Adiabatic compressibility factor of ethyl alcohol I am looking for a source where I can find the adiabatic compressibility factor of ethyl alcohol at approximately 15 degrees Celsius. I have found the compressibility factor at 20 degrees. Is there maybe a way to convert this to the desirable temperature?

A calculation using the NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, and assuming a temperature of $T=15.5\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$, yields an isothermal compressibility of ethanol of
$$\varkappa_T=-\frac1V\left(\frac{\partial V}{\partial p}\right)_T=1.0795\times10^{-9}\ \mathrm{Pa^{-1}}$$
and an adiabatic (or isentropic) compressibility of
$$\varkappa_S=-\frac1V\left(\frac{\partial V}{\partial p}\right)_S=9.1749\times10^{-10}\ \mathrm{Pa^{-1}}$$
The number of digits is the default setting of the software and does not necessarily correspond to the number of significant digits.

By way of comparison, the results for a temperature of $T=20\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$ are
$$\varkappa_T=1.1130\times10^{-9}\ \mathrm{Pa^{-1}}$$
and
$$\varkappa_S=9.4729\times10^{-10}\ \mathrm{Pa^{-1}}$$
The data in “Pressure and Temperature Dependence of Liquid Density”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL include a value of $\varkappa_T=1.119\times10^{-9}\ \mathrm{Pa^{-1}}$ for $T=20\ \mathrm{^\circ C}$. Therefore, it seems a likely supposition that the corresponding calculation result of $\varkappa_T=1.1130\times10^{-9}\ \mathrm{Pa^{-1}}$ is reasonable but the number of digits is excessive.


How to dispose of NaOH solution? I have a range of sodium hydroxide solutions with different concentrations. (from 1.0 mol dm^-3 to 0.5 mol dm^-3). The max volume of each is 100ml. How can you easily dispose of these? Can I pour them down the sink?

You may pour them down the sink. NaOH will be soon neutralized and destroyed by the CO2 from the atmosphere, or by the bicarbonate ions present in all drinkable calcareous waters.


Calculate ion concentrations in mixed buffers I have to state that I'm not a chemist, nor a chemistry student, but an electrical engineer with some chemistry knowledge. However, for my work I have to deal with a problem, from which my question arose.

There is a buffer solution made by dissolving $\ce{Na2HPO4}$, citric acid and KCl for a certain measurement. I would like to calculate what the ion concenctrations are in the solution.

This is where i got.

[$\ce{K+}$], [$\ce{Na+}$] and [$\ce{Cl-}$] are quite obvious.
Denoting the hydrogen phosphate anion as $\ce{BH2-}$, I have written the equation for the phosphate: $\ce{BH^2- (aq) + H2O <=>[K_b] BH2^- (aq) + OH- (aq)}$ and [$\ce{BH^2-}$]+[$\ce{BH2^-}$]=$M_{\ce{Na2HPO4},init}$
Denoting citric acid simply as $\ce{AH3}$ I have also written the equations for its three hydrolisation steps, such as $\ce{AH3 (aq) + H2O <=>[K_{a1}] AH2- +H3O+}$ and so on two equations more, plus [$\ce{AH3}$]+[$\ce{AH2^-}$]+[$\ce{AH^2-}$]+[$\ce{A^3-}$]=$M_{citric~acid,init}$
I have written the water autohydrolysis equation $\ce{H2O <=>[K_W] 1/2 H3O+ + 1/2 OH-}$
I have so far have the 9 unknowns [$\ce{H2O}$], [$\ce{H3O+}$], [$\ce{OH-}$], [$\ce{BH2-}$], [$\ce{BH-}$] [$\ce{AH3}$], [$\ce{AH2^-}$], [$\ce{AH^2-}$], [$\ce{A^3-}$] but only 7 equations. I could do the approximation that the water concentration is constant, which would be more than enough for practical purposes, but still there are 8 unknkowns remain. I could make the approx. that the dissolved citric acid concentration is equal to the final undissociated citric acid contentration, since hardly any dissociates, leaving 7 unknowns. But this is the point where it gets interesting for me, because I don't know how I should proceed without making these approximations, since I don't see clearly the physical and chemical processes in the system. Can you help me what other equations are there which (maybe at the cost of introducing some more variables) finally result in an exactly solvable equation system?

You can ignore K+, Na+ and Cl-. You have to deal with H2O, H3O+, OH-, Ap3-, HAp2-, H2Ap-, H3Ap, Ac3-, HAc2-, H2Ac-, H3Ac. Here H3Ap = H3PO4 and H3Ac = citric acid in neutral state. You have 11 unknowns.

(+1)    [$\Delta$H+] = 3*[citric3-]+2*[Hcitric2-]+1*[H2citric-]+[(PO4)3-] - [HPO4-]-2*[H3PO4]
(+1) [H(from_water)+ + $\Delta$H+][OH(from_water)-]=10^-14 (dissociation of H2O)
(+1) [H(from_water)+]  = [OH(from_water)-]
(+3) Three dissociation constants of H3PO4 are known
(+3) Three dissociation of citric acid are known
(+1) Concentration of [Ac3- + HAc2- + H2Ac- + H3Ac] is known
(+1) Concentration of [Ap3- + HAp2- + H2Ap- + H3Ap]is known

You have 11 equations.

OH(from_water)- corresponds to total OH-
H3O+ corresponds to H(from_water)+ + $\Delta$H+]

//(+1) Water concentration is assumed 1 in solutions

Forget about (PO4)3- and H3PO4. They will have negligible impact on overall pH.

You didn't tell the ratios of Na2HPO4 and citric acid and these ratios are critical. Given pKa of citric (3.15, 4.77, 5.19) and phosphoric (2.15, 7.20, 12.35) acids you will most likely deal with either of these regions:
2.15-3.15 H3Citric + H2Phosphoric-
3.15-4.77 H3Citric+H2Citric-
4.77-5.19 H2Citric- - HCitric2-
5.19-7.20 H2PO4- - (HPO4)2-

If you know which interval you are working with (check ratio of the reagents, aka stoichiometry) then you can pick the relevant equilibrium and set all other concentrations to zero for initial evaluation. 10^-15 M of ions will have zero impact on overall pH.

I regrouped some equations and expanded the one you asked about.

Can you solve it for a general case. Yes, you can. But it is like accounting for air friction, Special and General Theories of Relativity and quantum effects. It is harder, but it does cover more cases. In this case you will be dealing with x^11+a*x^10+...=0 instead of a square equation. You can solve it using Wolframalpha. But at the end of a day you will find that in a given region the impact of majority of members is negligible. Think of it this way. You can approximate the system with a spline (5 polinomials of second degree) or one polinomial of 11th degree.


State of aggregation of "ate" complexes I understand that ate complexes can result from the addition of carbanions to neutral aluminum alkyls.

I was wondering what the balanced equation for this reaction is/what the structure would be in solid state - like does it dimerize?

$$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -> ?}$$

The crystal structure of lithium tetraethylaluminate has been published (DOI)
and as far as I can tell from the abstract, there is no indication for a dimeric anion here. It might be save to assume that the balanced equation therefore can be written as:

$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -> 2 Al(CH2CH3)4^{\delta-}\,Li^{\delta+}}$


Why does shower gel / shampoo foam a lot more on hairy than non-hairy body parts? We usually use shower gel by rubbing it on our palms and then applying it to each body part in turn in order to clean oily substances that clung to it since our last shower. This procedure does not cause the gel to foam (if it is applied to a completely hairless place, such as a clean-shaven head or back, it will produce some small amount of foam (also depending on how much you rub it); if it is applied to a place with relatively small clusters of hairs, such as arms or legs, it might foam up a bit more, and so on).

However, when applying shower gel or shampoo to a hairy place, such as the head, the groin, etc. it usually foams up significantly and rather quickly. A clear shower gel or shampoo will become a white foam in a short time.

Why is that? Is it the air trapped among the hairs?

(Please do not confuse this question with this one, they are different enough).

It's mainly due to the surface area and the spacial distribution of hair compared to that of hairless skin.

A typical human hair width is around one micron.  Given the various hair length and thickness (space between hairs) on the different hairy regions of the body, there are orders of magnitude more surface area on which soap bubbles (foam) can adhere as compared to smooth (or even very rough), hairless skin.

The second part of that answer, the spacial distribution, just means that the hair gives support to sustain the bubbles so they don't immediately collapse.

Short answer, but I think this pretty much covers it.


Is it possible to use Vmax as Kcat in a Michaelis-Menten reaction? I am studying a biological system using ODEs. As part of my work I need to use the '$k_\text{cat}$' version of the Michaelis-Menten reaction, which is:

$$\frac{V_\text{max}\cdot [\ce{S}]}{(K_\mathrm m+[\ce{S}])} = \frac{k_\text{cat} \cdot [\ce{E}_\text{free}] \cdot [\ce{S}]}{(K_\mathrm m+[\ce{S}])}$$

I have found in a publication a value of a $V_\text{max}$ for one of my reactions:

$$V_\text{max}=\mathrm{\frac{11.4\;pmol}{min^{-1}{\;pmol^{-1}\;enzyme}} = \frac{0.19\;nmol}{s^{-1}\;nmol^{-1}\;enzyme}}$$

My question is, since this $V_\text{max}$ value is quoted in terms of 1 unit of enzyme (i.e. 0.19 nmol is the maximum one nmol of enzyme can catalyze in a second), can I use this $V_\text{max}$ value as a $k_\text{cat}$?

here is the abstract I found the information from :

  Retinoic acid (RA) is a critical signaling molecule that performs
  multiple functions required to maintain cellular viability. It is also
  used in the treatment of some cancers. Enzymes in the CYP26 family are
  thought to be responsible for the elimination of RA, and CYP26A1
  appears to serve themost critical functions in this family. In spite
  of its importance, CYP26A1 has neither been heterologously expressed
  nor characterized kinetically. We expressed the rCYP26A1 in
  baculovirus-infected insect cells and purified the hexahistidine
  tagged protein to homogeneity. Heme incorporation was determined by
  carbon monoxide difference spectrum and a type 1 spectrum was observed
  with RA binding to CYP26A1.We found that RA is a tight binding ligand
  of CYP26A1 with low nM binding affinity. CYP26A1 oxidized RA
  efficiently (depletion Km
  9.4  3.3 nM and Vmax 11.3  4.3 pmoles min1 pmole P4501 ) when supplemented with P450 oxidoreductase and NADPH but was independent of
  cytochrome b5. 4-Hydroxy-RA (4-OH-RA) was the major metabolite
  produced by rCYP26A1 but two other primary products were also formed.
  4-OH-RA was further metabolized by CYP26A1 to more polar metabolites
  and this sequential metabolism of RA occurred in part without 4-OH-RA
  leaving the active site of CYP26A1. The high efficiency of CYP26A1 in
  eliminating both RA and its potentially active metabolites supports
  the major role of this enzyme in regulating RA clearance in vivo.
  These results provide a biochemical framework for CYP26A1 function and
  offer insight into the role of CYP26A1 as a drug target as well as in
  fetal development and cell cycle regulation.

