would require insane amounts of water. That being said, your derivation (though impractical) is not meaningless; it is just slightly wrong.
Like J. Ari said in the comments, there is an equilibrium involving $\ce{H+}$ and $\ce{OH-}$, and it will be shifted by the dissolved acid (that is, the concentrations will shift so the equilibrium constant may stay the same). Therefore you can't assume that water will always contain precisely $10^{-7}$M protons from autodissociation (plus maybe some extra from acid). It doesn't work that way.
How does it work, then?
$$\ce{[H+]\cdot[OH-]=10^{-14}}\\
\ce{[H+]-[OH-]=}c(\rm{HCl})$$
(since the acid is strong, and hence fully ionized even in not-so-diluted solutions)
$$\ce{[H+]=10^{-6.995}}\\
\ce{[OH-]=10^{-7.005}}\\
c\ce{(HCl)=10^{-6.995}-10^{-7.005}=10^{-6.995}(1-10^{-0.01})}\\
V_{total}=V_0\cdot{c_0\over c}\\
V_{added}=V_0\cdot\left({c_0\over c}-1\right)
=V_0\cdot\left({10^{6.995-pH_0}\over1-10^{-0.01}}-1\right)$$
So it is only 4.3 million times the original volume, which is about 2 times less than your estimate.
Not that it made much of a difference, really.


text:
Why is only one part of buffer in the reactants side?
Given that buffers consist of either

a weak acid and its basic salt, or
a weak base and its acidic salt

and that the buffer consisting of CH3COONa and CH3COOH is added to water (H2O) ...
The reaction is:
$$\ce{CH3COOH + H2O <--> CH3COO- + H3O+}$$
(Equivalent reaction is $\ce{CH3COO- + H3O+ <--> CH3COOH + H2O}$)

Why aren't both parts of the buffer that is being added to the water represented in the reactants side of the reaction?
In other words, how is it that one part of the buffer is making another part of the buffer if both needed to exist in order to be a buffer to add in the water in the first place?

A buffer is a solution that is meant to prevent the pH from drifting away from its intended value in both directions (i.e. the solution should neither become more acidic nor more basic than desired). For the buffer to accomplish this, two components are added one of which is intended to counteract acidic additives the other to counteract basic additives.
Whether or not these two components are conjugate Brønsted acids and bases does not matter; a solution of e.g. ammonium acetate will also buffer. Every liquid inside the human body is buffered by multiple components to a very narrow pH range, so many different compounds together can also produce a buffer effect. However, the combination of an acid and its conjugate base is typically the most effective and narrow-range buffer system available in a typical lab context — and also the easiest to set up and use.
While in the buffered solution the two components of a buffer can indeed be said to react; this can be proven with isotope labelling. For example, if you had $\ce{^13C}$-labelled acetic acid and $\ce{^12C}$-labelled acetate, these two would react in the following way:
$$\ce{CH3^12COO- + CH3^13COOH <=> CH3^12COOH + CH3^13COO-}\tag{1}$$
This reaction does not, however, describe the buffering action.
Buffering itself can only take place if some acidic or basic compound is added to the solution as hinted above. Depending on what is added either equation $(2)$ or equation $(3)$ will take place.
$$\begin{align}\ce{CH3COO- + H3O+ &<=>> CH3COOH + H2O}\tag{2}\\
\ce{CH3COOH + OH- &<=>> CH3COO- + H2O}\tag{3}\end{align}$$
The first equation $(2)$ is the one if something acidic is added, equation $(3)$ takes place if a basic compound is added.
As you can see, one component of the buffer basically exists to quench basic additives (the acidic part) while the other only exists to quench acidic ones (the basic part). Therefore, in a buffering equation you will only see one of the two active at once.

Both parts of the buffer are not on the reactants side as they do not react. The equation you mentioned is simply the dissociation of ethanoic acid in water. This produces the ethanoate ion (which would be in the salt) and a hydrogen ion. When you make a buffer using a weak acid and its salt (or a weak base and its salt), the salt does not react with the acid. Essentially what happens is the acid dissociates (albeit, to a very small extent), producing the conjugate base and hydrogen in an equilibrium reaction. Adding more salt is the same as adding more of the conjugate base which shifts the equilibrium to the left, producing more acid. This creates the buffer. The actual dissociation produces very little ethanoate (roughly less than 1% of the acid is dissociated). At no point does the acid react with the salt directly as they are part of separate reactions. The acid dissociating is the forward reaction and the salt reacting with the hydrogen ions is the backward reaction.


text:
Which white powders form acids when mixed with water?
I have a mystery acid which isn't citric, boric, or oxalic acid. When mixed with water, it smells pungent, like hydrochloric acid--though I haven't compared them side by side.
Which other white powders form a strong acid upon being combined with water? Is there any powder that decomposes into HCl when mixed with water?
(Note: the mystery acid was sold as part of a set, along with sodium chlorite, for generating chlorine dioxide for use in aquariums.)

