What is the true meaning of Gibbs energy and chemical potential? After doing some research and reading I found some problems which I will try to state as clearly as possible.

The definition of Gibbs Free Energy says "the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition. (I have taken the definition from the wikipedia). The equation of the Gibbs energy $$ G= H -TS$$ is also clear.
But the way I understand it is this: the maximum non-expansion work , like transporting an electron, breaking a chemical bond, moving real life things etc.  which can be obtained from a system . These things made sense during the study of Second Law of thermodynamics but as I moved to Chemical Equilibrium these concepts began to tremble (for me).
In elementary classes it is said that Equilibrium is a state where the composition of reactants and products do not change over time, but in higher chemistry classes it is said that equilibrium is a condition corresponds to $$ \Delta G = 0$$ So, my first question is how do these two concepts mean same thing?  The next thing which causes problem is chemical potentials. If we adhere to the formal meaning of potential i.e. something which is stored and can be used when proper conditions are met, so chemcial potential would mean the potential of substance to react and again this is related to Gibbs energy whose definition I gave above. So, how do chemical potential and Gibbs energy can have any relation?
A question which is off topic over here but I want to mention it, why do we bother so much about standard  things like $$ \mu_\mathrm A = \mu_\mathrm A^\circ + RT \ln(p_\mathrm A)$$ why we wanted to express it in that standard (that little circle) form ?
I want to make myself clear that the conception of Gibbs energy was quite clear to me in the context of thermodynamics, we simply meant it to be the work which can be extracted from a substance, but it all-pervading use has made me to doubt myself, just like in mathematics the number $\mathrm e$ appears in odd places. Even if Gibbs energy (according to the understanding that gave above in bold) is appearing mathematically then also it would be having some  physical meaning because Thermodynamics and Equilibrium are natural sciences and not the mathematics.

Thank you, any help will be much appreciated.

Your first question:

Before the second law was understood it used to be thought that the maximum amount of work that could be extracted from a reaction was $-\Delta H$ but many experiments showed that this was not the case. It is true that, according to the first law, (the law of conservation of energy),  that external work done must be equal to the loss of energy of the system, unless some heat is given to or taken from the surroundings. This was first understood by Gibbs.  In an isothermal reaction working under  reversible conditions, the heat absorbed from the surroundings is $T\Delta S$, if this is positive then the work done will be even greater than the heat of reaction and so heat will be taken from the thermostat.

Second question:

The $\Delta G$ you mention is actually the gradient of the free energy with extent of reaction $(\partial G/\partial \xi)_{T,p}$ at constant temperature and pressure. This differential is sometimes written as $\Delta G'$. The equation in general is $\Delta G'=\Delta G^\text{o}+RT\ln(Q)$ where $Q$ is the ratio of partial pressures, for example, in a gas phase reaction.  At equilibrium a plot of $G$ vs $\xi$ reaches a minimum and the gradient is zero, i.e. $\Delta G'=0$ then $\Delta G^\text{o}=-RT\ln(K_p)$  where $K_p$ is the equilibrium constant and is used instead of $Q$ at equilibrium. This reaction tells us how the position of equilibrium can be determined in terms of standard state free energies of reactants and products at 1 atm pressure.

(The extent of reaction is zero when only reactants are present and is one when one mole of reactants have changed to product)


Stereoisomers of 1,3,4,6-tetramethylcyclohex-1,4-diene How many stereoisomers are possible for following compound (1,3,4,6-tetramethylcyclohex-1,4-diene)?

My Approach
the 1,4 methyl groups lie in the plane, whereas 3,6 are possible chiral centers. However it doesn't seem like enantiomers are possible due to centre of symmetry, one structure can be meso. Apart from this I don't see any possibility of Geometric isomers as well. However the answer given is:

3

What am I missing in this?

Your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, does not have line symmetry (plane of symmetry), and hence $\ce{C}$3 and $\ce{C}$6 are chiral centers. Thus, maximum possible stereoisomers are $2^2 = 4$. However, there is a symmetry element in the structure, which is order 2 ($180^\circ$) rotational symmetry (also called point symmetry). Therefore, one stereoisomer, $(3S,6R)$ would be superimposable with it's mirror image $(3R,6S)$, thus it is a meso-isomer:

To understand this, I put "$\bullet$" symbol to indicate relevant hydrogen is above the plane of paper. And, "$\bf{.}$" symbol to indicate relevant hydrogen is below the plane of paper. Thus, first two structures have both 3,6-dimethyl groups below the plane of the paper while the third structure has 3-methyl group below and 6-methyl group above the plane of the paper. Accordingly, all chiral centers have CIP nomenclature as indicated under each structure:

The first structure is $(3S,6S)$ and the second structure is $(3R,6R)$, thus they are non-superimposable mirror images as indicated by the mirror. Therefore, they are enantiomers. To make all the groups to coincide, you may need to rotate the second structure $180^\circ$ horizontally along the axis going through $\ce{C}$3 and $\ce{C}$6. However, after the rotation, the two methyl groups at $\ce{C}$3 and $\ce{C}$6 will be now above the plane. Therefore, these two structures cannot be superimposable.
On the other hand, the third structure is $(3S,6R)$ ($\ce{C}$3 methyl is below the plane and $\ce{C}$6 methyl is above the plane). To the mirror image of this molecule, if you follow the same operations as previous steep and do another $180^\circ$ rotation on the plane of the paper (or vise versa), you'd get the same compound.  That means, the third compound is not optically active and called meso-isomer. This can be found easily by looking at the two chiral centers: They are mirror images of each other, $(3S)$ and $(6R)$ or $(3R)$ and $(6S)$.

Therefore, your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, has three stereoisomers, but only two of them are optically active.


Can I disable "Display visual cues" by default in Avogadro? These huge arrows are very annoying to me. I disable them every time I open Avogadro. Is there a way to disable them permanently?

Sure. There's a big checkbox in the Navigate Tool settings:

When you quit, the settings should be remembered for next launch.

If not, please e-mail the Avogadro mailing list and we can figure out what's going on.


Why is Coupled Cluster not variational? It has been noted in several sources (e.g. J. Romero et al. Strategies for quantum computing molecular energies using the unitary coupled cluster ansatz. arXiv:1701.02691 [quant-ph]) that one of the disadvantages of the Coupled Cluster method for the description of electron correlation is that, since $e^T$ in
$$E_\mathrm{CC} = \langle0|e^{-T}He^T|0\rangle$$ isn't unitary, the method isn't variational (the energy obtained by solving CC equations isn't a rigorous upper bound to the exact energy).

Could someone please explain to me the connection between non-unitarity of $e^T$ and the method not being variational.

Variational coupled cluster (vCC)
First, you can make a variational CC with a non-unitary T operator in the following way.
Recall the variational principle (taken from Wikipedia):

$$\varepsilon[\Psi] = \frac{\langle\Psi| \hat{H} |\Psi\rangle}{\langle\Psi|\Psi\rangle}$$
The variational principle states that

$\varepsilon \geq E_{0}$, where $E_{0}$ is the lowest energy eigenstate (ground state) of the hamiltonian
$\varepsilon = E_{0}$ if and only if $\Psi$ is exactly equal to the wave function of the ground state of the studied system.

So all you have to do to make a variational CC is set $|\Psi\rangle = e^{\hat{T}}|0\rangle$ in the above expression. This is called "variational coupled cluster" or "vCC", because it literally uses the "variational principle" shown above.
Unitary coupled cluster (uCC)
Unitary transformations preserve eigenvalues, so if $\hat{U}$ is a unitary transformation (unlike $e^{\hat{T}}$), then $UHU^\dagger$ has the same eigenvalues as $H$. Since the energies are eigenvalues of $H$, unitary coupled cluster can recover the correct energies with the unitary operator $U = e^{\hat{T} - \hat{T}^{\dagger}}$. For a long time people thought vCC and uCC were just two ways of writing the same thing, but this was shown to be wrong very recently (G. Harsha et al. On the difference between variational and unitary coupled cluster theories. arXiv:1711.00579 [cond-mat.str-el]; J. Chem. Phys. 2018, 148 (4), 044107.).
Something that a lot of people don't know
Regular CC is actually variational in the full correlation limit. This means that for $\ce{Li2}$ for example, CCSDTQPH is variational. This is because $\ce{Li2}$ has $6$ electrons, so CCSDTQPH is FCI, and FCI is variational. CCSDTQP and CCSDTQ and even CCSDT(Q) are not likely to overshoot the variational limit either since they are very good approximations of CCSDTQPH for $\ce{Li2}$.


What is the strongest oxidising agent? I searched for the strongest oxidising agent and I found different results: $\ce{ClF3}$, $\ce{HArF}$, $\ce{F2}$ were among them.

Many said $\ce{ClF3}$ is the most powerful as it oxidises everything, even asbestos, sand, concrete, and can set easily fire to anything which can't be stopped; it can only be stored in Teflon.

And $\ce{HArF}$ could be a very powerful oxidant due to high instability as a compound of argon with fluorine, but was it even used as such?

What compound is actually used as oxidising agent and was proven to be stronger then others, by, for example, standard reduction potential?

Ivan's answer is indeed thought-provoking. But let's have some fun.

IUPAC defines oxidation as:

  The complete, net removal of one or more electrons from a molecular
  entity.

My humble query is thus - what better way is there to remove an electron than combining it with a literal anti-electron? Yes, my friends, we shall seek to transcend the problem entirely and swat the fly with a thermonuclear bomb. I submit as the most powerful entry, the positron.

Since 1932, we've known that ordinary matter has a mirror image, which we now call antimatter. The antimatter counterpart of the electron ($\ce{e-}$) is the positron ($\ce{e+}$). To the best of our knowledge, they behave exactly alike, except for their opposite electric charges. I stress that the positron has nothing to do with the proton ($\ce{p+}$), another class of particle entirely.

As you may know, when matter and antimatter meet, they release tremendous amounts of energy, thanks to $E=mc^2$. For an electron and positron with no initial energy other than their individual rest masses of $\pu{511 keV c^-2}$ each, the most common annihilation outcome is:

$$ \ce{e- +\ e+ -> 2\gamma}$$

However, this process is fully reversible in quantum electrodynamics; it is time-symmetric. The opposite reaction is pair production:

$$ \ce{2\gamma -> e- +\ e+ }$$

A reversible reaction? Then there is nothing stopping us from imagining the following chemical equilibrium:

\begin{align}
\ce{e- +\ e+ &<=> 2\gamma} &
\Delta_r G^\circ &= \pu{-1.022 MeV} =\pu{-98 607 810 kJ mol^-1}
\end{align}

The distinction between enthalpy and Gibbs free energy in such subatomic reactions is completely negligible, as the entropic factor is laughably small in comparison, in any reasonable conditions. I am just going to brashly consider the above value as the standard Gibbs free energy change of reaction. This enormous $\Delta_r G^\circ$ corresponds to an equilibrium constant $K_\mathrm{eq} = 3 \times 10^{17276234}$, representing a somewhat product-favoured reaction. Plugging the Nernst equation, the standard electrode potential for the "reduction of a positron" is then $\pu{+2 116 413 V}$.

