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How to explain two NMR spectra of the (allegedly) same compound with slightly different shifts?
For my synthetic lab work I synthesised a Horner-Wadsworth-Emmons phosphonate; for the sake of the question let’s assume it to be the one presented in the image below.

When first synthesising it, I of course recorded NMR data. I performed a Horner-Wadsworth-Emmons reaction using a tiny bit of the phosphonate and kept the rest in the fridge.
One month later I retried the HWE and got very bad yields, so I decided to record another NMR spectra. The two spectra themselves look identical and the signals (those that aren’t multiplets) also show the same coupling constants. However, when referencing the spectrum to the chloroform signal ($7.26~\mathrm{ppm}$), all signals are shifted by the constant value of $\approx 0.07~\mathrm{ppm}$.
I decided to check the old literature spectrum from 1986, but it was unhelpful: It only reported multiplets such that both of my spectra comfortably fall well within the multiplets’ limits.
Considering that the two spectra look identical and that the difference between two signals corresponding to the same hydrogens is identical as are the coupling constants, I want to assume that both spectra show an identical compound. However, the $0.07~\mathrm{ppm}$ shift for everything save chloroform still irritates me.
Both spectra were recorded at $500~\mathrm{MHz}$. The parameters of the earlier one state a temperature of $25~\mathrm{^\circ C}$, the later one was apparantly measured at $25.1~\mathrm{^\circ C}$. However, I don’t think that that could explain the shift.
What could explain this shift difference?
(My first thoughts were some changes to the compound that only affect the phosphorus atom, but aside from deoxygenation being chemically unlikely shouldn’t it also cause random chemical shift differences including slightly different coupling constants? In fact, shouldn’t any chemical modification cause the shifts to change in a seemingly random fashion depending on the proximity of hydrogen in question and entity modified?)
After checking because I was asked in the comments: The later spectrum (the one that has higher shift values) has broadened peaks. That includes the chloroform peak.

Do the shifts change for carbon and phosphorous?

For reference, the later spectrum included a dichloromethane peak that stemmed from the bulb of the pipette I used for pipetting chloroform. The former didn’t, so I wasn’t able to check if dichloromethane was shifted, too. A very minor peak at $\approx -0.02$ or $\approx -0.09$ was shifted.

@Lighthart I didn’t record carbon and phosphorus spectra to be quicker.

Do the peak widths change at all?

@Lighthart edited the answer in.

Maybe the solvent signal is shifted? You have not used an internal standard like TMS?

@Mart As far as I know today’s NMR relies on the solvent signal for referencing and TMS is almost never added … At least, I hardly ever added any.

If the peak widths are different, that means the machine's field has drifted a little.  And as a result, somewhat wider peaks (or worse shim) may be modestly adjusting the digital calculation of peak maximum (and hence chemical shift).  Wider peaks would provide evidence that this is a measurement problem, not a chemical purity issue.  That being said, you indicate you are not telling us exactly the compound synthesized, so we can't really judge if 0.1C will impact for reasons of conformational/rotomer changes.

@Lighthart I think that the first bit is the answer I am looking for and I am relieved that that may be the case. About the compound I actually used, it is very similar but not identical to the one I depicted here. And if I drew it for you, you would probably agree that conformation or rotamer changes would not be a logical conclusion. ;)

If you're goal is synthesis, I think you just botched your reaction, try it again.  If your goal is a rigorous description of the deviation, you've got some number crunching to do.

@Lighthart the former ;) I do also have another explanation on why the reaction failed but I was so confused at this spectrum today that I was too scared to redo the HWE ;)

Another possible reason is that the shift is caused because of aggregation. Not unreasonable to happen for a phosphonate but on the other hand the shift you observe is too wide to be explained by aggregation alone? Have a look at this paper  for example. I only know about this from its use as an assay in med chem so I couldn't help with more specific info.

How large was the DCM peak? Maybe I missed it, but which direction were the signals shifted: lower or higher ppm than before? Also, did you see any other signals that were new? Maybe some water or an alcohol that could hydrogen bond with some of oxygens here?

If all non-solvent signals are shifted by exactly the same amount, I would pick a shift of the solvent signal itself as the most likely source. If this were measured in H2O/D2O, I wouldn't be surprised at all, but the CDCl3 chemical shift is not as temperature sensitive as the H2O/D2O one.

I agree with @MadScientist. The solvent could shift due to presence of methylene chloride if there is a significant amount. Probably have to be a lot, though.

@Senders The DCM peak is really minute. You could publish the spectrum because it wouldn’t have been seen in normal scroll intensity. The new spectrum had slightly higher ppm values.

@MadScientist And that’s the reason why I always add methanol to $\ce{D2O}$ NMRs ;)

That just seems so strange. How could the magnetic properties of one nucleus change without any others changing? Even if the probe had drifted, I would expect everything to move, not just the solvent peak.

That's a pretty small chemical shift difference.  Keep in mind that chemical shifts depend on temperature, concentration, pH, and other factors.  The concentration difference is the most likely, but I wouldn't rule out T effects.  If you're running it in a different batch of CDCl3 (or even the same batch, but it's aged), the pH could be different as CDCl3 becomes acidic over time.
If you used identical settings when collecting the spectra (e.g. using a default Proton) then you can measure the signal-to-noise ratio of two identical peaks and get an estimate for how different the concentrations are.  You can also simply test how sensitive your chemical shifts are to concentration by making up a more concentrated sample (or diluting your current sample).  Or try changing the temperature with the VTC and ramp it up to 25.5 or 26.0 C and see how much the chemical shifts change (referencing everything to CHCl3 for consistency).
If your compound were reacting with something, there would be a much more dramatic change in the chemical shifts than 0.07 ppm.

Actually I never had a shift difference as large … but I never had phosphonates.


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Why stepwise dissociation for acids but not hydroxide bases?
My son asked this over the weekend and I didn't have a good answer: how come with polyprotic acids we assume that the release of each $\ce{H+}$ is progressively harder, but with bases like $\ce{Ca(OH)2}$ we write $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?
What I know about, say, sulfuric acid, seems to suggest that when we put calcium hydroxide in water we should get $\ce{CaOH+}$ and a single hydroxide, and it should be much harder to release that second hydroxide.  What am I missing?

The $\ce{H+}$ is bonded covalently to the rest of the molecule. So, when you take off the first $\ce{H+}$, the molecule fundamentally changes. It becomes another molecule. The second $\ce{H+}$ is being taken away from a different species and so it is governed by a different equilibrium.
In $\ce{Ca(OH)2}$, $\ce{Ca^2+}$ and $\ce{OH-}$ are held together via ionic bonds. When you dissolve that in water, all the ionic bonds are going to break at the same time. You cannot break some of the ionic bonds without also breaking the rest.

This is not a general rule. Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base.
In the case of $\ce{Ca(OH)2}$ you don’t have a molecule like $\ce{HO-Ca-OH}$ or something like you can assume for $\ce{HO-SO2-OH}$ (sulfuric acid). Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in $2~\mathrm{mol}$ of $\ce{OH-}$ per mole of $\ce{Ca(OH)2}$ dissolved.
As a different example take sodium oxide $\ce{Na2O}$. If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2 NaOH}$ first before a second set of protons will generate water. This is much the same mechanism as for diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)

It would be nice to tell about aquo complexes also @ortho

@Mithoron I don’t think that that belongs here …?

I think he meant it as a cc to me, since we basically wrote the same thing within 1 minute of each other (mine being a bit less expanded) :D @Mith, what about aquo complexes?

@orthocresol I was actually saying that aquacomplexes don't necessarily belong here ;)

In water there are equilibria of protonation involving complexes of Ca. Maybe I'll make an answer with reasoning involving it if you don't want ;)

If I would conclude from Jan's statement

Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in 2 mol of $\ce{OH-}$ ion per mole of $\ce{Ca(OH)2}$ dissolved. [$\ldots$]

then it would be not that difficult to give the following two reactions to answer the original question.
1.
$$\ce{Ca(OH)2 (aq) <=> Ca(OH)+ (aq) + OH- (aq)}$$
2.
$$\ce{Ca(OH)+ (aq) <=> Ca^2+ (aq) + OH- (aq)}$$

Water is already in an ionised-equilibrium state, so there may be availability of protons and hydroxides in the (universal) solvent

Why not try to rearrange the above equation by taking the help of protons contributed from water, let's do it.

Rearranging 1st equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{Ca(OH)2 (aq) + H+ (aq) <=> Ca(OH)+ + H2O}$$

Again, rearranging 2nd equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{Ca(OH)+ (aq) + H+ (aq) <=> Ca^2+ + H2O}$$

It would now be easy for us to conclude why the reaction must proceed harder progressively; because as soon as we see the modified the form of reaction 1&2: we notice the repulsive interaction between $\ce{Ca(OH)+}$ ions and $\ce{H+}$ ions. So one may conclude with this pointed statement:

In the case of polyprotic acids we assume that the release of $\ce{H+}$
ions is progressively harder.

said by Asker, has justified why must not "strong bases like Ca(OH)2", "directly form  $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?", further added by Asker.
Jan says

This is not a general rule.

Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base. In the case of Ca(OH)2 you don’t have a molecule like HO−Ca−OH or something like you can assume for HO−SO2−OH (sulfuric acid).
Rather, you have an ionic compound composed of two hydroxide ions per a calcium ion. Once we dissolve this, all the ions are released at the same time resulting in 2 mol of hydroxide per mole of $\ce{Ca(OH)2}$ dissolved. As a different example take sodium oxide $\ce{Na2O}$.
If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2NaOH}$ first before a second set of protons will generate water. This is pretty much the same thing that happens to diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)*
Jan statement is much correct and easily explains my asserted warning at last sentence under the conclusion.
Conclusion:
So let me again repeat the whole equation but with a slight change, this time for $\ce{H2SO4}$, a strong acid:

Rearranging 1st equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{HO−SO2−OH (aq) + OH- (aq) <=> O−SO2−OH- + H2O}$$
Again, Rearranging 2nd equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{O−SO2−OH- (aq) + H+ (aq) <=> O−SO2−O^2- + H2O}$$

Note that ${\rm p}K_{\rm a}$ and ${\rm p}K_{\rm a}$ are like constants whose value does not change until either temperature or some other thermodynamic variable changes (do not confuse with Kinetic Variables, acid dissociation constants have nothing to do with it) are changed (https://en.wikipedia.org/wiki/Acid_dissociation_constant)
TL;DR;

Firstly, determine the Acid Dissociation Constant constant of the given base, acid or mixed salt.

Secondly, determine the Thermodynamic conditions and values of its variables to help you go into detail of the reaction (not necessary, just knowing that it is constant or not, is enough)

Thirdly, write the whole dissociation reaction (only, because those that does not dissociate or whose dissociation constant is too small is generally ignored, i.e., Weak acids and Weak Bases and their mixed salts, can be ignored).