$V_{max}=k_{cat}[E_0]$ where $[E_0]$ is the total enzyme concentration and not just the free enzyme. See my answer here.

You cannot use the value of $V_{max}$ as $k_{cat}$ because while the latter is the property of the enzyme (actually the enzyme-substrate pair), the former is the property of the chemical sample i.e. it is a system parameter.

Obviously, the $V_{max}$ will be same as $k_{cat}$ for a unit-concentration of total enzyme (unit concentration does not mean units of enzyme. It means, for whatever you are using as the units of concentration in your equation, lets say unit-x, the concentration of the enzyme should be 1 unit-x)


I need Iron powder. What's the best way to produce it? I have Ferric Chloride and Aluminum...

I also have steel filings from grinding steel objects, though some Iron Oxide have be separated (I do have Oxalic Acid and Citric Acid). Though that would give me steel powder, not iron.

Which is the most economical and would yield purest samples?

Sigma-Aldrich has a product that is very pure but rather expensive. I suspect you can have high purity or low cost but not both.


Does the reaction of 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one with NaOH involve ring contraction? What is the product and mechanism when 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one is reacted with
  $\ce{NaOH}$?

The only reagent used is $\ce{NaOH}$ so I suppose it abstracts the acidic hydrogen from the $\ce{-OH}$ group. A nucleophilic attack at the carbonyl carbon is also possible, as it could form a tropylium like cation. But I am not sure as to how a ring contraction occurs. The final answer given is 3-methyl-4-nitrobenzoic acid. How does the formation of an acid group and the ring contraction take place?

Drawing on the suggestion of @user55119 I propose the mechanism shown below.

Whether the first intermediate bicyclic [4.1.0] heptanone arises from an intramolecular electrocyclic process or an intramolecular aldol is unclear but either seems possible. Once the highly strained bicyclic [4.1.0] heptanone intermediate is formed, the reaction proceeds with OH- attack on the carbonyl in a process analogous to a Favorskii mechanism here, with the second intermediate anion stabilised by the nitro group which also directs the direction of ring opening. Finally loss of OH- in a manner similar to base-catalysed aldol and driven by re-aromatisation gives the observed product.


What is the interaction between dissolved ions of opposite valence in a solution at rest? NaCl is dissolved in water. Ions sodium and chloride are sufficiently free from each other so that they may occupy different regions of the solution after an active transport process through a membrane, or a mere electric field is in place.

They still act upon each other, and a membrane voltage is set by oppositely charged ions lining its interior and exterior surfaces.

So my question is this: how free are those ions in a uniform solution at equilibrium? Aren't they actually forming pairs of watered ions, although weakly bound? Can they form even bigger clusters, be them of short lifetime?

Note: I originally asked this on Physics SE but after several weeks I have received no answer nor any other kind of reaction, so I am trying Chemistry SE now.

The answer is: it depends. Dissolution of a salt implies that the entropy gained exceeds the cost of breaking lattice interactions (the solution enthalpy, assuming it is positive). Electrostatic interactions compete with kT (thermal jostling). Under physiological conditions, long range interactions are strongly screened by intervening solvent molecules and other ions. At best ions are subject to an effective potential due to distant ions, in particular between segregated charges as in the cases you present (e.g. on opposite sides of a membrane).

However, ionic solutions are generally regarded as "non-ideal". This means that interactions between the ions and with the solvent cannot be dismissed. At high salt concentration ion pairing will be encouraged. Above the solubility limit salt precipitation will occur, preceded by formation of clusters (seeds). If an ion has multiple charges, then electrostatic interactions will be stronger and pairing will be encouraged. Some metal ions form permanent ligand complexes that can attenuate the overall charge or distribute it over the complex, and the ligands can form a "cage" about the central ion, altering the interaction potential with solvent and other ions. The effect of temperature is complex. The entropic cost of association increases with T, but the dielectric constant of the solvent tends to decrease also.

If you are interested in reading on ion-pairing in NaCl solutions you may want to start with Ref. 1. It explains that association is negligible in dilute solutions:

  For dilute NaCl, we predict a coordination number of 5.3 with only waters in the solvation sphere for Na+ at 25 °C, 1 bar. With increasing NaCl molality, we find increased Cl complexation of Na. At 25 °C and 1 bar the Na+ ions in an 8 m solution of NaCl have 4.3 waters and 1.0 Cl ligands in their first coordination shell. There is some evidence of solvent-separated Na–Cl ion pairs from the peak in the Na–Cl pair distribution function at 5 Å. However, this disappears with increasing temperature.

There is more on formation of NaCl complexes and clusters in that article.

References

Sherman, DM Collings, MD. Geochem. Trans., 2002, 3(11), 102–107


Why do we use external pressure when calculating work done by gas? For irreversible expansion and compression, why is dw= -pextdV but not dw= -pdV ?

Many of the texts and related questions on this site I encountered give a same general idea: because work done= opposing force x displacement. Using this reasoning, I cannot explain my following thought processes:

(a) In the case of compression, the opposing force is obviously caused by p but not pext. Based on Atkins' Physical Chemistry Ninth Edition page 50, we use pext even for compression.

(b) The explanation given by Atkins' Physical Chemistry (same page) is:

  'when a gas is compressed, the ability of the surroundings to do work
  is diminished by an amount determined by the weight that is lowered,
  and it is this energy that is transferred into the system.'

Sounds right. Now if I use this reasoning on expansion,

Imagine I have two pistons, two filled with different amount of gases at same volume and so different pressure.
Envisage the external pressure being a load resting on the piston.
When both gases expand by the same amount, the work done by both gases, based on the formula dw= -pextdV and Atkins' reasoning, are the same. This is because the load moved up by same vertical distance and the changes in volume of gases are the same for two piston.
However, logically, that piston with higher pressure should move faster and so does the load on it. This would mean that, that load actually has higher kinetic energy.
With both gases doing the same amount of work, two loads gain the same gravitational potential energy, but one of the loads gains greater kinetic energy, means that one of the loads actually gain greater amount of total energy. How is this possible?

(c)
We always say when p=pext during expansion, the work done by the system is maximum, otherwise 'work wasted'. How can you say that it is wasted when the work actually used to increase the kinetic energy of the load as a whole?

The ideal gas law (or other equation of state) is valid only when a gas is in thermodynamic equilibrium.  In a rapid irreversible expansion or compression, the pressure depends not only on the volume (and temperature) but also on the rate of change of volume.  This is because a gas has a viscous property which acts to dissipate mechanical energy and convert it to internal energy. In a reversible process, where the rate of change of volume is negligible, the ideal gas law can be used.  But the ideal gas law (or any other equation of state for that matter), cannot be used to determine the pressure (or more precisely, the compressive stress) in an irreversible process.

At the interface between the gas and its surroundings (say the inside face of a piston) where the gas is doing work, by Newton's 3rd law, the pressure of the gas matches the "external pressure," and, if this external pressure is known, the amount of work that the gas does on its surroundings can still be determined. In the "vertical piston" example, if one does a force balance on the piston during the process, one obtains:
$$P_{gas,I}=\frac{Mg}{A}+P_{atm}+\frac{M}{A}\frac{d^2V}{dt^2}=P_{ext}$$ where $P_{gas,I}$ is that gas pressure at the interface with the piston (the subscript I is present because, in an irreversible process, the pressure of the gas within the cylinder may not even be uniform), and the mass times acceleration of the piston is included. The work done by the gas on its surroundings during the process is obtained by integrating this equation between the initial and final equilibrium states to yield$$W=\int{P_{gas,I}dV}=Mg\Delta h+P_{atm}\Delta V=\int{P_{ext}dV}$$This equation acknowledges that, as a result of the viscous property of the gas, the kinetic energy of the piston has been damped out by the time the system reaches its final state.


Is vapour pressure of a liquid solution constant at a given temperature, no matter the size of closed container and amount of liquid taken? Does vapour pressure of a liquid solution depend on the size of closed container, amount of solution taken, given that, temperature is kept constant? Or will it change, if some extra gas is added to the container at constant temperature?

The equilibrium vapor pressure of a compound at a given temperature is an intrinsic property of the compound, meaning it does not depend on variables like volume of the container, quantity of the compound, etc.  Other examples of intrinsic properties include density and melting point. None of these intrinsic properties depends on the quantity of the material or it's physical surroundings.


Why do single, double and triple bonds repel each other equal amounts? I'm here to share with you something that totally confuses me, as I can't see the logic behind it, and my teacher doesn't know either.

Let's take a set of bonds that's trigonal pyramidal, with a lone pair of electrons and three bonds. The angle between the bonds is 107 degrees.

Now, the part where I get confused is if, say, we have one double and two single bonds (I know this makes the atom have 10 electrons, but just ignore that. It's not the point).

Now, in my brain, it would be logical that the double bonds, containing four electrons, would repel more than a single bond, containing just two. However, according to what I've been taught, it doesn't. What is that?