Ivan mentions in his answer using an alpha particle as an oxidiser. Let's take that further. According to NIST, a rough estimate for the electron affinity of a completely bare darmstadtium nucleus ($\ce{Ds^{110+}}$) is $\pu{-204.4 keV}$, so even a stripped superheavy atom can't match the oxidising power of a positron!

... that is, until you get to $\ce{Ust^{173+}}$ ...


What types of energies can be determined by molecular dynamics in gas phase or explicit solvents? For molecular dynamics in the gas phase I've seen that electronic energy can be determined and also decomposed in kinetic and potential energy. I've also seen that we can include ZPE. But can we determine (changes in) $\Delta H $, $\Delta S $ and $\Delta G$?

And how about explicit solvent systems? In particular I'm interested if it's possible to determine the free energy of activation and the entropy of activation for reactions in explicit solvent systems. Is it possible to determine such energies for the whole system or even for only the reacting molecules (if that makes sense to do)?

To add: I'm talking about using QM (DFT) for the molecular dynamics. Not molecular mechanics.

First, whether you use DFT or molecular mechanics to get your energies and forces doesn't really matter. If you have equations of motion, forces, and energies then you have the dynamics necessary to sample an ensemble and from there you should be able to get any physical observable you desire.

It is often very difficult to determine free energy changes using straight up molecular dynamics. The reason for this is that even very fast reactions are rare events when compared to typical molecular motions such as vibrations and rotations. For this reason, many methods have been developed to determine free energies without having to run simulations long enough to get good sampling of reactions. Usually, this is not possible anyways.

First, you may wish to read about umbrella sampling. Umbrella sampling, as a general idea, does not sample a trajectory on the potential energy surface determined the potential energy of the system (your output from DFT if you like), but rather, a harmonic potential energy is added to the system as well. This has the effect of lifting the molecule out of the potential well it may be stuck in so that sampling of rare events like reactions becomes much more common.

It is not obvious at all that any physically meaningful quantities should be able to be derived from this new system where the potential energy is constantly being modified. Yet, surprisingly and rather beautifully, it is possible. You should probably see ref. [1] and the citations therein.

The second way to do this I will mention is so-called meta-dynamics. The conceptual idea here is to first define the so-called "collective variables" of the system. These are literally any variables which can be used to describe the properties of a system. Typically one would choose some set of internal coordinates, particularly for a gas-phase system as you describe. However, my understanding is that it is also possible to choose various other coordinates such as intermolecular distances. One could imagine this being quite useful if you are interested in something like radial distribution functions in some hard-to-sample region.

Using these collective variables, again, a biasing potential is applied to the system, but this potential is said to be "history-dependent" in the sense that the form of the biasing potential might change based on what parts of the free energy landscape have already been seen. This differs from umbrella sampling where the potential is always harmonic and the rate at which the potential is biased and how extreme the bias is, is chosen by the user. You should see ref. [2] and references therein for more details on metadynamics.

So, yes, it is possible and even fairly routine to calculate free energies, enthalpies, entropies and pretty much anything else from molecular dynamics simulations. However, just because it is routing does not mean these methods are perfect. It can be very hard to measure error with these sorts of methods because one does not generally know the correct answer particularly for dynamical processes. That is, if your aim is to study details of dynamics which can't be observed experimentally, then it is very hard to know if the dynamics under your biased potential mean much beyond the fact they should give you pretty good ensemble averages. Also, there is the inherent subjectivity in metadynamics involved in choosing collective variables which has been known to cause problems. How many variables are enough and which ones?

Hopefully that is about what you were looking for.

I don't address the differences between explicit solvent and gas-phase much because these methods will work in any context (even implicit solvent) for which forces and energies can be supplied.

[1]: Kästner, J. (2011). Umbrella sampling. Wiley Interdisciplinary Reviews: Computational Molecular Science, 1(6), 932-942.

[2]: Barducci, A., Bonomi, M., & Parrinello, M. (2011). Metadynamics. Wiley Interdisciplinary Reviews: Computational Molecular Science, 1(5), 826-843.


Autodock Vina flexible metals and water molecules I have some problems with water and metals in binding site when using Autodock Vina.

Is it possible to make these flexible? They unfortunately remain where they are what leads to wrong interactions.

My other idea would be to delete it before docking, reinsert them and minimize everything.

Does anyone have a solution for metalloenzymes? (e.g. PDB:3NJY)

For the water molecules, you can try hydrated ligand docking in which they are added in the ligand file.
The documentation here: http://autodock.scripps.edu/resources/autodock-hydrated-docking.
I am not sure if the same approach can be used for the metal atom.


Differential equation for the rate of unimolecular reaction Given in my book is the following reaction mechanism for the unimolecular $\ce{A -> B + C}$,

$$\ce{A + M ->[k_1] A^* + M} \\ \ce{A^* + M ->[k_2] A + M} \\ \ce{A^* ->[k_3]  B + C}$$
Here $A^*$ is an $A$ molecule with enough vibrational energy to isomerize or decompose. In other words, part of kinetic energy of a bimolecular collision (between $A$ and $M$) has been used to raise an  $A$ molecule to higher vibrational energy.
Since $A^*$ is never present at very high concentration we can use steady state appoximination to obtain an expression for rate of reaction in terms of $[A]$ and $[M]$.
$$\dfrac{d[A^*]}{dt} = k_1[A][M] - [A^*](k_2[M] + k_3) = 0\tag{1}$$
$$-\dfrac{d[A]}{dt} = k_3[A^*]\tag{2}$$

I don't understand how the $(2)$ equation is obtained, in my opinion it should be $$-\dfrac{d[A]}{dt} = k_1 [A][M]- k_2[A^*][M].$$
I did not understand how we get $k_3$ instead of $k_1$ and why not $[M]$ in the rate law ?

To start with, the equation you wrote for the rate of change of $[A]$ is not correct. It should instead be

$$-\frac{d[A]}{dt}=k_{1}[A][M]-k_{2}[A^*][M].$$

Note that $[A]$ is consumed in the first reaction step and produced in the second. Now, rearrange equation $(1)$ to get

$$k_{1}[A][M]=[A^*](k_{2}[M]+k_{3}).$$

Plug in the above equation into the actual expression for $-d[A]/dt$ to get

$$-\frac{d[A]}{dt}=[A^*](k_{2}[M]+k_{3})-k_{2}[A^*][M].$$

This, of course, can be simplified to yield the desired equation $(2)$:

$$-\frac{d[A]}{dt}=k_{3}[A^*].$$


The effect of temperature on equilibrium If the forward reaction is exothermic, and the reverse is obviously endothermic. I getting rather confused. If the temperature is increased the equilibrium will want to decrease the temperature. However, (this is the bit im struggling with) if I want to decrease the temperature wouldn't you shift it to the left? As the reverse reaction is endothermic, and an endothermic reaction causes a decrease in the temperature or is this incorrect?

In order to understand an answer to this you need to be able to differentiate between system and surroundings
The system is the reactant mixture , now for an endothermic reaction the temperature of the reactant mixture will increase , however the temperature of the surroundings will decrease since heat is lost by the surroundings to the system , let's assume that an endothermic reaction took place in a test tube , if we put our hand around the test tube our palm will feel cooler since heat was transferred from our palm , however that heat was transferred to the reaction mixture hence the temperature of the mixture was raised . Hence this increase in temperature means we have altered the equilibrium conditions and thus excess heat needs to be lost therefore the eq will shift towards the exothermic reaction . However a decrease in temperature means the eq needs to gain heat which is in the form of an endothermic reaction


Can we predict Dextrorotatory and levorotatory character of an organic molecule? Can we predict in which direction a given molecule will bend plane polarized light?

If I have a molecule with R form in R-S nomenclature does it mean that the molecule will be dextrorotatory or is it to be found by experiments like finding out rate of a reaction by experiments?

There is no simple set of rules that one can use to determine the absolute configuration from the specific rotation, or the other way around.

A quick web search did dig up this reference that deals with calculating the specific rotation for arbitrary molecules, but it requires some fairly sophisticated algorithms. In principle, you could reproduce their methods, but it would require a substantial time commitment unless you are familiar with the software they describe. (Which is Gaussian, if you're interested.)


Hybridization in PH3 Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?

Arguments against hybridization:

$\ce{PH3}$ is less basic than $\ce{NH3}$.
This jibes with the supposition that $\ce{PH3}$ keeps its lone pair in what is essentially an unhybridized $\ce{s}$-orbital.
$\ce{s}$-orbitals are symmetrical and therefore do not concentrate electron density anywhere.
This lack of concentrated electron density as opposed to $\ce{NH3}$ which is $\ce{sp^3}$ hybridized makes $\ce{PH3}$ a poorer base.

However, my professor writes in his book that it can be argued that $\ce{PH3}$ exhibits $\ce{sp^3}$ hybridization.

Another professor told me that the wave function "blows up" when putting an electron pair in an s-orbital. What does that mean?

Pre: My views may be clashing with some people. I express them as far as I know.

To start with, hybridization is a hypothetical concept$^{*1}$ just to explain all facts and make our work easier, otherwise we would have to take into account every single interaction of electron-proton and also other phenomenons; many theories developed which explained all the major/important facts at a macro-level. And what I have been saying to take into account was all micro-level analysis. Attributing hybridization is purely personal choice because neither does it account all facts (for some molecules) nor it can be dropped altogether (for its $\cdots$), if then we would look for another theory$^{*2}$ to explain observed phenomenon. Parallels similarity between this and the situation of Ideal-Gas equation, in general no perfect theory has been developed, but the most easy, practical and useful are the hybridization$^{*3}$ and van-der-Waals equation.

As we move downwards in the periodic table the hybridization concept fails at many places due to overpowering of other factors. I wonder you didn't mentioned the bond angle$^{*4}$ ($93.5^\circ$) for a $\ce{sp^3}$ hybridized (ideally $109.5^\circ$).

In general atom/molecule doesn't seek to follow a theory, it does what it is best comfortable with, it is us, who mould the theory to the observations.

As a general rule, for sake of convenience, assume all molecules show hybridisation and the tendency to hybridise decreases down the group.

In conclusion, I would say it depends on the need of the situation and such dubious statements such as "Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?" should be replaced with "Does hybridization explain all (actually not all) properties of $\ce{PH3}$"

Also there are weights in $\ce{sp^\alpha}$ where $\alpha$ ranges from 1 to 3(even upto 4$^{*5}$ and maybe beyond) and is in between these for $\ce{PH3}$ and some may continue to treat it has a hybridised molecule with suitable weights.

