Buffer Solution - Sodium Benzoate A buffer with a pH of $\mathrm{4.79}$ contains $\mathrm{ 0.39\ M}$ of sodium benzoate and $\mathrm{0.10\ M}$ of benzoic acid. What is the concentration of $\mathrm{[\ce{H} ]}$ in the solution after the addition of $\mathrm{0.058\ mol}$ of $\ce{HCl}$ to a final volume of $\mathrm{1.6\ L}$?

My thought is that
$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]}{[acid]})}$$

When we added $\mathrm{0.058\ mol}$ of $\ce{HCl}$, the concentration of both acid and base decreased by $\frac{.058}{1.6}$, so I take that ratio and apply it.  I found $\mathrm{pK_a}$ by just applying what's given.  Not sure where I am going wrong.

In that equation, the acid and base are a conjugate acid/base pair.  We assume HCl quantitatively dissociates and protonates benzoates.  The concentration of the acid increases, and the base decreases.

Empirically, the pH should go down in this case.  Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:

$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$
From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.

Just applying the difference in the log values of the changed ratios, I get  $\mathrm{pH = 4.61}$


Gaussian calculation problem - Maxcycle values for Opt and SCF I've been trying to calculate vertical ionization potentials (IP) and electron affinities (EA) for a bunch of prostaglandin derivatives (23-25 heavy atoms). My calculation setup in Gaussian09 looks mostly like that:

# opt=(tight,maxcycle=1000)
freq=noraman cphf=noread
b3lyp/6-31g(d) geom=connectivity
integral=grid=ultrafine scf=maxcycle=1000

As you can guess, I have been struggling with converging the geometry and finding the stationary point for my molecules. Some were kind of stubborn and wouldn't converge, so in desperation I increased the Maxcycle values for Opt and SCF from a 1000 up to 5000, and I succeeded with some of them. Now, my question is if I need to recalculate all other prostaglandins that have already nicely converged with lower Opt(Maxcycle) or SCF(Maxcycle) options (I used them in different combinations like 1000, 3000 and 5000)? The thing is that I want to be able to objectivelly compare the IP and EA values between different derivatives.

I am concerned because I remember that someone told me that if I went for SCF=QC option I would need to recalculate the whole set of molecules with the same keyword option. However, as far as I know SCF=QC is like a substantial change in the setup as it uses another algorithm for calculations. Hopefully, changing the SCF(Maxcycle) value is not that dramatic, is it? How about SCF=Verytight - would it demand recalculating everything or not?

Another thing that bothers me is Opt=Tight. What if I decided to go for Opt=Verytight. Would I need then to recalculate all of the molecules with the same setting?

And lastly, what if I changed Integral(grid=ultrafine) to Integral(grid=ultrafine, acc2e=12) which is supposedly the most precise approach?

You are mixing two independent, but slightly related topics.

First is the SCF convergence, i.e. obtaining the wavefunction. This step is necessary for any calculation, without it the Gaussian crashes. For well behaved molecules, as the prostaglandines (organics) most probably are, the solution should be found fast (<30 SCF cycles). It also does not matter which converger you choose, i.e. QC or DIIS. How "tight" you decide for the SCF affects the final energy, but only the number of digits (precision). The SCF does not have any "verytight" option, according to manual, only the Conver=N.

On the other hand, the integral and grid settings do have effect on the energies, so all calculations have to be done with the same grid. I won't recommend changing the acc2e settings, the error is negligible.

Regarding the geometry convergence, you should obtain an optimized geometry in less than 100 steps for reasonable organic systems. If this does not happen, something is rotten in your setup and you should look deeper into the problem. In principle, you cannot converge the geometry to "tight" without sufficient accuracy of the SCF, namely the ultrafine grid, which you now do use.

How good the optimization was can be guessed by the frequencies which should be zero (translations and rotations). If they are significantly off, optimize the geometry more tightly. But as said above, you cannot optimize on noisy surface, where the noise is produced by the grid and SCF convergence.


Hell-Volhard-Zelinski Reaction What is the role of red phosphorus in hvz reaction?
Initially I think that red phosphurus is nothing but a catalyst. But it is not a catalyst.

TL;DR: When $\ce{PBr3/PCl3}$ is not available as reagent, red phosphorus is used which is reacted to bromine or chlorine to form the reagents in situ.
Long answer:
Hell–Volhard–Zelinsky halogenation reaction halogenates carboxylic acids at the α-carbon.  The reaction is initiated by addition of a catalytic amount of $\ce{PBr3}$, after which one molar equivalent of $\ce{Br2}$ is added. $\ce{PBr3}$ replaces the carboxylic -OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol, which will readily react with the $\ce{Br2}$ to brominate a second time at the α position.

Now, what happens if $\ce{PBr3}$ is not available?

You can use red phosphorus as a reagent. It is reasonably air stable (whereas white phosphorus combust in air almost spontaneously). Phosphorus reacts with bromine to give phosphorus tribromide which helps in the conversion of carboxylic acid into an acyl bromide.
$$\ce{2P_{(red)} + 3Br2 -> 2PBr3}$$

Why exactly $\ce{PBr3/PCl3}$ is needed and why an extra molar equivalent of
$\ce{Br2/Cl2}$ is needed?

When left out in the air, $\ce{PBr3}$ will readily combine with any water vapor present to form a new $\ce{P-O}$ bond along with $\ce{HBr}$. The strength of the $\ce{P-O}$ bond [about $\pu{140 kcal/mol}$] makes $\ce{PBr3}$ very useful for swapping out $\ce{C-O}$ bonds for $\ce{C-Br}$ bond and that's how carboxylic acids gets converted to acyl bromides.
However, $\ce{PBr3}$ by itself is not sufficient to perform the HVZ reaction. A molar excess of $\ce{Br2}$ is added to the already present $\ce{PBr3}$, which perform the halogenation of the alpha-carbon. That's why the reagent is always mentioned as ($\ce{PBr3 + Br2}$).
References

https://en.wikipedia.org/wiki/Hell%E2%80%93Volhard%E2%80%93Zelinsky_halogenation
https://www.organic-chemistry.org/namedreactions/hell-volhard-zelinsky-reaction.shtm
https://www.masterorganicchemistry.com/2020/08/26/the-hellvolhardzelinsky-reaction/

That generates another question: Can we use PI3 as reagent in HVZ reaction?


If I want to sample rate constants of chemical reactions, what the distribution will be appropriate? For example, I have a (chemical) reaction systems, I want to sample the parameter space. What will be the best and validated distribution for sampling? And why?

Since you have accepted my proposal of using a free-energy calculation, you can use that in a combination with classical Transition State Theory (think of it as an extension to the Arrhenius equation).

More practically, you can read-up, for example on

GROAMCS
NAMD

For more options, see this overview presentation.


Elements with no solid state Helium superfluid-3 has got at around 2K a special liquid state with 0 viscosity with interesting properties. I am really interested in what would happen at 0K. With no kinetic energy in the particles of the compound, would it be solid? Do elements with no solid state at all exist?

First of all, we cannot achieve temperatures of 0 kelvin in experiments, that is forbidden by quantum thermodynamics.

Currently, experiments can go down to seriously low temperatures of nanokelvins (source: a friend is working in that field, but spontaneously the only reference I can give  you is Wikipedia.)

At those temperatures the phases of 'solid' or 'liquid' don't really exist anymore. Rather collective quantum effects play a role, that make new phases appear, namely Bose-Einstein condensates and Fermi gases. What phase appears depends on the particles spin, which allows or disallows them to occupy the lowermost energy level.

Those exotic phases of matter count however still as 'condensed matter', which all elements at cold temperatures will have. The classical solid phase, which you're asking about, can be circumvented this way, but not at all pressures, as others correctly pointed out.


Carbanion Stability order Why is a 3° Carbanion unstable than 1° Carbanion? Carbon atom with negative charge will become little electropositive relative to a neutral carbon. So if a Carbanion is attached to three neutral carbons, essentially it should donate it's electrons distributing the negative charge of -1, thus stabilizing it. By this logic, more carbon atoms attached means more stability. But this does not happen. 1° Carbanion is most stable. Why?

Carbanions and Carbocations are perfectly opposite in order of stability because they are opposite in order of charge. The order is preferred of carbocations the same as acidity by Lewis' nucleophile/electrophile definition. Therefore, because a positive charge is stabilized better by a tertiary carbon, it is preferred to host a carbocation because the electron shells of its substituent carbons help stabilize the charge.
In the case of a carbanion, the negative charge is best stabilized when not in contact with neighboring nuclei that would resonate the negative charge away from the nucleophilic site. If you think about it like sterics, the further away the negative charge can be from other substituents, the better in the case of carbanions.


Are there proteins with permanently bound magnesium? Do you know something about the existence of a magnesium containing protein/enzyme in which magnesium is coordinated in the structure and the structure is stable? (like iron in haemoglobin or zinc in zinc fingers)

From the look of things you may actually be referring to a lot of different proteins, however I managed to
grab two that I found fascinating:

Enolase

A glycolytic enzyme which catalyzes the reversible dehydration of 2-phosphoglycerate to phosphoenolpyruvate:

Credits: Lehninger Principles of Biochemistry page 213 ("Enzymes")

In the enzyme active site, 2-phosphoglycerate undergoes strong ionic interactions with two bound Mg2+ ions causing the resulting intermediate to be easily abstracted.

GTPase specifically  ARF1 (ARF for ADP-ribosylation factor). These are important in priming phase in the formation of Clathrin-coated vesicles (CCVs).

Clathrin itself is  a protein that forms a polyhedral framework around vesicles that transport TM, GPIlinked, and secreted proteins from the Golgi to the plasma
membrane.

X-ray structures of ARF1.GDP and ARF1.GTP, were determined by Dagmar Ringe and by Jonathan Goldberg:

Credits Voet and Voet Biochemistry 4th ed page 435 ("Membrane Assembly and Protein Targeting")

X-ray structures of (a) ARF1.GDP (left) and (b)ARF1.GMPPNP (right) (GMPPNP is a nonhydrolyzable GTP analog in which the O atom linking GTP’s beta- and gamma-phosphorus atoms is replaced by an NH group.)

The bound nucleotides are drawn in stick form in white with their phosphorus atoms magenta and their bound $\ce{Mg^2+}$ ions shown as lavender spheres.

In ARF1.GDP, the protein’s N-terminal helix (red) together with its
covalently linked myristoyl group (not present in the X-ray structures) are bound in a shallow hydrophobic groove on the
surface of the protein formed in part by the residues of loop gamma 3.


The self-accepted answer given here is wrong.

Each line in a bond block of a MOL file represents one bond between two atoms.

A line in a bond block has the following compact format:

111222tttsssxxxrrrccc

With other words, each entry (=different number or letter) in such a line can take up to three spaces. If it needs less. e.g. when the atom number is smaller than 100, the data is right-aligned.