Finally, interpret the interactions between the filtered out dissociation-reactions, then cross out those that are against any of the 1st three rules. You may also filter out on the basis of what are you actually trying to obtain from these reactions. Just keep in mind, that the fundamental rules of Ionic Equilibrium in the context of whole Chemistry. must not be violated.
You will be having only the most important and relevant reactions. Just a few steps.

There is no reason to assume statements 1 and 2 have any physical significance.

I dunno, let's face it, it's bad. Putting the formatting and grammar aside, essentially everything here is irrelevant. Half of it is just quoting and claiming that the quoted sections prove some unknown "point". The other half is a disordered collection of assertions about pKa values, which are entirely irrelevant to the question to begin with. As Lighthart says, the first two equations have absolutely zero physical significance.

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Model of London forces and Hydrogen Bonding; the question is carried on forward, partially, as a personal question, from another question
Does Model of London forces and Hydrogen Bonding are themselves complicated and require multiple sub-models and assisted theories to explain its correctness.
I asked this question in the form (ironically) of a question, but I came up a new doubt later on. Just read below:
["
Lighthart says, "The actual magnitude is somewhat difficult to measure (and frequently dynamically changing due to molecular motion), but their existence is used to describe non-covalent bonding interactions between molecules such as London forces and Hydrogen Bonding." So, does it implies that the Model of London forces and Hydrogen Bonding are themselves complicated and require multiple sub-models and assisted theories to explain its correctness.
Can the statement "Now with partial charges the water molecule can have a dipole moment. ... So the gist is that ionic bonds and covalent bonds are labels that we put on molecules to predict behavior. Such models are useful in limited circumstances. But if we try to overextend the model it leads to wrongful conclusions." (quoted from text written as answer by MaxW.) can be used to strengthen my question's argument."]
The link to the site is: Partial charges in a covalent bond

What? Can you clarify what the actual question is here?

bon, please refer 'Partial charges in a covalent bond' - that's original question (of someone else) from where my doubts actually raised and I finally posted the question.

Yes but what is the question you are asking here? How is it different to the previous question?

If i underatabd correctly,
Now a days when in molecular dybamics simulation or docking studies, to calculate everything requires precise knowledge of polar, hydrophobic, ionic and non-bonding interactions. We have models to deal with each, and each has limitations. With polar groups the issues are not so much the models, which have distance dependance and orientation dependance, but computer power. This is really what limits the models. Often solvent effects and hydrophobic effects are ignored or estimated and simplified.
If there is a model missing is a hydrophobicity model. There isn't anything that is super.

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Mixing a Tagged Sample of Lead Iodide with another Iodide Salt
A sample of sparingly soluble $\ce{PbI_2{(s)}}$ containing radioactive $\ce{^{133}I}$ is added to 0.10 M $\ce{KI_{(aq)}}$ and stirred overnight, what happens?
I The radioactivity of the liquid phase increases  significantly.
II The concentration of the $\ce{I^{–}}$  ion in solution increases significantly.
The answer is I. I thought II would be true because some of the water evaporates, but I guess not to a "significant" extent. So that marks II off, and leaves II. I'm fairly new to nuclear chem so I don't know much about how/why radioactivity increases, could someone please explain this to me? How could radioactivity "increase?"

You are missing a basic aspect of how a precipitate in solution behaves. The precipitate isn't just statically immersed in the solution but rather some is dissolving and some of the ions (or molecules) in solution are precipitating out. So for a salt like $\ce{PbI2}$ there will be an exchange of ions which are in the precipitate with those which are in solution.


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Unique Solubility Expression for Sulfide Salts
While looking at aqeuous $\ce{K_{sp}}$ values in the back of my textbook, I noticed that there is a asterisk by all of the sulfide compounds (such as $\ce{HgS}$), and the asterisk says, "For a solubility equilibrium of the type $\ce{MS_{(s)} + H_2O_{(l)} <=> M^{2+}_{(aq)} + HS^{-}_{(aq)} + OH^{-}_{(aq)}}$." Now, why is it this expression, instead of the expected $\ce{MS_{(s)} <=> M^{2+}_{(aq)} + S^{2-}_{(aq)}}$?
Upon examination, I see that the given expression is just the expected expression and then tacking on the base equilibrium of the $\ce{S^{2-}}$ ion, which is $\ce{S^{2-}_{(aq)} + H_2O_{(l)} <=>HS^{-}_{(aq)} + OH^{-}_{(aq)}}$. But why do we involve the base equilibrium of $\ce{S^{2-}}$ when we don't involve other quite basic anions, such as $\ce{F^-}$ or $\ce{CO_3^{2-}}$?

$\ce{F^-}$ is not that basic. $\ce{CO3^2-}$ is, but still less so than $\ce{S^2-}$.

Why is it a new category in and of itself?

From Wikipedia article on hydrogen sulfide The sulfide dianion $\ce{S^{2−}}$ exists only in strongly alkaline aqueous solutions; it is exceptionally basic with a pKa > 14. https://en.wikipedia.org/wiki/Hydrogen_sulfide

To summarize the comments, the sulfide anion is a strong base. This site (sourced from the CRC Handbook of Chemistry and Physics) lists the second $K_a$ of $\ce{H2S}$ as $1\times 10^{-19}$, which is less acidic than water ($1\times 10^{-15.7}$).
Thus the $K_b$ of $\ce{S^2-}$, which is the equilibrium constant for the reaction with water, is $1\times 10^5$.
$$\ce{S^2- + H2O -> HS- + OH-}\ K_b= 1\times 10^5$$
After wrangling the algebra, if you start with a $1\ \mathrm{M}$ solution of $\ce{S^2-}$, at equilibrium $[\ce{S^2-}]=3.16\times 10^{-10}\ \mathrm{M}$ - almost all of it is converted to $\ce{HS-}$
Compare for some other anions (assume starting with total concentration $[\ce{A-}]+[\ce{HA}]=1\ \mathrm{M}$
$$\begin{array}{c|c|c|c|c}
\ce{A-} & K_b & [\ce{A-}]_{eq} & [\ce{HA}]_{eq} & \text{% protonated}\\ \hline
\ce{S^2-} & 1\times 10^5 & 3.16 \times 10^{-10}\ \mathrm{M} & \sim1\ \mathrm{M} & \sim 100\% \\
\ce{CO3^2-} & 2.1\times 10^{-4} & 0.99\ \mathrm{M} & 1.45\times 10^{-2}\ \mathrm{M} & \sim 1\% \\
\ce{F-} & 3.13\times 10^{-11} & \sim1\ \mathrm{M} & 5.59\times 10^{-6}\ \mathrm{M} & \sim 0\%
\end{array}$$

Very well presented.


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Laws of electrolysis and quantisation of charge relation

The quantisation of charge was first suggested by the experimental laws of electrolysis discovered by Faraday

I think about it a lot but could not find any relation, as the number of electrons involved is too large in electrolysis that the charge can be assumed to be continuous and discrete nature is negligible.
How did Faraday first suggested quantisation by this non microscopic process (electrolysis)?

Did you read https://en.wikipedia.org/wiki/Faraday%27s_laws_of_electrolysis? It's the number of  chemical valence electrons that is discrete, which is different for different metals. Some go into (or come out of) solution using one electron, others need two or three.

@CuriousOne      so discrete nature of valence electrons suggested quantisation of charge?

Yes. You are correct that one can't measure a single electron that way, but it's easy to distinguish 1 mole of electrons from 2 mole of electrons.


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Remove salt from dirt
I have a mixture that contains out of dirt/sand and salt. I want to extract and measure the salt. What would be the best way to extract the salt from this mixture?

Assuming you mean NaCl - the common "salt" (chemists call lots of things salt!).
If you extract the salt by physical means, it's a physics question.
Assuming that the "dirt" is not (or poorly) soluble in water, I would simply dissolve the salt in water, filter the liquid, then recrystallize (by evaporation of the liquid) and weigh the resulting crystals. Not sure if you call that chemistry or physics... it's physical chemistry. It is well described in text books and online - see for example this Scientific American page
Depending on how accurate you want to get, there are a slew of precautions to take. That would be the realm of the chemistry.stackexchange sister site.

The question is very vague.   Common road salt is typically $\ce{NaCl}$ or $\ce{CaCl2}$, both of which are soluble in water.  The problem is that "dirt" could be anything, including lots of things that are also soluble in water.
So the idea that washing the mixture with water to remove the solutes would isolate the salt is insufficient.   You would need to specify the type of salt you're dealing with and then ask on a chemistry forum how to chemically isolate just that compound.

You're right, the question is vague. But, that is mainly because I didn't know what salt this was and what materials where in the "dirt". Turns out that it was just common table salt with some sand and small stones. Thanks for your answer.

@bramhaag , how did you know it was just NaCl and no other water soluble salts?


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Why do the atoms in molecular models have the colors they have?
Molecular models are a great tool for visualizing a molecule in three-dimensional space. These models even appear on the web (practically every chemical's Wikipedia page), and there is a very specific color coding for atoms. After some searching I found that it was called the CPK coloring system. It seems rather intuitive for some elements, like for $\text{carbon}$, $\color{\green}{\text{fluorine}}$, and $\color{\yellow}{\text{sulfur}}$, which use the colors for the elemental forms of these atoms (diamonds aside).
On what basis, though, did Robert Corey and Linus Pauling choose the colors for $\color{\red}{\text{oxygen}}$, $\color{\turquoise}{\text{nitrogen}}$, and hydrogen (I'm not going to do white on white). Were these arbitrary choices that had to be made for these colorless elements, or is there some reasoning behind them?

There's an interesting article here - Illustrating Atoms and Molecules - that discusses atom colors:

In 1865, the chemist August Hoffman gave a Friday Evening Discourse at London’s Royal Institution on the “Combining Power of Atoms.” In order to demonstrate chemical bonding of atoms, he drilled holes in croquet balls and connected them with metal pipes (Figure 2). His choices were limited by the available colors of croquet balls and he relied on many of the same color conventions that had been adopted centuries earlier

These colors are listed as :

Carbon is colored black because it’s the color of charcoal.
Oxygen is red because it’s necessary for combustion.
Nitrogen is blue because it’s the most abundant element in the
  Earth’s atmosphere and the sky appears blue.
Hydrogen is white because it forms a colorless gas.
Chlorine is green because it forms a greenish gas.
Sulfur is yellow because that’s its color in mineral form.
Phosphorus is orange because it glows orange in a flame.
Iron is reddish brown because it rusts.

Which is interesting, but I've no idea where this list comes from...

It is interesting that it goes back so far (like quite a few decades before modern molecular structures were fully understood). But it is essentially arbitrary. But with a few links that make memorising the scheme slightly easier (nitrogen has nothing to do with the sky being blue but you are likely to remember the association because of the link).