$^{*1}$(as far as I know)
$^{*2}$(probably the MOT)
$^{*3}$(many may argue for other theories which is genuine, but in context this follows)
$^{*4}$(One may argue as we move down the group electronegativity decreases and atomic size increases. In case of $\ce{NH3}$ due to higher bond pair bond pair repulsion (since electronegativity of $\ce{N}$ atom is very high hence it attracts bonded electrons of $\ce{N-H}$ bond towards itself) bond pair moves away from each other and hence shows greater bond angle. This may be considered one of the major factors for less hybridization.)
$^{*5}$[$\ce{O}$ in $\ce{H2O}$ which means that they have 20% s character and 80% p character, but does not imply that they are formed from one s and four p orbitals]


Is Biotin water-soluble or fat soluble? Vitamin $\ce{B_7}$ or Biotin is a member of the vitamin B complex. Members of the vitamin B complex are usually known to be water soluble. However, this webpage mentions that biotin is water soluble. Again, according to another webpage, biotin is neither fat soluble, nor water soluble.

So, is biotin soluble in fat or water? As different sources are contradictory, please provide a reference.

Water solubility

The CRC Handbook of Chemistry and Physics, 97th ed., Physical Constants of Organic Compounds, p. 3-48, describes biotin solubility: soluble in water, ethanol; slightly soluble in diethyl ether and chloroform.

Descriptive terms for solubility, p. 1-38 (solubility – parts of solvent required for 1 part of solute (recalculated to mass%)):

very soluble – less than 1 (> 50%)
freely soluble – 1 to 10 (50% to 9.1%)
soluble – 10 to 30 (9.1% to 3.2%)
sparingly soluble – 30 to 100 (3.2% to 0.99%)
slightly soluble – 100 to 1000 (0.99% to 0.1%)
very slightly soluble – 1000 to 10000 (0.1% to 0.01%)
practically insoluble, or insoluble – >= 10000 (< 0.01%)

Aqueous Solubility and Henry’s Law Constants of Organic Compounds, p. 5-141, states that biotin solubility is 0.35 g kg−1 H2O (0.035 mass%) at 25 °C.

This solubility value would be described as very slightly soluble.

I don't know how to explain this inconsistency.

Octanol/water partition coefficient

.. However, low water solubility does not necessarily mean high fat solubility. A standard measure of lipophilicity is octanol/water partition coefficient (or its decadic logarithm, log P). Positive means more soluble in octanol (fats), negative means more soluble in water.

Experimental log P values for fat-soluble vitamins:lit

vitamin A +5.68
vitamin D3 +10.24
vitamin E +12.18
vitamin K0 +2.20

Typical water soluble vitamin:

vitamin C −1.85

Some B vitamins:

vitamin B1 −3.93
vitamin B2 −1.46
vitamin B4 −0.09
vitamin B7 +0.39
vitamin B8 −1.68
vitamin B13 −0.38

Biotin (vitamin B7) is interestingly slightly lipophilic, or little more fat-soluble than water-soluble.

Biological systems

Even though biotin solubility in water is quite low, it is still classified as water soluble. Originally, biotin (named vitamin H, originally) was discovered in extracts from egg yolk or liver, which are rather fatty tissues, in methyl ester form, practically insoluble in water.

On the other hand, biotin was originally not found in egg white, because it contains avidin protein, which binds biotin so strongly, that it's one of the strongest non-covalent interactions known.

Biotin is often bound in dietary sources to proteins as amide on lysine. By digestion, it's metabolised to biocytin, then to free biotin.

So the fat or water solubility view may be somewhat simplistic.


How do electrons travel through nodes I understand this is a basic question, but I'm having such a hard time wrapping my head around it. I'm trying to avoid thinking about it as an actual "particle" but as a wave, but that confuses me too.  Wouldn't a wave be the entire area the electron travels ? Or does the associated wave represent the path and not the electron? If so, where is the electron?
Here's a picture to refer to in your explanation if it helps.

From the image above and from what you're asking, I presume you are referring to a particle confined in a certain space region. You can find everything you need on the internet but I wanna clarify a few things about QM:

There is no concept in QM for "path of a particle": position and momentum are incompatible observables, which means that they cannot be measured simultaneously each with zero standard deviation: in fact the product of standard deviations for position and momentum must be greater than $\dfrac{\hbar}{2}$.
$$\sigma_{x}\sigma_{p}\geq\dfrac{\hbar}{2}$$
So it's meaningless in QM to talk about trajectories.
Meaning of the wave function: Let's assume that your system consists of a single particle in a one dimensional space. The wave function $\Psi\left(x,t\right)$
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared $\left|\Psi\left(x,t\right)\right|^{2}$
, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx
, gives you the probability for finding the particle between $x$
and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$
Of course you can integrate the previous eq. to find the probability over a certain region of space $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$
Stationary state: If the potential is time independent (in your case it's zero), the time dependence of the wave function can be factored out like
$$\Psi\left(x,t\right)=\mathrm{e}^{-i\omega t}\psi\left(x\right)$$
In this case it's easy to prove that the probability density is time independent.
The quantization: For this simple problem, you must solve the eigenvalue equation for the energy
$$-\dfrac{\hbar^{2}}{2m}\dfrac{d\psi\left(x\right)}{dx}=E\psi\left(x\right)$$
By solving it and by imposing that the particle cannot exists outside a certain region (outside the region $\left[0,3\right]$ in your picture), you find a natural quantization condition for the energy: the energy can assume only certain discrete values, which are called eigenvalues, $E_{n}$
. Of course, this means that for every eigenvalue there is an associated wave function, which is called eigenfunction $\psi_{n}\left(x\right)$
. The general solution to the eigenvalue equation is in fact a linear combination of eigenfunctions $$\psi\left(x\right)=\sum_{n}c_{n}\psi_{n}\left(x\right)$$
In the above picture, you can see the first three $\left|\psi_{n}\left(x\right)\right|^{2}$
. They of course represent probability density functions: note that where $\left|\psi_{n}\left(x\right)\right|^{2}$
is zero the probability of finding the particle is of course zero (for example in 0
and 3
, like we requested before).
So where is the electron? Like said before, in QM locating a particle it's a non-sense. All you can do is to compute the average position and momentum, which for a system in the state $\psi\left(x\right)$ are given by
$$\left\langle x\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)dx$$
$$\left\langle p\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]dx$$


pH of liquid hydrogen chloride vs hydrochloric acid Is hydrogen chloride in liquid state acidic like hydrochloric acid?
I know that hydrochloric acid is hydrogen chloride dissolved in water but I was wondering if liquid hydrogen chloride was also an acidic substance.
Is this the same for other acids?

The problem here is that the question is unintentionally loaded. As is, pH is a terrible measure of acidity or basicity for pure substances or when trying to compare solutions in different solvents. Furthermore, acidity and basicity are far more general concepts than pH. Better measures are pKa and proton affinity, but they too have their limitations.

That said, this is a clear-cut case. In truth, pure hydrogen chloride (gaseous or liquid) is a much stronger acid than aqueous hydrochloric acid. When hydrogen chloride is dissolved in water, it reacts quantitatively to produce hydronium ions $\ce{[H(H_2O)_{n}]^+}$. This is the actual acid present in hydrochloric acid solutions. The fact that this reaction between hydrogen chloride and water is quantitative, fast and very exothermic is a suggestion that the water is taming the acidity of the hydrogen chloride. This is a general phenomenon known as solvent levelling.

The exact factor by which the acidity of hydrogen chloride is reduced when dissolved in water is hard to measure precisely. There are attempts to form universal acidity/basicity scales, such as this article (overview here), which tries to define an absolute pH scale, as follows:

From the authors' data , gaseous hydrogen chloride at a pressure of $\mathrm{1\ bar}$ is an acid with an absolute pH ($\mathrm{pH_{abs}}$) of about 160, which is around 30 orders of magnitude stronger than even the most concentrated aqueous hydrochloric acid solution (pH in water approximately equal to -1), which would correspond to a $\mathrm{pH_{abs}}$ of about 190.


Would there be a solid residue left behind from a reaction between NaK and H2O? I gather the reaction produces Sodium Oxide and Potassium Oxide as well as Hydrogen. Would there be traces of Sodium Oxide and Potassium Oxide left of a cloth that was positioned near the reaction? Thank a ton.

According to this site:-

  NaK reacts explosively with water to form hydrogen, potassium hydroxide
  and sodium hydroxide.

This site says:-

  NaK is a strong reducing agent. It is corrosive and will react violently with water. The reaction with water will liberate extremely flammable gases potentially causing a fire or explosion.

This youtube video confirms the above fact.


What is more acidic: D3O+ in D2O or H3O+ in H2O and why? What is more acidic:

$\ce{D3O+}$ in $\ce{D2O}$

or

$\ce{H3O+}$ in $\ce{H2O}$

and why?

I think it's $\ce{D3O+}$ in $\ce{D2O}$ as I saw somewhere that this property is used in mechanistic studies (the inverse isotope effect), but I need a proper explanation.

EDIT: I found a statement about this in Clayden:  Water $\ce{H2O}$ is a better solvating agent for $\ce{H_3O^+}$ than $\ce{D2O}$ is for $\ce{D_3O^+}$ because O–H bonds are longer than O–D bonds hence $\ce{D_3O^+}$ in $\ce{D2O}$ is stronger than $\ce{H_3O^+}$ in $\ce{H2O}$. That is the origin of the inverse solvent isotope effect.

This is in contrast with all the answers given yet, so I am pretty confused.

For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.

The entire point of the above reference is that

$\ce{H3O+ + H2O <=> H2O + H3O+}$

(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.

$\ce{D3O+ + D2O <=> D2O + D3O+}$

would also correspond to an equilibrium constant of 1

So when Clayden and the OP write

  $\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$

it is wrong for the above reason.

Two genuine thermodynamic equilibriums are

$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$

Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).

But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$.  $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.

As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species.  See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.

Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles.  According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$."  My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.

Also the equilibrium constant for

$\ce{H2O + H2DO+ <=> HDO + H3O+}$

can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751

Explanation of Normal/Inverse Solvent Isotope Effect

For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state.  If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect".  Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.

Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.


Mozingo reduction mechanism I have been searching for the mechanism of the Mozingo reduction on Google but I haven't found anything relevant.

Is the mechanism known? If so, what is it; if not, what hypothetical mechanisms can we consider?

Here is the reaction scheme for the Mozingo reaction (or Mozingo reduction) taken from this Wikipedia link.

The reduction works for both aldehydes and ketones and involves two steps. First, the carbonyl compound is converted into a thioketal (or thioacetal). The mechanism for this step is analogous to the mechanism for ketal or acetal formation (see here) except sulfur replaces oxygen as the nucleophile attacking the carbonyl.

