text:
What is Markovnikov's addition?
The electrophilic addition of $\ce{HBr}$ across an unsymmetrical alkene follows Markovnikov's addition, but sometimes if the addition occurs via rearrangement then should we call it a Markovnikov addition?
Earlier, in an entrance exam, a similar question was asked and they have given the correct answer which states Markovnikov's addition for a reaction that occurs without a rearrangement.
In my opinion, if addition occurs without rearrangement then it should be Markovnikov's addition.

I always remembered Markovnikov addition as "the rich get richer" (in terms of protons). i.e. the carbon with more hydrogens will receive another, while the corresponding carbon will receive the halogen etc. As far as your question goes, I think it fair to say that mechanism is irrelevant. It's just a description based on observed products.

There is a difference between Markovnikov's addition and Markovnikov's addition product. The former always holds true as the carbocation formed initially due to an electrometric shift of electrons always happens at the most substituted carbon. However, Markovnikov's product does not account for rearrangements. If the carbocation rearranges to form a more stable carbocation, the product obtained usually won't be called a Markovnikov's product.

The idea of Markivnikov or anti-Markovnikov addition fundamentally isn't related to the mechanism by which a reaction took place, rather, the term is a way of describing the observed products obtained from certain kinds of addition.
The original statement of  Markovnikov addition is as follows, and applies specifically to the addition of hydrogen halides ($\ce{HX}$) across an alkene:

In the addition of hydrogen halides to unsymmetrically constituted [unsaturated] hydrocarbons, the halogen atom becomes attached to the carbon bearing the lesser number of hydrogen atoms.

The statement was a purely empirical one (i.e. it was based on the results of experiment, rather than any consideration of the mechanism/pathway that the reaction had taken). It can also be expanded to consider any addition of a polar reagent ($\ce{RX}$) across alkenes and alkenes. The following definition is taken from the IUPAC Gold Book (a standard reference for definition of chemical terminology):

In the heterolytic addition of a polar molecule to an alkene or alkyne, the more electronegative (nucleophilic) atom (or part) of the polar molecule becomes attached to the carbon atom bearing the smaller number of hydrogen atoms.

Thinking about the situation mechanistically, the Markovnikov rule basically just says that the less electronegative atom in the polar reagent $\ce{RX}$ ends up bound to the end of the alkene (or alkyne) most able to stabilise a developing positive charge (i.e. the most hindered end of the alkene/alkyne).
In terms of labelling the products of reactions, as discussed in your question, its perfectly valid to describe the products as those of Markovnikov addition (as stated previously, the term itself is saying nothing directly about mechanism, even if implied).

Good point, but it may still be useful to point out that in practice, Markovnikov additions usually proceed by carbocation or halonium Intermediates, whereas the two commonly taught anti-Markovnikov additions proceed by radical and concerted pathways (rather than cationic).

I think my point was that its not really saying anything about mechanism at all.  Being able to class reactions as Markovnikov vs anti-Markovnikov is simply a convenient correlation. I'd possibly go further and  actively advise for not putting reactions into boxes like that.

Wikipedia tells me:

The rule states that with the addition of a protic acid HX to an
  asymmetric alkene, the acid hydrogen (H) becomes attached to the
  carbon with fewer alkyl substituents, and the halide (X) group becomes
  attached to the carbon with more alkyl substituents. Alternatively,
  the rule can be stated that the hydrogen atom is added to the carbon
  with the greatest number of hydrogen atoms while the X component is
  added to the carbon with the least number of hydrogen atoms.

So, even with rearrangements, the halogen would still attack the most substituted site, so I'd say yes, you can consider it a Markovkinov addition.

If I take 3-phenyl propene then bromine goes to benzylic carbon which is having more number of hydrogen initially

@ArvindTiwari This might be a flaw in Markonikoff's rule. Nonetheless, don't use Markonikoff's rule. Simply compare carbocation stability.


text:
Why is the buckminsterfullerene the purest form of carbon?
Other websites say that $\ce{C60}$ doesn't have surface bonds that are attracted by other atoms as in graphite and diamond.
I understand that graphite may be attracted by other atoms because of its dangling electron. But why diamond? Each carbon in diamond is covalently bonded to $4$ other carbon atoms in a tetrahedral fashion.

Just to confuse you: I've read (albeit it has been a long while) that diamonds are somewhat resistant to impurities because it's difficult to get in its crystal lattice. Also if diamonds were pure, they would've been transparent. Diamonds are not transparent. I like bluish diamonds. Yellowish diamonds are also common.

@BerryHolmes I almost got confused. But maybe $\ce{C60}$ is relatively more pure than diamond but I don't know why. Maybe because you told that diamond contains impurities, and $\ce{C60}$ is artificial. But yeah, thanks for the info.

What happens at the edge of a diamond lattice...

@ReeshabhRanjan Buckminsterfullerene is not artificial - it is found in soot and has been detected in deep space.

@AndrewMorton I didn't know about deep space, but l it isn't like any soot. The soot is made by, as Wikipedia writes, in an 'arc-process between two graphite electrodes in a helium atmosphere where the electrode material evaporates and condenses forming soot in the quenching atmosphere.'

Diamond has dangling bonds on the outer surface of the crystal for pretty much the same reason as graphite. If you understood graphite differently, then you understood it wrong.
See, a molecule of oxygen contains 2 atoms, a molecule of sulfur has 8; but how many atoms are there in a "molecule" of diamond or graphite? Try drawing one to the end, so as to count them. You won't be able to do that. There is no end. The thing is infinite. But the real-world objects are finite, which means that at some point you have to say "Enough" and crop your ideal structure, and in doing so, you leave dangling bonds which attract other atoms. Fullerene lacks those, and hence is "more pure".
There is an altogether different dimension to the problem. Our thought experiment implied that we are able to produce a huge crystal without defects except maybe some on the surface. This is not true. Real-world compounds always contain impurities, and once you have a wrong atom built into the crystal lattice of graphite or diamond, it is stuck there forever. You'll never remove it, short of destroying the entire crystal. Fullerenes, on the other hand, are molecular compounds. They can be dissolved. They can be put through chromatography, sublimation, and other purification techniques. We can always remove any impurity (not that we can remove all of them, because nothing is ideal).
Either way, fullerenes win.

Thank you. Now I understand it well. The structure of $\ce{C60}$ clearly defines it's boundaries.

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text:
How do these cascade reactions work?

A. $\ce{Cr2O7^2-}$  B. $\ce{H+}$  C. $\ce{OH-}$  D.
  $\ce{Cr^3+}$  E. $\ce{Ba^2+}$
Which of the above is the appropriate ion for each blank in the
  following series of reactions?
$$\begin{array}{ccc}\ce{OH-} + (5) \xrightarrow{} & \ce{CrO4^2- + H+}
 & \\  & \ce{v} (6) & \\  & \ce{BaCrO4 (s)} & \ce{<=>[(7)][(8)]} &
 \ce{Ba^2+ + Cr2O7^2- + H2O}\\  & \text{(yellow precipitate)} & & \ce{v
 SO2 (g), H+} \\  & & & \ce{BaSO4(s)} + (9) + \ce{H2O} \end{array}$$

This question can also be viewed at https://i.stack.imgur.com/i08sR.jpg
So while studying, I came across this reactions' question, which I believe is called a 'cascade reaction'. I have no idea how to do it. I can figure out a few parts, like ($6$) is obviously $\text{E}$, but how do you do the others?

how do you do the others?

You do it like a detective. Even the slightest details matter, and you've to link a lot of concepts together.
The main theme you're missing in your question is this equilibrium:
$$\ce{ CrO4^2- <=>[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$
Dichromates are supposed to be stable in an acidic medium. So using this you can figure out that ($7$) is $\ce{H+}$ and ($8$) is $\ce{OH-}$. Again for ($5$) you're adding a base which is converting it to a chromate. So ($5$) should be a dichromate $\ce{Cr2O7^2-}$.
So we're only left with ($9$) and $\ce{Cr^3+}$ and they are bound to go with each other. Since this is not an exam here's a reasoning too,
$\ce{K2Cr2O7}$ is a strong oxidizer in acidic mediums. If it's going to oxidize others, it will itself get reduced. To be precise here's the reaction that's taking place:
$$\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$$

Thank you! The only question I have left is, how do you know that $\text{Cr}^{3+}$ goes in the spot $9$? I understand that you would have some Chromium, but why wouldn't the dichromate ion stay in place?

@Neal If you look at the oxidation states, $\ce{SO2}$ changes into $\ce{SO4^2-}$, i.e. sulfur gets oxidized. So, dichromate ion must undergo reduction. Thus, chromium changes from a $+6$ oxidation state to $+3$.

As noted by @Berry, hydroxide ion converts perchromate ion to chromate ion and a proton converts chromate ion to perchromate ion.
$$\ce{ CrO4^2- <=>[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$
So, in this case you figured out 7 and 8 to be proton(B) and hydroxide ion(C) and thus 5 would be (A) from above equation.
Since you want barium chromate from chromate ion, you just add any barium salt(E).
And last one would obviously would be left out option i.e (D). Unfortunately, I couldn't find any reaction for this. Only thing I can think of is that sulfur dioxide is acting as reducing agent(reduce perchromate to chromium(III)(D)).

Full reactions

$$\ce{K2Cr2O7(A) + 2KOH(C) → 2K2CrO4 + H2O}$$

If you use potassium ions as counter ions.

Potassium dichromate react with potassium hydroxide to produce
  potassium chromate and water. Potassium hydroxide - concentrated
  solution.(source)

$$\ce{Ba(OH)2(E) + K2CrO4 → BaCrO4 + 2KOH}$$

Barium hydroxide react with potassium chromate to produce barium
  chromate and potassium hydroxide.(source)

$$\ce{2BaCrO4 + 2HCl(B) → BaCr2O7 + BaCl2 + H2O}$$

Barium chromate react with hydrogen chloride to produce barium
  dichromate, barium chloride and water. Hydrogen chloride - diluted
  solution.(source).

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text:
Which reaction is more violent?
As far I know fluorine is the most electronegative element in the periodic table and caesium is the most electropositve element. I had a conception that the reaction between the both would be the most violent as they have extreme opposite natures and perhaps they would do it faster than other elements.
But recently, I came across a book where I read that the electron gain enthalpy of chlorine is a bit higher than fluorine. This means when a reaction would occur with chlorine it would liberate more energy than that of the reaction with fluorine and therefore more violent. But, chlorine is less electronegative than fluorine.
So which reaction would be more sudden or fast and will liberate huge amounts of energy?

