Calculate the concentration of the deprotonated adipic acid anion Adipic acid is a diprotic acid having $K_\mathrm {a1}=3.9\times10−5$ and $K_\mathrm {a2}=3.9\times10−6$. A saturated solution of adipic acid is about $\pu{0.10 M}~\ce{HOOC(CH2)4COOH}$. Calculate the concentration of the $\ce{−OOC(CH2)4COO−}$ in this solution. Express your answer to two significant figures and include the appropriate units

Lets represent adipic acid as $\ce{H2A}$ for the sake of simplicity.

$$\ce{H2A <=> H+ + HA-}\qquad   K_{\mathrm a1} = 3.9 \times 10^{-5}$$

$$\begin{array}{lccc}
\text{Initial}\quad &0.1 \quad &&0 \quad &&0\\
\text{Change}\quad &0.1-x \quad &&+x \quad &&+x\\
\text{Eq}\quad &0.1-x \quad &&x \quad &&x
\end{array}$$

$$3.9 \times 10^{-5} = x\times x/0.1-x$$

$x^2 = 3.9 \times 10^{-5} \times 0.1$ (neglecting $x$ in nominator because $x\ll0.1$)

$x = x_1 = 0.00197\ \mathrm M$

Let's designate $x$ as $x_1$. $\ce{HA-}$ dissociates again as

$$\ce{HA- <=> H+ + A-}\qquad K_{\mathrm a2} = 3.9 \times 10^{-6}$$

$$\begin{array}{lccc}
\text{Initial}\quad &0.00197 \quad &&0 \quad &&0\\
\text{Change}\quad &0.00197 -x \quad &&+x \quad &&+x\\
\text{Eq}\quad &0.00197 -x \quad &&x \quad &&x
\end{array}$$

$$3.9 \times 10^{-6} = x^2/0.00197$$

$x^2 = 0.00197 \times 3.9 \times 10^{-6}$ as shown above

$x = x_2 = 8.77 \times 10^{-5}$   (designate $x$ as $x_2$)

Concentration of $\ce{HOOC(CH2)4COOH}$ ($\ce{H2A}$) $= 0.1 - x_1 = 0.098\ \mathrm M$

But that was wrong. I then  $x_1 - x_2 = 0.00197 - 8.77 \times 10^{-5} = 0.00188\ \mathrm M$ but that answer is wrong.

I'll follow your lead and represent adipic acid as $\ce{H2A}$ for the sake of simplicity.

Equations we know:

(I) o.100 = $\ce{[H2A] + [HA^{-}] + [A^{2-}]}$

(II) $\dfrac{\ce{[H^+] [HA^{-}]}}{\ce{[H2A]}} = 3.9 * 10^{−5}$

(III) $\dfrac{\ce{[H^+] [A^{-2}]}}{\ce{[HA^{-}]}} = 3.9 * 10^{−6}$

(IV) $\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]}$

There are four equations in four unknowns so the system does have a solution. I could find any way to simply all the equations into anything which seems reasonable for an exact solution.

The way to solve this problem is to solve for the first ionization and assume that the second ionization has an insignificant contribution. A check can be done with equation (IV) to verify that  $\ce{[HA^{-}] \gg 2[A^{2-}]}$.

I am going to show an excessive number of significant figures in the answers so that various methods of calculation can be compared. The acid constants are only given to 2 significant figures so that is the limiting number of significant figures that can justified.

Quick and dirty

Let $\ce{[H^+] = [HA^{-}]}$ and let $\ce{[HA]} = 0.100$

then from (II)

$\ce{[H^{+}]^2 = [HA^{-}]^2 =} \sqrt{(0.100)(3.9E-5)}$

$\ce{[H^{+}] = [HA^{-}] =} 1.97E-3$

thus $[\ce{HA}] \approx 0.1 - 0.00197 = 0.09803$

from equation (III)

(V) $\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$

So indeed $\ce{[HA^{-}] \gg [A^{-2}]}$

This method can be made somewhat better by noting

                  x*(0.1/0.10197)
   HA    0.10000  0.09807
   HA-   0.00197  0.00193
         -------  -------
         0.10197  0.10000

Cleaner solution

The Quick and dirty is a bit dodgy since $\ce{[H2A]}$ was assigned a value of 0.100 to calculate, then reassigned a value of 0.09803. Since the pK constant is only given to two significant figures this isn't really wrong, but a more consistent approach can be easily calculated.

Let $\ce{[H^+] = [HA^{-}]} = x$ and let $\ce{[HA]} = 0.100 - x$, and also let's use $k_2$. Substituting into equation (III):

$x^2 = (0.100 -x)k_2$

$x^2 + k_2x -0.1k_2$

This is a quadratic in x so roots are:

$x = \dfrac{-k_2 \pm \sqrt{k_2^2 - 4(1)(-0.1k_2)}}{2(1)}$

The negative root doesn't work so the positive one is the only one. Also $k_2^2$ is so small that it can be neglected so;

$x = \dfrac{-k_2 + \sqrt{0.4k_2}}{2} =0.00186$

$\ce{[H^+] = [HA^{-}]} = 0.00186$

$\ce{[HA]} = 0.100 - 0.00186 = 0.09814 $

Again from equation (V)

$\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$

Exact Solution

Solve equation (II) for $\ce{[HA^{-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:

$\ce{[HA^{-}]} = \dfrac{k_1\ce{[H2A]}}{\ce{[H^+]}}$

and multiply (II) and (III) to solve for $\ce{[HA^{2-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:

$\ce{[A^{2-}]} = \dfrac{k_1k_2\ce{[H2A]}}{\ce{[H^+]^2}}$

Both of the above can be substituted into the left hand side of the following three equations to get the values shown on the right.

$\dfrac{\ce{[H2A]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}}$

$\dfrac{\ce{[HA^{-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k1[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $

$\dfrac{\ce{[A^{2-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k_1k_2[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $

Being a glutton for punishment I created a spreadsheet with the calculations. After some manual manipulation I was able to determine the "exact solution" to a totally ridiculous number of decimal places.

$\ce{[\ce{H+}]} \text{ }= 0.00195932639309$

$\ce{[\ce{H2A}]}~~= 0.09804455814272$
$\ce{[\ce{HA-}]} ~= 0.00195155732146$
$\ce{[\ce{A^{2-}}]}~~= 0.00000388453582$
$\text{sum} \quad = 0.10000000000000$

$\ce{2[\ce{A^{2-}}] = [\ce{H+}] - [\ce{HA-}]} = 0.00000776907164$

$\ce{K_{a1}}$ from data = $3.9E-5$
$\ce{K_{a2}}$ from data = $3.9E-5$

$\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]} = 0.00195932639309$


Why does graphite have a high melting point even though its intermolecular forces are weak? From what I've read online, when a substance undergoes a state change, the intermolecular forces are broken. The covalent bonds are not broken (http://www.bbc.co.uk/schools/gcsebitesize/science/add_ocr_pre_2011/chemicals/airmolecularrev2.shtml)

My textbook states that layers of graphite can slide over one another easily and can break of one another as they have weak intermolecular forces.

I thought that if an object has weak intermolecular forces, its melting point is lower. Why does graphite have such a high melting point if its intermolecular forces are weak?

My textbook also states that the strong covalent bonds are broken during state changes, isn't this incorrect?

It's not usual to consider graphite as a material composed of "molecules" in the typical sense, though it could be viewed as a kind of polymer with two-dimensional macromolecules. Regardless, it is fruitful to analyse the individual sheets in graphite as the limit of increasingly large polycyclic aromatic hydrocarbons (PAHs). The sequence goes: benzene ($\ce{C6H6}$) → coronene ("superbenzene", $\ce{C24H12}$) → $\ce{C54H18}$ → ...

The main kind of intermolecular interaction relevant to this sequence of compounds is pi-stacking. For the smallest example in the sequence, benzene, it appears that the strength of this intermolecular interaction is only about $\mathrm{10\ kJ\ mol^{-1}}$. This represents only a modest attraction; a hydrogen bond can easily be twice as strong even though it involves less atoms. In this sense, when "normalised" by the number of atoms taking part, pi-stacking is indeed a comparatively weak intermolecular interaction.

Furthermore, $\mathrm{10\ kJ\ mol^{-1}}$ is comparable to the average thermal energy of particles in ambient conditions ($\mathrm{k_BT_{amb}=2.5\ kJ\ mol^{-1}}$), so it doesn't take too much effort to pull apart benzene molecules. Indeed, benzene melts at $\mathrm{5.5\ ^oC}$ and boils at $\mathrm{80\ ^oC}$ under one atmosphere.

However, the next compound in the sequence, coronene, already melts at $\mathrm{437\ ^oC}$ and boils at $\mathrm{525\ ^oC}$. Larger PAHs would almost certainly have even greater values, eventually reaching the limit of graphite, which melts around $\mathrm{4000\ ^oC}$ under pressure. The type of intermolecular interaction hasn't changed, so why are these molecules suddenly so difficult to pull apart? The answer comes from realising that, though individual sections of each molecule interact weakly, the sum of many weak intermolecular interactions over an entire molecule leads to a very strong intermolecular interaction overall.

Crudely speaking, imagine that each individual aromatic ring contributes $\mathrm{10\ kJ\ mol^{-1}}$ worth of intermolecular attraction. Coronene contains 7 fused aromatic rings, which would lead to a total interaction of $\mathrm{70\ kJ\ mol^{-1}}$ between two molecules. As the molecules get larger, this value increases further and further. Eventually the total intermolecular interaction between two very large PAH molecules becomes enormous. For the substance to liquefy, it is only necessary to "break" a fraction of these intermolecular interactions (breaking all of them is turning the material into a gas), but even a small fraction eventually represents a huge amount of energy, so melting only happens at very high temperatures.

It is interesting to note how often chemists make the mistake of neglecting weak long-distance interactions (e.g. van der Waals), especially in the presence of stronger ones. For example, van der Waals interactions are fundamental to the stability of alkyl-substituted hexaphenylethane derivatives. In proteins, weak interactions are often overlooked in favour of hydrogen bonding ($\alpha$-helices and $\beta$-sheets), though they can be decisive in determining the correct conformation of an enzyme or how a protein interact with medicinal compounds.

As a last slight tangent, I just want to point out that in undergraduate-level chemistry, often one finds statements such as "boiling points for covalent compounds increase with their molecular weight. Now it should be evident this is not strictly true. It merely happens that compounds with higher molecular weights tend to be larger and allow a greater amount of intermolecular interactions per molecule, thus leading to higher boiling points.


