Why is vapour pressure not dependent on shape and size of the container? In my textbook, it's written that vapour pressure of a liquid does not depend on shape and size of a container but won't a container that provides less surface area to the liquid will have lower vapour pressure?

for example, consider 2 containers give in the diagram below, the volume of container and liquid is same in both the diagram just the shape is changed.

wouldn't the vapor pressure be lower in the diagram on the left side? and how does changing size not affect vapor pressure?

Vapor pressure does not depend on surface area because it is derived from the thermodynamic equilibrium between the liquid and gas phases of a substance. The closer the vapor pressure is to atmospheric pressure, the closer that substance is to boiling. At its boiling point, vapor pressure equals atmospheric pressure and the molecules that comprise the liquid now have enough thermal energy to overcome the intermolecular forces that are maintaining it in a condensed phase. So it might help to think about how container geometry does not affect the boiling point of a liquid; the same reasons are true for why it does not affect the vapor pressure.

The evaporation rate in terms of mass/time is absolutely a function of the container geometry, but equilibrium vapor pressure is an intrinsic molecular property.


Kinetic Energy Evaluation Integral Evaluation Program Ostlund I'm reading Ostlund's Modern Quantum Chemistry. In Appendix B, the kinetic energy integral is evaluated using the Gaussian Basis functions as

T11=T11+T(A1(I),A1(J),0.0D0)*D1(I)*D1(J)
T12=T12+T(A1(1),A2(J),R2).D1(I).D2(J)
T22=T22+T(A2(1),A2(J).0.0D0).D2(1).D2(J)

I understand T11 to be the kinetic energy integral using the first atom orbitals so it will give the kinetic energy of the electron of the first atom, and this is why it is passing zero as the bond length parameter. The same is T22 but for the second atom.

What is T12 used for, aren't T11 and T22 enough to calculate the total kinetic energy?

The off-diagonal elements $T_{\mu\nu}, (\mu \ne \nu)$ are just as necessary as the diagonal elements $T_{\mu\mu}$ to represent the kinetic energy operator in the chosen AO basis set. Substituting a few of Szabo/Ostlund's equations into each other (in my edition, 3.147-3.154), one finds
$$
F_{\mu\nu} = T_{\mu\nu} + V_{\mu\nu}^{\text{nucl}} + \sum_{\lambda\kappa} P_{\lambda\kappa}\left[ \left( \mu\nu \vert \kappa\lambda  \right) - \frac{1}{2} \left( \mu\lambda \vert \kappa \nu\right) \right]
$$
where $F$ is the Fock matrix, $V^{\text{nucl}}$ represents the nuclear attraction term, $P$ is a density matrix and $\left( \mu\nu \vert \kappa\lambda  \right)$ are two-electron repulsion integrals over the AO basis functions (denoted greek indices). Of course, your Fock matrix may be diagonal already, if you have a very special case and basis set, but in general and to perform the SCF calculation, the off-diagonal parts cannot be ignored.


Does an irreversible reaction have an equilbrium between reactants and products? Retrospective analysis 2/13/2017 -- The barium sulfate example is a poor choice. Equilibrium equations should really be defined using activities, and the activity of solid barium sulfate is by definition 1.

A previous question, Is every chemical reaction in equilibrium?, started much discussion. I objected to the answer that Curt F. gave and challenged him to derive the equilibrium reaction for a particular irreversible reaction. He countered with a reply indicating that I should make the specific case a separate question - so here it is. I'm going to change the reaction slightly. Given the following reaction between barium chloride and sodium sulfate, does "the" chemical equilibrium exist?

$$\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4 v + 2NaCl(aq)}$$

I contend that "the" equilibrum between reactants and products such as $$K_{\text{eq}} = \frac{\ce{[BaSO4][NaCl]^2}}{\ce{[BaCl2][Na2SO4]}}$$ doesn't exist since when the barium sulfate precipitates there could be a microgram or a kilogram as the product. Furthermore adding solid barium sulfate to the product will not shift the reaction to the left.

I'd agree that calling the reaction irreversible and saying that $K_{\text{eq}}$ doesn't exist is a tautology. So given reaction between
$\ce{aA + bB}$ to form products $\ce{cC + dD}$ then the reaction is a reversible reaction if the equilibrium such as $$K_{\text{eq}} = \frac{\ce{[C]^{c}[D]^{d}}}{\ce{[A]^{a}[B]^{b}}}$$ exists and if such an equilibrium doesn't exist then it is an irreversible reaction.

There are obviously "some" equilibrium in the reaction, but not "the" equilibrium between products and reactants.

Water has an autoionization equilibrium and $\ce{H2SO4}$ has two $\mathrm{p}K_\mathrm{a}$'s.
Barium sulfate has a $K_{\text{sp}}$.
Also the barium sulfate precipitate isn't static once formed. It is
dissolving and reprecipitating at the same rate, but the rates depend
on the surface area of the precipitate not the "concentration" (or mass) of the precipitate.

So, if I'm wrong, how do you calculate $K_{\text{eq}}$ for the overall reaction given? Assume you mix $500\mathrm{ml}$ of $0.1$ molar barium chloride with $500\mathrm{ml}$ of $0.1$ molar sodium sulfate. What is $K_{\text{eq}}$?!

The reaction you are interested in is the precipitation of barium sulfate, a relatively insoluble salt.

The reverse of this reaction is the dissolution of barium sulfate crystals.  The dissolution kinetics of barium sulfate have been studied in a variety of systems over the decades.  Here are a few links:

Kornicker et al. 1991 reported that barium sulfate dissolution into water reached equilibrium in less than 30 minutes at 25 °C and took about 5 minutes at 60 °C.  They also presented a rate law for $\ce{BaSO4}$ dissolution of $r=k A (C_{eq} - C)^2$, where $A$ is the specific area of the barium sulfate crystals, $C_{eq}$ is the equilibrium concentration of barium sulfate, and $C$ the instantaneous concentration of dissolved barium sulfate.  This rate law had been used historically (see refs.) but these authors advocate instead for a first-order rate law.
Dove and Czank 1995 also disagreed with second-order rate law of Kornicker et al. and proposed a first-order reaction for $\ce{BaSO4}$, i.e. $r = k A (C_{eq} - C)$.

I couldn't find any references for barite solubility in sodium chloride solution, but I think it is safe to say that the solubility is finite.  Probably the kinetics of dissolution would not be more than an order of magnitude different than in pure water.

Since this dissolution is the reverse of the precipitation reaction you are interested in, and since the literature is clear that dissolution happens at non-zero rates, then both "reverse" reaction and forward reaction must be happening in your system.  That sounds like the definition of a dynamic equilibrium to me.

Note the appearance of $A$ in the equations.  The surface area of barium sulfate solid involved in the equilibrium determines the kinetics.  Usually we assume that the affect of $A$ is the same for dissolution and precipitation kinetics, so that the effects cancel out.  This is why we say that the activity of a solid phase is 1 and does not vary.  However, the effects of $A$ do not always cancel out.  The solubility of nanoparticles of $\ce{BaSO4}$ is probably higher than the solubility of bulk $\ce{SO4}$ for this reason.

Experimentally, one could probe the dynamics of the equilibrium by mixing $\ce{^131BaSO4(s)}$, i.e. radiolabeled barium sulfate, with a saturated solution of $\ce{^138BaSO4(aq)}$.  This could be done even in a sodium chloride solution.  At regular time intervals, samples of the solution would be taken, which could be filtered to remove any traces of solid particulates, and the levels of the radiotracer that had entered solution could be measured.

I don't know, quantitatively, what the outcome of the experiment would be, especially in sodium chloride solution.  But I am 100% confident that at some time scale, probably within tens of minutes, radioactivity would appear in the liquid solution.  That indicates that there is an equilibrium.

The relevant equilibrium constant in this case is the $K_{sp}$ for barium sulfate, as was alluded to in both your question and in the comments.


Does benzene have isomers and resonance structures? Does benzene have both isomers and resonance structures?

I know that it has resonance structures because you can manipulate the position of the double bonds but what about isomers?

Well, it depends on what kinds of isomers you have in mind. It certainly has constitutional isomers: For example, the sum formula $\ce{C6H6}$ does describe benzene as well as hex-1,5-en-3-in $\ce{H2C=CH-C#C-CH=CH2}$.
There are also a lot ring structures that are isomers of benzene:

As far as I know there aren't any configurational isomers of benzene since a six-membered ring doesn't allow for trans double bonds (this would be sterically impossible).


Molecule or molecular ion that spontaneously breaks up in two different ways I teach physics, and I'm looking for simple, easy examples from chemistry to use in order to illustrate basic ideas about quantum mechanics for my students. A conceptually simple example is the $\text{He}^-$ ion, which has a metastable state with a mean lifetime of 0.36 ms. This is for the totally isolated ion, i.e., we're not talking about a gas or plasma in which there can be collisions or reactions. (How would chemists refer to this? "In vacuo?") This is easy to relate to in chemical terms (a noble gas needs an electron like a fish needs a bicycle) and provides a simple example of quantum mechanical concepts like exponential decay.

What would be more fun for my purposes would be an example of a metastable molecule or molecular ion that would spontaneously break up in two different ways. I would imagine that for the ions, one could measure the half-life and branching ratios by storing them in a circular accelerator. So we could have something like the decay channel

$$\text{XYZ}^+ \rightarrow \text{XY}^+ + \text{Z}$$

competing with, say,

$$\text{XYZ}^+ \rightarrow \text{X}^+ + \text{YZ}.$$

We can then discuss things like the probabilities (branching ratios) and the fact that if the ion starts out in a well-defined state, presumably its ground state, then we can only predict these probabilities, not the actual outcome.

Part of what's making it hard for me to find examples is that I don't know the terminology to google on. Is this "autodetachment," or is that only used for the loss of an electron? "Autodissociation?"

Note: I really am looking for the process described above, not something else like breakup induced by a collision, nuclear decay, electromagnetic decay of an electronic excitation, electron emission, or chemical reactions that require two reactants to come in contact. Those other examples might be fine for some educational purposes, but they're not what this question is about.

This happens all of the time in mass spectrometry following electron-impact ionization. Following ionization, each molecular ion decomposes (or fragments, to use mass spectrometry lingo) into other radicals and ions, usually following multiple pathways. As the pressure inside a mass analyzer is usually near vacuum, each fragmentation event is independent. No collisions or reactions between particles occur. The mass analyzer collects all of the fragments and generates signal that computer software converts to a mass spectrum - a frequency bar graph of the masses of species generated. For some very simple molecules, only one fragmentation pathway may predominate, but for most organic compounds, there are multiple fragmentation pathways.

For example, let's consider the fragmentation of but-3-en-2-one radical cation with the formula of $\ce{C4H6O+}$ and the structure shown below. The mass/charge ratio (m/z) is 70 for this compounds

There are at least two different fragmentation pathways.

