text:
Why are potassium organic salts ever preferable to the sodium equivalents?
$\ce{KOH}$, $\ce{KCN}$, $\ce{KNO3}$, $\ce{K2CO3}$, and plenty of other potassium salts are frequently used in industry.  But, as far as I can see, the sodium equivalents are also widely used, often interchangeably.  Since Na weighs only 60% of K it seems like the sodium salts should always be preferred because they carry more of whatever base is attached per unit weight (i.e., the sodium salts have "more of what you want"). $\ce{Na}$ is also slightly more abundant (on earth) than K.
So what other factors favor the industrial use of K-salts?  E.g., are K-salts just more abundant in natural sources?  Are they more reactive because the ionization energies of K are lower than those of Na?  Are the behaviors of these simpler salts actually different between the K- and Na- variants, and if so, is there some other elemental characteristic that suggests this?
Going the other way, it seems like if K is preferable to Na, Rb should be even more so (because it appears that in every elemental characteristic in which K exceeds Na, Rb exceeds K).  Would it be if it weren't so relatively scarce?

Sodium salts are used more often 'cause they're cheaper, but sometimes they aren't interchangeable.

@Mithoron: Maybe a strong example or two of where they aren't interchangeable  (but where one is still using the salt for the anion) would make a good answer.

Some cations are harder or softer than others; Na+ is harder than K+ and although the difference isn't huge, it's enough to make a difference sometimes.

For example, sodium t-butoxide is rather moisture-sensitive, whereas potassium t-butoxide can be handled under atmospheric conditions most of the time.

There's a couple of reasons with which I'm familiar:

Degree of hydration: Some sodium salts are readily available as their hydrates, but the anhydrous form is more expensive to produce due to how hygroscopic it is. On the other hand, the potassium salt of the same anion may not form hydrates; for example, sodium molybdate is commonly found as the dihydrate while potassium molybdate does not form hydrates. In these cases, the potassium salt may sometimes be preferred if a water-free procedure is required, or if the degree of hydration is high enough that the hydrated sodium salt would actually contain less of the attached anion(s) than the anhydrous potassium salt.

Working with sodium-sensitive organisms: For example, working with organisms that are used to growing in soil. Many soils have much higher potassium than sodium and therefore soil-based life is often well-adapted to potassium - many plants even require a lot of potassium as a major macronutrient. However, the same organisms will experience significant salt stress if placed into an environment with high sodium. Hence, potassium salts are often used in such setups; for example, phosphate buffer is often prepared with KH2PO4 and K2HPO4 rather than with the sodium equivalents to avoid the aforementioned salt stress (e.g. https://www.usbio.net/media/N1000/Nematode%20Growth%20Medium/data-sheet). Similarly, a lot of plant growth regulators are solubilized using potassium hydroxide instead of sodium hydroxide (https://phytotechlab.com/media/documents/ProductLiterature/PlantGrowthRegulators.pdf - notice the number of entries in the solvent column saying "KOH" but none saying "NaOH").

Double-displacement reactions where you want to precipitate the potassium salt of something else: As a very general rule of thumb, most potassium salts are less soluble than their sodium equivalents. If you were conducting a double-displacement reaction where you wanted to separate the two products, it might make sense to select the potassium salt as one of the reactants so the potassium byproduct can be precipitated.

Differing physical properties of the products: For example, saponification with sodium hydroxide typically results in a much firmer, solid soap bar while saponification with potassium hydroxide usually results in softer soaps or even liquid soap. I'd expect there to be other situations where use of the sodium compound results in a product with significantly different physical properties from the potassium compound.

As for rubidium - it varies. Rubidium could be used over potassium in 3 (precipitation of potassium side-product) if it were more common while it might also be used in 1 (degree of hydration) on a case-by-case basis. I'm a bit iffy on whether rubidium could replace potassium in 4 (physical properties) as it might be the case that the physical properties of the potassium-based product are exactly what is sought after (e.g. liquid soap) and use of rubidium would result in an unusable product. I doubt rubidium-based compounds would replace potassium in biological applications* (use 2); experiments where rubidium was used as a potassium substitute in lab animal diets resulted in the deaths of said animals (https://doi.org/10.1152/ajplegacy.1943.138.2.246).
*: Though, in an alternate timeline where rubidium was really common on Earth, any biological life that developed would probably have ways of using or detoxifying rubidium, the same way most terrestrial life has learned to utilise or detoxify elements like potassium, copper, or selenium when they're encountered at the levels which life is accustomed to. In that alternate timeline, rubidium might be used when working with organisms in the same way potassium is today, but it's unlikely rubidium would replace potassium in such applications e.g. if we came across a big chunk of rubidium in space and mined it.

The differences in the properties of the alkali cations are subtle, and the rationale for choosing one over the other will depend heavily on the application. I suspect you're right, that to some extent natural abundance of, e.g., potassium salts relative to sodium ones in certain mineral contexts explains the industrial preference for one or the other -- but, really, variations in abundance are just further examples of these subtle differences in the properties of the cations!
One specific example that I recently came across is the Hofmeister series (see also the Wikipedia article), where the choice of cation makes a demonstrable difference in the solubility properties of proteins.  Speaking also from personal experience, the choice of, e.g., $\ce{Na2SO4}$ versus $\ce{K2SO4}$ versus $\ce{(NH4)2SO4}$ as a component of the supporting electrolyte in electrodeposition of copper from acidic $\ce{CuSO4}$ matters significantly to the properties of the deposit obtained.
Addendum, 17 Apr 2023: Another example of a significant difference in properties is the aggressiveness of $\geq 1\mathrm{M}$ aqueous hydroxide solutions on aluminum and its alloys. In a $1\mathrm{M}\ \ce{KOH}$ solution, 6061 Al reacts slowly, releasing small traces of fine bubbles. In a $1\mathrm{M}\ \ce{NaOH}$ solution, the reaction is vigorous, with significant gassing and relatively rapid consumption of the metal.

Wow, that explanation of the Hofmeister series is illuminating!  I guess I had forgotten about other elemental characteristics that vary significantly within a group like ionic radius, polarizability, surface charge density....

Sodium and potassium salts often can be used interchangeably - it these cases the difference in price and accessibility can be important. For example one can easily buy cheap sodium chloride in shop, but not potassium chloride, which isn't as cheap or accessible as NaCl. Other differences can be however very important.
There are some sodium and potassium salts that strongly differ in water solubilities. For example antimonates - potassium hexahydroxoantimonate (V) is even used to detect sodium via precipitation. On the other hand solubility of potassium perchlorate is about 100 times lower than that of analogous sodium salt.
Difference between sodium and potassium salts is big for living organisms. For example potassium chloride is rather bitter than salty and it's very dangerous when injected - can easily cause heart attack.

One difference between $\ce){Na^+}$and$\ce{K^+}$ is the former, more compact cation is more likely to attract water molecules and form hygroscopic salts. This is a disadvantage in analytical applications, as Wikipedia reports in the case of sodium permanganate:

Sodium permanganate behaves similarly to potassium permanganate. It dissolves readily in water to give deep purple solutions, evaporation of which gives prismatic purple-black glistening crystals of the monohydrate $\ce{NaMnO4·H2O}$. The potassium salt does not form a hydrate. Because of its hygroscopic nature, it is less useful in analytical chemistry than its potassium counterpart.

Another difference, in the case of permanganates, is that the common method of producing potassium permanganate does not work with sodium as the cation:

Potassium permanganate is produced industrially from manganese dioxide, which also occurs as the mineral pyrolusite. In 2000, worldwide production was estimated at 30,000 tonnes.[1] The MnO2 is fused with potassium hydroxide and heated in air or with another source of oxygen, like potassium nitrate or potassium chlorate.[1] This process gives potassium manganate:
$\ce{2 MnO2 + 4 KOH + O2 -> 2 K2MnO4 + 2 H2O}$
(With sodium hydroxide, the end product is not sodium manganate but an Mn(V) compound, which is one reason why the potassium permanganate is more commonly used than sodium permanganate. Furthermore, the potassium salt crystallizes better.[1]).

So preparation of sodium permangnate requires more extensive and more indirect preparation giving, as noted above, a less desirable salt for analytical uses.
Cited Reference

Miessler GL, Fischer PJ, Tarr DA (2014). Inorganic Chemistry (5th ed.). Pearson. p. 430. ISBN 978-0321811059.


text:
Diethyl Ether reaction mechanism
I was trying to determine the reaction mechanism behind the production of diethyl ether. Looking at the ethyl alcohol + sulphuric acid method, I have come to the following conclusions for the reaction steps:
The ethanol nucleophilic oxygen gets protonated by the addition of sulphuric acid. This is more favourable because of the high density of electrons surrounding the oxygen. The bond broken between the carbon and oxygen is also needed to form the bond with hydrogen.
Water is then released and the ethyl alcohol conjugate base is formed (H3C-CH2+). The remaining ethyl alcohol molecules can either react with the HSO4- or the ethyl conjugate... Since HSO4- is more stable than the ethyl conjugate, ethyl alcohol reacts with the ethyl conjugate. Carbon is more electronegative than hydrogen and is an electrophile. It is attracted to the oxygen from the ethyl alcohol and replaces the H, thus forming diethylether, H+ and H2O where the sulphuric acid essentially acts as a catalyst!
Am I correct in presenting the mechanism this way? If the sulphuric acid still remains in the solution, why does it not protonate the diethyl ether? Is it because the stability of the diethyl ether conjugate base is unstable?
Thanks

Your mechanism would create a primary alkyl cation, namely the ethyl cation ($\ce{CH3CH2+}$). Primary carbocations are extremely unstable and only generated if there are stabilising effects present such as in the benzylic or allylic case — and these are then often not considered primary cations but rather allylic or benzylic ones.
Rather, the protonation of ethanol ($\ce{CH3-CH2-OH}$) creates an oxonium ion as you suggested in the first step:
$$\ce{CH3-CH2-OH + H2SO4 -> CH3-CH2-OH2+ + HSO4-}$$
The $\ce{CH2}$ carbon is then attacked from behind in a nucleophilic $\ce{S_{N}2}$ reaction liberating water in the same step. It is important to note that the attack of another ethanol molecule and the liberation of water happen synchronously; there is no first or second.
$$\ce{CH3-CH2-OH2+ + CH3-CH2-OH -> CH3-CH2-OH+-CH2-CH3 + H2O}$$
Finally, the oxygen which is now part of a protonated ether loses the proton to give the final product diethyl ether.

In the case of a nucleophilic substitution on a substrate like tert-butanol, or even better, tert-butyl bromide, one would consider the leaving group to be liberated first and a carbocation to be created; however, tertiary carbocations are orders of magnitude more stable than primary ones.

in case of  sn1 reaction will unsymmetric ether be formed?