In a second step, the thioketal is subsequently reduced to the corresponding methylene compound.  The reduction can be a standard Raney nickel hydrogenolysis or hydrazine can be used at higher temperatures. The mechanism of reactions on metal surfaces or at higher temperatures are usually not well understood.  Note however that a typical $\ce{C-S}$ bond is around 20 kcal/m weaker than a $\ce{C-C}$ bond.  It is also known that sulfur is readily bound at metal surfaces (catalyst poisoning by sulfur being an example).  The complexation of sulfur and the metal likely weakens the $\ce{C-S}$ bond further.  Whether a $\ce{C-S.}$ radical is formed that reacts with a hydrogen radical complexed at the metal surface, or whether the hydride anion is generated at the metal surface and displaces the sulfur at the weakened $\ce{C-S}$ bond is not known.

The Mozingo reaction is complementary to the Clemmensen (run under acidic conditions) and Wolff-Kishner (run under basic conditions) reductions in that it can be performed under neutral conditions.  So if acid- or base-sentive functional groups are present in the starting carbonyl compound, the Mozingo reaction is likely the pathway of choice to reduce the carbonyl.


Electrical and Thermal conductivity of electrum alloys I have tried to search for this to little effect - I am trying to find electrical and thermal conductivity values for Electrum (a gold and silver alloy). Most sites mention the values depend on the percentage of gold and silver in the alloy, but provide no actual figures.

Is there a table out there that shows approximate electrical and thermal conductivity for electrum under different amounts of component metals in the alloy?

This reference may be a good place to get started:

Thermoelectric power and thermal conductivity in the silver-gold alloy system from 3-300°K

https://doi.org/10.1080/14786437008228219


Can hemi-acetal rings form under basic conditions? I recently encountered a problem on a test.  It was a standard synth problem, "Provide reagents to produce y from x".  The starting material was DHA and the ending molecule was a branched pentose. (I included an image of the solution I came up with).  In my solution I use acetaldehyde under basic conditions to form the pentose.
My question is would pentose form under basic conditions or do I need to acidifying at the end?  Would acidifying just encourage the formation of side products?  Is there some other way to get this pentose as a primary product that I am not thinking of?

Using base means that the reaction will stop at the hemiacetal and acetal formation would not take place. Acidifying at the end might encourage acetal formation.


What is the number of electrons being transferred in electrolysis of aluminum oxide? $$\ce{ 2/3 Al2O3 -> 4/3 Al +O2}$$
As we know, to find the Gibbs free energy, we need to use this formula:
$$ G= -nFE^\circ$$
What would $n$ be in this case? After breaking it down into the half reactions $n = 12,$ but this doesn't seem right.

I guess the misunderstanding arises from the fact that you treat $n$ as the same entity as the lowest common multiple (LCM) used to balance the net equation using half-reactions for the electrolysis of molten aluminium(III) oxide:
$$
\begin{align}
&\text{cathode:} &\ce{Al^3+(melt) + 3 e- &-> Al^0(l)} &|\cdot4\tag{red}\\
&\text{anode:} &\ce{2 O^2-(melt) &-> O2^0(g) + 4 e-} &|\cdot3\tag{ox}\\
\hline
& &\ce{4 Al^3+ + 6 O^2- &-> 4 Al^0 + 3 O2^0}\tag{redox}
\end{align}
$$
or, finally, with the smallest whole-number coefficients:
$$\ce{2 Al2O3(l) -> 4 Al(l) + 3 O2(g)}\label{rxn:R1}\tag{R1}$$
as well as normalized to a single formula unit:
$$\ce{Al2O3(l) -> 2 Al(l) + 3/2 O2(g)}\label{rxn:R2}\tag{R2}$$
In the equation
$$Δ_\mathrm rG = -nFE$$
$n$ is the number that matches the amount of electrons (in mol — since we are using molar convention for Gibbs free energy) transferred in the balanced equation from a reducing agent to an oxidizing agent.
From the balanced reaction \eqref{rxn:R2} it is evident that the number of electrons $n$ transferred from $\ce{O^2-}$ to $\ce{Al^3+}$ is $3 \cdot 2 = 6,$ where $3$ is the number of oxygens per formula unit, and $2$ is the number of electrons each oxygen donates.


Is activation energy related to reactivity? I understand that activation energy is a property of a reaction.

If a substance is highly reactive, will it have low activation energy and vice versa ?
How is the activation energy calculated and what are the factors that affect it?

If by reactivity you mean, "does the reaction occur quickly", then yes, a highly reactive molecules do have a low activation barrier. The fastest molecules can react bimolecularly is if they do so on first encounter with one another. This effectively means that  there is a very small potential barrier between reactants and products so that they have enough energy, at the prevailing temperature, to always overcome this. In solution the rate constant is determined by how fast the molecules can diffuse together and in mobile solvents such as water or acetone the bimolecular rate constant is around $\ce{10^{10} dm^3mol^{-1}s^{-1}}$ depending on exact solution viscosity and far less in viscous solvents such as glycerol.

If by reactivity you mean that the reaction is very exothermic, i.e. free energy is large and negative, $\Delta G < 0$ then there is generally no way to precisely predict how big the activation barrier will be. It is usually measure using the Arrhenius equation and plotting log(rate constant) vs. reciprocal temperature. I'm not sure if ab initio quantum calculations can predict this in general. In fact many exothermic, thermodynamically spontaneous reactions occur slowly. Most organic substances react with oxygen exothermically, but are quite stable, because the activation energy to do so is so large. You and me for example :)


Why is a solution of PCl3 more acidic than MgCl2? Why is a solution of $\ce{PCl3}$ more acidic than $\ce{MgCl2}$? I thought that both $\ce{MgCl2}$ and $\ce{PCl3}$ would react with water to form $\ce{HCl}$ so both solutions should have a very low pH. But does $\ce{MgCl2}$ actually form $\ce{HCl}$ in water or it simply dissolves and forms $\ce{[Mg(H2O)6]^2+}$, but if it only forms this, how is it slightly acidic?

There is no such thing as a water solution of $\ce{PCl3}$. It reacts with water, and quite vigorously. Once the reaction is over, there is no more $\ce{PCl3}$ in the solution. Arguably, there is no $\ce{MgCl2}$ in the other solution either, but it would form (as a hexahydrate) if you evaporate the solution; $\ce{PCl3}$ would not.

The reason for this is that $\ce{Mg}$ is a metal and $\ce{P}$ is a nonmetal. They are different elements, about as different as two elements can be. They have nothing in common, except for a few sparse random traits, like being able to burn brightly in air when ignited. No wonder they are different in this regard as well. $\ce{MgCl2}$ hydrolyzes only very slightly (say, one part in a million). $\ce{PCl3}$ hydrolyzes completely.


Pauling's equation on the relationship between Electronegativity and Percentage Ionic Character In Paulings eqn:

$$ \%~\ce{ionic~character} = \left[ 1−e^{−(Δχ/2)^2}\right] \times 100$$

What is $e$?

I for the love of me can't find it anywhere

'e' is a natural exponential function, so not an actual measurement.

The equation is correctly written as:

% Ionic character $= (1 − e^{−(\frac{Δχ}{2})^2})×100$

where the $−(\frac{Δχ}{2})^2$ term is raised as a power of the natural exponent 'e'.

Source: The University of the West Indies page Lecture 3. Polarizability, you'll need to scroll down the page a bit to the section labelled 'Bond character based on electronegativity differences'.


Skipping the defatting stage in extraction of alkaloids I am attempting to extract certain alkaloids from green plant matter, which is very high in waxes and fats.

I understand that in this scenario it is common to first acidify and then use a non-polar solvent to de-fat the solution.

I am curious though: say I were to basicify and extract alkaloids without any de-fatting. If I were then to add HCl solution to the non-polar solvent which now contains the alkaloids, wouldn't this in theory be equivalent to the de-fatting stage? Wouldn't the alkaloids would now be in the aqueous layer, and the lipids would be stuck in the non-polar layer?

As the acid/base method is so common, I am wondering if there are any downsides to the method described above that I am unaware of. I only have enough material for one shot so I would just like to hear some input before I attempt this expedited method.

May main fear would be emulsions that wouldn't break up. Is this a valid concern that is remedied by a preliminary de-fatting stage? Are there other concerns I should be wary of?

One big risk with the method you propose is foaming and gellation turning your extractions into a huge mess.  Under basic conditions, the fats (e.g. triacylglycerols or TAGs) are likely to hydrolyze, forming carboxylate salts of fatty acids.  "Carboxylate salts of fatty acids" is a long way of saying soap.  These soap-like compounds could lead to excessive foaming or gellation of the sample, which will hinder phase separation and gunk up your apparatus.


Can aniline autodeprotonate? I had an argument with an assistant in my lab, as he stated that in a hypothetical reaction two aniline molecules could react in a way, that one takes a proton off of the other one, thus creating a $\ce{R-NH3^+}$ and a $\ce{R-NH^- }$

However, the $\mathrm pK_\mathrm a$ of aniline is 27, which means, that it is, if at all, a very very poor acid, making it in my eyes impossible to deprotonate. Maybe it could be possible with n-BuLi, but that was not part of the reaction. The main topic was a nucleophilic aromatic substitution to 1-bromo-4-nitrobenzene, with aniline.

Is it even hypothetically possible for such a deprotonation in pure aniline to occur?

I'm going to side with you @TheChemist.

Your TA might be suggesting this because you need often need a negatively charged nucleophile in order to carry out nucleophilic aromatic substitutions.

Having the nitro group in the 4 position makes the your electrophile much more receptive to attack, since the negatively charged intermediate will be resonance stabilized.

Even if the one aniline does deprotonate another, the concentration of the deprotonated anilide will be so low (on the order of 5 x 10-28 mol/L in pure aniline, if the pKa is 27) that this will have no impact on the reaction.


How can you change the proportions of reactants and products at equilibrium if Le Chatelier's principle is occurring? For example, how could you make a buffer with equal concentrations of acid and conjugate base? If you add more conjugate base, it will shift the equilibrium to form more acid, which will keep the proportions of each the same. So how can you achieve equal concentrations if this is happening?

The Henderson Hasselbalc equation is a good way to calculate both the pH and the concentration of acid/conjugate base in an equilibrium solution. According to it:

pH = pKa + log([A-]/[HA])

Since you want to know when [A-] = [HA], you are looking for when the pH is equal to the pKa. This means that you will have equal concentrations of the acid/conjugate base whenever your pH is equal to your pKa, which you can find by monitoring the pH. While adding more conjugate base will shift the equilibrium towards acid formation, it will normally increase the acid concentration less than the amount of base that you added.