The first fragmentation pathway is a loss of a vinyl group ($-\ce{C2H3}$, -27 m/z):
$$\ce{C4H6O+ -> C2H3O+ + C2H3 }$$

The second pathway is a loss of a methyl group ($-\ce{CH3}$, -15 m/z):
$$\ce{C4H6O+ -> C3H3O+ + CH3}$$

Each of these fragmentation pathways is independent. Here is a copy of the mass spectrum from the NIST Chemical Webbook, which is a frequency bar graph of all possible fragments. You can see in addition to the two pathways mentioned above, there are many other possible fragments.

We can estimate the relative rates of these processes from the frequencies of the peaks in the mass spectrum. The first pathway occurs with a relative frequency of 83.19 (based on the height of the m/z = 43 peak) , and the second pathway occurs with a relative frequency of 1.000 (from the height of the m/z= 55 peak). We can use these as proxy for the relative rates of decomposition. We can also see that the parent ion survives this process with a relative frequency of 0.57779 (from the m/z = 70 peak), which indicates that the parent ion has at least some stability.


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

My question is:

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

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

$$\begin{align}
\ce {Fe^{2+} {(aq)} + 2e^{-} &<=> Fe {(s)}} &E^⦵= \pu{-0.44V}\\
\ce {Fe^{3+} {(aq)} + e^{-} &<=> Fe^{2+} {(aq)}} &E^⦵= \pu{+0.77V}\\
\ce {Mg^{2+} {(aq)} + 2e^{-} &<=> Mg {(s)}} &E^⦵= \pu{-2.37V}\\
\end{align}$$

Thus

$$
\begin{align}
\ce {Mg {(s)} + 2Fe^{3+} {(aq)} &<=> Mg^{2+} {(aq)} +2Fe^{2+} {(aq)}}&&E^⦵_{\text{cell}}= 0.77--2.37=\pu{+3.14V}\\
\ce {Mg {(s)} + Fe^{2+} {(aq)} &<=> Mg^{2+} {(aq)} +Fe {(s)}}&&E^⦵_{\text{cell}}= -0.44--2.37=\pu{+1.93V}\\
\end{align}$$

Both reactions are feasible, and I'd imagine the first would proceed until the concentration of $\ce {Fe^{3+} {(aq)}}$ is very low, and then the second would start at a noticeable rate.
This assumes, of course, the iron solid is able to deposit on the carbon cathode, and I don't see why not.

Hope this was helpful.

(Thank you Dr. J for pointing out my typo)


Can dry ice explode? Can dry ice be used as a explosive chemical? I am intrigued by this chemical dry ice and seen various experiments with it, and seems to have potential to be a   explosive.

So can dry ice explode?

Chemicals used as explosives are those which, by some reaction/state change, can produce a large pressure (and thus usually volume) change very quickly.

While dry ice (solid carbon dioxide) is much more dense than in its gaseous phase, at room temperature the change happens very slowly, and is just unstable rather than stable but reactive, so can't be left and triggered later.

This makes it pretty useless as a working explosive.

Since it can exert a significant pressure while it sublimates, however, it can make other things explosively fail -- like the dry ice in the soda bottle demonstration. A quick search suggests that soda bottles are rated to withstand a pressure difference of around 100 psi (~6.8 atm, 0.69 MPa), and the increasing pressure won't stop the carbon dioxide subliming/boiling (since at that point it'll be a liquid) until ten times that.

Now, there is a market for things that slowly expand with a lot of force ("non-explosive demolition agents"), but they're solids that you mix into a slurry and pour into a cavity, since these can trivially exert orders of magnitude more pressure -- a common brand-name one, Dexpan, is rated at 18000 psi -- and they also turn from a slurry or solid into another solid. If you tried this operation with something that turned into a gas, you'd need to seal the cavity to stop the gas just escaping, whereas with solid-to-solid reactions, gravity and the solid itself will do the job for you.


Why is solid sodium hydride a base and not an acid when reacted with water? When solid $\ce{NaH}$ is mixed with water, the products are not $\ce{Na-}$ and $\ce{H3O+}$. The reaction yields $\ce{NaOH}$ and $\ce{H2}$ gas.

Why does solid $\ce{NaH}$ not act as a Brønsted acid in this case?

The answer lies in electronegativity. Sodium has a significantly lower electronegativity ($\approx 1.0$) than hydrogen ($\approx 2.1$) meaning that hydrogen pulls electron density towards itself, away from sodium to generate a sodium cation and a hydride anion.

For a compound to be a Brønsted acid, you need to draw electron density away from the hydrogen, i.e. connect it to an electronegative atom such as oxygen, fluorine, nitrogen etc. Only then can it formally be described as $\ce{H+}$ and can dissociate as such.

A hydride is much better described as $\ce{H-}$ and has a free lone pair. As such, it is a Brønsted base, not an acid. In fact, if you extend the Brønsted acid/base definition in the way Lewis did, you will arrive at the sodium ($\ce{Na+}$) being the acidic species here.

The Brønsted acid/base reaction product of the base $\ce{H-}$ and the acid $\ce{H+}$ happens to be $\ce{H2}$, and because the acidic hydrogen is abstracted directly from water (hydrogen gas can bubble away, shifting the equilibrium even if the reaction were not thermodynamically favoured), $\ce{OH-}$ ions are left which can be written with the remaining $\ce{Na+}$ cation to give sodium hydroxide.


Functional group naming order? I'm currently studying some organic chemistry at highschool level. I understand that when naming an organic compound, the names for substituents should be done in alphabetical order. For instance 3-bromo-3,4-dimethylhexane (B in bromo before M in methyl), or 3-chloro-3-ethyl-4-methylhexane (C in chloro before E in ethyl before M in methyl).

However, it doesn't seem to be the same for functional group suffixes. For instance, for one compound I wrote but-1-amin-3-ene, while the correct answer was apparently but-3-en-1-amine. What's more, these are referred to as suffixes, while what I thought was prop-3-amin-1-ol was in fact 3-aminopropan-1-ol.

So is there some kind of pattern or rule as to the order of writing functional group "suffixes" in organic compounds? At first I thought perhaps it follows the priority the groups have over each other in numbering the parent chain carbons (i.e. alcohol > amine > alkene, etc), but this doesn't seem to be it either. It's possible that it'll be explained further on in the course, but I want to make sure I haven't missed anything.

Generating the systematic names of organic compounds follows the rules for IUPAC nomenclature.

In this system, a name has several parts:

Prefix (meth-, eth-, prop-, but-, pent-, etc.) - which indicates the number of carbon atoms in the parent chain.
Infix (-an-, -en-, -yn-) - which indicates the presence or absence of carbon-carbon multiple bonds in the parent chain.
Suffix (-e, -ol, -amine, -oic acid, -al, etc.) - which indicates the highest priority functional group in the parent chain.
Substituents (bromo-, methyl-, hydroxy-) - indicate the presence of other things on the parent chain.
Locants (the numbers) - indicate where all of these things happen.

You are confusing the roles of infixes and suffixes.

  Infixes

The base infix is -an- indicating no carbon-carbon multiple bonds. The presence of alkenes or alkynes in the parent chain changes the infix to -en- or -yn-, respectively. No other functional group can change the infix, as there are no carbon-carbon bonds of higher order than 3 or lower order than 1. It is also possible to have a complex -enyn- infix if you have both, but let us not worry about that now. In your example compounds:

3-bromo-3,4-dimethylhexane - the infix is -an- because there are no carbon-carbon double or triple bonds.
but-3-en-1-amine - the infix is -en- because there is a carbon-carbon double bond.
3-aminopropan-1-ol - the infix is -an- again because there are no carbon-carbon multiple bonds. The amine and alcohol functional groups do not consist of carbon-carbon multiple bonds so they cannot change the infix.

  Suffixes

The base suffix is -e, signifying a hydrocarbon with some substituents. Many functional groups change the suffix of a molecule's name, but the name can have only one suffix. If two or more suffix-changing functional groups are found on the parent chain, then the one with the highest priority is chosen to be the suffix and all others become substituents. The common functional groups by decreasing priority are:

functional group     suffix         substituent name
carboxylate anion    -oate
carboxylic acid      -oic acid      carboxy
ester                alkyl -oate    alkoxycarbonyl or alkylcarboxy
acid halide          -oyl halide    halocarbonyl
amide                -amide         amido
nitrile              -enitrile      cyano
aldehyde             -al            oxo
ketone               -one           oxo
alcohol              -ol            hydroxy
thiol                -ethiol        mercapto
amine                -amine         amino
hydrocarbon          -e             varies by group

The name 3-aminopropan-1-ol is correct. Both the alcohol and the amine cannot change the suffix. Only the higher priority group (alcohol) changes the suffix. The other (amine) becomes a substituent. As mentioned before, neither group changes the infix.

  Substituents

In addition to the substituents listed above, several functional groups are always substituents (which are listed alphabetically in front of the parent name):

functional group     substituent name
alkyl                alkyl
alkyl halides        halo
ethers               alkoxy
nitro                nitro
sulfides             alkylthio

There are some other common substituents that deal with exception cases, like phenyl, aryl, vinyl, methylene, formyl, allyl, and acetyl.


Calculate pH of a buffer solution Calculate the pH in a buffer prepared from $\pu{50 mL, 0.30 M}$ formic acid ($\ce{HCOOH}$) and $\pu{30 mL, 0.40 M}$ sodium formate ($\ce{HCOONa}$).

My way of solving:

$n_{\ce{HCOOH}} = 0.05 \times 0.3 = \pu{0.015 mol}$

$n_{\ce{HCOONa}} = 0.03 \times 0.4 = \pu{0.012 mol}$ (which means $\ce{NaOH} = \pu{0.012mol}$)

$\ce{HCOOH + NaOH <=> HCOONa + H2O}$

After doing a BCA-table, I got that the remaining moles of HCOOH = $\pu{0.003 mol}$

and HCOONa = $\pu{0.012 mol}$

Calculating the new concentration:

[HCOOH] = $\ce{0.003 / 0.08 = 0.0375M}$

[HCOONa] = $\ce{0.012 / 0.08 = 0.15M}$

Using the Henderson–Hasselbalch equation:

pH = pKa + log [HCOONa]/[HCOOH] = $\ce{3.75 + 0.60 = 4.35}$

So my answer is that the pH of the buffer is $\ce{4.35}$

Answer key

However, in our answer key, they found the new concentration of HCOOH and HCOONa by simply taking:

[HCOOH] = $\ce{(0.05 *0.3)/0.08 = 0.1875M}$

[HCOONa] = $\ce{(0.03 *0.4)/0.08 = 0.15M}$

and when using those concentrations in the Henderson–Hasselbalch equation they get the answer pH = $\ce{3.65}$

Question

Is the answer key correct? If we are adding a base, then the pH should increase to become more basic, not decrease as in the answer key. If the answer key is correct, why does the pH decrease?

Is the answer key correct?

Yes, they have a very good approximation of the pH, and their way of solving it is straightforward and conventional.

  After doing a BCA-table, I got that the remaining moles of HCOOH = 0.003moles

The sum of formic acid and formate has to remain 0.027 mol. You could start out with 0.027 mol of formic acid and turn some of it into formate by adding NaOH. Or you could just mix them as described in the problem. However, you are claiming that there is a total of 0.015 mol of formic acid plus formate, so you lost some atoms there. As the comments state, there is no reason to think about where they got the sodium formate (stock room?).