Does it dissolve in water completely?

@IvanNeretin It dissolves completely except for some chalky looking impurities. (The other powder, the sodium chlorite, also leaves impurities visible in the water when dissolved.) I'll measure the solubility when I get home.

Using some of "red cabbige pH indicator" (http://www.discoveryexpresskids.com/blog/test-ph-levels-with-red-cabbage), is the solution of X in water rather acidic, or basic? In other words, "mystery acid" is based on knowing that it is an acidic material?

@Buttonwood I used full-range pH paper (not that precise), but it looks like it's pH 2 or lower. Certainly 2.5 or less. I'll have narrow-range acid pH paper next week, and that'll let me estimate whether it full disassociates. (Is that right? If not, it'll get stuck at some pH, and if so, it'll get more acidic the more I add.)

Is it sodium acetate? Because it is very soluble in water to form acetic acid. It has pungent smell and has ph of 2.4.

@NilayGhosh Thanks, that's good to know, but no vinegar smell. Plus it reacts with NaClO2 to immediately (almost violently) form ClO2, and vinegar doesn't have that effect.

@IvanNeretin What were you getting at regarding complete disolution? I found it leaves quite a bit of white "scum" floating on the top when it's dissolved. The solubility is high--at least 40g per 100g of water at around 25 deg C.
Are acids ever supplied as meta-materials with an acid gas adsorbed over another substance? I think at the very least, my mystery acid is a mix, because I think I also smell a hint of chlorine.

Is it sodium hypochlorite aka bleach (NaOCl)? It has a very pungent chlorine like smell, soluble in water to form hypochlorous acid (pH = 4.0).

@NilayGhosh I wasn't aware hypochlorite could be packaged as a solid. If so, that's probably one component of the powder, unless the powder contains/yields HCl. (Unless my nose deceives, it definitely contains chlorine. But the smell is more "aggressive" than bleach. More like HCl.) NaOCl can't be the main acid, though--if I could trust my color sense more, I would say the pH was 1.

@piojo I was thinking of some salts that hydrolyze easily.

@NilayGhosh Don't confuse people. Sodium acetate solution does not have pungent smell; more importantly, it has alkaline pH rather than acidic. True, it does contain some minuscule amounts of acetic acid resulting from hydrolysis, but that doesn't make it an acid.

@IvanNeretin From a chemical engineering perspective, this powder should ideally contain (or yield) either straight HCl, or HCl and NaOCl, because either combination is used to produce chlorine dioxide when mixed with NaClO2. I'm not sure whether a salt of HCl would behave the same way. Does HCl have (solid) strong acid salts?

Did I say "salts"? I meant "chlorides" (including covalent chlorides). $\ce{PCl5}$ would behave like that, but it is most likely too aggressive to be actually used.

@IvanNeretin Thanks! I have a feeling this is heading in the right direction. The material is probably moisture sensitive, due to being packaged in an extra plastic bag. It's not PCl5, since I can put it in water or drip water onto it (sorry, shouldn't even have tried that) without any apparent reaction. Next, I'll try dissolving a granule and letting it dry. If this is a hydrolyzable chloride, its size will decrease after evaporation. I can also probably tell that by seeing whether it's infinitely "soluble" in a drop of water.

From the pH and solubility information alone, I would propose sodium bisulfate: cheap, readily available, safe in the powerfully oxidizing solution. (Possibly also potassium bisulfate.)
One way of manufacturing $\ce{HCl}$ produces sodium bisulfate as a byproduct:
$$\ce{NaCl + H2SO4 -> HCl + NaHSO4}$$ The odor of $\ce{ HCl}$ could be a minor impurity in this mystery acid. The other impurities also suggest it is not a reagent grade product.