Black Smoke Produced from Regular Lighter but Not Torch Lighter Applied to Copper I have a piece of copper mesh from which I would like to remove all coatings/oils. The copper comes from a pot scrubber like this. In order to remove any residue, I held a standard bic lighter up to it. It gave off black smoke. I moved the lighter around the mesh, waiting for it to stop giving off any such smoke, but seemed to do so indefinitely. Even when I just held the lighter in one place, it would not stop giving off a steady stream of black smoke.

I then did the same thing with a torch lighter on a fresh piece of copper from the same source. This time it was hard to tell if there was any smoke at all-- and if there was, it stopped being produced very quickly.

With both the normal lighter and the torch lighter, the surface of the copper turned gray-black. I assume part of this is from the formation of copper oxides, but part of it may also come from the deposition of incompletely combusted carbon on the surface. In both cases I was able to restore the shiny surface by putting the mesh in $\ce{HCl}$-based toilet cleaner for a few minutes.

Why is the normal lighter producing black smoke where the torch lighter is not? What does this smoke consist of?

Common copper wire, especially used in electronic components like inductors and transformers, is usually covered with insulating varnish. This is the common source of black smoke, sometimes it may even burn on its own. I doubt, however, that kitchenware is coated with such varnish, there is no point in it.

There are two types of devices with flames: ones where the fuel is premixed with air, that usually produce blueish flame, and ones where fuel is not pre-mixed with air, but burns and mixes with it in same time, that may produce bright yellow flame.

For many organic compounds, especially ones that require large amount of oxygen per molecule to fully burn, in case the fuel is not premixed with enough air, the flame usually contains oxygen-deficient region where tiny particles of carbon and molecular hydrogen are formed. These carbon particles, since they are heated, give thermal light emission and are responsible for bright non-transparent orange-yellow organic flames. A typical example of such flame is a flame of a candle. It is a flame of heavy hydrocarbons or fatty acids (c20+), that require tens of molecules of oxygen per molecule of the fuel to fully burn.

In case gases from this yellow part of the flame are cooled without mixing with more air, this carbon particles may be obtained in form of deposits or black smoke. For this, a cold effective conductor of heat (i.e. metal, especially copper and silver) or a solid surface with high heat capacity introduced into yellow part of the flame may be used. In some cases, when too many of such particles are produced and the flame is not particularly hot, there is no need even to cool the flame. This is the usual case of hydrocarbons with big amount of carbon, like aromatics, and chlorinated hydrocarbons, like PVC (as long as they burn at all, which often is not the case even if there is no much chlorine in the compound).

Propane torches usually have a mean to regulate amount of air premixed with fuel. Torch lighters are small propane torches, usually without such means. When propane in such torch is premixed with excess of air, a very hot blueish short flame is formed. However, when air is not premixed with propane, the flame is organized in layers with more oxidized outer layers and more reduced inner layers. In inner layers, the dominating reactions are thermal decomposition

$\ce{C3H8 = 3C + 4H2}$

and incomplete combustion

$\ce{C3H8 + 2O2= 3C + 4H2O}$

(this is an oversimplification, as numerous other intermediates are formed, but that's advanced stuff)

In outer layers, propane combusts completely, just as if it was premixed with air

$\ce{C3H8 + 5O_2= 3CO2 + 4H2O}$

and carbon particles (and other reduced products) from inner layers also combust here.

Now, to get carbon particles from the flame in form of smoke the key is to cool them before they mix with more air. An ideal solution would be to suck them with tiny pipe and cool with water. However, it seems, that skein of copper mesh, at least until it is heated enough, works good for this purpose, both cooling the inner layers and hindering their mixing with air.

Since flames of fuels that are not pre-mixed with air have very efficient emitter of visible and and infrared light in it (many carbon particles), they tend to be cooler, especially the inner part. The flames with fuel pre-mixed with air (like in torch lighters with blueish flame like this), the flame is thin, quite short and much hotter.

Copper wire, when heated in air, also oxidizes. If done properly, it is a good way to produce pure black copper (II) oxide. However, it usually stay on the surface of the wire and indeed, can be washed with hydrochloric acid. If there are black particles that are not dissolved in hydrochloric acid, they are carbon particles.


Does sulfur and phosphorus expand their octet? There are many compounds in which the stability of a molecule is not governed by the presence of octet configuration in central atom. In most of the cases the central atom is generally sulfur or phosphorus. In these cases either the central atom has 6 electrons, or sometimes the electron count exceeds 8 and 10 or 12 electrons in its valence electron shell.

This makes the concept of octet somewhat incorrect. My question is how can they increase their octet to such an extent?

Could someone explain this in relation with sulfur trioxide?

Whether sulfur or phosphorous actually expand their octet is contested within the chemistry community. Another term for this octet expansion is "hypervalency." You can find many works of research regarding hypervalency. The consensus, according to Wikipedia, is that both can expand their octets, but not to a significant extent. In other words, d-orbital utilization is low at best.

Regarding sulfur trioxide, this article may be of interest.

  We find no evidence to support notions of p π –d π back-donation from oxygen to sulfur.

Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: $\ce{C, N, O}$, and $\ce{F}$. And note that for carbon, there are well known exceptions in organic chemistry - carbocations (which have less than an octet of electrons). Carbocations however tend to exist mainly as reactive intermediates. Therefore, a more nuanced statement might be $\ce{C, N, O}$, and $\ce{F}$, when involved in neutral and thermodynamically stable compounds, generally have 8 valence electrons.

  this makes the concept of octet somewhat incorrect.


Does the solubility of iron (iii) oxide increase when the pH of the water it is in decreases? I did an experiment and I got these results:

pH 4: 3.46
pH 3: 3.77
pH 2: 2.18
pH 1: 3.69
The numbers on the right is the amount of iron oxide remaining that was not filtered.

Do these numbers make sense? (The measurements are in grams.) The original amount of iron oxide in each solution (4 different solutions as there are 4 different pH's) was 3.81 grams

I basically put iron (iii) oxide in different pH's, left them, and filtered out the water after. I measured how much iron (iii) oxide remained. I am not sure if my numbers make sense though because I thought that if the pH drops it increases in solubility. However, my pH 2 was the highest in solubility. Do these numbers make sense chemically?

Iron (III) oxide is insoluble in water. If you add an acid (e.g. $\ce{HCl}$) it might react with the oxide and form a soluble iron salt.

How much of the oxide gets dissolved depends, amongst others, on what acid is used, its concentration and amount, and the time of reaction.


Testing for manganese in water Is there a decent way to test for solvable manganese in water without purchasing a 150USD test kit? I have a fair chemistry set available (I use it to teach my kids chemistry), and would rather pay 50USD for something to add to it that can also be used for other things later than pay for some test that may not be that accurate anyway.

Our city water supply to our house produces a black slime that is likely from bacteria that feeds on iron or manganese. I’d like to start getting to the bottom of it. This is a first step.

Best way is sending a sample for Atomic Adsorption Spectroscopy (AAS). You can get the concentration for $\ce{Mn, Pb, Fe, Cr, Ni}$, and other water goodies. Contact a local university or lab and get a price quote for a test.
Or get a flame spectroscope and DIY https://web.pdx.edu/~atkinsdb/teach/427/Expt-AtomicSpec.pdf


Difference in titration curves In my titration curve for HCl being titrated with NaOH, the initial pH-value was plotted as 2.1 according to a pH-meter. When I found the equivalence point and calculated the concentration of HCl, it was 0.075 M. I calculated how much the initial pH-value should be from that by -lg(0.075) which is around 1.12. That is way off from the original pH-value I got.
Why did this happen? Is it because the risk of me calibrating the pH-meter incorrectly? If that is the case, will the x-value of the equilibrium point also be wrong?? For the latter, I feel like the answer would be no, because what is wrong is that if the pH-meter was calibrated wrong, then all of the values would be 0.9 pH higher than what they should be, so the curve would only differ in the y-direction, but would be exactly the same in the x-direction. Is that correct?
Thanks in advance!

Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of $\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. Just for the fun try this experimentation : Take your $\ce{HCl}$ solution with an electrode showing  pH $2.1$. Add an equal volume of saturated $\ce{NaCl}$ solution (which is neither acidic nor basic). You will not believe your eyes by reading your electrode : The pH values goes down to $1$, as if diluting the initial solution produces an increase of concentration !
The activity is sometimes described as the amount of $\ce{H3O+}$ ions divided by the volume of the so-called "free water" (and not by the volume of the solution). In concentrated solutions the "free water" is this small amount of water that is free to move, and not attracted and fixed around the ions. One Liter $1 M$ $\ce{HCl}$, should contain much less that $1$ Liter free water, maybe about $0.1$ Liter free water, or less.


Generally available fluid with good heat transport properties I have designed a container which is heated up and has to heat up some tube-coils of water inside of it. This container is filled with a fluid that can help store and transfer the heat received from the container to the tube-coils.

My plan was to find some material that I can easily buy and which is not very expensive to fill the $636cm^3$ container.

It has to have a high coefficient of heat transfer, high heat capacity and an operating temperature of at least the range of 10-400 degrees Celsius. I will also need to calculate how much the volume changes when it goes from around -35 to 500 degrees Celsius.

I also think a high density and if the material is non-toxic might be a plus.

Does anybody know any fluid that could be used for this or somewhere where I could find such information?

400 C is beyond long term use of any organic heat transfer fluid.  Dow Corning boasts 427 C for one of its silicone fluids.  Don't bet on that.  Inorganic eutectic salts will long term decompose into corrosive volatiles.  Best bet for a red hot closed system is fusible alloys of indium, gallium, and bismuth - beware of metal boiling points.  Open systems will air oxidize.   Accommodate sealed inert gas fill (argon) with a metal bellows add-on, preferably on a pipe extension away from the hot zone.  Fill composition must be compatible with container composition.  Mercury is out for toxicity and boiling point, 357 C.  Thermal coefficient of linear expansion, solid vs. liquid alloy, must be looked up.  The increment is cubed for volume change.

500 C (red heat) will change the metallurgy of stainless steels over time.  They will harden (carbide precipitation), among other things.  Don't try to drill a hole or tap a thread in stainless that has been soaked at red heat.


In a multi equilibrium system, how does change in one equilibrium affect the other? Suppose in a reaction vessel of fixed volume we have the following equilibria established (existing simultaneously in the vessel),
\begin{align}
\ce{A (g) &<=> B (g) + C (g)}\tag{i}\label{i}\\
\ce{D (g) &<=> E (g) + C (g)}\tag{ii}\label{ii}
\end{align}

Frankly speaking I do know that when we increase, say, $\ce{A (g)}$ concentration in $\eqref{i}$ then indeed the concentration of $\ce{C}$ or $\ce{B}$ is going to increase as per Le Chatelier's Principle. However if we increase the concentration of $\ce{E (g)}$ in the vessel what impact will it have on the concentration of $\ce{C}$ and $\ce{A}$?