@starunique2016 In the case of $\mathrm{S_N1}$ reactions you are well in the area where sterics play an enormous role and you may not get any ether formation at all. But if you do and you have two different alcohols then you will likely get a distribution symmetrical and unsymmetrical ethers. If you have a case like tert-butanol and isopropanol, the unsymmetrical ethers will probably be the main product by a significant margin.

You got it all wrong. First of all the reaction written by you concisely is
$$\ce{diethyl~ ether + H2SO4 -> CH3-CH2-OH + CH3-CH2-HSO4}$$
Your first step is correct that oxygen gets protonated due to acid. Then you said that  water is then released and the ethyl alcohol conjugate base is formed $\ce{H3C-CH2+}$. This is the point where you got it wrong. The reaction goes under SN2 mechanism( I am not going into the details of it) and there is no formation of intermediates as you gave it as $\ce{H3C-CH2+}$ So no carbocation inter mediates are formed only transition stage is formed.
Your next question was for what if $\ce{H2SO4}$ is in excess.
Here it goes like this

Ethanol formed is again protonated by $\ce{H2SO4}$ by the same reason you stated.
Hence now there are $\ce{CH3CH2OH2+}$( where +ve charge is on oxygen) and $\ce{HSO4-}$ anions.
These $\ce{HSO4-}$ again attack on $\ce{CH3CH2OH2+}$ by SN2 mechanism.
Now you will get salt of the 3rd step and water as by product.

I don't see how the HSO4- would attack the protonated ethyl alcohol. If that were true, you wouldn't have diethyl ether.

According to wiki, the unprotonated ethyl attacks the protonated ethyl. From the looks of it, even though the protonated ethyl alcohols are positive, the oxygen still forces the C to be the electrophile. Thus the SN2 attack is of unprotonated  ethyls on the carbon back with an H2O and H+ leaving group.

I am not getting what is your question. Please send me that link and I think that I will help you out

@NukulParmar, your "You got it all wrong" comment isn't exactly constructive to the original poster of the question. The reason people post their questions on stack exchange is because they want answers/clarification, not so someone can tell them how wrong they are... this entire community works because people feel able to do this.  If you take issue with the original question, you're free to down vote it, just as people are free to down vote on your 'answer'.

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text:
Justification for Freezing Point Depression & Boiling Point Elevation in Solutions?
I was wondering if the following justification for freezing point depression and boiling point elevation are conceptually correct. The reason why I ask this question is because I have been self studying chemistry for a course I will be taking this fall, and I don't have a human reference to check with. My book is also very confusing on this subject, and I don't have a world renowned memory (meaning I'd rather understand than memorize the equation), so I tried to think about it and came with the following conclusions:
-Freezing Point Depression: When a solvent is not pure and has particles dissolved in it, there are constituents (be it molecules, ions, etc.) that take up volume in the solution. For freezing to occur, these constituents need to aggregate to form a lattice, bonds, etc. Say we have a pure solvent X. This process occurs at a temperature Y. However, when this solvent is not pure and has dissolved solutes, there is "blockage" impeding these constituents from aggregating and bonding. Therefore, the temperature Y needs to be reduced further to slow down the kinetic energy of all constituents in solution (solvent and solute), which gives a greater statistical probability of intermolecular forces getting an opportunity to bind solvent-solvent constituents.
-Boiling Point Elevation: Boiling, if I'm not mistaken, occurs when the partial pressure of a vapor exceeds atmospheric pressure. For a gas to enter the vapor phase, molecules in solution need to have a certain kinetic energy (as specified by a Boltzmann distribution) to free themselves of the intermolecular forces in the solution phase and become a vapor. When a pure solvent is now diluted with other particles, the new solute-solvent bonding energy needs to be overcome to allow a solvent molecule to exist as a vapor. One way to overcome this energy is to increase the average kinetic energy of all molecules, and this is done through a temperature increase.
This is for an introductory chemistry course, and I'm not sure if my explanations are sound. I just want a yay or a nay in terms of my logic, because I don't have the most intimate background in chemistry and I'm trying to conceptually understand things.

I think you are looking at the problem from slightly the wrong angle. The central quantity when dealing with colligative properties is entropy and not solute-solvent or solvent-solvent molecule interactions. Of course the interactions are important in the sense that they affect the entropy but since we are dealing with thermodynamics here the way you think about it at the moment clouds the main effect at work here imo.
So, when trying to picture how colligative properties work on a microscopic level think about the effect of the solute on the entropy. Boltzmann's entropy formula tells you that the entropy $S$ is proportional to the logarithm of the number of microstates available to the system. The solute is present in the solution in small quantities and the molecules are really dissolved and don't form a different phase. By their presence the solute molecules disturb the local order of the solvent molecules and introduce new available microstates into the system. As an extremely simplified example you can think of it like this: If you want to place 10 identical blue balls (solvent molecules) on 10 fixed spots there is only 1 way (microstate) to do this, since all the balls are alike. But if you replace one of the blue balls with a red ball (solute molecule) you suddenly have 10 possible distiguishable ways to arrange your balls on the 10 spots. So, when comparing the initial situations of a solution of pure solvent and one of solvent plus solute, the second solution will have a higher entropy to start with, i.e. $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$.
Now, what happens when you want to freeze both solutions? In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{s}} \leq G_{\mathrm{l}}$ or put differently $\Delta_{\mathrm{m}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{l}} - G_{\mathrm{s}} \geq 0$. The Gibbs free energy is given by
\begin{equation}
 G = H - TS
\end{equation}
where $H$ is the enthalpy and $T$ is the temperature, so that
\begin{equation}
 \Delta_{\mathrm{m}} G = (\underbrace{H_{\mathrm{l}} - H_{\mathrm{s}}}_{\Delta_{\mathrm{m}} H}) - T (\underbrace{S_{\mathrm{l}} - S_{\mathrm{s}}}_{\Delta_{\mathrm{m}} S})
\end{equation}
Since there is only very little solute present in the second solution you can assume that $\Delta_{\mathrm{m}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{m}} H^{\mathrm{pure}}$. So, the difference between $\Delta_{\mathrm{m}} G^{\mathrm{impure}}$ and $\Delta_{\mathrm{m}} G^{\mathrm{pure}}$ will mainly be caused by the entropical term. Let's have a look at that: the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules, and thus $S_{\mathrm{s}}$ must be the same in both cases. Furthermore, you generally find that the entropy of a solid phase is lower than the entropy of a liquid phase, i.e. $S_{\mathrm{s}} < S_{\mathrm{l}}$. Above I established that $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$. It follows that $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Putting all this together gives you the following picture: For $\Delta_{\mathrm{m}} G$ to become equal to zero the entropical term $T \Delta_{\mathrm{m}} S$ has to balance $\Delta_{\mathrm{m}} H$, which is assumed to be about the same for the pure-solvent and the solvent-plus-solute case. Since $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$ the temperature at which freezing becomes possible must be lower in the solvent-plus-solute case, i.e. adding the solute leads to a freezing point depression.
An analogous argument can be made for the Boiling point elevation. The focal point is again that the solute raises the entropy of the liquid phase, thus affecting the entropical term in the Gibbs free energy equation for the phase change. For the boiling process to happen the Gibbs free energy of the gaseous phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{g}} \leq G_{\mathrm{l}}$ or $\Delta_{\mathrm{vap}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{g}} - G_{\mathrm{l}} \leq 0$.
Requiring the Gibbs free energy of vapourization
\begin{equation}
 \Delta_{\mathrm{vap}} G = (\underbrace{H_{\mathrm{g}} - H_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} H}) - T (\underbrace{S_{\mathrm{g}} - S_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} S})
\end{equation}
to be equal to zero gives
\begin{equation}
 \Delta_{\mathrm{vap}} H = T \Delta_{\mathrm{vap}} S
\end{equation}
Again, we can assume that $\Delta_{\mathrm{vap}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{vap}} H^{\mathrm{pure}}$. Since the entropy of a gaseous phase is generally higher than the entropy of a liquid phase, i.e. $S_{\mathrm{g}} > S_{\mathrm{l}}$, and since $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$, it follows that $\Delta_{\mathrm{vap}} S^{\mathrm{impure}} < \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Thus the boiling temperature must be higher in the solvent-plus-solute case, i.e. adding the solute leads to a boiling point elevation.

"the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules” Can you explain why you’ve made this assumption?

This is an excellent answer. I wish more people had seen and upvoted it. Dang it, I wonder why the O.P didn't accept it. @lightweaver: The solid phase only consists of the solvent molecules. Since we're looking at the situation from an entropic point of view, it is safe to say that in both cases, the entropy of the solid phase doesn't change due to this (correct) assumption that the solid phase is pure.

@Kaumudi.H But why is the solid phase pure? Where did the solute disappear to? Did it crystallize into its own phase?

Philipp, in almost every textbook that I look into, and online notes indicates that colligative property is only and only dependent on a solute's concentration in a solution. And my doubt has been, is it temperature dependent as well? because concentration of a solution is highly temperature dependent. Colligative properties on their own, thrive on "points" (freezing point - a temperature, boiling point - a temperature, vapour pressure - is temperature dependent). Is colligative property temperature dependent or not?

@bonCodigo Well, if you look at the derivation of the freezing point depression, e.g. here, you'll find that it is dependent on the freezing temperature of the pure solvent. In that respect it certainly is temperature dependent. Apart from that: most statements about colligative properties are made for ideal solutions and only for those they will hold. In real solutions you get interactions between the molecules and those lead to (sometimes sizeable, sometimes very small) deviations from the predicted behavior...

@bonCodigo ... Furthermore, in some equations you have certain approximations or assumptions, such as $\Delta_{\mathrm{m}} H$ being independent of temperature, in order to simplify integrals. All this can lead to finding a slight temperature dependence in practice, where theory doesn't predict it.

@lightweaver I'm a novice in chemistry, but I will try to answer your question. Since free energy, as a function of state, in a solution ( mix of substances ) is dependent on X ( molar fraction ) of the solvent and  the solute, probably he made the assumption that in the solid phase the contribution of the solute X is so much smaller ( in Entropy computation ) with respect to liquid phase ( due to less possible macrostates configurations ) that could be neglected. This is what I can assume from all the answer he gave above.

@Philipp Nice answer. I understand that it is reasonable to approximate $S$(impure solid) = $S$(liquid) But I cannot understand that why $S$(pure gas) = $S$(impure gas) . Why in gaseous phase also, you assumed both solutions to have equal entropy?