  If the answer key is correct, why does the pH decrease?

Decrease or increase depends on your frame of reference. If you compare it with a 1:1 buffer where the pH = pKa, it decreased because there is a higher concentration of acid than conjugate base.

If you compare it to the pH of 0.3 M formic acid (roughly pH = 2.1), then it increased because you added conjugate base (neutralized and diluted the acid).


Empirical formula when given percentages? A binary compound of lead and oxygen contains $90.66 \% \ce{Pb}$. What is the empirical formula for the compound? How do I use the percentage to get to the normal formula, then how do I change the normal formula to empirical?

You will get the empirical formula straight from this data. In order to convert to a molecular formula, you need more information.

By way of another example:

  A binary compound of tin and chlorine contains 62.89% tin by mass. Determine the empirical formula.

Our first goal is to figure out the ratio of moles of tin to chlorine in this compound. Once we know the ratio of moles, we can determine the formula.

What do we know? We know what percentage of tin is in this compound by mass. What is the percentage of chlorine?

In order to get to moles, we need at least a mass of the sample. We were not given that. Since percent by mass is a ratio of masses, it turns out we can choose any mass of sample we want and get the same formula (Try it yourself!).

If we assume that we started with 100.0 g of this stuff, then how many grams of tin would you have? How many grams of chlorine would you have? 100.0 g is a nice starting mass since you are working with percents.

Now, as Ari's answer says, you need to convert mass of tin and mass of chlorine into moles of tin and moles of chlorine by dividing each mass by the atomic molar mass of each element (look it up on your periodic table).

Now that you are in moles, determine the lowest whole number ratio of the two elements by dividing each number of moles by the smaller number of moles. If your numbers are close to whole numbers, you can round. For example, round 1.97 to 2. If your numbers are not close to whole numbers, you may need to do a little more arithmetic:

If you get a number like 1.48, this is close to 1.5 which is 3/2. Multiply both numbers by 3.
If you get numbers like 1.34 (close to 4/3) or 1.67 (close to 5/3) multiply by 3.

Now you have the empirical formula. As Ari's answer says, you need a molar mass of the species to get a molecular formula.

The answer for my example is $\ce{SnCl2}$. Work through the steps and see if you can get to this answer. Then repeat for your lead and oxygen compound.


What is the molecular weight of bilirubin? There is a discrepancy among different sources on the molecular weight of bilirubin, $\ce{C33H36N4O6}$:

584.673 (Fisher Scientific);
584.66 (Sigma Aldrich).

What is the correct value and why?

The discrepancy is simply due to round-off error.  The atomic masses of these elements are usually given as follows.  These are known as the standard conventional form, or standard formal short form, since they are given without an interval indicating the uncertainty in their values (see  https://en.wikipedia.org/wiki/Standard_atomic_weight):
C = 12.011
H = 1.008
N = 14.007
O = 15.999
Hence 33 C + 36 H + 4 N + 6 O = 584.673
However, if someone were to round off all the masses to four significant figures, but then report the result to five significant figures, we would have:
C = 12.01
H = 1.008
N = 14.01
O = 16.00
And now 33 C + 36 H + 4 N + 6 O = 584.658 = 584.66


Calculating approximate pH of polyprotic acids When I took up ionic equilibria and titrations after a long break, I found it hard to solve the questions regarding pH calculations of polyprotic acds. Consider these two questions as examples:-

  Find the approximate pH of a solution with $0.4M\text{ }\ce{HCO3-}$ and $0.2M\text{ }\ce{CO3^2-}$$K_{a1}=4\times 10^{-7}$,
  $K_{a2}=4\times 10^{-11}$
  Find the approximate pH of the resulting solution formed after mixing $20ml$ of $0.1M\text{ }\ce{H3PO4}$ and $20ml$ of $0.1M\text{ }\ce{Na3PO4}$. Express in terms of $pK_{a1},pK_{a2},pK_{a3}$.

I have mentioned 2 questions here because I want to know the method to solve such type of questions and not these two questions here in particular. The problem arises when there is a polyprotic acid with hydrolysable salts. For the second case, the ongoing equilibria are the dissociation of the three acids and the hydrolysis of their respective salts. For the first one, there are just two dissociation equilibria(and the dissociation of water). Even by writing down the equilibrium expressions of these reactions, I am unable to reach the final expression with pH or pOH. And, all the solutions have to be approximate (using approximations like the negligible contribution of hydrogen from the dissociation of water, low value of the successive ionization constants) and not at all accurate.

I would appreciate if someone explained how to do such numericals, taking one of the above as an example (I already have the answers for the aforementioned questions. I just want to use them to illustrate the general principle behind these questions).

Let me know if my question looks too homework-like and off-topic.

For the first scenario, the Henderson-Hasselbalch equation is probably good enough. In this case, we assume the major species to be $\ce{HCO3-}$ and $\ce{CO3^{2-}}$ and we will use $K_2$ in the equation.

$$\text{pH}=\text{p}K_2 + \log_{10}{\frac{[\ce{CO3^{2-}}]}{[\ce{HCO3-}]}}$$

I was working on a more general approach using systematic treatment of equilibrium, and it was driving me crazy. I will try to come back to it for the second part of you question.


What is double titration? I an currently studying acid-base titrations. There is a concept of double titration. Can someone provide me an idea on what actually it is? Also what resources can I refer to?

Double titration also often refers to Gilson's titration method (also known as "Gilman color tests" or "Gilman double titration") for determining the concentration of organometallic compounds in hydrocarbon solvents.
At the first stage the total concentration of the base in the solution is determined.
At the second stage the residual amount of alkali after the organometallic compound reacts with the halide is established, allowing to calculate the concentration of metalloorganic compound. [1]

For the double titrations, a 3 ml aliquot of the solution to be analyzed was
withdrawn by pipette using a suction bulb and added to 10 ml of ether.
The solutionwas hydrolyzed with 10 ml of distilled water and titrated with standard acid, using phenolphthalein as indicator, to give the total present.
A second 3 ml aliquot was withdrawn and added 10 of ether, containing 1 ml of the halide, under an atmosphere of nitrogen.
The solution was swirled gently and allowed to stand for 2 min, after which it was hydrolyzed with distilled water and titrated immediately with standard acid, again using phenolphthalein as indicator.
In titrating the latter solution the end point is easily overstepped, since the
aqueous layer decolorizes before the ether layer. Vigorous shaking near the end point
is recommended.
There is the possibility of interference from hydrolysis of the halide, so that the titration should be carried out within a reasonable length of time.

An updated cumulative summary for the method can be found in Wakefield's Organolithium Methods [2, pp. 17–18]:

For detecting the presence of an organolithium compound (for example for checking whether any remains at the finish of a reaction) Gilman's Colour Test 1 is convenient:
$$
\begin{align}
\ce{(4-Me2NC6H4)2C=O + RLi &-> (4-Me2NC6H4)2C(R)OLi}\\
\ce{ &->[(i) H2O][(ii) I2, AcOH] [(4-Me2NC6H4)2CR]+ I-}
\end{align}
$$
A sample (0.5—1 ml) of the test solution is added to an equal volume of a I solution of Michler's ketone, 4.4'-bis(dimethylamino)benzophenone, in dry benzene.
A few drops of water are added, and the mixture is shaken.*
A few drops of an 0.2% solution of iodine in glacial acetic acid are added,
resulting in the formation of a blue or green colour.
Gaidis proposed two modifications, to distinguish between alkyl- and
aryllithium compounds (15).
(a) Following hydrolysis, the solution is buffered to pH 9 by the addition of 20% aqueous catechol (1—2 ml).
A few drops Of iodine in benzene are added and the solution is shaken.
A green to blue colour in the organic layer indicates the presence of an aryllithium compound.
The mixture is then acidified with acetic acid to reveal the presence Of an alkyllithium compound.
(b) Following hydrolysis the solution is acidified with acetic acid (5—15
drops).
20% aqueous sodium bisulphite is added; aryl- but not alkyllithium
compounds give a green colour.
Gilman's Colour Test 2 (16) which also distinguishes alkyl- from aryllithium compounds, depends on the rapid metal—halogen exchange of the former with 4-bromo-N,N-dimethylaniline:
$$
\begin{align}
\ce{4-BrC6H4NMe2 + RLi &<=>> 4-Li2C6H4NMe2 + RBr}\\
\ce{4-Li2C6H4NMe2 &->[(i) Ph2CO][(ii) H3O+] [Ph2CC6H4NMe2]+}
\end{align}
$$
The solution to be tested (0.5-1 ml) is added to an equal volume of a solution of 4-bromo-N,N-dimethylaniline in dry benzene.
A 15% solution of benzophenone in dry benzene (1 ml) is added.
After a few seconds, water is added and then the mixture is acidified with concentrated hydrochloric acid.
A positive result is indicated by a red colour in the aqueous layer.

* Aerial oxidation sometimes leads to formation of the blue-green at this stage.

References

Gilman, H.; Cartledge, F. K.; Sim, S.-Y. The Analysis of Organic-Substituted Group IVB Lithium Compounds. Journal of Organometallic Chemistry 1963, 1 (1), 8–14. DOI: 10.1016/S0022-328X(00)80039-X.
Wakefield, B. J. Organolithium Methods; Best synthetic methods; Academic Press: London; San Diego, 1988. ISBN 978-0-12-730940-8.


Why is phenol soluble in sodium carbonate but not in sodium bicarbonate? The original question was basically to predict which of sodium carbonate or sodium bicarbonate can be used to distinguish between phenol and benzene.

I was sure that phenol neither reacts nor dissolves in $\ce{NaHCO3}$ but I was not sure about the $\ce{Na2CO3}$.  By searching on Google I found that phenol is soluble in sodium carbonate. I am curious to know why this is, even though phenol doesn't liberate $\ce{CO2}$.

The liberation of $\ce{CO2}$ from $\ce{Na2CO3}$ requires two protonation steps.

$$\begin{align}
\ce{CO3^2- + HA &<=> HCO3- + A-}\\
\ce{HCO3- + HA &<=> CO2 ^ + H2O + A-}
\end{align}$$

Neglecting the concentrations, the position of these equilibria will depend on the acid strength of $\ce{HA}$. If $\ce{HA}$ is a stronger acid than $\ce{HCO3-}$, for example, the first equilibrium will favour the products.

This is the case for phenol, which has a $\mathrm{p}K_\mathrm{a}$ of $9.95$, whereas $\ce{HCO3-}$ has a $\mathrm{p}K_\mathrm{a}$ of $10.32$. Therefore, as long as you don't have too much phenol, it will exist in the soluble phenolate form:

$$\ce{PhOH + CO3^2- <=>> PhO- + HCO3-}$$

The solubility of phenol does not necessitate that it releases carbon dioxide.