Nice suggestion! It looks the same as the bisulfate photo on wikipedia, and based on its packaging in a plastic bag within a bottle, I suspect the mystery acid is also hygroscopic. The one wrinkle in this theory (and it's minor because I don't trust my perception) is that I thought I also smelled chlorine (like a pool smell). Still, you're likely right. I'll try to think of a way to confirm.


text:
In which pH range are silica-based TLC plates stable?
I am trying to understand why the experiment asks for raising the pH of a fairly acidic solution. The analytes should not be affected by it, so I thought the silica plate might be. Are silica-plates vulnerable in acidic conditions and, if yes, what happens?

I should add that the goal pH was between 6 and 7.

The pH value of an aqueous solution of $\ce{NH4Cl}$ is mildly acidic, and with an aqueous solution of $\ce{NaHCO3}$ you catch much of acidity, too.  If you prepare a long chain alcohol, for example, the quench with aqueous $\ce{NH4Cl}$ is meant to have no longer the alcoholate generated by the Grignard reaction, for example; simultaenously, to remove remains of salt ($\ce{MgBr2}).  In this perspective, the then extracted organic phase is pure enough to attempt the TLC, regardless if the pH of an aqueous solution in contact with the organic phase were 7.00 or about 6.

At the stage of monitoring the progress of the reaction, or to monitor the quench and (extractive) workup of the reaction, chromatography does not yet deal with purification; which may be done subsequently by column chromatography, or other means like distillation, recrystallisation.

Assuming "normal" TLC with silica as stationary phase, and a mix of ethyl acetate / hexane as eluent, charged compounds (i.e., salts) do not elute well; their adsorption is too tight to move forward.
Hence, if your reaction is performed under acidic conditions (sulfonation reaction, for example), drop a small sample of your reaction into a test tube or small vial, gently quench it against an aqeuous solution of $\ce{Na2CO3}$, extract it with some ether / MTBE / methylene chloride (well...) and spot the extract on the TLC plate.  With some training, this works easily around o.5 mL of total volume.
Conversely, if your reaction proceeds under basic conditions (like a Grignard reaction, for example), bring the sample back to neutral, for example with dilute HCl (or, with more fragile products like tertiary alcohols, with an aqueous solution of $\ce{NH4Cl}$), and then spot the organic extract on the plate.

I referred to "normal" TLC as there are other stationary phases than silica, too.  Alumina (basic, neutral, acidic) works with the same principle as silica.  These two contrast to applications of RP-TLC, were the stationary phase is coated with alkyl chains, and much more polar solvents are deployed as eluents, hence with much lesser issues of free salts / free acids. But the price tag for these plates is higher than for normal phase TLC.


text:
Counting the degree of unsaturation
What is the difference between a ring and an aromatic ring here (degree of unsaturation)?
For a benzene molecule, they count it as four, so it seems like benzene contains a ring...
 but doesn't it contain an aromatic ring? I am very confused.

Ignore that line altogether, as if there were no such thing as aromatic ring. Benzene contains a ring (1) and three double bonds (3*1), thus making a total of 4, that's it, period.

Well, I would suggest you to ignore these lines in your textbook and simply use some formulas for finding the degree of unsaturation of compounds.
For a hydrocarbon,
DU = ((2×C+2) - H)/2
For a compound having carbon, hydrogen and nitrogen,
DU = ((2×C+2) - (H - N))/2
For a compound having carbon, hydrogen and oxygen,
DU = ((2×C+2) - H)/2
For a compound having carbon, hydrogen and any halogen,
DU = ((2×C+2) - (H + X))/2
where, DU is degree of unsaturation
             C is no. of carbon atoms
             H is no. of hydrogen atoms
             N is no. of nitrogen atoms
             X is no. of halogen atoms


text:
Why do they flush people with water after chemical attack?
I recently viewed a video (dated April the 6th 2017), which purports to show the aftermath of a chemical attack (where Sarin was employed).
Could anyone explain why the victims in the video were doused with water? To what degree does such measure help?

Not much. It could have been of some help against blister agents, but not against nerve agents.

Probably because water is all they have immediately to hand and in the hope of physically flushing it off the skin. Chemically it would not be very effective at deactivating sarin.

A standard protocol after unintended chemical exposure is to quickly douse the victim in large quantities of water to wash away chemicals that could remain and cause further harm, such as powders and other types of solids that could spread over your body as you moved. Also, the water will dilute any chemicals to the point of least harm that soak into clothing (provide a large amount of water is used). However, as Ivan Neretin pointed out in the comment, sarin gas does not do physical damage to the external parts of the body, but enters the body and causes nerve damage extremely quickly. In this type of situation, the water dousing probably did little to nothing to help the victims.

so basically they are doing it for nothing (the watter spraying). Since all of them a paralyzed, shaking etc. There aren't any visible burns on their bodies

But if sarin sticks to your hair and cloth etc washing does help otherwise you would inhale stuff hanging in your clothes and hair...