Although fluorine is more electronegative than chlorine it has less electron gain enthalpy because of its small size.
The electrons in the outer shell of fluorine repel the incoming electron more effectively than the electrons in chlorine. More closer the electron comes to the nucleus more energy is released. In case of chlorine, the electron comes more close to the nucleus.
Thus, chlorine has more electron gain enthalpy.
Size matters in chemistry. It is more dominating factor than electronegativity in most cases.
You would not expect enormous difference in enthalpy of formation of $\ce{CsF}$ and enthalpy of formation of other alkali halides. In case of $\ce{CsF}$, The bond length of the resultant salt will be larger than other alkali halides due to large size of cesium atom and small size of fluorine atom. The bond between them would not be very strong. Consequently, the reaction will not be strongly exothermic.
Putting the theory part aside, here is the data :
$\Delta{H^{\circ}_f}$  -  $\ce{CsF}$ $\rightarrow$ $-550\frac{KJ}{mole}$,
$\Delta{H^{\circ}_f}$ - $\ce{NaF}$ $\rightarrow$ $-575\frac{KJ}{mole}$,
$Electron\space gain\space enthalpy\space - Chlorine$ $\rightarrow$ $-349 \frac{KJ}{mole}$
The comparisons in your question are very different and they can only be compared by looking at the data. The prediction will be very inaccurate. If you want to compare two properties, you will at least need some similarity between the two properties. Like, you can compare enthalpy of formation of two different alkali halides or electron gain enthalpies of Halogens or chalcogens. Electron gain enthalpy and enthalpy of formation of alkali halides are very different things.

"When caesium reacts with fluorine, you would not expect enormous release of energy." - I beg to differ

@NicolauSakerNeto, I edited my answer.

@BhavyaSharma, thanks for the answer, all I can now infer from the data is that the reaction between Cs and Cl is the most violent reaction that can probably occur in the whole periodic table?

Not at all, the reaction between aluminium and chlorine is more violent than CsCl. Check this out : https://en.m.wikipedia.org/wiki/Standard_enthalpy_of_formation?wprov=sfla1


text:
Are there non-brittle silicon-based alloys?
Requirements:

Silicon part is greater than 50%
Has plasticity comparable to iron.

Could you elaborate a little more? Mentioning why you want such an alloy  may help a potential answerer. Also, even though metallurgy isn't my area of expertise...if an alloy with a 50%+ silicon content isn't brittle, then I don't know what it is ;)

Yes, please elaborate on the use case. There are many silicides with >50 at.% Si, and I'm not going to sift through them all looking for mechanical property info...

The only alloys I can think of where silicon would be present in that high a quantity would be ferrosilicon, silumin (where you very rarely get 50% Si), and things like silicon-germanium which are usually used as semiconductors for very small scale applications. Otherwise, you're looking at ceramics like silicon carbide.
All of those materials tend to be very brittle on a macro scale: on a nano scale, some ceramics show ductility. I'd be very surprised if you can find anything that matches your criteria though.


text:
What theoretically is the smallest cyclic trans hydrocarbon to correspond to a local potential energy minimum?
Trans-cyclooctene is isolatable at room temperature.
Trans-cycloheptene is also known at low temperature.
Ab Initio Conformational Analysis of trans-Cyclohexene (1995) says that there is spectroscopic evidence of the existence of trans-Cyclohexene, but that trans-cyclopentene hasn't been studied.
So I'm looking for any further information on trans-cyclopentene, trans-cyclobutene or trans-cyclopropene, particularly whether any of these have been theoretically calculated as (or ruled out as) corresponding to a local potential energy minimum.


text:
Confusion regarding sigma bond and pi bond
I was recently told that a sigma bond between two $\mathrm{3p}$ orbitals is possible but a $\pi$ bond between the same orbitals is not possible.
Although I was not told the reason, I am speculating that a $\pi$ bond between them is not possible due to a comparatively larger size in comparison to $\mathrm{2p}$ orbital.
Is it the same reason why $\ce{S2}$ does not exist?
Can someone confirm the theory and give reason for the same ?

Pi bond is not impossible, just very weak.

What about the reason for no S2?

How can I predict whether a bond is strong or weak ?

Can you please [edit] your question to include a more descriptive title?

I wonder what kind of logic produces this size argument over and over again. Are you saying that two 2p orbitals form a $\pi$ bond just fine, but two 3p orbitals can't do the same, because one of them is too large for the other, even though the other is exactly as large?

@Anmol there is diatonic sulfur, https://en.m.wikipedia.org/wiki/Disulfur.

Meant d i a t o m i c sulfur, device wrongly autocorrects my words.

@Ivan Neretin ,  I have been taught that the electron probability in 3p decreases as compared to 2p.Just wanted to confirm.

Your statement is unclear. Total (that is, integrated all over $\mathbb R^3$) electron probability on a 3p orbital is 1, as it is for 2p or any other orbital. Probability values at particular coordinates are higher at some places and lower elsewhere.

$\ce{S2}$ does exist above 1000K.

The most common explanation for the poor multiple bonding in 3rd row main group compounds is poor orbital overlap due to the radial node of the 3p orbital.
Since the 2p orbitals have no radial nodes, the lobes can overlap in-phase to form a reasonably favorable bonding interaction. The radial node of 3p orbitals makes it so that when the outermost (most overlapping) parts of 3p orbitals are overlapping in-phase (stabilizing), the part that penetrates into the inner parts are by definition out of phase (destabilizing). Hence the overall interaction is much, much weaker. Bonding through 3p orbitals are much more favorable when they are more end-on than when they are side-on like in multiple bonds. Hence, $\ce{P4}$ is much more stable than $\ce{P2}$ and $\ce{S_2}$ is less stable than larger sulfur cycles.


text:
How to add a color to vapors?
I want to add a color to vapors, like the vapor that comes out of glycerine when heated.
Is there a chemical compound I can put into glycerine that makes the vapor colorful? Or is there a chemical compound over which I can pass the vapors and they get colored?

Just remember that all that colored fog will "dye" whatever it comes in contact with...

By means of physics, just vaporise the glycerine and shine light on the vapours -- like it is done for water curtains.  Similar results may be achieved with fog machines (here).  The advantage: your light source defines when and what kind of colour is visible behind a dark background / in a darker environment.  To quote from a lecture by French chemist Hervé This: "Egg white appears only white because of the white light normally used in the kitchen, as soon we use red light it were egg red."

(picture source)
If, on the other hand, you want to use chemical means, mix glycerine, water and appropriate food colours, and evaporate it in a fume machine. A how-to like this may be a good starter.


text:
Why is it necessary to heat an esterification reaction?
Why do we heat the mixture acid and alcohol?
The reaction of esterification is exothermic so it is not affected by increasing temperature, then what is the use of heat ?

reaction of esterification is exothermic, so it is not affected by increasing temperature This is most certainly not true. Increasing the temperature increases the rate for a start and also drives off water which pushes the equilibrium towards the products.

Basically the use of heat is to liberate the water molecule from the mixture of acid and alcohol that is $\ce{OH-}$ ion from alcohol and $\ce{H+}$ from acid that results in formation of ester.Moreover it helps in evaporation of water since esterification reaction forms an equilibrium.The removal of water from mixture forces the reaction to move forward.It helps us to get a large amount of product.Alternative to heating some dehydrating agents like conc. $\ce{H2SO4}$ can also be added.

We do not get hydrogen ions from the acid and hydroxide from the alcohol.  The mechanism is more complicated, http://www.chemguide.co.uk/physical/catalysis/esterify.html.

$\ce{OH-}$ is a poor leaving group. You don't get hydroxide ions. A better alternative would be to convert the bad leaving group $\ce{OH-}$ into a good leaving group $\ce{H3O+}$. This isn't the true mechanism, however. As far as I know, the $\ce{H+}$ attacks the carbonyl oxygen to produce a geminal alcohol.


text:
Finding sulfate contamination in crystallized silver nitrate
What is the best way to check for a small amount of sulfate contamination (if any) in silver nitrate? Usually, I would use dilute barium chloride to precipitate out the resulting insoluble white barium sulfate in an unknown aqueous solution that contained sulfates. I tried this and got a white precipitate, but now I believe the precipitate I'm getting is insoluble silver chloride.
Is there a better/different method to check for sulfates in silver nitrate? I was researching for a better solvent to use, but haven't been able to find any conclusive answers so far.

When you suggest "sulfate contamination" how much sulfate would you expect in wt%? // Barium nitrate wouldn't ppt AgCl, but would ppt BaSO4 (assuming a reasonable amount of sulfate).

In this case a very small amount of contamination.

Your choices are restrained as the precipitation of $\ce{SO4^{2-}}$ in $\ce{BaSO4}$ is the classical way to quantify the former and an electrochemical determination (in aqueous solution) is not practical.

Electing $\ce{Ba(OH)2}$ may lead to the formation of silver hydroxyde, equally poorly soluble in water.
$\ce{Ba(PO3)2}$ itself is very poorly soluble, as were the $\ce{Ag3PO4}$, too.
As correctly stated by you, an aqueous solution of $\ce{BaCl2}$ is not suitable, as $\ce{Ag+}$ will form the precipitate of $\ce{AgCl}$.
Hence I suggest to give $\ce{Ba(NO3)2}$ a try.  The anion is the same as in $\ce{AgNO3}$. The solubility of this salt in water is reported to equal $\pu{4.95 g / 100 mL }$ ($\pu{0 ^\circ{}C}$); or even $\pu{10.5 g/ 100 mL}$ ($\pu{25 ^\circ{}C}$), respectively, according to the English entry in wikipedia.

Addition:
Please note, the solubility of $\ce{Ag2SO4}$ ($\pu{0.83 g / 100 mL}$ water) at $\pu{25 ^\circ{}C}$ (ref.) is substantially lower than the one of $\ce{AgNO3}$ ($\pu{256 g / 100 mL}$ water) at $\pu{25 ^\circ{}C}$ (ref).

Barium nitrate has a water solubility of $\pu{10.5g/100mL}$ at $\pu{25^oC}$. It isn't specified in the question what concentration of sulfate you suspect might be present, but given that you are trying to check for sulfate by precipitating with barium, barium nitrate should be the way to go.

Interesting, I had thought about using barium nitrate instead of the chloride, but did not have the reagent on hand and was hesitant to prepare any because I was unsure if that would work. I'll try that out and see what happens. In this particular case I suspect a very small amount of sulfate contamination.