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What is the hybridisation of trifluoromethyl free radical?
Having three σ bonds in a similar manner to $\ce{CH3^·}$ free radical, $\ce{CF3^·}$ should also have $\mathrm{sp^2}$ hybridisation. However, if we look at its shape, it is pyramidal and not planar like $\ce{CH3^·}$ free radical (which is $\mathrm{sp^2}$-hybridised), which signifies that $\ce{CF3^·}$ should have $\mathrm{sp^3}$ hybridisation.
But how is this possible because the three σ-bonds will bond with three hybrid orbitals? Where does the third p-orbital come?
If its hybridisation is $\mathrm{sp^3},$ then why is it?

Geometry drives hybridization, not the other way around.  2. An odd of consequence of #1 is that non-integer exponents in the sp/sp2/sp3 notation have physical relevance.

As Lighthart suggested, it is neither sp2 or sp3. It will be somewhere inbetween. Exactly where is an interesting question which I don't have the answer to right at the minute.

This question is very relevant, although not a dupe IMO.

Hyperconjugation between the orbital that contains the lone pair and the $\ce{C-F}$ antibonding orbital contributes to the pyramidalization of $\ce{CF3^·}$ radical:

In the planar geometry the orbitals are perpendicular and no overlap occurs.
References

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th edition.; Springer: New York, 2008. ISBN 978-0-387-68346-1.

This is a very popular, albeit very organic chemistry, view on that matter. The explanation in terms of Bent's rule might be much easier. Also that hand-wavy argument 'repulses more' cannot be proven, and is likely as wrong as spd hybridisation in hypercoordinate molecules. The second part is actually quite nice though

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Hock rearrangement of hydroperoxides
Which hydroperoxides undergo a Hock rearrangement?
e.g. Cumene hydroperoxide undergoes a Hock rearrangement to yield acetone and phenol

Would the following hydroperoxide of ethylbenzene also undergo any similar rearrangement? How does one decide if a Hock rearrangement is likely?

The main parts of the mechanism that will affect the rate of the reaction are the following steps (full mechanism on Wikipedia): abstraction of a hydrogen to form a benzylic radical and the rearrangement of the of the protonated peroxide.
These steps will be slower as they involve the formation of the higher energy species when ethylbenzene is used.

Secondary benzylic hydrogen have a higher bond bond dissociation energies ($357.5\: \mathrm{kJ\: mol^{-1}}$) than tertiary benzylic hydrogen ($353.3\: \mathrm{kJ\: mol^{-1}}$)$^{[1]}$. This will diminish the rate at which the initiation step proceeds, though a difference of $4.2\: \mathrm{kJ\: mol^{-1}}$ shouldn't make or break the reaction.

The formation of a primary carbocation would otherwise be highly unfavorable, except for that it is bonded to an oxygen atom. A lone pair on the oxygen can therefore be shared with the carbocation, thereby stabilizing it. This step will have a higher activation energy, but likely all that will be necessary for this reaction to occur is a little more heat and time.

[1] Baciocchi, E.; D'acunzo, F.; Galli, C.; Lanzalunga, O. Tertiary : Secondary : Primary C–H Bond Relative Reactivity in the One-Electron Oxidation of Alkylbenzenes. A Tool to Distinguish Electron Transfer from Hydrogen Atom Transfer Mechanisms. J. Chem. Soc., Perkin Trans. 2. 1996, 133–140.

Thanks! But isn't this addressing the formation of the Hydroperoxide? I'm wondering if, once formed, the hydroperoxide of EB would split up or not via the Hock Rearrangement.

@curious_cat I think the second half covers what you're looking for specifically.

@curious_cat The hydroperoxide will form nearly instantly once the benzylic cation is formed. It is not part of the rate-determining step.

@ringo He is talking about the step to form the ether (the alkyl shift), after the hydroperoxide formation, I believe.

If the image below is correct (I believe it is), then higher substitution = more stable carbocation = more likely to react this way.

That being said, a carbocation adjacent to an oxygen atom like that is pretty stabilized, so it may not require a tertiary carbon. I'm not sure what side reactions would be promoted by a non-tertiary carbocation, though.

Actually, this picture is wrong. It applies, but not perfectly. Will fix when I'm back to my computer.


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Redox reactions: identifying what is oxidized vs reduced
$$\ce{SO4^2- + H2O2 + 2H+ -> SO2 + O2 + 2H2O}$$
In the above chemical reaction, I'm having trouble identifying what is oxidized and what is reduced.
My guess is: sulfur has reduced but I can't work out what has oxidised. My guess is the middle oxygen as it has lost two hydrogen atoms to $\ce{2H2O}$.

The sulphate ion and the hydrogen peroxide are reduced , the hydrogen is oxidised ....

I think there would be more steps to the actual reaction tho ....

When the species cannot be found in redox tables (or the reaction scheme is just for illustration as in the question, e.g. SO$_2$ does not exist in solution as its a gas), using oxidation numbers is the only simple way. These values can be zero, positive, negative or fractional, e.g S in S$_4$O$_6$ or Fe in Fe$_3$O$_4$. In the question S has a value +6 (6-8*2=-2) in SO$_4$ and +4 in SO$_2$, as the total must be zero in a neutral species. As a decrease in oxidation number corresponds to reduction the sulphate ion is reduced and the peroxide is oxidised.

Yes sulfur of $\ce{SO4^2-}$ has undergone reduction as the oxidation number reduces from +6 to +4 and Oxygen of $\ce{H2O2}$ has undergone oxidation as its oxidation number increases from -1 to 0.
$$\ce{\overset{\color{red}{+6}}{S}O4^{2-}\+H2\overset{\color{red}{-1}}{O2}\+2H+->\overset{\color{red}{+4}}{S}O2\+\overset{\color{red}{0}}{O2}\+2H2O}$$
NOTE: Oxidation sate of oxygen in peroxide is -1. This because in the neutral molecule hydrogen peroxide the oxidation state of $\ce{H}$ is +1 and as the sum of oxidation numbers of the two hydrogen atoms and the two oxygen atoms must be zero so each oxygen in $\ce{H2O2}$ has -1 oxidation state.

You can use formal charges, but that's exactly what they are: a formality, not actual charges.
A redox reaction is a redox couple, consisting of an oxidation half reaction and a reduction half reaction. I would consult a table of reduction potentials (e.g.), which lists half-reactions and their standard reduction potentials.
On this table, we find:
\begin{align}
  \ce{O2(g) + 2 H+ (aq) + 2 e– &-> H2O2 (aq)},&
  E^{\circ} &= \pu{+0.682 V}\\
  \ce{SO4^2– (aq) + 4 H+(aq) + 2 e– &-> SO2 (g) + 2 H2O},&
  E^{\circ} &= \pu{+0.2 V}\\
\end{align}
If I invert the first reaction, I have the following oxidation:
\begin{align}
  \ce{H2O2(aq) &-> O2 (g) + 2 H+ (aq) + 2 e–},&
  E^{\circ} &= \pu{-0.682 V}\\
\end{align}
Summing this with the second half reaction (reduction) gives the total [redox] reaction provided in your question.
This tells us two things. Hydrogen peroxide is being oxidized (that system is losing electrons), since it is the reactant in the oxidation half reaction. Sulfate and hydrogen ion are participating as reactants in the reduction half reaction; they are reduced.
We didn't actually need to do anything with potentials here, but you can easily see that the total reaction is non-spontaneous.

The oxygen atoms on sulfate ion are not normally redox-active, whereas the oxygen in hydrogen peroxide could be redox-active given that they act that way in other reactions.  So I would select the hydrogen peroxide oxygens as the reducing agent here.


text:
What makes a species "fluorescence quencher"?
I just came across an article where they said that sodium fluoride is not quencher (in the context of fluorescence). Why is NaF not a quencher but NaBr is one?

quencher is any species that decreases the intensity of fluorescence when photochemically atoms collide

A fluorescence quencher is any species that causes your fluorophore to stop fluorescing. There are several ways this can happen, but there are two broad categories: chemical reactions and energy transfer. I will give simple examples of each category below. However, which method is operating is going to depend on the fluorophore. Bromide and fluoide will not behave the same way with every fluorophore. Some might be quenched by fluoride and not bromide. Some might be quenched by both. Some might be quenched by neither.
However, in the case of the specific fluorophore in the article, it is quenched by bromide and not be fluoride. The article likely provides an explanation or hypothesis about this behavior, or it cites something that does. Likely this article then is about a selective sensing application. Since bromide turns the sensor off (quenches fluorescence) and fluoride does not, the sensor is selective for bromide over fluoride.

Quenching by chemical reaction

Any chemical reaction will change the relative energies and energy spacing of the frontier molecular orbitals, which could lead to quenching. This chemical change could be something irreversible like oxidizing or reducing the fluorophore (probably not what is happening here). Other chemical reactions are possible, and it could be that bromide reacts and fluoride does not.
This change could also be a reversible one, like protonation/deprotonation, ligand binding, etc. If the fluorophore in your article is a transition metal complex, than it could be quenched by complexation with bromie anion (which displaces some ligand that has charge-transfer absorption/fluorescence with the metal). Fluoride might not do this.
This behavior could be more complex. Perhaps the fluorophore only fluorescences in the presence of some analyte. Maybe bromide reacts with the analyte (and fluoride does not) or bromide competes with the analyte but does not promote fluorescence.

Quenching by energy transfer

Fluorescence works as follows. The fluorophore absorbs a photon and is excited from its ground state to its excited state. When the excited state relaxes back to the ground state it emits another photon.
Fluorescence can be quenched if the excited state can be intercepted. An example is as follows. You have a system with your fluorophore and another molecule. Your fluorophore has a smaller HOMO-LUMO gap than this other molecule, so the fluorophore selectively absorbs photons of a certain frequency range and is excited. However, if the other molecule has a lower energy excited state than your fluorophore, then energy transfer can occur between them if they are close. Now your fluorophore is back in its ground state and the other molecule is in its excited state. If this molecule does not fluoresce, then fluorescence is quenched. This is probably not happening in your example. Note this phenomenon can be given various names like "resonance energy transfer" or "Forster energy transfer."
This type of quenching can get more complex as well. The energy acceptor could go and do chemistry or it could be so eager to be reduced that it not only steals the energy from the fluorophore, but it grabs an electron as well. This is not likely what is going on with bromide and fluoride.
The other type of behavior that can go on here is analyte-induced aggregation. Aggregation of the fluorophore causes the electronic properties to smear out over the aggregate, which likely means no more fluorescence (and perhaps no more absorbance) at the frequencies you care about. This phenomena could be happening in your example. Bromide could induce aggregation while fluoride does not.