111 is the number of the first atom

222 is the number of the second atom

These atom numbers refer to the position of the atom in the atom block in the same file, where the cartesian coordinates and the atom type (element) are given.

ttt is the number decoding the bond type: 1=single, 2=double, 3=triple, etc.

sss describes the stereochemistry, 1 is a single bond pointing up, 6 is a single bond pointing down, etc.

The latter are used in the flat 2D MOL file of morphine cited in the question, e.g. for the only hydrogen atom in the file: a bridgehead H-atom.


Estimate Partition Function from a simulation I have a system and I've carried out a long molecular dynamics simulation over it. I would like to estimate the partition function $Z.$ Theoretically, one would compute:
$$Z=\dfrac{1}{N!h^{3N}}\int \exp \left(-\beta\frac{-p^2}{2m} \right) \exp\left(-\beta V(r) \right) \ dp \ dr, $$
but if the system I'm dealing with is, say, a protein in water or something like that, this integral is of course absolutely untractable. When thinking of how to estimate the partition function of susch a system from a molecular dynamics simulation, I guess the first thing that comes to mind is the following: Consider all the snapshots obtained during your simulation, label them with the index $j$ and call $E_j$ to the energy of each snapshot. For giant thermalized systems, I guess we can neglect the probability of a concrete configuration repeating, so I believe a nice try would be:
$$\tilde{Z}=\frac{dV_{\text{ps}}}{N!h^{3N}}\sum_{j} \exp\left( -\beta E_j \right),$$
where $dV_{\text{pc}}$ is the volume of small cube in phase space (its volume would have to be adjusted manually depending oh how many snapshots one has taken).

My question, briefly put, is, how can I estimate the error of estimating $Z$ by $\tilde{Z}$?

It seems intuitive that typically $\tilde{Z}$ should be a good aproximation to $Z,$ but one can easilly think of situations where this isn't true: Imagine that my initial configuration lies in, so as to speak, a valley in the energetic landscape, surrounded by a chain of "high mountains" (high-energy configuration) and that at the other side of these "mountains" lie low-energy configuration. It is clear that in such a case there could be a considerable error when estimating $Z$ from $\tilde{Z}.$ But since one does not know the aspect of the energetic landscape beforehand, it seems difficult to deal with this.

The question stops here, but just to provide context I'll explain briefly what I'm thinking about: I want to study a chemical reaction, and to that end I would like to know the free energy profile along a certain reaction coordinate. For that, an Umbrella Sampling technique is used: you drag the reaction coordinate along the reaction path, forcing it to be at certain positions at different steps (otherwhise, the simulation wouldn't take you over the energetic barriers in short simulation times). I am aware of several methods to compute free energies, like FEP or WHAM, which is quite sophisticated, but I was guessing that if the technique I mentioned above to estimate the partition function $Z$ was correct, then the goal of getting the free energy profile would be fairly simple: you just drag along the reaction coordinate, perform molecular dynamics simulations for different restricted values of the reaction coordinate (you force the reaction coordinate to be at a concrete position by means of a harmonic potential), and at each value of the reaction coordinate you estimate the partition function as above. That's the idea and that's why I am interested in having some idea on methods to estimate the error of approximating $Z$ by $\tilde{Z}.$

Note: this question was originally in Physics.stackexchange. It didn't receive any attention at all and I asked for it to be migrated to Chemistry.stackexchange, but the flag was ignored, so I've decided to delete my question on Physics.stacexchange and post it here

What you're saying does make sense. You can, in fact directly estimate the partition function using nested sampling, with error bounds. See for example this paper http://journals.aps.org/prb/abstract/10.1103/PhysRevB.93.174108 . In your case without PBC it's actually even simpler. Also check out Livia's blog here https://liviabpartay.wordpress.com/publications/


What is the oxidation state of Calcium in teeth? What is the oxidation state of Calcium found in human teeth?

As a group II metal, you will be hard pressed to find calcium in the 1+ oxidation state (a quick google search yields an isolated report of calcium(I) chloride). Calcium's willingness to oxidise is evinced by its vigorous reaction with water. As you have noted, the calcium in tooth enamel is hydroxyapatite, often with varying degrees of fluorination. Which brings me to your next question, which hopefully you or a mod can slipstream into the main body -

I would definitely warn against eating chalk without good reason as it will evolve $\ce{CO2}$ in your tummy and raise your stomach pH, and is associated with a variety of negative health outcomes:

Milk-Alkali Syndrome
Hypercalcaemia
Antacid side effects

As always, all things in moderation. On a personal note eating chalk sounds like exactly the thing certain sectors of the 'alternative health' crowd would be into (namely, those who assert that the body is 'too acidic' (there's another crowd who claim the opposite and will recommend, say, drinking $\ce{HCl}$ (through a glass straw, natch)), ignoring for a moment the fact that the human body is pretty good at maintaining its pH and will up and have a cardiac arrest or something if the pH drifts more than a bit)


Structure of aminoglycosides Attached is the structure of one aminoglycosides, kanamycin. I understand that the ring binds to 4 position of 2-DOS is numbered with prime (') and sugar that links to 4 or 5 position of 2-DOS is numbered with double prime ('') but my question is how to do the numbering for 2-DOS (center ring). Can anyone teach me how? I have searched through Google and still have no idea on it.

Like @orthocresol said in his comment, biomolecules are often numbered by convention. The 'traditional' numbering for 2-deoxystreptamine is shown below on the left, and IUPAC numbering (not used) is shown on the right:

The most likely rationale for the traditional numbering is that the compound is a diamine so the numbering starts from one of the amino substituents and goes towards the other one.

EDIT:
OK, now I understand exactly where lies your confusion. The molecule of 2-deoxystreptamine (2-DOS) has a plane of symmetry. If we are to number the molecule alone, it doesn't matter from which amine group we will start. But as you noticed, the kanamycin is asymmetrical and there are indeed two ways to number the ring of 2-DOS. To solve this problem, we need to look on the structures of other aminoglycosides and their biosynthetic pathway. Many of the antibiotics from this class are biosynthesized through an intermediate compound known as paromamine (I prefer Haworth projection):

The numbering is logical - if we would start from the other amine then the glucosamine substituent would be at atom 6 (against the usual rules of numbering where we prefer lower numbers for substituents). Most aminoglycosides can be divided into two groups: 4,5-disubstituted (neomycins, paromomycins) and 4,6-disubstituted (kanamycins among others). Therefore the red numbering in the original question is the right one, if we assume the synthetic pathway described in this article:

Park J.W. et al, "Discovery of parallel pathways of kanamycin biosynthesis allows antibiotic manipulation" Nat. Chem. Biol. 2011,  7(11):843-852 (DOI: 10.1038/nchembio.671).


How to determine the boiling point of substances? Which of the substances has the lowest boiling point?
a) $\ce{H2O}$ b) $\ce{H2S}$ c) $\ce{H2Se}$ d) $\ce{H2Te}$

I've been searching for any formula or rule to determine this, but I read that it didn't exist.
Do I have to remember some boiling points by heart?

While it may be difficult to determine the exact boiling point of a substance, many factors play to compare if a substance has a higher boiling point than the others.

1) Intermolecular force. Notice that it is "intermolecular force" instead of
"intramolecular force". The reason being when you boil something you want to transform the substance from liquid state into gas state, and that can only be achieved by weakening the connection of molecules, not by breaking the intramolecular bond of a molecule. There are generally three types of intermolecular force : London dispersion force [weakest], dipole-dipole, and hydrogen bonds [strongest]. The stronger the force, the larger the amount of energy needed to break off the connection between the molecules, thus the boiling point is higher.

London dispersion force (LDF/Van der Waals) usually occurs when there exists no significant dipole in the molecule (propane, hexane)
Dipole-dipole occurs when there exist an observable amount of charge difference between atoms in a molecule. This can be seen in the case of hydrogen chloride. The chlorine atom is more electronegative, thus it attracts the more electropositive hydrogens of other molecules too (but doesn't bond with them).
The hydrogen bond is the strongest one. It occurs if a molecule has a hydrogen atom bonded to either F (fluorine), O (oxygen), or N (nitrogen). These atoms are highly electronegative. This happens in the case of water $H_2O$. I reccomend you to think why the hydrogen bond is the strongest and Van der Waals (London dispersion force) is the weakest intermolecular force.

2). Molecular mass. Suppose you have two substances, both of which happen to interact through LDF. Another thing to consider is the mass of the molecule. If the molecule is bigger, then the surface area is larger, which results in a bigger LDF. This leads us to the conclusion that the mass of the molecule is proportional to the boiling point.

3). Branches. In alkanes (consisting of C and H only), generally a straight chain alkane has a higher boiling point than similar branched alkanes due to the surface area between two adjacent molecules. This would be harder if the alkane has branches. Compare : 2,3-dimethylbutane and hexane. Both have 6 carbons, but the boiling point of 2,3-dimethylbutane is 331.15 K whereas hexane is 341.15 K.

Take a look at your case. $H_2O$ has oxygen in it, so hydrogen bond. Although the electronegativity of hydrogen, selenium, sulfur, and tellurium are around 0.1 - 0,48 difference, it doesn't really contribute much to the overall dipole of the molecule. So these three must interact with LDF. We can see that the molecular mass of sulfur < selenium < tellurium (S < Se < Te). From what we know, the bigger the mass, the larger the boiling point.

From this point we can tell that :
$H_2S < H_2Se < H_2Te < H_2O$

And a quick Google search will show us that the boiling point of these molecules are :
$H_2S (-60) < H_2Se(-41,25) < H_2Te(-2.2) < H_2O$ (100)


Reaction of haloalkane with hydroxide ions My teacher gave me the following question:

What are the possible products of these reactions? Then, what is the mechanism of those reaction (elimination (E1 or E2) or subtitution (SN1  or SN2)? Explain in as much detail as possible.
$$\ce {CH3Cl + OH- ->} ?$$
$$\ce {(CH3)3CCl + OH- ->}?$$

I suggest that the first reaction will produce methanol and chloride:
$$\ce {CH3Cl + OH- -> CH3OH + Cl-}$$
and it is a SN2 reaction, because $\ce{OH-}$ attacking from behind and this makes $\ce{Cl}$ leave the chloromethane. It becomes $\ce{Cl-}$ because it brings one electron from the carbon atom.
Then I suggest the second reaction is a SN1 reaction because there are some barriers behind the $\ce{C-Cl}$ (the barrier is $\ce {(CH3)3}$). But I don't know the product that can be formed.

You are on the right track with the hydroxide substituting the chloride, so you can predict the alcohol product in each reaction. In order to determine whether the reactions will be SN1 or SN2, we must look at the mechanism of each reaction:
SN1

SN2

Wikipedia:Nucleophilic_Substitution
So we can see that in an SN1 reaction, a stable carbocation intermediate must be formed, while in an SN2 reaction there is no intermediate. We can determine the stability of our carbocation intermediate by looking at the following stability diagram:

Wikipedia:Carbocation
From the diagram, you can see that carbocation stability increases with the number of alkyl groups attached to the ionized carbon. If the carbocation is relatively stable, the reaction will likely be SN1, if the carbocation is relatively unstable, the reaction will likely be SN2.
Another important idea, which you have mentioned, is that a central carbon surrounded by bulky groups will be more difficult to attack due to steric hindrance (thus not likely an SN2 reaction), and becomes more easily attacked when dissociated to a carbocation.
You can use these concepts to find/explain the answers to your question.