I think the water dousing is not completely useless, see the possibility Fl.pf. posed, but it has very limited positive effect. The water will not help alleviate the problem compared to a situation like spilling a corrosive liquid on yourself.

The dousing may also help any emergency responser not get contaminated.


text:
How well can triglycerides be resolved using TLC?
The only part of triglycerides that significantly interacts with a TLC plate are the ester moieties which are present in all triglycerides. Other than that, they do not differ much, except in mass (dependent of the chain lenght of FAs) and shape (saturation degree of FAs). How much do these differences alter the interactions causing the resolution?

If you work with normal phase TLC, you may lower the polarity of your mobile phase, and instead of mixing ethyl acetate with petrol ether, you may give diethyl ether / petrol ether (like 5:95, v/v) a try.
Some of the TLC manufactures even provide databases, pointing towards publications were compounds of interest were successfully separated.  Often, these references refer to a separation on TLC-scale.  CAMAG's CCBS Database is one noteworthy example.


text:
How does this oxazolone behave as an acid?
how can you explain the increased acidity (pka around 14) of this oxazolone compared to a standard cyclic amide/lactam)?

It's a cyclic carbamate so not really comparable to a lactam. The O has a major electron'withdrawing effect. The enolate/anion is also stabilised by delocalisation.

so would the most acidic proton be @ N? Is both inductive and mesomeric delocalisation at work in the conjugate base?

Yes, the NH is most acidic and both factors contribute to make the conjugate base stable.

Better question may be why it would be only that weakly acidic.

It's a cyclic carbamate so not really comparable to a lactam. The O has a major electron withdrawing effect. The enolate/anion is also stabilised by delocalisation.


text:
Selectivity of m-CPBA
I have a question about the selectivity of m-CPBA in this reaction. In my experience, I have used m-CPBA for epoxidation of alkenes. It surprised me that in this reaction, the ketone reacts to give an ester and the alkene remains unchanged? Is it due to the presence of the overhanging ether group? Does this provide a sort of steric crowding so that the peroxyacid can't get in? Cheers.

Did you make two accounts? I may cause problems...

The transformation taking place in your question is a name reaction known as a Baeyer-Villiger.
The Baeyer-Villager transforms ketones into esters/lactones (cyclic esters) via an alkyl migration in which the group most able to stabilise a positive charge is observed to migrate.
A general mechanism is presented below (taken from the wonderful name reactions book cited). The key steps are (1) attack of the peroxide into the carbonyl followed by (2) the alkyl migration to form the ester grouping.

Source: Strategic Applications of Name Reactions, 2005.

That is great-thank you for the reference and mechanism of tranforming the ketone to an ester. However, I am still unclear why the peroxyacid selects the ketone functional group over the alkene-is it simply more reactive?

Its not always clear cut which will happen. Fundamentally it depends on the specific alkene and the specific carbonyl.


text:
Can atoms other than hydrogen be reduced to a bare nucleus?
In acid-base theory, hydrogen is able to become the $\ce{H+}$ ion, or a bare proton. I'm curious if other, larger atoms can have their electrons stripped to the point of being only a nucleus. I know helium can in the form of an alpha particle, but are there any other examples? For example, could you have $\ce{Li^3+}$?

While technically it is true, in real chemical environment the H+ is never a solo proton going around: it is always attached to something (solvent or other molecules). The reason is the a bare proton (or other nucleus) would be very unstable energetically, and strongly polarize its environment.

http://chemistry.stackexchange.com/questions/41997/does-the-hydrogen-ion-actually-exist

Only in the low pressure gas phase it is possible to strip all the electrons off atoms, and as pointed out by @Greg ions in solution are always going to be solvated, which true also of the electron.

The fact that ionization energies have been measured for many single-electron ions is clear proof that bare nuclei, other than the examples mentioned in the question, have been produced. However, some might say that the process of ionization is more of a physical than a chemical one. This distinction is left to the reader.

In the atomic physics community, they are 'fully stripped' nuclei. Sending energetic ions through thin foils under the right conditions will result in some fraction of fully stripped ions, which can then be magnetically separated out from the other charge states. But, yes, it isn't chemistry.