Fluorescence (or phosphorescence) quenching is the process by which an excited state looses its energy to another species during a collision (in solution or vapour phase) and it always occurs in competition with fluorescence (or phosphorescence) and non-radiative transitions of internal conversion and intersystem crossing. Thus quenching always results in reduced fluorescence yield and lifetime.
Let S$_1$ be the excited singlet S$_0$ the ground state, T the triplet  and Q the quencher. The mechanisms are (a) heavy atom quenching (b) energy transfer (c) electron transfer and (d) chemical reaction.
In (a) The heavy atom, bromide ions, iodide ions, krypton, xenon etc. causes increased non-radiative transitions so  S$_1$ crosses to T with a greater rate constant and so the fluorescence yield falls as does the fluorescence lifetime.The quenching is caused by increased spin-orbit coupling from the quencher. The quencher does not need to remove any energy from the fluorophore since this is left as vibrational and rotational energy in the triplet, which then transfers it to solvent. The triplet can then suffer quenching from another quencher molecules to form S$0$ the ground state. This also reduces any triplet lifetime and phosphorescence yield.
The extent of spin-orbit coupling is proportional to the quenchers mass so is far greater for heavy species hence the name. This explains the effect of Br vs F in your question. Paramagnetic species such as O$_2$ also quench by causing extra spin-orbit coupling but O$_2$ can also quench by energy transfer as it has two low lying excited states.
(b) Energy transfer. This has two forms (i) Dexter transfer and (ii) Forster or dipole dipole transfer. Dexter is generally short range but Forster can be long range transfer in which S$_1$ can be a few nanometers away from Q for quenching to occur. The condition for transfer is one of energy conservation, the emission spectrum of the donor (S$_1$) must overlap in energy with the absorption of the acceptor. In dipole-dipole transfer the acceptor is electronically excited. This is usually written as D$^*$ + A -> D + A$^*$
(c) In electron transfer the donor and acceptor have to be stable in their ground states. Either the donor or acceptor has to be electronically excited and thus its redox potential is changed to be favourable ($\Delta G$ negative) and reaction occurs. D$^*$ + A -> D$^+$ + A$^-$ or A$^*$ + D -> A$^-$ + D$^+$. After this the D$^+$ and A$^-$ can recombine to D and A in their ground state. The rate constant of energy depends exponentially on their separation and only occurs at close to contact.
(d) In a reaction (other than electron transfer) the excited state is deactivated as bonds are broken or made. The subsequent mechanism depends on the exact chemical nature of the excited molecule and quencher.


text:
How can a nonspontaneous reaction occur?
$$\frac{-\Delta G}{T}=\Delta S_{universe}$$
(This equation applies under isobaric and isothermal conditions.)
I understand that if $\Delta G$ is positive, the reaction is nonspontaneous, and adding that amount of energy to a system can drive that reaction to occur. However, if you’re increasing the Gibbs free energy of the system, then by that equation, the entropy of the universe would be decreasing, which would violate the Second Law of Thermodynamics. Can anyone explain this?

Can you give an example of such are reaction? I think I might know where you confusion is but I'm not entirely sure.

@bon How about an electrolysis reaction? I don’t know; I’m just thinking about nonspontaneous reactions in general...

In electrolysis you are putting in energy to drive the reaction forward. That energy has to come from somewhere and the process that creates in has a large, negative $\Delta G$.

@bon I understand that, so that’s not really the focus of my question. I’m confused about the formula I posed above, really. My central confusion is that if you’re adding energy, then you’re increasing free energy, but you can’t decrease the universe’s entropy!

@lightweaver: The equation is true only when non-compression work is zero.

When you add energy from somewhere, let's say a battery, you are creating a greater amount of entropy there (in the battery or whatever). That's all there is to it. The Second Law is not violated.
The electrolysis process is not non-spontaneous. Yes, it would not occur in the absence of the battery. But the fact that it does occur with the battery there must tell you that overall it is spontaneous. The fact that you are coupling a strongly spontaneous reaction (discharge of the batter) with a non-spontaneous reaction (the electrolysis) means that overall, the electrolytic process is spontaneous.
It's incorrect to treat the electrolysis as if it occurs without the battery.

So you’re saying that when we’re adding energy to drive a nonspontaneous reaction, the equation for $\Delta G$ changes to incorporate the source of energy into that system?

You could look at it that way. To be more precise, the "system" that you are talking about should really include the battery as well. The equations follow.

So, just to verify your answer, in order for reactions that decrease entropy to occur, we need to supply them with energy. By doing that we increase entropy much more somewhere else and thus the overall entropy of the system is increased?

@ΚελεσίδηςΑντώνιος That depends on your definition of "system". The Second Law only applies to an isolated system. If you consider the reaction vessel to be the "system" (in this case a closed system, not an isolated system), and everything outside the reaction vessel to be the "surroundings", then the system and its surroundings make up an isolated system, and the overall entropy of the isolated system increases. However the entropy of the reaction vessel, a closed system, decreases. The entropy of the surroundings increases.

So the system, in which the entropy must and is increased, consists of the beaker with the reactants and the enviroment that supplies the necessary energy. I understand now. Thank you.

@ΚελεσίδηςΑντώνιος That's right!

Is it possible to drive any non-spontaneous process, like melting of ice at -10°C, by coupling  it with a strongly spontaneous reaction?

Any reaction depends on two factors: enthalpy and entropy, and the summation of which is the change of the Gibbs energy by equation (for isobaric and isothermal process):
$$\Delta G = \Delta H - T \Delta S \qquad(1)$$
or, after manipulations we have
$$\frac{\Delta H-\Delta G}{T} = \Delta S \qquad(2)$$
The difference between equations (2) and one you've represented is the last one describes process of the hypothetical system (or real -> Universe) which always become to the initial state and there's no change of enthalpy: $$\Delta H = 0 \qquad(3)$$
If you will try to change Gibbs energy: $$\Delta G = G_2 - G_1 \qquad (4)$$ where$$G_i =  H_i - TS_i \qquad (5)$$ or $$ G_i =  U_i + PV_i - TS_i\qquad (6)$$ you have to change internal energy or volume of the system which will effect on enthalpy and the change of it:
 $$\Delta H \neq 0 \qquad(7)$$ and you will get the equation (2)
Answering on your question: handle the change of enthalpy (7) of the reaction.

A non-spontaneous reaction does not proceed without the continuous supply of energy:  A spontaneous reaction is one that, once started, continues without assistance. Under a particular set of conditions (e.g. room temperature), the reverse of a spontaneous reaction will always be non-spontaneous. For example, if A to B is spontaneous at 300 K, B to A is non-spontaneous at 300 K. However, a substantial increase in temperature (achieved by adding energy) often reverses spontaneity, so that (in our example) B to A is now spontaneous at (say) 1000 K and A to B non-spontaneous at 1000 K.
Electrolysis is an unusual tactic for forcing some non-spontaneous reactions to occur at room temperature. For example, the decomposition of water at room temperature is non-spontaneous, but a continuous supply of electricity causes continuous production of hydrogen and oxygen gases. Part of the electricity provides the energy to break the O-H bonds in water and part of the electricity is used to warm the water reservoir and the laboratory; the latter is essential for the entropy of the universe to increase without which the non-spontaneous reaction could not occur.

As one of the commenters near the top correctly said, the formula $-\Delta G/T = \Delta S_{total}$ only applies when non-compression work is zero. That is, the system is set up so that the only way it can exchange energy with the environment is to expand or contract against some ambient pressure, and to absorb or release heat.  In that case, the enthalpy change of the system is equal to the heat absorbed from the environment: $\Delta H = Q$.  This comes from the fact that the work done on the system is just $W = -p\Delta V$ at constant pressure. So $\Delta H = \Delta(U+pV) = \Delta U + p\Delta V = Q + W + p\Delta V = Q -p\Delta V + p\Delta V = Q$.
Then you can put that into the formula for total entropy:
$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T $$
$\Delta G$ and $\Delta S$ are the Gibbs energy change and entropy change of the system only, while $\Delta S_{total}$ is the entropy change of "the universe", which figures into the most common way of stating the second law (i.e. total entropy never decreases).
In the electrolysis example, however, there is electrical work $W_e$ being done on the chemical system in addition to compression work.  So total work is $W = W_e - p\Delta V$.  When you put this into the formula for enthalpy change, $\Delta H = Q + W + p\Delta V$, the p-V terms cancel again and you're left with $\Delta H = Q + W_e$.  Then substitute $Q = \Delta H - W_e$ in the formula for total entropy, and you get the equation that applies in this situation:
$$ \frac{W_e- \Delta G}{T}  = \Delta S_{total}$$

$$\frac{-\Delta G}{T}=\Delta S_\mathrm{universe}$$
(This equation applies under isobaric and isothermal conditions.)

Another requirement for this equation to hold is that the non-PV work is zero. When deriving it, you use the $\Delta H = q$, which is not the case when there is non-PV work.

I understand that if $\Delta G$ is positive, the reaction is nonspontaneous, and adding that amount of energy to a system can drive that reaction to occur. However, if you’re increasing the Gibbs free energy of the system, then by that equation, the entropy of the universe would be decreasing, which would violate the Second Law of Thermodynamics. Can anyone explain this?

The Gibbs energy is a state function, so no matter how you run a process, for the same starting and ending state you get the same change in Gibbs energy. "Adding energy" will not drive the reaction, but doing work on the system will. The battery does electrical work on the system if you place it outside of the system. In that case, you have an example of a process that goes forward even though $\Delta G_\mathrm{sys}$ is positive.

authors: Αντώνιος Κελεσίδης


text:
Do both NH2 groups in semicarbazide have the same properties?
Would it be correct to consider both the hydrazine and amide $\ce{-NH2}$ groups in semicarbazide (hydrazinecarboxamide) as derivatives of ammonia?
I thought about this possibility while writing the products obtained by reacting a carbonyl compound with semicarbazide. I was taught that only the hydrazine $\ce{-NH2}$ will be treated as a derivative of ammonia, so addition will take there and not at amide side.
Is there any possibility to add at amide? If not why?

Amide lone pair is delocalised into the C=O system (just google it), so it does not react with carbonyl compounds.

@orthocresol but at the same time, the amide hydrogen is more acidic because of conjugate base stabilization by resonance. Im not very sure about this. Can you please comment on it?

@DatBoi I don't see what the contradiction is?


text:
Reversibility of an acid's reaction with water depending on its strength
This is how my chemistry professor compared reaction of a weak and a strong acid with water:

$$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$
  In this case $\ce{H2O}$ breaks the bond between $\ce{H}$ and $\ce{Cl}$ in $\ce{HCl}$ which results in he formation of $\ce{H+}$ and $\ce{Cl-}$ ions. The $\ce{H+}$ ion combines with $\ce{H2O}$ to form $\ce{H3O+}$ (also known as hydronium). Since adding water to an acid causes its hydration, which is an exothermic process, the reaction also results in the emission of heat. This reaction favors going to the right hand side of the equation.
  $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$
  In this case, the size of the acid is greater than the size of $\ce{HCl}$. It means that the process of hydration will be slower. This will lead to a comparatively less number of $\ce{H+}$ ions being combined with water. So, considering Arrhenius' definition of acids, $\ce{CH3COOH}$ is a weaker acid than $\ce{HCl}$. Also, this reaction is going to be reversible.