Now, the second protonation. If you want to quantitatively determine whether $\ce{CO2}$ will actually be liberated, you need more data than the $\mathrm{p}K_\mathrm{a}$ values, since quite a significant amount of $\ce{CO2}$ can exist in the aqueous form. However, one thing is certain: if you want to produce $\ce{CO2 (g)}$ you first have to protonate $\ce{HCO3-}$ to form $\ce{H2CO3}$ (which dissociates to give $\ce{CO2 (aq)}$).

This is not possible with phenol, since $\ce{H2CO3}$ has a $\mathrm{p}K_\mathrm{a}$ of $6.3$ (even after correcting for dissolved $\ce{CO2}$).


Nucleophilicity vs Basicity Can something be both a nucleophile and a brønsted base?
Also , I'm actually quite confused about the difference between these two terms.

by definition a nucleophile, is a lone pair donor. phile meaning 'loving'. i.e its attracted to species which are electron deficient. the term is used to describe a species in an organic mechanism.a  bronsted base, is a proton accepter. Mostly in organic chemistry, you draw curly arrows, going from the nucleophile to electrophiles,most usually a electron deficient carbon. so it acts more as a Lewis base. which is a lone pair donor. however there are protons floating around in most organic reactions, also very electronegative atoms such the halogens(leaving groups). which pick up protons so i guess by defintion you could say they act as bonsted bases as well as nucleophiles. However the term nucleophile is mostly reserved for the species acting on the electrophile in an organic mechanism.


When DNA degrades, does any chemical environment preserve sequence information? Small bits of DNA information can be found in ancient remains.   In analogy to
old bones becoming mineralized, is there a chemical environment in which the
nucleic acid sequence of the original strand might remain readable though in other, more
stable, chemical form?

A mineralized form might be possible (maybe DNA can form a pattern of missing atoms in stone) but reading it will be next to impossible. We have many techniques to amplify and read tiny bits of DNA, doing this for imprints in stone would be a huge challenge, especially because many (all?) techniques for reading DNA depend on the amplification of DNA.

As for conservation, as far as I know most of the damage making the DNA unreadable is hydrolysis of the backbone. If the DNA is in an environment without water this will not happen. Of course you need an extremely dry environment for DNA to last millions of years.


Standard Reduction Potentials and Electropositivity I was looking through a table of standard reduction potentials and the trend seemed like it correlated with electropositivity. Is it reasonable to qualitatively compare two metals to see which has a higher standard reduction potential based on electropositivity? If not, why is this unreasonable?

Your reasoning is reasonable. However, although we can make the association between electropositivity and reduction potential qualitatively, relying on the former one to imply about the reduction potential is not always safe. For instance, following the trend of electropositivity on the periodic table, we would imply based on the reasoning that you proposed, that Zn which has a greater electropositivity than Cu (check the table bellow) has also a greater reduction potential, which is not the case. When it comes to transition metals, there are actualy some other effects that play a role on the reduction potential of the atom, ex.: shielding effects, final electron configuration, etc. The bottom line is, although electropositivity could help to justify a certain trends in reduction potential, it should not be used to make rigorous predictions of these trends since it could easly lead to mistakes mainly when considering transition metals.

P.S. The table shows the trend according to electronegativity. You can visualize the electropositivity trend by inverting the color scale at the bottom.


Why don't the solute molecules in a volatile solvent also escape to contribute to vapor pressure? I'm currently studying relative lowering of vapor pressure with the addition of a  solid non volatile solute.

The book explains it with the notion that some of the non volatile solute also appears at the surface between the liquid solution and the vapor. However, i don't understand why don't the molecules of the solute at the surface just escape? Obviously, a solid can't vaporize (Unless you raise the temperature, (And maybe reducing the pressure?)) however the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid.

Hence, if i understand this correctly, they, just like the water (Or other solvent) molecules should be constantly moving around in the solution and sometimes 'bump' into other molecules to gain enough velocity to escape the solution. The few molecules that are are the surface between the solution and the air should be getting frequently hit with high energy molecules from below, which should constantly knock such molecules into the vapor. This would mean two things:

The solute also has a small partial vapor pressure
The vapor pressure should still lower, as once solute particles are knocked out into vapor, a few of the solvent particles should escape through the 'opening' before another solute can take it's place. Rinse and repeat, like a valve being opened and closed. Hence, while there would still be a lowering of vapor pressure, there shouldn't be any permanent layer being formed between the solution and the vapor.

But both of these are in contradiction to what's in the book. So where did i go wrong in my reasoning?

It is true that "non-volatile" is a relative term and that virtually all compounds have some, possibly infinitesimal, vapor pressure, but the idea is that this is negligible. And you are also correct that "...the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid."

Concerning that latter statement however, there are many reasons that a compound might maintain very low volatility as a dilute solute, even though the intermolecular forces within the bulk solid are not at play. For example, compounds having a large molecular weight tend to be non-volatile, and these will still require much more kinetic energy to be "bumped out" of the solvent, meaning this will be a statistically rare event compared to the transfer of solvent molecules to the vapor phase.

Ionic solids are another type of solute that are typically quite non-volatile. In order for an ionic species occupying a solute surface site to escape into the gas phase, in the case of an aqueous solvent for example, a rather large amount of energy would be required to dissociate the ion from it's solvation sphere and kick it out into the gas phase as a free ion.

I hope this answers your question "where did I go wrong in my reasoning". Don't hesitate to ask for clarification in the comments if I missed something.


The natural equilibrium of water Why can the equilibrium reaction which exists in water be simply represented as

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

rather than

$$\ce{2 H2O <=> OH- + H3O+}$$

Doesn't this change both the equilibrium constant AND the enthalpy change?

Why would this still give the correct answer when you use the ionic product of water in solution?

The former equation is just a simplified version of the latter.

The important factors are the concentrations (activities) of the positive and negative species.

However, since these values are small ($10^{-7}\,\mathrm{mol\cdot l^{-1}}$) compared to the concentration of water ($55,56\,\mathrm{mol\cdot l^{-1}}$) this distinction would not affect the value of the dissociation constant either.


Enol form of acetone is treated with D2O Could find the mechanism in brief online. Still have few queries. What makes the D2O not attack the the central carbon (to which OH is attached) and choose other the carbon instead (both have a double bond). Again, in the 'repeat' part of mechanism, there's only one carbon with double-bond (carbonyl group). Why does D2O attack the second carbon again and not the carbonyl group one? I'm really confused.

The oxygen of the water will attack the carbon of the C=O forming the hydrate, but this is in equilibrium and completely reversible and a dead end so of no consequence when considering the enol reaction.


Why is is this molecule called ketene? The marking scheme says that the molecule has both alkene and ketone.
So there is carbon to carbon double bond.ok.check there is alkene.But what about ketone.Ketone has this structure:

So can we technically say that there is ketone in it?

This molecule is a ketene because it contains the ketene functional group:

Whilst the ketene appears to contain an alkene and a ketone functional group, which as you have suggested is where its name comes from, its chemistry is notably different from both of its 'constituent groups'.


Does migraine medicine Topiramate work by supplanting Pyridoxal phosphate in enzymes? I have seen it said that the precise mechanism of action of migraine medicine Topiramate is not known. But I certainly see a resemblance between that molecule and PLP (Pyridoxal phosphate, the activated form of vitamin B6).

A significant similarity is that both have an "offshoot chain" (Topiramate's sulfamate taking the place of PLP's phosphate) where PLP's phosphate is what binds it to enzymes in the first place, and both have a more electronegative element in the middle of what would otherwise be a six-carbon ring, apparently important in the typical PLP enzyme reaction.

I have been wondering if Topiramate supplants PLP in the enzyme glutamate decarboxylase, changing the reaction into a deamination. This would deplete glutamate, causing the pill's migraine relief, and the deamination could even cause Topiramate hyperammonemia.
If it does the same or similar to serine hydroxymethyltransferase, which is also PLP dependent, it could cause glycine deficiency, thus collagen deficiency - and this would even explain Topiramate hair loss.
When you look at it this way, it's frankly hard to imagine the mechanism of action is anything different. Have I stumbled upon it? Or has this been suggested already?

Background
Topiramate (TPM) is an anticonvulsive and antimigraine drug that helps prevent or reduce the frequency of epileptic seizures and that is used as a prophylaxis of migraine. (Shank and Maryanoff, 2008)
TPM not only inhibits kainate subtypes and α-amino-3-hydroxy-5-methylisoxazole-4-propionic acid (AMPA) subtypes of glutamate activated ionotrophic receptors, the modulatory effects on GABAA receptors and voltage-gated Ca2+ and Na+ channels (Gibbs et al., 2000), but also inhibits carbonic anhydrase enzymes (demonstrated in electrophysiologic and biochemical studies). see drugs.com
OP question
It is an intriguing possible correlation that “Topiramate supplants PLP in the enzyme glutamate decarboxylase [GAD]”, notably based on structural similarity of the alkyl-sulfamate to the alkyl-phosphate as well as the linked rings.
However, the literature does not seem to align with this premise:

Shank and Maryanoff et al.(2008) noted that “[Topiramate’s] sulfamate moiety is essential, but not sufficient, for its pharmacodynamic properties.”
Whereas PLP is a substitute pyridine (planar) derivative, TPM is a sulfamate, substituted derivative of β-D-fructopyranose (Maryanoff et al., 1987) where its substituted ring is in the boat conformation and its ring is puckered. See the crystal structure as entry SEQKAA02 at CCDC

TPM does not seem to inhibit glutamate decarboxylase. “Single-dose and repeated treatments with TPM were without effect on the activities of GABA-T and GAD (Fig. 2) in mouse brain at 4 h after administration.” (Sills et al., 2000)
GAD inhibition and Topiramate treatment have opposite effects. “Clinical evidence shows that lowered GAD activity is associated with several forms of epilepsy which are often treatment resistant.” (Zhang et al. 2017; Lloyd et al. 1986) Furthermore, Zhang et al. synthesized an ethyl ketopentenoate (EKP), a lipid-permeable GAD-inhibitor, and used it to induce refractory seizures in zebrafish larvae. (yes, that’s mean.. but informative here.)

Molecular basis of carbonic anhydrase inhibition
Topiramate is a carbonic anhydrase inhibitor (multiple classes), a Zn-dependent metalloenzyme. Below is how Topiramate (green) binds to human carbonic anhydrase II. The zinc ion is shown as a silver sphere. RCSB entry 3HKU.pdb DOI: 10.2210/pdb3HKU/pdb and Lopez et al. (2009):

Conclusion
Given the above, I wouldn’t think the apparent structural similarity of Topiramate to PLP is relevant to epilepsy and/or migraine treatment.
References
Lloyd, KG, Bossi, L, Morselli, PL, Munari, C, Rougier, M and Loiseau, H (1986) “Alterations of GABA-mediated synaptic transmission in human epilepsy” Adv. Neurol. 44:1033–1044. source
Lopez, M, Paul, B, Hofmann, A, Morizzi, J, Wu, QC, Charman, SA, Innocenti, A, Vullo, D, Supuran, CT and Poulsen, SA (2009) “S-Glycosyl Primary Sulfonamides−A New Structural Class for Selective Inhibition of Cancer-Associated Carbonic Anhydrases” J. Med. Chem. 52:6421-6432. DOI: 10.1021/jm900914e
Maryanoff BE, Nortey SO, Gardocki JF, Shank RP, Dodgson SP (1987) “Anticonvulsant O-alkyl sulfamates. 2,3:4,5-Bis-O-(1-methylethylidene)-β-D-fructopyranose sulfamate and related compounds” J. Med. Chem. 30:880–887. DOI: 10.1021/jm00388a023
Shank, RP and Maryanoff, BE (2008) “Molecular pharmacodynamics, clinical therapeutics, and pharmacokinetics of topiramate” CNS Neurosci. Ther. 14:120-142. DOI: 10.1111/j.1527-3458.2008.00041.x
Sills, GJ, Leach, JP, Kilpatrick, WS, Fraser, CM and Brodie, MJ (2000) “Concentration-Effect Studies with Topiramate on Selected Enzymes and Intermediates of the GABA Shunt” Epilepsia 41 (Suppl. 1):S30-S34. DOI: 10.1111/j.1528-1157.2000.tb02168.x


Rate of solvolysis of allyl and alkyl halides Suppose I have 2 compounds, 3-chloropropene and 1-bromo-1-methylcyclohexane. Without knowing the values, how can we compare the rate of solvolysis?