From my perspective, when $\ce{E}$ is increased in concentration, it will now combine with $\ce{C}$ to give $\ce{D}$ so that reaction goes backwards, and so more of $\ce{C}$ needs to be formed as a result $\ce{A}$ concentration in equilibrium $\eqref{i}$ goes down. However I am not convinced with my own explanation as I feel I am missing out on something. Please feel free to help me out.

Increasing $\ce{[E]}$ lowers $\ce{[C]}$ by a simple application of Le Chatelier's Principle. Regarding the effect of increasing $\ce{[E]}$ on $\ce{[A]}$, your understanding is correct. To make your observation more obvious, we may reverse reaction (ii) and add it to reaction (i): $$\ce{A(g) + E(g)<=>B(g) + D(g)}$$
So we can, once again, simply apply Le Chatelier's Principle and see that increasing $\ce{[E]}$ lowers $\ce{[A]}$.


How does addition of AgCl affect concentration of iodide in KI solution? In a mixture of $\ce{AgCl(s)}$ and initial $\pu{0.10 M}$ $\ce{KI}$, what is the $[\ce{I-}]$ after the reaction proceeds?
Disclaimer: This is not a graded assignment, but rather a problem I'm trying to figure out for practice purposes from my textbook.

I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it.
Given the reaction:
$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}$$
there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is to say that if the initial concentration of $\ce{I-}$ is $\pu{0.10 M}$, then the final concentration of $\ce{Cl-}$ will be  $\pu{0.10 M}$. (This is a significant figure simplification. The $\ce{[I-]}$ isn't given to 10 decimal places.)
Now Wikipedia gives the $K_\mathrm{sp}$ of silver chloride as $1.77 \times 10^{-10}$ and for silver iodide as $8.52 \times 10^{−17}$.
Since the final concentration of $\ce{Cl-} = \pu{0.10 M}$, the final concentration of silver is given by:
$$\ce{[Ag+]} = \dfrac{K_\mathrm{sp, AgCl}}{\ce{[Cl-]}} = \dfrac{1.77 \times 10^{-10}}{0.10} = \pu{1.77\times 10^{-9} M}$$
So the final concentration of $\ce{I-}$ is given by:
$$\ce{[I-]} = \dfrac{K_\mathrm{sp, AgI}}{\ce{[Ag+]}} = \dfrac{8.52 \times 10^{−17}}{1.77 \times 10^{-9}} = 4.8\times10^{-8}$$
Since the $\ce{[I-]}$ starts out as $\pu{0.10 M}$ and ends up as $\pu{4.8 \times 10^{-8} M}$, it is better to say that the reaction proceeds quantitatively rather than completely.


Determine the concentration of hydrogen cations to prevent precipitation of monocalcium phosphate? Determine the $\ce{[H+]}$ needed to just prevent precipitation by $\ce{H2PO4-}$ in a $\pu{0.25 M}$ $\ce{H3PO4}$ solution that has $\ce{[Ca^2+]} = \pu{0.15 M}$.
  The $K_\mathrm{a1}$ of $\ce{H3PO4}$ is $7.1 \times 10^{–3}$. The $K_\mathrm{sp}$ of $\ce{Ca(H2PO4)2}$ is $1.0 \times 10^{-3}$.

For this problem, I first found the $\ce{[H2PO4-]}$ present in the solution when it becomes saturated, which is $\pu{8.2 \times 10^-2 M}$. So I interpreted this as at equilibrium, there is at most $\pu{0.082 M}$ $\ce{H2PO4-}$ present in the solution. When I went to set up my ICE chart for $$\ce{H3PO4 <=> H2PO4- + H+},$$ I put $\pu{0.082 M}$ as the equilibrium value of $\ce{[H2PO4-]}$. However, the answer key wrote

  $
\displaystyle
K_\mathrm{a1}
= \frac{\ce{[H+][H2PO4-]}}{\ce{[H3PO4]}}
= \frac{x (0.082 + x )}{(0.25 - x )}$

meaning that $\pu{0.0082 M}$ is the initial concentration and the equilibrium concentration of $\ce{H2PO4-}$ is $0.082 + x$. But if this was the case, then wouldn't $\ce{Ca(H2PO4)2}$ precipitate since there's more than $\pu{0.082 M}$ at equilibrium?

I think I'm probably misinterpreting the question, so any clarification on what's happening here would be appreciated.

edited - MaxW
Added image (cut and paste job for needed parts...) of problem from 2002 U. S. NATIONAL CHEMISTRY OLYMPIAD NATIONAL EXAM—PART II

I was looking for something else and stumbled across this old problem. I noticed Martin's comment about the source, and I looked for one. Noticing that this was sponsored by ACS made me wonder if my initial analysis had been correct.

The answer key, the OP, and I agree that the $\ce{H2PO4^-}$ concentration must be $8.2\times10^{-2}$ molar to cause $\ce{Ca(H2PO4)_2}$ to start to precipitate.

Problem 1 - The problem statement lists $K_{\alpha1} = 7.1\times 10^{-3}$ but the answer key uses $7.3\times10^{-3}$

My faith in the infallibility of the ACS is shaken...

Problem 2 - There are really two ways to solve the problem.

A. Neutralizing the acid

The solution starts off as 0.25 molar $\ce{H3PO4}$ and 0.15 molar $\ce{Ca^{2+}}$. You add $\ce{NaOH}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Mass balance for all phosphate species is:

$0.25 = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$

The solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 = \ce{[H3PO4] + [H2PO4^-]}$

$\ce{[H3PO4] = 0.25 - [H2PO4^-] = 0.25 - 0.082 = 0.168 }$

Now using the equilibrium expression for the first ionization of phosphoric acid

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO3^-]}}{\ce{[H3PO4]}}$

Rearranging and using $7.3\times10^{-3}$ from the answer key

$\ce{[H^+]} = \dfrac{K_{\alpha1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = \dfrac{(7.3\times10^{-3})(0.168)}{0.082} = 0.014956 \ce{->[Rounding]} 1.5\times10^{-2}$

Fortuitously using 7.1 from the problem statement yields the same rounded value.

$\dfrac{(7.1\times10^{-3})(0.168)}{0.082} = 0.014546 \ce{->[Rounding]} 1.5\times10^{-2}$

B. Making a buffer

To the 0.25 molar solution of $\ce{H3PO4}$ you add $\ce{NaH2PO4}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Let $x$ be the nominal molarity of the $\ce{NaH2PO4}$.

Again the solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 +x = \ce{[H3PO4] + [H2PO4^-]}$

But $\ce{[H2PO4^-] = 0.082}$, so

$\ce{[H3PO4]} = 0.25 + x - \ce{[H2PO4^-]} = 0.25 + x - 0.082 = 0.168 + x$

The charge balance is given by

$\ce{[Na^+] + [H^+] = [H2PO4^-] + 2[HPO4^{2-}] + 3[HPO4^{3-}] + [OH^-]}$

Since the solution is very acidic we can neglect $\ce{[HPO4^{2-}]}$, $\ce{[HPO4^{3-}]}$, and $\ce{[OH^-]}$. So

$\ce{[Na^+] + [H^+] = [H2PO4^-]}$

$\ce{[H^+] = [H2PO4^-] - [Na^+]} = 0.082 - x$

Now substituting into the equilibrium expression for the first ionization

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO4^-]}}{\ce{[H3PO4]}} = \dfrac{(0.082-x)(0.082)}{0.168+x}$

$0.0073(0.168+x) = 0.082^2 - 0.082x$

$(0.0073 + 0.082)x = 0.0893x = 0.082^2 - 0.0071(0.168)$

$x= 0.06194$

So the buffer solution would be 0.25 molar $\ce{H3PO4}$ and 0.062 molar $\ce{NaH2PO4}$.

$\ce{[H^+] = 0.082 - 0.6194 = 0.02006 ->[Rounding] 2.0\times 10^{-2}}$


Deriving an alternative expression for the Van der Waals equation using given parameters The question is to find out an alternative expression for the Van der Waals equation of corresponding states for 1 mole of gas using the parameters given as $$A=\frac{P}{P_c},\quad B=\frac{V}{V_c},\quad C=\frac{T}{T_c} ,$$ where $P_c,T_c,V_c$ represent the critical pressure, temperature and volume respectively.

I am aware of the simple form of the Van der Waals equation which is $$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT .$$ The values of $P_c , T_c \text{ and } V_c$ can also be written down in terms of the Van der Waals constants but I have no idea on how to proceed after this. I tried to substitute the known values but it made things complicated. Is there any other approach to this problem?

$\def\p{\partial} \require{enclose}$Start with the van der Waals equation $(1)$.

$$\left(P + a\frac{n^2}{V^2}\right)\left(V - bn\right) = nRT\tag1$$

Note that you have a slightly different form of the equation, namely you have used molar volume $V_m$. Divide both sides of $(1)$ with $n$, use $V_m = V/n$, and $n^2/V^2 = 1/V_m^2$ to reach your variant.

Values for critical parameters in terms of $a, b, n, R$

The critical point $(P_c, T_c, V_c)$ is a critical point and an inflection point of the isotherm by definition.$^{[a]}$ Therefore we arrive at two equations;

$$\left(\frac{\p P}{\p V}\right)_{T\ =\ T_c} = 0,\tag2$$
$$\left(\frac{\p^2 P}{\p V^2}\right)_{T\ =\ T_c} = 0.\tag{2'}$$

Equation $(2)$ is a sufficient condition for a critical point (mathematically, a derivative's non-existance works too). Equation $(2')$ gives a necessary condition for an inflection point.

So our first step should be to find the equation $P = P(T, V)$. Just use equation $(1)$.