The depression of freezing point and elevation of boiling point are, as has been described by @phillip, due to entropic effects.
The basic idea can be seen in the change in free energy with temperature as shown below. The phase change occurs where two solid lines cross. However, rather that use free energy G, which is extensive it is more general to use the molar free energy $G_m$ which is intensive (just like T and p) and is usually given the symbol $\mu$ and called the chemical potential.
Just as in mechanics where a ball will spontaneously roll down hill to the lowest potential so in a chemical system the direction of spontaneous change is that which lowers the chemical potential.
Free energy is defined as $\mu = G_m = H_m-TS_m$  where the subscripts m indicate molar quantities. The most stable state of a substance is that with the lowest chemical potential.
At low temperatures $\mu$  is determined by the enthalpy $H_m$ because the entropy  changes only slowly on increasing the temperature. This is because a solid has strong intermolecular interactions and molecules remain in their relative places  as the temperature increases.
At high temperature $\mu$ is dominated by $TS_m$ and so the chemical potential of a gas decreases rapidly with temperature. In the liquid the situation is such that neither term dominates entirely on its own. But the entropy does increase from that of the solid simply because molecules can now exchange places with one another, which is not possible in a solid, and have more freedom to move in other ways.
The chemical potential of a solution of an involatile solute is lower than that of a pure solution because it has a higher entropy due to the increased number of arrangements that the two types of molecules can take compared to a pure liquid. Entropy is $ S=R\ln(\Omega)$ for $\Omega$ different arranegments. The figure shows plots of chemical potential vs temperature. It can be seen that compared to the pure liquid the solution has a lower melting point and a higher boiling point.

Notes:
The dotted lines show only that it is possible to  super-heat and super-cool at a phase change.
The change in free energy vs temperature at constant pressure is calculated using $dG=Vdp-SdT$. As $dG$ is an exact differential ($dG$ depends only on starting and ending values, not the path taken) differentiating at constant pressure gives
$$\left (\frac{\partial G}{\partial T}\right)_p=-S$$
or
$$\left (\frac{\partial \mu}{\partial T}\right)_p=-S_m$$
which shows that the slope of the change of free energy (or chemical potential) with temperature is $-S$. The third law  means that entropy $S$ is always a positive quantity, and thus we can conclude that at constant pressure the chemical potential always decreases with increase in temperature and the rate of decrease is greatest for systems with the  greatest entropy.

In case of solutions we can say that $G=G(T, p, n_1, n_2)$ where $n_1$ is the number of solvent's particles and $n_2$ the number of solute's particles. Is it valid to divided by $n$, where $n=n_1 + n_2$ in order to get $G_m$? Is it because $G$ is an extensive quantity with respect to $n_1$ and $n_2$?

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text:
Chemical Equillibria Comparisons
In which of the following reactions is the equilibrium farthest to the left?
a) $\ce{P4O10 + $x$ H2O <=> 4 H3PO4(aq)}$
b) $\ce{AlCl3 + $x$ H2O <=> [Al(H2O)6]^3+ + 3 Cl- (aq)}$
c) $\ce{Li2O + $x$ H2O <=> 2 Li+(aq) + 2 OH- (aq)}$
d) $\ce{H2S + H2O <=> H3O+ + HS- (aq)}$
e) $\ce{CaC2 + $x$ H2O <=> Ca^2+(aq) + 2 OH- (aq) + C2H2(g)}$
The correct answer is d, but I'm not sure how to compare these equilibria.  I know that the answer can't be e, since the evolution and purging of gas will drive the equilibrium forward.  Also, the dissociation of ions in water is favorable due to solvation, so that rules out c and b.
But I'm not sure how to compare the remaining two reactions.
Any help would be greatly appreciated.

By farthest to the left, do you mean which one will have the smallest equilibrium constant?

Yes, I believe that is how the question is to be interpreted.

You have already ruled out the equations b, c, and e.

But I'm not sure how to compare the remaining two reactions.

Have a look at equation a again.
The reaction of $\ce{P4O10}$ with water generates phosphoric acid ($\ce{H3PO4}$), which is a fairly strong three-protonic acid ($\mathrm{pK_{a1} = 2.16; pK_{a2} = 7.21}$).
As a result, subsequent deprotonation of phosphoric acid
$$\ce{H3PO4 + H2O <=> H3O+ + H2PO4-}$$
will pull the equilibrium in equation a to the right.
Hydrogen sulfide ($\ce{H2S}$), on the other hand, is a rather weak acid with $\mathrm{pK_{a} = 7.0}$. Consequently, you will find a significant amount of undissociated hydrogen sulfide in aqueous solution. Therefore, the correct answer in the comparison is indeed d.

Isn't the hydrolysis of phosphorpentoxid irreversible just because of the thermodynamics? I doubt this has anything to do with the phosphoric acid getting deprotonated. Stated otherwise: I highly doubt that I would be able to generate phosphorpentoxid from phosphoric acid just by removing the water from it - Phosphorpentoxid is just energetically unfavored thus driving "equilibrium" to the product side in a quasi irreversible fashion. Besides, this argument ignores intermediate formation of meta and ortho phosphoric acid that is already irreversible.


text:
Oxidation of Carbons

In which of the reactions below is the first compound in the equation NOT oxidized?

The correct choice is (D), which can be seen through process of elimination, but I was wondering how to do this through the carbon oxidation number?
I feel that to know the oxidation number of nitrogen, I need the oxidation number of carbon, but to know the oxidation number of carbon, I need to know the oxidation number of nitrogen.
Or is the correct way to think about this problem that: (D) didn't gain any oxygen atoms, therefore, wasn't oxidized and leave it at that?

There are several ways to determine oxidation number/state. The most useful method for organic (covalent) compounds uses the following three rules, and works for most elements engaged in covalent bonding (just replace carbon with some other element):

Every bond between a carbon atom and a more electronegative atom increases the oxidation number of that carbon atom by 1.
Every bond between a carbon atom and a less electronegative atom (hydrogen is most common) decreases the oxidation number of that carbon atom by 1.
Every bond between a carbon atom and another carbon atom has no effect on the oxidation number of the carbon atom (although we could get more complicated since carbon atoms with different "hybridizations" have different electronegativities).

This approach is equivalent to assigning all of the electrons in a covalent bond to the more electronegative atom for purposes of determining oxidation number. This page on Master Organic Chemistry has a number of examples.
Here's an example from the Wikipedia article on the matter, using acetic acid:

Atom            Bonds to   Contribution to Ox#
H (methyl)       C         +1 (2.5 > 2.1)
          TOTAL Ox#        +1
C (methyl)       H         -1 (2.1 < 2.5)
                 H         -1
                 H         -1
                 C         +0 (same atom)
          TOTAL Ox#        -3
C (carboxyl)     O         +1 (3.5 > 2.5)
                 O         +1 (double bond counts as 2 bonds)
                 O         +1
                 C         +0
          TOTAL Ox#        +3
O (carbonyl)     C         -1 (2.5 < 3.5)
                 C         -1
          TOTAL Ox#        -2
O (hydroxyl)     C         -1
                 H         -1 (2.1 < 3.5)
          TOTAL Ox#        -2
H (hydroxyl)     O         +1 (3.5 > 2.1)
          TOTAL Ox#        +1

Or is the correct way to think about this problem that: (D) didn't gain any oxygen atoms, therefore, wasn't oxidized and leave it at that?

This is the simplest way of thinking of the problem.
Many introductory organic chemistry texts define oxidation as carbon forming additional bonds to oxygen, and reduction as carbon forming additional bonds to hydrogen. So if we have a reaction converting ethanol into acetaldehyde, this would be an oxidation; the carbon previously connected to the $\ce{-OH}$ group is now connected to oxygen with two bonds instead of one.
We can generalize these statements to say that oxidation involves the formation of bonds to a more electronegative element, and that reduction involves the formation of bonds to a less electronegative element.
With this in mind ...
In your problem, specifically, choice (D), carbon previously has two bonds to oxygen and ends up with two bonds to nitrogen. This is not considered an oxidation because it started with two bonds to a more electronegative element and ended up with two bonds to a more electronegative element.


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Where did the description that acids are sour and bases are bitter come from?
Where did the description that acids are sour and bases are bitter come from? With Arrhenius definition, those descriptions may have fit acids and bases, but with the Lewis definition, I guess they would not really apply.
For example, $\ce{AlCl3}$, a known Lewis acid, does not taste sour but rather "gives water a bitter astringent taste" as stated by PubChem in this link.

AlCl3 in water hydrolyzes and becomes Bronsted acid.

possible duplicate of Do acids without hydrogen taste sour?

Ok, I edited it so that it won't be the same with that question. :)

Modifying the content of your question to say something else is prohibited in SE, but I reckon since your new revision was a part of your last question this is allowed.

authors: زكرياء البوش


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Why does hydrogen fluoride have a boiling point so much lower than that of water?
$\ce{F}$ has more unshared electron pairs and is very electronegative, so $\ce{H}$ of another $\ce{HF}$ molecule can $\ce{H}$-bond with it.
$\ce{HF}$ has normal boiling point of $\pu{19.5^oC}$ while $\ce{H2O}$, as you know, has normal boiling of $\pu{100^oC}$.

I think we may have been focusing too much on $\ce{HF}$ and $\ce{H2O}$. If we look at the bigger picture of the boiling points of the hydrides of the elements of their respective groups, we can see that group 17 hydrides really have lower boiling points than that of group 16 hydrides. I still don't know why is the trend this way within each period, but the figure definitely tells us that:

the deviation is more probably due to the differences between the groups
the H-bond concept still works the same way for HF; it really has higher BPs than the other hydrogen halides

So I think now the main question is: why do hydrogen chalcogenides have higher BPs than hydrogen halides (and is actually the group of halides that have the highest BPs in each period)?
(figure is from http://www.vias.org/genchem/kinetic_12450_08.html)

In addition, HF and H2O both can make the maximum number of H-bonds per molecule. HF has three H-bond acceptors (three lone pairs of F) and one H-bond donor (from H of HF) while H2O has 2 H-bond acceptors (2 lone pairs of O) and two H-bond donors (from 2 H of H2O).