My attempt: Resonance is quite dominant character thus alkyl chloride resonance I think should dominate over hyperconjugation but then we have an exception that 3 degree carbocation is more stable than benzyl carbocation hence there might be some disparity in orders.

Resonance effect is more stronger than hyperconjugation irrespective of the number of αH. In first case allylic carbocation is formed which is stabilised by resonance. So solvolysis rate is more for the first compound


Identifying Lewis acid What is the Lewis acid on the reactants side?

$$\ce{HNO2(aq) + HPO4^{2-}(aq) <=> NO2-(aq) + H2PO4-(aq)}$$

The book says it is the $\ce{H+}$ only, not the whole molecule of the nitrous acid, how so?

What we have, in the Lewis theory, is an acid-base displacement reaction.  The Lewis acid, $\ce{H^+}$, is initially combined with one Lewis base, $\ce{NO2^-}$.  Then another Lewis base, $\ce{HPO4^{2-}}$, takes the $\ce{H^+}$ away displacing the $\ce{NO2^-}$.

Similarly when $\ce{HCl}$ is dissolved in water and dissociates, the water is displacing the chloride ion.  The Lewis acid involved in this displacement is again $\ce{H^+}$.


Help understanding the following conversion I am given the substrate and reagents in steps (1,2,3). I am asked to figure out the product! I am not sure how the reagents in 2 (aq. Na2CO3) and 3 (NaH, MeI/DMF) work to give that answer. Why is the answer what it is?

You can work backwards here. Step 3 converts cyclohexanol to the product by substitution reaction of methyl iodide and alkoxide. Sodium hydride is the base that deprotonates the alcohol.

Step 2 is a bit dubious. They want to substitute a secondary chloride with hydroxide. In principle this will work, though I am suspicious about whether or not it will actually take place. For cyclohexyl chloride (product from step 1), the substitution is probably only going to happen from the axial conformer; the equatorial conformer has the flag pole hydrogens blocking the approach to the $\sigma^{*}$ orbital. But from the axial conformer, I suspect that elimination is also a competitive pathway. Hard for me to say exactly what the relative rates are for sure.

Step 1 you've likely figured out. It's radical chlorination of an alkane.


Diprotic Basic Buffer So, as part of an experiment, I plan to make a diprotic basic buffer. For a monoprotic basic buffer, I used the standard combination of Ammonia and Ammonium Chloride. However, how can I make a diprotic basic buffer using commonly available reagents? If possible, could you also explain how the buffer behaves in the presence of an acid/base?

For a typical diprotic base buffer system you can use sodium bicarbonate. Adding hydrochloric acid to sodium carbonate solution

The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is:

$$\ce{Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + CO2(g) + H2O(l)}$$

If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point - because of the 1 : 2 ratio in the equation

Take a look at he sample titration curve:

The sense of diprotism of a base, is revealed through its ability to accept the two protons from hydrochloric acid to form water.

Referencing and diagrams from

pH (TITRATION) CURVES


Zero dipole moment of p-dinitrobenzene Why is $\mu=0$ for p-dinitrobenzene ? How can the dipole moments of the two Nitro groups exactly cancel each other ? Shouldn't it have a non-zero, albeit feeble, dipole moment ?

This paper (J. Sep. Sci. 2011, 34, 1489–1502) lists 1,4-Dinitrobenzene as having a very small dipole moment (0.01 Debye). No error is given on the dipole moments, but they do list compounds with exactly zero dipole moment (naphthalene for example).


Why doesn't open-chain glucose react with water readily in aqueous solution? Though the exact number given varies from 0.02% to 3%, it's well agreed that glucose spends a very small percentage of its time in aqueous solution as a linear aldehyde. Glucose's overwhelming preference as a cyclic pyranose is commonly explained by the cyclization being intramolecular; while the aldehyde group in its linear form can react with other molecules like protein and lipids during rare encounters, it far more readily reacts with an alcohol group that is always around by virtue of being on the same molecule.

My problem with that explanation is that there is indeed another molecule that is always around in aqueous solution: water. All sources I could find said that the common instability of hemiacetals and geminal diols are both explained by oxygen repulsion. Therefore, I would expect there to be a glucose-geminal diol concentration comparable to the cyclic hemiacetal concentration in aqueous solution, yet I cannot find any mention of a glucose-geminal diol, as if it didn't exist. What am I missing?

Why doesn't water react with the aldehyde to form a linear geminal diol?

It does, and depending on the nature of the aldehyde, this will be the major species. You are right in saying that the high concentration of water helps to form the diol. In other solvents, the diol form will be present at lower fraction, even when they contain a bit of water. The Wikipedia article on geminal diols has some equilibrium constants for various aldehydes and ketones in aqueous solution: https://en.wikipedia.org/wiki/Geminal_diol#Reactions

  Glucose's overwhelming preference as a cyclic pyranose is commonly explained by the cyclization being intramolecular; while the aldehyde group in its linear form can react with other molecules like protein and lipids during rare encounters, it far more readily reacts with an alcohol group that is always around by virtue of being on the same molecule.

Yes, that is the special thing about glucose: the aldehyde can form an intramolecular hemiacetal without strain. It can also form a geminal diol, but that is no different from other aldehydes, so it is usually not mentioned in that context.

Is the geminal diol ever important for the reactivity?

There is an enzyme called aryl sulfatase which has an aldehyde (a cysteine modified post-translationally to a formyl glycine) in its active site [review]. X-ray crystal structures show that the main form is the hydrated species, i.e. the geminal diol. In the suggested mechanism, one hydroxyl attacks a sulfate ester, displacing the alcohol. Then, the second hydryoxyl group reforms the aldehyde and releases the sulfate in an elimination reaction.

Are there other instances where we don't explicitly mention reaction with water?

The hydrogen ion occurs in aqueous solution in various hydrated states. Carbon dioxide in aqueous solution occurs either hydrated (carbonic acid) or not. In either case, it is easier to lump the species into one than to talk about them separately.

What is the equilibrium constant for the hydration of the aldehyde in glucose

I did not find data for glucose, but did find some for glyeraldehyde 3-phosphate (Petersson and Petersson 1999). [In the HTML version of the paper, some equations and schemes don't show, but the PDF is an improvement.] In aqueous solution, only about 3% of this phosphorylated sugar is present in the aldehyde form. In the reaction they were studying, only the aldehyde form reacts in the enzymatic reaction, so there is a slow non-enzymatic first step to dehydrate the diol and a fast second step.

Wikipedia has an article on glycoaldehyde, with 4% of it in the aldehyde form.
If glucose behaves in a similar manner, most of the linear form would be in the diol form. Equilibrium constants given in the literature probably use the total concentration of linear glucose (hydrated or not) in the equilibrium constant expression rather than the actual concentration of aldehyde.


pH and materials selction I working on the Navy study guide for their nuclear engineering programs and I am not a Chemist.  Thus, I have come here to try and develop a better understanding of the subject matter.

Why is pH important in materials selection?

The pH scale is merely a way of keeping track of the concentration a species (the hydrogen ion $\ce{H^+}$, $pH = -log\  a_{H^+} \simeq -log\  [H^+]$) in a medium (usually liquid, not necessarily aqueous). You can define arbitrary scales for other species, too. Analytical chemists are more familiar with scales such as pCl, with $pCl = -log\  a_{Cl^-} \simeq -log\  [Cl^-]$, and that's fine, but they're much rarer. The reason chemistry pays so much attention to hydrogen ions specifically is because the pH scale keeps track of what is arguably the most reactive species in all of chemistry (one could make a case for alpha particles, i.e. bare helium-4 nuclei $\ce{^4_2 He^{2+}}$ instead, but hydrogen ions are nevertheless comparable in strength and far more common).

If you stop to think about it, a (non-solvated) hydrogen ion, $\ce{H^+_{(g)}}$ is a bare proton. Let me explain why this is an important realization. In chemistry, we talk a lot about how electrons get shoved around when different molecules meet (also known as chemical reactions), and the exact way in the electrons shuffle about is dependent on complicated electromagnetism at the quantum level. However, to a simple but enlightening approximation, one can imagine how substances behave based on how much electrical charge they concentrate in a region; simply, concentrated charges will display stronger electromagnetic effects.

In most of chemistry, we deal with charges of order unity relative to the fundamental electrical charge ($e=1.6022 \times 10^{-19}\ C$) spread in the volume of a few atoms (of order $10^{-3}\  nm^3$). The resulting charge density can be found by dividing the charge by the volume. Now here's the thing - the nucleus of an atom is tiny compared to the size of the entire atom (with its electrons). A lone hydrogen atom has a radius of approximately $0.1\ nm$, but a lone hydrogen nucleus has a radius of about $1\ fm = 0.000001\ nm$, a difference of five orders of magnitude. The volume difference is thus fifteen orders of magnitude. Therefore, the bare nucleus which comprises a $\ce{H^+}$ ion has a charge density 15 orders of magnitude greater than a similarly-charged atomic ion, for example $\ce{Li^+}$. In effect, this means that a bare $\ce{H^+}$ has such a huge positive charge density that it will tear opposite charges (electrons) from absolutely anything. Helium and neon are notoriously unreactive, but both react quite happily (and rather exothermically) with gaseous $\ce{H^+}$ forming hydrohelium(+) and hydroneon(+) as the respective products.

So why doesn't water (or any ready source of $\ce{H^+}$ ions, i.e. protic substance) immediately dissolve absolutely everything it touches? It so happens that, in condensed phases, whatever molecules close to a hydrogen ion are so attracted that they effectively encase the hydrogen ions into solvated hydrogen ions, forming a shell which protects everything else from the aggressive reactivity of the hydrogen ion. For example, in water, $\ce{H^+}$ is actually not a very correct description of reality. Rather, one finds $\ce{H(H_2O)_{n}^+}$ (often alluded to by formulae such as $\ce{H3O^+}$, $\ce{H5O2^+}$, $\ce{H9O4^+}$ and others), or more concisely (but less distinguishably) $\ce{H^+_{(aq)}}$. After being surrounded by water molecules, the hydrogen ion loses almost all of its reactivity, as the original charge is now effectively dispersed over several molecules, greatly dimishing the charge density of the solvated hydrogen ion compared to the non-solvated hydrogen ion. Even after the massive handicap, solvated hydrogen ions are still quite reactive. Thus, it is important to keep track of their amount, to figure out if certain materials are capable of withstanding the onslaught.