Solubility vs Dissociation of HF Ok, so I understand that solubility is a measure of how well a compound is solvated by the solvent. And dissociation is a measure of how much of a compound splits into its ionized forms.
Does HF have high solubility, but low dissociation (weak acid)? How can that be?

Is it because it forms strong hydrogen bonds in solution and thus forms stable solvation spheres (solubility), but its high electronegativity prevents H+ atoms from leaving its vicinity (dissociation)?

I think your last sentence answers your own question very well:

  It is because it forms strong hydrogen bonds in solution and thus forms
  stable solvation spheres (solubility), but its high electronegativity
  prevents $\ce{H+}$ atoms from leaving its vicinity (dissociation).

Dissociation is one mechanism of dissolution, but not the only one. Also, however, the $\ce{H-F}$ bond strength is not solely determined by electronegativity.


Does tetraoxidovanadium (VO4+) ion exist? In my book (NCERT India), it is mentioned that: (page 224)

  ...$\ce{Mn2O7}$ gives $\ce{HMnO4}$ and $\ce{CrO3}$ gives $\ce{H2CrO4}$ and $\ce{H2Cr2O7}$. $\ce{V2O5}$ is, however, amphoteric though mainly acidic and it gives $\ce{VO4^3–}$ as well as $\ce{VO2+}$ salts. In vanadium there is gradual change from the basic $\ce{V2O3}$ to less basic $\ce{V2O4}$ and to amphoteric $\ce{V2O5}$. $\ce{V2O4}$ dissolves in acids to give $\ce{VO^2+}$ salts. Similarly, $\ce{V2O5}$ reacts with alkalies as well as acids
  to give $\ce{VO4^3-}$ and $\ce{VO4+}$ respectively. The well characterised $\ce{CrO}$ is basic but $\ce{Cr2O3}$ is amphoteric.

The sentence under scrutiny has been boldened. I wonder how the $\ce{VO4+}$ ion can exist.

Going by the usual oxidation numbers, four oxide anions should contribute a total of -8 charge. To balance it, vanadium needs to be in +9 oxidation state. But, vanadium has only five valence electrons $(\ce{[Ar] 3d^3 4s^2})$, so how can it even attain the +9 oxidation state? Though, it may be possible that this ion has a different structure (possibly a peroxide linkage?) that allows it to exist with a +5 oxidation state on vanadium, but I am not sure.

I have read relevant documents on the internet, including the Wikipedia page for the vanadate ion and a Scholar search, but neither yield any relevant results. I can't search it on ChemSpider Structure Search either because I don't know its structure, and a textual search yields only unrelated partial matches.

So, is it true that the $\ce{VO4+}$ ion dost not exist, or have I missed something? Or is it that there is a misprint in the book, and they intended to write another ion in place of $\ce{VO4+}$?

Since the comment by Ivan pointing to a possible misprint got four votes and no answer yet, I'll post it as an answer. In short:  this is a misprint in the book, and the book probably meant $\ce{VO2+}$.

Here's a citation to back it up: Bertrand, G. L.; Stapleton, G. W.; Wulff, C. A.; Hepler, L. G. Thermochemistry of Aqueous Pervanadyl and Vanadyl Ions. Inorg. Chem. 1966, 5 (7), 1283-1284. DOI: 10.1021/ic50041a048.

  ...it does appear that below $\mathrm{pH} \approx2$ the principal species are the pervanadyl and vanadyl ions represented by $\ce{VO2+(aq)}$ and $\ce{VO^2+(aq)}$ (respectively).

Also worth noting that the formation of the pervanadyl ion is more favourable with a free energy of formation of $\pu{-140.3kcal/mol}$, as opposed to $\pu{-106.7kcal/mole}$ for the vanadyl ion (as mentioned in the source above)


Thermodynamics: Relation between different types of energy: My Analogy I have always been confused between the different kind of energies in thermodynamics so today I thought of finally getting it right. Here's my analogy(which is wrong and I need help in improving it, please read it once):

Let us suppose we take a gas at temperature $T_1<T_{room}$ in a beaker fitted with a piston and conducting walls.

Its present energy can be represented by the following diagram:

When the temperature increases both the pressure volume energy and internal enrgy increase by fixed amounts, independent of path(being state variables), according to the equation $\Delta(PV)=nR\Delta T$ and $\Delta U=3/2\ nR \Delta T$

As enthalpy is the sum of both these energies, even it is a state variable, and its change is independent of the path followed.

Now event though $\Delta (PV)$ is a state function and there is a fixed change in its value, there are 2 ways of changing (increasing in this case) it. One by increasing volume and the other by increasing pressure.
The problem is that in the process of increasing pressure-volume energy by increasing volume, the system has to do some work, which means it has to throw out some energy. But we can't let that happen. So we need to provide the system with some extra energy that it can throw out to do work. We haven't increased the energy of the system by doing that as it has come back to the surroundings in the form of work.
There is this fundamental change in energy of the system that is absorbed by the system from the surroundings and there is this extra energy $P\Delta V$ that is also absorbed but then thrown out in the form of work. The total energy absorbed is termed as heat $q$ and the energy thrown out is called work $(-w)$. They depend on the path as it depends on the way we increase the $PV$ energy.

Problem:
The biggest problem with this analogy is it cannot explain $\Delta U=q+w$. According to my analogy $\Delta U+\Delta (PV)=q+w$. This means pressure-volume energy is a part of internal energy and not something different. But then what would be the significance of enthalpy $H=U+PV$?
Moreover I don't know where to put:
1. gibbs free energy as I don't have an intuitive understanding of it, I only know the mathematical formula and the definition: maximum useful work that the system can do and
2. energy lost due to entropy ($T\Delta S$).

It would be great if someone could help me improve my understanding of thermodynamics as I am tired of getting confused each time I open this chapter. An approach coinciding with this one (something like a venn diagram) would be appreciated but anything is welcome. Please help, I really need it!

This is not an attempt at a complete answer but I think its relevant to share this great explanation from "An Introduction to Thermal Physics" by Schroeder:


Data repository of organic compounds Does there exist any data repository of chemical compound that gives the structure of the compounds with the help of coordinate geometry. It would be like each atom has a coordinate and the list of bonds it has with other atoms so that the 3D structure of the compound can be created from this very theoretical description?

I'm not entirely clear on what you want - I guess a database of a bunch of 3D coordinates for many organic compounds?

By far the largest such database is NIH PubChem with over 30 million compounds - mostly organic. You can search by structure or name or identifier. Most of these have 2D and 3D geometries you can visualize or download and many computed properties.

A similar database is the RSC ChemSpider with over 34 million compounds. Much like PubChem, names, 2D and 3D geometries and other properties are available.

I'll put in a plug for the Pitt Quantum Repository which contains ~64,000 molecules and growing, all with geometries computed using quantum chemical methods. One benefit of PQR is that you can easily access the site using smartphones and tablets and use a QR code or DOI to link to the compound.


Why does trihaloacetaldehyde formed during the haloform reaction not undergo a Cannizzaro reaction? We know trihaloacetaldehyde is formed during the haloform reaction of acetaldehyde with a halogen in strong base. Since it has no alpha hydrogens, I expected that it might undergo a Cannizzaro reaction, as we are using a strong base.   Why not?

Albeit haloform reaction is primarily for methyloketons, aldehydes which have relatively good leaving groups connected with carbonyl (like chloral), or get it during halogenation, indeed also react.

Here we have another factor limiting usage of Cannizzaro reactions, as although intermediate tetrahedral anion (6) could in principle undergo this reaction if $\ce{R = H-}$, it doesn't if there's a better leaving group then $\ce{H-}$. The point of haloform reactions is to make them if they aren't present already.

(source of picture)


Cohesive Energy Density table request Quick question:  Any one knows a resource for Cohesive Energy Density of some organic materials ?

I am not looking for any specific molecule, actually I want to plan my project based on available data. I have found some tables for internal cohesive pressure but could not find any thing for CED.

Anything including tables/papers and books are fine.

Some resources to get you started:

Free Volume, Cohesive Energy Density, and Internal Pressure as Predictors of Polymer Miscibility (White and Lipson, 2014) (paywalled).

Where the authors use cohesive energy density as part of their research to develop

  strong correlations between bulk behavior and theoretical pure component character, and use those correlations to motivate a discussion of what drives polymer miscibility.

The Materials Design webpage Organic Materials Properties: Densities, Cohesive Energies, and Heats of Vaporization contains a couple of lists and information pertaining to a computer based modelling tool.  Also included is a bibliography of further resources.
The Tampere University of Technology in Finland's webpage Exploitation of CED provides a list of CEDs of several organic compound, as well as an explanation and further information on the previous and next pages.


Calculate the wavelength of the radiation released when an electron moves from n= 5 to n=2 Calculate the wavelength of the radiation released when an electron in a hydrogen atom moves from $n = 5$ to $n = 2$.

How can I find this wavelength?

The Rydberg formula (for the hydrogen atom) is usually given as

$$\tilde{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_\mathrm f^2}-\frac{1}{n_\mathrm i^2} \right)$$

Here, $R$ is the Rydberg constant; take care of the unit ($\mathbf{m^{-1}}$  or $\mathbf{cm^{-1}}$) here! Wikipedia quotes a value of $\pu{1.097 \times 10^7 m-1}$ for a nucleus of infinite mass. In principle, the value for a nucleus of finite mass (as is the case here) is slightly different, but for the purposes of this question, we can ignore that.

$n_\mathrm f$ is the final state, and $n_\mathrm i$ is the initial state.

So, if you substitute $R = \pu{1.097 \times 10^7 m-1}$, $n_\mathrm{f} = 2$, $n_\mathrm i = 5$, you can arrive at $$\tilde\nu = \pu{2.304 \times 10^6 m-1}$$

and the wavelength is $\lambda = 1/\tilde\nu = \pu{4.341 \times 10^-7 m}$, or if you convert to nanometers, $\lambda = \pu{434.1 nm}$. Double-checking against the Wikipedia page on the Balmer series, this value is indeed correct.


What is the bonding type in hydrides of group 2 elements? In one of my MCQ questions, there was a statement saying all group 2 elements form covalent hydroxides. I know it is incorrect. I know that $\ce{BeH2}$ is a covalent hydride. What about the rest? Are all of them metallic hydrides? Is it true that metallic hydrides produce the hydroxide and give off $\ce{H2}$ with water?

There is a quite extensive chapter about hydrogen and its compounds in Holleman and Wibereg's book "Inorganic Chemistry". As I only have the german version, I will briefly summarise what I read.