I think there are a variety of qualitative ways of looking at this:

Perhaps the most obvious is that $\ce{H2O}$ can form a greater number of hydrogen bonds due to having an equal numbers of hydrogen bond acceptors and donors. Each of the hydrogen atoms can be hydrogen bond acceptors; each of the lone pairs on the oxygen can be donors. In $\ce{HF}$ however there is only one hydrogen bond acceptor and theoretically three hydrogen bond donors. This imbalanced ratio inevitably leaves some acceptors without a donor. Imagine this as two dance parties; one party has two men and two women but the other has one man and three women. Obviously fewer bonds can be successfully formed in the latter, assuming that everything is monogamous.
As mentioned by the other poster, both O and F are very electronegative elements. F however is more EN than O, which means that F stabilizes electrons better than O. As a result, electrons localized on F are weaker hydrogen bond donors; they are already fairly well stabilized. Remember, electrons "want" to be stabilized by nuclei. If they already are fairly well stabilized then they will "feel" less of a need to be associated (and therefore stabilized) with (by) other nuclei. This suggests that the hydrogen bonding present among $\ce{HF}$ molecules might be weaker than the hydrogen bonds present among $\ce{H2O}$ molecules.

Experimental results are the gold standard, of course, as opposed to qualitative reasoning, and qualitative reasoning here can also lead us in the opposite direction; one might argue that by having three lone pairs, F has a lot on its plate, so to speak; it might be able to stabilize one lone pair of electrons very well but three lone pairs is a bigger ordeal, and perhaps big enough that the hydrogen bonds among $\ce{HF}$ molecules are stronger than those among water molecules. Again, this is all qualitative, but this is the sort of reasoning introductory chemistry teachers desire.
If we limit our thinking to just electrostatics then we might surmise that $\ce{HF}$ should have the stronger hydrogen bonds because F is more electron withdrawing and therefore hydrogen should be more positively polarized in $\ce{HF}$ as opposed to hydrogen in water. However, hydrogen bonding is more than just electrostatics. Hydrogen bonding actually has a covalent component; this, however is usually ignored by introductory treatments of chemistry. The bond angle of elements involved in a hydrogen bond is critical. The closer the elements involved in a hydrogen bond are to 180 degrees, the stronger the bond (this specific angle is the case with hydrogen bonding in water; not necessarily other molecules). If hydrogen bonding were purely electrostatic then this would not be the case; angles would not matter - only distance would.
Other issues must be explored as well, such as the number of viable hydrogen bonds and the electron-donating/releasing tendencies of the involved elements.
After scouring the web some common "explanations" that would not explain why water has a higher boiling point than $\ce{HF}$ would be:

water can form 4 per molecule while HF can only form 2.

Incorrect because if we only look at hydrogen bonding as involving the polarity of atoms, then how can water form four hydrogen bonds per molecule? It has two positively charged hydrogens, and a negatively charged oxygen. Seems like it should only form three hydrogen bonds. One must understand that lone pairs can each be hydrogen bond donors.

So, should we accept the reasoning that goes with the experiment and disregard the others? Well, I am asking this because I am wondering what percentage of my high school book is true!

It doesn't really detract from your point, but you have acceptor and donor the wrong way round.

@orthocresol It is known that HF shows hydrogen bonding in gaseous phase too. It simply means that while going from liquid to gaseous phase, the hydrogen bonds don't have to be broken in the case of HF while in the case of water, the hydrogen bonds need to be broken. This simply explains why boiling point of water is greater despite HF making stronger hydrogen bonds. Is this reasoning wrong because it hasn't been discussed in the answer?

We don't only have to explain why the boiling point of water is higher than that of HF, we also have to explain this "huge difference" in their boiling points. The factors discussed above cannot alone account for this.

Water can form 2 hydrogen bonds, as oxygen has a 2δ− partial charge, therefore it attracts to hydrogen atoms with a δ+ bond to it. What you found on the internet wasn't entirely incorrect because they may have meant that water has 2 covalent bonds and can have 2 hydrogen bonds (4 bonds).

The answer lies in hydrogen bonding.
Hydrogen bond energy depends on the electronegativity of a highly electronegative atom which is bonded to hydrogen. Electronegativity of hydrogen is $2.2$, for oxygen it's $3.44$ and it's $4$ for fluorine.
Difference in electronegativity between $\ce{F}$ and $\ce{H}$ is $1.8$ and between $\ce{O}$ and $\ce{H}$ is $1.24$. Hydrogen bond energy of $\ce{H-F}$ is $\pu{41.83 kJ/mole}$ and that of $\ce{O-H}$ is $\pu{23 kJ/mole}$.
$\ce{H-F}$ bond is stronger in comparison to $\ce{O-H}$ bond. In the case of $\ce{H-F}$, there exists hydrogen bonding even in vapour state, 4 to 7 $\ce{HF}$ molecules together form one unit in vapour state. However, in the case of water, there is no hydrogen bonding in vapour state; each water molecule exists independently.
So, to boil liquid water, all hydrogen bonds have to be broken and it requires a large amount of energy. This isn't the case in $\ce{HF}$; all hydrogen bonds need not to be broken, and therefore a lesser amount of energy is required. So $\ce{HF}$ boils at a much lower temperature as compared to water.

Hydrogen fluoride is known to have a way stronger hydrogen bond than water; but, the question asks why the seemingly opposite view is true.

Both fluoride and oxygen are very electronegative. When they bond with hydrogen the hydrogen becomes slightly positive and the electronegative atom slightly negative. Because of this, attraction occurs between the negative atoms and hydrogen in different molecules, called hydrogen bonding.
In water there are two hydrogen making more charge dipoles for stronger and more numerous hydrogen bonds.
Another factor is water disassociates (i.e. liquid water is partially made up of $\ce{H3O+}$ (three hydrogen and an oxygen, which is positively charged) and some $\ce{OH-}$ (one hydrogen and one oxygen, negatively charged). This increases inter-molecule interaction, making the boiling point higher.

Well, [HF2]-, the bifluoride anion, exists in (concentrated) HF solutions. Wouldn't this have some effect on intermolecular interactions as well? Also HF definitely ionizes just like water and actually HF ionizes to a greater extent than water since HF is a stronger acid than water.

Intermolecular forces (IMFs) are the key here. IMFs are directly related to boiling point.
Both $\ce{HF}$ and $\ce{H2O}$ have hydrogen bonds ($\ce{H}$ attached to $\ce{N}$, $\ce{O}$ or $\ce{F}$). But $\ce{H2O}$ has two hydrogen bonds whereas $\ce{HF}$ only has one. Thus, $\ce{H2O}$ should have a much higher boiling point.

please note that there are exactly zero hydrogen bonds in H₂O or HF


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Iron chemistry: acetates for ebonizing wood
I have recently been “brewing” what is commonly called “iron acetate” for ebonizing wood, and I'd like to understand the chemistry of the iron/acetic acid reaction, which should be fairly simple, but turns out to be more complex than I thought.
One previous question relates to the tannin/mordant reaction, a later stage in the process; another asks for feedback on a recipe for iron (II) acetate, which only adds to my confusion; while a third asks about decomposition of iron (III) acetate.
Overview
The basic idea is that you put fine steel wool in white vinegar, the iron reacts with the acetic acid to form an iron (II?, III?) acetate solution; you then paint this on the wood where the iron acetate reacts with naturally occurring tannins (or introduced by way of a previous application of a tannin-rich solution, such as black tea) in the wood, yielding iron tannates, which are dark-colored, thus staining the wood.
Problem
There are many websites, blogs, and articles online describing the ebonizing process and how to make “iron acetate” with varying degrees of utility and accuracy. None of them contain a detailed and accurate description of the chemistry, which I would like to understand.
From the existing online articles, there is a great deal of conflicting information (even to the point of direct contradictions). It's so bad that it is not at all clear to me what the goal of producing the solution is: do we want iron (II) acetate or iron (III) acetate? Does it matter?
Procedure
I put one US quart of warm distilled white household vinegar in a mason jar (5%, ~0.83 mol CH3COOH).
I added two 0000 wire wool pads (~23g, 0.41 mol Fe) that were unrolled, cut into strips, washed in acetone (to remove protective oil), and dried.
I placed the mason jar inside my vanilla Weber barbecue grill in the hot sun, reasoning that the somewhat elevated temperature might speed up the reaction.
I lightly agitated the contents a couple of times per day with a bamboo skewer, squeezing out bubbles trapped in the wire wool to keep it from floating on the liberated gas (presumably hydrogen).
Throughout this process, the liquid remained clear, although some reddish crud was deposited on the mouth of the mason jar.
As some of the solution evaporated, it was topped up with vinegar. This was done twice or three times over a week. Total additions were at most 4 fl.oz. (100 ml).
After a week or so (maybe 8 days), all of the steel wool appeared to have been consumed, and there was a dark grey or black sludge in the bottom of the mason jar, which could just have been unreacted fragments of steel wool.
There was some (predominantly) charcoal grey “foam” on top, but with a large bright orange-brown spot.
I filtered the liquid through a coffee filter into a transparent plastic container (a washed-out 40 oz. Lysol bottle).
During filtering, the liquid appeared dark, with a distinct green tinge.
The last few ounces were filtering slowly (lots of crud in the filter), so I left it for a couple of hours.
On returning, the solution did not appear green but had a definite purple tinge (my g/f even said “Why is it purple?”).
Over the last few days, the solution has lost its purplish color and stratified — there are a few millimeters of dark brown precipitate at the bottom (probably not as much as it seems due to the concave bottom of the Lysol bottle), the bulk of the liquid is almost clear orangey-brown, there is a narrow but distinct dark band about 35mm from the top, then above that, it is cloudier and slightly lighter-colored than the bulk. Unfortunately, the narrow band and top stratum are too subtle to be seen in the photographs.

On top, there is an orangey-brown deposit around the edge and crystalline flakes floating on the surface.

The filter paper has dried, leaving a fine reddish-brown powder.

Other Observations
The colorless liquid in the mason jar, prior to filtering, was much more effective than expected. From all online descriptions, iron acetate is not terribly effective on low-tannin woods like pine, yet only a few hours after putting the steel wool into the mason jar, a drop of the solution placed on an untreated pine strip produced a very noticeable graying effect (much like I'd expected from the final result); the next day, the solution produced a mid-gray/brown; and after a few days, a dark brown.

On pine treated with tea, a jet black color was produced by all but the weakest solution, which was dropped on while the pine was still damp with tea (hey, I'm impatient).