Why can't helium nuclei (alpha particles) react with fluorine? Alpha particles are basically just helium nuclei, so it will accept an electron pair to become stable.

Will fluorine, being highly electronegative, not just donate an electron pair but form a bond with helium?

Why, many atoms would readily create a bond with helium, when it comes in a form of alpha particle (just cooled down enough to chemically interact with). There is comparatively stable $\ce{HeH}^+$ and other similar particles, too. But a positive particle is not a compound yet. And when you try to form a compound out of it, that is, to balance it with some negative ions - well, that's where the problem arises. It would violently react with absolutely anything else, form some compound, and happily release the neutral helium atom.


Trifluoroiodomethane in basic medium $$\ce{CF_3I + NaOH(aq) -> Major Product}$$

The Major product, according to me, should be ditrifluoromethyl ether $\ce{CF3-O-CF3}$. The hydrogen on the $\ce{-OH}$ group is quite acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism:

However, the textbook gives the following reaction mechanism and claims that the product is trifluoromethane $\ce{CHF_3}$

Which one of these is correct?

You didn't give a mechanism for your expected product, so I can only speculate as to how you intended to form the final product.

If you thought to use an Sn2 mechanism, I'd say that the reaction is highly improbable as the halide is attached to a trisubstituted carbon, i.e. a carbon with a lot steric hinderance, and therefore, it would be rather difficult for an Sn2 reaction to take place.

If you chose to use an Sn1 mechanism: with the first step, when iodine leaves, you would create a primary carbocation, which is unstable. Additionally, being that fluorine is an electron withdrawing group, the three fluorines would likely serve to further retract electrons from the already electron-poor carbon, and not serve to stabilize the primary carbocation. Therefore, I doubt that this reaction would occur as well.

In the textbook provided mechanism, the oxygen attacks the accesible iodine, which is next to a very psuedopositive carbon. And although this places a  negative charge  on the carbon atom, the electron-withdrawing fluorines serve to help stabilize the carbanion via their inductive properties. In the second step, the carbons rids itself of its negative charge, by taking a hydrogen atom from a nearby hypoiodous acid (pKa 11).

References:
McMurry, J. (2015). Organic chemistry: with biological applications. Stamford, CT: Cengage Learning.


Energy absorbed by a system in isothermic process in which the phase changes from liquid to gas A system contains $x$ mole of material, with surface area $S$ is pressed by a mass $M_1$ (no fraction and no any other outer forces involves). Let $\Delta H_\mathrm{l\rightarrow g}$ be the transition enthalpy from liquid to gas. The system is position in $(T,P_1)$ such that the material phase is liquid for those temperature and pressure values ($T$ and $P_1$). Now instantaneously the mass $M_1$ is decreased to to an unknown value which we denote $M_2$ so the pressure decreases. In case that the temperature stays constant until the inner pressure is equal to the outer pressure $(T,P_2:=\frac{M_2g}{S})$ system gets to back to equilibrium, and the material phase in $(T,P_2)$ is gas, I wish to calculate how much energy was absorbed by the system till it gets to pressure equilibrium $(T,P_2)$.

In the question is says that the volume of the liquid can be neglected when compared to gas' volume. And that the volume of the water is independent of the pressure.

What I did so far:

The chemical equilibrium pressure (i.e the pressure in the phase transition point) can be calculated because the exercise gives the triple point $(T_\triangle , P_\triangle)$ so using the transition enthalpy constant and the constant $T$ of the process, one can calculate $(P_\mathrm{l\rightarrow g, eq},T)$ (using Clausius–Clapeyron relation). And by $PV=nRT$ one may calculate $V_\mathrm{eq} = \frac{nRT}{P}$ the volume in the chemical equilibrium point. By the unstruction $V_\mathrm{eq}$ is the total volume of the material (we neglect the liquid volume)

We know the starting inner pressure which is $\frac{M_1g}{S} \text{[Pa]} $ and by the given instruction the pressure doesn't change the liquid volume we learn that until the phase transition point, the volume of the system stays constant $V_\mathrm{eq}$ which we calculated above.

When the system gets to pressure equilibrium point $(T,P_2=\frac{M_2g}{S})$ we don't know the volume or $P_2$ because $P_2$ is unknown how every we not that its larger then $V_\mathrm{eq}$.

I'm pretty much stuck here, I don't see how to translate this information to evaluate the amount of energy absorbed. It seems to me that maybe one more piece of information is needed, but the exercise doesn't give one.

As indicated by @Buck Thorn,the work done by the substance on its surroundings is $P_2(V_2-V_1)$.  So, from the first law, $$U_2-U_1=Q-P_2(V_2-V_1)$$Solving for Q then gives:
$$Q=(U_2+P_2V_2)-(U_1+P_1V_1)-(P_2-P_1)V_1=H_2-H_1-(P_2-P_1)V_1$$Here, $$H_2=H_1+(H^{sv}_1-H_1)+(H_2-H^{sv}_1)$$where $H_1$ is the enthalpy of saturated liquid in State 1, and $H_2$ is the enthalpy of the superheated vapor that exists in State 2 (i.e., at a pressure lower than the equilibrium vapor pressure at the constant temperature T).  For an ideal gas, the change in enthalpy at temperature T from the saturated vapor in State 1 to the superheated vapor in State 2 is zero.  So the previous equation reduces to $$H_2=H_1+(H^{sv}_1-H_1)=H_1+x\Delta h^{vap}$$where $\Delta h^{vap}$ is the heat of vaporization at temperature T.  Substitution of this into the equation for the heat added gives:
$$Q=x\Delta h^{vap}+(P_1-P_2)V_1$$where $V_1$ is the volume of the saturated liquid at T. The second term on the rhs will typically be negligible.


Why do the field of study "Chemical biology" often return papers that are in general anything disease related? For example, if one tries to look up research opportunities or browse papers in the field chemical biology, most often return research fields related to biological sensing, drug activity etc. and which the ultimate goal of all these researches are related to treating diseases of some kind.

http://pubs.acs.org/journal/acbcct

and only occasionally, do the other non disease treatment related sector of chemical biology (those related to proteins, environmental biochemistry, microbial reactors etc.) can be found

http://www.ch.cam.ac.uk/rig/biological/about

http://chemistry.mit.edu/research/biological

http://www.kemi.uu.se/research/molecular-biomimetic/microbial-chemistry/

  Why are most chemical biology research dominated by disease treatment goals despite the chemistry of biological systems are not limited to the medical sector, or is it because the non disease related sector of chemical biology are often lumped into the environmental science, biotechnology and engineering sectors?

Because when we try to sell some new molecule to the chemical biology journals' editors and reviewers we think health and drug like actions sell better than a mere signalling molecule or so ?


Why is AlCl₃ is known as aluminium chloride while PCl₃ is known as phosphorus trichloride? Why is $\ce{AlCl3}$ is known as aluminium chloride while $\ce{PCl3}$ is known as phosphorus trichloride?

I think it is due to variable valency if phosphorus. Is it true?

Names of binary molecular compounds (which contain only two different elements, normally nonmetals, and do not contain positive and negative ions) can contain the prefixes mono, di, tri, tetra, penta, etc.
Normally, names of ionic compounds do not use these prefixes to indicate the number of positive or negative ions. $\ce{CaCl2}$ is calcium chloride, not "calcium dichloride".

However, aluminum chloride, $\ce{AlCl3}$, is sometimes called aluminum trichloride which is not incorrect in this case because it actually is a molecular compound (it has very polar aluminum-chlorine covalent bonds) even though it looks like it should be ionic since it contains metal and nonmetal elements typical of ionic compounds.
 $\ce{MnO2}$ is often called manganese dioxide rather than manganese(IV) oxide for the same reason. Usually, it is best to play it safe with compounds like these and use the ionic names:

$\ce{FeCl3}$ is ferric chloride or iron(III) chloride rather than "iron trichloride." $\ce{PCl3}$, a molecular compound (expected since only nonmetallic elements are present), is commonly called phosphorus trichloride, but is sometimes called "phosphorus(III) chloride."


Use of fractional distillation in organic chemistry Can fractional distillation be used to distinguish between:
1) d/l form of a compound
2) geometrical isomers of a compoun (Cis and Trans)

The boiling point of an (R)-isomer is the same as the one of the enantiomer, the (S)-isomer.   Since the condensor does not offer a chiral environment to the chiral molecules, there is no discern of them and consequently, neither "single" nor "fractional distillation" separates them.   (You may expose the molecules to a chiral surface to adsorb, though, and separate the enantiomers by chiral gas chromatography; a specialized approach in analytical chemistry and -- because of the scale of amount typically accessible -- much less frequently deployed on a preparative one (a review).)
A successful separation of (cis)-(trans)-isomers by distillation of course depends on how different the boiling points of the pure compounds are.  Looking up the data for (trans)-stilbene (bp 305 C reference) and (cis)-stilbene (307 C at 1 atm reference) suggests other separation techniques may be much more suitable since less demanding.  This especially becomes relevant if you need a technique easy to scale, to manage and to be run continously (e.g., distillation).


Is it stupid to put used acid into a pool after cleaning the chlorinator? I have a salt pool, and when the chlorinator salts up with calcium, I clean it with a diluted acidic solution.

What happens to the acidic solution after the reaction happens? I.e., does the pH change?

Would it be stupid to put the used solution into the pool afterwards? Am I correct in thinking the following sentence? "We are trying to get the calcium out of the pool, but putting it back in just means the chlorinator has to get it out again." Or, it is stupid for a different reason, e.g. the acidic solution is too weak after the reaction anyway?

The bicarbonate buffer capacity of the pool  water probably would deal with the remaining acid, but the pH would IMHO get lower.

But returning calcium to pool is not wise, as it will cause calcium accumulation and progressively shortening periods of calcium precipitation.


How come electrons in the d-subshell are considered valence electrons in transition metals? Similarly, why are f-subshell electrons considered valence electrons in the lanthanides and actinides?

I was under the impression that the outermost electrons, and therefore the electrons with the highest principal quantum number, made up the valence electrons.

Also, if the d-subshell electrons are considered valence electrons in transition metals, how come they're not considered valence electrons for elements in Groups 13-18?

I understand that the highest d-subshell is of higher energy than the highest s-subshell but I fail to see the logic in how it would be considered a valence electron.

The valence electrons are the ones responsible for engaging in chemistry and corresponds to the orbitals which are highest in energy. These electrons are the most easily accessed because they are not buried in the energy well of an atom. For main group elements, the highest principal quantum number p and s orbitals are the highest in energy, as you said, but for the transition metals, the partially filled d orbitals are the highest in energy. Generally, whichever orbitals are being filled last as you go across the periodic table are the ones that should be considered the valence electrons.