The way you have learnt to predict molecular geometries is called the VSEPR theory. As all theories, it makes assumptions and has a certain number of limitations. The issue you are asking about is one such limitation.
When predicting the geometry of molecules with VSEPR, one should treat multiple bonds as a single entity. However, the geometry obtained is just a first approximation, or “idealized” geometry. In fact, multiple bonds induce distortions from this ideal geometry, pretty much like lone pairs do (I assume you know that already. If not, well, check the above-linked Wikipedia page for “lone pair”.) Citing from this online course:

Multiple bonds contain a higher electronic-charge density than do single bonds, so multiple bonds also represent larger electron domains ("fatter balloons"). Consider the structure of phosgene, Cl2CO, which is shown below.

From the Lewis structure of phosgene, we might expect a trigonal planar geometry with 120°-bond angles. However, the double bond seems to act much like a nonbonding pair of electrons, reducing the ClCCl bond angle from 120° to 111°. In general, electron domains for multiple bonds exert a greater repulsive force on adjacent electron domains than do single bonds.

So: yes, lone pairs and multiple bonds give rise to deviations from the ideal VSEPR geometries. It's nice that you figured it out on your own!


What side reactions are suppressed by the use of TMSCl in the acyloin condensation? I was reading about the acyloin condensation and came across a statemant according to which, the use of TMSCl obviates competing reactions. My question is what kind of competing reaction would happen if it was not used and how? I've read about the possiblity of some sort of Claisen condensation but I fail to see how would that be possible.
https://en.wikipedia.org/wiki/Acyloin_condensation

The acyloin condensation of diethyl adipate (1) on your Wikipedia citation generates ethoxide that can cause a Dieckmann condensation (intramolecular Claisen condensation, structure 6). TMSCl scavenges ethoxide and silylates the enediolate 3 formed in the acyloin condensation. The process is a 4-electron reduction (see details below). Mild hydrolysis of 4 gives the acyloin 5.


Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia? I know that soft bases cleave diborane symmetrically but hard ones don't. However, what makes ammonia harder than trimethylamine? And why does hardness/softness affect this?

Excerpts from my answer here.

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}B}$ nmr.

It can then be interpreted that hard bases like ammonia don't experience the steric effects which softer bases like trimethylamine do. Sterics inhibit the attachment of the second donor at the same boron atom.
Source: Boron Hydride Chemistry, Earl Muetterties


Explaining the oxidation states of palladium Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr]  4d^{10} 5s^0}$

The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$.
How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$?

Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment.

  Sodium nitrate, $\ce{NaNO3}$

When $\ce{NaNO3}$ dissolves in water, the two ions dissociate.

$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$

Both ions are solvated by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.

  Palladium nitrate, $\ce{Pd(NO3)2}$

Transition metal ions are different. They tend to form complex ions by coordinating solvent molecules. These dative interactions are far stronger than the ion-dipole interactions in solvated sodium ions.

Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the electron configuration of that complex ion.

$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$


Is it true that the better the trial function, the better approximation for the energy in variational principle? While studying the variational principle in McQuarrie's Quantum Chemistry, I came across the following problem: to relate the difference between an approximation $\phi$ and the exact ground-state wave function $\psi_0$ with the difference between the approximate energy $E=\int\phi^* \hat H \phi \, \text{d} \tau$ and the exact energy $E_0=\int\psi_0^* \hat H \psi_0 \, \text{d} \tau$. For instance, if I know that $\phi$ and $\psi_0$ differ by at most, say, $O(\epsilon)$ over all space, can I say that $E-E_0 \approx O(\epsilon)$?
It is intuitive to think that the better approximation $\phi$ is, the better the energy approximation gets, but I cannot prove this.
Without loss of generality, suppose $\phi$ and $\psi_0$ normalized. By writing $\phi$ as a linear combination of $\psi_n$, we get
$$
\phi=\sum_{n=0}^{\infty} c_n\psi_n
$$
such that $\sum_{n=0}^{\infty} c_n^* c_n=1$.
Therefore
$$
\hat H \phi=\sum_{n=0}^{\infty} c_n\hat H\psi_n = \sum_{n=0}^{\infty} c_nE_n\psi_n
$$
Hence $E=\int\phi^* \hat H \phi \, \text{d} \tau=\sum_{n=0}^{\infty} c_n^* c_nE_n$.
From here, the author shows that $E \geq E_0$, but how can I estimate the difference between the approximate energy and the exact energy?

TL;DR
No this is not true. Although a "better approximation" might yield a lower energy, this is not always the case.
Full answer
There is a major issue in the formulation of your problem: You want

to relate the difference between an approximation $\phi$ and the exact
ground-state wave function ψ0 with the difference between the
approximate energy $E=\int\phi^* H\phi \mathrm{d}\tau$ and the exact energy

The second part of that statement includes a clear mathematical definition of the target property (energy $E=\int\phi^* H\phi \mathrm{d}\tau$). However, there is no clear definition of what a better approximation of the wave function actually is. How do you want to define or measure it, so it can be expressed in a mathematical way?
There a various options to define such a measure, which would allow comparing two trial wave functions and judge their quality. The most obvious and commonly used one is the total energy. But this is the property you want to compare with, so we need something else. Examples for further options are:

overlap with the exact wave function (if available) $\langle \phi|\psi_0\rangle$
expectation values other than total energy, e.g. $\langle r\rangle$ or $\langle r^2\rangle$
any molecular properties of the chosen model system (e.g. dipole moment, excitation energies, etc.)

The next problem is: the choice of the measure for a good approximation might influence the result. For example the overlap $S_i=\langle \phi_i|\psi_0\rangle$ for two different trial wave functions $\phi_1$ and $\phi_2$ (possibly each with a different variational ansatz) might yield $S_1 > S_2$ (meaning $\phi_1$ would be the better approximation), while the total energy of the second trial wave function might be lower (and therefore close to the true ground state): $E_2 < E_1$.
The different measures on how good of an approximation a given trial wave function is do not correlate, therefore there is no general answer which one is actually better. It depends on the property you want to calculate.
There is a even paper available covering this issue and including a more detailed example. Sadly, it is paywalled although it is primarily intended for educational purposes: https://pubs.acs.org/doi/10.1021/acs.jchemed.8b00959

how can I estimate the difference between the approximate energy and
the exact energy?

In essence, this is the central question quantum chemistry is concerned about. And the field developed a whole lot of different methods to achieve reasonable approximations. Thus, I am afraid there is no simple answer to this question.


What is the difference between Dielectric and Capacitive Polarizable Continuum Model? I have good background knowledge on Quantum Chemistry calculations but my knowledge on Solvation Models is not that amazing.

Many times I read a paper that uses TD-DFT, they also assign some solvent to their calculations. Some papers use Dielectric-PCM and some others use Capacitive-PCM. I know that D-PCM assumes the solvent as Dielectric media and C-PCM assumes as Capacitive Field and I have some very basic knowledge on them but what I can not understand is when we use which?

For example what is the criteria for selecting DPCM or CPCM for Methanol solvent?

Short answer: It matters most that you are using a modern solvation model than which one

Slightly longer answer: If you have the time, try a few solvation models, compare with experiment and see which gives you the best results. In my experience, modern models do fairly well for TDDFT calculations. Be careful about the atomic radii used and whether the computational program generates a reasonable solvation surface.

Also keep in mind that not every program supports both D-PCM and C-PCM. Many packages just implement one solvation model.

There are a number of excellent reviews on polarizable continuum solvation models. As you're probably aware, they are a really efficient way to handle molecular solvation in calculations, since the details of the solvent are largely ignored. Instead, the solvent is treated as a dielectric environment, and you handle polarization between the molecule and the solvent self-consistently. That is, the solute molecule is treated at normal atomistic detail, creates a solvation shell (usually treated by a set of overlapping spheres) and the electrostatics of the molecule influences the solvent, and vice-versa.

Now, I've never seen "C-PCM" as a "capacitive" PCM. Usually people refer to the "C" as a conductor-style model. That is, in a C-PCM model, the dielectric constant is set to $\infty$ but then you re-scale the charges on the molecule later.

There are more than just these techniques. The main problem is that it's very hard to bench-mark solvation energies.

Some great recent reviews:

"Thirty years of continuum solvation chemistry: a review, and prospects for the near future." Theoretical Chemistry Accounts (2004) v. 112(4) pp. 184-203
"Quantum Mechanical Continuum Solvation Models." Chem. Rev., 2005, 105(8), pp 2999–3094
"Perspective on Foundations of Solvation Modeling: The Electrostatic Contribution to the Free Energy of Solvation" J. Chem. Theory Comput., 2008, 4 (6), pp 877–887
"A Universal Approach to Solvation Modeling" Acc. Chem. Res., 2008, 41 (6), pp 760–768
"The COSMO and COSMO-RS solvation models" WIREs Comput Mol Sci, (2011) 1: 699–709
"Polarizable continuum models." WIREs Comput Mol Sci (2012) 2: 386–404.


Does mercury chloride (ii) evaporate at room temperature? if you leave mercury chloride (ii) exposed to air in a room does it evaporate to any significant level which can cause toxicity?

According to this 1911 Journal of the American Chemical Society article (Johnson, F. M. G, JACS 1911, 33, 777), the vapor pressure of $\ce{HgCl2}$ is 3 mmHg at 152 degrees Celsius. Since the article is behind a paywall, I will reproduce the data below:

$$\begin{array}{|c|c|}
\hline
T/ \mathrm{^\circ C} & P/ \mathrm{mmHg} \\ \hline
152 & 3\\
195 & 20 \\
231 & 82 \\
256 & 198\\
262 & 237 \\
275 & 325\\
278 & 421 \\
283 & 481\\
302 & 754\\
\hline \end{array}$$

The authors did not report values for lower temperatures since their method focused on setting a pressure in the system and then measuring the temperature of sublimation/boiling point. However, using the data and some help from the Clausius-Clapeyron relationship, an estimate for the room temperature vapor pressure is $2.4\times 10^{-4}\ \mathrm{mmHg}$, which assuming you are at 1 atm, is 0.32 parts per million when the atmosphere is saturated. Consult a safety data sheet or other source for toxicity considerations at that exposure level.


Equations for UV degradation of condensation polymers I'm an A-level student studying the F324 unit of OCR GCE Chemistry 'A'.

The exact question is

  Explain how condensation polymers may be photodegradable. Equations needed!