Assumptions
Ferrous (iron II) acetate is soluble in water. It is a white crystalline solid. Its tetrahydrate is light green.
Ferric (iron III) acetate is insoluble in water, but soluble in ethanol. It is an orangey-brown solid.
The effects of impurities can be ignored.
The distilled vinegar did not have much dissolved oxygen.
Speculation
In the mason jar, the iron in the wire wool reacted with the acetic acid, mostly yielding ferrous acetate tetrahydrate (and hydrogen, which just bubbled off), which accounted for the green tinge of the solution observed during filtration. There was also some ferric acetate created in the presence of the limited dissolved oxygen and that available at the surface; this precipitated out into the gray sludge, which also contained unreacted fragments of wire wool. The orange spot in the surface foam was, most likely, ferric acetate and (?) ferric oxide hydrates.
In the Lysol bottle: on filtration, atmospheric oxygen was introduced to the solution, which enabled some ferrous acetate to transition to ferric acetate (pathway?), as this is what has precipitated out. This ferrous->ferric reaction (?) is continuing at the surface. Some of the precipitate has been deposited around the edge of the bottle and some has crystallized to form the flakes on the surface.
In the filter paper, the sludge was residual unreacted wire wool, ferric acetate, and ferric oxide hydrates, dampened with ferrous acetate and acetic acid solution. These have reacted and dried out to a mixture of ferric acetate and ferric oxide hydrates, accounting for the orangey-brown color.
Questions

Is the above speculation broadly correct?

If the end result of putting wire wool in vinegar ultimately becomes ferric acetate (insoluble precipitate, soluble in ethanol), why don't the instructions ever tell you to filter off the precipitate and dissolve it in denatured alcohol? Surely that would make more sense. If that is the goal, why doesn't the usual recipe call for some hydrogen peroxide?
OTOH, if the desired result (for ebonizing) is ferrous acetate, why don't the instructions tell you to avoid introducing oxygen? Tell you to brew it with an airlock? Warn you that the solution will ultimately “go bad” when the ferrous acetate becomes ferric acetate and precipitates out? Tell you to keep it in an airtight container, use it quickly, etc.

What caused the purple tinge?

Immediately after filtration, the solution appeared purplish. Was this just the mixture of an orangey color and a greeny color as the greenish ferric acetate tetrahydrate turned to orangey ferrous acetate or was there something else going on?

If an aqueous solution of ferrous acetate becomes ferric acetate, which precipitates out, what is the most likely reaction/pathway (under these conditions)?

I'm sure I could work something out, but I'm not sure it'd be correct :)

Are there any simple diagnostic tests I could do?

There may be very simple (backyard/household chemical) tests that would enable me to confirm/refute parts of my speculation or estimate relative concentrations. I'm thinking along the lines of “a drop of that in two drops of ammonia will turn a banana blue” rather than “put 10 ml in a test tube with Ludicrously Expensive's reagent in an argon atmosphere”.

In my own experimentation the reaction is quite temperature sensitive. It barely proceeded (>1wk most wool unreacted) in my minimally heated shed (~50F), proceeded slowly indoors (~67F) and proceeded rapidly (~24 hours most wool reacted) in a 150F water bath. Sorry for commenting on such an old question but I hope this helps the next person who comes across it like I did!

Ok, without going through all of this experimentally I'll shoot from the hip knowing some chemistry.
(1) With this coloration you don't want a paint that sits on top of the wood.
(2) The desire is to get the iron to wick into the cellulose fibers.
(3) An iron-acetate complex must wick better than a iron ion (or else you could use iron chloride).
The gist here that an iron ion isn't floating around in a water solution as a "free" ion. It really has water molecules which are "bonded" to the atom to form a "complex." The acetate complex must wick better, probably because it is overall uncharged. So you basically can't wick $\ce{Fe^{2+}}$ into the wood and leave $\ce{Cl^{-}}$ in the solution.
(4) If the pH of the solution gets too low, then iron hydroxides form.
(5) The overall reaction of the acetic acid to dissolve the steel wool is:
$$\ce{Fe} + \ce{H^+}\ce{OAc^+} \ce{->} \ce{Fe^{2+}}\ce{OAc^{-}_2} + \ce{H_2}$$
The point in (5) is that the reaction consumes $\ce{H^+}$ and the pH will increase (i.e. the solution is less acidic).
I suspect that if the pH gets too low then nasty things happen. I think you want to use pickling vinegar with is 10% acetic acid, not household vinegar which is only 3%. For consistency you also don't want to leave excess steel wool in the solution forever. I'd guess figure on about a 75% reaction. So figure out much steel wool would react with 75% of all of the acetic acid and use that much steel wool. That will insure that you have excess acetic acid which will dry off the ppt.
The staining solution is probably unstable, so just mix what you want to use within some "short" period of time.
The dry solid should be stable if stored in a "sealed" container.

You have the direction of pH change backwards.  The reaction consumes $\ce{H+}$, which means the pH will rise, not drop.  At high pHs, the oxidation of iron(II) ions to iron(III) ions occurs more readily.  At high pHs, the solubility of iron(III) is much lower and iron (hydr)oxides will form.  But you are right when you say the solution is to use excess vinegar.  That will keep the pH low.

tl;dr. But the object of the vinegar process is to produce soluble iron(III) to react to ferric tannates. (ferrous tannates are more soluble, see iron gall ink, rust converter, etc. in Wikipedia). The easiest way to do that with household chemicals is vinegar. Tannins are a large family of polyphenolic compounds. Their color ranges from orange to brown to black to purple to almost blue (navy blue, say). If this were done industrially, I'd expect iron sulfate would be used instead of iron acetate (check out the solubility of iron(III) salts. Note that in most cases, Fe(III) is more stable (under environmental conditions) than Fe(II), so that you could use the more soluble Fe(II) salts and wait for the Fe to oxidize. Which might take minutes, hours, days, years, IDK. The one thing I'd worry about with sulfates is the fate of the sulfuric acid left behind after the fe tannate precipitates...an acetate would be a LOT gentler on the wood....so perhaps acetate is actually the best choice...IDK.

My knowledge of chemistry is too non-existent to provide you with the precise answer to your questions but I believe that I have all the data needed to infer it. Since I’m myself interested in understanding what is going on, I’ll share them here hoping that competent people will interpret my findings. It seems I managed to get the staining substance in pure form. Let me explain how I got it.
First of all, the reaction I have is much cleaner than yours and all the others’ I’ve seen on the web. This is probably due to the fact that I use not white vinegar but what is called the “vinegar essence” and is obtainable from the most of the “Russian groceries” shops one can find throughout Europe and, I presume, everywhere where there are enough Russians. The label says it’s the 25% solution of acetic acid. At all the stages of the reaction the liquid is transparent and clear (apart from bubbles reminiscent of sparkly water); there are no layers, no muddiness. After the steel wool has completely dissolved you are left with a little bit of black sediment. As days go by the solution darkens from weak black tea colour to strong black tea colour (it is visually indistinguishable from tea) and the black crystals grow on the surface. First time I made this stain, I assumed that this was carbon from the steel.
As a side note, the others expressed some misconceptions. It is not entirely true that the end colour of wood is always the same. The more diluted solution gives a lighter grey colour which will darken further if subsequent coatings are applied until all the tannins have reacted. However, the more concentrated solution, providing the same final level of darkness, will have a brown tint. Since I was after the pure grey colour, I spent some time experimenting with proportions. The final ones were: one piece of steel wool dissolved in 250ml of 25% acetic acid — it took a couple of days to dissolve but then it sat for another week before I filtered it. The resulting concentrate was diluted with 10 parts of water. It has the pH of about 4. In my experience, the final solution does not “go bad” as others suggested. In three years it hasn’t change its appearance nor the effect. But I was close to running out of it. I decided to repeat the process, and here is when the interesting things happened.
I followed my recipe to the letter with the exception that I left the concentrate to sit for almost 4 weeks before filtering. I took a sample of the filtered concentrate, diluted it with 10 parts of water and found out that the new solution has a much lighter colour than the old one and a much weaker effect on wood. I took another sample and diluted it until it matched the colour of the old solution (to achieve this I added only 2 parts of water). Correspondingly, it produced exactly the same wood colour. It became clear that somehow I lost much of the active substance.
And then I realised that in the last batch I produced a whole lot more of the black crystals that grow on the surface and that I wrongly assumed to be carbon and discarded the first time — probably because the concentrate sat still for four weeks instead of one. Under the microscope, I saw that the crystals only appear black when they are in large enough chunks. When small, they are brown and transparent, like the beer bottle glass. I took some and put them into water to verify they are not soluble. But I was wrong and after about half an hour they completely dissolved (they also turned out to be not soluble in ethanol). The resulting solution had the same colour as my old solution (by pure chance of me putting just enough crystals in just enough water); I covered a piece of wood with it and got the same colour as the old stain produces.
So, lo and behold these crystals are the substance that does the job. I hope it is clear how to get them: follow my procedure and let them grow on the surface of the concentrate (on the second image in your post I see them — “the crystalline flakes”, as you said — floating on the surface). I got a ton of them from one piece of steel wool (see images). Since the crystals do not dissolve quickly, it is easy to rinse them with water and then make the final solution without any acetic acid in it. It is definitely nicer to work with because it smells like water.
In addition to the questions you initially asked, I’d like to know what is this crystalline substance and why it grows on the surface (the crystals actually sink if stirred and float only when they have a dry top). Hope this helps.
(To moderators: I've read the guidelines and I understand that I'm in violation, strictly speaking, but I judged that the contributed information outweighs the offence. If not, sorry)

Since I had some rusty scrap steel (mostly nails) that were not going away anytime soon, I put them into a mason jar with 5% white wine vinegar; but there may have been some contaminants like dust or oil on the nails.
The solution I made slowly but steadily went from clear to orange-brown in color with a white powder forming on top of what remained of the nails.  There was never a time when my solution turned green or purple even though I was checking the solution about every half hour in the first several hours of starting and a couple times each day afterwards.  My solutions were about a week old before I tried filtering one at which point both solutions were staying dark orange/brown, with the what looked like a 'crust' starting to form on top.
I put my solution through a coffee filter which caught the crust, but after a couple weeks, the solution did not change color; it instead formed a new dark brown crust on the top.  This new layer of crust seems to be getting thicker as the water level decreases, but it's hard to tell how thick because some of the crust stays on the side of the jar where the water level used to be.  The filtered solution was inside another mason jar during this part of the setup.
From what I read about Iron (II) acetate, the hydrate can be made by the reaction of ferrous oxide or ferrous hydroxide with acetic acid.
Iron (II) hydroxide can be made by mixing hydrogen peroxide with acid and iron which would make a green solution that can then react with excess acetic acid to form Iron (II) acetate.
I went with the easier method of using iron that oxidized in the presence of oxygen and water over time.  I don't think that the nails I used were galvanized, but it is a possibility that some of the zinc came of and allowed rust to form on parts of the nails.
I would have thought that there might be a slight green color if I mostly had iron (II) acetate, so maybe much of the Iron (III) acetate was suspended in the water and managed to make it through the filter somehow.  Maybe the iron (II) acetate changes into iron (III) acetate over time or under the right conditions?
After awhile (a little over a month), I got back to my a mason jar of filtered solution and found a brown powder that looked sort of like coffee grounds.  I broke it apart into a powder and added some tap water.  My powder seemed the get absorbed or at least suspended in the water.  A small amount of the solution was poured onto a piece of wood that I had on hand (I think that the wood was spruce).  The solution had some suspended particles, but seemed to mostly be liquid.  The wood visibly started changing color in a minute; after a few minutes, the spruce was almost the same color as walnut.  I think either both types had formed alongside each other, or maybe I only had iron III acetate which was merely suspended in water, but helped the iron acetate ebonize the wood which was awesome (sorry, I have no pictures of this).
In conclusion, my altered version of this investigation failed to produce both the green solution and the purple one.  I wish I knew where the purple color came from; I am completely stumped on that point.  In any case, the iron acetate worked, regardless of whether iron II or iron III acetate was created.  Also, it does not seem to matter what the source of iron is; my rusty nails worked wonderfully and they were free.
This was an easy solution to make; I am going to make my own homemade stain using old nails and vinegar from now on.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

Honestly, the question is so long that I no longer know what the question is and whether this answers the question. If the observations are not reproducible, then the chances of having a definite answer to "what chemistry is going on" are automatically reduced, anyway. I'll undelete it in the meantime...