The most common bare-nucleus species on Earth is actually not $\ce {H+} $.  It's $\ce {He^{2+}} $, better known as  $\alpha $ particles produced by radioactive decay.  These $\alpha $ particles do not have a very long shelf life, as they will rapidly take up electrons to become just plain helium.  But they are real enough for household smoke detectors to rely on them or at least their (ionized) reaction products obtained by taking up said electrons.


text:
Entropy concept
Entropy change does not depend on the heat that is actually absorbed,but it depends on the heat absorbed reversibly.
How?
Since the surroundings emit irreversible heat then how the system which is also a part of universe absorb heat reversibly? I can't get the concept.

To determine the change in entropy of a system that has experienced an irreversible process between an initial and final thermodynamic equilibrium state, you must first use the 1st law of thermodynamics to establish the final equilibrium state. After that, you can totally and entirely forget about the irreversible process.  It is of no further use.  Instead, you need to devise a reversible process between the initial and final states.  You can even apply different reversible processes to the system and to the surroundings, as long as each of them passes through a continuous sequence of thermodynamic equilibrium states between the two end points. There are an infinite number of reversible processes for the system and the surroundings that can take them each between their initial and final states; each of these paths give exactly the same entropy change, as determined by the integral of dq/T.  The reversible paths do not need to bear any resemblance whatsoever to the real irreversible path, as long as they start and end at the same initial and final states.  We need to follow a reversible path to determine the entropy change since this is the only way we know of to determine $\Delta S$.

Maybe it would be useful to add that although heat is a path dependent variable, entropy is a state function.

Chester Miller does 1st law is applicable to only reversible process?

The 1st law is applicable to all processes, both reversible and irreversible.  Have you never applied the 1st law to an irreversible process?


text:
Alkaline Water Electrolysis with RbOH
In industry, it appears the alkaline water electrolysis is largely done with solutions of NaOH or KOH - presumably due to low cost.
I presume that alkaline water electrolysis is possible under RbOH, however, I can not seem to find any evidence in the literature showing the use of RbOH for water electrolysis. Can water electrolysis be done with RbOH to give oxygen and hydrogen gas evolution at the working (TiO2) and counter (Pt) electrode respectively?
Thanks!

Of course! Rb is more electropositive than hydrogen, so hydrogen is released.
As you mention, $\ce{RbOH}$ is a bit more expensive than $\ce{NaOH}$. At present, $\ce{RbOH}$ is ~US\$500/kg and $\ce{NaOH}$ is ~US\$500/tonne, three orders of magnitude cheaper.
You might use $\ce{CsOH}$ or $\ce{FrOH}$, too. The Fr will even illuminate the reaction with Cherenkov radiation.


text:
Does triphenylphosphine show resonance?

The lone pair of phosphorus and the double bonds in benzene looks to be in conjugation. Do the molecule show resonance? If so, how?

The final arbiter will probably be some computational study, but phosphorus 3p overlap with carbon 2p is not great to begin with.

Somewhat related, since the lone pair of the $\ce{P}$ atom is addressed: http://chemistry.stackexchange.com/questions/38599

@orthocresol,What do you think about the answer below?

There's probably no question that some degree of resonance exists; and indeed it's not possible for donation into all 3 rings to happen at once. However, the devil is in the details: how much of it is there? I personally don't know, but I don't think it's a lot, at least simply based on overlap arguments.

The triphenylphosphine molecule would show resonance because there is a lone pair on phosphorus and it is in conjugation with the double bonds in benzene. But the lone pair would be in resonance only with one ring at a time because the molecule is non-planar (tetrahedral) having sp3 configuration. Also, we should consider the steric effects on the molecule. The steric hindrance produced by other two benzene rings would cause less resonance effect and the lone pair would be inhibited from going into the other two rings.
So, the resonance in this molecule would be simply like that in aniline.


text:
Cyclic Voltammetry - HOMO and LUMO levels
I'm a physicist, so I apologize if these are obvious questions.
I've carried out CV measurements on a few different types of material (with a ferrocene reference). I am interested in determining the HOMO and LUMO levels of the material.
My CV measurements result in a current-voltage curve (IV curve). It has been explained to me I just need to measure the mV difference between the peak of the ferrocene and the oxidation or reduction curve, then subtract/add that onto the reference for ferrocene (4.8eV) to get the HOMO/LUMO levels.
Sounds simple, right?
But:

Do I measure to the onset of the oxidation/reduction curve or the
peak?
What justification do I use to convert the voltage to an energy? How can I say that a difference in mV corresponds to eV? Maybe this might make sense if it was a one electron process, but how do I know that?
Are these peaks even the HOMO/LUMO levels at all? In this eminent paper it warns me

" Often cyclic-voltammetry-based ionization potentials and electron
  affinities are inappropriately referred to as "HOMO" and "LUMO"
  energies"

Also relevant: http://chemistry.stackexchange.com/a/16225/5017

Yes, you can "convert" this way, but you're correct to be skeptical.
Let's start with interpreting the cyclic voltammetry curves themselves.