Another thing to note, as elements are ionized, the orbitals with higher orbital angular momentum drop in energy faster than orbitals with lower angular momentum. So, in the transition metals, the n d orbitals drop much more than the (n+1)s and p orbitals. Since, transition metals have <10 electrons (the number it would take to fill up all the d orbitals), all of them reside in d orbitals (and none in the s and p) and so these are the highest occupied orbitals and therefore the ones that engage in chemistry.


Is 1% m/m the same as 1% w/w? Hydrocortisone cream seems to be sold as 1% w/w, but my cream is 1% m/m, which is confusing. Are they the same units or the same concentrations?

“[M]ass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity.” (Wikipedia quoting [de Silva, G.M.S. (2002), Basic Metrology for ISO 9000 Certification, Butterworth-Heinemann, 214p.]
Well-known relationship:
$$
W = m \times g
$$
where W is weight, m is mass and g is the acceleration of gravity.
Because g is a constant that cancels out in a ratio, mass/mass and weight/weight are identical.
Likewise, the ratios % m/m and % W/W are identical.


Why is the rate of a reaction proportional to the concentrations of reactants raised to their stoichiometric coefficients? Consider a gaseous state elementary reaction $$\ce{aA(g) + bB(g)} \overset{k_\mathrm{f}}{\underset{k_{\mathrm{b}}}{\ce{<=>}}}\ce{ cC(g) + dD(g)}$$

I know that for this reaction, $$\Delta G = \Delta G^{\circ} + RT \ln \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) \tag1$$

Now, if $$K = \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) ,$$

then $\Delta G = \Delta G^{\circ} + RT \ln K$.

Now, how can we say that $$K = \frac{k_{\mathrm{b}}}{k_{\mathrm{f}}}$$

and $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$  and $\, r_\mathrm{b} = k_{\mathrm{f}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$
$\,$where $r_{\mathrm{f}}$ and $r_{\mathrm{b}}$ are rates of forward and backward reaction as assumed in this answer.

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$  and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

A late answer...

In a molecular approch to chemical reaction rates we asume that the molecules move randomly in solution and have a velocity randomly assigned according to the Boltzmann distribution and that the chance of being in one position in solution is the same as being at any other, i.e. the molecules are in thermal equilibrium at some temperature T.

The reaction rate for the reaction $\mathrm {A+B \rightarrow}$ can be found by considering the number of collisions per second and then this number reduced by the chance that a reaction occurs on collision. To arrive at a rate the chance that a randomly chosen combination of molecules of type A and B at time t will react in the next infinitesimal time interval $t+\delta t$ and has not reacted up to t has to be calculated. As a result the total number of collisions /second between $N_A$ molecules of type A and $N_B$ of type B is found to be $\sim N_AN_B\sigma_{AB}$, where $\sigma_{AB}$ is the cross sectional area for a collision, and this expression produces the rate  when  multiplied by the probability $p_{AB}$ of A and B reacting, i.e. rate $\sim N_AN_B\sigma_{AB}p_{AB}$. The probability of reaction $p_{AB}$ depends critically on the activation energy.

The important point is, however, that the  rate is proportional to the product of the numbers of molecules of type A and B which in constant volume is the same as being proportional to the product of concentrations. That it also is proportional to the collision cross section and the chance of reacting is also physically realistic. The rate constant is proportional just to the probability of reaction and the cross section for collision.

Thus the origin of the product of concentrations in the rate expression is found in the chance that a molecule of type A will collide with that of type B in a small time interval $t, t +\delta t$.  Thus if we ignore the fact that there is variation due to changes in experimental conditions, such as temperature, the electronics of the detectors used to measure the reaction etc. perfect kinetics would be measured, i.e $dA/dt$ or $ dB/dt$ vs. time would be noise free.

If we now consider the random nature of the collisions between molecules a slightly different approach has to be taken. If a small volume of the solution is considered and the rate calculated in that region, because the number of species entering and leaving in any given time period varies, the rate expression is also potentially subject to variation. This happens simply because of the random nature of the motion of molecules in a solution or gas. In this case the rate of change of a species X should be written in terms of its average (or mean) value represented as $\langle X(t)\rangle$;

$$ \frac{d\langle X(t)\rangle}{dt} = \langle f\times X(t)\rangle $$

where $\langle f\times X(t)\rangle$ represents some rate expression such as $k_1\langle X(t)\rangle + k_2\langle X(t)\rangle^2$ etc. except that the number of  $X$ are represented as their average values. Clearly the $X$ are always positive values. These  values have a variance (square of standard deviation) and so we would expect that different experiments give slightly different results because of this. (This is the approach taken by Gillespie, see 'Markov Processes' publ. Academic Press 1992)

In a bimolecular reaction such as $X + X \rightarrow $ with rate constant k the rate expression is

$$ \frac{d\langle X(t)\rangle}{dt} = k\langle \;X(t)(X(t)-1)\;\rangle/2 $$

The term $\langle X(t)(X(t)-1)\rangle /2$ arises becaus of the number of distinct X-X pairs in the reaction volume at time t. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that X is large compared to $1$ and so $\langle X(t)(X(t)-1)\rangle \rightarrow \langle X^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle X^2(t)\rangle \sim \langle X(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $X(t)=\langle X(t)\rangle$ which means that $X$ is a deterministic process, so that the randon nature of the numbers of $X$ entering into the reaction volume is always the same, which is not what was assumed at the outset and seems to fly in the face of what we know about how molecules move in solution, i.e. their random diffusional nature.

However, in any normal kinetic experiment the number of molecules we measure is vast, $\langle X(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the  number of molecules is tiny and the approximation made above is valid.

$$ \frac{\sigma_X}{\langle X \rangle} \sim \frac{\sqrt{\langle X \rangle}}{\langle X \rangle} \sim \frac{1}{\sqrt{\langle X \rangle}}\sim 10^{-9}$$

Even if there are as few as 1000 molecules the realtive fluctuation  is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.

Notes :

(A)  To see that the variance is zero use the definition $\sigma^2_X = \langle X^2 \rangle -\langle X \rangle^2$ where $\sigma$ is the standard deviation)

(B) There are other situations where the rate equation approach breaks down in macroscopic systems and that is when they are sufficiently non-linear, one example is the Schlogl system of equations, which for certain values of concentrations and rate constants becomes bi-stable;

$$B_1+2X \rightleftharpoons 3X; \;\; B_2 \rightleftharpoons X$$


Why is benzyne an intermediate? I've always seen benzyne (benzene with a triple bond) classified as an "intermediate". I really don't see why it needs to be an intermediate, though. The possible reasons I can come up with are flimsy:

It has an unstable sp2–sp2 $\pi$ bent bond: we have bent bonds with more angle strain in cyclopropane, which is not an intermediate. Though those bonds are $\sigma$ bonds.
It immediately reacts with itself to form a dimer, biphenylene, in the absence of other reagents: there are many other molecules which spontaneously dimerize (eg $\ce{NO2}$ at certain temperatures), yet they are not classified as "intermediates".

So, what makes benzyne an "intermediate"?

More generally (but possibly too broad), what makes a molecule an "intermediate"?

Intermediate is a middle step in the chemical process: reactants -> intermediate(s) -> products. Benzyne is formed from benzene and is highly reactive. Thus, it is never the first or last step of the process. To my knowledge benzyne is rarely observed as a product, because its reacting fast with other molecules because its triple bond.

What makes a molecule an intermediate? It is a highly reactive "middle product" of a reaction.
See here and there.


Why is only monochromatic light used in polarimeter? While using the polarimeter to determine optical rotation, why can't we use polychromatic light? What would change?

The reason is partly historical and partly scientific. Most of the polarimetric data exists with the sodium yellow emission at 589 nm because sodium lamps were among the most convenient ones and remember electricity is rather new. Early light sources were either flames or sunlight. It was easy to generate sodium emission by introducing salt in the flame. If you are familiar with Raman spectroscopy, it was discovered using sunlight (heliostats) not lasers in the 1920s-30s. Today nobody can think of using sunlight for spectroscopy.

At a more fundamental level, optical rotation is a function of refractive index and refractive index is a strongly dependent on wavelength. You have to see the optics of a polarimeter, the question of using polychromatic light does not arise. You will see a number of prisms are used. Impossible to use polychromatic light.


Reduction and oxidation in organic chemistry What is the meaning of reduction and oxidation in organic chemistry?
In general chemistry I learned an OIL RIG mnemonic, i.e. oxidation is loss and reduction is gain (of electrons).

But what do we say about their usage in organic chemistry? I have seen their usage a lot of times in the form of reducing agents and oxidizing agents but never really understood it. What really makes a substance a good oxidizing or a good reducing agent?

In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.

Reduction of alkenes (gain of hydrogen):

Oxidation of secondary alcohols (loss of hydrogen):

The strength of an oxidising or reducing agent can be evaluated by measuring its reduction potential. Good oxidising agents have highly positive $E^\circ$ values, meaning they are 'more keen' to gain electrons, whereas good reducing agents have highly negative $E^\circ$ values. For example:

$$
\begin{align}
\ce{MnO4- + 8 H+ + 5 e- &<=> Mn^2+ + 4 H2O} &\quad E^\circ &= \pu{+1.51 V} \\ \ce{Cr2O7^2- + 14 H+ + 6 e- &<=> 2Cr^3+ + 7 H2O} &\quad E^\circ &= \pu{+1.33 V}
\end{align}
$$

Since the redox potential for the $\ce{MnO4-}$ system is more positive, $\ce{MnO4-}$ is a stronger oxidising agent than $\ce{Cr2O7^2-}$ under acidic conditions.

The reduction potential is affected by the $\mathrm{pH}$ (along with many other factors — temperature, concentration of reactants — see the Wikipedia article for more discussion). Under neutral or alkaline conditions, $\ce{MnO4-}$ undergoes a different reaction with a less positive reduction potential, meaning it is a weaker oxidising agent under alkaline conditions that under acidic conditions.

$$\ce{MnO4- + 2 H2O + 3 e- <=> MnO2 + 4 OH-} \quad E^\circ = \pu{+0.59 V}$$

Reactive metals, such as zinc, tend to be good reducing agents. Zinc amalgam and hydrochloric acid are used in the Clemmensen reduction, to reduce ketones to alkanes.

$$\ce{Zn^2+ + 2e- <=> Zn(Hg)} \quad E^\circ = \pu{−0.7628 V}$$

Many commonly used reducing agents in organic chemistry, such as $\ce{LiAlH4}$ and $\ce{NaBH4}$, serve as sources of the hydride ion $\ce{H-}$. As an example, in the reduction of ketones using $\ce{NaBH4},$ hydrogen is added across the $\ce{C=O}$ double bond though a tetrahedral intermediate.


How high in (mmH2O) is the column of water in the tube? A long glass tube, sealed at one end, has an inner diameter of 23.4 mm. The tube is filled with water and inverted into a pail of water. If the atmospheric pressure is 735 mmHg, how high (in mmH2O) is the column of water in the tube (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL)?