I understand:

What a condensation polymer is.
Photodegradation may be possible by either a) Light-sensitive additives which catalyse a (slow) reaction in the presence of UV OR b) carbonyl bonds (C=O) breaking from the light energy.

What I don't understand:

What my teacher wants me to do for option 'a' in terms of equations. I'm inclined to leave that out, as it seems to be well outside the scope of the course.
Once the carbonyl bond breaks, what happens then? My best guess is that the oxygen atom receives the electron pair from the bond, then $H^{+}$ ions from 'somewhere' arrive to form water and finish the reaction. If I didn't know better, I'd say you needed a reducing agent for things to photodegrade.

So, what I'd ideally like is an equation (and explanation, if possible) showing what happens after the carbonyl bond breaks from UV.

Having spoken to my Chemistry teacher, the idea is that the UV causes the molecules to split; after the fission, the molecules may rejoin or react with 'other radicals' to break the polymer down, hence degrading it.

E.g. for a polyester:

$R_1COOR_2 \xrightarrow{hv} R_1CO\boldsymbol\cdot + R_2O\boldsymbol\cdot$


When dispensing liquid nitrogen, why does the surface get wet rather than frozen? When I go to dispense liquid nitrogen after someone else has just used it (say in the last 30 minutes or so), the copper pipe used to transfer the liquid is covered in frost/ice.

This makes sense, as the pipe gets incredibly cold, condenses water out of the air which freezes extremely quickly due to the vast difference in temperature (77K for liquid nitrogen vs. water freezing at around 273K).

What I find confusing is that once I start dispensing the nitrogen, this frost/ice appears to 'melt' into a liquid, this seems less intuitive to me. My first thought is that so much new water is condensing out of the air that the frost/ice gets 'wet', however I'd expect this to freeze very quickly and the ice to eventually get so thick that this issue of wetting became irrelevant. Alternatively, one could imagine condensing something else out of the air which is indeed a liquid at these temperatures such as oxygen, but in this case I don't observe the characteristic blue colour.

  An example of the type of dewar in question

Your "first thought" is basically right, with one clarification. The addition of "so much new water" to the existing ice results in the formation of liquid water due to the latent heat released upon condensation. This results in melting of the existing frost and thus the transitory existence of liquid water.

Also you can rule out oxygen condensation.  Firstly there is some insulation between the 77 K liquid nitrogen and the atmosphere, so the surface is not likely to cool all the way to the 90 K boiling point of oxygen.  More importantly, oxygen would not condense from the air at 77 K anyway.  You have to think in terms of saturation vapor pressures, and the vapor pressure of oxygen in our air is only about 0.21 atmospheres.

By analogy, albeit an extreme one, the temperature in the coldest regions of our atmosphere drops below the sublimation point of carbon dioxide.  Does that cause the formation of dry-ice clouds?  Not even close. The sublimation point of carbon dioxide in this example, like the boiling point of oxygen in your example, is only relevant at 1 atmosphere of pressure of that compound, and in neither case does our atmosphere contain 1 atmosphere of either of those compounds.


How is octane rating defined for negative values and those over 100? Based on my reading, the octane rating for a given fuel is calculated by finding the compression ratio at which it knocks and finding a mixture of $X\%$ isooctane and $(100 - X)\%$ n-heptane which knocks at the same compression ratio. The $X\%$ isooctane is then that fuel's octane rating.

But how does this work with negative numbers or those over 100?

I recommend the @Stanley Yu own answer, as I admit I was not particular familiar with the shape of the ON number curve.

It is done by an extrapolation of the critical compression ratios of mixtures 0-100% of isooctane and 100-0% of n-heptane for ON values <0 and >100.

If, just illustratively, 100% isooctane had the critical compression ratio 1:20, 90% isooctane 1:19 ,... 90% n-heptane 1:11, 100% n-heptane 1:10, then the fuel with the critical compression ratio 1:21 would have the octane number 110.

The octane number/rating differs for different procedures. See Research octane number RON, Motor octane number MON,  Anti-knock index AKI=(RON + MON )/2 and Observed Road octane number RdON.

As seen in the OP own answer, the extrapolation of the octane number (ON) scale  above 100 is done by linear  extrapolation by the empirical aviation formula for performance number $$PN=100 + 3 * ( ON - 100 )
where on the graph is coordinate ON substituted by tetraethyllead addition.


Why do elements change their colour in reactions? We are almost common with the reaction:
$\mathrm {CuSO_4+Fe = FeSO_4+Cu }$. In this reaction initially Copper sulphate is bluish in colour, which gradually changes to pale green. I want to know, why these elements have so and so specific colour, OR what determines the colour of these elements?  Further, why there is such specific change of colour as in copper sulphate observed?

Elements don't have inherent colors at all; substances do. Furthermore, while there are situations when some compounds (and even copper sulphate) change their colors, this is most definitely not one of them. Here copper sulphate is blue and never changes to pale green. It's just the reaction... wait, didn't you write the equation yourself? See, in the end there is no more copper sulphate. It is gone, and so is its color. What you see now is the color of $\ce{FeSO_4}$, or more precisely, of $\ce{Fe}^{2+} (aq)$.

As to why compounds have certain colors... well, this is a really huge body of knowledge. Long story short, color is caused by absorption of light with some specific wavelength. What wavelength is it, is determined by possible electron transitions within the substance in question. The transition may be between some molecular orbitals, if the compound is molecular. Or it may be between the orbitals of some metal ion, as is the case in your example. In many cases you may be able to predict the color of a compound you have never seen, using simple techniques.

Come to think of it, this question must have been asked before, and surely there is already a nice answer with lots of pics and stuff.


Is an SN1 reaction possible in halobenzene? Is an SN1 reaction possible in halobenzene?

Will the phenyl carbocation be stabilized via resonance? Suppose the leaving group is very electronegative and the environment is protic. What then would be a suitable nucleophile?

Yes, that reaction occurs, but generation of the phenyl carbocation was much more difficult than anyone might have guessed. Here is a drawing of the phenyl carbocation.
First note that the 6 p orbitals making up the aromatic pi system are all still intact and overlapping - the aromatic nature of the benzene ring has not been tampered with.  Then notice that the cationic center exists as an empty $\ce{sp^2}$ orbital where the

$\ce{C-H}$ bond used to be. The empty (cationic) $\ce{sp^2}$ orbital is orthogonal to the aromatic pi system; as a consequence there is no resonance interaction or stabilization of any sort between the two systems.  Further, being constrained by the 6-membered ring, the cationic $\ce{sp^2}$ orbital cannot rehybridize.  This results in considerable (low energy) s-character being used to stabilize an empty orbital, and is not available to stabilize other orbitals that do contain electrons! The vinylic carbocation, the ethylenic analogue of the phenyl cation, is much more stable and common (it can even be observed by nmr) because it is not constrained by a ring and can rehybridrize, thereby lowering its energy. It is both this total lack of stabilization and relatively high (higher than need be if rehybridization could occur) energy of the molecule that has made the phenyl carbocation such an elusive target.

To stack the deck in their favor, investigators have run experiments using extremely good leaving groups such as nitrogen and the triflate anion. Magic acid and its variations, a very powerful acid that stabilizes carbocations, was often employed as well.  The nitrogen leaving group route, using thermal decomposition of aryl diazonium cations has been one of the few methods that has allowed access to this transient intermediate.

A few references if anyone would like to read more:

On Attempts at Solvolytic Generation of Aryl Cations - 1976
Attempts at Solvolytic Generation of Phenyl Cations - 2004
A one paragraph review - 2000


Making sulfate from elemental sulfur I have $\approx\pu{2 g}$ of elemental sulfur that I need to convert to dissolved sulfate in DI water. What's the easiest lab chemistry that will get me there?

You burn the sulfur and lead the fumes through a solution of $\ce{H2O2}$. Afterwards you boil that solution for a moment to remove the rest of the peroxide.

Use a vertical tube reactor or something so you can lead a soft stream of oxygen gas (or pressurised air, in a pinch) over the sulphur.


Are there still "prairie dog refineries"? In the old days (meaning the 1940s) there still many small "prairie dog" oil refineries, that were little more than a bunch of hooch stills soldered together, but from what I understand these are all gone now and there are only a handful of oil refineries in the United States.

Are there still small refineries in the United States or have they been regulated out of existence?

Any collection of 'hooch stills' that were soldered (?) together would probably have been uneconomical almost as soon as they were built, because welding together a series of serviceable fractional distillation towers is not that technologically advanced and much more efficient.  However, there are certainly a number of relatively small refineries.

For instance, about the largest refinery in the U.S. (for comparison) is Exxon's in Baytown, TX that has a capacity of 560,000 barrels per day.

A couple of the smallest ones are the 5000 bbl/day refinery in Wyoming (Silver Eagle, in Evanston) and the 2000 bbl/day Foreland Refinery in Ely, NV.  The latter (if run all day) is equivalent to around 1.5 barrels a minute.

Also (to respond to a comment) there are still some refineries that don't do catalytic cracking (probably just straight distillation of the petroleum into fractions, which is viable for small refiners because they can save on expensive (and often proprietary) cracking units and are small enough that they can sell all the fractions that they produce.  The recently closed Flint Hill refinery in Alaska used another approach, reinjecting unused fractions back into the TransAlaska pipeline.)

For a list of refineries, the EIA is always a good place to look (pdf warning).

A wiki on refineries is "A Barrel Full" which has detailed information on individual refineries.


Why does buffering capacity increase with the pKₐ of the acid component? For example, the $\mathrm{p}K_\mathrm{a}$ values of ethanoic, propanoic, and butanoic acid are related in the following way:
$$\text{ethanoic} < \text{butanoic} < \text{propanoic}$$

And their buffering capacities are related in exactly the same way. What causes this?

Buffer capacity depends on the $K_\mathrm{a}$ of an acid. I will use the explanation that can be found in David Sheehan: Physical biochemistry: principles and applications. John Wiley & Sons, 2013.