(figure from Wikipedia)
Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.

Do I measure to the onset of the oxidation/reduction curve or the peak

You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.
You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.

What justification do I use to convert the voltage to an energy? How can I say that a difference in meV corresponds to eV?

In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.
The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.
So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.
You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.

Maybe this might make sense if it was a one electron process, but how do I know that?

Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.
You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.
You can change the scan rate and use the Randles-Sevcik equation
$$i_p = 268,600 \ n^{\frac{3}{2}} AD^{\frac{1}{2}} Cv^{\frac{1}{2}}$$

Are these peaks even the HOMO/LUMO levels at all?

They are not. There should be a correlation between oxidation and reduction potentials and HOMO and LUMO energies.
But step back and think - in one case, you have electrochemistry in the electrolyte solution at an electrode surface.

Can the solvent change the observed redox potential? Yes.
Can the electrode change the observed redox potential? Yes.
Could the molecules aggregate and change the observed redox potential? Yes.
Could the oxidized or reduced species undergo reactions that change the observed redox potential? Yes.
Can the sweep rate change the observed redox potential? Yes.

Okay - Thank you for your detailed answer, that seriously helps to answer most of my questions. Apart from one regarding the dimensions (sorry for the physics): I'm measuring volts [kg $m^2$ $s^−3$ $A^−1$] but without (rigorous?) justification, I'm taking these volts for energy peaks and saying they correlate to an energy  [kg $m^2$ $s^-2$] in eV. I.e. I'm measuring the energy difference (between the peak and ferrocene) in mV.

@Tomi - as I now indicate above, the difference is that in most cases, one electron is transferred in the redox couple, so the conversion is "swept under the rug" so to speak.

Actually, do you not think it would be best to measure the onset of the oxidation/reduction peak? - The HOMO according to the theory you quoted should be the "first" electron allowed into the "band", so it would make sense that we consider the onset. That would match literature "Therefore, the oxidation and reduction potentials are estimated by the onset potential,  defined as the potential at which the initial injection of holes or electrons to the HOMO and LUMO..." Buzan et al. http://onlinelibrary.wiley.com/doi/10.1002/adma.201004554/abstract

While you can make arguments along those lines, many electrochemists prefer to take the average of the current peaks to define $E^0$ - for practical purposes. It can be tricky to define the onset, depending on the baseline and the peak. (Imagine a sloping baseline when the capacitance is changing and a small peak.) The peak position is much easier to define accurately. So I understand Bazan and Kaifer's paper, but I don't think this is widespread practice.

'ere we go! https://chemistry.stackexchange.com/q/72106/402


text:
What is the area that 1 litre of gasoline would spread above the sea?
I am studying for a general chemistry test with Atkins and Jones "Principles of Chemistry" 5th edition.
In the Chapter 18, page 743, section 18.3, there is this funny question:

How would you estimate the area that 1 litre of gasoline would spread
  above the water surface?

Observation: I translated it from the Brazilian Portuguese edition. The English version may have different units, but the idea should be the same.
Let's say that the water body is the sea. But it could be a big lagoon, too. It doesn't really matter.
It got me thinking and now I'm stuck. How can we do that?

Estimate the thickness of the layer, take 1 litre and divide by the thickness.. otherwise you probably need more data

If we had more data, what is the best way to do this? I thought about using surface tension properties and equations, but I am not sure if it is the easiest way...

You can calculate the surface area per molecule, using surface tension measurements

For an estimation I might start with this table regarding the dispersal of oil spills on water. I would assume that gasoline would spread at the very thin end of that table, and guess something on the order of 50 nm thickness and say that's the amount you can see or some such thing. There is no reason it couldn't spread much thinner, too thin to see even.

This site offers a hand rail: http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/phillip4.html  Be careful, as it mixes SI and non-SI units; but the principle is legible.

These articles were helpful! I think I got it now. Thanks!