This problem really confuses me, I know I have to use the ratio of heights of the liquid columns is inversely related to the ratio of the densities of the liquids:
hH20/hHg = dHg/dH2O. I just can't seem to put it together. I would appreciate any help on this problem.

Consider the air pressure in mm of mercury given. Is water more or less dense? Would an amount of water of the same mass as the mercury in the column take up more or less room?

So what would you need to do to change the 735 mm of mercury to an equivalent mass of water, given the densities cited in the problem?

The diameter of the column is red herring; don't go off on the kipper.


Stirring the reaction between hydrochloric acid and a powdered substance For an experiment, I'm adding several substances (as powders) to hydrochloric acid, and measuring the temperature changes to calculate enthalpy. However when adding the powders, the mixture requires some stirring to get all of the substance reacting.

As I am measuring temperature via a temperature probe hooked up to a data logger (to obtain a temp. against time graph), I need a way to briefly stir the mixture in a controlled way that won't affect the change in temperature (reactions are both endo and exothermic). I have tried stirring with the probe itself, but this is not ideal as the probe is pretty sensitive and friction/bumping with the side of the vessel may affect the recording.

I initially looked at a magnetic stirring to obtain uniform and controlled stirring, but then I realized the magnet would absorb heat and potentially increase the temperature of the mixture through kinetic energy. So I'm after a stirring method that will not affect temperature, so that the temperature change I record is solely due to the reaction.

Any suggestions are welcome, because I can't think of any tool or method that would suit.

To minimize thermal coupling between the stirring apparatus and the solution, the best choice is a small, slowly stirred, PTFE or equivalent coated magnetic stir bar. Heat generated by friction should be negligible.

A potential source of heat transfer into the system from this configuration is heat from the stirrer motor. To mitigate this, the surface of the stir plate should be covered with a strip of Styrofoam as thermal insulation. Furthermore, the reaction vessel could be physically separated from the stir plate by using a ring stand and clamp or equivalent. It may be necessary to adjust the thickness of the Styrofoam and distance of physical separation in order to keep the stir bar magnetically coupled to the stir plate magnet.

The setup should first be tested by doing several full runs without adding the acid or the powders. This is essentially an analytical blank. If the temperature changes can be considered negligibly small throughout the duration of an experiment, then you are ready to go. If necessary, any temperature change measured from the blank runs can be subtracted from the actual experimental measurements.


Mechanism for basic hydrolysis of α-chloronitrile to ketone? In Corey's 1969 prostaglandin synthesis,1 compound 2 is formed from compound 1 by:

  [...] treatment with 2.5 equiv of potassium hydroxide (added as a hot saturated aqueous solution) in dimethyl sulfoxide for 14 hr at 25-30°.

Can anybody explain the mechanism of this reaction? I thought of an $\mathrm{S_N2}$ reaction, followed by nitrile hydrolysis and elimination of $\ce{CO2}$. But $\mathrm{S_N2}$ there seems impossible, and the elimination of HCl should be more reasonable than decarboxylation.

Reference

(1) Corey, E. J.; Weinshenker, N. M.; Schaaf, T. K.; Huber, W. Stereo-controlled synthesis of dl-prostaglandins F2α and E2. J. Am. Chem. Soc. 1969, 91 (20), 5675–5677. DOI: 10.1021/ja01048a062.

Your analysis is slightly incorrect, actually. If you do a $\mathrm{S_N2}$ on the chloride, hydrolyse the nitrile, and decarboxylate, you end up with the wrong oxidation state:

Instead, if you actually get to the cyanohydrin, you could (in theory) simply lose $\ce{CN-}$ to form the ketone:

However, a $\mathrm{S_N2}$ reaction at that centre is going to be difficult, to say the least.

The reaction has been studied by Shiner et al.,1 who determined that the primary site of attack is likely to be the nitrile. They wrote:

  No mechanistic studies of the hydrolysis step have previously been reported, although pathways involving nucleophilic displacement of chloride or cyanide, affording intermediate cyano- or chlorohydrins, have been proposed. A priori we regarded both $\mathrm{S_N2}$ and $\mathrm{S_N1}$ displacements as improbable in these systems.

The immediate product of nitrile hydrolysis is the amide, and the α-chloroamide intermediates were actually isolated:

The mechanism after this involves a cyclisation of the amide onto the $\ce{C-Cl}$. The exact steps are not entirely clear. However, the following was proposed:

Reference

(1) Shiner, C. S.; Fisher, A. M.; Yacoby, F. Intermediacy of α-chloro amides in the basic hydrolysis of α-chloro nitriles to ketones. Tetrahedron Lett. 1983, 24 (51), 5687–5690. DOI: 10.1016/S0040-4039(00)94173-X.


Order for acidic strength of hydrides across a period I have learnt that in a group , say group $15$, the acidic strength increases on going down the group as

$NH_3 < PH_3<AsH_3<SbH_3<BiH_3$

This is due to decrease in the thermal stability on going down the group. But we can also reason it as follows : on going down the group , the electronegativity of the central atom decreases , and the bond between the central atom and hydrogen becomes less polar , so the bond strength also decreases hence it becomes easier to lose a hydrogen atom thus increasing the acidity.

Now if we compare the hydrides of a period, that is

$PH_3 , HCl , H_2S$ , according to the above logic , the order should’ve been $PH_3>H_2S>HCl$ because the electronegativity increases across a period. However , the correct order seems to be the reverse. Could you please explain why ? Is there a flaw in my reasoning?

The reasoning you gave for the increasing acidity of hydrides of group $15$ down the group is not the best one. You claim that less polar bonds are easy to break which is absolutely wrong. Non-polar covalent bonds are very difficult to break (think of beaking  $\ce{C-H , N-H , O-H }$ bond, which is easier to break ?)The logic can be modified as follows.

Actually the basicity of those hydrides decrease down the group, which in turn results in the increase of acidity of them. The reason for decrease in their basicity down the group is the increase in inability to donate their lone pair to accept the proton. In $\ce{NH3}$, the lone pairs recide in the $sp^3$ hybrid orbitals, whereas for other hydrides like  $\ce{PH3 , AsH3 , BiH3}$ the lone pairs recide almost in pure $s$ oritals (i.e. the hydrogen atoms are bonded to almost pure $p$ orbitals). So, as the $s$ character of the orbital where the lone pair is residing increases, it becomes closer and tightly bound to the nucleus, and thus difficult to protonate. Thus the basicity of group  $15$ hydrides decreases down the group, alternatively, the acidity increases.

Coming to the case of  $\ce{PH3 , H2S}$ and  $\ce{HCl}$, two factors govern the acidity of them. One is the  ease of bond-dissociation, and other is the solvation of the medium. When you consider the gas-phase acidity, i.e. no solvation is present, the bond-energy will only be the determining factor and bond energy is given as,  $\ce{P-H < S-H < Cl-H}$ (easily justifiable by  decreasing size of central atom), which tells us in gas phase the acidity of  $\ce{PH3}$ should be the highest. But normally when we talk about acidity, we talk about $pK_a$ in water. Whenever there is a medium, it stabilises the dissociated species via solvation (in water , this is hydration). As. the size of the conjugate base decreases across the period, the extent of hydration energy released will dominate over the bond-dissociation energy. Chloride is the smallest ion among them having higher electronic charge density, and $\Delta H_{hyd} \propto \frac{1}{r} $, It is highest for hydrolysis of  $\ce{HCl}$, making it a stronger acid among three.


Interference and absortion filters I am learning a bit about Uv-vis spectroscopy filters. As far as I know there are two commonly used: interference and absotion filters.
The first one works with the interference of light principle and the second is a glass with some pigments which absorbs light (of some specific frequency).

Question

Have you got any idea if there is any further classification of absortion filters? Any help will be greatly welcome.

There are a great many of each type of these filters.

Interference filters are optimized for different wavelengths, different bandwidths, transmission vs. reflection etc. by controlling the materials and the thicknesses of the layers. A commercial supplier, Edmund Optics, shows how to specify and order filters.

In a similar manner, absorptive filters depend on the chemicals used to make them. For example, a solution of $\ce{CuSO4}$, copper sulfate, in water removes wavelengths between ~600-750 nm, but letting blue light through..


Acid-base equilibrium of NH4CN I'm having difficulties understanding this problem.

We have in the problem that $K_\mathrm a$ for $\ce{HCN} = 6.2\cdot10^{-10}$ and $K_\mathrm b$ for $\ce{NH3} = 1.8\cdot10^{-5}$.

Write chemical equations to represent the dissociation of $\ce{NH4CN}$ in water, and the acid-base equilibrium equation.
Will the salt react as an acid or a base?

I know this is part of the solution:

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$$\ce{CN- + H2O <=> HCN + OH-}$$

I understand Le Chatelier's principle and I know that the reaction will shift to the side of less energy.

Fun!  weak acid versus weak base!
Ammonium cyanide, $\ce{NH4CN}$, is a solid where the atoms are grouped into the same ions that are generated in solution: $\ce{NH4+}$ and $\ce{CN-}$.  Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN.  Dangerous stuff!
The first part of the answer to 1. is the decomposition of $\ce{NH4CN}$ into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so.  So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is no $\ce{NH4CN}$ as such in solution.
The acid-base equilibrium equations can be the ones you wrote.  Again, you may want to add more information, like this:
$$\ce{NH4+ (aq) + H2O <=> NH3 (aq) + H3O+ (aq)}\quad K=5.5\times10^{-10}$$
Here the equilibrium constant $K$ is the acid dissociation constant of the acid $\ce{NH4+}$, the protonated form (aka conjugate acid) of ammonia.  I got its value by dividing $K_\mathrm w$, the dissociation constant of water ($1.0\times10^{-14}$), by the $K_\mathrm b$ of ammonia, given in the problem.
I encourage you to add similar information for the  equilibrium involving cyanide ion.  Hint: if you keep the reaction as you wrote it, the appropriate $K$ is not the $K_\mathrm a$ of $\ce{HCN}$ but the $K_\mathrm b$ of the deprotonated form (conjugate base), $\ce{CN-}$.
(But you can instead consider the alternate reaction $\ce{HCN + H2O <=> CN- + H3O+}$, and then the equilibrium constant is the $K_\mathrm a$ of $\ce{HCN}$.)
How about part 2?  The only correct answer to

Will the salt react as an acid or a base?

is "It can do either, depending on what it's reacting with".  But maybe what they mean is whether the solution of the salt is acidic or basic.  If so, one way to find out is to combine the two equilibrium equations and their $K$s.  I don't want to give problem away completely, but think of it this way:

What is the pH of a (say) 1 M solution of $\ce{HCN}$, based on the $K_\mathrm a$ of $\ce{HCN}$?
(I'm assuming the calculation of the pH of a solution of a pure weak acid or a pure weak base has been covered in the course before getting to the question you posted today.)

What is the pH of a 1 M solution of $\ce{NH3}$, based on the $K_\mathrm b$?