Now, my questions are as follows:

Why the first reaction favors going to the right hand side of the equation? I think that it has something to do with heat as the heat is lost to the surroundings and the energy level on right hand side decreases. I am still not sure.
Why is the second reaction reversible? I have no idea of why it is reversible so I can't provide my opinions.
Why does water causes ionization of acids? This question might look dumb to some but I am wondering what special water has got that it ionizes acids (or maybe other species too).
Why is there a release of heat? I have read that breaking of bonds is an endothermic process.

$$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$
This reaction owes its ability to go to completion to the great stability the ions it forms when dissolved in water ($\ce{H+}$ and $\ce{Cl-}$). This ionization is caused by stronger attractive forces between the water and the atoms in a molecule (or ions in a salt) than between the atoms within the molecule have themselves, an is an endothermic process. In the dissociation of salts like $\ce{KNO3}$, the solution becomes notably cooler. When an acid dissociates, however, the hydration of $\ce{H+}$ to $\ce{H3O+}$ releases a great deal of heat.
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$
This reaction also results in the formation of $\ce{H+}$ ions, though I wouldn't say that $\ce{CH3COO-}$ is less stable because it is larger than the $\ce{Cl-}$ ion. Consider that $\mathrm{p}K_\mathrm{a}(\ce{HCl})=-7$, $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=4.75$, and $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=0.75$. In case you are unfamiliar, the $\mathrm{p}K_\mathrm{a}$ is a way of expressing how much an acid dissociates in water, and a drop in $\mathrm{p}K_\mathrm{a}$ by $1$ means an acid releases $10$ times as many $\ce{H+}$ ions. $\ce{HCl}$ releases $10^{11.75}$ times as many $\ce{H+}$ ions as $\ce{CH3COOH}$, but when acetic acid's $\ce{C-H}$ bonds are swapped out for $\ce{C-F}$ bonds to make $\ce{CF3COOH}$ (an even larger molecule), its acidity increases by $10^4$. This shows that ion stability has more to do with bond strength and the inductive effect than ion size.
Both of these reactions are reversible. This just means that the forward reaction is occurring as well as the backward reaction (when occurring at the same rate this is known as equilibrium), but the first reaction lies more towards the formation of the products due to its greater stability. All reactions proceed in the backward direction, unless they are irreversible. Though the dissociation of $\ce{HCl}$ is often considered irreversible due to how shifted its equilibrium is, it is not truly irrversible because it has the potential to reform a molecule of $\ce{HCl}$. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, which does not happen in this case.

I got your point about why the rate of reversibility is more in case of acetic acid. But I still don't understand why does the reaction go backward?

@unknownCoder Addressed.


text:
How to calculate pH of a solution when H3PO4, NaH2PO4, Na2HPO4 and Na3PO4 are mixed together in certain amounts to form a solution?
How to calculate pH of a solution when $\ce{H3PO4}$, $\ce{NaH2PO4}$, $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ are mixed together in certain amounts to form a solution?
I know how to calculate pH when they are present individually but not when all of them together. I searched the net also but couldn't find any such example.

Let's start from a simpler one: do you know what to do with a solution of $\ce{CH3COONa;/;CH3COOH}$?

Yes its a buffer.Henderson equation. @IvanNeretin

So you may do the mixture of (say) $\ce{Na2HPO4/NaH2PO4}$ in the same manner, I suppose?

But there are 4 mixtures isn't it ?? And I will get a different pH for each combination of two mixtures.I'm confused @IvanNeretin

Wait a sec. You can deal with any neighboring two of them, right?

Yes I can.Then ? @IvanNeretin

Usually doing calculations of this kind is not hard.  Roughly, you start from some idea of where the pH is going to end up.  For example, if you are only adding these phosphates, count the number of phosphate ions and the number of sodium ions.  Divide the number of sodium ions by the number of phosphate ions.  You (hopefully) get a number $x$ between $0$ and $3$, say $x=2.5$.  Almost all phosphates will then be either $\ce{NaH2PO4}$ or $\ce{Na2HPO4}$.  Pretend that these are the only ions and treat the problem like an ordinary two component buffer.  Then you are done.  This is an approximation but it will work rather well, as long as the $\mathrm{p}K_\mathrm{a}$ of the various acids are different enough.  If this is not good enough, you just treat this as a bunch of acids, with a bunch of $\mathrm{p}K_\mathrm{a}$ and use the acid-base equilibrium equation over and over, that is
$$\frac{[\ce{H}][\ce{B}_i]}{[\ce{A}_i]} = \mathrm{p}K_\mathrm{a}$$
where $[\ce{A}_i]$ is the concentration of an acid "$i$" and $[\ce{B}_i]$ that of the conjugate base.  Given the pH, this gives you the ratio of the concentration of each base to each acid.
But, really, for phosphate buffers, you are very well off if you just use one acid base pair.  For acids like terephthalic acid or isophthalic acid, which have two $\mathrm{p}K_\mathrm{a}$ values that are rather close to each other, that is they differ by less than $1.5$ or so, finding and solving such polynomial equations makes possible sense.  Otherwise, other approximations we routinely make are more important.
For an explicit example, suppose that we consider a compound $\ce{BH2}$ that can give up two $\ce{H+}$ ions, making both $\ce{BH-}$ and $\ce{B^2-}$.  Suppose that we have a total molar concentration of $\ce{B}$,
$$B_t=[\ce{BH2}]+[\ce{BH-}]+[\ce{B^2-}],$$
and that we have also added strong bases or strong acids to change the $[\ce{H+}],$ so that the total possible molar concentration of $[\ce{H+}]$ is
$$H_t=[\ce{H^+}]+[\ce{BH^-}]+2[\ce{BH_2}].$$
We then also have the relationships
\begin{align}
[\ce{H+}][\ce{BH-}] &= K_{\mathrm{a},\ce{BH_2}}[\ce{BH_2}]\\
[\ce{H+}][\ce{B^2-}] &= K_{\mathrm{a},\ce{BH-}}[\ce{BH-}]\\
\end{align}
If we now call $H=[\ce{H+}]$ and $B=[\ce{B^2-}]$ we find the equations
\begin{align}
H_t &= H + B (K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH2}} H))\tag{1},\\
B_t &= B (1 + K_{\mathrm{a},\ce{BH^-}} H (1 + K_{\mathrm{a},\ce{BH2}} H))\tag{2}.\\
\end{align}
We need to solve these equations for $B$ and $H$. Equation $(2)$ is easy to solve for $B$ and we can then substitute into equation $(1)$ to find
$$H_t = H + B_t
  \frac{K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH_2}}H)}
  {(1 + K_{\mathrm{a},\ce{BH-}} H(1 + K_{\mathrm{a},\ce{BH_2}}H))}\tag{3}.$$
We need to solve this formula for $H$, and (of course) $\ce{pH}=-\log_{10}(H)$.  Some algebra from this point gets us to a cubic equation for $H$.  Of course this has three possible solutions.  But, only one is sensible, provided we insist on "physical sense", e.g. all the equilibrium constants and concentrations are positive real numbers.  Somehow, this must be easy to see from the rules about roots of polynomials.  I am not currently willing to do that proof.
Physically, however, the total amount of hydrogen bound to the base will increase with increasing (positive) $H$.  So the right hand side of $(3)$ increases with increasing $H$ - from $H$ for $H=0$ to $2B_t + H$ as $H\to\infty$.  As the right hand side of $(3)$ increases with increasing $H$ there is exactly one positive solution.  However, this right hand side is also not a very "nice" function - the rational fraction is approximately constant except when we are near a "buffer" condition, where it changes quickly.  So, probably an effective way to solve the equation would be to use a computer and "regula falsi" or something like that.  A little thought will allow sensible guesses for any set of $H_t$ and $B_t$.  Then regula falsi will converge quickly.
Or, from a good enough initial guess even Newton-Ralphson.
Or, possibly it would be better in many cases to figure out which acid/base pair is closest to being in the "buffer" region.  And then, suppose that you have $H$ in that buffer region.  Solve for $H$ with only that acid-base pair.  Then, figure out what you did wrong - there can be $\ce{H+}$'s that are not bound to the base in the buffer pair you have chosen or $\ce{H+}$'s that are bound to stronger acids.  Assume that $H$ calculated from a single buffer pair is correct.  Calculate how many $\ce{H+}$'s are unexpectedly bound or unbound from your assumed buffer pairs.  Use this to calculate a new $H_t$.  If $H_t$ has not changed much, which will be the case unless there is another acid-base pair with very nearly the same $\mathrm{p}K_\mathrm{a}$ - differing I suppose by less than an integer from that of the pair you are considering, then use this value of $H_t$ to solve again for $H$, and recurse.
For the stated problem, all the $\mathrm{p}K_\mathrm{a}$ are very different so the $K_\mathrm{a}$ values differ by orders of magnitude.  And, really, all you need do to is figure out which buffer region you are near.  In principle, you can then find tiny errors using the recursion.  But, this is silly in most circumstances.  This will get a clear, continuous curve which always has the correct slope.  But, it will vastly overstate what you know about the problem.  The changes will be smaller than (I expect) lots of stuff: deviations from ideal solutions, errors in $K_\mathrm{a}$ values, etc..
).

I can provide an example which is useful:

$1,2,3$ and $4$ moles are taken of $\ce{H3PO4}$, $\ce{NaH2PO4}$, $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ respectively with $K_{a.1}=10^{-3},K_{a.2}=10^{-7},K_{a.3}=10^{-13}$.
Find the resulting $pH$ when these are mixed together to form an aqueous solution.

First write the $3$ ionic equations that are relevant here for the given $K_a$ values and then pair them up in such a way that you get rid of $\ce{H^+}$ terms on both sides and the $K$ value is of a very high magnitude such that the reaction shift almost completely in the forward direction.
For eg. in the given question, we can pair up the ionic equation for $K_{a.1}$ and reversed equation for $K_{a.3}$ such that the net $K=10^{10}$ and then we put in the value of the moles of each substance:
$\ce{H3PO4 + PO4^{3-}<=>H2PO4^{-} + HPO4^{2-}}$
 1        4            2          3
 _        3            3          4

Similarly we can pair $K_{a.2}$ and reversed $K_{a.3}$ ionic equations, and are finally left with $10$ mole of $\ce{HPO4^{2-}}$ for which pH can be evaluated with the given data as $10$.


text:
Does Finkelstein's reaction proceed through an SN2 mechanism?
I was dealing with a question whose answer was justified only when it was assumed that the Finkelstein reaction proceeds through an $\mathrm{S_N2}$ mechanism. So, does it really follow an $\mathrm{S_N2}$ mechanism or is there some other mechanism at play?

What brings you to assume any other mechanism?

Why do you think it wouldn't proceed through $\mathrm{S_N2}$?