After all I said, it seems like it would be a good idea to minimize the amount of $\ce{H^+}$ in water, which can be done by going to high values in the pH scale. However, by doing so you begin to significantly increase the amount of hydroxide ions $\ce{OH^-}$ present, in order to obey the autodissociation reaction equilibrium of water. In some sense, hydrated $\ce{OH^-}$ ions in solution are about as reactive as solvated $\ce{H^+}$ (even though non-solvated $\ce{H^+}$ is much, much more reactive than non-solvated $\ce{OH^-}$). Thus, going to either end of the pH scale can be a bad idea. One could say that water solutions are least reactive* when their pH is close to 7 (including pure water), as one minimizes the sum of $\ce{H^+}$ and $\ce{OH^-}$ present.

* Least reactive in the specific sense that if a material resists acidic aqueous solutions or basic aqueous solutions, then it must resist pure water and neutral aqueous solutions, assuming that whatever else anything else in the water incapable of altering the pH value is also unreactive; a pH 7 sodium chloride solution for example can be quite corrosive to metals, but not because of the water itself.


How does water deionization process work? What are the technical/scientific details behind the process of water deionization? How does deionization machines like those manufactured by Millipore work?

Most cost/energy efficient method to make deionized water is reverse osmosis. The mechanism is to apply external pressure to water and force it to go through a layer or layers of semipermeable membrane.


Why is osmium the densest known element? Why is osmium so dense despite there being heavier elements after it in the periodic table?

The density of an element is related to how many atoms can be placed in a given volume and the weight of the nuclei.  Therefore, the smaller the atomic radius of an atom and the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation.  This small atomic separation along with osmium's relatively high atomic number gives rise to osmium's high density.

The small atomic radius can be attributed to the following factors:

f orbitals are very diffuse and therefore result in poor screening of the electrons further out.  In the case of osmium ($\ce{[Xe] 4f^{14} 5d^6 6s^2}$) the poor screening by its 4f orbitals leads to a contraction in the n=5 and n=6 orbitals.
Due to osmium's high atomic number, relativistic effects come into play.  These relativistic effects have been discussed previously on SE Chem for gold and mercury.  A general discussion as well as the mathematics describing the relativistic effect can be found in these earlier posts.  Basically, for heavy nuclei the s electrons must circulate at relativistic speeds in order to maintain a stable orbit (not fall into the nucleus).  Therefore the electron mass increases and the s orbital radius decreases (the p orbital radius also decreases, but to a lesser degree).

The orbital contraction caused by these two effects leads to a small atomic radius for osmium.  Therefore, short metal-metal bonds result.  This is reflected in the small unit cell volume for osmium (27.96 cubic angstroms).  For comparison the unit cell volume for lead is 121.3 cubic angstroms! As a result, more osmium atoms can pack into a given volume compared to other elements

Osmium's relatively high atomic number along with its small atomic radius, as explained above, gives rise to osmium's high density.


Titration of sulfuric acid with sodium hydroxide $55.0~\mathrm{mL}$ of $0.250~\mathrm{M}~\ce{NaOH}$ is used to titrate $35.0~\mathrm{ml}$ of $\ce{H2SO4}$.
  What is the molarity of $\ce{H2SO4}$?

I know that the equation for this reaction is:
$$\ce{H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$

I know that I have to use $$i_\mathrm a\cdot M_\mathrm a\cdot V_\mathrm a = i_\mathrm b\cdot M_\mathrm b\cdot V_\mathrm b$$ formula where $i$ is the number of ions, $M_\mathrm a$ and $M_\mathrm b$ are the molarities (of the acid and base, respectively), and $V_\mathrm a$ and $V_\mathrm b$ are the volumes of the acid and base (in milliliters).

I starting doing this problem and got up to this point:

$$i\cdot M_\mathrm a\cdot 35~\mathrm{ml} = 2\cdot 0.250~\mathrm{M}\cdot 55~\mathrm{ml}$$

So my question is:
What should I put for the number of ions on the left hand side? Is it three or two?

That is one way of doing it.

In reality, this is a stoichiometric question where you are trying to find the molarity of an unknown polyprotic acid by titrating it with a strong base (woo-hoo!) of known molarity. I prefer using this method as it is easier to see the relationship between the units as well as use chain-ink conversion which most people are familiar with.

We know that

$$ \text{M} = \frac{\text{mol}}{\text{L}} $$

Therefore,

$$ 55 \ \text{mL} \ \ce{ NaOH} * \frac{250 \ \text{mmol} \ \ce{NaOH}}{1000 \ \text{mL} \ \ce{NaOH}} * \frac{1 \ \text{mmol} \ \ce{H2SO4}}{2 \ \text{mmol} \ \ce{NaOH}} * \frac{1}{ 35 \ \text{mL} \ \ce{H2SO4}} = $$

Which equals

$$ 0.19643 \ \frac{\text{mol}}{\text{L}} $$


Why nature favours lower potential energy? It is very abstract but I was asked this question recently and unfortunately I was not able to answer it decently enough.

I've seen it coming up in various forums of physics, chemistry or biology. Most of the answers I've read give examples from everyday life or approach the matter more philosophically. I'd appreciate any clarification in this matter.

It doesn't.  Nature favors higher entropy, and in many situations higher entropy can be achieved by turning potential energy into heat, so it looks like the general proposition is that systems evolve to states of lower energy. In fact, that was considered as a possible general law of nature by early philosophers.

However, there are plenty of counter-examples.  For example, consider the smoke from a fire.  It rises spontaneously, so the atoms and molecules in it are spontaneously ascending to a higher state of gravitational potential energy.  Or a steam engine, in which the flow of heat from a fire can easily do work (compress springs, move objects uphill, run a generator to charge a battery, and otherwise increase the potential energy of things).  Very early philosophers rationalized these distinctions by saying that things tended to "seek their own level," with certain things seeking states of low energy and others seeking states of higher.  This isn't very satisfactory, of course, because there's no way to determine ahead of time which thing seeks which state.  One of the greatest triumphs of 19th century science was to discover the underlying principle of entropy, and that all spontaneous changes can be understood as tending to increase the entropy of the universe.

The question of why nature favors states of higher entropy is much harder to answer, because such a preference is not built into the laws of physics.  Basic physics tells us states of higher entropy are favored only when you generally begin in states of low entropy.  So far as anyone can tell, the universe generally began in a state of very low entropy, and it has been increasing ever since, and that's the only reason why we observe an evolution to state of higher entropy.  Why the universe started off in a state of low entropy, nobody knows.


Shortcut method to find atomic term symbols ignoring the spin orbit coupling Is there any shortcut method by which we can find the atomic term symbols ignoring the spin-orbit coupling, i.e. ignoring the values of J.

For instance, when writing the d-d transitions for any complex, we write the atomic term symbols and them split them further into other states. I know the lengthy procedure to find the term symbols by considering the values of Ml, and finding the exact states including the spin-orbit coupling but is there a shortcut way by which we may find the atomic term symbols?

Like, in the above picture, how did we find the terms for d2 to be 1S, 1G, 3P, 1D, and 3F

After searching on the internet, and going through textbooks, I understand there is no "shortcut-method" to get the atomic term symbols in atoms with a large number of electrons.


Production of hydrogen through the reaction of aluminium metal and sodium hydroxide If I have a solution of 25% $\ce{NaOH}$, and I simply add a cut-up aluminium can or some aluminium foil, will it react into hydrogen and sodium aluminate?

In principle this reaction should happen, but it didn't work when I attempted it. Perhaps an oxide layer prevented it, or a protective layer of sodium aluminate was formed. (Perhaps my experiment was faulty.) Should this happen, and what can I do to fix it? (My goal is to generate hydrogen gas.)

Oxide layer would not hold against the strong alkali, and sodium aluminate does not form protective layer either. I suppose cans and foil may be covered by some coating, to make them suitable for food grade applications. Try burning it off, or rather use other sorts of aluminium (cut-up wire, maybe). Normally it would react quite vigorously.


Why Molecular Solids can be treated as Monoatomic in Reactions I'm trying to figure out what is the precise rule for the formula of elemental substances in reactions. I'm afraid I can't word the question properly, so I'll try to give some examples.

1)Oxidation of some metal, e.g. Calcium

$2Ca+O_2 \rightarrow 2CaO$

Since Calcium is a metal I understand that each atom reacts independently and the formula is plain $Ca$

2)Oxidation of some molecular gas, e.g. Hydrogen

$2H_2+O_2 \rightarrow 2H_2O$

Now atomic hydrogen in gas form is not what one usually expects, but rather the diatomic gas, so the formula must be $H_2$.

3)Oxidation of Sulfur

$S+O_2\rightarrow SO_2$

Now, here I get confused. I just copied the reaction from a textbook, but everywhere I looked in the internet says that this is the correct formula. On the other hand the textbook says that the most common allotrope of sulfur has formula $S_8$. And now I wonder why the reaction should not involve molecular sulfur and be

$S_8 +8O_2\rightarrow 8SO_2$.

I do understand that either way one would get the correct mass proportions, but I'm sot sure if that wold be the case in other reactions or with other substances.

So, I get that for metals the formula for the elemental substance that appears in reactions is just the atom. I get that for carbon and other network covalent materials the same applies. And I get that for molecular gases you need the diatomic formula, because each molecule participates as a whole. What puzzles me is what is the rationale beyond plain atoms appearing in molecular solids like sulfur, phosphorous, etc.

I apologize for not being able to state the problem in a more succinct fashion, feel free to edit the question or ask for further clarifications.

Generally, other than the seven diatomic gases at STP ($\ce{H2}$, $\ce{O2}$, $\ce{N2}$, $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$... though  $\ce{Br2}$ & $\ce{I2}$ have fairly low vapor pressure at room temperature), for convenience, other elements are considered as if they reacted as individual atoms, though, as you state, that is not actually the case. Phosphorus, for example, can be in allotropes with formulas of $\ce{P2}$, $\ce{P4}$, or even a graphitic form, but in a reaction, it's usually considered to be monatomic (unless one must take into account the energy to break the bonds of a specific form).

BTW, iodine vapor is diatomic only at moderate temperatures. From Wikipedia: "1% of a sample of gaseous iodine at atmospheric pressure is dissociated into iodine atoms at 575 °C." Furthermore, at STP, iodine solid behaves as a semimetal, so perhaps it shouldn't be considered diatomic... but it's easier to just say "all the halogens are diatomic", ignoring iodine and fleeting astatine.

In brief, then, this is the conventional list.


Electron's Energy In The Atom I know that the exact (OK most of the times not the exact) energy of an electron can be calculated by solving Schrodinger's differential equation, but can we explain an electron's energy of a specific orbital using the Penetration and Shielding effect?