You are of course right, hydrogen forms salt-like structures with alkaline and earth alkaline elements, except for beryllium. Compounds like $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{RbH}$ and $\ce{CsH}$ crystallise in a sodium chloride structure. Every cation is coordinated by six hydride ions and vice versa. The compounds $\ce{CaH2}$, $\ce{SrH2}$ and $\ce{BaH2}$ crystallise in a fluorite structure. The hydride ions are tetrahedrally coordinated by four earth alkaline cations. The cations are cubically coordinated by eight hydride ions. This β-structure is persistent at higher temperatures. Below $780~^\circ\mathrm{C}$ ($\ce{CaH2}$), $855~^\circ\mathrm{C}$ ($\ce{SrH2}$), and $598~^\circ\mathrm{C}$ ($\ce{BaH2}$) it changes to the α-form, i.e. the lead dichloride structure. The magnesium compound exists in a rutile structure. Hydride ions are trigonally coordinated by magnesium ions. The cations are octahedrally coordinated by hydride ions.

However, according to the difference in the electronegativities, you would expect a predominant covalent character. The maximum is $1.3$ for $\ce{CaH2}$ and the minimum $1.0$ for $\ce{MgH2}$, which translates to a maximum $30\%$ and minimum $18\%$ ionic contribution. Maybe this is where the statement of the question originates from. Experimental findings rather support the more ionic nature of the compounds though, i.e. the found lattice energies were supported by calculations assuming a purely ionic structure.

The conclusion of this according to Holleman and Wiberg is that the elements are fused by ionic bonds with covalent contributions. The latter increases from $\ce{CsH}$ to $\ce{LiH}$ and from $\ce{BaH2}$ to $\ce{MgH2}$. The magnesium hydride is already borderline covalent.

The question about their reactivity towards water is mainly answered in the chapters for the elements itself. Magnesium reacts readily with water and releases hydrogen. Because of this it is sometimes used as a drying agent. $$\ce{MgH2 + 2H2O -> Mg(OH)2 + H2 ^ }$$
The other elements react similar. The reaction is highly exothermic, the reaction of calcium hydride releases $228~\mathrm{kJ/mol}$. It is often used as a drying agent for gasses because of this feature. Strontium and Barium also react exothermic, i.e. they release $177~\mathrm{kJ/mol}$ and $172~\mathrm{kJ/mol}$, respectively.


Is it possible to blow out a candle flame using pure oxygen? This sounds like a stupid question perhaps, but if a stream of pure $\ce{O2}$ was directed at the flame as a concentrated jet would it blow off the candle?

It depends on the wick.   A flammable wick, exposed to oxygen, will burn, but
a non-flammable wick, saturated with liquid wax, could be chilled by the gas flow,
and when the evaporated wax is blown clear of the wick, the candle flame
could extinguish for lack of fuel.   Basically, the oxygen flow might
(at sufficiently high velocity) remove the vapor fuel so that the flame
becomes a detached fireball, and the wick and wax of the candle cool.
The oxygen flow just has to be faster than the flame propogation in the
fuel-oxygen mixture.

A solid flammable wick, though, won't be displaced, just burn faster,
and on discontinuing the oxygen, the candle will still be lit.


Good resource for stereochemistry of more than 4 substituents? I have several classes of problems I'm facing in which I'd like to be able to discuss stereochemistry of non-tetrahedral centers. Specifically, I'd like to be able to check that the absolute configuration around a (for instance) trigonal bipyramidal transition state matches the absolute configuration that I've calculated.

For instance, these (chemically meaningless) trigonal bipyramidal structures are not superimposable:

What's a good resource discussing "higher order" stereochemistry? I'm particularly interested in classification of stereochemistries and in exchange rules for different orderings (analogous to the parity rules for tetrahedral centers). I'm coming from a computational/algorithmic slant.

The best (only) resource I've found so far is the opensmiles specification: http://opensmiles.org/opensmiles.html#_trigonal_bipyramidal_centers. I'm happy to purchase a book, or academic paper, or read program source code.

I'm not sure how easy this is to implement algorithmically, but for what it's worth, the official IUPAC recommendations are as follows.
Section P-93.3 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) describes the guidelines for assigning configuration at non-tetrahedral atoms. The portions of this answer will be quoted from there.
These guidelines, and further details on certain aspects, are also discussed in Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) (link), section IR-9.3.

Stereodescriptors for nontetrahedral configurations are composed of three parts.
(1) A symbol indicating overall geometry called the 'polyhedral symbol';
(2) A symbol called the 'configuration index' that describes the configuration of ligands around the central atom;
(3) A symbol indicating the 'absolute configuration' associated with the central atom called the 'chirality symbol'.

For the first example you provided:

which is a trigonal bipyramid, the appropriate 'polyhedral symbol' is TBPY-5 (P-93.3.2, Blue Book). The numbers in red are the priorities according to the Cahn–Ingold–Prelog system, which will soon be required.
The 'configuration index' is

a series of numbers to identify where each atom or group is located. It is based on the CIP priority order of the atoms attached. The atom or atoms with highest priority are given priority number one '1'; the next priority number two '2', and so on. [...] (P-93.3.3.1)

For a trigonal bipyramid, this "consists of the priority numbers of the two apical atoms (lower number first if different), representing the reference axis of the system" (P-93.3.3.5.1). In this case, the two apical atoms have priority numbers 1 and 2, so the configuration index is simply 12.
The 'chirality symbols' available for a trigonal bipyramidal system are 'A' and 'C'.

The atoms in the plane perpendicular to the reference axis are viewed from the side with the atom or group of highest priority on the reference axis. If the direction from the atom in this plane with highest priority to the atom with the next higher priority is clockwise the chirality symbol is 'C', if anticlockwise 'A' (P-93.3.4.2.1).

Looking along the reference axis (which is the axis containing the axial ligands), from the direction of the higher priority atom (oxygen), the equatorial ligands decrease in priority going in a clockwise order. So, the chirality symbol is C. (Essentially, this is analogous to assigning an R-configuration for an ordinary chiral carbon.)

The three parts of the stereodescriptor are enclosed in parentheses and separated from each other by hyphens and cited at the front of the name separated from it by a hyphen. (P-93.3.1)

The full stereodescriptor is therefore (TBPY-5-12-C).


Can you write (with and ink pen) on hydrophobic paper? Would the ink be repelled from the paper or prone to smearing?

This depends on what your ink is based on. If it can be washed off ordinary paper using water, it will be repelled on hydrophobic paper. However, if it can't, it has a chance of staying on.

Note that the paper in particular might repel all or most liquids, not just water. In this case, the ink has a very low chance of staying on the paper.


How can I reach the two minima connected by a first-order saddle point (transition state) using the Gaussian Software Currently, I have several transition state structures for a reaction of the kind A + B -> AB. These TS structures were calculated using Gaussian's own Berny-Optimization procedure and validated as first-order saddle-points by performing frequency calculations. Each of the structures has one, and only one, negative frequency. (It really doesn't matter for this question, but these are DFT calculations and they don't take too long as the system is moderatly small).

To validate and analyze these transition-states in the realm of the bigger picture of the whole PES some more, I would like to perform optimizations starting from them to yield the two minima connected by the transition-state.

Using the Gaussian software the most primitive idea to do this is to perform a regular optimization. I of course did this and, oddly, for every transition state the structure of the product of the reaction is yielded. Conceptually, a true transition-state on the PES should have zero gradient. This means that starting a primitive, gradient-based optimization from a hypothetical, 100% converged transition state should not yield another minimum anyway. I assume that the Gaussian software is in this case too "smart" and "user-friendly": The optimization might not be performed along the imaginary mode or using only the gradients, but most likely all convergence-accelearting black magic available is used. For most cases, this is a reasonable default scenario as it speeds up the optimization towards a truely relevant structure. However, I am also interested in the "path" of the optimization downward the imaginary mode. So this does not work for me.

It seems like the "EigenvectorFollowing" (EF) option for Opt in Gaussian is exactly what is needed to solve my problem. However, I cannot seem to get it to work for a regular optimization starting from the TS structure. Keywords that have failed include:

#p opt(z-matrix,EF) freq b3lyp/aug-cc-pVDZ
#p opt(cartesian,EF) freq b3lyp/aug-cc-pVDZ
#p opt(EF) freq b3lyp/aug-cc-pVDZ
#p opt(calcall,EF) freq b3lyp/aug-cc-pVDZ
#p opt(calcfc,EF) freq b3lyp/aug-cc-pVDZ
#p opt(readfc,EF) freq b3lyp/aug-cc-pVDZ

In every case, the error is the same:

 (Enter /apps/gaussian/g09_D/g09/l113.exe)

 EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-
 EIGENVECTOR FOLLOWING MINIMUM SEARCH
 INITIALIZATION PASS

 ************************************************
 ** ERROR IN INITEF. NUMBER OF VARIABLES (  0) **
 **   INCORRECT (SHOULD BE BETWEEN 1 AND 50)   **
 ************************************************

Searching for this error on the internet yields many results, however they are all realted to MP4 / CCSD(T) optimizations and not to my scenario. I have also tried to specify the molecule as Z-Matrix or Cartesian coordinates.

Note that - speaking plainly - the system in concideration is two organic molecules and previous investigations showed that it is quite boring from a quantum-chemical viewpoint. By this I mean that the structures are "well-behaved" and not biradicals or other quantum-chemical-corner-cases where DFT fails.

I hope that this issue can be resolved by an experienced gaussian user. I am  using Gaussian 09 Rev. E. The first part of my question therefore is how can I use the EF option of Gaussian to perform an optimization along an imaginary mode starting from a transition structure?

Secondly, I still do not have any idea how to reach the "second" minimum connected by the TS (assuming the reaction is concerted, this would be the seperated educts). The only possible approach to tackle this which I can think of currently is to perturb the system manually along the imaginary mode and perform EigenVector-Following optimizations from these perturbed structres. Is there a more elegant way, circumventing manual structural modifications, to yield the two minima connected by the TS?

how can I use the EF option of Gaussian to perform an optimization along an imaginary mode starting from a transition structure?

I have never used the EF mode of Gaussian 09, but...

  Is there a more elegant way, circumventing manual structural modifications, to yield the two minima connected by the TS?

I believe the standard way of proving that your TS is the TS you want is by an intrinsic reaction coordinate calculation:

http://gaussian.com/irc/

Make sure to check out the "options" tab. A decent write-up on a small system is here:

http://www.cup.uni-muenchen.de/ch/compchem/geom/irc5.html

Basically, you need to restart the calculation using the checkpoint (chk) file of your TS calculation, and as you calculated the vibrational spectrum, you can use "irc=rcfc" to read the force constants from the restart file. By default the IRC calculation does 10 points in both directions of the "reaction coordinate" starting from the TS, you can change this using MaxPoints=n, and you can also specify which direction you want to study using other keywords listed in the "options" tab on the Gaussian website.