Consider this reaction: (CH3)3-C-Br + NaI (dry acetone) ---> (CH3)3-C-I                      I have every reason to believe this reaction will proceed through Sn1 mechanism as there is stearic hindrance to the incoming nucleophile and moreover I'm getting a very stable tertiary carbocation.

The mechanism should depend on what type of haloalkane you use in the reaction i.e. primary, secondary or tertiary. For primary haloalkanes, the reaction must proceeed through ${S_{N}2}$ mechanism, since the carbocation formed will not be stable, and there is very little steric hindrance while ${S_{N}1}$ mechanism must be used for a tertiary haloalkane sue to stable carbocation formation and steric hindrance. Not really sure about secondary haloalkanes, though.

The Finkelstein reaction is a nucleophilic substitution reaction. As such, it generally has the choice of proceeding via the $\mathrm{S_N1}$ or the $\mathrm{S_N2}$ mechanism.
The reaction is typically conducted in acetone; a polar but not very protic solvent. The solvent is used because all the sodium halides except for sodium iodide are not very soluble in it and precipitate out. It does not show any real preference for either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.
Therefore, we should study the nucleophile and leaving group. Iodide is both a good nucleophile and a good leaving group; the former is important as it allows us to assume a viable $\mathrm{S_N2}$ mechanism. The other halides that you are substituting with iodide are not as good leaving groups but still rather okay. This again does nothing really to tilt the scales towards either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.
The final choice we have is that it depends on the substrate. Probably, if you are replacing a primary halide chances are that the reaction will proceed according to the $\mathrm{S_N2}$ mechanism while an $\mathrm{S_N1}$ mechanism should predominate for tertiary halides.


text:
What reactions occur when pyridine, actic anhydride and citric acid is mixed?
The reaction of pyridine, acetic anhydride, water and citric acid is used as a colorimetric identifier of citric acid. What reactions occur and why does the color change from colorless to greenish? I have been looking everywhere for the answer and have, so far, come up with nothing.
The reaction is made from 1 ml of water containing citric acid, 1.3 ml of pyridine which is then mixed and then 5.7 ml of acetic anhydride is added. Then it is left in a water bath with a temperature of 32 degrees celsius.

@PhMgBr - I am not so sure. If it were a homework question, I should have been able to learn more about it using popular search engines. I cannot. This sounds like chemistry that happens in a proprietary colorimetic assay kit, and the chemistry in these assay kits is generally obfuscated by the vendor.

Right @Ben, my bad. Weren't paying enough attention.

@BenNorris The chemistry here was reported in J. Bio. Chem. in the late 40s. It's not proprietary anymore. The reaction with citric acid, from what I see, gives a red color, not green. Green, supposedly, comes from tartaric acid. [This article] is probably behind a paywall for those not hooked up to the proper IP address, but it's all I've really found so far.

Okay, then it might be behind a paywall for the OP. Regardless the question can be improved.

Sorry. This article. Should be able to see an abstract (or something, not sure if SD does first page for old articles).

This one worked with my VPN connection off.

We have been doing this experiment as a part of a school project where we produced citric acid by fermentation by Aspergillus niger. We have made several reactions with pure citric acid and some have turned green while others have turned red, most of them have been green. We don't know what factors causes the experiment to turn either green or red. We have used a spectrophotometer and measured at 420 nm as is also done in the aforementioned article if I'm not mistaken.

I don't need a specific answer, i only need some idea of what happens and why, so if anyone has any idea why the color change occurs i would appreciate it very much

Pyridine as a base and acetic anhydride might facilitate the decarboxylation and elimination of water from citric acid, leading to citraconic (and itaconic) anhydride. These anhydrides might form charge transfer complexes with pyridine. If I remember correctly, similar complexes have been observed for pyridine with maleic anhydride.


text:
The chlorofluorocarbon reaction in the lab?
The putative reason for banning Freon and its class of chemical (chlorofluorocarbons) is that they will rise to the top of the atmosphere, to the ozone layer and "catalyze" the reduction of ozone to ordinary $\ce{O2}$.
The objection that the amount of chlorofluorocarbons would be tiny in comparison to the amount of ozone, is answered by the idea that the chlorofluorocarbons, do not do a simple reaction, but catalyze at high rate the conversion of $\ce{O3}$ into $\ce{O2}$. In other words, the idea is that a single molecule of chorofluorocarbons could convert millions of ozone molecules into $\ce{O2}$ every minute or something like that.
Unfortunately, this supposed reaction (which would seem to be unlikely on the face of it) seems to have never been duplicated in a laboratory, which I would expect would be easy to do.
Would it be possible to easily set up and measure this reaction in a laboratory? The goal would be to determine the rate of reaction. For example, given a certain molar concentration of chlorofluorocarbon in the reaction vessel, how fast is a given molar concentration of $\ce{O3}$ converted to $\ce{O2}$.

I guess it's possible, maybe already done.

The catalytic stratospheric ozone destruction by CFC's and other compounds is almost certainly too complex to fully replicate in a laboratory with current technology.  The only laboratory of proper scale and conditions is the earth's atmosphere as a whole (no pun here, seriously ;).
Even the 1995 Nobel prize winners Crutzen, Rowland and Molina had made serious omissions from their original work in the 70's that are now known to be critical to the mechanisms of stratospheric ozone depletion.  While Crutzen is pretty much a pure theoretician, Molina and Rowland did much of their work in the laboratory and combined that work with atmospheric measurements.  The bottom line is that only pieces of the whole process can be controlled in a single or set of experiments.  One important factor, for example, that was omitted from their original work was the role played by condensed-phase surface process in mediating catalysis by active chlorine.  Specifically, without the different types of polar stratospheric clouds (i.e. water ice, nitric acid trihydrate,  liquid supercooled solutions of sulfuric acid, nitric acid and water, and other complex matrices) catalytic ozone destruction does not occur. So, although the net of their predictions in the 70's was correct, it was the real-world observations decades later, not laboratory confirmations, that sent them to Stockholm for their prize.
In 2010, an article was published called "Key compound of ozone destruction detected; Scientists disprove doubts in ozone hole chemistry.", Karlsruhe Institute of Technology (KIT).  ScienceDaily. ScienceDaily, 22 July 2010, excerpt available here.  The summary states:

For the first time, scientists in Germany have successfully measured
  in the ozone layer the chlorine compound ClOOCl, which plays an
  important role in stratospheric ozone depletion. Doubts in the
  established models of polar ozone chemistry expressed by American
  researchers based on laboratory measurements are disproved by these
  new atmospheric observations.

This represents an even more recent case in which rigorous laboratory attempts to simulate stratospheric ozone depletion processes failed in the face of experiments and measurements made in-situ in the "real world" laboratory.
The article further points out:

The ozone hole above the Antarctic and the destructive role of
  chlorofluorocarbons (CFC) and their decomposition products have become
  a synonym of both global environmental problems and their solution by
  concerted agreements worldwide. Scientific fundamental research into
  ozone chemistry of the atmosphere was the basis of international
  agreements, such as the Montreal Protocol of 1987, which has put
  limits on CFC production. The success of the political implementation
  of these scientific findings is reflected by the fact that the
  chlorine content of the atmosphere and, hence, the ozone destruction
  potential recently started to decrease slowly.

So, the upshot here is that it was primarily in-situ experiments and measurements that led to predictions of catalytic chlorine mediated ozone destruction being later confirmed by measurements. The same was true for formulation of a solution to the problem, which again in-situ measurements confirmed.
In short, laboratory experiments have repeatedly been outperformed by in-situ experimentation and observation in the case of stratospheric ozone depletion studies.  The same is true for many complex, global chemical and physical process; it simply is not always feasible to re-create a system of large scale and complexity in a laboratory.


text:
How to make sense of the crystal systems names
I cannot fix the names and structures of the seven crystal systems in my head. There's a thousand books explaining their geometry and it's totally fine and understandable, no problem. But what do the names mean, etymologically?

Cubic
Tetragonal
Orthorhombic
Monoclinic
Hexagonal
Rhombohedral or Trigonal
Triclinic

OK, cubic is simple enough, hexagonal too. I know that "-eder/-hedral" means "side" and "-gon" means "angle". Why is a system with three different edges and angles triclinic? What are the rationales behind each of the names?
A rhombus has sides of equal length. Orthorhombic has only right angles (orthogonal), but the sides a,b,c have different lenghts. So why "rhombic"?

If you're looking for one cohesive system of nomenclature, then there isn't one. If you want the etymology of each word then look the words up in a dictionary.

I want the etymology (thanks for the word ;)). Sadly, the dictionaries (http://www.merriam-webster.com/dictionary/triclinic ,http://www.oxforddictionaries.com/de/definition/englisch/orthorhombic) are not helpful. They just repeat the definition.

AFAIK, "-clin" is a cognate of "inclined", as in "not straight", so triclinic and monoclinic both make perfect sense: they have three and one non-$90^\circ$ angles, correspondingly.

We're  getting somewhere, I think. "n-gonal" means "one Cn axis, angles are 90°, resp. fixed by the symmetry"

"Trigonal" and "rhombohedral" come from different features of that system.  The former comes from the presence of a threefold rotation axis (the principal unit-cell axes are distributed around this threefold axis, so they have the same lengths and angles between them).  "Rhombohedral" comes from a polyhedron with this ($D_{3d}$) symmetry whose faces are all rhombi.

I've found that oed.com is the best source for word origins.  Unfortunately, it's behind a paywall.  Barring that, you might try dictionary.com  E.g., here's what the latter says about "triclinic":  ORIGIN OF TRICLINIC
1850–55; tri- + Greek klī́n(ein) to lean, slope + -ic

Monoclinic and triclinic
Ivan (in the comments) is spot on about these two. Among all the systems, these are the only ones with angles that are not fixed by symmetry, so you actually have to measure them.
Rhombohedral or Trigonal
These are two different settings of the same lattice. The unit cell can either be a prism (with angles of 90, 90 and 120 degrees, blue lines in the figure) or a rhombus (with all three angles equal, and all three sides equal, black lines in the figure).

In general, there are infinite choices for unit cells (you just need three translation vectors that are not linearly dependent, so pick an origin on a lattice point and three other lattice points and its a unit cell). The conventional choices usually have more of the symmetry elements along the axes, face diagonal or body diagonal, and keep the volume of the unit cell fairly small.
Orthorhombic
This used to be called pseudo-orthorhombic.

Again, you have a choice of unit cell, and the more uncommon one is a rhombohedral prism. When you switch from all right angles to a rhobohdral prism setting, face-centered turns into primitive and primitive turns into face-centered.
Here is a quote from Wikipedia:

In the orthorhombic system there is a rarely used second choice of crystal axes that results in a unit cell with the shape of a right rhombic prism;2 it can be constructed because the rectangular two-dimensional base layer can also be described with rhombic axes. In this axis setting, the primitive and base-centered lattices swap in centering type, while the same thing happens with the body-centered and face-centered lattices.