Any aqueous solution containing both $\ce{A-}$ and $\ce{AH}$ is, in principle, capable of resisting change in $\mathrm{pH}$ according to
$$\ce {A^- + H^+ <=>> AH}.$$
If alkali is generated in the solution (that would tend to remove protons) we have
$$\ce{AH + OH- <=> A- + H_2O}.$$

Most buffers consist of mixtures either of a weak acid and its salt or of a weak base and its salt. The ability to resist change in $\mathrm{pH}$ is finite, especially if the number of protons involve is especially large. This limit is represented by the buffering capacity of the buffer $\beta$.
This is defined as the amount of substance in moles of $\ce{[H^+]}$ which must be added to a liter of the buffer to decrease the $\mathrm{pH}$ by one unit. It can be mathematically calculated by
$$\beta = \frac{2.3 K_\mathrm{a}[\ce{H+}][C]}{(K_\mathrm{a}+[\ce{H+}])^2},$$
where $[C]$ is the sum of conentrations of $\ce{A-}$ amd $\ce{AH}$. This relationship means that buffering capacity increases with the buffer concentration. Buffers work best at $\mathrm{pH}$ values around their $\mathrm{p}K_\mathrm{a}$  most pH are effective one $\mathrm{pH}$ unit above an one below their $\mathrm{p}K_\mathrm{a}$.
I hope that the formula previously reported help in clarifying how the buffer capacity is related to $\mathrm{p}K_\mathrm{a}$ (since they are indeed related).

Example (based on ZUMDAHL, Steven. World of chemistry. Cengage Learning, 2012.)

A chemist needs a solution buffered at $\mathrm{pH}~4.30$. choosing from ethanoic,  propanoic, butanoic acid. We can calculate the ratio  $\ce{[HA]/[A^-]}$ required for each system. Considering that a $\mathrm{pH}~4.30$ corresponds to $\ce 5\times 10^{-5} M$ using the equation
$$[\ce{H+}]=K_a \frac{[\ce{HA}]}{[\ce{A^-}]}$$
we substitute the required $[\ce{H+}]$ and $\ce{K_a}$ for each acid to calculate the ration $[\ce{H+}]/[\ce{A-}]$ needed in each case is:

\begin{array}{llr}
\text{Acid} &
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &
5.0 \times 10^{-5} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &
2.88 \\[1ex]
\text{Propanoic} &
5.0 \times 10^{-5} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
3.749\\[1ex]
\text{Butanoic} &
5.0 \times 10^{-5} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
3.98  \\ \hline
\end{array}

In this case the choice would be ethanoic acid since it has the ratio closes to 1.

If we need a buffer near $\mathrm{pH}~5.5$ the best choice is butanoic acid:

\begin{array}{llr}
\text{Acid} &
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &
3.2 \times 10^{-6} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &
0.18 \\[1ex]
\text{Propanoic} &
3.2 \times 10^{-6} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
0.23 \\[1ex]
\text{Butanoic} &
3.2 \times 10^{-6} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
0.25 \\ \hline
\end{array}


Confused by Alternate Names I am highly allergic to propylene glycol and polyethylene glycol.  I am finding many hidden sources of both and MANY synonyms for PG and PEG.  Is there a practical way to tell how closely related another chemical might be to my allergens using such resources as the U.S. Household Products data base?  For example:  glycerin's alternate name is propandiol.  Propandiol is also an alternate name for propylene glycol.  I'm very confused about sodium laureth sulfate in particular.  It has PEG listed as an alternate name, but is not an alcohol like the glycols.  Thank you for any clarification you can give me!

You'll probably be better served asking an allergist about this.

While many things contain PG, that is not the same as a compound similar to PG, which could have PG (or part of its IUPAC name, propane-1,2-diol, not to be confused with propane-1,3-diol, which would be its possibly harmless-to-you although structurally similar isomer) as part of its name or could interact with your body in a similar way. Case in point, your confusion over sodium laureth sulfate, which is nothing like PG, although that's an irritant for everybody. However, it looks like PG allergy is sometimes co-sensitized with corticosteroids - which are themselves structurally nothing like PG!

Thus, the confounding variable here is the way your body decides what will or will not set off your immune system, which can have very little to do with the chemical behavior of the allergen and everything to do with what mechanisms your body is using to make that distinction.


For chemical reactions occurring in the interstellar medium, which involve metal-ions, will the metal ions be solvated or not? Usually, metal-ions in terrestrial chemistry will be solvated - either by gas-phase solvent molecules or by liquid-phase solvent molecules. But the interstellar medium (ISM) is never in local thermodynamic equilibrium. Moreover, the solvation of any system is driven by the system's proclivity to attain thermodynamic stability. But given the lack of local thermodynamic equilibrium in the ISM, solvation should not at all be an important factor in the ISM. Is it right or wrong? Any reference from the Astrochemistry/Astrophysics community will also be highly appreciated.

Per Wikipedia, the upper bound on the particle density in the interstellar medium is approximately $\pu{E6}$ molecules per cubic centimeter, in "cool, dense regions."

In the context of terrestrial aqueous solvation, where the concentration of water is approximately $\pu{55M}$, the particle density is on the order of $\pu{3.3E22}$.

So, even if every single particle in the vicinity of a metal ion in the ISM were a water molecule, the ISM would be roughly $\pu{E18}$-fold more dilute in neighboring water molecules as compared to aqueous solution.

No, metal ions in the interstellar medium will never be solvated.

Note that this does not imply that it is impossible for aquo-complexes of transition metal ions to occur in the ISM, given their significant affinity for ligands in the first coordination shell and for the appreciable organizing influence (thanks, MaxW!) such ions have on more distant coordination shells. It's just highly improbable that one would find a metal ion associated with a sufficiently statistically large number of water molecules to be considered solvated.

Further, to address the portion of the question about thermodynamic equilibrium: I don't think that's a particularly significant factor, no. On Earth, it's frequently the case that ions are solvated but not in thermodynamic equilibrium: the proton gradient that is sustained in the aqueous milieu of biological organisms to drive ATP synthase is one of the most significant examples that comes to mind. Thus, an absence of thermodynamic equilibrium does not necessarily entail a lack of solvation.


Why use Km in catalytic efficiency? The catalytic efficiency of an enzyme is given by $k_{cat}/k_M$ where $k_{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme.
$K_{M}$ is a measure of the affinity of the enzyme with the substrate, or the likelihood of binding.
Why bother dividing the $k_{cat}$ by $K_{M}$? Isn't the affinity of the enzyme already encoded into the quantity of $k_{cat}$? How could you be an enzyme that has low affinity, but still have a huge turnover? To me this doesn't seem possible, and thus it is redundant to divide by $K_{M}$. Likewise, could there be a situation where $k_{cat}$ is low, but $K_{M}$ is high?

The most basic kinetic scheme for enzymes is represented as
$$\ce{E + S <=>[K_m] ES ->[k_{cat}] E + P}$$
As should be clear, the $k_{cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The resulting rate (kcat[E]tot) is only achieved when every molecule of enzyme essentially always is in the act of converting substrate to product. That is, every time a product molecule is released, a substrate molecule immediately binds. Even if the substrate molecule dissociates before reacting, another immediately takes its place. We describe this situation as the enzyme being "saturated" with substrate.
The Km is a measure of how tightly the substrate binds to the enzyme, approximately equal to the equilibrium constant for the dissociation of the substrate from the enzyme. If the substrate is at a low concentration relative to this Km value, then many of the enzyme molecules will not have substrate molecules bound to them and will be unproductive as a result. The overall rate will be substantially below the maximum  kcat[E]tot.
All of this is captured in the basic Michaelis-Menten kinetics equation:
$$Rate = \frac{k_{cat}[E]_{tot}[S]}{K_m + [S]}$$
You can see that the impact of $K_m$ on the rate increases substantially as [S] decreases relative to $K_m$.
Since most substrates exist physiologically at concentrations below what is required for maximum rate, it is the combination of $k_{cat}$ and $K_m$ relative to [S] that determines the in vivo rate of reaction (along with the amount of enzyme of course).


Why can't we use 30 mL of water and 70 mL of alcohol to make a 70% (v/v) solution of Isopropyl Alcohol? Volume percentage (V/V): Solutions that are formulated from two liquids are often expressed as volume percent
with respect to the solute. The volume percent is the volume of a liquid in 100 mL of solution. The label on a bottle
of ordinary rubbing alcohol reads “isopropyl alcohol, 70% by volume”. Such a solution could be made by mixing
70 mL of alcohol with water to make a total volume of 100 mL, but we cannot use 30 mL of water, because the
two volumes are not necessarily additive:

What exactly does "not neccessarily additive" mean?

Assuming you are talking about either isopropyl alcohol or ethyl alcohol (ethanol) and not both. They are different beasts, they are both used as disinfectants and solvents, but isopropyl is way more toxic to humans (who more or less regularily drink etanol solutions).
70ml ethanol and 30ml water, whem mixed, do not make 100ml mixture, but somewhat less.
That's what "non additive" volume means.
I don't have the numbers at hand, but the "contraction" is pretty much measurable and in particular important in the mass production of desinfectants and drinks.


Should I have a pair of shoes that I wear only in the lab? I bought some very nice heavy leather boots when I was working in a solid state synthesis lab, mainly with selenium and sulfur compounds, a few years back. I probably did spill small quantities of some very basic/acidic solutions with some selenium/selenate content on them a few times.

I don't work in a synthetic lab anymore, but I wore them again this weekend. It feels strange though, perhaps just from ingrained lab safety habits, to touch my face after putting them on etc. My gut feeling is that a) rain has washed away anything which was on them b) most chemists I knew as a research student wore their normal day shoes in the lab, so it should be fine. But what are the standard safety rules on this? Is it generally okay to wear the same shoes in the lab and in daily life?

Standard safety rules (from a pharma and semiconductor processing POV):

To follow industrial hygiene practice absolutely to the letter, you should have dedicated outer layer garments that remain in a quarantined area that would not pose a gross cross contamination risk to other things that cannot be controlled for exposure risk. This means that you should not have been able to take those boots home with you after you wore them in the lab even one time. I do not think any company enforces such a rule, in fact, it is the employer's responsibility to provide PPE to workers and the worker's responsibility to use and maintain PPE (if applicable.) If the risk of cross contamination was assessed and determined to be moderate or higher, you should have been provided shoe covers at your solid state systems lab or dedicated work shoes that had to be kept in a quarantined area.