Opening the logfile in Gaussview, you can see the IRC profile - the same way you can look at the optimization profile of an "opt" calculation. I'd follow up on the IRC by optimizing each of the end-points of the profile using a standard "opt" calculation, just to make sure.

Hope that helps.


Wood burning and carbon dioxide or monoxide? I am building greenhouse and i want to operate greenhouse at winter time. For heating i ll use wood. And also i am thinking to give back  carbon monoxide which ll come from burning process of wood. There is a question, how can avoid to get carbon monoxide from burning wood, do i must give more air flow (or oxygen flow) to burning process? Or when you burn wood there is just one gas output that is carbon dioxide?

A good reference to read is Laboratory and field investigations of particulate and carbon monoxide emissions from traditional and improved cookstoves Atmospheric Environment February 2009, Pages 1170–1181.

Unforntunately, there is a significant amount of carbon monoxide, from 29 to 118 grams of CO per kg of wood.

They find that the drier the wood, the less CO, but don't expect to completely eliminate CO.


Are orbitals always filled in from closest to nucleus to farthest away? In which orbital, 4f or 6s, would an electron have a greater likelihood of being near the nucleus.

I figured that the 6s would be closest, since it is filled in first when using the Aufbau principle.
However, does that necessarily imply that it is closer to the nucleus? Or would 4f be closer, since its principal quantum number is smaller?

As you have studied for the quiz, you should know the expected answer.

In full-scope reality, the question has nothing to do with the two explanations you offered, but with the shape of atomic orbitals:

Radial distribution of selected orbitals in $\ce{Sm^{3+}}$, http://pubs.rsc.org/en/content/articlehtml/2004/dt/b407173e


How can you infer that density decreases when ice melts from the water phase diagram? I know the phase diagram shows negative slope for water which is an exceptional case...  But how do you read that graph and get to the conclusion that density is increasing?
It's the one with pressure plotted against temperature
Thanks ..

sorry i couldn't attach the graph here...

Draw an isotherm at a low enough temperature (< the "triple point temperature of 0.01°C), so that it passes through "all three" phases (ignoring the high-pressure ices that are not relevant to the question).

Note that as pressure goes up you go from solid to liquid.  Increasing pressure always favors a denser phase, so liquid is denser than solid for water.

Water is not quite unique in this respect. Silicon and germanium are two elements that do the same thing.


History of acid-base reactions Basic and acidic materials exist in nature, so it makes sense that our forefathers could know what happens when they react. I would like to know when and how did they learn about this. In particular:

a. What is the earliest time in which people could witness acid-base reaction that causes fire or explosion (as in this video)?  What are the earliest written accounts of such reactions?

b. What are the earliest written accounts of neutralization reactions?

c. When were acid-base reactions intentionally used, e.g. for neutralizations or to create explosions?

There's a bit of a misconception on what actually happens in the video linked in your question:

The reactions between alkali metals, namely lithium ($\ce{Li}$) and sodium  ($\ce{Na}$) and water, diluted acid and concentrated sulfuric acid are not neutralisations, but redox reactions, that is transfer of an electron from one reaction partner to the other.

\begin{align}
\ce{2H2O &<=> 2H+ + 2OH-\\
2Na &-> 2Na+ + 2e-\\
2H+ + 2e- &-> H2 ^}
\end{align}

If you take this three reactions as a (mathematical) equation system, with which you're probably more familiar with, and sum it up, the total reaction is
\[\ce{2Na + 2H2O -> 2NaOH + 2 H2 ^} \]
To put this into words:

The metal is dissolved and forms metal hydroxide.
The resilting solution is alkaline (basic), hence the name alkali metals.
The reaction releases a lot of heat!
Hydrogen gas is formed.

Since the reaction releases a lot of heat, the hydrogen gas is burnt with the oxygen from the air, yielding water, This is the explosion that you noticed.

Taken these vigorous reactions into account, it is not astonishing that in nature, sodium and and lithium do not exist as metals, but in the form of various salts. It wasn't until 1807 that sodium metal was actually made in the laboratory, followed by lithium in 1818.

Turning back to your question on neutralisations, let's define what it actually means.

(I apologize if I'm getting to colloquial here!)

The term neutralisation is typically used to describe that two effects don't sum up to result in something stronger but rather incapacitate each other. In the case of acid-base reactions, this translates to:

\[\ce{effective\ acid + effective\ base -> something\ ineffective + water}\]

In water, a typical example would be the reaction of of a strong acid (such as hyrochloric acid) with equimolar amounts of strong base (such as sodium hydroxide):

\[\ce{HCl + NaOH -> NaCl + H2O}\]

Adding the right amount of a strong and corrosive acid to a solution of also corrosive and harmful caustic soda a harmless solution of sodium chloride (table salt) in water.
The only side effect of the reaction is that the solution can get pretty hot, but an explosion is not to be expected here.

Typically, usable explosions do not result from neutralisation reactions but from fast combustion reactions in with large volumes of gases are formed, possibly (but not necessarily) accompanied by the release of heat.

Taking into account that strong acids had to be made by men, I'd be surprised to find that these materials were wasted in simple neutralisation reactions.

You will however find historical accounts on the use of the acids or the bases. Nitric acid was used to separate silver from gold (Scheidewasser). Calcium oxide (burnt lime), the precursor of calcium hydroxide e.g. is used to make cement.


How do I solve such questions on paramagnetism and ferromagnetism? Match the type of magnetism given in Group I with the material given in Group II:

  $$
\begin{array}{ll|ll}
\hline
 & \textbf{Group I} & & \textbf{Group II} \\
\hline
\text{P} & \text{Ferromagnetic} & 1 & \text{Nickel oxide} \\
\text{Q} & \text{Ferrimagnetic} & 2 & \text{Sodium} \\
\text{R} & \text{Antiferromagnetic} & 3 & \text{Magnetite} \\
\text{S} & \text{Paramagnetic} & 4 & \text{Cobalt} \\
\hline
\end{array}
$$

  A) P-4, Q-3, R-1, S-2
  B) P-4, Q-1, R-3, S-2
  C) P-1, Q-2, R-4, S-3
  D) P-3, Q-2, R-1, S-4

It was taken from a mock test sheet for GATE in India.
With regard to whether a material is ferromagnetic or paramagnetic etc. All that I know is to check the magnetic domains and decide. However, with regard to such compounds I have been facing extreme difficulties. Is this something that is totally experimental and that I need to know by information? Or is there a certain intrinsic property of the elements which can help me to decide? If yes, please explain how to solve such questions.

I believe GATE is a university entrance exam in India, so they will not expect you to solve an extremely complicated equation to predict ferromagnetism or to memorize an infinite list of substances. Sadly, there is some element of rote memorization still lingering like a pest in the educational testing system.

Good news for you is that relatively few compounds and metals show ferromagnetism or ferrimagnetism. Find out the list of ferromagnetic metals from the web, you might end up with very few elements and oxides, such as iron compounds, iron oxide containing materials. Later you can attempt the multiple choice questions by the process of elimination. For example, in the given choices, choose the answers which show that magnetite is ferrimagnetic.

Lo and behold, there is only ONE choice which matches with Q-3!


Work done in cyclic process I have been told that when $\Delta V=0$, work done will be zero $(W=0)$. I know that, in cyclic process $\Delta V$ will be zero, since it's a path function and $\mid W \mid$ i.e. magnitude of work done will be the area enclosed by the curve. But my doubt is when $\Delta V=0$, shouldn't $W$ also be zero in such processes?

The reversible work is $W=\int{PdV}$ over the path of the process.  This integral is not zero for the cyclic process (involving V changing) even through $\Delta V$ from beginning to end of the process is zero.  So, whoever told you that $W = 0$ when $\Delta V=0$ is wrong.


Tanabe-Sugano diagram x-axis question While reading up on Tanabe-Sugano diagrams I noticed that many diagrams spread over the internet that all seem to come from this Wikipedia article: https://en.wikipedia.org/wiki/Tanabe–Sugano_diagram. Those graphs are all derived from this data: http://wwwchem.uwimona.edu.jm/courses/Tanabe-Sugano/TSspread.html
When I myself create the same graph from the same raw data the x-axis is not the same in regards to scale. My question is, since the creator of the Wikipedia graph left no further documentation, what step am I missing?

The step that was missing was basically just human error in the creating of the graph (wrong column for x) on top of a 10Dq/B to Dq/B conversion. The Wikipedia graph was correct and the raw data was correct, the creator of the graph on the right (me) made a mistake. My bad.


How to compare the boiling point of water, ammonia and hydrogen fluoride? According to the values of boiling points that I found on internet the order is as follows:

  $\ce{H2O}$ >  $\ce{HF}$ >  $\ce{NH3}$

I was expecting $\ce{HF}$ to have highest boiling point because F is the most electronegative atom hence should form stronger H Bonds.

Now, I think the factor here might be number of H Bond rather than strength of H Bond.

$\ce{H2O}$ have 2 lone pair and 2 H atoms, hence it can form 4 H Bonds with surrounding molecules.
Similarly  $\ce{NH3}$  is also capable of forming 4 H Bonds with surrounding molecules.
$\ce{HF}$ on the other hand is only capable of forming 2 H Bonds with surrounding molecules.

Still the above hypothesis doesn’t explain why the boiling point of $\ce{HF}$ is greater than $\ce{NH3}$. I’m not sure what am I missing here.

Though ammonia has three H atoms, it is not capable of forming four H-bonds

An important difference in terms of hydrogen bonding between ammonia and water, lies in the ratio between how many partial positive hydrogen atoms and how many lone pairs of electrons each have.

A water molecule has two partial positive hydrogen atoms and two lone pairs of electrons located on the oxygen atom. This implies that each water molecule can potentially have both its hydrogen atoms and both its lone pairs involved in hydrogen bonding.

But in ammonia, there is a shortage of lone pairs on the central nitrogen atom.
Hence it is capable of forming only two H-bonds.

And the difference in electronegativity is higher in $\ce{HF}$ than in $\ce{NH3}$.

Hence boiling point order is $\ce{H2O} > \ce{HF} > \ce{NH3}$


Rate of electrophilic bromination of toluene and ethylbenzene My book says that electrophilic bromination of toluene (methylbenzene) is faster than that of ethylbenzene. How is that possible?