Historically, the first data available to crystallographers (and minerologists) were the shapes of crystals. Orthorhombic crystals can have faces that are at 90 degree angles or not, depending on the kinetics of growth. Pictured below are three crystal forms with underlying orthorhombic symmetry, with the 2nd one showing is a shape with (+/- 1, +/- 1, +/-1) faces labeled orthorhombic pyramid:

Cubic, Hexagonal and Tetragonal
See OP's comments for those.
History
The German Wikipedia article on crystal systems has the terms Bravais used in 1886:

Assemblages terquaternaires (cubic)
Assemblages sénaires (hexagonal)
Assemblages quaternaires (tetragonal)
Assemblages ternaires (trigonal)
Assemblages terbinaires (orthorhombic)
Assemblages binaires (monoclinic)
Assemblages asymétriques (triclinic)


text:
Is poison still poisonous after its 'expiration date'?
As we all know that,

Any poison is nothing but a chemical compound.

And as discussed in the question: Chemicals-do-have-an-expiry-date!
So, my question is: Is poison still poisonous after its 'expiration date'? and Is every poison always as poisonous as just after it was synthesised  ?
please feel free to update with the suitable tag(s).

The problem here is tagging compounds with "expiration date" often is very unscientific.

Yes you're right, but do they have expiry date? As in afterwards, they won't be as much effective as they were before??

You see, there is no physical phenomenon as "expiry date". Unless something happens to the toxin, i.e. it degrades, reacts with moisture, air etc. that eventually kicks in, there's no clock ticking counting towards the 'expiry date'.

Ok, that is explained well in the tagged question. And more importantly, why do medicinal drugs have 'the' expiry date?

Commercial chemical products carry an expiration date for legal reasons: It's the date past which the company no longer guarantees the product to work properly, remain safe, etc.

@Brian oh, i now understand. Thank you, you too Grignard reagent ;D

Please check the updated question. Does it make proper sense?

An expiration date is merely a guarantee that the product will still have a certain potency when stored as recommended given known ways it can degrade. It doesn't suddenly change at the date: it is just a statistical average. So, for things that do degrade (not all poisons will, like elemental lead), the date simply means the poison might be slightly less effective that it was before the date not that its poisonous effect will suddenly disappear.

Protip: Poisons are not the opposite of food. Just because you can eat food before it's expiration date doesn't mean it's a good idea to eat poison after it's expiration date

@PyRulez Yes that's a very good point.

It's very common to see bottled water with an expiration or best by date. I've even seen salt sold on being in the ground for 'millions of years' with a 4 year expiration date. So, the question is, if it did, would you believe it?

Yes, we shouldn't believe everything that we're forced to read.

It depends on what the poison is.
If we take the colloquial use of the word and include toxins and venoms, many are things like proteins that will certainly denature or otherwise degrade, eventually becoming harmless. e.g. tetrodotoxin, ricin, botulinum, etc.
I would expect that type of poison to have the shortest shelf-life as they are relatively fragile.
Many other poisons are small organic molecules.
These can often be degraded by oxidation in air, exposure to UV, hydrolysis etc. and would include things like nicotine and nerve agents like sarin and VX.
Many nerve agents, have shelf lives of a few years and research has actually been done to extend them for use in munitions.
Several metals are known to be poisonous (like lead, mercury, and cadmium) and are problematic because they are toxic in not only their elemental forms, but also in inorganic and organic compounds.
There may be a great difference in toxicity of the different forms, (see elemental mercury vs methylmercury), but most forms remain at least somewhat toxic.
These may last a very long time because reactions likely to occur under normal conditions may not render them safe, e.g. a chunk of cinnabar ($\ce{HgS}$ mineral) sitting on your desk will not undergo any significant change to render it safe, even on a geological timescale.

@ABcDexter: The existing example (cinnabar mineral) wasn't synthesized at all ! You assume poisons are synthetic, when you say "just after it was synthesised". But many are naturally occurring.

So many see the word "chemical" and get the heebie jeebies. Few folks stop to consider that if you're allergic to strawberries they can be just as deadly as cyanide. What is a strawberry? Just a sack of chemicals.

Yes, agreed. But again, 'some' naturally occurring chemicals like strawberries do have an expiry date after which they rot, right?

@ABcDexter: No, they pretty much start to rot as soon as they're harvested. (But that's biology more than chemistry).

Ok, so that comes under botany or boichemistry :-)

Cinnabar (or mercury sulphide in general) is actually one of the few mercury compounds that is practically nontoxic if sitting on the desk because it has no noteworthy vapour pressure, does not dissolve in water to any meaningful extent and therefore is not taken up under ambient conditions. Even if it is taken up (orally), I expect most of it to come back out on the other side unchanged. So I there should be much better examples out there (gut feeling tells me arsenic would be one).

Chemicals can also decompose due to contained impurities.  I would generally and universally agree that deterioration does occur over time. The extent of which varies.  I would also say that business reasons,  aside from legal reasons, also play into the expiration date.  See also concepts of shelf-life and pot-life.
Also,  the decomposition or reaction product(s) may or may not be relatively benign compared to the reactant(s).

"Not up to quality" for a poison could mean anything. It could be more toxic; it could be less. It could be differently toxic, so that it's no longer as effective against its target but could cause severe harm to other species. It only means that one chemical may have started to degrade, and the end product of that degradation could be very different.
So don't get too hung up on the category "poison". Poison are just chemicals, and time affects them in the same way other chemicals change over time. And don't get too hung up on expiration dates, either: they're just guesses given the wide range of storage conditions.


text:
How does molecular orbital theory explain CH4?
I have understood the formation of $\ce{CH4}$ by valance bond theory, but I'm having trouble understanding it through molecular orbital theory.
The energy level diagram of molecular orbitals of $\ce{CH4}$ is not clear to me.

What specifically are you having trouble with?

Methane doesn't illustrate the difference very well, but the difference is nicely illustrated with benzene. Valence bond theory would have the electrons in the "double bonds" isolated between atoms as so:

But the electrons in the "double bonds" are not isolated in three double bonds but distributed equally over all 6 carbon atoms. Valence bond theory tries to wiggle out of this dilemma by posing the Kekule resonance structure:

which still belies the situation. The double bonds don't "resonate." The extra 6 electrons are distributed in one aromatic molecular bond (with three degenerate orbitals) evenly over the 6 carbon atoms.

Did you read the question? The OP wasn't asking about the different approaches that MO and VB theory take to double bonds.

I've edited the question to try to point out the difference better. I don't really think the OP wants to know why MO and VB theory give difference values for the bonding energies when you crunch the numbers. I think the OP is missing the idea that VB theory isolates chemical bonds as only being between two particular atoms. To me this problem is best illustrated with conjugated double bonds.

@MaxW The bonds in $\ce{CH4}$ are not conjugated, though.

@Ringo - I know that methane doesn't have conjugated bonds but I'm trying to show why VB isn't enough and why MO are more general. The difference between the two in methane just seems much much harder to explain. I don't see how you'd explain the difference without crunching the numbers which is very complex.

There is an explanation in this weblink that directly answers the question, but I really don't think that it will solve the OP's dilemma. http://www.users.csbsju.edu/~frioux/h2bond/MethaneMOBonding.pdf // I really should have waited for the OP to clarify bon's query about what wasn't understood rather than guessing...

Thank you sir @MaxW for all the explanation and for the link. I finally got my problem resolved.

@MaxW You can still edit accordingly.


text:
Why is solid sodium hydride a base and not an acid when reacted with water?
When solid $\ce{NaH}$ is mixed with water, the products are not $\ce{Na-}$ and $\ce{H3O+}$. The reaction yields $\ce{NaOH}$ and $\ce{H2}$ gas.
Why does solid $\ce{NaH}$ not act as a Brønsted acid in this case?

It's actually one of strongest bases...

@Mithoron Of course, since it produces the very strong base NaOH.

The answer lies in electronegativity. Sodium has a significantly lower electronegativity ($\approx 1.0$) than hydrogen ($\approx 2.1$) meaning that hydrogen pulls electron density towards itself, away from sodium to generate a sodium cation and a hydride anion.
For a compound to be a Brønsted acid, you need to draw electron density away from the hydrogen, i.e. connect it to an electronegative atom such as oxygen, fluorine, nitrogen etc. Only then can it formally be described as $\ce{H+}$ and can dissociate as such.
A hydride is much better described as $\ce{H-}$ and has a free lone pair. As such, it is a Brønsted base, not an acid. In fact, if you extend the Brønsted acid/base definition in the way Lewis did, you will arrive at the sodium ($\ce{Na+}$) being the acidic species here.
The Brønsted acid/base reaction product of the base $\ce{H-}$ and the acid $\ce{H+}$ happens to be $\ce{H2}$, and because the acidic hydrogen is abstracted directly from water (hydrogen gas can bubble away, shifting the equilibrium even if the reaction were not thermodynamically favoured), $\ce{OH-}$ ions are left which can be written with the remaining $\ce{Na+}$ cation to give sodium hydroxide.

The hydrogen is in the form of a $\ce{H^-}$ ion, negatively charged instead of the usual partial positive. Ths hydride ion strips a $\ce{H^+}$ from a $\ce{H_2O}$ molecule to become $\ce{H_2}$ gas and leaves behind $\ce{OH^-}$.

While it's correct, I think you should elaborate to provide better answer.


text:
Why Does Coordination of Metal Ions Happen Anyways?
I've been studying coordination chemistry, but I still have a fundamental question about the area: why do transition metals even form coordination compounds, while main group metals do not? Why can't they just form simple ionic compounds, treating their d electrons as valence electrons?
On that note about d-orbitals as valence electrons: when do they act as valence electrons and form "simple" ionic (such as $\ce{CrCl_3}$) or covalent ($\ce{MnO_4^-}$) bonding, and when does coordination happen instead? Lately, I've began to think that transition metals don't even form simple ionic bonds or covalent bonds. It used to seem to me that in the chromate ion, $\ce{CrO_4^{2-}}$  (by drawing the Lewis structure) the d-orbitals in chromium participate as valence electrons in covalent bonding, but then how is it colored without d–d transitions? Sulfate, with $\ce{S}$ instead of $\ce{Cr}$, is colorless $\ce{SO4^2-}$. Thus, there must be coordination chemistry involved in $\ce{CrO4^2-}$ (is $\ce{O^2-}$ even a good ligand)? And with $\ce{CrCl_3}$, is it indeeed a simple ionic bond like I thought, or is there coordination going on with that, too?
Basically, why can transition metals exhibit coordination in addition to (at least what seems like) ionic and covalent bonds?

I wouldn't say main group metal do not. I would actually say they almost certainly do, and you're not likely to see an unsolvated metal in solution. They do tend to be more labile bonds, however, due to the more metallic character of the alkali and alkaline earth metals. Look at the mechanism for the Weinreb Ketone Synthesis for an example of a synthetically useful of main group metal coordination.