One simple way is to estimate the molecular size and assume the layer is one molecule thick. A check on the thickness estimate is the interference patterns on the surface which mens the layer must be of the same order as the wavelength of visible light. Either way, you will get a crude but useful estimate.

Benjamin Franklin tried this experiment on the pond in Clapham Common in London  in the 1770's.  There is an account in the link below which also contains his own account.  http://www.rsc.org/learn-chemistry/content/filerepository/CMP/00/000/687/isms-9.pdf?v=1399160447243  or search for Franklin Oil Drop Experiment.

The hint here assumes a layer thickness of (crude) oil of about 1 mm. If I remember the pictures in the News around Deepwater Horizon correctly, and compare what I saw with own eyes at the local petrol station, petrol has a lower viscosity; likely spreads out wider than crude oil.
From here we learn the layer thickness is much lesser than 1 mm (millimetre), namely 168 nm (nanometre); hence all these colourful interference patterns in visible light.
$$ 1 \text{ L} = 1 \text{ dm}^3 =  10^{24} \text{ nm}^3$$
Deviding by 168 nm provides with googles built-in calculator an area of about $$5.95 \times 10^{21} \text{ nm}^2 = 5952 \text{ m}^2$$
To set this into a context, UEFA and FIFA agreed on the size of European Football / Association (soccer) field ($ 105 \times 68 \text{ m}$) which equals $7140 \text{ m}^2$.
Addendum:  According to this, an American Football field measures "$120.0 \text{ yd } (109.73 \text{ m}) \times 53.3 \text{ yd } (48.74 \text{ m})$ or $6400 \text{ yd}^2 (5351.2 \text{ m}^2$)".


text:
Effect of nuclear spin statistics on the IR spectrum of CO2
Specifically, I am talking about $\ce{C^{16}O2}$. With a nuclear spin of zero meaning, it is a boson - it's total wavefunction must be totally symmetric. The total wavefunction is made up of an electronic, vibrational, rotational and a nuclear spin contribution. For this molecule the electronic and vibrational (in ground state) wavefunctions are symmetrical. There is no antisymmetric nuclear spin wavefunction so only the even J rotational levels exist.
Now consider the degenerate bending vibrational mode. Excitation of this mode leads to excitation into the excited vibrational state of this mode and the IR spectrum has a corresponding P, R, and crucially a Q branch (due to additional angular momentum from the degenerate bending mode).
I have been asked whether there are similar J restrictions in the excited state as there are in the ground state. I initially jumped into assessing the symmetry of the contributions to the wavefunctions and concluded that the vibrational mode is antisymmetric and thus the corresponding excited state wavefunction is antisymmetric leaving the possibility of only odd J values in the excited state. However, there is a Q branch so this must be incorrect since the Q branch indicates the presence of J=0 in the excited state.
Why is there seemingly no restriction on the excited state?

Do you mean electronic excited state ? If not look at this answer http://chemistry.stackexchange.com/questions/70715/rotational-constant-for-co2-from-an-ir-vibrational-rotational-spectra/70722#70722

Thank you but that does not answer my question. I am talking about the excited vibrational state and why nuclear spin statistics appear not to affect this state (i.e they only effect the ground state).

@RobChem As the energy scales responsible for either IR spectroscopy and NMR spectroscopy are so remote from each other, and routine ATR-FTIR spectrometers have a resolution of about 0.5 wavenumbers, I am at least surprised other IR spectroscopes were able to discern the (in comparision) tiny differences of energy levels, caused by the coupling with different nuclear spins of the sample. – At least as the original question does not mention an external magnetic field excersing the Zeeman splitting on the later.  What did I miss here?

This has absolutely nothing to do with NMR.

@RobChem Thank you, this lowers / relaxes my eyebrow "back to ground state".

You should follow the same argument for rotational levels built on v=1, v=2 levels as for the vibrational ground state; account for symmetry product of electronic, vibrational, rotational and nuclear wavefunctions. As the symmetry of v$ \gt 0$ wavefunctions (asymmetric stretch) alternate the pattern should be clear.