I am doubting it because ethyl is a better $+\ce{I}$ group than methyl, so ethyl would increase the electron density in the benzene ring more than methyl. As a result, bromine, which is an electrophile, should react with ethylbenzene faster than methylbenzene.

So, why is the bromination of methylbenzene faster than ethylbenzene and please give the reaction mechanism, if possible?

First, I think it's probably important to note that the difference in $\text{+I}$ effect of a methyl vs. ethyl is very small. In principle, the ethyl is more electron-rich, however the strength of the inductive effect decays very rapidly with increasing distance. Ultimately, I suspect variance in steric factors, solvation, and electron-donating ability to the $\pi$-system of the ring by hyperconjugative overlap are the decisive factors, not the difference in induction.

If we consider sterics, the ethyl is obviously the bulkier group, and that may partially explain the difference in reactivity. The ethyl, moreso than the methyl, will hinder the approach of the electrophile, and may potentially raise the energy of the transition state and $\sigma$-complex intermediate slightly by:

(a) electrostatic repulsion toward the electrophile, particularly in the case of ortho- attack, and

(b) inhibit potential stabilization of the $\sigma$-complex intermediate by solvent molecules.

From the standpoint of electronic effects, I suspect the differences are marginal. To the extent of my understanding, the energies of $\ce{C-H}$ and $\ce{C-C}$ bonding $\sigma$ orbitals are very similar, and the efficiency of hyperconjugative overlap should consequently be similar as well. I've occasionally encountered claims that $\ce{C-H}$ bonds are slightly better $\pi$-donors, but I'm not aware of any conclusive evidence to that effect. If that is true, however, it might contribute to the observed disparities in reactivity.

For a quantitative approach, you may want to take a look at the substituent constants derived for various alkyl groups from the Hammett equation. The equation was originally developed with the aim of analyzing substituent effects on the thermodynamic equilibria of benzoic acid derivatives. However, since the reactivity of substituted benzoic acid is also affected by the mesomeric and inductive properties of ring substituents (albeit in the opposite sense from a benzene ring in an EAS reaction), I think the Hammett equation furnishes a reasonable proxy measure of at least the relative strengths of substituents as either donors or acceptors of electron density. I found one table of said substituent constants. (Keep in mind that those table values are for ionization of benzoic acid, so a negative value is actually indicative of an electron-donating effect, which is favorable for an EAS reaction but destabilizing to an acid.)

Finally, as to the reaction mechanism, I'm not going to reproduce it here. Every EAS reaction follows a very similar, schematic mechanism. For the mechanisms, take a look, for example, at the Wikipedia articles on electrophilic halogenation and electrophilic aromatic substitution more generally.


Understanding reason behind existence of chemical equilibrium Many chemistry reactions are reversible in nature.
In such reactions initially reactants react to form products but after some time products also react to give reactants back again. And an equilibrium is established where forward and backward reactions occur with equal rates.

If products are more stable, why will they react again to give reactants?

It is best to think of reversible reactions statistically

A good understanding about equilibrium can be had by thinking about the detailed molecular level processes involved at the molecule level.

Take, for example, the simple equilibrium between liquid water and water vapour in a closed vessel. At room temperature, the equilibrium clearly favours liquid water. But there is always some water vapour present (that's what we call humidity in the wider atmosphere). Why is there any vapour if the equilibrium favours liquid? The reason is that the water molecules don't all have the same amount of energy. some are travelling fast and some some are slow. A small percentage of the fast ones can escape the liquid and become vapour (ie gas-phase water molecules not liquid phase water molecules). In the gas above the liquid, the same is true. Not all the gas molecules have the same speed. Some are travelling slowly and, if they hit the liquid surface, stay in the liquid. Both processes are dynamic and are happening fairly fast. So molecules are constantly exchanging between the vapour and the liquid.

Equilibrium is not reached when everything stops: at room temperature there is a lot of kinetic energy shared out among the molecules of the system. Equilibrium happens when the rates of evaporation and liquefaction are the same. So the number of molecules per second going into the vapour from the liquid equals the number going the other way.

The same is true for other, more complex, reversible reactions. Equilibrium is not stasis but an average state resulting from dynamic reactions in both directions.

This point of view makes it clear why, even when one product is far more stable, the reverse reaction can still happen.


Why do we insert a numeric multiplier after group positions when naming an isomer? Question: Why do we insert a numeric multiplier after group positions when naming an isomer?

Take 2,3 dimethylbutane for example. Why do we add the numerical multiplier di when we've already explicitly mentioned two locations? Why don't we just use 2,3 methylbutane? This just doesn't feel intuitive.

take 2,3-dimethylbutane

In the case of different substituents on different or the same atom, we use a combination of the position and the name of the substituent, such as in

2-chloro-3-methoxy...
3-cyano-4-bromo..., etc.

Following this rule, 1,2-dimethoxyethane would be 1-methoxy-2-methoxyethane. Using the multiplier di, together with the positions, just shortens and simplifies the name and still is unambiguous.

The multiplier di explicitly denotes that there are two identical substituents at the given positions and not just one that bridges two centres.

An example for such a situation would be 1,4-methanonaphthalene (1). Note that even in this context, di can be used, as in 1,4:5,8-dimethanonaphthalene (2).


How would I find the molarity/concentration of a gas using pV = nRT? The problem I'm working on is:

At $\pu{1100 K}$ and a total pressure of $\pu{1 atm}$, $\ce{SO3}$ is $75.0\%$ dissociated.
$$\ce{2SO3(g) <=> 2SO2(g) + O2(g)}$$
Assume $\pu{1.00 mol}$ of $\ce{SO3}$ was present before dissociation.
Calculate the molarities of $\ce{SO3},$ $\ce{SO2},$ and $\ce{O2}$ at equilibrium.

I've already found the moles at equilibrium to be $\pu{0.25 mol}$ for $\ce{SO3},$ $\pu{0.75 mol}$ for $\ce{SO2},$ and $\pu{0.375 mol}$ for $\ce{O2}.$ I'm just not sure how to go about finding their molarities.

Supposing the dissociation process is isochoric, the total volume $V$ required for calculation of molarity at equilibrium of $i$th component $c_{\mathrm{eq},i}$

$$c_{\mathrm{eq},i} = \frac{n_i}{V}$$

can indeed be found from the ideal gas law using initial amount of undissociated sulfur trioxide:

$$V = \frac{n_0(\ce{SO3})RT}{p} = \frac{\pu{1.00 mol}\cdot\pu{8.21e-2 L atm K-1 mol-1}\cdot\pu{1000 K}}{\pu{1 atm}} \approx \pu{82.1 L}$$

You have found amounts of substances at equilibrium correctly, the rest is just math:

$$c_\mathrm{eq}(\ce{SO3}) = \frac{\pu{0.25 mol}}{\pu{82.1 L}} \approx \pu{3.0e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{SO2}) = \frac{\pu{0.75 mol}}{\pu{82.1 L}} \approx \pu{9.1e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{O2}) = \frac{\pu{0.375 mol}}{\pu{82.1 L}} \approx \pu{4.6e-3 mol L-1}$$


Where does magnetic inequivalence come from? As the question says, I'm having trouble understanding where does magnetic inequivalence in $\ce{^1H}$ NMR come from. All the sources I've found say that two protons are magnetically equivalent if all $J$ values to all other spins in the molecule are identical. So for example, in 1,2-dibromobenzene, the $\ce{H^3}$ and $\ce{H^6}$ are not magnetically equivalent because $J(\ce{H^3, H^4}) \neq J(\ce{H^6, H^4})$ and $J(\ce{H^3, H^5}) \neq J(\ce{H^5, H^6})$. I understand how to apply the rule, but I don't understand why does this affect the spectra. For example, assuming a not unusual $J_\mathrm{ortho} = \pu{7 Hz}$ and $J_\mathrm{meta} = \pu{2 Hz}$, then:
$J(\ce{H^3, H^4}) = \pu{7 Hz} = J(\ce{H^5, H^6})$, and
$J(\ce{H^3, H^4}) = \pu{2 Hz} = J(\ce{H^5, H^6})$. So they are both split the same way, but in reality they also split each other because of their magnetical inequivalence. This is what I find hard to grasp: if they are being in overall affected the same way, by the same type of atoms, in the same positions, how do they "know" that they are, in fact, different.

I guess related to this would be, why do all protons in ethene give the same signal? If they are also related differently between each other? ($\ce{H^1}$ and $\ce{H^{$1'$}}$ have different coupling constants to each $\ce{H^2}$ and $\ce{H^{$2'$}}$, cis and trans, respectively).

PS. Maybe related to this, although I can open a new question if it's not, of course, is why are diastereotopic protons different in NMR – i.e., how does the molecule know that "left" and "right" are different in, for example, (2R)-2-bromobutane, where the only difference between the protons in the $\ce{CH2}$ centers is that they "feel" the substituents to their left or right (looking at all its possible staggered Newman projections). Also, correct me if I'm wrong, these are not even chemically equivalent. Why would left and right make it so different (yet not so for enantiotopic protons)? I realize this might be a completely different questions, but something tells me they might be related?

Firstly, thank you for asking a very interesting question that certainly pushed my limits of knowledge (and others' as well). I apologise if I have made mistakes, it is a real possibility. All references are to Levitt's Spin Dynamics, 2nd ed. (an excellent book, by the way).

From a quantum mechanical point of view, the complexity in the spectrum of magnetically inequivalent nuclei should be considered the "norm". Fundamentally, in any strongly coupled system where the difference in Larmor frequency between two spins $\Delta \omega$ is much less than the coupling constant $J$, we expect to see complex spectra (see second half of this answer for more discussion).

Magnetic equivalence is a special case which leads to simplifications in the Hamiltonian and hence in the spectrum. As I mentioned in a comment, it can be shown that any coupling between magnetically equivalent nuclei "disappear". The word "disappear" merits further explanation, though. It is not true that the coupling constant $J$ goes to zero, for example. Neither is it true that it is averaged to zero by some sort of physical process, e.g. molecular rotation. It only means that the coupling term in the Hamiltonian has no impact on the expectation value of any quantum mechanical observable, or its evolution. Since an NMR spectra is simply us humans reading out some expectation values, the NMR spectrum itself is therefore unaffected by the coupling term between magnetically equivalent nuclei.

Having established that it is magnetic equivalence, not inequivalence, that is the special case, I would like to rephrase your question to: "Is there a physical explanation for why magnetic equivalence leads to simplification of the spectrum?"

The mathematical reason is given in Appendix A.9 of Levitt. I won't reproduce it here, so I will just mention it has to do with some operators which commute with others, and these only commute under very specific circumstances, i.e. the rules for magnetic equivalence that you have surely been taught (they must be chemically equivalent, plus either they have the same coupling constant to all other nuclei, or there are no other spins in the molecule).