If you have enough of your unknown solution and a good microbalance, you can reasonably quantify down to 10's of mg/L sulfate by precipitation with barium. The limiting factor becomes the solubility of barium sulfate, which is ~2-3 mg/L at room temperature. Anyway, best of luck ;)


text:
How to correctly prepare a standard solution?
Having just spoken to my teacher about the creation of a standard solution, and she suggested one way of doing it, however a previous chemistry teacher had told us a different way of doing it. My question is, (of the below methods) which is the "correct" way of doing it, and what are the pros/cons of doing one over the other? If you're interested in the full procedure of making this/any other standard solution (with the names of compounds, it is page 24 here).
Method 1:
The solid was weighed accurately in a weighing bottle, and then transferred to the volumetric flask, and then the bottle is re-weighed to calculate the amount of solid transferred to the flask.
Method 2:
Rather than re-weighing the solid in the weighing bottle, the weighing bottle is rinsed with DI water, and then poured into the volumetric flask. (this is repeated a few times).
If I'm not including enough detail in the above methods please let me know and I'll try to expand. For reference the 2 methods are Part 1, step 3 of the above link.

It somewhat depends what you're weighing and how soluble it is.

If working with small volumes and masses,  I usually weigh the solid directly in to the volumetric flask. Otherwise, I would go with method 2. But I feel that this is one of those things that is very much a matter of opinion. If you think about it, the "leftover" solid on your weigh boat is probably going to be in the microgram range, and may or may not be significant. Really depends on the purpose of the standard.

The reason you are using a weighing glass is because it is a lot smaller and it has a much smaller mass than a volumetric flask. These glasses usually are more stable, less affected by temperature, and pressure changes, or drafts in the laboratory. The smaller surface also tends to not accumulate much dirt due to static, your hands, etc.
If you were using method one, then you could skip the part with the weighing glass altogether. This method suffers from getting a stable reading as due to the size it is more affected by the conditions in the laboratory.
Method two is certainly the better method, because the total error tends to be much smaller. When you are using such equipment, then your concentrations are typically smaller than 1 mol/L and you are working with small quantities of solid substances. Continuously rinsing out the weighing vessel will ensure to transfer almost all of your solute.
If you were to produce a saturated solution, then the mass of the solute would not matter in the first place.
If your compound does not readily dissolve then the attempted concentration is too high to be used for analytical chemistry and you should rather start from a saturated solution. All primary standards are easily dissolved. If you are not dissolving a primary standard, then your should check the concentration against one.
Therefore, method one can be used if accuracy is of a minor concern (leave out the weighing glass) because it has to be determined again with a primary standard. Method two should be used when preparing accurately concentrated solutions of primary standards.

This is rather interesting as when I approached my teacher she described method 2 as the "quick way" and we are also told to do method 1, not method 2, but thank you :)


text:
Between 3-chlorophenol and 3-fluorophenol, which one is more acidic?
Between 3-chlorophenol and 3-fluorophenol, which one is more acidic?
In this case, I know that both F and Cl are electron-donating groups. Furthermore, I know that F is more electron-donating than Cl. But I can't figure out whether that makes 3-chlorophenol more acidic, or 3-fluorophenol.

Experimentally, the $\mathrm pK_\mathrm a$ values for 3-chlorophenol and 3-fluorophenol appear to be (perhaps unexpectedly) equivalent.
The table below1 shows the measured $\mathrm p K_\mathrm a$ values of substituted phenols in dimethyl sulfoxide (DMSO):
$$\begin{array}{ccc} \hline
\text{substituent} & \sigma & \mathrm p K_\mathrm a \\ \hline
\ce{H} & 0 & 18.03 \pm 0.04, \color{silver}{16.4} \\
\ce{p-CH3} & -0.170 & 18.86 \pm 0.05 \\
\ce{m-CH3} & -0.069 & 18.23 \pm 0.01 \\
\ce{p-Cl} & 0.227 & 16.74 \pm 0.03 \\
\color{red}{\ce{m-F}} & \color{red}{0.337} & \color{red}{15.88 \pm 0.07} \\
\color{red}{\ce{m-Cl}} & \color{red}{0.373} & \color{red}{15.83 \pm 0.05} \\
\ce{m-CN} & 0.56 & 14.76 \pm 0.05 \\
\ce{m-NO2} & 0.710 & 14.39 \pm 0.12, \color{silver}{13.75} \\
\hline
\end{array} \\
\small\text{(Greyed out values are taken from ref 2.)}$$
In general, halides have slightly unusual effects on aromatic rings. Despite being highly electronegative (and deactivating), they actually donate into the pi system via their lone pair.
This donation means that the phenoxide generated upon deprotonation (i.e. the dissociation being measured by the $\mathrm pK_\mathrm a$ value) isn't as stabilised as it would be with a purely withdrawing group (such as the nitrile or the nitro groups).
References

Bordwell, F. G.; McCallum, R. J.; Olmstead, W. N. Acidities and hydrogen bonding of phenols in dimethyl sulfoxide. J. Org. Chem. 1984, 49 (8), 1424–1427. DOI: 10.1021/jo00182a020.
Chantooni, M. K., Jr.; Kolthoff, I. M. Comparison of substituent effects on dissociation and conjugation of phenols with those of carboxylic acids in acetonitrile, N,N-dimethylformamide, and dimethyl sulfoxide. J. Phys. Chem. 1976, 80 (12), 1306–1310. DOI: 10.1021/j100553a009.

Fluorine can donate its lone pairs to the conjugated ring. Chlorine cannot do that because of ineffective 3p-2p orbital overlap.
For more acidic compound, the electron density on the -OH group must be taken away from it. $-I$ effect (inductive effect) makes a compound more acidic.
Since fluorine and chlorine are both $-I$ groups, the acidic strength of 3-fluorophenol should be higher but due to its contribution to resonance ($+R$ effect) in the aromatic ring, 3-fluorophenol becomes less acidic.

Please, don't use mathjax like that, save it for actual mathematical formulae etc.


text:
Why the transition of diamond to graphite is so slow?
Why the transition of diamond to graphite is so slow? Do I have to worry that one day I'll wake up and see graphite instead of my diamond?

The transition is so slow because chemistry. If that doesn't convince you, think of the prominent diamonds which have been known for centuries; they didn't turn to graphite, not a little bit.

There's another way to look at it. Diamond has an adamantane structure, while graphite has a hexagonal sheet structure. Imagine trying to change the structural arrangement of $6.023×10^{23}$ atoms. Even nature can't do such a difficult task so fast!


text:
Removing sulfur in Barton-Kellogg reaction without triphenylphosphine
I'm trying to understand how the sulfur was removed in the following reaction, taken from a paper by Ben Feringa:1

After some research I found out that this is a Barton-Kellogg reaction. Based on this I went through the mechanism as follows:
I'm stuck on the final step of how to remove the sulfur. I looked over multiple papers and all of them used $\ce{PPh3}$ to open the episulfide but the reaction from the paper I'm considering does not have $\ce{PPh3}$. I looked at all the sources and supporting information the paper had linked but couldn't find anything to explain how the double bond forms. My only guess is that something happens when they purify it using $\ce{SiO2}$ flash column chromatography. Here is the excerpt from the SI (L2 is the alkene product):

Freshly prepared diazafluorene 1 ($\pu{350 mg}$, $\pu{1.80 mmol}$) and thioketone (±)-2 ($\pu{383 mg}$, $\pu{1.80 mmol}$) in THF ($\pu{10 mL}$) were heated at reflux for 14 h under a $\ce{N2}$ atmosphere. The solution was cooled to rt, concentrated and purified by FC ($\ce{SiO2; EtOAc}$) to afford (±)-L2 ($\pu{327 mg}$, $52\%$) as yellow solid [...]

I would like any suggestions on what people think may be happening.

Reference

Wezenberg, S. J.; Chen, K.-Y.; Feringa, B. L. Visible-Light-Driven Photoisomerization and Increased Rotation Speed of a Molecular Motor Acting as a Ligand in a Ruthenium(II) Complex. Angew. Chem. Int. Ed. 2015, 54 (39), 11457–11461. DOI: 10.1002/anie.201505781.

In almost every example I've seen, the episulfide has been isolated and then subjected to some kind of phosphine reagent to cause the elimination.  I'd be inclined to think, given the nature of the paper (not really discussing this reaction apart from the SI), that its simply an oversight in the preparation.  Are you looking to do this reaction (in which case there are other more detailed/explained preparations)? Or is it simply as mechanistic challenge?

There were other papers by Feringa that does the same reaction and none of them used PPH3 which makes me think that this is not simply an oversight. I'm not planing to do the reaction, I'm only trying to understand the mechanism.

@MathHelpNeeded The paper's site states both Mr Nobel Laureate Professor Feringa, as well as Dr Wezenberg as corresponding author.  Why not putting a foot forward, and writing an e-mail in tune of "I read your publication 2015ACIE11457, and by comparision with other's work on the Barton-Kellog reaction I noticed the lack of $\ce{PPh3}$ in your protocol. The comparision with earlier papers A, B, and C by your group suggests to me this is not a mere oversight.  May you assist me in the rationale how the episulfide forms the alkene?" And, for the benefit of ChemSE, to share the rationale here?

This is a very heavily substituted episulfide. This suggests the possibility of removing the sulfur just under reflux. I've never seen this before, but that's how I would rationalize the mechanism.

@Zhe do you have a possible mechanism you can show me.

@Buttonwood I sent an email; I'll post whatever answer if he replies.

Your curved arrows in the step after the equilibrium suggest a C-C bond is forming; did you intend this and you're skipping some rearrangement step?

Can you just extrude elemental sulfur?

After contacting the authors of the paper, one author, Prof. Wezenberg said that a conclusive mechanism is not yet available for this case but he believes that the unstable episulfide undergoes desulfurization by itself under reflux.


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Does copper react with HCl when current is present?
Copper does not react with hydrochloric acid, according to many sources.  However, I found an experiment posted in a video, which I will try to repeat, that shows a copper pipe dissolving into HCl and plating a nail when the copper pipe and the nail are connected to a 12-volt battery, becoming electrodes.  No other chemicals, according to the video, are present.  How does the electrical current cause the copper to react with the HCl when otherwise it would not?

Electric current is stronger than anything in chemistry. It would make pretty much anything react with anything else.