Is $\ce{HCN}$ a stronger acid than $\ce{NH3}$ is a base, or is $\ce{NH3}$ a stronger base than $\ce{HCN}$ is an acid?  Based on this, which would you expect to "win out" if you mix the two in equal proportions?

(The point here is that what you get by dissolving $\ce{NH4CN}$ is indistinguishable from what you'd get by mixing equal amounts of equimolar solutions of $\ce{HCN}$ and ammonia.  In other words, if you mix 1 L of a 2 M solution of $\ce{HCN}$ with 1 L of a 2 M solution of ammonia, the result is very nearly  2 L of a 1 M solution of $\ce{NH4CN}$.   Footnote: "very nearly" because volume is not always preserved exactly, unlike mass.)


What happens to an acid when it completely dries out? What happens to an acid when it completely dries out?  Does it stop being acidic?  And if there is a base there also, does it give back it's H's?

When an aqueous solution of an acid is dried, water evaporates and the $\ce{H+}$ ions crystallize with the anionic part of the acid and form a crystal.

If the acid used to make the solution was not of crystalline form then after evaporation of water the anion and cation would have combined to form a compound of some other type (may be a solid formed by co-ordinate bonds or a liquid or a gas like $\ce{CO2}$)

Initially, because of presence of $\ce{H2O}$ molecules, the $\ce{H+}$ ions were bonded with them to form $\ce{H3O+}$ molecules and the anionic part of the acid were solvated by these $\ce{H3O+}$ ions. This solvation happened because it liberated more energy than consumed in dissociation of the acid molecule (i.e. was thermodynamicaly favorable)

But, now since the water molecules are absent, the $\ce{H+}$ and $\ce{A-}$ (anions) will combine to form $\ce{AH (s)}$ as its formation will be favorable because there is no more water to compensate the energy required in there dissociation.

If there had been a solution of acid and base then the $\ce{H+}$ ions from the acid and $\ce{OH-}$ ions from the base would have already reacted to form $\ce{H2O}$ and hence the scenario will be the same as above instead we will also notice crystallization of the cation $\ce{(B+)}$ from the base $\ce{(BOH)}$ and the anion ($\ce{A-}$) from the acid ($\ce{AH}$) to form $\ce{BA (s)}$.

If it would had been an aqueous solution of only a base then also the water would have evaporated and crystals of the corresponding base would have formed.

To talk a bit more (though not necessary in the answer): Acidity is a measure of concentration of $\ce{H+}$ ions in water and now since there is no solution present but only the crystals, the crystals won't behave like an acidic solution until and unless solvent is added to it.


Possible to Calculate Electronic Charge Distributions/Densities for Organic Molecules? I was wondering if there are any programs that can calculate the electronic charge distributions around an organic molecule, possibly in the form of a volumetric charge volume density? I am reading up about using DFT programs that (correct me if I'm wrong) I believe may be able to tell us something about finding the charge density.

I'm working on a project that involves obtaining the electric field values surrounding an organic molecule. My current approach involves finding a function for the molecule's electronic volume charge density and doing a volume integral to get the electric field.

Is this a potentially valid method? If not, what else should I look into?

If I understand correctly, you want the total electric field at some point $\mathbf{r}$ due to a molecule. This will have at least three components. The first due to the electric field from the nuclei. This can be calculated classically from the field due to point charges. The second due to the dipole moment of a molecule. This can also be calculated completely classically (though you will have to use some electronic structure calculation to get the dipole and direction). Finally, you will need to find the total charge density of the electrons.

The charge density can be found using either DFT or any post-Hartree Fock method such as MP2 or CCSD(T), etc. When using something like MP2, you will get the charge density from the molecular orbitals. You can do this manually, but one code I know of that will give you the relevant information from the HF orbitals is NWChem. You will want to use the dplot command. This allows you to print out electron densities for each orbital using a grid. You will want to use a fine enough grid to capture the actual spatial changes in density. Notice that you also provide limits i x, y, and z. This is because, in principle, the charge density extends to infinity in all directions. Thus, you will want to choose a cutoff where the density is sufficiently small as to be negligible. Then, you can again calculate the field due to the electron density from classical equations which can be found online or in an E&M textbook. You can either calculate the field discretely from the points on the grid, or fit it to some functional form and then do a true integral over the density.

Another way to do this is to use a code which will give you approximate charges per atom from some type of population analysis such as Mulliken charges. You can do this with Gaussian using population keyword. This will aggregate the nuclear charge and the electronic charge into one field. Fair warning, however, Mulliken charges (and all population analyses) are approximate and not even necessarily physical.

There may be other ways to do this from DFT, but I surprisingly cannot find any software packages that do this directly.


Most dense hydrogen containing composition I am searching for a composition which could host the most hydrogen atoms in a specific volume. Note that this density is sought for justifiable conditions in terms of pressure, temperature, ... and should be realizable in a reasonably equipped lab in quantities of a few kilograms. As for hydrogen itself, besides its very low melting point (which is a problem in certain respects), the density itself is 0.086 g.cm-3 which is lower that water (nearly 0.110 g.cm-3), so that it is evident that a composition should be sought after. Any help would be appreciated.

A few binary metal hydride candidates:

Magnesium hydride gives about 0.11 grams of hydrogen per cubic centimeter, and unlike water it can release its hydrogen leaving the element behibd on simple heating.  It does, however, require 287°C to decompose, limiting reversibility.

Lithium hydride offers nearly the same hydrogen density as the magnesium compound but decomposes at a much higher temperature, limiting its utility.

Beryllium hydride approaches approaching 0.12 grams hydrogen per cubic centimeter.  But it has several drawbacks; beryllium is expensive, it's a serious health hazard and the element does not react directly with hydrogen, making an element/hydride cycle impossible.

Aluminum hydride, with a favorable stoichiometry, comes inat nearly 0.15 grams hydrogen per cubic centimeter and also releases its hydrogen at 150°C.  Like magnesium hydride it has been considered for hydrogen storage.  However, reversible cycling between the element and the hydride is again inhibited because aluminum and hydrogen do not react directly except at very high pressure (10GPa for aluminum hydride versus 20 MPa for a direct synthesis of magnesium hydride).  As with beryllium hydride, getting the aluminum back to hydride form requires indirect methods with other hydride sources.

A ternary compound with 0.15 grams of hydrogen per cubic centimeter is $\ce{Mg2FeH6}$, which can be made from the elements at 2-12 MPa pressure after compressing the metals into pellets.  This compound can also release its hydrogen thermally and is one of many ternary hydride candidates for hydrigen storage.


How to define the atom numbers in unsaturated hydrocarbons correctly according to IUPAC? I would like to address the compound below.

This compound on elaborate elongation would result in this structure.

But is the below given numbering correct? I am just wondering that what if the same compound is given numbering, starting from that carbon which is assigned 7th number in the first elaborate diagram (second diagram), by doing so, triple bonded substituent will also get the least number, as shown in the below given diagram.

I have a similar problem with this compound:

What if you just reverse the numbering? By reversal I mean, the carbon assigned 1st number should be given the last number, am I violating any rule? If I do that, by numbering the compound in reverse order, we get least number for the methyl chain substituent, so is it necessary that methyl group substituted in the side chain should get the highest possible number?

In short, I mean, is the below given numbering valid or not?

Your first compound should be (3E)-5-ethynylhepta-1,3,6-triene. The 2 conjugated double bonds at the end determine the first carbon.

The second compound should be (4-methylpentyl)cyclobutane. The aliphatic group has less precedence than the cyclobutane group.


Elimination from Vinylic Halides Vinylic halides do not show nucleophilic substitution at all, because of their stability due to resonance. But elimination to form alkynes is possible. No nucleophile at all can displace the halide ion in a nucleophilic substitution, but a strong base suffices for elimination.

Why is this?

In the $\ce{sp^2}$ transition state for an $\mathrm{S_{N}}2$ reaction

the attacking nucleophile, the central carbon and the leaving group prefer a linear arrangement
the 3 remaining groups around carbon are arranged with 120° bond angles

(image source)

Now, let's look at ethylene (as a model for a vinyl halide) and let's use the bent bond description of ethylene because it makes it easy to visualize what's happening (the bent bond model has been shown to be equivalent to the pi bond model in all aspects, see below for the same explanation phrased in terms of the pi bond model).

(image source)

Replace one of the ethylene hydrogens with halide and imagine what happens as we approach the transition state.  The bond angle in the two-membered ring wants to be small, but as we approach the transition state it needs to open up wider to 120°!  This would create a lot of strain.  It is this large strain energy that causes the $\mathrm{S_{N}}2$ reaction at $\ce{sp^2}$ hybdridized vinylic carbons to be a very high energy pathway and therefore less likely to occur.

On the other hand, the elimination reaction at a vinylic carbon is quite facile.  Here as the base removes a proton we build up negative charge on an $\ce{sp^2}$ hybridized carbon.  An $\ce{sp^2}$ orbital has more s-character than an $\ce{sp^3}$ orbital; the more s-character in an orbital the lower the energy of an electron in that orbital.  Therefore, negative charge is more stable in an $\ce{sp^2}$ orbital than a $\ce{sp^3}$ orbital.  Consequently it is easier to remove a proton when the negative charge will reside on an $\ce{sp^2}$ hybridized carbon compared to an $\ce{sp^3}$ hybridized carbon (and that's why it is even easier [lower energy] to remove a proton from an $\ce{sp}$ hybridized carbon).

Explanation Based on Pi Bond Model

As expected for equivalent models, the bent bond explanation can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state. It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons are in higher energy unhybridized p orbitals and this is not offset by the energy lowering experienced by the electrons as some bonds rehybridize  from $\ce{sp^2}$ to $\ce{sp}$) thereby increasing the energy required to reach the transition state.


What dictates cathode vs anode nomenclature use? what was the justification for the use the cathode and anode terminology in IEF?

There is no redox taking place in IEF, just proteins interacting with an electric field

Who was responsible for this naming system and how can we change it?

Michael Faraday was responsible for the terms anode and cathode more than hundred years ago.

All the confusion regarding the nomenclature will vanish if you do not associate electrostatic signs with these two terms. One should identify the electrode labels with the redox processes rather than the signs.

Cathode: The electrode where reduction occurs
Anode: The electrode where oxidation occurs

These definitions are true whether you have a electrolytic or galvanic cell. This eliminates the need to call the negatively charged electrode as a cathode in an electrolytic cell or the positively charged electrode as the cathode in the galvanic cell. These definitions are self-consistent and hence nobody bothered to changed them.

Isoelectric focusing is "sort" of an an electrolytic "cell" hence its cathode is has a electrostatic negative sign because if we were to remove this gel and replace it with just salt water, we will see hydrogen evolution (= reduction of water) at this electrode.


Na Cl battery within Zn Cu battery? https://youtu.be/iHBz77E625o

In this video there are pieces of matboard that are soaked in water saturated with NaCl. The pennies themselves are made of zinc and copper. With the electrical energy from the zinc and copper would the chloride let go of its electron and the sodium gain back its electron producing a Na Cl battery within a Zn Cu battery?