Given the (relative) complexity of the mathematical explanation, I suspected that a physical explanation would not be particularly instructive. Levitt, though, does offer us something (p 461):

  The apparent disappearance of J-couplings between magnetically equivalent spins is quite surprising. A physical explanation runs as follows. For magnetically equivalent spins, the local magnetic fields, whatever their source, are exactly the same on the two spins. It follows that the motion of the two spin polarizations is identical. Whatever happens, the polarizations of the two spins are locked in the same relative orientation. There is, therefore, no need to incorporate a term in the spin Hamiltonian that takes into account the dependence of the energy on this relative spin orientation.

What this means is: it does not matter what the relative orientations of the spins are. For example, in ethene, it does not matter whether two spins are up and two are down, or whether all four are down, or whether all four are somewhere between up and down (i.e. a superposition of $\alpha$ and $\beta$, which is in fact the most likely case by far). The relative orientations can be anything, but the point is that the time evolutions of all four spins are synchronised. Therefore, the coupling terms between them – which are terms, scalar products, that depend on the relative orientation of spin vectors – may be omitted from the Hamiltonian without any impact on the observables of the system.

Returning to a mathematical interpretation, another sign of magnetic equivalence between nuclei $i$ and $j$ is when the labels $i$ and $j$ can be permuted in the spin Hamiltonian without changing it. This implies that $[\hat{\vec{I}}_i,\hat{H}] = [\hat{\vec{I}}_j,\hat{H}]$. From the standard formula

$$\frac{\mathrm d}{\mathrm dt}\langle Q \rangle = -\frac{\mathrm i}{\hbar}\langle[\hat{Q},\hat{H}]\rangle$$

it is immediately obvious that $\mathrm d\langle\vec{I}_i\rangle/\mathrm dt = \mathrm d\langle\vec{I}_j\rangle/\mathrm dt$, thus proving that the spins $i$ and $j$ move in concert.

Regarding ethene: you are correct that the cis and trans (and vicinal) couplings in ethene are different. However, all four of the protons in ethene are chemically equivalent and must be considered together as a group. It is not correct to look at two out of the four protons and call them magnetically inequivalent, without also taking into account the remaining two. In this case, it turns out that if you do the maths, the relevant commutators don't cancel out if you only consider two out of the four nuclei. However, when you consider all four nuclei together, the relevant operators fully commute.

Now, none of the simplifications above apply to magnetically inequivalent nuclei. These experience very tiny instantaneous differences in magnetic fields – hence the name – and therefore all the couplings in the system end up being reflected in the spectrum, as we should have expected a priori. Returning to your analysis ("the coupling constants are the same, so shouldn't they be the same?"), the issue is that the coupling constant $J_\ce{AX}$ is only one part of the coupling term in the Hamiltonian, which is $2\pi J_\ce{AX}(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$. Although $J_\ce{AX}$ is indeed equal to $J_\ce{A'X'}$, the dot product $(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$ is in general not equal to $(\hat{\vec{I}}_\ce{A'}\cdot\hat{\vec{I}}_\ce{X'})$.

Warning: as with the ethene case, one cannot look at individual A and A' nuclei, because they are symmetry-related. You have to look at both of them together as a group. Once you do that, though, the spectra become possible to analyse.

Can magnetic inequivalence be explained without any QM?

This was the original crux of my answer, but I realised it is slightly tangential. I leave it in, however, since there is a possibility it may be useful.

The Hamiltonian of a spin system can be written as:

$$\hat{H} = \sum_i \omega_i \hat{I}_{iz} + \sum_i\sum_{j<i}2\pi J_{ij}(\hat{\vec{I}}_i\cdot\hat{\vec{I}}_j) \tag{1}$$

where $i$ ranges from $1$ to $n$, the number of spins in the system, $\omega_i$ is the Larmor frequency of spin $i$, and $J_{ij}$ is the coupling constant between nuclei $i$ and $j$ in $\pu{Hz}$ (hence the factor of $2\pi$).

Under a particular condition - when $|\omega_i - \omega_j| \gg J_{ij}$, or "weak coupling" - the term $$(\hat{\vec{I}}_i\cdot\hat{\vec{I}}_j) = \hat{I}_{ix}\hat{I}_{jx} + \hat{I}_{iy}\hat{I}_{jy} + \hat{I}_{iz}\hat{I}_{jz} \tag{2}$$ may simply be approximated by $\hat{I}_{iz}\hat{I}_{jz}$. This is called a "secular" approximation, cf. Appendix A.6 of Levitt. The reduced Hamiltonian is then

$$\hat{H} = \sum_i \omega_i \hat{I}_{iz} + \sum_i\sum_{j<i}2\pi J_{ij}(\hat{I}_{iz}\hat{I}_{jz}) \tag{3}$$

Since the Hamiltonian only consists of $\hat{I}_z$-type operators, the eigenstates of the Hamiltonian will simply be products of the eigenstates of $I_{iz}$:

$$\Psi = \psi_1\psi_2\cdots\psi_n \tag{4}$$

where $\psi_i$ is either $\alpha_i$ (up spin) or $\beta_i$ (down spin). The energies of such systems are then found by using $\langle\alpha_i|\hat{I}_{iz}|\alpha_i\rangle = +1/2$ and $\langle\beta_i|\hat{I}_{iz}|\beta_i\rangle = -1/2$. In general, therefore, the first term in the Hamiltonian will give some linear combination of $\omega_i$ (with each coefficient being $\pm 1/2$) and the second term in the Hamiltonian will give another linear combination of $2\pi J_{ij}$ (with each coefficient being $\pm 1/4$).

The allowed transitions correspond to a difference between these energies, and we regain the simple picture that we learn in introductory NMR: peaks at $\omega_i$ (or their equivalent in the chemical shift scale), and splitting between peaks of exactly $2\pi J_{ij}$ (or $J_{ij}$ in the Hz scale). Often, this is taught using a simplified model of considering each spin to be like a bar magnet, i.e. it is either up or down spin. With this model one may achieve some understanding of first-order spectra, and can rationalise the $n+1$ rule, for example.

However, this is an assumption which leads to precisely equation $(4)$. By assigning a distinct "spin" to each nucleus, we discard the possibility of quantum entanglement, i.e. we discard any possible quantum states that are not simple direct products. For example, consider the two-spin wavefunction

$$\Psi = \frac{1}{\sqrt{2}}(\alpha_1\beta_2 + \beta_1\alpha_2) \tag{5}$$

This is a perfectly valid wavefunction, but it is not possible to assign an up or down spin to either of the two nuclei. Equivalently, by assigning up or down spins to each nucleus, we are ruling out the possibility of any such states being important.

Now, this is not a problem as long as we are in the weak coupling regime. It may be helpful to think of the discarded term $(\hat{I}_{ix}\hat{I}_{jx} + \hat{I}_{iy}\hat{I}_{jy})$ as a perturbation to the simplified Hamiltonian in equation $(3)$. The unperturbed eigenfunctions of the simplified Hamiltonian are given by equation $(4)$. If the perturbation is small, i.e. if we are in the weak coupling regime, then the eigenfunctions of the full Hamiltonian are very close to that of the simplified Hamiltonian, and hence our simplified model (with discrete up and down spins) works. However, if the perturbation is large, i.e. strong coupling, then the simplified model fails. The eigenstates are no longer the same, the energy eigenvalues are modified, and hence the transition energies (which depend on the energy eigenvalues) and intensities (which depend on the eigenstates) are totally different.

Having two nuclei which are chemically equivalent, i.e. having the same chemical shift, is a case which simply not play into our simplified picture above. Since they have the same chemical shift, $|\omega_i - \omega_j| \equiv 0$ and hence weak coupling is never true. Note that chemically equivalent nuclei still possess a coupling constant $J_{ij}$ to each other and hence, in fact, chemical equivalence represents the most extreme case of strong coupling.

Fortuitously, when two chemically equivalent nuclei are also magnetically equivalent, it can be shown that the coupling is simply not manifested in the spectra (Appendix A.9, Levitt). Therefore, we can still use our simplified model, but simply with the caveat that "equivalent protons do not couple to each other". Strictly speaking the model is already incorrect, because the eigenstates of the system are no longer products of individual spin functions (see my answer here).

This is, however, not the case for magnetically inequivalent nuclei. Therefore, we have to throw our simplified model out - it just does not work here. It is therefore – to the best of my knowledge – not possible for one to explain the physical origin of magnetic inequivalence based on simplified arguments such as how "spin A sees spin B".


Can nuclear shielding affect relaxation time? Can nuclear shielding affect the transverse relaxation time, i.e extend the relaxation time the more the nuclei is shielded?

I'd say yes, but not by any detectable amount. In some relaxation mechanisms (e.g. dipolar relaxation, a major relaxation pathway for $\ce{^1H}$), the rate of relaxation varies with $\langle B_\mathrm{loc}^2\rangle$, i.e. the magnitude of the random fields that the spin experiences due to the magnetic momenta of all the other spins. The magnitude of any field that a nucleus experiences will be attenuated by the induced magnetic field arising from its own electrons, so it stands to reason that – all else being equal – a more shielded nucleus will have a smaller $\langle B_\mathrm{loc}^2\rangle$, and a longer relaxation time, than a less shielded nucleus. This applies to both $T_1$ and $T_2$.

Bear in mind, however, the most obvious effect of nuclear shielding, which is to change the resonance frequency of the nucleus. The spread of frequencies is on the order of $10^{-6}$ relative to the strength of the external magnetic field. If we crudely estimate that the spread in $T_1$ and $T_2$ arising from sheilding differences is also on the ppm level, that corresponds to ~µs levels of difference, which I strongly doubt is detectable.

In any case, for two protons that have different chemical environments, there would be other factors which affect the relaxation times to a much greater extent.


Way to establish the acid/base properties of a compound from its chemical formula? Is there a way to figure out the acid/base properties of a compound just by looking at the chemical formula?

$\hspace{4.3cm}$

Consider the two compounds above. Both have the formula $\ce{C5H10O2}$, yet one is about 20 orders of magnitude stronger an acid than the other. Without knowing the general structure of a molecule (what functional groups it is comprised of), suffice it to say no, you cannot know how acidic the molecule is.


When to disconnect tubing in vacuum filtration? The UCLA site says in step 6 to disconnect the vacuum tubing before turning off the vacuum and the second link from Leeds says in step 11 to turn off the vacuum source first and then disconnect the tubing.

I've been taught both ways and am confused about the difference in methods.