Related: Copper Doesn't React with Hydrochloric acid

A reaction between copper metal and hydrochloric acid would go something like this: $$\ce{Cu + 2HCl -> CuCl2 + H2}$$
...which we can write as: $$\ce{Cu + 2H+ -> Cu^2+ + H2}$$
If we break this up into half reactions, we get: $$\ce{Cu -> 2e- + Cu} ~~~E°=-0.34~V$$ $$\ce{2H+ + 2e- -> H2} ~~~E°=0.00 ~V$$
If we add the voltages together, we get the cell potential of the combined reaction is $-0.34+0.00=-0.34~V$. Since the cell potential is negative, the reaction is not spontaneous, so the reaction is not going to occur without any artificial interference ($\Delta G°=-nFE°_{cell}$).
However, if you connect a battery (galvanic cell) to this reaction, you can supply the necessary electrons for this reaction to occur (electrolytic cell). Since you are connecting a 12 V battery, there is more than enough voltage to cause this reaction to occur.

The copper oxidation half reaction needs to have the copper ion’s charge on the right hand side.

When I want to plate small parts with copper, I submerge lots of waste copper and the steel part into muriatic(HCl) acid.  The oxides on both materials dissolves quickly and overnight there deposits a layer of copper on the steel suitable for forming a passivation layer.  No electricity involved.

The short answer is because the electric circuit provides the energy needed to do those reactions. You should know that you could try the same experiment but instead of using HCl, try table salt dissolved in water. That should work (12 V is a whole lot of voltage), but I can't say I actually looked it up to verify it. If that doesn't work, add some vinegar (acid) to the salt solution. Be sure to use the right connections (the copper will only dissolve near one wire). Waiting 5 or 10 minutes should be long enough to see if its working or not.
Dr. Who said "Time makes liars of us all." and he was right. It is just not true that HCl is unreactive towards Cu. I found dozens of YouTube videos and testimonies which contradicted that claim. The real world is complicated. Most general statements of the form "A doesn't react with B" are likely to be true in some circumstances and false in others. "Water doesn't react with sugar" for instance. Well, that seems to be true; I can dissolve sugar in water and then evaporate the water and get back the sugar. Well, wait a minute: it turns out that sucrose reacts (very slowly with water at room temperature, it takes years) and converts to a mixture of glucose and fructose. Not only that, but if I heated the water to 2000° I know that dumping some sugar into it would decompose the sugar. So, we need to qualify statements of fact (either implicitly or explicitly). In the case of your question, we need to make sure that the copper and the acid are pure. We need to prevent contamination with any oxidants (such as O2 from the atmosphere) and we need to specify the conditions of temperature and pressure (and protect the system from other interferences, such as radiation, gravity, etc. etc. (admittedly, Earth's surface gravity isn't strong enough to have much effect on reactions at lab scale.). Most copper is coated with various oxides, so cleaning a penny with HCl (which makes it shiny) doesn't prove that the HCl is reacting with the metal, but demonstrates that the HCl is either reacting with the oxide, with the metal, or with the copper alloy. (Copper wire is quite pure, 99.99%, pennies not so much). So, let's say we condition the HCL solution to eliminate all oxygen and other impurities, abrade the surface of a (pure) copper wire, and place the wire into the solution and leave it there at room temperature, 25°C. Will the copper react? Well, let's say that it won't. We can assume this because supposedly the "standard potential" tables indicate it won't. (Be careful here!, what concentration of acid is used to determine "standard potential"??) 1M HCl is ~4% but you can easily buy 30% concentrations. (The devil is in the details.) So, assuming your "sources" are trustworthy, then they are referring to specific conditions, and not some "universal truth". If you doubt it, you could try (the dangerous experiment of) carefully heating the diluted HCl with a scraped-clean-of-oxide piece of Cu wire in it, and seeing if the copper starts to react. It will but it might not be an experiment you should do. HCl is volatile, and can vaporize. The problem with that is that it is also corrosive and toxic. If you breath it, you could die. If it condenses on various metal surfaces around where you're heating it, they might corrode (and lets not even think about what it will do to electronics!).======== OK. So, we've diluted and cleaned the system and if your sources are right, the copper just sits there. But if you connect the metal to an the right side of a 12V transistor battery, you can indeed drive the reaction. Why? Well the flow of current indicates that energy is being used, just like the flow of water. You can use some of that energy (from the battery) to do work. (Incidentally, you could also use a battery charger and do the same thing, even a dc charger or power source which you probably have dozens of (you know, those little black cubes you have to plug into your AC wiring and then plug into your cell phone to charge it, or into thousands of other things that require DC power)). Anyway, a battery by definition works by ionizing something (usually a metal) and guiding the electrons that flow from one electrode to the other out of the battery (if a circuit is made and closed). The opposite of M → M+ + e- which is what powers most batteries (where M is a metal like lead or lithium) is M+ + e- → M. In other words you can use the electrons to ionize some metal, M, if there's enough energy. We measure energy in a circuit by voltage, the higher the voltage, the 'stronger' the energy (voltage, or potential, is like the height of a tank of water - the higher it is, the more work it can do). A 12V battery has a LOT of potential (relative to what's needed to ionize metals) but generally doesn't have much current. That means it can ionize most (I think, all, actually) metals but not very much of them. Electrochemistry is a bit of a strange area of chemistry, because it deals with two different concepts of "potential": the electrical term, same meaning as the word as voltage, more or less and the chemical term, meaning the ability (or potential) to react. Most metals prefer to ionize to the M+ ion, the cation. (By "prefer" I don't mean the atoms are choosing what to do, they do what the Laws of Physics and Chemistry tell them to do, no exceptions) But metals losing electrons is called "oxidation", metal ions gaining electrons (and so going back to the neutral (uncharged) atom, is called reduction. In order for a + ion (a cation) in solution to gain an electron, it needs to take it from something. The easiest place to take it is from a wire which has a negative charge and is "full" of extra electrons (from the battery). It happens at the wire's surface, and this allows the metal to "plate out" on the wire because uncharged (neutral) metal atoms are generally not very soluble in water. The ions are soluble, they are changed to the neutral metal at the electrode surface and plate out (precipitate out) onto that surface. Oh, one last thing. 12 Volts is more than enough to break apart water. In gentle conditions, (not working with explosive mixtures, using small batteries, etc.) the reaction which will occur is the one which uses the least energy. This means that even if you have enough energy to, say, ionize gold or platinum or silver, that they won't dissolve in a dilute acid if the reaction 2H2O → 2H2 + O2 uses less energy. (You can see this by watching bubbles form at the wires. If bubbles form at one wire, you know that something is going on at the other. The water reaction, above, makes hydrogen at one wire (electrode) and oxygen at the other. One explodes, the other we breathe. (Collecting the H2 is mostly safe unless sparks, flames, or heat (or platinum) is present.)

Wallet of text. Consider making paragraphs for readability.


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How to solve the Schrödinger equation of 1D hindered rotor?
I have the following equation:
$$-\frac{{{\hbar }^{2}}}{2I{}_{r}}\frac{{{d}^{2}}\psi }{d{{\theta }^{2}}}+V\left( \theta  \right)\psi =E\psi, $$
with
$$V\left( \theta  \right)=a+\sum\limits_{n=1}^{N}{{{b}_{n}}\cos \left( n\theta  \right)}+\sum\limits_{m=1}^{M}{{{c}_{m}}\sin \left( m\theta  \right)}$$
where the notations are straightforward ($\theta \in \left[ 0,2\pi  \right)$). I would like to get $K$ eigenvalues of the spectrum. How to integrate it (with some derivations)? Is there any non-commercial program which is able to provide them? I used MSTOR by Truhlar's group, but the potential does not contain the sine-terms.

If I remember correctly, Gordy and Cook (Microwave Molecular Spectra) discuss the solution in the case where the potential consists of one sine term. They expand the solution as a Fourier series with appropriate symmetry properties which then results in the evaluation of an infinite determinant. Truncating this determinant and diagonalizing it results in the eigenvalues. I believe this should also be applicable to this case.  There is also a review by Lin and Swalen (Rev. Mod. Phys. 31, 84 (1959)) where they discuss the solutions.

I found this article which shows how to solve your problem. Solution of the Schrödinger Equation for One-Dimensional Anharmonic Potentials: An Undergraduate Computational Experiment. J. Chem. Educ., 2011, 88 (7), pp 929–931 DOI: 10.1021/ed1000137. The abstract says  'A method of solving the Schrödinger equation using a basis set expansion is described and used to calculate energy levels and wavefunctions of the hindered rotation of ethane and the ring puckering of cyclopentene.'


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How do water molecules overcome the van der waals distance and form hydrogen bonds?
I read in Voet & Voet that the hydrogen bonds are at least 0.5 angstrom shorter than the calculated van der waals distance. As I understand, van der waals distance is the distance of closest approach between two atoms/molecules. How are two water molecules able to break this to get closer to each other and form hydrogen bonds?
We also know that ice is less dense than liquid water. In ice, water molecules are hydrogen bonded to each other and the distance between them is shorter than van der waals distance. but it liquid water not all of them is hydrogen bonded and it should be ideally at a minimum distance of van der waals and the density should be less. What is making the water molecules to interact closer to make liquid water more dense than ice?


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Finding a substance which has a continuous gradient in refractive index
I need to find a substance whose refractive index is constantly changing and directly proportional to the square root of the depth. I know I'm looking for a lot, but I'd really appreciate any ideas or advice from anyone. Thanks a lot

Square root or whatever, there is no such compound. In effect, you are asking for a metamaterial with peculiar properties; whether or not it can be made is a separate question.

There are commercially available optical fibres with a  gradient in refractive index, presumably lenses also, often called GRIN lenses.

It's not entirely true that no such substance exists, though you won't find this property in a homogeneous substance.
It's well known that changing the concentration of solute in a solution can change the index of refraction; a concentration gradient would then provide a changing refractive index, though having this mimic a square root function would be difficult. Typically, the refractive index changes linearly with the concentration of solute, so a nonlinear function may be difficult to achieve.
A similar approach could be applied to solids, forming a solid solution, though again there would be some difficulty in finding two compounds with similar lattice parameters that would allow to a continuous gradient between compositions to product the refractive index gradient.
Your best bet may be with polymers, where changing proportions of the monomers used could change the refractive index; organic dopants can also be used to effect* this gradient in polymers (see, e.g., Polym. Adv. Technol. 2014, 25, 1099). Again, the difficult part would be matching a square-root function, but you could at least have a varying gradient.
*https://xkcd.com/326/


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Graph between PV & P, Boyle's Law and its interpretation
My text book says:

At higher constant temperature, the volume increase and value of product PV should increase due to increase of volume at same pressure, but PV remains constant at this new temperature & a straight  line is obtained parallel to pressure axis in the fig

I could not get the idea if, this is a Boyle's law then why pressure remains same instead of temperature?

What is your exact question?

That's the graph you get under isothermal conditions.
Recall the ideal gas equation of state:
$$PV = nRT$$
At isothermal conditions, temperature is constant (duh!), so as a result the $PV$ term is going to be constant as well. You may as well try changing the volume of your gas container, but pressure will just adjust itself to make sure $PV$ remains the same.