Would the water in the process separate into hydrogen and oxygen through electrolysis? If so since hydrogen is very flammable would a combustion reaction occur in the air giving you water again?

The electrolyte in that clip is a solution of $\ce{NaCl}$ and acetic acid (vinegar). The zinc anode is oxidized, resulting in $\ce{Zn^2+}$ ions going into the electrolyte. The electrons went through the red LED, lighting it, and then on to the copper cathode. From there, they reduced some $\ce{H+}$ to hydrogen gas. So two $\ce{H+}$ ions were reduced to one $\ce{H2}$ molecule (which left: it's a gas) and the $\ce{Zn^2+}$ ion simply replaced the two $\ce{H+}$ ions.

Acetic acid is a weak electrolyte, so a solution of it (vinegar) is not a great conductor of electricity. The $\ce{NaCl}$ just makes the electrolyte a better electrical conductor: movable ions are necessary for a steady flow of electrical current in this battery. Without the $\ce{NaCl},$ most of the generated voltage would have been wasted on the electrolyte's resistance.

No sodium is produced (by reducing sodium ions) because the $\ce{H+}$ ions get the electrons. And no chlorine gas is produced (by oxidizing chloride ions) because the $\ce{Zn}$ anode is more readily oxidized.


Calculate the cathode electrode potential in this redox reaction In this problem I'm asked to calculate the electrode potential of the magnesium electrode ($E_{Mg^{2+}/Mg}$) given the red-ox reaction:$$\text{Mg(s)}+2\text{Ag}^+(10^{-2}\text{M})\leftrightarrow 2\text{Ag(s)}+\text{Mg}^{2+}(10^{-2}\text{M})$$

The information I'm given is:

Temperature equal to $278\,K$ (the standard).
The molar concentration of the $\text{Mg}^{2+}$ ions is of $0.01\,\text{M}$.
The standard reduction potential of the $Mg$ electrode is: $E^{º}_{Mg^{2+}/Mg}=-2.34\,V$.

I know I have to use the Nernst Equation, but my solution and the book's answer are very different.

My solution:

$$E_{Mg^{2+}/Mg}=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#ff3300}{\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}}=\ldots=\boxed{-2.28\,V}$$

My book's solution:
$$E_{Mg^{2+}/Mg}=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#00ff00}{\mathbf{\log\frac{1}{\left[\text{Mg}^{2+}\right]}}}=\ldots=\boxed{-2.40\,V}$$

($n=2$ because 2 moles of electrons where exchanged per 1 mole of reaction)

I've colored the part where I go wrong (basically the logarithm). I don't know exactly why the book put the concentration in the denominator, since $\text{Mg}^{2+}$ is a product (not a reactant), therefore following what I studied, its concentration should be in the numerator.

Any help? Sorry for asking such an elementary question, but I'm still learning the basic stuff in Chemistry!

Note
I attach an image of the solution provided by the book

The equation that corresponds to the standard reduction potential is:

$$\ce{Mg^{2+} + 2e- -> Mg}$$

Notice that this is not the actual half reaction in your total equation, where magnesium is oxidized. But the book is determining the reduction potential, so the magnesium ion is a reactant not a product. In the reaction quotient, the concentration/activity of magnesium ion should be in the denominator.

I'm not sure I agree that we should choose the standard reduction potential as the electrode potential for the electrode where oxidation occurs, but at least the book is consistent.


Can water be liquefied or solidified just by adjusting the temperature, regardless of the pressure? technically the line between liquid and solid phase would go a long way before hitting the y-axis, but the point is it will eventually.

So the question remains, can any liquids, including water (with its unique phase diagram) be liquefied regardless of the pressure?

If I actually understood your question correctly, yes there are regions above which there will be only solid state (rather than no solid state). This is limited to the molecular region.

The Roman numerals indicate various ice phases.

From the above phase diagram (which can be found here) the solid-liquid equilibrium line doesn’t touch the y-axis as you suggest. But there will be only solid phase above the $10\ \mathrm{GPa}$ region(approximately).

In the phase diagram you drew, if (a big if) such a PD exists for some element/molecule, then there will be no solid above the point where solid-liquid line meets y-axis.

As pointed out in the comments, when we reach high temperatures and pressures we won’t be seeing the molecular water. It will dissociate into atoms/ions as shown in the figure.


Which is a stronger acid: H3O+ or HCl? I know as a fact that HCl is a stronger acid than H3O+, but I want to know the reason behind it. The definition of an acid is a substance which dissociates to give H3O+ ions. In that case H3O+ should be the strongest acid available because it does not even need to dissociate it give itself!!

Here is my opinion, which may be subject to errors and corrections are very welcome:

Using the definition of Brønsted, acidity is the tendency to give off protons. This, unlike the definition of Arrhenius, is not limited to aqueous solutions. However, if you do have an aqueous solution of an acid, something interesting happens: any acid HAc (or base B) stronger than H$_3$O$^+$ (or OH$^-$) completely dissociates via:

$$\ce{H2O + HAc <=> H3O+ + Ac-}$$ or
$$\ce{H2O + B <=> OH- + BH+}$$

So the maximum acidic compound in water is the hydronium ion H$_3$O$^+$; you couldn't, for example, differentiate aqueous hydrochloric and perchloric acid solutions by pH. To rank very strong acids by their acid strength, non-aqueous solutions are used to determine the dissociation constants and the data obtained is then transferred approximately onto water as a solvent. Also note that we are talking about solutions of acids.

Specifically for HCl, we can denote the pairs of conjugated acids and bases:

$$\ce{H2O ~(~base~1) + HCl ~(~acid~2) <=> H3O+ ~(~acid~1) + Cl- ~(~base~2)}$$

So we can compare the strengths of the two acids: $\ce{HCl}$ and $\ce{H3O+}$. We know that this equilibrium is shifted to the right, so hydrochloric acid is stronger.

Very late edit to try and answer comment by MukulGoyal (unfortunately, he didn't elaborate on his questions), by number:

Is the definition of acidity the tendency to give H+ ions, or
H3O+ ions? That seems to matter here because in first case H3O+ will
not even be acidic in water as the equilibrium H2O + H+ ---» H3O+
has a very high equilibrium constant.
Similarly even if the solvent is NH3, the equilibrium constant of
NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason,
and hence it would actually be a very weak acid.
Does your answer imply that any acid which has a Ka greater than
1 will be a stronger acid than H3O+?

First of all, please remember that equilibria are dynamic.

1) As I've mentioned, acidity, according to Brønsted, is the tendency to give off $\ce{H+}$. In $\ce{H2O}$, $\ce{H+}$ are solvated, on average, by 6 water molecules, giving rise to species like $\ce{H3O+(H2O)6}$ (for more information, check out Wikipedia and this).

1) and 2) you seem to be talking about the dissociation of $\ce{H3O+}$. However, you're using terms like "very high" and "very low", so that, in my opinion, becomes kind of a misleading discussion.

When you mention

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

you need to remember that the proton has to come from somewhere. In fact, what you are talking about there is

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

which is the autoprotonation of water. Let's just calculate that equilibrium constant you talk about all the time:

$$K = \cfrac{c(\ce{H3O+})~ \cdot~ c(\ce{OH-}) }{c^2~(\ce{H2O})}$$

Well, guess what? The concentration of water (c = n/V = m / (V * M) = $\rho$ / M = 55,51 mol/L) remains about constant for pure water and diluted solutions (because not many water molecules form ions), so it is brought into the constant, yielding

$$K_w = c(\ce{H3O+})~ \cdot~ c(\ce{OH-})$$

$K_w$ depends on temperature and equals $10^{-14} $ mol$^2$ / L$^{2}$ at 25 °C. It remains constant in pure water and also in diluted solutions.

In pure water where $c(\ce{H3O+}) = c(\ce{OH-})$, we then get

$$K_w = c^2(\ce{H3O+})$$

and

$$c(\ce{H3O+}) = 1.0 * 10^{-7} ~mol~ / ~L~ $$

which essentialy quantifies what you, as I suspect, were talking about, in that it tells you the concentration of $\ce{H3O+}$ in the autoprotonation equilibrium for pure water. By the way, because pH is defined like so:

$$pH = -lg ~c~(\ce{H3O+})$$

at pH = 7 (neutral), $\ce{H3O+}$ and $\ce{OH-}$ are equal. In diluted solutions, when they are not equal, you can calculate one using the other.

Also, remember that in the Brønsted–Lowry theory acids and bases are defined by the way they react with each other, as mentioned in Wikipedia. So far, we have only looked at water and its autoprotonation, no solvents and other molecules, which have their own acidies and basicities.

2) As for the reaction you mentioned in ammonia as a solvent, well, first of all, $\ce{H3O+}$ needs to come from somewhere (let's imagine we drop some aqueous solution of hydrochloric acid in there). Second, I am not sure that constant would actually be as you describe it - I think that here, ammonia (which behaves ampholytic) would act as a base and pretty happily accept a proton. If anyone can provide insight into this, that would be great.

3) Acid strength is defined using $pK_a$, which is given by $pK_a$ = -lg $K_a$. From my experience, if $pK_a$ is defined via concentration, then $pK_a$ of hydronium ions is 0, and if it is defined via activity, it is -1.7 (-1.74, to be more exact; similarly, for $\ce{OH-}$, the analogous $pK_b$ is 14 via concentration and 15.7 via activity).

Coming back to your question about "Ka greater than 1", that seems about right, because -lg (1) = 0.

I refer you to an acid strength table, for instance, here. Also, I'd like to refer you to this excellent discussion on the matter, where, by the way, orthocresol lent a hand.

Also, just to add this, strong acids have $pK_a$ < 0, while weak acids have $pK_a$ > 0 (again, this is via concentration - for example, here in Wikipedia they use -1.74).

Everyone, corrections are very welcome!


Primary carbocation in addition of HBr to ethylene? All mechanisms of the addition of a halogen acid to ethylene (ethene) I can find involves a primary carbocation as the intermediate ion.

However, primary carbocations are generally unstable and unlikely to form. This can indeed explain why the first step is the rate-determining step, but does primary carbocations really form in the process? Is there no other plausible explanations?

There is quite an evidence that carbocations are involved in this process, and that no concerted pathways exist for this transformation.

While the case of ethyne may be difficult to analyse, lets consider 3-methylbut-1-ene:

If you consider the regular Markovnikov's rule to predict the product, you would probably get this:

While in actuality, the major product is actually this:

You can't really explain why that happens without accepting the existence of carbocations and the following mechanism:

This shows that carbocations are involved in the addition of $\ce{HBr}$ to an alkene.

Note: Since this question is about carbocations, I'd like to mention George A. Olah, a reputed chemist who studied carbocations by preparing them in superacid environments, where they were stabilized and studied using spectroscopid techniques. It was his research that led to the conclusion that carbocations have a planar structure. Have a look at his wikipedia page for more information.


Bromate forming in open water According to this article, the city of Los Angeles (in the USA) is facing a problem where bromate, a suspected carcinogen, has been forming in their open water supplies.  They believe that the formation of bromate has to do with exposure to sunlight, and are now distributing millions of "shade balls" into water supplies in order to block the sunlight.