Because the degenerate bending mode has vibrational angular momentum, it is a vibrational $\Pi_\text{u}$ state and just as in electronic $\Pi$ states there exist levels with $+$ and $-$ parity. As a consequence, the degenerate bending mode has $\Pi_\text{u}^\pm$ symmetry and the vibrational wave functions can be both symmetric and antisymmetric.
See also Herzberg Vol 2 (p372 and Fig 99 on p373).


text:
Why do fluoride ions "win" over thiocyanate ions in a solution with iron ions?
When you mix aqueous iron(III) nitrate and potassium thiocyanate together you get a red solution. Then if you add sodium fluoride, the red colour disappears.
This is basically because the $\ce{Fe(SCN)^2+}$ ions are red, but $\ce{FeF^2+}$ ions are colourless.
I get Le Chatelier's principle in that, the more iron ions that are stolen by the fluoride, the less $\ce{Fe(SCN)^2+}$ ions there will be because they keep dissociating etc.
The thing I wonder about is why the fluoride atoms get all the iron. I can think of a couple of potential reasons for this, but I don't know which, if any of them, are true, even after googling a fair bit.
Could it be the case that iron is a lot more attracted to fluoride than thiocyanate? Or that the fluoride reaction isn't reversible? And if either of those is true, can this experiment be seen as evidence for that?

Well, I guess Fe(III) (aq) is "hard" enough to prefer F- from SCN-

I found this on the MEL science website (https://melscience.com/US-en/articles/fake-chemical-cut/)

"Adding sodium fluoride destroys the iron(III) thiocyanate complexes, and colorless hexafluoroferrate(III) [FeF₆]³¯ ions form. Fe(CNS)₃ + 6NaF→Na₃[FeF₆] + 3NaCNS"

Seems that it may create this hexafluoriferrate ion instead... just thought I'd add to this discussion. I was actually just trying to find out a bit more about this reaction myself. AP Chem teacher here :)


text:
Decreasing magnitude of peaks in cyclic voltammogram?
I'm in an instrumentation course and we covered cyclic voltammetry earlier in the semester and will be doing an experiment in CV next week.
My question is this:
If you have a reaction and sweep the potential, will the new peaks be entirely within the previous peak? (Meaning: lower magnitude in the subsequent peaks?) It seems that with each sweep, it should have lower electrical potential.
Or would the peaks trace over the previous ones, nicely?
Pardon me for not fully understanding (or if my question is wrong.)
Enjoy a poorly drawn image of what I mean:

Perhaps your observation is an indication of irreversible reaction conditions.  More specifically, a tight deposit on the electrodes hampering the advancement of the electrochemical reactions. This happens sometimes with some less soluble organic dyes used organic electronics, too.  Take a look if the electrodes are still shiny, maybe they need to be cleaned again.  If this is the case, try out a different solvent/electrolyte system.
One aiding companion to trouble shoot is this.

The oxidation and reduction peaks on the cyclic voltammogram should be reproducible if done under the same conditions and for a reversible reaction, i.e. no side reactions depleting species. The difference in voltage between the peaks can be used as a test of reversibility. Most probably as @Buttonwood writes you have an irreversible reaction where the peak shapes are not the same in shape or size as time progresses due to various other species being formed. Sometimes, such as with quinones, it is possible to prepare an electrode so that it can behave reversibly.


text:
How do you identify parent chain of molecule?
Imagine a molecule that has aldehyde and alkene group, then how do you choose the longest chain to name as parent chain. Does the parent chain have to include both aldehyde and alkene group even if there is a larger one just with an aldehyde group?
My TA says that you choose the parent chain that has both aldehyde and alkene chain, if possible, even if it is the shorter chain, without giving any further reasons for that. So then my question is that is it possible for parent chain to include more than 2 functional groups? Like I know the likelihood of seeing such an example may be rare, but if you have a molecule that includes a chain, among other candidates, with more than 2 functional groups, like a carboxylic acid, alkene, alcohol or aldehyde and so on all in one chain… then can you choose such a chain as a parent chain over any other chain anytime when that other chain has fewer functional groups or is even longer or shorter? If yes, then would you say that when finding the parent chain we should try to find parent chain that not only has the highest priority substituent but also the one that has most high priority substituent as possible?
Also, one more question, let's say you have a parent chain that has ketone at second carbon atom from one side of the linear chain and an alcohol group on carbon 1 when viewed from the other end? Even though ketone has higher substituent than alcohol, where would you start numbering the molecule? Would carbon with aldehyde be numbered 2 and carbon attached to alcohol 6 or 7? Or would carbon with alcohol be 1 and ketone be numbered 5 or 6?

Possible duplicate of How to identify the parent chain if two chains have the same number of skeletal atoms and the same number of double bonds?

These are reasonable questions, but I'm voting to close because you're asking too many things in one post. Please edit this post down to just one question, and create new posts for each of the others.

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?