Can we explain an electron's energy of a specific orbital using the
  Penetration and Shielding effect?

Yes, to some extent it is possible. Look no further than Slater's rules which specifies the values of the effective nuclear charge $Z^*$, and the effective principal quantum number $n^*$ that can be used to replace the nuclear charge $Z$ and the principal quantum number $n$ in the hydrogen-like atom orbital energy expression. The energy of each and every orbital can be approximated then by
$$
    E = - \left( \frac{Z^*}{n^{*}} \right)^{2}.
$$
But I would be careful calling this an explanation, since it is rather a rationalization. I wrote about the difference between those two here.


Electronegativity Considerations in Assigning Oxidation States I have never seen anything other than a set of rules like these when textbooks present how to assign oxidation numbers. Such as these:

However, if we keep in mind that oxidation numbers are simply imaginary numbers which suppose all bonding to be ionic - i.e. not electron sharing - and if we keep in mind simply relative electronegativity, we can easily work out the oxidation state of any element in any compound.

For example, take water. Bonding order: $\ce{H-O-H}$. Oxygen has two lone pairs. Oxygen is more electronegative than hydrogen. So we suppose that oxygen takes both electrons in both bonding pairs. Oxygen has 8 electrons. Its valence is 6. 6-8 is -2; oxygen has two more electrons assigned to it than what it has in its valence, so naturally its oxidation state is negative 2. No rules needed.

Now hydrogen peroxide. $\ce{H-O-O-H}$. Here we have an oxygen-oxygen bond so in this case, neither element wins the electronegativity battle. Electrons are split between the oxygen and the oxygen. However, since the oxygens are much more electronegative than the hydrogens, we assign both electrons in both the $\ce{O-H}$ bonding pairs to oxygen. Oxygen has 7 electrons assigned to it; oxygen has a valence of 6; oxidation state: -1. No need to memorize exceptions as stated in the table above.

We could go on. However, I am curious:

1) Has anyone been taught to assign oxidation states this way?
2) If not, what do you think of this method?

In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted in the earlier version of this answer, parts of which are included below.

The electronegativity battle scheme is most helpful for all kinds of compounds since it is one of the most generic ways to derive oxidation states. The table represents just a cheat sheet that might be very helpful in the beginning. If you are spending most of your time with chemistry, this table will be present as some sort of muscle memory - usually referred to as chemical intuition.

The IUPAC gold book now defines oxidation state as follows:

  oxidation state
  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
  OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

  The underlying principle is that the ionic sign in an $\ce{AB}$ molecule is deduced from the electron allegiance in a LCAO-MO model: The bond’s electrons are assigned to its main atomic contributor. Homonuclear $\ce{AA}$ bonds are divided equally. In practical use, the ionic-approximation sign follows Allen electronegativities (see Source). There are two general algorithms to calculate OS:

  Algorithm of assigning bonds, which works on a Lewis formula showing all valence electrons in a molecule: OS equals the charge of an atom after its heteronuclear bonds have been assigned to the more electronegative partner (except when that partner is a reversibly bonded Lewis-acid ligand) and homonuclear bonds have been divided equally:

  Algorithm of summing bond orders: Heteronuclear-bond orders are summed at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum, yielding the OS. This algorithm works on Lewis formulas and on bond graphs of atom connectivities for an extended solid:

  NOTE:

  Specific uses may require modified OS values: Electrochemical OS is nominally adjusted to represent a redox-active molecule or ion in Latimer or Frost diagrams. Nominal OS values may also be chosen from close alternatives for systematic-chemistry descriptions.
  Some OS may be ambiguous, typically when two or more redox-prone atoms enter bonding compromises and nearest integer values have to be chosen for the OS.
  The caveat of reversibly bonded Lewis-acid ligands originates from the simplifying use of electronegativity instead of the MO-based electron allegiance to decide the ionic sign. Typical examples are the transition-metal complexes with so called Z ligands in the CBC ligand-classification scheme (see Source).
  When used in chemical nomenclature as a symbol, the OS value is in roman numerals.
  At the introductory teaching level, prior to the bonding-based definition and algorithms: OS for an element in a chemical formula is calculated from the overall charge and postulated OS values for all the other atoms. For example, postulating $\text{OS} = +1$ for $\ce{H}$ and $−2$ for $\ce{O}$ yields correct OS in oxides, hydroxides, and acids like $\ce{H2SO4}$; with coverage extended to $\ce{H2O2}$ via decreasing priority along the sequence of the two postulates. Additional postulates may expand the range of compounds to fit a textbook’s scope.

  Source:

  Karen, P.; Mcardle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure Appl. Chem. 2014, 86 (6), 1017–1081 DOI: 10.1515/pac-2013-0505.
  Karen, P.; Mcardle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure Appl. Chem. 2016, 88 (8), 831–839 DOI: 10.1515/pac-2015-1204.
  Nomenclature of Inorganic Chemistry. IUPAC Recommendations 2005. p. 34. Available as pdf from old.iupac.org

I recommend having a look at the sources cited above, they contain a lot more information. (Note that the link from the gold book to the red book on the IUPAC website is broken.)

Prior to 2016 the IUPAC defined oxidation states in their gold book (via the Internet Archive) differently. I am including this definition, since it will be still present in prominent text books and it is what lead to the postulates given in the table of the original question.

  oxidation state (deprecated, pre-2016 definition)
  A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

  the oxidation state of a free element (uncombined element) is zero;
  for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
  hydrogen has an oxidation state of $1$ and oxygen has an oxidation state of $-2$ when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of $-1$ in hydrides of active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of $-1$ in peroxides, e.g. $\ce{H2O2}$;
  the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$, $\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and $\ce{H2SO4}$ are, respectively: $-2$, $04, $+4$, $+6$ and $+6$. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction.

What we see is, that the table in the original question actually reflects some of these rules. However, this set is not generic at all and it lacks a definition for compounds that do not contain oxygen or hydrogen, are elemental or monoatomic ions. With only these rules it is impossible do determine the oxidation states for $\ce{BF3}$ and many, if not most, compounds.

The lack of the definition is very well known, but it took the IUPAC until 2016 to actually change the official set. Hans-Peter Loock proposed a much simpler concept in Expanded Definition of the Oxidation State, which resembles the currently used definition quite well.

  The oxidation state of an atom in a compound is given by the hypothetical charge of the corresponding atomic ion that is obtained by heterolytically cleaving its bonds such that the atom with the higher electronegativity in a bond is allocated all electrons in this bond. Bonds between like atoms (having the same formal charge) are cleaved homolytically.

So he basically came to the same conclusion as Dissenter in the question. Another statement from this article is

  This is not a new definition, but it predates the IUPAC rules by several decades. For example, Linus Pauling provided a similar definition of the oxidation state in his 1947 edition of General Chemistry (3).

  (3):Pauling, L. General Chemistry; Freeman: San Francisco, CA, 1947. Republished by Courier Dover Publications, 2012.


How Can an Amino Acid Be Non-Polar? Looking at the structure of an amino acid (as a zwitterion), it seems that the part of the molecule with the amino group and the carboxylic acid group are quite polar (the amino group has a positive hydrogen and the carboxylic acid group has a negative oxygen). However, my textbook says that the character of the R group completely determines the polarity of the molecule (i.e. if R is non-polar, the molecule is non-polar). Why is this?

There are two things at play here and you need to mentally separate them.
When seen as molecules, all amino acids are polar to at least some degree. Most of them are very polar, especially when compared to other organic molecules. This is (obviously) due to the carboxylic acid and amino groups both of which are very polar functional groups and both of which are ionised in neutral aquaeous solution.
However, it is desirable to categorise the different amino acids into various groups, to better explain their individual contributions to an overall peptide or protein structure. As all amino acids have a carboxylic and amino group and as these groups are connected in peptide bonds to form the peptide or protein backbone, these two groups alone are not able to distinguish between the different amino acids.
For this reason, the amino acids have been classified according to their residue R. If the residue contains only carbon and hydrogen atoms (e.g. Gly, Ala, Val, Leu, Ile), it is a non-polar residue and thus these amino acids are classified as ‘non-polar amino acids’ in simplistic terms. In a peptide or protein, these amino acids’ non-polar side chains will usually act as non-polar domains within the tertiary structure
A similar case can be made for the ‘charged’ amino acids (Glu, Asp, Lys, Arg): obviously and trivially, all amino acids have charged groups when dissolved in water but only those four have a charged side chain. So only these four are able to build up salt bridges to cross-stabilise tertiary structures and are thus classed as charged.


Does the carbocation rearrangement in a SN1 reaction specify a change in degree? I am asked to determine whether the following process is likely to involve a carbocation rearrangement or not:

I think it is possible through the rearrangement given above (though the rearrangement cannot become a secondary or primary carbocation). Can someone please clarify if the carbocation rearrangement specifies a change in degree?

The most common carbocation rearrangements involve [1,2] hydride or alkyl shifts. As a typical result a secondary carbocation is converted to a more stabilized tertiary carbocation.

In the course of these reactions, the migrating group and the positive charge change place, they move to opposite directions.

This is not the case in your example.

I fail to see how your transformation can be described as a cationic rearrangement!


Qualitative analysis of the anionic part of a salt using it's soda extract While testing for anions using dilute $\ce{H2SO4}$, we prefer using the soda extract for tests rather than the actual salt. But we don’t use it for tests of carbonates, bicarbonates and sulphites. Why is this so?
I get that soda extract itself is a carbonate so no point to use it for carbonates but why not bicarbonates and sulphites?
Also I’m told that we use soda extract because sodium salts are mostly soluble. Then why specifically sodium? Isn’t there any other metal we can use?

If a solution of $\ce{H2SO4}$ is added to a some carbonate, bicarbonate or sulfite, it will produce some $\ce{CO2}$ or $\ce{SO2}$ effervescence with gas bubbles filling the container where this test is done, and maybe cause it to overflow. With sodium bicarbonate the equation is :
$$\ce{H2SO4  + 2 NaHCO3 -> Na2SO4 + 2 H2O + 2 CO2}$$
With sodium sulfite, it is :
$$\ce{Na2SO3 + H2SO4 -> Na2SO4 + H2O + SO2}$$


Why is telluric acid dibasic? $\ce{Te(OH)6}$ has an octahedral geometry with $6$ $\ce{-OH}$ groups.

Is it a protonic acid (like $\ce{HCl}$) or a Lewis acid [like $\ce{B(OH)3}$]?

What makes $\ce{Te(OH)6}$ dibasic?

Two may be as many protons that a Bronsted-Lowry base as strong as hydroxide ion can remove.  Building up more than two negative charges than that on the ion is a tough job.  Possibly a stronger base, such as amide ion in ammonia, could do that.

That the basicity of the medium could impact how many protons can dissociate is nicely demonstrated by citric acid.  Ordinarily it is triprotic in aqueous solution, but spectroscopic evidence indicates that at high alkali concentration (pH>14.4) a fourth, alcoholic proton can be removed.  There would be no doubt with a base such as amide in ammonia.