Given that an overwhelming majority of water on this planet is exposed to sunlight, why are there not high levels of bromate present in all water, especially shallow bodies of water where the surface area is maximized?  Or is it?

The water referred to in the article was being stored in a reservoir after being treated with chlorine.  It is the combination of chlorine and sunlight that facilitates oxidation from bromide to bromate.

So, no, this is not applicable to natural bodies of water, only bromide containing water which has been artificially treated with chlorine.

See Bromate ion formation in dark chlorination and ultraviolet/chlorination processes for bromide-containing water for more information.


What is the correct formula for diamminechloridonitrito-n-platinum(II)? Is the correct formula for diamminechloridonitrito-n-platinum(II) $\ce{[Pt(NH3)2Cl(NO2)]}$ or $\ce{[PtCl(NH3)2(NO2)]}$?
I was taught that the sequence of symbols within the coordination entity should be written in the following order:

The symbol of the central metal atom is written first.
The ligands are written afterwards and should be mentioned in alphabetical order according to the first symbol of their formulae.
For example, for the coordination compound $\ce{[Co(H2O)2(NH3)4]Cl2}$,
$\ce{H2O}$ is cited first and then $\ce{NH3}$. (As H comes before
N)

However, it is given in my textbook that the formula for diamminechloridonitrito-n-platinum(II) is $\ce{[Pt(NH3)2Cl(NO2)]}$, where $\ce{NH3}$ is followed by $\ce{Cl}$.
I assumed that the correct formula would be $\ce{[PtCl(NH3)2(NO2)]}$ ($\ce{Cl}$ followed by $\ce{NH3}$) instead. What am I missing here?

So I did some digging, and here's what I found:
According to the current version of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005,

IR-9.2.3.1 Sequence of symbols within the coordination formula
(i) The central atom symbol(s) is (are) listed first.
(ii) The ligand symbols (line formulae, abbreviations or acronyms) are then listed in
alphabetical order (see Section IR-4.4.2.2).Thus, CH3CN, MeCN and NCMe would
be ordered under C, M and N respectively, and CO precedes Cl because single letter
symbols precede two letter symbols. The placement of the ligand in the list does not
depend on the charge of the ligand.
(iii) More information is conveyed by formulae that show ligands with the donor atom
nearest the central atom; this procedure is recommended wherever possible, even for
coordinated water.

So in this case, $\ce{Cl}$ should come before $\ce{NH3}$ and the correct formula would be $\ce{[PtCl(NH3)2(NO2)]}$.


Could this solution be resolved by titration? I have a mixture of $\ce{NH_4NO_3}$ and $\ce{NH_3}$ at $\ce{pH}$ of about 13.

It is possible to determine the concentration of both on them by acid-base titration?

I was told it's possible, and that you need first titrate with $\ce{HCl}$ and then with $\ce{NaOH}$.

Thanks.

Edit: In the lab were I was working, there were an auto titrator, that had a program that could tell the concentracion of $\ce{NH_4NO_3}$ and $\ce{NH_3}$ in a solution, just by titrating it and controlling the $\ce{pH}$ with a pH meter. I simply don't understand how it does it, because depending of the $\ce{pH}$, $\ce{NH_4+}$ and $\ce{NH_3}$ would be in the same form and I don't know how you could tell them apart.

The ammonium nitrate dissociate in water:

$$\ce{NH4NO3 -> NH4+ + NO3-}$$

The $\ce{NH4+}$ and $\ce{NH3}$ form together the ammonia $\mathrm{pH}$ buffer with $\ce{NH4+}$ acidity constant $\mathrm{p}K_\mathrm{a}=9.25$:
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log { \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}}$$

The titration curve as $\mathrm{pH}=f(V)$ is the most flat at $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=9.25$, where the buffer has the biggest capacity.

For a given solution, containing $\ce{NH4+}$ and $\ce{NH3}$, titration with $\ce{HCl}$ or $\ce{NaOH}$ causes  respective acid-base reactions:

$$\begin{align}
\ce{NH3 + H+ &-> NH4+ \\
NH4+ + OH- &-> NH3 + H2O} \\
\end{align}$$

The ammonia buffer causes   $\mathrm{pH}$ being quite stable during titration, until one of the 2 ammonia forms is practically spent.

At that point, the $\mathrm{pH}$ titration curve gets very progressively bent up or down, respectively and the titrator stops the titration.

The content of $\ce{NH4NO3}$ and  $\ce{NH3}$ is then obtained by obvious routine calculation.

$$\begin{align}
m_{\ce{NH3}}&= M_{\ce{NH3}}\cdot c_{\ce{HCl}}\cdot V_{\ce{HCl}}\\
m_{\ce{NH4NO3}}&= M_{\ce{NH4NO3}}\cdot c_{\ce{NaOH}}\cdot V_{\ce{NaOH}}\\
\end{align}$$

Note that at $\mathrm{pH}=13$, $\ce{NH4+}$ ions practically do not exist, as $ \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}=5620$

If we consider $\mathrm{pH}$ of ammonia solution itself, it is
$$\begin{align}
\mathrm{pH}=14 - 0.5 \cdot ( \mathrm{p}K_\mathrm{b}-\log c) \\
13=14 - 0.5 \cdot ( 4.75 -\log c) \\
\end{align}$$

$c_{\ce{NH3}}$ would be 563 mol/L, what is nonsense. Such $\mathrm{pH}$ is not achievable by ammonia alone, but a mineral hydroxide would be present.

If there was initially any ammonium nitrate, it would be converted to ammonia:

$$\ce{NH4NO3 + NaOH -> NaNO3 + NH3 + H2O}$$

In such a case, determination of both compounds by titration is not possible, as there are no ammonium ions and ammonia would be titrated together with hydroxide.


What would be the product of following elimination reaction? Question

Doubt

The answer is given B. But why cant it be C after all formation of cyclopropane would increase angle strain?

The first thing you notice is an alcoholic base reacting with a molecule. The base looks for the most acidic hydrogen in the molecule. Since the no.of moles of $alc.KOH$ is not mentioned, we go for both the acidic hydrogens at the adjacent positions to the carbonyl functional group. Once the acidic hydrogens are abstracted, it's a simple elimination of $Cl$ that will happen.

I agree with the comment made by @Waylander as there are plenty of examples with three-membered rings. Well, (c) is not possible but you can say (a) is a possibility. Of course (a) forms too by a simple elimination reaction but the most probable one is the abstraction of acidic hydrogen in the molecule. The stability of [3]rings can be brought about by banana bonds.


Visualizing s and p orbitals from contracted basis set I was thinking about basis sets used in Quantum Chemistry programs and thought of why not try to visualize them.
I started with GTO from EMSL website for $\ce{H2}$, specifically cc-pVDZ basis
which is the following,
! s functions
H    5    3
     13.0100000            0.0196850              0.0000000              0.0000000
      1.9620000            0.1379770              0.0000000              0.0000000
      0.4446000            0.4781480              0.0000000              0.0000000
      0.1220000            0.0000000              1.0000000              0.0000000
      0.0297400            0.0000000              0.0000000              1.0000000
! p functions
H    2    2
      0.7270000           1.0000000              0.0000000
      0.1410000           0.0000000              1.0000000

I framed the basis functions for $s-$orbital as following,
$$0.019685\times exp(-13.01  r^2) + 0.137977\times exp(-1.962  r^2) + 0.4446\times exp(-0.478148  r^2)$$
$$1\times exp(-0.12200 r^2)$$
$$1\times exp(-0.02974  r^2) $$
This is only the radial part (the associated spherical harmonic is not here), but still I can plot the above against $r$.

The above plot is for the $s-$functions.

I am able to do the same thing for $p-$functions, and I get similar Gaussians (not shown here) with different widths, and centered at zero.
Shouldn't the lobe of $p-$ be centered larger than zero (which implies that the functional form I am using is not correct/incomplete for $p-$function). Or the lack of explicit spherical harmonic functions does not allow the $p-$functions to be similarly visualized. Could someone clarify.

Your suspicion is correct - you need the angular part, that is, a spherical harmonic (sometimes it may be cartesian instead) to get the shapes of the orbitals and thus, actual lobes of $p$-functions.

Also note that projection to $(x)$ from $(x,y,z)$ may not be straightforwardly the same for all angular momenta. Additionally, the angular part also has influence on the normalization of the basis functions, which may or may not be included in the basis set format at hand. The last point, however, can be checked by numerical integration.


Which factors determine the keto:enol ratio for aldehydes, ketones, and 1,3-dicarbonyl compounds? Arrange the following compounds in order of increasing enol content:

I think the order is D > C > B > A considering conjugation, hyperconjugation and hydrogen bonding. The enol formed in case D will have a substituted double bond (that is, more hyperconjugative structures) than the enol of C.
Both enols A and B lack hydrogen bonding. Also, enol of B is more stable because of more substituted alkene.

Your reasoning is along the right lines, but it is incomplete.

enol of B is more stable because of more substituted alkene.

The methyl group in acetone does stabilize the carbon-carbon double bond in the enol form for the reasons you suggest. However, the methyl group also stabilizes the carbonyl double bond in the keto form (see here for example, this is why a carbonyl in a ketone is slightly stronger than a carbonyl in an aldehyde). In fact, in simple carbonyl compounds, the methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond. As a result, simple aldehydes generally have a higher enol content than simple ketones. Hence, the enol content in acetaldehyde is greater than the enol content in acetone.
$$\ce{carbonyl <=> enol}$$
$$K_\mathrm{eq}=\frac{[\text{enol}]}{[\text{carbonyl}]}$$
$$
\begin{array}{lc}
\hline
\text{compound} & K_\mathrm{eq} \\
\hline
\text{acetaldehyde} & 6 \times 10^{-7} \\
\text{acetone} & 5 \times 10^{-9} \\
\hline
\end{array}
$$
The two β-dicarbonyl compounds have a higher enol content than the two monocarbonyl compound because hydrogen bonding and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above.
Therefore, the overall order of increasing enol content is C > D >> A > B.
Unless you've studied this before, you might not know that a "methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond". So, a general rule that might come in handy in these situations is that the enol content increases with the acidity of the enolic hydrogen. A more acidic α-hydrogen implies a weaker $\ce{C-H}$ bond. Since the position of a keto-enol equilibrium is dependent on the relative stabilities of the keto and enol forms (the compound with the highest bond strengths overall will be more stable and will predominate at equilibrium) a weak $\ce{C-H}$ (lower $\mathrm{p}K_\text{a})$ generally implies a higher enol content.
$$
\begin{array}{lr}
\hline
\text{compound} & \mathrm{p}K_\text{a} \\
\hline
\text{acetone} & 19.3 \\
\text{acetaldehyde} & 16.7 \\
\text{2,4-pentanedione} & 13.3 \\
\hline
\end{array}
$$
Finally, this earlier answer provides a review of the key factors controlling keto-enol equilibria.