At the lab bench level, it is my experience that people wear their daily shoes unless you work at a larger site that has a safety shoe/shoe covering policy.


Which of these solution have the lowest vapor pressure in 298 K? Which of these solution have the lowest vapor pressure in 298 K? Please give a brief explanation.

0.01 m $\ce {CaCl2}$
0.01 m $\ce {NaCl}$
0.02 m $\ce {NaCl}$
0.015 m $\ce {CaCl2}$
0.01 m $\ce {Na2SO4}$

(This problem is taken from an Indonesian Chemistry Olympiad)

All examples are mixtures of not-volatile (here: salts, but it would be the same for a teaspoon of sugar) and volatile compounds.

The vapour pressure above these mixtures is proportional to the mole fraction of the volatile component. With other words:

The higher the concentation of the salt, the lower the vapour pressure.

But as user80551 pointed out in his comment, we should not look at it this way. Water isn't "just the solvent" here.

It is the volatile compound in the mixture that matters! The higher the mole fraction of the volatile compound, the higher the vapour pressure!

With (again) other words: Raoult's Law ;)


When using Gaussian to calculate NMR, what's the default solvent and frequency? If I choose GIAO HF/STO-3G, is the default solvent $\ce{CD3Cl}$? Is the default frequency for $\ce{^1H-NMR}$ 500 MHz? What about the frequency for $\ce{^13C-NMR}$?

While Greg's answer describes sufficiently what the program does, I would like to extend a little on his remark. (This should have been a comment, but it was too long.)
The HF/STO-3G level of theory is what you call the minimised ab initio approach. It is unreliable in many ways, but it is fast. However, even geometries of well known and/or calculated molecules may be wrong. In short, it fails more often than it is correct. The problems start with HF, since it does not describe correlation sufficiently. (It only describes Pauli correlation, but this is exact.) The next problem is the basis set: It substantially lacks basis functions. It is called the minimal basis for a reason. Using this setup for NMR calculations is worse than guessing.
The first thing to do would be to increase the basis set. Maybe move away from Pople basis all together. A few basis set families come to mind: Dunning's augmented family, aug-cc-pVDZ and larger, Ahlrichs and Weigend's def2 family, def2-SVP and larger, and many others, see EMSL.
Calculating NMR tensors is always an accuracy versus time tradeoff. And it might be complicated to find the best (in the sense of most efficient) setup. The best way is to start a geometry optimisation on quite a low level. For reasonably large molecules a semi-empirical model like pm6 can be used to scan for configurational space.
Increase the level of theory to pure DFT. You can use density fitting here to speed up your calculation: e.g. #P BP86/Def2SVP/W06 DenFit.
Validate your geometries with a higher basis set, e.g. #P BP86/Def2TZVPP/W06 DenFit and different functionals, e.g. TPSS, PBE1PBE, M06, etc.
Validate your geometries with the dispersion correction of Grimme, e.g. BP86 EmpiricalDispersion=GD3BJ, PBE1PBE EmpiricalDispersion=GD3BJ, B2PLYPD3, etc.
Apply solvent corrections for your system, e.g. SCRF=(Solvent=Chloroform). Make sure to run single point calculations first and try optimisations later.
If you have fully converged geometries at DFT level, you might want to check basic properties with ab initio techniques. Most appropriate would be perturbation theory, e.g. MP2.
After you have done all the above you can go ahead and calculate NMR tensors. It may be combined with solvent approximations. It requires a very well converged structure, a local minimum on the potential energy hyper surface. (Check that with a frequency calculation, e.g. FREQ.) You can set it up like this sample input:
%oldchk=filenameOfPreviousOptimization.chk
%chk=newCheckpoitfile.chk
#P MP2/def2TZVPP
SCRF(solvent=chloroform)
NMR(GIAO)
geom=AllCheck guess=read

Be advised that that computing NMR data with only one method can lead to many misinterpretations. It is very important to verify your structures and tensor on a variety of methods. If you cannot afford MP2 with a large basis, then you should at least check 4 or 5 DFT methods. It is also important to know, that including solvent effects is still something where the theory is not exact. If you have a system that changes a lot with the solvent, then you should consider explicit treatment. Then you might want to dive into molecular modelling, molecular dynamics, MM/QM, ONIOM, etc. pp.


How do I isolate calcium from calcium carbonate? I have the ability to use many kinds of reactions, and the use of most common acids/bases.

The answer is you cannot do it almost for sure.

To obtain $\ce{Ca}$ metal from $\ce{Ca (+II)}$, you need to perform a reduction process. For this purpose, you would need a very strong reducing agent, because calcium is very close to the bottom in the Ellingham diagram (used in metallurgy for this same purpose: obtain metals from its salts).

So you will need to perform an electrolysis of molten calcium chloride, which is not cheap, and by any reason not doable at home or at an standard lab (just a quick view of calcium salts melting point will give you a hint about why is this...)


Heating paper to a yellowish smoke I put some papers into a biscuit tin and put it over fire and baked it. Then I made a small hole at the top and saw yellowish smoke coming out. Any idea what the smoke could be?

What you've performed is, technically, "thermal decomposition of cellulose", or pyrolysis. During heating, there are large number of changes producing many different chemicals, so the smoke you saw was a mixture of different chemicals.

Pyrolysis is used to produce many industrial chemical, such as "wood alcohol" (methanol), and during gasoline shortages, to run internal combustion engines.


Why pyranosides do not mutarotate? For mutarotation, at least one hemiacetal group must be present in the
  sugar. Therefore, non-reducing sugars do not show mutarotation.

Why?

Consider the following reaction:

$$\ce{\beta-D{-glucose} + CH3OH + HCl -> \alpha-D{-glucopyranoside} + \beta-{D-glucopyranoside}}$$

Even though the pyranoside has an acetal group, the reaction is still reversible? Therefore, shouldn't pyranoside show mutarotation? Isn't the mechanism for the reversible reaction for acetal same as the one for hemiacetal?

In principle, the acetalisation reaction is indeed reversible. Typical conditions to reverse it would be to use $\ce{TsOH}$ or $\ce{PPTS}$ and water.

However, both full acetalisation and liberation of the aldehyde require acidic conditions. Hemiacetalisation — the prime mechanism of mutarotation — proceeds without acidic catalysis, i.e. the inherent (low) acidity of water is sufficient. And to observe mutarotation, you typically dissolve the pure sample in a pure solvent (with no acid added).

If you dissolve e.g. methyl α-ᴅ-glucopyranoside in water and add catalytic amounts of acetic acid, you will observe mutarotation but not if you simply dissolve the O,O-acetal in water.


Why would CuSO4-EDTA drops be used in Pb titrations? We use CuSO4-EDTA drops (4) when preparing our lead titration samples for final titration by EDTA.  What mechanism would there be for adding this to a titration?  What benefit does it produce on Pb titration?  We digest galena with nitric and sulfuric, clean up samples with a 5% sulfate wash, and remaining sample is washed into original beaker with 50% ammonium acetate solution (ammonium acetate/acetic acid buffer).  Samples are boiled then cooled before final steps.  The samples have ~1.5g ascorbic acid, 4 copper-EDTA drops, and 4 xylenol orange drops added to each sample prior to titration with EDTA.

I'm not an expert on coordination chemistry, and in particularly, metal-indicator complexes in EDTA titrations for heavy metals. But, I assume, the addition of $\ce{[Cu(EDTA)]^2-}$ solution to your unknown $\ce{Pb}$ sample is to get a sharp end-point (example of Replacement or Substitution Titration). Since, the extra metal ion added ($\ce{Cu^2+}$) has already complexed with EDTA, it won't interfere with the calculations. Meantime, both $K_{\mathrm{f}}$s of $\ce{[Cu(EDTA)]^2-}$ and $\ce{[Pb(EDTA)]^2-}$ are so closed ($5\times 10^{18}$ and $2\times 10^{18}$, respectively) that following equilibrium exists:
$$\ce{[Cu(EDTA)]^2- + Pb^2+ <=> [Pb(EDTA)]^2- + Cu^2+}$$
Where, $$K_{\mathrm{eq}} = \frac {[\ce{(Pb(EDTA))^2-}][\ce{Cu^2+]}}{[\ce{(Cu(EDTA))^2-}][\ce{Pb^2+]}} = \frac{K_{\mathrm{f}(\ce{(Pb(EDTA))^2-})}}{K_{\mathrm{f}(\ce{(Cu(EDTA))^2-})}} = \frac{2}{5} = 0.4$$

Thus, released $\ce{Cu^2+}$ cpmplexed with Xylenol Orange to give stable red color at operating pH (e.g., Figure 1A, Ref.2). My assumption is based on EDTA titration of $\ce{Pb^2+}$ and $\ce{Bi^3+}$  ion mixtures in the presence of Xylenol Orange as an indicator. In that titration, pH of a test sample solution containing $\ce{Pb^2+}$ and $\ce{Bi^3+}$ is adjusted to $3$  and add the indicator (Xylenol Orange). The test solution becomes red due to the complex formation of $\ce{Bi^3+}$ and Xylenol Orange (Figure 1B). When titrate with EDTA, at the first end point, the solution becomes yellow, the color of the free indicator at that pH. To titrate for $\ce{Pb^2+}$ at the same solution, it is needed to adjust pH to 5 - 5.5 until the appearance of red color again (indicating the complexation of $\ce{Pb^2+}$ with Xylenol Orange). Now, the slution can be titrated with EDTA until the turning of red to yellow. It is important to note that solution of $\ce{Pb^2+}$ ions does not become red with Xylenol Orange until pH is above 5. That means, the [$\ce{Pb^2+}$ + Xylenol Orange] complex is unstable below pH 5.