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


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Which of the following represents the steps needed to produce crystals from a solution?
From a solution of zinc sulfate, we have excess zinc in a solution and need to form pure crystals.
Possible methods:
A- Filter, concentrate the solution, evaporate to crystallization point.
B- Filter, evaporate to crystallization point, distill.
My thoughts:
In choice A, how can we concentrate the solution (without heating it) if there already is some excess zinc? And in choice B, the last step (distillation), do they mean that we will heat the mixture of crystals and un-crystallized solution so the solution will evaporate and we will be left with the crystals? Wouldn't that remove some of the water of crystallization of the crystals too?

You could concentrate without heating, depending on the solvent. Apply vacuum maybe? It doesn't really matter, you just need to decide which set of options gives you the pure product. In option B, you would have to consider what happens to any impurities. If you work that out, it should help with your decision.


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Can metal oxides react with water to form a hydroxide, then that with HCl to make a salt?
Lets say I have a metal oxide and I put it in water, and it reacts. As far as I know, only metal oxides from group 1 and 2 react in water, making hydroxides.
Now here is my question: If I add HCl to the hydroxide, can I be sure that it will make a salt and water? Or are my previous assumptions wrong?
An example that works would be Sodium Oxide which reacted with water and then with HCl produces NaCl and water. Does it work the same way with all metals from this group though?

Some salts are dangerously toxic, not every salt is goody-goody $\ce{NaCl}$ you know. If you still have your wits with you, Do Not Drink.

I'm not actually going to drink anything. Are my other assumptions correct though? Are there any specific salts from group 1 and 2 I should know about (in the sense that they are very toxic)? Also, why are people downvoting the question?

I can't say why the DV's for sure, but in part it probably had to do with the "would I die if I drank it" phrase, which I removed; otherwise the question would have been closed as personal medical. Also, you could improve the quality of the question by giving a specific example. That's my 2 cents anyway.

tl;dr: Yes, depending on extent of reaction
The group 1 and (group 2) hydroxides for example: $\ce{NaOH,  KOH, RbOH, CsOH, Ba(OH)2, Ca(OH)2}$ are strong bases, being essentially fully ionized in aqueous solution; (although some are weaker $\ce{LiOH}$, $\ce{Mg(OH)2}$) . These should react with $\ce{HCl}$ - a strong acid forming water and a neutralisation product a typical corresponding salt $\ce{MCl}$ or $\ce{MCl2}$ .
Being soluble in aqueous envronment implies the salt is also obtained in this phase, so the question will be on extration method to use, in order to get this salt product in solid form.
Why I said depending on extent of reaction, is beacuse some metal oxides e.g $\ce{MgO}$ hardly form the hydroxide in the first place, and also insolubility with render the second reaction effectively out of question.

If the chloride is soluble, but the hydroxide isn't, the reaction should still proceed, but more slowly (since it would only occur at the surface of the solid). The chloride thus formed will dissolve away, exposing more hydroxide. And some oxides can react with $\ce{HCl}$ (and other acids) directly.


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Covalent Compounds In Written Form
How do you determine which non-metal is written first in a covalent compound? This has been a confusion for me for quite some time.

Why is methane's molecular formula conventionally “CH₄”, while water is “H₂O” (among others)?

Why do we write NH3?

Why is arsenous acid denoted H3AsO3?

It's just a convention.

Typical convention for organic molecules is to list carbon first, than hydrogen, and then other elements in alphabetical order. The Merck Index, for example, does it this way.

The way my teacher taught me was C,H,O first, and then in order of decreasing electronegativity.


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pH change with pressure in a bicarbonate buffer system
Looking at small scale bioreactors with no added pressure (1atm) compared to a commercial bioreactor which is pressured to 5 psi, where both have the same concentration of sodium bicarbonate, if both reactors are at pH of 7.2, should the pCO2 be the same as well? I've been trying to figure this out using Henry's law as well as the Henderson-Hasselbach equation but feel that the added pressure means that the equilibria for solubility and also acid dissociation should be altered and make the previous question false. Thoughts?


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Why does AsF3 boil at 60.4 Celsius and AsF5 at -52.8?
I don't understand why the boiling point of $\ce{AsF3}$ is at 60.4 degrees Celsius and $\ce{AsF5}$ boils at -52.8 degrees Celsius. I understand that it has to do with the bondings, but I can't see what would cause such difference?

I guess AsF3 has a huge dipole moment, while AsF5 has none.

This has to do with the dipole moment of those compounds. As the difference in electronegativity between As (2.18) and F (3.98) is very large, those bonds are highly polarized.
Using the VSEPR theory, $\ce{AsF3}$ is an $\ce{AX3E}$ molecule. As such, it has a pyramidal structure. Therefore, there is a separation of charge between $\ce{F^\delta-}$ and $\ce{As^\delta+}$ in the molecule, leading to a dipole moment. This implies dipole-dipole interactions between those molecules, which greatly increases the boiling temperature.

However, $\ce{AsF5}$ is an $\ce{AX5}$ molecule and has a trigonal bipyramidal structure. The centers of charges coincide and the net dipole moment is null. There is no dipole-dipole interaction between the molecules and the boiling point is much lower.


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What does iron (III) get reduced to in a galvanic cell: solid iron or iron (II)?
Recently, I built a galvanic cell in the lab with iron (III) nitrate, magnesium sulfate, and carbon and magnesium electrodes.
My question is:

When the iron $3+$ ions are reduced, do they reduce to iron $2+$ ions or to solid iron?

I know the reduction potential for iron $3+$ to iron $2+$ is greater, indicating that it is more likely to occur.

Magnesium electrode? And what, non-aqueous solution?

$$\begin{align}
\ce {Fe^{2+} {(aq)} + 2e^{-} &<=> Fe {(s)}} &E^⦵= \pu{-0.44V}\\
\ce {Fe^{3+} {(aq)} + e^{-} &<=> Fe^{2+} {(aq)}} &E^⦵= \pu{+0.77V}\\
\ce {Mg^{2+} {(aq)} + 2e^{-} &<=> Mg {(s)}} &E^⦵= \pu{-2.37V}\\
\end{align}$$
Thus
$$
\begin{align}
\ce {Mg {(s)} + 2Fe^{3+} {(aq)} &<=> Mg^{2+} {(aq)} +2Fe^{2+} {(aq)}}&&E^⦵_{\text{cell}}= 0.77--2.37=\pu{+3.14V}\\
\ce {Mg {(s)} + Fe^{2+} {(aq)} &<=> Mg^{2+} {(aq)} +Fe {(s)}}&&E^⦵_{\text{cell}}= -0.44--2.37=\pu{+1.93V}\\
\end{align}$$
Both reactions are feasible, and I'd imagine the first would proceed until the concentration of $\ce {Fe^{3+} {(aq)}}$ is very low, and then the second would start at a noticeable rate.
This assumes, of course, the iron solid is able to deposit on the carbon cathode, and I don't see why not.
Hope this was helpful.
(Thank you Dr. J for pointing out my typo)

@Tamir-Did you mean to say that the first would proceed until the concentration of Fe3+ is very low, and then the second would start.  Or did you actually mean to say  Fe2+?

yes, my mistake, I'll edit it


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Can benzyne be trapped inside a polycyclic hydrocarbon?
I came across an article in my organic chemistry textbook, which told about how benzyne was trapped within a molecular container and stabilized for short periods of time before undergoing a Diels-Alder reaction with the molecular container.
Obviously the container molecule must have had double bonds, which caused the powerful dienophile benzyne to react with it. While thinking about fully saturated molecular containers, dodecahedrane came into my mind:

The volume enclosed inside a dodecahedrane molecule was found to be approximately $2.8 \times 10^7 \text{pm}^3$. If I approximate the dodecahedrane to be a sphere, I get the diameter of the dodecahedrane to be approximately $5970 \text{pm}$.
Benzyne was conveniently prepared by diazotizing o-anthranilate, followed by heating simultaneously removes both $\ce{N2}$ and $\ce{CO2}$, and creating the benzyne molecule:

I am unsure on exactly how to find how much transverse space an anthranilate ion would occupy when stuffed into a dodecahedrane, but benzene itself is $496 \text{pm}$ wide, so I can suspect anthranilate maybe a bit wider, but nontheless can easily fit inside dodecahedrane.
And another advantage is, since both $\ce{N2}$ and $\ce{CO2}$ are linear molecules, they can easily escape through the pentagonal faces, leaving the benzyne trapped inside.
Edit: I made a mistake in calculating the diameter of the dodecahedrane. I now get it as $377 \text{pm}$, which is slightly smaller than a benzene ring width, so this compound won't work.  Still if we used bigger molecule, it could fit in. Can this be done? Will this work? Are there any complications involved?

All this calculation, and you could find out it's too small just by counting bonds.

I don't understand what you're saying @Mithoron.

Dodecahedrane is way too small to contain anything.

I'm also not sure how you did your calculation for the diameter. Did you square root instead of cube root? $V = 4\pi r^3/3$

@orthocresol, whoops, that's my mistake. I used $πr^2$ instead of $\frac{4}{3}πr^3$. Time to learn the difference between circles and spheres.

So, this question can be deleted I guess.

Happens to the best of us.

@orthocresol, should I delete this question or keep it for future readers? (though I doubt anyone would think of a crazy idea like this)

Well, it's still better then most questions. Maybe you could change this to asking what molecule it would fit?

@Mithoron Ahh that's a nice idea, but with all these calculations done, we could simply just plug the values of diameter or width of other molecules and see if it's shorter than the diameter of dodecahedrane. Thank you anyway.

@IvanNeretin Actually it can contain He https://en.wikipedia.org/wiki/Dodecahedrane#Encapsulating_atoms

@Mithoron OK, I stand corrected.

@IvanNeretin I guess it's one of smallest things which could be trapped though, so almost anything would be too big ;)

@Mithoron Yeah, I'm of the same opinion.

@Mithoron Yup I just saw a YouTube video about it. Helium barely fits inside it. They called it "The world's smallest Helium Balloon".


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Compare concentrations using LC/MS
Suppose I run a reaction and then pass the results through LC/MS to find the conversion. One idea could be to compare the peaks of the reactants to the peak of the product. But:

MS peak will depend on the levels of ionization of each of the compounds
UV spectra depends on the molar extinction coefficient of the compounds

So it looks like I can't compare the peaks. Is this right and if so what are the possible ways of finding the conversion?