SEARCHES : iron buff, wood vinegar and steel wool stain, giggle juice (found manly by people looking to add paints i have found) and ebony wood.
WHAT'S THE GOAL : iron acetate II (reddish in color after filtering and may have to sit a few days to obtain color)
HISTORY : Iron buff is one of 3 main chemical stains. It goes back to the 1890's where they used it to turn maple gun grips gold by heat treating it after iron buff was applied for the flintlock guns and was used commonly up to the 1960's.
BREWING : While there are many ideas on how, what is clear is that a 1 qt jar of glass or plastic is best, with a plastic top, or one that wont rust. It's not a big deal if you have to use a mason jar lid. Lid can't be airtight do to the hydrogen gases created by the chemical reaction of the acid in the vinegar and the iron in the wool. This also makes temperatures of 100+° from my understanding and experience. Oxygen also helps with oxidizing. One qt of vinegar (distilled white is best) to 1 steel wool (the finer the better, or faster). Also you can rinse the steel wool to clean off the protective oiltiand fluffed or cut up, which all will speed up brewing as well. The brewing time varies, but average is 24 hours to 1 week (I believe that a controlled environment at room temperature, witch I have seen on sites as a suggested temp for reliable results). Also, those sites don't talk about heating up the vinegar or adding peroxide. Filter with coffee filter or cheese cloth to stop brewing by taking out any steel wool left behind. I seen it come out apple cider color after filtering and reddish if cider is what you get let sit sealed up over night or until reddish (purplish is OK but close to being burnt. Yes, over time the solution will go bad since there's no way to fully stop the reaction. Yes you could filter more but I have found there is a thing as to much filtering).
NOTE: You can burn the solution by brewing to long. The color will be black. I just now read that you can renew the solution by adding peroxide to. However, I dont know how much.
OPTIONAL ADDITIVE : Hydrogen peroxide can be used to speed up brewing. (Anything else, or why, I don't know and never looked into it). You can also heat up the vinegar to speed up brewing.
ADDITIONAL INFORMATION: I have found that over all, no matter what you really do to make it, the end color is the same. I have even found a project site were they used black tea and iron buff ended up with the same color in the end, with pine. What I have found is if its too light when applied to wood, wait a day or two, plus when clear coats applied it darkness it. It doesn't fade in sun so it's good indoors or out. Now there are many ways to age wood using things at home. I hear stove cleaner is great for graying wood, so the different questions you have could work just as well, but it's all up to how you make it and the results your looking for.
BACKGROUND: I helped make a solution for my job a few years ago. We make about 30 gallons at a time. I have since moved departments so I don't play with it anymore but lately I started experimenting again and always enjoy learning more about it
Hope this helps if your still working with it.

The question is not about ebonizing wood, it's about understanding the chemistry as thoroughly as possible.

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


text:
Could sodium fluoride or aluminum accumulation in soils contribute to increased amounts of palmitic acid in food?
I am interested in the effects of plant environments on their biochemistry in relation to their effects on the environment and animals(us).
Could sodium fluoride or aluminum  accumulation in soils contribute to increased amounts of palmitic acid in food?
Thus increasing the inflammatory effects of certain foods?

This article explains the effects of fluoride ion concentrations to Prosopis juliflora.

Both catalase (3.2 folds) and peroxidase (2.7 folds) enzymes activity increased with increase in F concentration.

Checking out Wikipedia articles about these enzymes, neither catalase not peroxidase are involved in the fatty acid biosynthesis. So as far as I know, I don't think there's any effect of fluorides in the biosynthesis of palmitic acid.
Also, regarding aluminum ions, this article explains aluminum toxicity as a hindrance to cell division and elongation, but nothing is mentioned about fatty acid synthesis.
References:
Saini, P, et al. “Effects of Fluoride on Germination, Early Growth and Antioxidant Enzyme Activities of Legume Plant Species Prosopis Juliflora.” Journal of Environmental Biology., U.S. National Library of Medicine, 34 Mar. 2013, www.ncbi.nlm.nih.gov/pubmed/24620580.
Mossor-Pietraszewska, T. (2017). Effect of aluminium on plant growth and metabolism.. [online] Psjd.icm.edu.pl. Available at: http://psjd.icm.edu.pl/psjd/element/bwmeta1.element.bwnjournal-article-abpv48i3p673kz [Accessed 23 Aug. 2017].

Good answer...for half of the question. But the OP asks about fluoride and aluminum, the latter of which is not mentioned in your answer.

@airhuff Fixed!

Laterite soils contain significant aluminum : You could look up if they have any effect on crops/forage.


text:
Liquid nitrogen to cool liquid?
Is it possible and safe to use liquid nitrogen to quickly cool a 5 gallon tank of liquid from room teperature to say 34 °F? Or is there another way that can produce really fast results?

Which liquid are thinking of?

Liquids are coffees, fruit juices, and water.

Liquid nitrogen will freeze them to a little bit lower than 34 degrees Fahrenheit. Like, 50 degrees colder. And this question has a big lack of data. What is the specific heat of that liquid? How will you use or pour liquid nitrogen? Do you even have access to it?

Sorry for the lack of information, I am using this in a restaurant environment and have heard that this is possible but just cant find the specific details on how to do it or if it is the best way. The starting temp would be about 23°c(73°f) and want to cool to 0°c(32°f). I am wondering if this is possible, I understand that pouring the liquid into the liquid nitrogen and mixing is the better way. I do have access to it, I just wanted to know if this is a good way to do it or if there are better alternatives before I try it, as it seems to be potentially dangerous.

Dippin' Dots.... Made with LN2

https://en.wikipedia.org/wiki/Liquid_nitrogen  see paragraph 3

I don't know whether liquid nitrogen can be used to cool a volume as large as 5 gallons safely (that is a lot to do at once in a kitchen or restaurant) but it can and has been used in cooking.
Here is an example from youtube.
But it can be dangerous if not done carefully. This BBC story has a warning of what happens if it isn't.

Is it possible and safe to use liquid nitrogen to quickly cool a 5 gallon tank of liquid from room teperature to say 34 °F? Or is there another way that can produce really fast results?

No, at least not in a safe way.
For rapid cooling of such volumes, intense stirring is needed. Liquid nitrogen tends to stay on top of the heavier water and evaporates very easy. It shall form an ice layer with rest of the liquid relatively unaffected.
Possible solutions is food-grade dry ice that is denser than water and causes stirring if submerged thanks to bubbles formed from it. It shall still require good ventilation, but might work.
BTW, both options likely are extremely cost-inefficient.

Pouring LN2 into a volume of warm water would produce a VIOLENT effect of converting the LN2 into a large volume of gas:
https://youtu.be/rVi58N25-Zc
I treat LN2 with the same respect that I do gasoline.  It has an expansion rate of 1:694 at 20 °C (68 °F) and it is quite dangerous to wet human flesh and especially the eyes which are always wet.  Dry skin doesn't freeze instantly but wet surfaces conduct heat at a 25X higher rate.
So you must consider playing with LN2 the same as as playing with fire.  If you must use LN2 to cool water I suggest a heat exchanger system where the exhaust is routed to a well ventilated room.  Better yet use another refrigerant in a closed cycle arrangement.
I work for a large chemistry department that routinely uses ~1000 Liters of LN2 a week.  I am also the safety officer and if anyone suggested such a plan to me they would be shutdown immediately.

One safer way would be to put water ice into the LN2 and transfer the super-cooled ice cubes using a strainer into the bucket of water you wish to cool.  They will cool the water Much Much faster than normally cooled ice.

Also you should have cryogenic gloves, face-shield and apron to be on the safe side.

I used to do this often for a chemistry demonstration for an organization I was in.
We would mix up a fruit punch with kool aid, soda, maybe some fruit cocktail, etc and pour liquid nitrogen on top to flash freeze into slushies.
Honestly, it didn't work well since it basically just froze the top and left the bottom undisturbed. The key was to pour it incredibly slowly, stir vigorously and take breaks (because of the massive cloud of nitrogen that would result).
Edit: Let me also make it perfectly clear that all possible precautions were taken including goggles, cryo-gloves, lab coats, well-ventilated area, etc.


text:
Kinetics and Half Lives Using Pressure

For a certain first-order gas-hase reaction
$$\ce{A \rightarrow B + C}$$
the time required for half of an initial amount of A to decompose is 8.8 min. If the initial pressure of A is 400 mmHg, the time required to reduce the partial pressure of A to 50 mmHg is:
(a) 1.1 min
(b) 17.6 min
(c) 26.4 min
(d) 35.2 min
(e) 70.4 min

The correct answer is (c).
When I approached this problem, I started off thinking about it as a zeroth order reaction for simplicity.
$$\mathrm{400~mmHg \rightarrow 200~mmHg \rightarrow 100~mmHg \rightarrow 50~mmHg}$$
So, if this was a zeroth order reaction (not dependent on the amount of A remaining), then time would be $\mathrm{3 \cdot 8.8~min = 26.4~min}$. But, since this is a first order reaction, then [A] does factor into the half life and thus the time to reach $\mathrm{50~mmHg}$ should be slightly higher than $\mathrm{26.4~min}$.
However, this doesn't seem to be the case. Perhaps it's because we're using pressures instead of concentrations? But isn't pressure synonymous with concentration in the gas phase?

Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.
What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.
For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is
$$\ln\left({\ce{[A]}\over\ce{[A]_0}}\right) = -kt$$
and the usual expression for half-life (where I've rearranged to solve for $k$) is
$$k = -{\ln(1/2)\over 8.8\;\mathrm{min}} = 0.0787\;\mathrm{min}^{-1}$$
So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:
$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$
where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.
As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.