Also note that use of ammonium acetate ($\ce{NH4OAc}$) as a neutral buffer is misleading. $\ce{NH4+}$  and $\ce{AcO-}$  are not a conjugate acid/base pair, which means that they do not constitute a buffer at pH 7. However, dissolution of ammonium acetate salt in water results in pH 7, but this pH is highly labile (Ref.1). Ammonium acetate does provide buffering around pH 4.75 (the $\mathrm{p}K_{\mathrm{a}}$ of acetic acid) and around pH 9.25 (the $\mathrm{p}K_{\mathrm{a}}$ of ammonium). Thus, advantage for using $\ce{NH4OAc}$ is a drop from pH 7 to around pH 4.75, which is considerably less dramatic (e.g., addition of $\pu{1 mM} \: \ce{H+}$ to a $\pu{10 mM}$ neutral pH solution will decrease the pH to 5.8) than doing the same to pure water. Remember, during the titration, with out "real" buffer, pH of the solution changed due to release of $\ce{H+}$ ions by complexation.

References:

L. Konermann, “Addressing a Common Misconception: Ammonium Acetate as Neutral pH “Buffer” for Native Electrospray Mass Spectrometry,” J. Am. Soc. Mass Spectrom. 2017, 28(9), 1827–1835 (https://doi.org/10.1007/s13361-017-1739-3).
Interesting reading: O. J. V. Belleza, A. J. L. Villaraza, “Ion charge density governs selectivity in the formation of metal–Xylenol Orange (M–XO) complexes,” Inorganic Chemistry Communications 2014, 47(1), 87–92 (https://doi.org/10.1016/j.inoche.2014.07.024).
Relevant: S. Murakami, “Semi-Xylenol Orange Complexes with Bivalent Metal Ions,” J. inorg, Nucl, Chem. 1981, 43(2), 335–343 (https://doi.org/10.1016/0022-1902(81)90020-8).


Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones In the reactions above, taken from Tetrahedron Lett. 1982, 23 (23), 2355–2358, the choice of protecting group on the α-hydroxyl group has a large influence on the diastereoselectivity of reduction.

Why does the benzyloxymethyl protecting group lead to the formation of the anti diol (consistent wiht the Cram chelate model), whereas the tert-butyldiphenylsilyl protecting group gives the syn diol (consistent with the Felkin–Anh model)?

I think Jan did an excellent job of answering your question, however I am writing a slightly longer answer largely with more details concerning the Felkin–Anh rule, just so that someone who is not already familiar with them can learn something too.

Assuming that there isn’t an unusually electronegative atom on the carbon next to the carbonyl, the largest group prefers a conformation where it is perpendicular to the carbonyl moiety. This gives two relatively competitive lowest energy conformations, with the medium and smallest groups differing in their proximity to the carbonyl oxygen.

For the TBDPS-protected substrate shown above, the OTBDPS group would be considered the largest group. The medium and small groups are methyl and hydrogen respectively. This placement is also consistent with a modification to the Felkin–Anh model, which places electronegative substituents (such as –NR2 or –OR) perpendicular to the carbonyl (an explanation is given at the bottom of the answer). In any case, the two conformations of interest are the two shown below:

Moreover, the nucleophile doesn't prefer to attack the carbonyl carbon at a 90° angle; rather it prefers something closer to the tetrahedral angle (Bürgi–Dunitz trajectory). Naturally, the nucleophile prefers attack from the opposite side to the large group. This leaves two possibilities: for the nucleophile to attack over the small group, or for it to attack nearest to the medium group. The former is preferred, and the major diastereomer will be the syn diol:

On the other hand, Lewis basic groups such as $\ce{-OCH2OBn}$ are also capable of forming a chelate with the metal together with the carbonyl oxygen. This feature forces the $\ce{OR}$ group to be held parallel to the carbonyl. Nucleophilic attack still occurs along the Bürgi–Dunitz trajectory, and the nucleophile attacks opposite the larger of the two remaining groups. This will invert the stereochemical outcome as compared to the Felkin–Anh model previously described, and hence leads to the formation of the anti diol.

Now, to answer your question of why does the choice of protecting group matter. The previous discussion illustrates that the reason we get the Cram Chelate product is the chelation of the metal by the $\ce{OR}$ oxygen and the carbonyl. This chelation can take place easily if the $\ce{R}$ group is small, but is greatly hindered with a bulky alkyl/silyl substituent, in which case we get the product determined by the polar Felkin–Anh relationship.

The bulk of the TBDPS group does indeed prevent chelation, and you get the observed stereochemistry in accord with the Felkin–Anh rule. However, chelation is possible when BOM is the protecting group (Jan already offered an excellent explanation), and you end up with the Cram chelate product.

When there is an electronegative atom or group α to the carbonyl, there is a slight change to the Felkin–Anh model which may be applied. This ‘slight change’ is that this electronegative group becomes the equivalent of the ‘large’ group in the Felkin–Anh model.

The reason for this that the energy of the C–X σ* antibonding orbital is rather low, and so when the C–X bond is perpendicular to the carbonyl group, this can overlap with the π* MO of the carbonyl group to make a new, lower energy LUMO.

In "human" speak, this means that this conformational arrangement is more
reactive than any other. With this change in mind, the Felkin–Anh type addition now proceeds as before. However, the stereochemical outcome will generally be different from the original Felkin–Anh model.

For more information, see: How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?


How does the HCl-KCl Buffer work? I have just been studying the $\ce{HCl}$-$\ce{KCl}$ 'buffer', but there are still quite a few things I am uncertain about. I would appreciate any help in clearing up some questions I have.

What I understand (possibly incorrectly) from my research so far:

Water is the weak acid acting as a buffer in this system. The $\ce{HCl}$ and $\ce{KCl}$ in the system are there to increase the solution's ionic strength, which somehow improves water's buffer capacity.
In high ionic strength solutions, the 'standard' equilibrium equation has to be changed to include the activity coefficients. For water's equilibrium, as the ionic strength increases, the activity constants decrease and so the K value decreases.

If anything above is incorrect, please point out my misunderstanding. There are still a few questions I can't find answers to:

Why does the increased ionic strength of the solution improve water's buffer capacity?
Isn't a higher equilibrium constant required to make water a better buffer? Then isn't it bad that the ionic strength decreases K?
Could any other salt have been used to increase the ionic strength, or is there some specific reason $\ce{HCl}$-$\ce{KCl}$ is used?

I'm finding this 'buffer' system to be extremely confusing, so any help answering these questions would be greatly appreciated.

Introduction

Let's define buffer capacity quantitatively as

$$\beta=\cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}=-\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})}$$

that is, the relationship between concentration (in equivalents) of strong base ($c_\mathrm{b}$) or acid ($c_\mathrm{a}$) added to a solution and its change in $\mathrm{pH}$. From now on I'll assume we're adding a monoprotic base or acid, so "equivalents" and "moles" (and their concentrations) can be used interchangeably.

Pure water

Pure water has a very low buffering capacity - its $\mathrm{pH}$ is very sensitive to the addition of acids or bases. For instance, imagine we add a concentration $c_\mathrm{b}$ of base to pure water:

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

The self-ionisation equilibrium of water will be displaced:

$\ce{[OH-]} \approx c_\mathrm{b}$

and therefore

$ c_b \approx \ce{[OH-]} = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}}$

so, taking the derivative,

$\beta_{\ce{OH-}} = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})} = \cfrac{\mathrm{d}}{\mathrm{d}(\mathrm{pH})} 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \ln{10}$

it's easy to show that, similarly, for the addition of acids,

$\beta_{\ce{H+}} = -\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})} = 10^{-\mathrm{pH}} \ln{10}$

and, combining the buffer effect of both semi-systems, we get the total buffer capacity of water:

$\beta_{\ce{H2O}} = \left( 10^{-\mathrm{pH}} + 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \right) \ln{10}$

Weak acid/base pair

1:1 solutions of weak acids/bases are a typical buffer system around $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ for the buffer. To understand why, let's imagine we add a weak acid/base pair, $\ce{HA/KA}$, with total concentration $C_\mathrm{A}$, to which we later add a certain concentration of base, $c_\mathrm{b}$. This weak acid/base pair will be described by the equilibrium

$K_A = \cfrac{\ce{[H3O+] [A-]}}{\ce{[HA]}}$

which, taking into account that $C_\mathrm{A} = \ce{[HA]} + \ce{[A-]}$, implies

$\ce{[A-]} = C_\mathrm{A} \cfrac{K_\mathrm{A}}{\ce{[H3O+]} + K_\mathrm{A}}$

From charge balance:

$\ce{[H3O+]} + \ce{[K+]} = \ce{[OH-]} + \ce{[A-]}$

Taking into account that $\ce{[K+]}$ is equal to the formal concentration of $\ce{KA}$ we added, so $c_\mathrm{b} = \ce{[K+]}$, and the previous expression for $\ce{[A-]}$:

$c_b = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} - \ce{[H3O+]} + C_\mathrm{A} \cfrac{K_A}{\ce{[H3O+]} + K_\mathrm{A}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} - 10^{-\mathrm{pH}} + C_\mathrm{A} \cfrac{10^{-pK_A}}{10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}}}$

So the buffer capacity of the solution will be:

$\beta = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}= \left(10^{\mathrm{pH}-\mathrm{p}K_w}  + 10^{-\mathrm{pH}} + C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \right) \ln{10}$

Note that the first two terms are the buffer capacity of water, so the contribution of the acid/base pair is

$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \ln{10}$

Additive buffer capacity contributions can be calculated this way.

For instance, for a typical acetate buffer ($\ce{HAc} \ 0.2 \mathrm{M} $, $ \ce{NaAc} \ 0.2 \mathrm{M} $; $\mathrm{p}K_\mathrm{a} = 4.76$, $C_\mathrm{A} = 0.4\mathrm{M}$), the buffering capacity looks like this:

Ionic strength

Since buffers are moderately concentrated electrolyte solutions, the assumption that activity coefficients in them are approx. 1 shouldn't be automatic. How does this affect the calculations of buffer capacity we've performed until now?