You;re completely right, its one of the 'issues' with trying to monitor reactions by LCMS. If its a reaction you plan on running a lot, its not too much effort to measure the UV absorption and then calibrate the LC results, but failing that, GCMS or NMR are both viable options

Even in GC one has to calculate something akin to absorption factors which I don’t know the name of right now.

You can indeed find the conversion (not the absolute concentrations), even without calibrating the system.
At time t = 0, take an exactly measured volume of reaction mixture (e.g. 1 mL) and dilute it in a volumetric flask with water and/or acetonitrile to an exact volume (e.g. 100 mL).
Inject an exact volume (e.g. 50 uL) of the diluted solution in the LCMS and note down the (UV-based) peak area of the limiting reagent. This is your ${A_0 = \epsilon \cdot C_0 \cdot k}$, where ${C_0}$ is the concentration of your limiting reagent at this time, and ${k}$ is a constant related to the sample size and dilutions.
At any time t, do exactly the same as above, and you get an area ${A_t = \epsilon \cdot C_t \cdot k}$. Under these conditions, ${k}$ is the same, and of course so is ${\epsilon}$.
As you know, the conversion is ${y = 1 - \frac {C_t}{C_0}  = 1 - \frac {A_t}{A_0} }$.

Could you describe what A and ϵ are? And if this is a well-known equation, could you please name it and/or give the references?

It's based on Beer-Lambert. ${A}$ is the the LCMS peak area from a UV detector, which is what I wrote, and is proportional to absorbance. ${\epsilon}$ is the molar extinction coefficient. Not sure why my answer got a-1 mark, considering that it answers the question (how to measure conversion by LCMS).

Thanks! I don't think we'll be able to do the measurements at t=0 in our case, but still this clears some things up for me.

In some special cases you need no t=0 measurement. E.g. if your reaction is 1st-order (or at least pseudo-1st order), plotting ${log(A_t)}$ vs t should give you a straight line, from which you can extrapolate ${log(A_0)}$. Ordinary reactions (zero order, 2nd order, etc) should all be amenable to such treatment. With different functions of course.


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selective hydrolysis problem
I am aware that esters can be selectively hydrolysed depending on the type of alkyl group present e.g. Me ester vs t-butyl ester, the latter is reactive to acidic conditions and not basic because of its inability to form tetrahedral intermediate. But what reagent could I use to carry out the following transformation?

I see it is the O-C bond that breaks, so it is acyl cleavage (Ac). Acidic conditions would destroy my Me ester as would basic, so how can I get around this? Could a big bulky base, say LDA, take it down a B(Ac)2 route (generate CO2?)

If you use methanol as a solvent and add a suitable base, you should end up with the correct product, since any MeOH attacking the protecting group you want to keep will simply re-form the methyl ester, whereas the acetate will cleave to give the unprotected alcohol.
You could potentially also achieve this using sodium methoxide/methanol to speed the reaction up, but standard potassium carbonate/methanol is usually just fine.

this makes sense as it would reversibly add/leave. how does cleavage take place mechanistically? Is it attack of methanol at the carbonyl then leaving group is the "rest of the ester"

Pretty much. Standard addition elimination mechanism

You could try refluxing in NaOMe/MeOH. The presence of methanol would preserve the methyl ester, while transesterification would produce methyl acetate as a by-product.


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Interhalogen with alkene

Due to electronegativity differences, the iodine will be electrophilic and the chloride nucleophilic. Correct me if I am wrong here.
My question is one of regioselectivity: will it proceed through an iodinium ion (like a bromonium ion which is observed when Br2 is added) where the iodine and Me groups are on opposite faces. Then, chloride attack at the most substituted carbon? Or would it be a more akin to a planar tertiary carbocation, so attack at either face almost equally probable? **chlorine should be "back" in first structure

If I recall correctly, iodoniums are much less likely to form than bromoniums or chloroniums.

ok, so we're looking at a significant build up of positive charge at the tertiary carbon. Would you expect a racemate then?

AFAIK iodoniums are common enough, it's the fluoroniums that are rare. Now, there's no chirality present in any of your reagents, so you shouldn't expect any enantioselectivity. You can't get an enantiomeric excess from nothing.

i am confusing myself with regioselectivity

Iodine monochloride is usually highly specific for anti addition to alkenes. While bromonium ions form more readily than do iodonium ions, the latter do form. Additionally, the large size of the iodine atom further reduces the chance of syn addition. To answer your question in particular, it is unlikely that a planar carbocation is formed, and the anti-addtion would be expected to be the main product.
The more typical issue with iodine monochloride is that while the chloride is typically added to the Markovnikov carbon, the regioselectivity is sometimes relatively mediocre. For example, the addition of ICl to propene gives 2-chloro-1-iodopropane and 1-chloro-2-iodopropane in a 69:31 ratio; the yields of the expected regioisomer are better when styrene is used, giving approximately 95% of the expected 1-chloro-2-iodo-1-phenylethane. (J. Chem. Soc., 1931, 2742) This is likely due to the greater stability of the benzylic stabilization of the cation.


text:
How can I filter dissolved PVA from water
I'm working with a 3D printer that uses PVA filament as a support material.  After a 3D printed part is made, it is put in water for some time to dissolve away the PVA supports.
I'm trying to speed up the process by using a heated magnetic stirring plate with a pump connected to a filter.  The filter is for removing the already dissolved PVA from the water solution so that the solution doesn't get saturated.
I plan on using a 2L container of water and heating it to about 60-80 degrees C.  There would be about 10-50 grams of PVA being dissolved at a time.
I have two main questions:

What kind of filter(s) could I use for PVA in a water solution?
Do I really need a filter?  I.e. would 50 g of PVA have issues dissolving in 2L of 80C water?

PVA?

I don't have a ready answer for you, but I don't think that a 2.5% w/w would pose much of a problem and the whole filtering process is probably completely unnecessary. This publication DOI 10.1007/0-306-46915-4_3 might give you more information, I don't have access to it.

I tend to agree with @Martin-マーチン, a 2.5% w/w solution isn't going to be a problem, even with a moderate amount of repeat uses.

Since you mentioned 3D printing, have you considered there's a separate stackexchange site for it?

You might be able to add a brine solution and crash out the PVA.

Rig up a column that you can fill with granular activated carbon (GAC). Throw batting on either end to prevent carbon fines from entering your dissolution chamber. Replace the carbon when you reckon it is no longer working. If you can afford the heat, throw a cooling coil in front of the filter.
Another method is indicated in this patent: US4078129. They add 0.25g sodium borate and 0.6g sodium sulfate per liter to precipitate between 0.03 and 3% PVA.
This may not be a good solution if you are prone to overdosing, as it would say, goop your soup, when you go to do the next batch.
A third method seems to be already popular at the large scale: ultrafiltration. There are many papers discussing its usage for waste water discharge applications. Use the smallest pore UF you can find (50kDa is what I've seen in the literature). The pressures required to operate it are pretty low. Be sure the temperature of the solution is <50 degrees C prior to entering the UF (almost all membrane applications fail >50 degrees C). Run it outside-in. Don't dead-head it. You will generate a small concentrate stream. Perhaps try method 2 on that to reduce waste generation.

The first thing that comes to my mind is Reverse Osmosis, since it can exclude large molecules dissolved in water. I don't think that a typical filter (eg: paper filters for buchner apparatus)  would work if the polymer is actually dissolved in water and it's not a precipitate.
An alternative (maybe cheaper) idea is the following: since your mixture is heated at 60-80°C, you could attach a side apparatus, separed by a valve from the main one, and gradually distill at 100°C the water from it, with supplemental heating (or even at 60-80°C, if you use a vacuum system). The solid residue will be your polymer. The clean, distilled water could be driven back to your mixture. The valve will then be opened again, and the procedure repeated.
This would allow you to perform the process without using any filter.

distillation could work but would not be practical as it would be as slow as the PVA dissolving and thus wouldn't provide a significant speed up.

Why all the way to RO? A UF membrane should reject a polymer of significant size (>50kDa). A carbon filter might even do the trick.

Back in the day we used carbon filters on process water that was used to wash circuit boards. We also used PVA solder masking dots. The PVA would accumulate on top of the carbon filters and increase the pressure drop until we had to change them. These were 12 Cu ft filters and you could see the carbon media compress over time as a white film (PVA) would build up on top of the filter bed. On a small scale I think that the carbon filters would be readily available & semi inexpensive to try.


text:
Is there anything like "Saturated Fullerene"?
This question arose from my previous question: Can benzyne be trapped inside a polycyclic hydrocarbon?. While dodecahedrane was too small to fit an anthranilate ion or even a benzyne ion, there are lots of other cage molecules, notably fullerenes. Unfortunately fullerenes themselves have partial double bond characters in them, making them liable to undergo a Diels-Alder reaction with benzyne.
However, if it were possible to "saturate" the fullerene, it could serve as a molecular container to hold the benzyne molecule.
Is it possible to saturated fullerenes? After being saturated,  can we use it as a molecular container?

That's really two questions: can fullerenes act as a cage-like container; and can they be saturated. They don't necessarily depend on each other.

Yes, benzyne can be trapped. See: R. Warmuth
"o-Benzyne: Strained Alkyne or Cumulene? NMR Characterization in a Molecular Container"
Angew. Chem. Int. Ed. 1997, 36, 1347-1350
http://onlinelibrary.wiley.com/doi/10.1002/anie.199713471/abstract

https://en.wikipedia.org/wiki/Buckminsterfullerene#Hydrogenation

I'm limiting my answer to $\ce{C60}$-fullerene, but the same idea applies to the higher fullerenes as well.
In short, it's not possible to fully hydrogenate $\ce{C60}$ to $\ce{C60H60}$. This is a result of the bond strain that would result upon hydrogenation, as the addition of hydrogen across an alkene in $\ce{C60}$ produces carbon atoms that would optimally be tetrahedral with angles near $109.5^\circ$, while the bond angles in $\ce{C60}$ are $108^\circ$ and $120^\circ$ for the pentagonal and hexagonal rings, respectively.
Partial hydrogenation of $\ce{C60}$ with zinc and hydrochloric acid has been done to form $\ce{C60H36}$,1 but this compound degrades quickly in the presence of oxygen to form hydroxylated groups; degradation of the fullerene cage and formation of ketones also occurs. Efforts to hydrogenate further under high pressure hydrogenation conditions results in complete breakdown of the fullerene cage, with pretty interesting results, including ejection of carbon atoms and $\ce{C2}$ loss, producing hydrogenated species with fewer than 60 carbon atoms.2 Collapse of the fullerene cage was also noted, forming hydrogenated compounds with 25 - 45 carbon atoms.