Your approach was perfect! Infact thanks for reminding me this method. But your conception is wrong regarding order of reaction.
For your kind information,
Zero order reaction has half life directly proportional to its initial concentration
If [A]° (initial concentration) ,  K (rate constant)
Then
T (1/2)= [A]° /  2K
But
For first order reaction,
T (1/2)=  0.693 / K
And you can get now that in first order reaction the half life is totally independent of the initial concentration.
So the shortcut you applied for Zero Order reaction was actually valid for the First Order reaction.
Now I request you to go through the textbook once again thoroughly and get the right idea about Rates and Order of reactions.
Cheers


text:
How is sulfuric acid with a concentration of 12 M also 72% w/w?
According to LabChem, $12~\mathrm{M}$ of $\ce{H2SO4}$ is $72~\mathrm{w/w\%}$.
When I try to convert $12~\mathrm{M}$ of $\ce{H2SO4}$ to $\mathrm{w/w\%}$ I get $117\%$ which is obviously wrong.
$$\begin{multline}
12~\mathrm{M}\cdot\left(\frac{1~\mathrm{L}}{1000~\mathrm{ml}}\right)
\left(1~\mathrm{\frac{ml}{g}}\ \ce{H2O}\right)
\left(98.09~\mathrm{\frac{g}{mol}}\ \ce{H2SO4}\right)\\
= \frac{1177~\mathrm{g}}{1000~\mathrm{g}} = 1.177=1.177\cdot100\%=117.7\%
\end{multline}$$
I must be doing something wrong. How do you convert it correctly?

Your problem is that you've assumed the density is 1 gram/mL.
Remember that a molar is defined as a mole of solvent per liter of solution, not solvent. Usually, in introductory chemistry classes, we skip over the fact that adding solute to a solution increases its density, because it makes life more complicated.
As you just found out though, sometimes you really need to account for the change in density that occurs as you add solute. This webpage claims that the density of $12~\mathrm{M}$ sulfuric acid is about $1.634~\mathrm{g/mL}$.
Using this updated calculation, we still have the original $1177~\mathrm{g}$ of sulfuric acid. However, the solution mass is now $1000~\mathrm{mL}\cdot 1.634~\mathrm{g/mL} = 1634~\mathrm{g}$. Finally, we obtain that
$$
\frac{ 1177~\mathrm{g} \, \ce{H2SO4}}{1634~\mathrm{g} \, \mathrm{solution}} = 72.03\% \; \mathrm{w/w} $$
A note about this calculation: it's possible we're begging the question, since I don't know which values on that linked page are measured and which are calculated. It's possible that the density was calculated from measured data. The important thing here isn't how the data was obtained though, it's that the density of solution changes as you add more solute, and so you have to change how you calculate things from the molarity.

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


text:
What happens if a buffer is made up with only half as much NaH2PO4 compared to Na2HPO4?
A solution of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$, when provided in equal amounts at the same concentration, form an excellent buffer. What happens is that the two sodium-phosphate compounds each dissociate independently:
\begin{align}
\ce{Na2HPO4 &<=> HPO4^2- + 2Na+}\tag1\\
\ce{NaH2PO4 &<=> H2PO4^- + Na+}\tag2
\end{align}
What would happen if you made up a solution that had only half of the reactant in (1)?
Am I right in saying that Le Chatelier's principle is not involved, and that shifts in equilibrium don't occur?
Will the solution just have a lower buffering capacity for bases relative to acids?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

You are not right, Le Châtelier's principle plays a very important role.
Let's have a look at the reactions involving disodium monohydrogen phosphate, as it is an amphoteric substance and therefore a buffer on its own:
\begin{align}\ce{
Na2HPO4 + H2O &<=> 2Na+ + H3+O + PO4^3- \\
Na2HPO4 + H2O &<=> 2Na+ + {}^{-}OH + H2PO4^- \\
}\end{align}
To a small extent, there will also be the formation of phosphoric acid, but we'll ignore that for the moment.
Let's ignore the counter ion and formulate the equilibrium constants:
\begin{align}
\ce{HPO4^2- + H2O &<=> H3+O + PO4^3-} &
K_1 &=\frac{c(\ce{PO4^3-})\,c(\ce{H3+O})}{c(\ce{HPO4^2-})\, c(\ce{H2O})} &
= K_\mathrm{a}\cdot c(\ce{H2O})\\
\ce{HPO4^2- + H2O &<=> {}^{-}OH + H2PO4^-} &
K_2 &=\frac{c(\ce{H2PO4^-})\,c(\ce{{}^{-}OH})}{c(\ce{HPO4^2-})\, c(\ce{H2O})}&
= K_\mathrm{b}\cdot c(\ce{H2O})\\
\end{align}
The whole reaction can be described with a coupled equilibrium constant:
$$K = K_\mathrm{a}\cdot K_\mathrm{b}\cdot c^2(\ce{H2O})$$
At all times in a pure solution there will be phosphate, monohydrogen phosphate, dihydrogen phosphate and maybe even phosphoric acid in solution. Changing any of the concentrations of these species will change the equilibrium.
The same equations can be applied to the sodium dihydrogen phosphate system. The initial concentrations of course will be different.
Depending on the ratio of the two solutions, the pH will also change.
The buffer capacity is of course not only dependent on the overall concentration, but also if it is near the ideal buffer point.
Overall, polyprotic acids are very complicated systems and it solely depends on your understanding of the word "excellent" how the buffer behaves. There are many fine tuned buffer systems for various purposes, this website has quite a few neat tables.

To make the above simple (1) pH will change to be more acidic (2) Buffer capacity at that pH will be less than if the ideal solution was made.

@Max, this is exactly what I tried to avoid with my answer. It does not make things simple, but lazy.


text:
Most stable conformation of aldehydes and ketones
Why is the most favorable conformation of ketones and aldehydes the one where the alkyl group is anti to the other alkyl group (or hydrogen atom in an aldehyde) as attached? Why isn't more stable the conformation where the bond between the alpha and beta carbon atoms can participate in hyperconjugation with the carbonyl pi antibonding MO?

By contrast, the slightly more favorable conformation for butene is the right one:

You have to bear in mind that the acidity of the \alpha-hydrogens/enolisability of the ketone. There will be an equilibrium between the keto form and the enol form which will also play into the preferred conformation. Admittedly the equilibrium lies very much to the keto form...

I would have expected the conformation shown to be the most stable for ketones - to minimize steric interactions between R groups. But for aldehydes, I would have expected the opposite, since H is smaller than O. What is the evidence?

http://pubs.acs.org/doi/abs/10.1021/ja01091a016

It could be because of some weak hydrogen bonding between a hydrogen atom attached to one of the carbons of the R group and the carbonyl oxygen?

I found in Ian Fleming's Molecular orbitals and Organic Chemical Reactions, page 93-94 the answer (the aldehyde is propanal):

The polarisation of the H-C bonds in the methyl group leaves a weak
  positive charge on the outside and this is attracted to the partial
  negative charge on the oxygen atom, lowering the energy of this
  conformation

(the difference in energy of the two eclipsed conformations of propanal is 8 kJ/mol).

That is of course assuming R = Me. If you add a page number with your reference, I'll upvote you.

Of course, for a C(Me)3 group, the other conformer of the aldehyde would be more stable.


text:
Something about the idea of Molarity is confusing me
Okay, so molarity is defined as moles of solute per liter of solution. It is used to describe the concentration. So in the lab, when I see a bottle labeled something like $\mathrm{3.5~M}~\ce{H2SO4}$, is it saying that the only chemical in the bottle is this $\ce{H2SO4}$?
Because the way I think of it is "there are 3.5 moles of $\ce{H2SO4}$ per liter of solution". The word "solution" is kind of throwing me off because it makes me feel like they are saying that there is more than just $\ce{H2SO4}$ in the bottle. Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?
I don't know why I'm having such a hard time grasping this. But any help would be awesome!

@michael I dont know whether this will answer your query but I had a similar doubt long time back. You should look into it http://chemistry.stackexchange.com/questions/671/the-mole-is-used-extensively-in-chemistry-why-not-elsewhere . This is how i remembers a mole 1 mole=6.2*10^23 of something. something includes atom,molecules, cow, aliens, stars etc :)

Related: What is the difference between molarity and concentration?

Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?

You are, unfortunately, incorrect. In wet chemistry, you need just a solvent$^\dagger$ and at least one solute to make a solution.  While this does mean that you need at least two constituents to make up a solution, you do not need at least two solutes.

... is it saying that the ONLY chemical in the bottle is this $\ce{H2SO4}$?

It depends on your definition of a chemical:

If you consider water to be a chemical, then no: there are two chemicals in the bottle$^\ddagger$, $\ce{H2O}$ and $\ce{H2SO4}$.
If you are not considering water to be a chemical, then yes: the only chemical in the bottle$^\ddagger$ is $\ce{H2SO4}$.

$^\dagger$ If not specified, the solvent is almost always water.
$^\ddagger$ Except for small amounts of inevitable impurities.

Brian has the correct answer.
I'd like to add, however, that in the most general sense, we don't have to take a molar to mean a mole of solute per liter of solution. I've actually seen it used to mean "a mole of stuff per liter of space" for particular values of stuff.
Since the above sentence isn't exactly self-explanatory, let's take a look at a few examples.
Example 1: Water
At 4 °C, water has a density of 1000 g/L. Water has a molar mass of 18.02 g/mol. This means that in a liter of water at 4 °C, there are:
$$
1~\mathrm{L} \cdot 1000 ~\frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{18.02 \mathrm{g}} = 55.5 ~\mathrm{mol}
$$
Since there are 55.5 moles of water per liter of water, we can actually say that, at 4 °C, water is 55.5 molar. (See Example 3 on Wikipedia).
Note how this works: there is a certain amount of stuff (in this case, water) per amount of space (one liter, or one cubic decimeter). It doesn't matter that we don't have a solution--since there's stuff, and it takes up space, we can calculate a molarity for it.
Example 2: Ideal gas
An ideal gas takes up 22.4 L per mole, at STP. Taking the inverse of this, we find that it is 0.0446 moles/L. Therefore, the concentration of an idea gas at STP is 0.0446 molar.
Again, there's a certain amount of stuff (one mole of ideal gas) and a certain amount of space (22.4 liters), which allows us to calculate a molarity. The fact that the substance is pure doesn't stop us from writing a concentration for it.
A note of caution
This will NOT work for a measure of concentration like molality, because molality is defined by the amount of solvent used. It also won't usually work for things like normality, since you (usually) need to be in solution to generate ions.