$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$

Apply conditions $(2)$ and $(2')$ on $(3)$.

$$\left(\frac{\p P}{\p V}\right)_T = -\frac{nRT}{(V - bn)^2} + 2a\frac{n^2}{V^3} \overset{(2)}{=} 0\tag4$$

$$\left(\frac{\p^2 P}{\p V^2}\right)_T = \frac{2nRT}{(V-bn)^3} - 6a\frac{n^2}{V^4} \overset{(2')}{=} 0\tag{4'}$$

Both $(4)$ and $(4')$ are equal to zero. Thus we may multiply either equation by non-matching real numbers, and still maintain equality: $(4) = (4')$. We will use this to our advantage by multiplying $(4)$ with $2/(V-bn)$ and $(4')$ by $-1$.

$$-\frac{2nRT}{(V - nb)^3} + 4a\frac{n^2}{V^3(V- nb)} \overbrace{=}^{(4)\ =\ (4')} -\frac{2nRT}{(V - nb)^3} + 6a\frac{n^2}{V^4}\tag5$$

$$\frac{2}{V - nb} = \frac{3}{V} \implies 3V - 3nb = 2V \implies \enclose{box}[mathcolor="green"]{V_c = 3nb}\tag{5'}$$

Plug result $(5')$ into equation $(4')$.

$$-\frac{nRT}{(3nb - nb)^2)} + 2a\frac{n^2}{27n^3b^3} = 0 \implies -\frac{RT}{4} + \frac{2a}{27b} = 0 \implies \enclose{box}[mathcolor="red"]{T_c = \frac{8a}{27bR}}\tag6$$

The outcomes $(5')$ and $(6)$ are to be placed into equation $(3)$.

$$P_c = nR\frac{8a}{27bR} : 2nb - a\frac{n^2}{9n^2b^2} \implies \enclose{box}[mathcolor="blue"]{P_c = \frac{a}{27b^2}}\tag7$$

Values for $a, b, R$ in terms of critical parameters

$$b \overset{(5')}{=} \frac{1}{3n}V_c\tag{8a}$$
$$a \overset{(7)}{=} 27b^2P_c \overset{(8a)}{=} \frac{3P_cV_c^2}{n^2}\tag{8b}$$
$$R \overset{(6)}{=} \frac{8a}{27bT_c} \overset{(8a, 8b)}{=} \frac{8P_cV_c}{3nT_c}\tag{8c}$$

Reduced van der Waals equation

Replace the values of $a, b, R$ from equations $(8a)$$-$$(8c)$ into the original equation $(1)$. Immediately after this substitution multiply both sides by $3/(P_cV_c)$. You should have the result

$$\left[\frac{P}{P_c} + 3\left(\frac{V_c}{V}\right)^2\right]\left(3\frac{V}{V_c} - 1\right) = 8\frac{T}{T_c}.\tag9$$

Define $P/P_c = \pi, V/V_c = \varphi, T/T_c = \tau$. Finally, enjoy the reduced van der Waals equation

$$\enclose{box}[mathcolor="orange"]{\left(\pi + \frac{3}{\varphi^2}\right)(3\varphi - 1) = 8\tau}.\tag{10}$$

What does equation $(10)$ imply by there being only reduced quantities $\pi, \varphi, \tau$?

   Hint: theorem of corresponding states.

$[a]$ While I believe this to be technically true, I also think it deserves a question of its own. The fact that a phase critical point physically always has to (?) be a inflection point confuses me a bit, so is best left to another answerer.


What tools are used to predict the products of a chemical reaction? I was wondering if there is a decent and non-complex way to predict the outcome of mixing two chemicals together.

I know that molecular dynamics is used in computational chemistry to understand some of these types of problems, but in many cases I have noticed that they seem to be restricted to classical physics and therefore cannot accurately predict any real reactions.

My point in asking is such that I can have a broader understanding of this; However, at the moment, I am interested in the difference between this reaction:

and the same reaction while replacing the triflic acid ($\ce{CF3SO3H}$) with chlorosulfonic acid.

Would there be any differences if these two superacids were switched?

The paper from which I obtained this reaction from is here:

Klumpp, Douglas A., Rendy Rendy, Yun Zhang, Alma Gomez, and Aaron McElrea. "Dicationic Intermediates Involving Protonated Amides: Dual Modes of Reactivity Including the Acylation of Arenes." Organic Letters 6, no. 11 (2004): 1789-1792.

I don't know for sure, but I suspect these two reagents would lead to the same product.  They are both superacids, and extremely harsh.

Whichever is a better Bronsted acid (proton donor) in benzene probably leads to a faster reaction, or better conversion.  But who knows?  Maybe the reactions are so acidic (two plus charges in a small organic molecule!) that they are just about the same.

There may be some superacid expert lurking on chemistry.se who knows this answer, but I think the only way to know is to do the reaction.

You may also want to write the corresponding author of the paper. They may already know the answer. Or, they might be flattered and may find the question interesting enough to go to the lab and do the experiment.

(I don't think computational chemistry will help here.)


Sodium carbonate titration in presence of phenolphthalein Why phenolphthalein cannot show the endpoint for the complete titration of Na2CO3 with only phenolphthalein?

What's the reason that we consider only half of the moles of sodium carbonate when titrating with phenolphthalein indicator?

Suppose you want to find out the amount of sodium carbonate  in a solution of sodium hydroxide. Let's use phenolphthalein as the indicator and HCl for titrating the mixture.

The first reaction that takes place is $$\ce{NaOH + HCl -> NaCl + H2O}$$
The pH at this point isn't alkaline enough for phenolphthalein to become colourless, since there's still base ($\ce{Na2CO3}$) present in the solution.

Then $\ce{Na2CO3}$ reacts. When all the $\ce{CO3^2-}$ becomes $\ce{HCO3^-}$, the pH is given by:

$$\mathrm{pH}=\frac{\mathrm{p}K_{\mathrm{a}1} + \mathrm{p}K_{\mathrm{a}2}}{2}$$

which is approximately $8.3$. (The above expression is for amphiprotic salts.) Phenolphthalein changes colour at approximately pH $8.2$.

So when the colour of phenolphthalein disappears, it means that all the carbonate is converted to bicarbonate. So if $x$ is the volume of HCl required to neutralize NaOH and $y$ is the volume of HCl required for complete neutralization, of carbonate, then phenolphthalein changes colour when volume of HCl used is $\displaystyle x+\frac{y}{2}$.

Remember that for complete neutralization, two moles of HCl is required for every mole of carbonate. For half neutralization only one mole of HCl is required. (Which is half the amount required for complete reaction).


Formation of (Gamma) Methylparaconic acid I came across this question recently,

I tried to devise a mechanism but I'm struck here,
My try -

Moreover, I think my mechanism is somewhat inefficient because of formation of some byproducts, which we don't want.

In short in my mechanism,

First we convert succinic acid to its ester (by esterification using ethanol).

Then, we use this ester to react with acetaldehyde in basic medium (dilute NaOH), intramolecular magic occurs to yield an ESTER of the desired product.

Now, here's where i'm struck, how do I convert ester of the desired product into acid, Can I do Hydrolysis, but I think then the ring might get cleaved!

Moreover, I think during this reaction, some acetaldehyde may undergo self condensation to yield a byproduct, this may be true for ester of succinic acid.

Any help will be appreciated!

The mechanism looks OK. Though catalytic NaOEt would be a better base than NaOH as the EtO- is regenerated as the reaction proceeds.

The opening of the lactone under basic conditions is reversible but the reformation of the lactone is favoured. A mild basic hydrolysis - 1eq LiOH in 19:1 THF/H2O - should give you your target.

Self condensation is always a possibility with acetaldehyde - this is why few reactions go with 100% yield in real life


Why isn't hybridization done in America? Warning: This is a question from my teacher.

In America hybridization isn't done, while in Europe it is done around the 10th grade. Is there any reason or advantage for doing(not doing) it as part of the school program?

In Russia hybridization of carbon learns at the beginning of the 10th grade, before the main organic chemistry. Hybridization in the atoms of metals (complexes) are studied in grade 11. These two topics are difficult for students to understand. But they can be on the exams.

Advantages:

It is easier to understand the spatial structure
Explanation of coordination numbers in complexes
For an explanation of the facts, when the atom forms a large number of bonds than the number of unpaired electrons in the ground state

Disadvantages:

Difficult to understand
There are many different theories
Rarely used in the study follow-up program


How to prevent fuel cell gas leak? My research is fuel cell. I need to do a open system fuel cell, but in cathode side always have gas leak. Have any one have experience to tell me, how to prevent gas leak?

Since it's the cathode where the leak is I presume it's the hydrogen you can't hold. Hydrogen is very hard to contain because of it's small molecules. I use plastic glass as its bubbler and a small hose to move it elsewhere. If it's your bubbler where the leakage is insulate it using silicon around the connections/openings.


How are these two organic resonance structures different? I've been working through some questions online (self study, not homework) predicting the various forms for molecules. One of the questions involved the following molecule (see image below). It had four structures listed in total, but the two below have me baffled. They appear to be identical if rotated by 180 degrees along the horizontal axis.

Please explain the difference between these two structures.

Resonance structures do not describe the conformation (=energy state) of a molecule, but are didactic. They show how you can just flip schemactic $\pi$ electrons around the structure, without breaking any connection. The bond lenghts and angles are constant in the actual molecule, at values somewhere between those indicated by the resonance structures.

Between resonance structures, you don't look at a possible symmetry of the location of atoms, but regard them as unique and fixed (see also https://en.wikipedia.org/wiki/Born-Oppenheimer_approximation). Resonance structures can have different energy or be symmetric (=conformationally identical) to each other. If one resonance structure has higher energy than the others, that is just a hint that the actual structure might by more closely described by the other resonance structures.


Is there an otherwise liquid material that has angle of repose? I wonder whether there is a material which consists from individual molecules (that is, fluid), which has its angle of repose thus exhibiting propertes of a powder?

Such materials are called Bingham plastics.  An angle of repose means that the material must withstand a non-zero shear stress (from its own weight) while still exhibiting zero shear rate.

Examples include toothpaste, ketchup, mayonnaise, latex paint, and concentrated suspensions of coal particles in a otherwise newtonian liquid.  I'm not sure if any of these materials meets your requirement that it consist "of individual molecules".  I don't think a single-component, single-phase, non-polymeric substance will ever exhibit Bingham plastic behavior, although (I might be wrong and) I can't cite any theoretical reason why this is so.


Why does radical chlorination and bromination of propane occur at different positions? When propane reacts with chlorine, we will mostly (theoretically) get an isomer with a chlorine on the end. ($\ce{CH_3-CH_2-CH_2Cl}$). However, when propane reacts with bromine, we will mostly get an isomer with bromine attached to the second carbon ($\ce{CH_3-CHBr-CH_3}$). Could anyone please explain why this happens?

The Hammond Postulate (HP): Exothermic reactions have early transition states (TS). Endothermic reactions have late TS's. [The HP may easily be demonstrated with a retractable, metal tape measure. Open several feet of the tape with the left hand holding the case and the other hand at the end of the tape. Bend the tape upward so that it looks like a TS. Raise your left hand above the right. This is an exothermic reaction. Note how the high point (TS) moves to the left. Make an endothermic reaction. Watch the TS move to the right.]
To compare the radical chlorination and bromination of propane, the following chart is instructive. Only the first propagation steps need to be examined because this is where the 1o vs. 2o choice is made. Chlorination (A & B) is exothermic (earlier TS) for both C-H bonds with the secondary more favorable by virtue of the weaker 2o C-H bond. Bromination (C & D) is endothermic (later TS) and more reversible than chlorination. The difference in the heat of reaction in both chlorination and bromination is 2 kcal/mol with the secondary radical formed in favor of the primary radical. But this equal difference does not explain the difference in the greater selectivity in the bromination vs. chlorination. The difference in activation energy (Ea) in the TS between secondary C-H bond cleavage vs. primary is greater in the bromination than in chlorination. [ADDENDUM: The ΔΔH‡ (ΔEa) for chlorination is 1 kcal/mol; for bromination, 3 kcal/mol. Therefore, bromination is more selective.][2] [continued]

The data for methane in the chart below is instructive. Fluorination has a very low Ea and it is more exothermic than chlorination. A fluorine atom is virtually indiscriminate in choosing among primary, secondary and tertiary C-H bonds. The less reactive chlorine atom has a higher Ea and shows greater discrimination. While these two halogenations are readily conducted at ambient temperature, bromination is conducted at elevated temperature to obtain useful rates of reaction. With a high Ea selectivity is the greatest with bromination.
To put a folksy spin on the discussion, a bromine atom has more "time" to make choices as to what C-H bonds it is capable of breaking while a fluorine atom is virtually a bull in a china shop.