For the colour of chromate thing, please search - it's been answered before more than once. If chromate doesn't work search for permanganate. Furthermore I would strongly advise against this black-and-white approach of classifying things so strictly into one category or another. Some approaches to bonding are more instructive for certain compounds and less useful for others. For example I hardly expect CrCl3 to be completely ionic, either in solution or  even in the solid state. There are numerous examples of solid state structures being driven by coordination preferences.

Check out this question about a main group metal complex. They certainly do exist and some of them are stable enough to be crystallised as salts with all sorts of weird anions like this.

Your approach is fundamentally wrong. Main group elements form complexes all the time. It is just that these complexes are typically kinetically labile and quickly decompose, rearrange and so on.
Main group elements’ complexes are also typically colourless due to the lack of d-electrons or due to the fully populated d-subshell meaning that all electron transitions are in the UV range, so you cannot easily study them via colour changes.
The state of the d-subshell (full or empty, nothing in-between) also means that no additional stability of certain geometries exist; the s and p orbitals can more or less rearrange to whatever is required.
And finally note that ‘simple’ ionic compounds are, in fact, huge coordinative compounds. A single sodium ion in a $\ce{NaCl}$ crystal could be described as a $\ce{[Na(Cl)_{\frac{6}{6}}]}$ complex.


text:
Calculating the mass of water removed by air conditioning condensation
I cannot figure out how to calculate the mass.
The question is:

"Suppose the percent relative humidity is $\mathrm{80.0\%}$ at $\mathrm{91.4\ ^\circ F}$ ($\mathrm{33.0\ ^\circ C}$) in a house with a volume $\mathrm{245\ m^3}$. Then the air conditioner is turned on. Due to the condensation of of water vapor on the cold coils of the air conditioner, water vapor is also removed from the air as it cools. After the temperature has reached $\mathrm{77\ ^\circ F}$ ($\mathrm{25\ ^\circ C}$), the percent relative humidity is measured to be $\mathrm{15.0\ \%}$. What mass of water vapor has been removed from the house? (Reminder: take into account the difference in saturated water vapor pressure at two temperatures.)"

The answer is $6.00\ \mathrm{kg}$.
Relative humidity= (actual pressure of $\ce{H2O}$ pressure/partial pressure of $\ce{H2O}$ if saturated) * 100
So, my first step was to find the actual pressure of the initial and final conditions.
$X/355.1 = 0.80$
$X = 284.08~\mathrm{Torr}$
Then $X/23.76 = 0.15$
$X= 3.564$
$\mathrm{284.08~Torr- 3.564~Torr = 280.516~Torr}$ (water removed, I think)
I attempted to use the ideal gas formula ($pV= nRT$) but kept getting the wrong answer. How do I approach the rest of the problem?

Are you using the right gas constant since you're in torr? Did you convert the moles to grams  by multiplying the mols by 18.02 g/mol? Which temprature did you use? When did you include the relative humidity?

Step 0) Realize that sometimes the answer given for a homework problem is wrong.
In this case it appears that the correct answer is $\pu{6.11 kg}$ of water, not $\pu{6.00 kg}$ of water.
The approach to the problem given below is to simply calculate the total water mass in the house before and after cooling, then subtract the latter from the former to get the mass of water removed by condensation.

Step I) Assemble parameters applicable before and after cooling:
$R =$ gas constant $= \pu{8.31m^3~bar~mol^{-1}~K^{-1}}$
$V = \pu{245m^3}$
$MW_{water}=\pu{18.0g~mol^{-1}}$
${n=\frac{PV}{RT}}$

Step II) Calculate water mass before cooling:
$T = \pu{33^oC = 306~K}$
$P_{saturation} = \mathrm{50.3~mbar^{(1)}}$
$P = 80\%$ of $P_{saturation} = \mathrm{40.2~mbar}$
${n=\mathrm{\frac{0.0402~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 306~K}=387~mol~\ce{H2O}}}$
$\mathrm{387~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=6.97~kg~\ce{H2O}}$

Step III) Calculate water mass after cooling:
$T = \mathrm{25^oC=298~K}$
$P_{saturation} = \mathrm{31.7~mbar^{(1)}}$
$P = 15\%$ of $P_{saturation} = \mathrm{4.76~mbar}$
${n=\mathrm{\frac{0.00476~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 298~K}=47.9~mol~\ce{H2O}}}$
$\mathrm{47.9~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=0.862~kg~\ce{H2O}}$

Step IV) Calculate total water removed from the house:
$$\mathrm{6.97~kg~\ce{H2O} - 0.862~kg~\ce{H2O} = 6.11~kg~\ce{H2O}}$$
Reference:
1) Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.
....


text:
Chemisorption vs Physisorption?
What is the fundamental difference between chemical adsorption and physical adsorption.
Why do some compounds chemisorb and others physisorb?

It's the difference between binding to an adsorbent chemically (undergoing a chemical reaction) vs simply binding through attraction, i.e. electrostatic attraction.

It appears to be one of degrees of bonding, without a clear distinction. As stated in this Wikipedia article:

Physisorption, also called physical adsorption, is a process in which
  the electronic structure of the atom or molecule is barely perturbed
  upon adsorption.$\mathrm{^{[1][2][3]}}$

A more thorough description is given later in the article:

"The fundamental interacting force of physisorption is caused by van
  der Waals force.   [...]   In comparison with chemisorption, in which
  the electronic structure of bonding atoms or molecules is changed and
  covalent or ionic bonds form, physisorption, generally speaking, can
  only be observed in the environment of low temperature (thermal energy
  at room temperature ~26 meV) and the absence of the relatively strong
  chemisorptions. In practice, the categorisation of a particular
  adsorption as physisorption or chemisorption depends principally on
  the binding energy of the adsorbate to the substrate."

1) K. Oura; et al. (2003), Surface Science, An Introduction, Berlin: Springer, ISBN 978-3-540-00545-2
2) M. C. Desjonqueres; et al. (1996. Corrected printing 1998), Concepts in surface physics (2nd ed.), New York: Springer-Verlag, ISBN 3-540-58622-9
3) Hans Luth; et al. (1993), Surfaces and interfaces of solids, Springer-Verlag, ISBN 978-3-540-56840-7


text:
Electrocatalysis of Hydrogen Evolution Reaction with metal oxides
I'm new to the topic of electrocatalysis so I have a few questions. Also, I don't have much knowledge about thermodynamics, but I will try to ask anyway.
I'm looking for a material to catalyze the Hydrogen Evolution Reaction for $\ce{H2}$ production. So, for pure metals we have the Volcano curve:

Then, with this curve, I found in literature that we can mix metals from the different branches (positive and negative slopes) and get a new alloy with an intermediate M-H energy strength (which is desirable to improve the catalyst activity.)
But, I want to know if there exists the same curve (and the same logic) for metal oxides, because I need to predict if an alloy of rhenium oxides with copper oxides are going to be a better catalyst than only rhenium oxides.
Is there some method of theory to predict this before going to the lab and doing some experiments?

Presence of Titanium on this diagram is hilarious. It's good there is no aluminium here, or I'd laughed my ass off.

Is there “oxide alloys”? I am not sure the two oxides can freely mix. 2) There can similar volcano curves depending on the mechanism of reaction.

You can use a similar, DFT-based approach to estimating the electrocatalytic activity of oxides, as has been done with the oxygen evolution reaction, but since most oxides are not stable at reducing conditions present in HER, there's not as much literature on that particular material-reaction pairing.  As an aside, a popular theory of metal-adsorbate bond formation, the d-band model was formulated specifically for metals, so the question as to whether or not there's a model with comparable simplicity that includes oxides (or other binary compounds) is still somewhat of an open question.


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Why do compounds with similar properties have different Stitch IDs?
I'm not from the Pharmacy or Chemistry background. I have been wondering why certain drug compounds have similar properties but each of them have different Stitch IDs in PubChem?
For example:

Stitch ID 9782 Betamethasone
Stitch ID 5743 Dexamethasone

Bethamethasone and dexamethasone are different chemicals, hence they behave different, therefore they have different identification numbers. The numbers you cite are not STITCH IDs, and as far as my research goes these do not exist. Instead they are CID - PubChem Compound Identification numbers.

As you can see the molecules are diastereomers. The stereo centre of the steroid differs at carbon-16. When you look them up with STICH 4.0 you will see differences with their predicted functional partners. While the first one is nearly identical, the second partner for betamethasone is proopiomelanocortin, while for dexamethasone it is ABCG2.
For more information on STITCH (‘search tool for interactions of chemicals’) I recommend: Michael Kuhn, Christian von Mering, Monica Campillos, Lars Juhl Jensen, and Peer Bork, Nucl. Acids Res. 2008, 36 (Database issue): D684–D688. (Open Access); PMCID: PMC2238848.
Image sources:

National Center for Biotechnology Information. PubChem Compound Database; CID=9782, https://pubchem.ncbi.nlm.nih.gov/compound/9782 (accessed Apr. 20, 2016).
National Center for Biotechnology Information. PubChem Compound Database; CID=5743, https://pubchem.ncbi.nlm.nih.gov/compound/5743 (accessed Apr. 20, 2016).


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Gibbs Free energy Process v/s Reaction
How is Gibbs free energy for a reaction different from that in a process? Is it because a reaction generally occurs at a constant temperature and pressure, and so dG = dH - TdS, while for a process temperature may or may not be fixed for a process, say expansion of a gas, and so dG = dH - TdS - SdT?
Also, how do we define H at non-constant pressure? Is it dH = dU + d(PV)?
Also, why do we say that The G at any point is the max non-mechanical work that can be drawn out? Why does this not include mechanical work too?

The second part has been addressed multiple times already - just a search for "gibbs work" will bring up several questions asking the same thing. http://chemistry.stackexchange.com/questions/32873/regarding-gibbss-function-why-is-there-non-expansion-work

We look at dG at constant temperature and pressure only as a vehicle for determining the relationship we need to use to calculate the equilibrium constant for the reaction.  That relationship can be used for all temperatures and pressures.


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Does the plastic cup PETE 1 material leak toxic material if subject to boiling water even for part of a minute?
Sometimes a cafe may pour boiling tea into a plastic cup made of PETE 1 and then add lemon and cold water / ice to it.  When the cup is handed to the customer, the cup looks ok.  And I suppose even if a person drinks such drinks for a few years, and gets cancer, what proof do you have that it is due to the cup?   The cafe doesn't really care, because money, profits, and speed is of the utmost importance in the country.
For PETE 1 plastic, when it is subject to boiling, 100°C, water or tea for 30 seconds or 1-2 minutes, will it already leak out toxic chemicals?

PETE? Do you mean PETP (sometimes PET)?

I saw the PETE 1 and PET 1 are the same

#1 is the recycling number for the plastic.