Cataldo, Franco. Fullerenes, Nanotubes and Carbon Nanostructures 2003, 11 (4), 295-316. DOI: 10.1081/FST-120025852
Talyzin, Alexandr V., Yury O. Tsybin, Jeremiah M. Purcell, Tanner M. Schaub, Yury M. Shulga, Dag Noréus, Toyoto Sato, Andrzej Dzwilewski, Bertil Sundqvist, and Alan G. Marshall. J. Phys. Chem. A 2006, 110 (27), 8528–8534. DOI: 10.1021/jp0557971

Wait a minute. A hexagon in a flat surface would have a 120° angle, but a hexagon in a curved surface should have an angle less than 120°, no?

@PrittBalagopal The hexagons and pentagons themselves remain flat and regular; it's the junctions between them that gives curvature to the fullerenes. Note that at a vertex in a repeating hexagonal tessalation on a flat surface, the angle sum at the vertices is 360º. However, in a truncated icosahedron (the shape of $\ce{C60}$), two regular hexagonal faces and a regular pentagonal face form each vertex, giving an angle sum of 348º; this leads to the intrinsic curvature.


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Strength of Dipoles, Covalent bonds and H-Bonds
My textbook says:

Dipole-dipole forces are 1% as effective as covalent bonds and the strength of a hydrogen bond is generally twenty times less than that of a covalent bond.

What do these statements mean?

It means that of the three forces, dipole-dipole forces are the weakest, hydrogen bonding is the second weakest, and covalent bonds are the strongest. Now, dipole-dipole attraction happens between an ion and a polar molecule. Due to a net dipole movement, this is weak. In comparison to a covalent bond, it has only $\frac{1}{100}$ the efficacy. A hydrogen bond is a force between H and either N, O, or F, the three most electronegative elements. These are classified as d-d forces, but due to the differing electronegativities, they are a tad stronger. Covalent bonds are much stronger than either, $20$ times more than the H-bond and $2000$ times stronger than the d-d force. A covalent bond involves the sharing of electron pairs between atoms, generally two non-metals. There are also ionic bonds and metallic bonds, an ionic bond involving the transfer of electrons, such as $\ce{NaCl}$ and a metallic bond being between metals.


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What is the point of using molality when we have molarity?
Molality and molarity are both concentration terms. Given molarity's popularity molality seems rather antiquated.  Why is molality still discused in modern chemistry classes?

This is by no means an answer (you'll just have to wait for one), but the way I see it: "Molarity" is convenient, "Molality" is invariant ;)

IMHO I think molality should be more fundamental than molarity. Molarity is used simply for convenience, eg: pH is defined considering molarity, but a lot of actual chemical properties depend on molality, like boiling point rise and lowering of freezing point.

@PrittBalagopal The proper way of defining pH is through molality. What do you mean by more fundamental? Such notions tend to be rather subjective (in my opinion).

The answer is that molarity requires you to measure the final volume of the solution and varies with temperature. While molarity is great for labs as you can easily determine how much of a substance you are adding, it is limited to small batches.  In industry it is not practical to measure the final volume of 5000 gallons precisely but mass is much more easily measured. So rather than use molarity, you can use molality to produce repeatable results in a production facility.  In essence is the chemical engineers' fault.

One of the key attractions of molality is that changing the temperature of a solution does not change the molality, while it may change the molarity. This is because the volume of the solution changes as a result of expansion or contraction of the solvent upon changing the temperature, and thus the molarity changes, since $$M=\frac{n}{V}.$$
The downside, of course, is that the amount of solute required to reach a given molality depends on the nature of the solvent; ambiguity may also arise when a mixed solvent system (e.g. a water-ethanol mixture) is used, as the choice of which compound is the solute changes the molality of the resultant solution.1 Molarity avoids this problem by requiring only the volume of solvent.
As for why it's introduced in school, that's a pedagogical question beyond the scope of this answer, but both molarity and molality have their uses in the laboratory.

1 Consider, for example, a solvent made from 95% water and 5% ethanol. Molality could be determined with respect to only the water, or the combined water-ethanol solution, leading to different calculated concentrations.

In addition to JSK's excellent answer, I'd like to point out that there's a common pitfall related to molarity (which JSKs answer might have slipped on): molarity is defined as amount of substance of solute per liter of solution, while molality is amount of substance of solute per kilogram of solvent.
This might not seem like a huge difference, but if you have a highly concentrated solution, you can sometimes get pretty big discrepancies between the volume of solvent and volume of solution. Look at this question for an example of where the difference comes into play: the mistake would likely not have happened if molality was used as a measure, since doubling the volume of solute in a fixed volume of solvent doubles the molality, while it might do weird things to the molarity.

Thanks for point this out; it is certainly worth mentioning. I also thought about including some of the other potential pitfalls when making solutions in my original answer, such as the potential for volume contraction, e.g. when making methanol/water solution (see J. Chem. Ed. 1929, 6(6), p 1144),. This, of course, further complicates the situation, but I omitted it as it was only peripherally related to the question at hand.

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


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Relation between chemical kinetics and chemical equilibrium
In my chemistry book, the law of chemical equilibrium is derived from the law of mass action:

For a reversible chemical reaction $$\ce{aA +bB\rightleftharpoons cC + dD}$$ where $a$, $b$, $c$ and $d$ are stoichiometric coefficients in the balanced equation, we have, from the law of mass action, the rate of the forward reaction $$r_\mathrm{f} = k_1 \ce{[A]^a[B]^b}\label{1}\tag{1}$$ where $[\ce{A}]$ represents the concentration of $\ce{A}$ and so on. The rate of the backward reaction is $$r_\mathrm{b} = k_2\ce{[C]^c[D]^d} \label{2}\tag{2}$$
At equilibrium, $r_\mathrm{f} = r_\mathrm{b}$, so, we have $$\frac{k_1}{k_2} = \frac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$ Replacing $\displaystyle \frac{k_1}{k_2}$ by $K_\mathrm{c}$, we have
$$K_\mathrm{c} = \ce{\frac{[C]^c[D]^d}{[A]^a[B]^b}}\tag{3}\label{3}$$
which is the law of chemical equlibrium. $K_\mathrm{c}$ is supposed to be constant, for any concentration of the reactants, at a particular temperature.

However, in chemical kinetics, they tell us that the rate equation, derived from the law of mass action is not always correct. That is, the exponents of the concentration terms in the rate equations may not be equal to their stoichiometric coefficients. If this is true then, the above two rate equations ($\eqref{1}$ and $\eqref{2}$) may not be correct for a particular reversible reaction.
Then, how can we apply the law of chemical equilibrium on every reaction, without experimentally verifying the rate equations? If the rate equations derived from the law of mass action are wrong, then the value of $K_\mathrm{c}$ obtained by the final equation will not be constant.

You are essentially correct in your reasoning.

Just because the kinetics are not mass action does not mean that the thermodynamic equilibrium constant is not given by the usual relationship.

A cursory reading of https://en.wikipedia.org/wiki/Chemical_equilibrium suggests that the section cited by the asker is the historical derivation (which is also easier to understand). There is a derivation based on the Gibbs / free enthalpy.

The above deduction is only a simplified way which must be accompanied by the observation: "The equilibrium constant can only be obtained by the kinetic way if the chemical equations are elementary reactions."

The derivation that you cite only holds true if both forward and backward reactions are elementary steps. In this case, the rate of the forward reaction is indeed given by
$$r = k[\ce{A}]^a[\ce{B}]^b   \tag{1}$$
and so on, and hence the derivation is logically sound. If they are not elementary steps, then even though the conclusion is correct, the method of arriving there is not, for the reasons you have already pointed out. See also: hBy2Py's answer and Ben Norris's answer where the assumption of an elementary step is explicitly mentioned.

In general the equilibrium constant does not need to be derived in this manner; a more conventional method of arriving at it is by using (pressure here is taken to equal an arbitrary but specified standard pressure $p^\circ$)
$$\mu_i = \mu_i^\circ + RT\ln a_i \tag{2}$$
where $\mu_i$ is the chemical potential of species $i$, $\mu_i^\circ$ is the standard chemical potential of species $i$, and $a_i$ is the chemical activity of species $i$. We now have, for a reaction
$$\ce{a A + b B -> c C + d D},$$
the following equation:
$$\begin{align}
\Delta_\mathrm r G^\circ &= c\mu_\ce{C}^\circ + d\mu_\ce{D}^\circ - a\mu_\ce{A}^\circ - b\mu_\ce{B}^\circ \\
&= c(\mu_\ce{C} - RT\ln a_\ce{C}) + d(\mu_\ce{D} - RT\ln a_\ce{D}) - a(\mu_\ce{A} - RT\ln a_\ce{A}) - b(\mu_\ce{B} - RT\ln a_\ce{B}) \\
&= c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B} - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \\
&= \Delta_\mathrm r G - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]
\end{align}$$
However, at equilibrium, $\Delta_\mathrm r G = 0$ and hence
$$\Delta_\mathrm r G^\circ = - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \tag{3}$$
The term in square brackets can then be identified as the equilibrium constant, $K$. As you can see there is no need for any recourse to kinetic rate laws.
Why is $K$ a constant? Well, the standard chemical potentials $\mu_i^\circ$ are thermodynamic constants, just like standard enthalpies of formation, standard entropies, etc. So $\Delta_\mathrm r G^\circ$ - a sum of constants - is itself a constant, and it follows that $K = \exp(-\Delta_\mathrm r G^\circ/RT)$ must be a constant.

Note for high-schoolers like me: The chemical potential $\mu_i$ is nothing but the partial molar Gibbs energy of the species $i$.

Just a question, where does the equation of chemical potential come from? How is it derived?

@ShoubhikRajMaiti See this.