[2]: Streitwieser and Heathcock, Introduction to Organic Chemistry, 4th ed., pg. 114.


Is Linus Pauling's General Chemistry still readable and up-to-date today? I've found a Dover edition copy of Pauling's General Chemistry in the bookstore near me. I'd like to know if it's still readable today, I mean, is it still up-to-date? I have no background in chemistry, but the book was cheap and it seems to be a nice read.

I currently have the Dover copy of Pauling's book on my shelf.  It indeed is a nice read but it is in no way 'up-to-date' as it was written many years ago.  That being said, a lot of conceptual information presented in the book still apply just as much today as they did 6 decades ago (I would stray away from using the book as a source for benchmark measurements and such).  It is a book that should find its way onto any chemist's bookshelf, and it makes for a fascinating read into many areas of general chemistry.


Extrema in free energy of a solution? I am a high school student reading about the second law of thermodynamics, and one of the equations given was the free energy change in dilution.
$$\Delta G_{\text{dil}} = -RT \log(C_1/C_2)$$
This implies that as you make a solution more dilute, the free energy decreases. However this equation does not take into account the entropy change in breaking a solid crystal, which I would expect to be highly positive as the crystal is a highly symmetric arrangement. Also it assumes ideal solution.
If we take into account the entropy change in breaking up the solid crystal, we see that the free energy increases while diluting, and also increases in breaking up the solid , that is, concentrating. So I would expect a minima in the free energy vs concentration curve. My guess is it might be the saturation point?
Is my reasoning correct? If it is, then is it possible to calculate the minima?

For Gibbs Free Energy, the standard equation is ΔG=ΔH−TΔS. The overall change in Gibbs Free Energy will depend, at least in part, on the enthalpy of dissolution for the specific substance. As for entropy, increasing entropy by breaking up the crystal decreases Gibbs Free Energy. The saturation point is where ΔG for the dissolution equals zero: where ΔH = TΔS. As ΔH and ΔS vary depending on the particulars of the solute and solvent, the saturation point also varies.
The equation you cite is specifically for when you have two solutions, so the dissolution of the solid is not described by it. Since the saturation point depends on the enthalpy of dissolution, you can't use the log equation to find it.


Do carboxylic acids form hydrates like aldehydes and ketones? Having learned that aldehydes exist in equilibrium with their hydrate form -

Do carboxylic acids do the same in a water solvent and form a hydrate with 3 $\ce{OH}$ groups and one alkyl group?

Let me go against the previous answers, and say that they "do" form hydrates.
What you should consider is that the equilibrium towards such hydrates is extremely unfavorable, making them virtually non-existent.

There have been, anyway, clever ways of "blocking" those forms as orthoesters. If I recall correctly, Corey used a protecting group of the OBO family in the synthesis of a natural product.

A more recent paper, which describes a protecting group of that nature, is this:
https://pubs.acs.org/doi/abs/10.1021/ol047493n?journalCode=orlef7

Anyway, in normal mechanism-writing, that hydrate is not considered, being highly unfavored and unlikely to form.


Comparison of C-Cl bond length in CH3Cl and CF3Cl How can one compare the $\ce{C-Cl}$ bond lengths in chloromethane ($\ce{CH3Cl}$) and chlorotrifluoromethane ($\ce{CF3Cl}$)?

Bent's rule is also consistent with Gillespie's VSEPR model, and may provide alternative rationalisation for effect of electronegativity. So it is restated as: 'more electronegative atom not only prefers to stay in the orbital having more $\mathrm{p}$ character but also can increase the $\mathrm{p}$ character in its attached orbital from the central atom depending on the circumstance.'
With increase in $\mathrm{p}$ character in an orbital, bond length will increase while with increase in $\mathrm{s}$ character in an orbital, bond length will decrease. For example, $d_\ce{(C-Cl)}$ in $\ce{CH3Cl}$ ($\pu{1.78 \unicode{xC5}}$) $\gt d_\ce{(C-Cl)}$ in $\ce{CF3Cl}$ ($\pu{1.75 \unicode{xC5}}$).

Source: Concise Inorganic Chemistry for JEE (by J.D. Lee and Sudarshan Guha)
I understand this alternative statement to Bent's rule.
Can anybody explain how have the $\ce{C-Cl}$ bond lengths been compared here? I don't get this. I understand that fluorine is more electronegative than hydrogen and might induce small positive charge on the carbon atom. Also, the $\ce{C-F}$ bond will have more $\mathrm{p}$ character than the $\ce{C-H}$ bond. How do I proceed further to reach the conclusion?

The paragraph you cite does not define Bent's rule (see What is Bent's rule?, Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?).

If you compare $\ce{H3CCl}$ and $\ce{F3CCl}$, then the $\ce{C-H}$ bond has less carbon-p character than an idealised sp³ orbital, while the $\ce{C-F}$ bond will have more carbon-p character than the idealised sp³ orbital. This in turn means that in $\ce{H3CCl}$ the $\ce{C-Cl}$ bond will have more carbon-p character as an idealised sp³ orbital and is therefore longer, while in $\ce{F3CCl}$ the $\ce{C-Cl}$ bond will have less carbon-p character and will therefore be shorter.

Note of caution: I have read many questions about the concepts introduced in J. D. Lee's Concise Inorganic Chemistry; many of which revolve around outdated theories. I do not have a copy of that book, but from what I have experienced so far, I do not think it is a good book. Alas, in this instance it appears to be correct though.


How to standardise NaOH? My aim is to "To determine the concentration of citric acid in lemon juice using titration technique." and as following i need to standardise the NaOH in order to titrate it with validity. Is there a procedure and step to standardising NaOH. Thank you

You should find it in any analytic chemistry textbook, considering volumetric methods.
We did so in the high school.

In short, a solution of known molar concentration of well defined acidic substance is made, that get titrated by the $\ce{NaOH}$ solution.

Typically oxalic acid $\ce{(COOH)_2 . 2 H2O}$, weighted at good scales, preparing solution $\pu{0.05 mol / l}$.

When you Google "acidimetric titration sodium hydroxide oxalic acid" , you get tons of references, from illustrating videos to PDF documents.


How does adding acids (H+ ions) to limited soluble salts help it further dissolve? For example, adding $\ce{HCl}$ to the very insoluble $\ce{CaF2}$ will help it further dissolve. How?

I believe the concept deals with equilibrium and Le Chatelier's Principle:

$$ \ce{ CaF2 <=> Ca^2+ + 2F^-} $$

Does adding $\ce{H+}$ react with the $\ce{F-}$, which shifts the equilibrium to the right-hand side and therefore further dissolve the salt?

In general, the solubility of a sparingly soluble salt having a basic anion will increase with the addition of a strong acid.  Given:

$$\ce{X+A- <--> X+ + A-}$$

where $K_{sp}$ is small and A- is a weak base, the addition of a strong acid will form HA, driving the reaction to the right and increasing $K_{sp}$.

In the specific case of

$$\ce{CaF2 <--> Ca^2+ + 2F−}$$

the addition of $\ce{HCl}$ gives

$$\ce{CaF2 + 2HCl <--> Ca^2+ + 2Cl- +2HF}$$

driving the reaction to the right, thus increasing the solubility of $\ce{CaF2}$.  Note that $\ce{HF}$ itself is a weak acid, having a pKa of $3.17$, and will be in equilibrium with $\ce{H+}$ and $\ce{F-}$.


Extended Hückel method: what are the atomic orbital energies usually used? I was checking out an extended Hückel (non-iterative) code, and realised that for s-block elements it would be impossible to simply use atomic ionisation energies as an approximation to the orbital potential energies (due to the p orbitals being empty). In this case, what values are used?

Please see the following reference:

  A. Herman, Modelling Simul. Mater. Sci. Eng. 2004, 12, 21.

It has $H_{vv}$ terms for different orbitals of all elements of the table.


What does pO2 of blood mean and why do we use it? I understand the basic Dalton's law of partial pressures in gases. Also, Henry's law of diffusion, says, the concentration of gas dissolved in a fluid is proportional to the partial pressure above it.

So if we say that the $p(\ce{O2})$ of oxygenated blood is $\pu{100 mmHg}$, where is the free gas existing in equilibrium with dissolved gas? Does it mean that the blood has a concentration of oxygen equal to that when placed in a surrounding of $p(\ce{O2}) = \pu{100 mmHg}$? If yes, why don't we directly report in concentrations instead? Is it easier to measure?

Wikipedia also says that the Henry's law doesn't stand if the gas is reacting. But isn't oxygen reacting with the Haemoglobin?

There is a good explanation in Relating oxygen partial pressure,
saturation and content: the
haemoglobin–oxygen dissociation
curve Breathe 2015; 11: 194–201

  The partial pressure of oxygen (also known
  as the oxygen tension) is a concept which often
  causes confusion. In a mixture of gases, the
  total pressure is the sum of the contributions
  of each constituent, with the partial pressure of
  each individual gas representing the pressure
  which that gas would exert if it alone occupied
  the volume. In a liquid (such as blood), the partial
  pressure of a gas is equivalent to the partial
  pressure which would prevail in a gas phase in
  equilibrium with the liquid at the same temperature.
  With a mixture
  of gases in either the gas or
  liquid phase, the rate of diffusion of an individual
  gas is determined by the relevant gradient of its
  partial pressure, rather than by its concentration.
  While in a gas mixture, the partial pressure
  and concentration of each gas are directly proportional,
  with oxygen in blood the relationship
  is more complex because of its chemical combination
  with haemoglobin. This allows blood
  to carry an enormously greater concentration
  (content) of oxygen than, for example, water
  (or blood plasma). Measurement of $p_\ce{O_2}$, therefore,
  does not give direct information about the
  amount of oxygen carried by blood.

So blood $p_\ce{O_2}$ does not correspond to a particular concentration of oxygen, because the concentration of haemoglobin can vary, and most of the oxygen is bound to the heme iron.

$P_\ce{O_2}$ is the partial pressure of oxygen in a hypothetical gas phase which would make the blood oxygen and gas phase oxygen be in equilibrium.