When should one actually disconnect the tubing and what is the rationale for each method?

If you were to turn off the pump while leaving the object attached then if you have an oil filled pump or a water pump then you will have a suck back which will either damage the pump or contaminate the thing you are filtering, or if you are very unlucky both.

If you have a house vacuum system then you should close the valve / tap between your filtration and the vacuum manifold before you remove the hose to your flask. If you do it the other way around then you will interfere with the use of the house vacuum by others.

I think you need to know if you have your own vacuum pump or if you are using a house vacuum system where a centralised pump serves many people.


Enolizablity of pyruvate and lactate I understand that to create an enol, you simply eliminate a proton alpha to the relevant $\ce{C=O}$ and then tack the proton onto the $\ce{O}$ and there you have it — an alkene and an alcohol — or enol.

Given this, between pyruvate and lactate, which is more enolizable?

I'm thinking that pyruvate doesn't even have an enol tautomer … there are no protons alpha to a carbonyl group. Sure, there is the carboxylate group … but that doesn't have anything to do with keto-enol tautomerism (does it?)

You're right and wrong.  Your right that pyruvate anion has a higher enol content than lactate anion, at least that would be my guess (pyruvate anion has 3 enolizable hydrogens, while lactate anion has one and needs to destroy the carboxylate resonance in order to create the carbonyl required for enolization...read on).

But I think that lactic acid could enolize, after all esters can enolize.  I suspect that the enol content in the lactate carboxylate anion would be non-measurable - you would need to disrupt the carboxylate anion resonance in order to generate the carbonyl required for enolization.  If we use 36 kcal/mol as an estimate of the carboxylate resonance energy and say that for every 1.4 kcal/mol we can shift an equilibrium by a power of 10, then we might expect 1 enol molecule in every $10^{26}$ lactate carboxylate molecules.  In the lactic acid itself, we have no carboxylate resonance to disrupt, the enol content would be higher.


Hydration, Coordinate Bonding, and Reduction Potential in Groups 1 and 2 Elements Whenever an alkali metal such as Lithium is hydrated/solvated, coordinate bonding occurs and it is the high hydration enthalpy of lithium which makes it the strongest reducing agent in aqueous solution. My question is, if lithium is hydrated and becomes surrounded by water molecules in coordinate bonding, then how does the overall reaction of adding lithium to water produce LiOH and H2(gas) ?  Is the coordinated ion an intermediate product?

I also am wondering what happens to the "lost" electrons from the metal in coordinate bonding.  Take Beryllium dissolving in water as an example. As it is hydrated it will lose the 2 valence electrons in its outer orbital and then it hybridizes its orbitals into 4 empty orbitals so that 4 molecules of water, supplying 1 lone pair each, can bond with the positive beryllium ion. Where did the original 2 electrons go?  How do these electrons vanish in order to allow the hybridization of the 4 empty orbitals for water molecules to bond with?

And if Lithium is the strongest reducing agent in aqueous solution I assume it means it is reducing water itself, in order to create lithium hydroxide and hydrogen gas, but again, how does that come about from the coordinate bonding and the hydrated lithium? Coordinate bonds preserve the charge of the cation but I am not quite sure how this is so.

I have included an image below from a website that attempts to explain it, but I am confused about to where the original electrons go.

The website from which I took the image is:  https://www.chemguide.co.uk/inorganic/group2/beryllium.html#top

This answer addresses your question regarding the reaction

$$\ce{2Li}(s) + \ce{2H2O}(l) \rightarrow \ce{H2}(g) + 2\ce{Li+}(aq) + 2\ce{OH-}(aq)$$

  Whenever an alkali metal such as Lithium is hydrated/solvated,
  coordinate bonding occurs and it is the high hydration enthalpy of
  lithium which makes it the strongest reducing agent in aqueous
  solution. My question is, if lithium is hydrated and becomes
  surrounded by water molecules in coordinate bonding, then how does the
  overall reaction of adding lithium to water produce LiOH and H2(gas) ?
  Is the coordinated ion an intermediate product?

The reaction mechanism for this reaction involves some radical chemistry:

Lithium hydroxide is soluble in water, and will exist as separate ions as long as the solution is not saturated. As soon as the ions are generated, they will be solvated by nearby water molecules. The solvated ions are not intermediates - they might better be viewed as products.

The lithium metal is very reactive due to its free electron, and will react quickly. Besides, Li metal is neutral, and will not immediately form any strong interactions with the polar water molecules, I would think.

  And if Lithium is the strongest reducing agent in aqueous solution I
  assume it means it is reducing water itself, in order to create
  lithium hydroxide and hydrogen gas, but again, how does that come
  about from the coordinate bonding and the hydrated lithium? Coordinate
  bonds preserve the charge of the cation but I am not quite sure how
  this is so.

Indeed, in the course of the reaction water is reduced and Li is oxidized. But I don't think the Li metal (neutral atomic Li) is solvated by water. The reaction with water takes place extremely quickly. And as mentioned above, it is not intuitive to me why water would want to solvate a Li atom in the first place, due to it being non-polar.


Why is distillation not a viable way to separate ammonia from water? Since the steam pressure of ammonia is higher than that of water, I would expect distillation to be a reasonable way of separating a mixture of both.

However, in industrial applications known to me there's always a step of stripping the ammonia chemically from the vapor. Why is this necessary?

Having embarrassed myself with my first answer, let me try this again. Ammonia and water hydrogen bond exceptionally well. In fact, the $\ce{O-H...N}$ value is 29 kJ/mol while the $\ce{O-H...O}$ value is 21 kJ/mol. (Ref) So, if there is any water around, ammonia would be happy to become solvated as they recondense.

Also, ammonia is a weak base and reacts with water (which is not usually the case when you're doing a distillation). $$\ce{NH3 + H2O <=> NH4+ + OH-}$$ So, if there's any water around as the ammonia is distilling and they recombine, they would also react which would make it difficult to remove the last of the water. (Some reaction would occur because the lowest energy point of the equilibrium is not at reactants or products.)

So, as noted in the comments, distillations to prepare dry ammonia use a scrubber to remove the water.


How do you explain pKa to non-professional? Let's say we have substance $\ce{A}$, which is mixed with substance $\ce{B}$ to improve shelf-life because $\mathrm{p}K_\mathrm{a}$ of the substance $\ce{A}$ is $7.9$ and in mix the $\mathrm{pH}$ is $5.2$.

Does this mean that in the solution with $\mathrm{pH}$ near $8$ the substance $\ce{A}$ has multiple molecules in neutral state and not dissociate, thus it forms precipitates?

Does $\mathrm{p}K_\mathrm{a}$ mean that most of the molecules of the substance $\ce{A}$ are in nondissociated state?
 Does lowering $\mathrm{pH}$ in this case causes substance $\ce{A}$ to dissociate?

Are bold parts right? Please explain in plain English without the use of external links.

pKa and pH are related concepts but often confused. pKa is a property of a compound that tells us how acidic it is. The lower the pKa, the stronger the acid. pH is a property of a particular solution that depends on the concentrations and identities of the components.

For this discussion, I'm going to use the terms protonated and deprotonated to mean that a compound is associated or dissociated with a proton.

Based on the relationship between the pKa of a compound and the pH of a solution, we can predict whether a compound will be protonated or deprotonated. If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated.

A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.

Given the information you provided, if compound A (with pKa 7.9) is in a solution of pH 5.2, compound A will be in the protonated state.

Without knowing anything about the identities of A and B, the following is speculation. Most likely compound A is a base and compound B is an acid. The protonated version of A (its "conjugate acid") is more stable for long term storage than A itself.

Another possibility is that compound A is an acid and B is a stronger acid. If the deprotonated version of A is unstable, then compound B is added to ensure that if the mixture came in contact with a base, something more acidic than A would be present to react with the base.

EDIT BASED ON NEW INFORMATION:
As Mateus B said, lidocaine will be protonated by HCl at the amine nitrogen. This is likely done for two reasons. First, as its hydrochloride, it will be more water soluble. In solution, the compound will dissociate into protonated lidocaine cation and chloride anion. Second, as the nitrogen is protonated, it is less reactive to the environment. Over time, amines can react with oxygen to form the corresponding N-oxide and/or can absorb carbon dioxide.


Are the energies of orbitals that are not filled in an atom, equal? In a hydrogen atom only 1s orbital is filled. Does this mean that the energies of all other orbitals are equal? If so,why?

Sure, if they are unoccupied they contribute nothing to the total energy of the system. Remember that orbitals are mathematical constructs and only have meaning with regard to their energy when being physically occupied by an electron that adheres this mathematical description. The energy required to fill such an orbital however differs (vastly) from orbital to orbital and nucleus to nucleus (see for example http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf).


How or why can carboxylic acids be protonated? In the wikipedia article "Step-growth Polymerization/Kinetics", it is stated that:

  The simple esterification is an acid-catalyzed process in which the
  protonation of the acid is followed by interaction with the alcohol to
  produce an ester and water

My question is: How are acids even protonated? Aren't they the ones supposed to be giving out protons? How are they the ones accepting the protons?

A general acid-base reaction is one of the kind $(1)$.

$$\ce{\underset{\text{acid}}{HA} + \underset{\text{base}}{B} <=> \underset{\text{conj. acid}}{HB+} + \underset{\text{conj. base}}{A-}}\tag{1}$$

All that the acid needs to react as an acid is some hydrogen atom bonded to something, and all that the base needs to react as a base (although it is not shown here) is a lone pair which can accept the acid’s proton. Depending on the relative strength of the acid and the conjugate acid, this equilibrium will be shifted. If the acid is stronger, it will be shifted towards the product; if the conjugate acid is stronger, towards the reactant. If the acidity are within reasonable range of each other, you can always expect a catalytic amount of the stronger acid to be protonated.

The above was the general introduction; now how does this apply to your question?

The ‘acid’ you are talking about is a carboxylic acid. These have the functional group $\ce{-COOH}$ and each of the oxygens in there has two lone pairs. Thus technically, a carboxylic acid could be protonated up to four times.[1] We only need one further protonation, so we can define the carboxylic acid to be compound $\ce{B}$ above — the base in the acid-base reaction.