NaHCO₃ + HNO₃ → NaNO₃ + H₂O + CO₂ ... but something weird happens to me in practice I put 98% white fuming nitric acid and then I pour some sodium bicarbonate expecting $\ce{NaNO3}$ being formed, I'm not really using much water because it will make evaporation takes ages, even so $\ce{NaHCO3}$ should give $\ce{NaNO3}$ when in contact with $\ce{HNO3}$ and it seems so, $\ce{CO2}$ is formed very quickly so I suppose the reaction is happening... and I can see those white crystals (as the solution is over-saturated of $\ce{NaNO3}$) I shake, shake and shake, and pour more Sodium bicarbonate tills $\ce{HNO3}$ gets a pH around 7 (usually a bit less like 5 or 2) when I notice that it doesn't want to react anymore, however, I try to add mol per mol.

I know there's some unreacted $\ce{HNO3}$ but it should decompose into $\ce{H2O} + \ce{NO2}$ so I put it to start drying in a hot plate, and then the solution becomes yellow (decomposing) it's suddenly brown... and then my "supposed" $\ce{NaNO3}$ is yellowish/pink... and I wonder that seems like a $\ce{NaNO3} + \ce{NaNO2}$ mix but it makes no much sense, I tried with $\ce{NaOH}$ instead $\ce{NaHCO3}$ and it's the same ending color; it's not working... what did I do wrong? how to do it properly?

Let's look at the equation first:

$$\ce{NaHCO3 + HNO3 \to NaNO3 + H2O + CO2}$$

In other words, when added in equal parts, you'd ideally expect the bicarbonate anion to be protonated by the nitric acid, forming weaker carbonic acid (which will dissociate into water and carbon dioxide under temperature or even agitation), and the remaining ions will pair up into sodium nitrate (soda niter) solution in water.

However, nitric acid is tricky, and doesn't like to act "ideally". First, you already know that nitric acid decomposes into nitrogen dioxide and water; this happens whether you want it to or not. Highly exothermic reactions (most nitric acid reactions, including this one, are so) ramp up this inevitable side reaction with the extra heat. This will reduce your yield, but I'd expect it to go the other way than you apparently did; 1:1 mixtures of acid and bicarbonate will inevitably leave unreacted bicarbonate in the solution (not acid) along with some dissolved nitrogen dioxide (aka nitrite hydrate, $\ce{H2NO3}$).

In addition, "pure" WFNA is defined as anywhere north of 98% nitric acid, so pretty much any quantity of nitric acid you could get is likely to have impurities. So, I think your first mistake was to keep adding sodium bicarbonate; if you mixed a mass of "pure" (>98%) WFNA (even if you made it from scratch) with equivalent molar weight of sodium bicarb, you'll have a small excess of bicarb due to impurities in the nitric acid (pre-existing and created), but the solution will still read acidic (because of dissolved $\ce{CO2}$ and $\ce{NO2}$) after you've added more bicarb than will dissolve in the solution.

Now, heating the equilibrium solution to obtain the dry solid salts will remove the water from the dissolved nitrogen dioxide, and it will start acting like the nitrite ion it is, reacting with any unreacted bicarbonate to form yellow sodium nitrite, water and more $\ce{CO2}$. This sounds like exactly what happened to you, resulting in a more yellowish solid than the pure white soda niter you were expecting.

If you're not careful with the heat while evaporating off the water, parts of the reaction, namely the precipitating solid collecting on the bottom of the flask, can be superheated beyond boiling, and reach the temperature at which both of the niter salts will decompose into sodium oxide and more oxides of nitrogen. This is typically accomplished only at temperatures in excess of those capable with the average benchtop rig, but it becomes more thermodynamically favorable in the presence of excess sodium (which you have in the sodium bicarbonate). I'm not saying this happened - it's more likely that the nitric acid decomposed more than you might have wanted - but it's a theoretical possibility.


Estimating the absolute entropy of water at room temperature I thought it would be fun to get an idea of how enormously large $W$ in $S = k\ln W$ is. I'd specifically like to estimate this for a mole of water at $\pu{1 atm}$ and $\pu{298 K}$.  To do this, I need an estimate of the absolute molar entropy of water at those conditions, enabling me to calculate W using:
$$W= e^{S/k} $$

Are there tables of absolute (as opposed to standard) molar entropies of water at various temperatures?  I searched, and was unable to find one.

Alternately, is there an expression (or series of expressions) that provides an estimate for $C_p(T)$ of ice at $\pu{1 atm}$, between $\sim\pu{0 K}$ and $T_\mathrm{fus}$?  If there were, I could use this to estimate the absolute entropy of ice at $\pu{1 atm}$ and $\pu{273.15 K}$ by substituting it into the following:

$$ S_T-S_0=\int_0^T \frac{C_p(T)}{T}\mathrm{d}T$$

It would then be an easy matter to add the difference in entropy between ice at $T_\mathrm{fus}$ and water at $\pu{298 K}$.

To evaluate $S=Nk\ln(\Omega)$ it is necessary to use statistical mechanics to calculate $\Omega$.

In quantum mechanics, for any independent system in a potential there is generally a set of permissible energy levels $\epsilon_1,\,\epsilon_2,\cdots$. The energy of the whole assembly of $N$ similar systems is $E$ and the volume $V$. The number of microstates or configurations (complexions) is $\Omega$ and this depends on $E,\,V,\, N$. It is assumed that all of the different microstates with the same values of $E,\,V,\,N$ are equally probable.

There are $n_1$ systems with energy $\epsilon_1$, $n_2$ systems with energy $\epsilon_2$ and so on. The total number of different states of the assembly is the number of ways of dividing $N$ things into groups with $n_1, n_2\cdots$ in each which is $N!/(n_1!n_2!\cdots)$. The total number of different states is then the  total of all of these and is

$$\sum_{\text{all possible sets }n_i} \frac{N!}{\prod_i n_i}$$

As the total energy $E$ is fixed as is $N$, then

$$\Omega = \sum \frac{N!}{\prod_i n_i}  \quad \text{with constraints} \quad \sum_i n_i =N, \sum_i \epsilon_i n_i=E$$

This is a difficult expression to evaluate but it happens that the largest term makes such an overwhelming contribution that only this term need be considered. The mathematical method to find the maximum is an interesting use of Lagrange's method of undetermined multipliers and is given in texts on stat mech but covers several pages so is too long to include here.

The result is that

$$S=k\left(N\ln\left(\sum_i e^{-\epsilon_i /(kT)}  \right) +\frac{E}{kT}\right) $$

where $\displaystyle Z=\sum_i e^{-\epsilon_i /(kT)}$ is the partition function. The entropy thus comes down to calculating the partition function. In fact all other thermodynamic properties can be found once $Z$ is known.

In a molecular system the $\epsilon_i$ are the energies of translation, vibration and rotation and electronic excited states (if any). Knowing the energy levels, say from spectroscopy, then allows each of these terms to be calculated. The translational partition function is calculated using the Sakur-Tetrode equation and a value $\ln(Z)\approx 70 $ is typical. For rotational motion $\ln(Z)\approx 10$ and for vibrations $\ln(Z)\approx 1.5$ for each vibrational mode. The vibrational contribution is small because at room temperature $kT\approx 210$ wavenumbers and vibrations are typically several hundred to a few thousand wavenumbers.  In the case of liquid water you have a particular problem which may be mitigated somewhat by adding H bonds to a molecule as an extra vibration.


Mass change during phase transition When you fill up a balloon about a quarter way with liquefied butane fuel and let it sit at room temperature it will turn into gas. But why does the gas weigh the same as the liquified butane?
The liquefied butane liquid weighs just about as much as water.

Assuming you fill the balloon only with liquid butane, the answer is very simple - the gas in the balloon at room temperature IS the liquid in the balloon initially. If you put some amount of liquid butane into the balloon and seal it at t=0, then allow it to warm until the liquid has evaporated, there is essentially no transport of other matter into or out of the balloon during that time. The only thing that has happened is that the initial charge of liquid has absorbed thermal energy from the surroundings and undergone a phase change from liquid to gas.

Because there is no change in the amount of material in the balloon - there are the same number of butane molecules in the gas phase as there were initially in the liquid phase - there can be no change in mass. Liquid and gas phase contain the same number of molecules, at some fixed mass per molecule.

Note that while the mass does not change, the density does.


Calculating the x, y and z distances between two molecules - AMBER (Molecular Dynamics) I am interested in calculating the x,y and z distance between two molecules over a given simulation time. I know how to calculate the total (r) distance between them, but I want the distance between them along all three dimensions.

For the total distance, cpptraj can be used as:

parm prmtop-file
trajin netcdf-file
distance DIST :1-4@heavy-atoms-from-molecule-1 :5-8@heavy-atoms-from-molecule-2 out dist.out
run
quit

But like I said before, I do not want the r-distance. Is there anyway to calculate the x,y and z-distances in cpptraj? If not, what other alternatives do I have?

I ended up realizing that if you run:

parm prmtop-file
trajin netcdf-file
vector v0 :1-4@heavy-atoms-from-molecule-1 :5-8@heavy-atoms-from-molecule-2 out v0.out magnitude
run
quit

You will get the total distance as well as x, y and z distances between your two molecules. The output file will contain 8 columns. The AMBER manual states that they are identified as follows:

Frame | x-Vector-Coordinate | y-Vector-Coordinate | z-Vector-Coordinate | x-Origin-Coordinate | y-Origin-Coordinate | z-Origin-Coordinate | Magnitude

The vector coordinates are your x, y and z distances. Magnitude is the total distance.

Decided to share this as someone might run into the same problem one day.


How to capture the gas produced by combusting alcohols I am burning various alcohols and am attempting to capture the carbon dioxide produced. I could burn the alcohols in a conical flask with a cork on the top, with a tube directing the gas into a gas jar. However, due to the small volume of a conical flask, there would be almost no oxygen to burn and the flame would extinguish quickly. Is there any experimental setup I might have access to with high-school lab equipment that allows me to burn alcohols in an environment with plenty of oxygen, while capturing the gas produced? Thanks

If you look at the overall reaction, there is actually no gas produced, but there is a consumption of gas:

$\ce{CH3(CH2)_n-OH + \frac{3}{2} (n + 1)O2 -> (n + 1)CO2 +(n + 2)H2O}$

So 1.5 mol of oxygen are consumed, while only 1 mol of carbon dioxide is formed, so the volume will be less if the temperature is maintained.

To perform this experiment, you must collect the carbon dioxide in a solution of alkali and then titrate the solution.
The carbon dioxide will be converted to carbonate:

$\ce{2NaOH + CO2 -> Na2CO3 +H2O}$

A constant flow of air must be put into the system and the gas that comes out must be bubbled with a diffuser through the alkaline solution.
The air flow must be slow, so that all the carbon dioxide is washed out of the gas stream.

In the titration there will be two equivalence points, the first one of the excess of alkali and the second one of the carbon dioxide.
This one has a much smaller pH change.

Be sure to make a careful blank of the alkali, because it normally has some carbonate that will add to the value of carbon dioxide.