The big issue here is that there is confusion between the observed rate law, which is not the same thing as the actual rate law.
For example, there are pseudo zero order reactions, but it is impossible for a reaction to be truly zero order (what is reacting then?).
Consider the following conversion of $\ce{A}$ and $\ce{B}$ to $\ce{C}$, via a two step process involving intermediate $\ce{A'}$.
$$\ce{A + B \overset{k1}{->} A' + B \overset{k2}{->} C}$$
Suppose that the first step is slow ($k1 \ll k2$). This means that the observed rate law would be $k\ce{[A]}$ because as soon as $\ce{A'}$ is formed, the reaction to the product is fast, so the rate is essentially determined by the first step. But it doesn't make any sense that the rate law would be independent of $\ce{B}$. What if $\ce{[B]} = 0$?
If you write out the full rate law without making any assumptions, every species is involved.
Let's see if I can do this properly.
The rate of the reaction is $\ce{k2[A'][B]}$. I'll apply the steady state approximation to $\ce{A'}$ here: because $\ce{A'}$ is a fleeting intermediate that is quickly consumed, we assume that $$\ce{[A']} \approx 0 \Rightarrow \frac{d\ce{[A']}}{dt} \approx 0 $$
$$\frac{d\ce{[A']}}{dt} = -k_{-1}\ce{[A']} - k_{2}\ce{[A'][B]} + k_{1}\ce{[A]} = 0$$
$k_{-1}$ is the rate of the reverse reaction to form $\ce{A'}$.
$$\ce{[A']} = \frac{k_{1}}{k_{-1}+k_{2}\ce{[B]}}\ce{[A]}$$
$$\mathrm{rate} = \frac{k_{1}k_{2}}{k_{-1}+k_{2}\ce{[B]}}\ce{[A][B]}$$
There are two cases to consider.
First, in most scenarios because the second step is fast, $k_{2}\ce{[B]} > k_{1}$. This means that $k_{-1}+k_{2}\ce{[B]} \approx k_{2}\ce{[B]}$, and this $\ce{[B]}$ will cancel the other one in the numerator. It's really important to note that this is a pseudo first order reaction with two component, but the real rate law contains both species.
The other regime is  $k_{2}\ce{[B]} < k_{1}$. In other words, there is very little $\ce{B}$ present. Here, $k_{-1}+k_{2}\ce{[B]} \approx k_{-1}$, so the $\ce{[B]}$ in the rate law cannot be canceled out, and we see that the reaction is regained its true second order form, which was concealed by the specifics of the reaction before.

Just a comment: "This means that the observed rate law would be..." I had a slight problem understanding things after here. But when I wrote the two reactions separately: $\ce{A \overset{k1}{->} A'}$ and then $\ce{A' + B \overset{k2}{->} C}$ your post made much more sense.

"First, in most scenarios because the second step is fast, k2[B]>k1" I need help understanding this. The rate of second step is $r_2=k_2[B][A']$ but that of first step is $r_1=k_1[A]$. So, shouldn't always $r_1>r_2$ since you said that $[A']\approx0$?

@GaurangTandon This has to be true. Imagine a case where the forward reaction weren't faster than the reverse reaction. Then you can't argue that the first step is slower and that the intermediate is fleeting. Note that approximately zero is not the same thing as zero. You might want to read up on some physical organic chemistry for some of these topics.

@Zhe Is there any way to do this without the steady-state approximation?

@ApoorvPotnis Sure, but you might not be able to use this approach at all. I'd look at orthocresol's solution using chemical potentials instead.

Then, how can we apply the law of chemical equilibrium on every reaction, without experimentally verifying the rate equations?

You can derive the equilibrium constant from the rate constants even if the rate law is different than the stoichiometry of the net reaction suggests. It just gets a bit more complicated than in your example.
You do need to know all the elementary reactions. At equilibrium, rates for each elementary reaction will be equal to the rate of the reverse reaction, giving one equation about rates for each elementary reaction.
If you multiply all these equations, you get an expression that contains all the rate constants and lots of concentration terms which will all cancel out unless they are reactants or products of the net equation. After cancelling and sorting rate constants on one side and concentration terms on the other side, you will arrive at the equilibrium constant expression.
Here is a simple example with A reacting to form B with the help of a catalyst E (E for enzyme):
$\ce{A(aq) <=>[E] B(aq)}$ (net reaction)
First elementary reaction:
$\ce{A + E <=> I}$
Second elementary reaction:
$\ce{I <=> B + E}$
First set of equal rates:
$ k_1 [A] [E] = k_{-1} [I] $
Second set of equal rates:
$ k_2 [I] = k_{-2} [B] [E] $
Multiply:
$ k_1 [A] [E] k_2 [I] = k_{-1} [I] k_{-2} [B] [E] $
Cancel:
$ k_1 [A] k_2 = k_{-1} k_{-2} [B] $
Sort:
$ \frac{k_1  k_2}{k_{-1} k_{-2}} =  \frac{[B]}{[A]} $
Compare to the expected equilibrium expression
$ K = \frac{[B]}{[A]} $
For the specific example, this shows that while the enzyme influences the rates (and some steps depend on the enzyme concentration), it does not change the equilibrium (it speeds up both the forward and the reverse reaction). If you think of the common energy diagram, the enzyme (or in more general terms - the path of the reaction) influences the rates and activation energies, but not the position of the equilibrium, i.e. the Gibbs energy of reaction.

If the rate equations derived from the law of mass action are wrong, then the value of Kc obtained by the final equation will not be constant.

The equilibrium constant expression is correct even if the reaction has a mechanism more complex than a single elementary reaction. One more aspect is the unit of the equilibrium constant. Rate equations are based on concentrations (units mol/L) while equilibrium constants are based on activities (dimensionless). If the number of reactants is different from the number of products in the net equation, you have to consider this by dividing concentrations by the respective standard states when using rate constants to calculate K.

I’m stating this sentence from the book ‘Modern approach to chemical calculations’ by R C Mukerjee (p. 581 footnote)

“In chemical kinetics, reactions which occur in a single step are termed as simple or elementary reactions and those which take place in two or more steps are termed as complex reactions.
  For elementary reactions, the law of mass action and the rate law expressions are generally same.”

In other words, for elementary reactions, the exponent is basically the stoichiometric coefficient.
The reason why we don’t use stoichiometric coefficients as the exponents for complex reactions is as follows :
When we derive rate law of complex reactions, we combine the rate laws of the elementary reactions (actually, we combine the laws of mass action, but they are the same for an elementary reaction) by getting rid of many concentration terms (practically, we can only measure the concentration of reactants of the rate determining or slowest step, and the final product). So the stoichiometric coefficient of the reactants may or may not appear directly in the final product.
Complex reactions comprise of many elementary reactions and finding out the actual path of every reaction is not very easy. Even if we can find out every intermediate and every reaction involved, it is very tiresome to write the law of mass action for each reaction and get rid of the concentration terms we can’t measure.
So we use experimental data and define order of the reaction (with respect to the reactants) which appear as the exponents in the rate law.


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On the activity of liquids and solvents in equilibium constants
This is a confusion I've had for quite sometimes, first when learning about equilibria (due to contradicting teachers), then when trying to teach it (due to contradicting sources, in books and on the Internet).
When facing an equilibrium (let's be original and take $\ce{a A + b B <=> c C + d D}$ as an example), the equilibrium constant is
$$K = \frac{a(\ce{C})^c \cdot a(\ce{D})^d}{a(\ce{A})^a \cdot a(\ce{B})^b}$$
where $a(\ce{X})$ is the activity of X.
For pure solids, the activity is 1; therefore solids "do not appear" in the expression of the constant.
For dilute solutions, the activity of the solute can be taken as its molar concentration. Thus, supposing A, B, C and D are not solids, the equilibrium constant can be written as
$$ K_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$
But what happens when facing liquids (e. g. water)? In that case, I have seen two things :

Just putting the "concentration" of the liquid into the equilibrium constant.
A pure (or nearly pure, such as the solvant in a dilute solution) liquid has an activity of (or nearly of) 1. Therefore, the liquid should not appear in the constant.

Let's apply both cases to acidity constants, for the reaction $\ce{HA + H2O <=> A- + H3O+}$.

As $[\ce{H2O}] \approx 55.55$ mol/l in dilute solutions,
$$K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}][\ce{H2O}]}
\textrm{ and }
K_a = \frac{K_c}{[\ce{H2O}]} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
As $a(\ce{H2O}) \approx 1$ in dilute solutions,
$$K_a = K_c = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

Which way is correct? I'd be more inclined to follow the option #2, but I've been stared at wide-eyed like a heretic when talking about that.

Does this answer your question? Acid dissociation constant and equilibrium constant

I concluded on this question after discussion with the physical chemistry professor at my home university, and reading some physical chemistry books.
Although many references include the concentration of water in the expression of Ka, this is incorrect: the activity of pure water is 1. In dilute solutions, $a(\ce{H2O}) \approx 1$.
Therefore, the second option is correct:
$$K_a=K_c=\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$
This has a direct consequence on the values of pKa for the two couples of water: $pKa(\ce{H3O+}/\ce{H2O})=0$ and $pKa(\ce{H2O}/\ce{OH-})=14$, instead of the values $-1.7$ and $15.7$ that are often found in textbooks (which result from including $[\ce{H2O}]=55.55$ mol/l).
A good article on this matter can be found here: What is the pKa of water?.
References:

B. Leyh, Department of Chemistry, University of Liège, Belgium.
Atkins, P., & De Paula, J. (2006). Atkins’ physical chemistry (8th ed.). London, England: Oxford University Press.
Reid, P., & Engel, T. (2006). Physical Chemistry. Philadelphia, PA: Pearson Education.

You should go through the derivation of the equilibrium constant from thermodynamics and see exactly what is meant by activity, and the importance of defining a "standard state". The activity of water in a dilute solution is very nearly 1 because our standard state assumes pure water. We use the concentration of solutes because the standard state typically involves 1M solutions. Using the concentration of water is wrong unless you define your standard state with 1M water (in...something?).
If you want to be extra careful, you might write the activity of, for example, C in your first equilibrium as
$$\left( \frac{[C]}{[C]°} \right)^c$$
Where the little ° means "in the reference state".

I am well aware of the definition of the activity, and that's where my problem is: there seems to be no consensus on what the value of $[\ce{H2O}]°$ is. Shouldn't there be a standardized standard state for water?


text:
Potassium phosphate buffer + EDTA
Given 10 mM potassium phosphate buffer (pH = 7.0), 50 mM potassium propionate and 20 mM EDTA (acid). After addition of all 3 components pH becomes 4 since EDTA has 4 $\ce{OH}$ groups.
How to calculate the ionic strength if I am pHing it with $\ce{KOH}$ up to 7.1?
I can calculate ionic strength for binary mixtures, but 4 different components makes me stuck.

Are you sure you are using EDTA (the acid) and not any of its salts?
The water solubility of EDTA is quite limited and I think it would be difficult to prepare a 0.020 M solution. Please check, because it might affect your calculations.

yep, it was acid and it dissolved completely upon bringing pH to 7.1

Ok. I suggest you state the charge balance for all protolytes that are present in significant amounts at pH 7.1. Then you calculate the concentration of each protolyte by using the Henderson-Hasselbalch's equation. Finally, you multiply each protolyte concentration with the charge of the protolyte squared. Sum up all products and divide by two and you will get the ionic strength of the solution. Show me!