In fact, acid-catalysed esterification is typically catalysed by the addition of a small drop of sulfuric acid. An average carboxylic acid has a $\mathrm{p}K_\mathrm{a}$ value of $\approx 5$, sulfuric acid $-3$. The hydronium ion $\ce{H3O+}$, that forms if water takes the place of the base in $(1)$, has a $\mathrm{p}K_\mathrm{a}$ of $-1.7$ or $0$, depending on who you ask. Other alcoholic oxonium ions have similar $\mathrm{p}K_\mathrm{a}$ values. The lower the $\mathrm{p}K_\mathrm{a}$, the stronger the acid. Thus, we can assume that most carboxylic acid molecules remain as $\ce{-COOH}$, while the catalytic amounts of sulfuric acid (practically completely deprotonated once) protonated a few solvent molecules at equilibrium. The stronger acid — sulfuric acid — practically governs that none of the weaker acid — the carboxylic acid — can lose its proton.[2]

Even though it is equilibrium, a few of the acid molecules — the ‘next-weakest links in the chain’ — will also be protonated at a given point in time, even though we can expect the $\mathrm{p}K_\mathrm{a}$ value of a protonated carboxylic acid to be rather low. This small proportion is enough to get the reaction going.

Notes:

[1]: However, each successive protonation introduces a positive charge which always makes further protonation less favourable. A positively charged compound is, generally, also more inclined to transfer hydrogen to something else, i.e. is more acidic.

[2]: This can actually be extended. There are a number of superacids with even lower $\mathrm{p}K_\mathrm{a}$ values — in fact, those values are so low that they can no longer accurately be calculated and instead a different value, the Hammett acidity is used. Some of the strongest superacids are even able to protonate methane which is unsuspiscious of providing a lone pair an acidic proton could interact with. Hence Ivan’s comment of:

  Everything can be protonated, given enough effort.


Why do electron-rich aryl groups have greater migratory aptitudes? In the following reaction there are three possible aryl groups that can migrate: an unsubstituted phenyl group, a p-methoxyphenyl group, and a p-nitrophenyl group:

Why does the electron-rich p-methoxyphenyl group migrate instead of the phenyl or p-nitrophenyl group, especially when it leads to a carbocation that is less stabilised (since it will no longer be conjugated with the stabilising p-methoxyphenyl group, cf. the crossed-out arrow)?

The answer varies slightly depending on whether the migration is one-step or two-step via an intermediate.
In either case, the respective transition state or intermediate looks like this:

You can see that the middle structure would be stabilized most by a group that is able to stabilize the ring via resonance, for example, an alkoxy substituent.
If the mechanism is single-step, then that substitution would stabilize the transition state and make this migration the fastest.
If the mechanism is two-step, then we can invoke the Hammond postulate to say that the transition state leading to the higher energy intermediate will resemble it in structure and thus also be stabilized by the same kind of substitution, leading to the same conclusion.
We therefore suggest that the propensity of migration is X=OMe > X=H > X=NO₂.


Which allotropes of sulfur exist naturally? I'm writing a paper on chemistry and I would like to know which allotropes of sulfur exist naturally. I've searched on the Internet but I haven't found anything about natural sulfur allotropes except this book(Ref. 1). Does anyone know the answer?
Reference

Steudel R., Eckert B. () Solid Sulfur Allotropes. In: Steudel R. (eds) Elemental Sulfur and Sulfur-Rich Compounds I. Topics in Current Chemistry, vol 230. Springer, Berlin, Heidelberg, DOI: doi.org/10.1007/b12110

The alpha and gamma form of cyclo-S8 are the only forms to occur naturally. Besides the reference provided in the question, you can also find more information in the wikipedia article of "sulfur allotropes".


How is the central carbon in 3‐bromopentane‐2,4‐diol chiral? My teacher told us the middle carbon in 3‐bromopentane‐2,4‐diol is chiral:
$$\ce{CH3-CH(OH)-\overset{\star}{C}H(Br)-CH(OH)-CH3}$$
Since there are two identical groups around it, I would guess it should not be chiral.
It was a pre-recorded class and I was not able ask this question. Has my teacher made a mistake?

Carbons #2 and #4 are chiral centres themselves. There are four possible diastereoisomers:

If C2 and C4 have different configurations, which is the case in molecules A and D, then C3 is termed a pseudoasymmetric centre and labelled with a small 'r' or 's'. These diastereomers are meso compounds: although they contain chiral centres, they are not chiral, because they have an internal plane of symmetry.
On the other hand, in molecules B and C, C2 and C4 have the same configuration and C3 is not a chiral centre (and not a pseudoasymmetric centre either). These diastereomers are chiral, though: the chirality comes from C2 and C4.


Can a reference electrode be inserted into an electrochemical cell horizontally? If I have an opening on the side of my electrochemical cell, I'm wondering if I can use this to insert a reference electrode. It is normally used to insert an L-shaped tube where you place the reference electrode, and is connected to luggin capillaries. I'm wondering if we can take out the entire connection with the capillaries and replace it with only the reference electrode (sans capillaries). This is in a 3-electrode system, so if I could insert the counter electrode in this way, that would work too. In my setup, I cannot do this with the working electrode.

From what I've read, it's good practice to not have the reference electrode vertical because of bubbles on the bottom of the frit (so most tilt it slightly), but I haven't read anything about a sideways electrode.

As long as there's nothing about the design of the reference electrode (RE) that will make it malfunction when oriented horizontally, it should be fine.  Obviously if there's free liquid inside the RE that will slosh away from the frit or the electrode internals (or spill out onto the lab bench!) when you tip it, then that won't work well at all. The concern about bubbles collecting on the bottom surface is very real, especially if you're not stirring very vigorously.

One practical downside to inserting the CE directly like this is that the frit of the RE will be liable to clogging/degrading much more rapidly than if you use a Luggin. Also, the composition of the internal electrolyte of the RE will change more rapidly if directly exposed to your working solution without a Luggin.

Of course, depending on your setup, this will allow you to get your reference much closer to your working electrode, reducing the $iR$ drop in your measurement.


Safe endothermic reaction that gets below freezing? I have a project/idea I am working on but I am only 14 years old and I need some help.

I'm looking for an endothermic reaction that gets below freezing, but the chemicals before the reaction and after the reaction is over are safe to touch and safe for the environment. The reaction can be harmful to plant life, but only for a short time before it decays or whatever needs to happen for it to be safe. I do not want it to have long term effects.

Additionally, two important parts are that these substances need to be easy to obtain and need to not to be a fertilizer.

The reaction of baking soda and citric acid (both are easy to come by and safe) in water does cool the water noticeably. I have tried it multiple times, and while it does not reach freezing, it will reliably reach $6\ ^\circ \mathrm{C}\ \left(43\ ^\circ \mathrm{F}\right)$, which is cold to the touch.

  Add 4 teaspoons each of sodium bicaronate (baking soda) and citric acid to a cup. Then add the smallest amount of water you can and start stirring!

You can tell a chemical reaction is occurring because a gas is produced:

$$\ce{3NaHCO3 + H3C6H5O7 -> Na3C6H5O7 + 3CO2(g) + 3H2O}$$

But, wait! We want a temperature below freezing! It might seem a little like cheating, but the reaction still works. Carbon dioxide bubbles are still produced, and the temprature gets down to around $-6\ ^\circ \mathrm{C}\ \left(21\ ^\circ \mathrm{F}\right)$.

If you do not have a thermometer, you can prove the second version is colder than the freezing point of water by doing the following:

Get a small, flat piece of wood and dampen the top surface of it.
Do the reaction in a glass, using 4 teaspoons each baking soda and citric acid and replacing the water with crushed ice.
As the temperature drops, the water between the wood and the glass will freeze, freezing the glass to the wood. If you lift the glass, the wood will stick to it.


Acidity of o-halobenzoic acids Rank the following compounds in decreasing order of acidity:

2-Fluorobenzoic acid
2-Chlorobenzoic acid
2-Bromobenzoic acid
2-Iodobenzoic acid

My approach
Since the ortho effect is proportional to the size of the substituent, it will be the greatest in the case of iodine and smallest in the case of fluorine. So, the molecule will be more non-planar in 2-iodobenzoic acid.
So resonance will be inhibited in all the above compounds. But what will happen to the acidity?

Acidity data:

The true order of acidity is $\ce{Br = I > Cl >> F}$. Relative to unsubstituted benzoic acid, all of the ortho-halogen substituted benzoic acid derivatives have a greater acidity, clearly having some stabilising effect on the resulting benzoate anion.

  Data retrieved from Journal of Research of the National Bureau of Standards, 1958, 60, p569 - research paper 271. $\mathrm{p}K_\mathrm{a}$ values measured in water.

Rationalisation:

Polar effects All of the halogens exert a negative inductive (-I) effect, removing electron density from the ring. This makes the benzoic acid derivatives more acidic in all 4 cases since the resulting conjugate base is stabilised to a certain extent by the presence of the electron withdrawing halogen.
Steric effects As the halogen gets bigger, strain may force the carboxylate out of planarity with the aromatic ring. Whilst fluorine is roughly the same size as the hydrogen atom, bromine and iodine are both sterically more demanding.
Internal H-bonding This is suggested by another answer below, whilst this has been studied and shown to operate for protic groups such as alcohols and amines, in halogens it has little impact on stabilising the negative charge.

Overall, steric and polar effects operate in tandem to give an overall acidity for the benzoic acid. It's perhaps not intuitive which one will dominate (e.g. fluorine is far more electronegative, but also far smaller than iodine, so which is more acidic).

The table below, taken from the cited paper, attempts to separate out steric and polar effects through calculations on both the acid and its conjugate base. Sadly, they only look into $\ce{F}$ and $\ce{Cl}$ (no $\ce{Br}$ or $\ce{I}$), but it can be clearly seen that:

Hydrogen bonding is making no significant contribution to the acidity of the substituted benzoic acid
Polar and steric effects are both in operation, with sterics being far more important in the ground state (the carboxylic acid) and polar effects dominating for the conjugate base (the carboxylate).

  Table copied from New. J. Chem., 2004, 28, p67


Is atomic weight universal? Are we advanced enough to calculate the abundance of an element throughout the universe or is our atomic weight only based on the occurrence of an element in our planet alone?

There is no "or".

Yes, we are advanced enough to calculate the abundance of elements throughout the universe, albeit somewhat hypothetically. (But then again, the estimates here on Earth are also largely indirect. The radius of Earth is 6400 km, and the deepest borehole pierced only 12 km.)
Yes, if you look at the Periodic table on the wall of your chemistry class, the atomic weights listed there are only based on the isotope abundances in our planet alone.

True, the "earthly" and "Universe-based" atomic weights rarely differ by much. One notable exception is argon (earthly product is mostly $\rm^{40}Ar$, cosmic version is mostly $\rm^{36}Ar$). More subtle discrepancies are numerous. They are well-known, documented, and routinely used to prove that a particular rock is a visitor from space (even if nobody saw it fall).

So it goes.