Excellent points. I hate normality -- it doesn't even have to be ionic. For example, for EDTA, typically its normality is equal to its molarity, since it binds metals $1:1$. If you don't the 'customary' conversion from molarity to normality, its impossible to identify logically. Small wonder the term is obsolete!

I think this is the best way to clear your doubt. Have you ever prepared tea? What are the main ingredients of tea?
$$\ce{water~(or~milk) + tea~leaves -> tea}$$
It's the addition of tea leaves which changes water to tea. Depending on how strong you want a tea you can add one teaspoon, two teaspoons or n number of teaspoons of tea leaves.
Similarly, in $\mathrm{3.5~M}~\ce{H2SO4}$ solution it contains:
$$\ce{H2SO4~(solute) + H2O~(solvent) -> H2SO4~solution}$$
To my knowledge there is no 100 percent pure $\ce{H2SO4}$ available commercially. The commercially available $\ce{H2SO4}$ is 96 percent $\ce{H2SO4}$ with a concentration approximately of 18 molar. See this link.


text:
How to convert ethanol to 1,1-dichloroethane?
How to convert ethanol to 1,1-dichloroethane?
My approach: Convert ethanol to ethene followed by $\ce{HCl}$ addition. But that's no use. Well if it is possible to create gem-diol from here, then maybe it can be done. But not getting a suitable reaction. Can't seem to figure out anything. Please help!

Well you have to at least tell us what your thoughts on this is.

Addition of chlorine gas with UV light to methane forms polychlorinated methane, so something alone these lines is nearing your goal.

convert ethanol to 1,1-dichloroethane?

A standard route for producing gem-dichloro compounds involves treatment of a ketone or aldehyde with $\ce{PCl5}$ (see the second to last paragraph in the "Chlorination of organic compounds" section here and page 18 here).

(image source)
Therefore, you could carefully oxidize your ethanol to acetaldehyde and treat it with $\ce{PCl5}$.


text:
Why don't electrons pair in [Fe(NH3)6]2+?
Why doesn't electron pairing take place in $\ce{[Fe(NH3)6]^2+}$ though $\ce{NH3}$ is a strong ligand field producer?

Because $\ce{NH3}$ isn’t a strong-field ligand.

Good thing I found this page, for an entire year I thought $\ce{NH3}$ was a strong ligand.

(If you know the basics, go to last paragraph and skip the introduction part)
There are two types of ligands. Strong field ligands and weak field ligands. Those which only cause a small extent of crystal field splitting is called weak field ligand and those which cause large splitting is called strong field ligand.
Why does crystal field splitting take place?
Let's take an octahedral complex. The $d_{x^2-y^2}$ and $d_{z^2}$ point along x,y,z axis and the $t_{2g}$ orbitals point between the axis. When the 6 ligands approach along the three axes, the energy of the $e_g$ orbitals increase more than the increase in the energy of $t_{2g}$ orbitals. Hence the d-orbital splits into two. Filling each electron in $t_{2g}$ orbitals decrease energy by $0.4\Delta_0$ and filling electrons in $e_g$ increase energy by $0.6\Delta_0$
Coming back to your question, you know how electrons are filled in the orbitals (according to hund's rule of course). But how does electron pairing take place in your case?
Let the energy required to pair an electron be P.
Let's say there are 4 electrons. The two arrangements are one electron in each sub-orbital (no pairing) and 4 electrons all in $t_{2g}$ itself (instead of 4th electron going to $e_g$, its paired up in $t_{2g}$).
In the first case, the total energy is $-0.6\Delta_0$ and in second case, its $-1.6\Delta_0+P$. So, only strong field ligands can form the second arrangement. $P$ value is constant for a given metal ion.
Finally, ammonia is weak field ligand but in certain cases, it acts as a strong field ligand (example: for Cobalt). For Fe, ammonia is a weak field ligand. Hence no pairing takes place.

There are not two types of ligands. At the very best, there are two extremes and a very broad range of in-betweens between them.

@Jan actually, you can classify them based on whether $\Delta$ is greater or less than pairing energy for the electrons.

@PrittBalagopal No you cannot. A certain combination of metal and ligand can exist in both a high and low-spin configuration. One of my [tag:coordination-chemistry] answers has the data which I am too lazy to dig up right now. All I remember is that the ligand is nitrogen-based.

I know I'm quite late for this party, but is there a specific reason why $\ce{NH3}$ is a weak ligand toward $\ce{Fe}$ but strong toward $\ce{Co} ?$

The dichotomy between weak- and strong-field ligands that I have often seen constructed here is false and physically inaccurate.
Complexes can be in a low-spin state or in a high-spin state. The high-spin state is the one according to Hund’s rule where electrons are filled into the five d orbitals one after the other. A low-spin state is any complex in which the Hund rule is not strictly observed, but some orbitals are filled with two electrons while others remain empty. Classic examples for high-spin complexes are, e.g. $\ce{[FeCl6]^3-, [CoCl4]^2-}$ and others. Examples for low-spin complexes are, e.g. $\ce{[Fe(CN)6]^3-, [Cr(CN)6]^3-}$ and others.
While these examples may seem to exemplify your stated dichotomy (with chloride being on the weak-field side and cyanide being on the high-field side), reality is much more complex and the spin state of a complex depends on ligands, metal, oxidation state, counterions and so on. In fact, for 2-picolylamine ($\ce{2{-}pic}$) as a ligand to iron(II), both the low-spin and the high-spin $\ce{[Fe(2{-}pic)3]^2+}$ complexes have been isolated and characterised by crystal structures. The $\ce{Fe-N}$ bond length differs by $19~\mathrm{pm}$ with the high-spin complex obviously having longer bonds.
Complexes have a tendency to be in the low-spin state if

the metal is not a 3d metal
the metal’s oxidation state is higher
metal to ligand π-backbonding can take place.

Especially that final point is important. Ammonia is very bad at π backbonding (and amide is even π-donating). So calling it a high-field ligand is misappreciating physicochemical reality. It is just ammonia’s strong σ bonding ability that causes a slightly larger field split than water. As such, ammonia is best termed a medium-field ligand. (The same is true for most neutral nitrogen donors.)

Nice thank.! So for third point, N does not have empty orbital in Nh3 is that reason for no pi back bond. Thank generous soul

$\ce{NH3}$ is placed in the middle of the spectrochemical series.
It mostly acts as SFL but sometimes may act as WFL.
When a metal ion having higher charge is interacting with ligand NH3 it acts as a SFL.
When the metal ion has lower charge it acts as WFL.
EG: $\ce{Co^2+}$ and $\ce{Co^3+}$
In $\ce{Co^2+}$ $\ce{NH3}$ acts as WFL.
SFL:Strong field ligand
WFL: weak..


text:
Naming convention for graphs?
I always have trouble naming my graphs in class. They never seem too scientific or professional.
For example, we are reviewing density now, and had to graph the mass (Y) of different volumes (X) of water. So, very simply, my title was "Mass of Different Volumes of Water". Is this appropriate?
Are there any naming conventions, such as "X vs. Y", or "Effects of X on Y" or "Relationship of X and Y", etc.? Sorry if this is a bit stylistic and opinion based, but I would like some feedback. Thanks.
Note: I'm in the tenth grade.
Note 2: I'm also not sure what site to post this under (there's no "science" site) nor what tags to post this with. Feel free to improve it.

Label your graph with a title of what it shows.
For example, if you have plotted a graph of the temperature recorded each day at one location for a year, you could title it:
Temperature at Observation Site A during 2014-15
Essentially, you want to summarise the X and Y variables and what your sample is in the title.
In this case, X = date, Y = temperature, and the sample is Observation Site A.
If you had measured the mass and volume of water at 25°C for numerous samples, you could title it:
Mass vs Volume of Water at 25°C
In this case, X = mass, Y = volume, and your sample is water at a fixed temperature.
I hope this makes some sense?

As for naming conventions, they tend to be dictated by whoever is going to assess it. If you write a paper to submit to a journal, they have certain styles. A different journal, different styles. Whatever you do decide on, be consistent.

Agree with your post except for one small bit (chances are it's a typo), the mass is the dependent variable here (ie plotted on y-axis) and the volume should be the independent variable (x-axis).

Probably a good catch! Obviously it will depend if you have measured the mass after changing the volume, or measured the volume after changing the mass. (One is inherently easier...)

The most general way of naming a graph is by writing,
Y vs X graph
An example is, Distance vs Time graph. In which the Y axis represents Distance and X represents time.
In order to know which comes in which axis you have to have a clear understanding about independent and dependant. Thus is easy to understand. Once you have clearly understood the axes you have to make the Y axis the dependant variable and X the independent one.
Then the name of the graph becomes ,
Dependant variable vs Independent variable.
For instance consider the Distance vs Time graph.
We see that Distance depends on how long has the time passed. Thus D becomes the dependant variable

Another way to think of it is "Y" as a function of "X".  For example, "Distance as a function of time" or "Distance vs. Time".

A x vs y graph would ensure that the x axis is filled with fixed variables (quantities that do not change throughout the experiment, eg Time intervals) whereas in the y axis, you would intend to fill in  the values that change during the experiment, eg number of observations. I hope this helps

Do you really say x vs y? I'm more familiar with y vs x, as outlined in the answer of slhulk.


text:
Why is benzyl alcohol a slightly stronger acid than cyclohexanol?
Cyclohexanol has a pKa = 16, while benzyl alcohol has a pKa = 15.4
The given explanation is that the "polarity of the benzyl group stabilizes the negative charge on oxygen".
My understanding is that when resonance is not a concern, benzyl groups are primarily electron donating by the inductive effect... which would destabilize charge on oxygen.
I know the inside of the aromatic ring is more negative, and the outside is more positive. Is it this positive charge that stabilizes the negative charge on oxygen?

(This CW answer is adapted from an earlier comment by ron.)
In both cases, the carbon directly attached to oxygen is $\mathrm{sp^3}$ hybridised. However, the next carbon in the two compounds have different hybridisations. In cyclohexanol, it is also $\mathrm{sp^3}$ hybridised; but in benzyl alcohol, it is $\mathrm{sp^2}$ hybridised.
In general, $\mathrm{sp^2}$ hybridised carbons are more electronegative than $\mathrm{sp^3}$ hybridized carbons. This is because the electron density in an $\mathrm{sp^2}$ orbital has more s-character (50%) compared to one in an $\mathrm{sp^3}$ orbital (33%), and is consequently more strongly attracted to the carbon nucleus. Therefore, the carbon attached to oxygen in benzyl alcohol will be more electron deficient (more positive) and will be better able to stabilise the negative charge on oxygen in the alkoxide conjugate base, via the inductive effect.