Let's define an effective equilibrium pseudo-constant, $K'$, operating on concentrations, that is mathematically equivalent to the actual equilibrium constant, $K$, operating on activities:

$K_\mathrm{A} = \cfrac{a_{\ce{H3O+}} a_{\ce{A-}}}{a_{\ce{HA}}} \iff K'_\mathrm{A} = K_\mathrm{A} \cfrac{\gamma_{\ce{H3O+}} \gamma_{\ce{A-}}}{\gamma_{\ce{HA}}} = \cfrac{\ce{[H3O+]} \ce{[A-]}}{\ce{[HA]}}$

So reflecting the impact of ionic strength on buffer capacity becomes a question of evaluating $K'_\mathrm{A}$ and using it to replace $K_\mathrm{A}$ in our previous expressions. For instance, we can use the Debye-Hückel relationship:

$\mathrm{p}K'_\mathrm{A} = \mathrm{p}K_\mathrm{A} + A \left( 2 z_{\ce{HA}} - 1 \right) \left( \cfrac{\sqrt{I}}{1 + \sqrt{I}} - 0.1 I\right)$

where $A$ is a constant ($\approx 0.51$ at room temperature), $z_{\ce{HA}}$ is the charge of the conjugate acid and $I=\frac{1}{2}\sum c_i z^2_i$ is the ionic strength of the solution. Note that $\mathrm{p}K'_\mathrm{A}$ has to be solved recursively, however, as it depends on $I$, which depends on the concentration of the different ions, which depend on $\mathrm{p}K'_\mathrm{A}$.

A similar correction can be applied to $K_\mathrm{w}$ to obtain $K'_\mathrm{w}$.

Applying this ionic strength correction to our acetic/acetate $\pu{0.4M}$ buffer above, we can see the impact of ionic strength:

The $\ce{HCl/KCl}$ buffer

So what about the $\ce{HCl/KCl}$ buffer? Unlike the example of acetic/acetate, $\ce{HCl}$ is a strong electrolyte - so it dissociates completely at all values of $\mathrm{pH}$. So we can treat it in two ways: as simply a water buffer with a different starting $\mathrm{pH}$, or as an acid/base conjugate pair buffer with a $\mathrm{p}K_\mathrm{A} < 0$. Both are equivalent, as the acid/base contribution is negligible compared to the water contribution due to the $\mathrm{p}K_\mathrm{A}$ value.

For instance, for a typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer ($\mathrm{p}K_\mathrm{A} = -6.3$, $\mathrm{pH} \approx 0.7$, $I \approx 0.8$), the contributions to buffer capacity are as follows:

$\beta_{\ce{H2O}} = \left( 10^{\mathrm{pH}-\mathrm{p}K'_\mathrm{w}} + 10^{-\mathrm{pH}} \right) \ln{10} \approx  10^{-\mathrm{pH}} \ln{10} = 0.4594$

$\beta_{\ce{HCl/Cl-}} = C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K'_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K'_\mathrm{A}} \right)^2} \ln{10} \approx C_\mathrm{A} 10^{\mathrm{p}K'_\mathrm{A}-\mathrm{pH}} \ln{10} = 5.865 · 10^{-8}$

So, for all effects and purposes we have a buffer that behaves just like water. It is, however, a buffer that behaves just like water in a region of the $\mathrm{pH}$ scale which has a relatively large buffer capacity by virtue of being close to $\mathrm{pH}=0$. For instance, this is what buffer capacity looks like around our typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer (the red dashed line marks the initial $\mathrm{pH}$):

Note that buffer capacity is highly asymmetrical around this point - adding acid increases the buffer capacity of this system, requiring increasingly more acid to decrease $\mathrm{pH}$, while adding base decreases the buffer capacity, making the system's $\mathrm{pH}$ more sensitive to further base additions. This can be easily visualised if we realise that the area under the $\beta$ curve represents the concentration that needs to be added to move in the $\mathrm{pH}$ scale.

Note also that, unlike in weak acid/base pairs, we cannot simply increase concentration to increase buffer capacity while retaining the same $\mathrm{pH}$ - as the concentration-dependent term, corresponding to the $\ce{HCl/KCl}$ base, is negligible. Ionic strength is irrelevant for $\beta$, as well, as $K'_\mathrm{w}$ and $K'_\mathrm{A}$ is not involved - $\beta_{\ce{H3O+}}$ is the only relevant contribution to $\beta$ in this very acidic region, and, by definition, it is not affected by activity concerns (unlike $\beta_{\ce{OH-}}$ or $\beta_{\ce{HCl/Cl-}}$).

Does that mean that $\ce{HCl/KCl}$ buffers with different concentrations all have the same buffer capacity? Not at all: but their different buffer capacity is mediated by $\mathrm{pH}$. In other words, different concentrations produce different $\mathrm{pH}$ which, in turn, produces a different $\beta$ - which is unresponsive to changes in $I$ that do not change $\mathrm{pH}$.

Two final plots to illustrate this. Imagine we prepare solutions of $\ce{HCl/KCl}$ (1:1) at different total concentrations $C_{\ce{Cl-}}=c_{\ce{HCl}}+c_{\ce{KCl}}$. If we compare the two contributions to $\beta$, it's clear that $\ce{HCl/KCl}$ increases buffer capacity almost exclusively through $\mathrm{pH}$; $\beta_{\ce{H3O+}}$ is proportional to $\ce{[H3O+]}$ which increases linearly with $c_{\ce{HCl}}$ as $\ce{HCl}$ is a strong electrolyte:

Conversely, what if we control for $\mathrm{pH}$ and change only $I$? Let's imagine a series of solutions containing a fixed concentration of $\ce{HCl}$ and varying amounts of a non-reacting, strong electrolyte without common ions - such as $\ce{NaBr}$ - so we can change the ionic strength of the solution without affecting neither the $\mathrm{pH}$ nor the $\ce{HCl/Cl-}$ equilibrium. The impact of $I$ on $\beta$ would look like this (note that the minimum $I$ corresponds to a solution with $\ce{HCl}$ and no $\ce{NaBr}$):

This is because although the $\beta_{\ce{HCl/Cl-}}$ and $\beta_{\ce{OH-}}$ terms do depend on $I$, these terms are many, many orders of magnitude lower than the $\beta_{\ce{H3O+}}$ term, which doesn't.

So, after this discussion, we can now directly respond to your questions.

TL;DR

1) Why does the increased ionic strength of the solution improve water's buffer capacity?

It doesn't. The buffer capacity of solutions of strong acids are dominated by the $\beta_{\ce{H3O+}}$ term, which (by definition of $\mathrm{pH}$) is not affected by ionic strength.

There are, however, other reasons to desire an ionic strength in that range - in biological systems, many molecules, particularly proteins, have a privileged ionic strength stability gap - above or below which they tend to denaturalise or precipitate, and outside of which their biological activity can be inhibited or poisoned. Although different systems will require different ionic strengths, buffers are typically prepared with ionic strengths in the range $0.1-1\mathrm{M}$ for that reason.

2) Isn't a higher equilibrium constant required to make water a better buffer? Then isn't it bad that the ionic strength decreases $K$?

The $\beta_{\ce{H3O+}}$ term, by construction, does not depend on $K'_\mathrm{w}$, so it is unaffected by ionic strength. The $\beta_{\ce{OH-}}$ term does, and as you mention, it is decreased at moderate ionic strength - although it is increased at higher $I$. In general, the impact of ionic strength on buffers is as follows: increasing ionic strength begins by lowering $\beta$, then, as $I$ keeps increasing, $\beta$ raises again.

3) Could any other salt have been used to increase the ionic strength, or is there some specific reason $\ce{HCl-KCl}$ is used?

Yes, any other salt could have been used to increase the ionic strength. Adding $\ce{KCl}$ affects the acid-base pair contribution to $\beta$, but as we've seen, that contribution is completely negligible - as will be the case with strong acids, which operate mainly through the $\ce{H3O+}$ contribution.


How do the stored chemicals in batteries drive electronic current from their terminal to outside devices? I have got around to at least read on Wikipedia, and it does not clarify how this works in any sensible manner.

I wondered how the chemicals make electrons "move". How is the basic understanding of this?

A battery, at its heart, is what we call a redox cell: some compounds are stripped of their electrons, and others gain those electrons. The movement of electrons between these two compounds can be harnessed to provide electricity, but that's taking us a step too far. What we want to understand here is why the electrons move. To do this, we'll have to understand the redox cell. As much as I hate to assume things, you're going to have to understand a bit about ions and equilibrium to understand this explanation.

(Note: the explanation I'm about to use is heavily taken from the excellent Chemguide site, run by Jim Clark. I'll include a link at the end.)

Let's say I place a bar of magnesium in a water bath. I'll take a look at the magnesium bar a short while after I place it in the water. If I could visualize atoms and electrons, this is what I'd see:

Some of the atoms of magnesium leave the bar and become ions in the solution, solvated by water (and other things, but that's not crucial to this explanation). Since they become ions, they leave electrons on the metal bar. Because the ions are positively charged, they are attracted to the negative charge on the metal bar, and tend to hang around there.

If I wait a little longer, I'll see more and more ions coming off the bar. As they start to build up in the water around the metal, every once in a while, one of the ions will hit the bar and reclaim its electrons. This is what we'll see then:

If I wait long enough, eventually, the number of atoms coming off the metal will be equal to the number of ions reabsorbing onto the metal. If I take out the "bites" in the metal bar to simplify the situation, this is what I see:

Notice how there's quite a buildup of electrons on the magnesium. This is because magnesium is a very active metal, like lithium. What if we used a much less active metal, one that doesn't have a large propensity to lose atoms into solution? An example of such a metal would be copper. Here's a side-by-side of the two:

Because the magnesium is a lot more active, it tends to lose ions more, and so it has more electron buildup. If we connect the two with a wire, we give the electrons a way to flow from the high concentration to the low concentration. We can then place something along the wire in the middle (say, a lightbulb) and harvest some energy!

This, of course, is a hilariously oversimplified scheme. For one, as the electrons go to the copper, more and more magnesium ions break off to replace them. At some point, there are so many positive ions in the solution that they draw the electrons back toward them, through electrostatic attractions (which is a fancy way of saying that positive and negative charges attract each other). To solve this, we'd put a salt bridge between the solutions that lets ions migrate back and forth.

There are other problems too: for example, this setup wouldn't work in real life because there are no ions in solution: real cells use salt solutions of the metal e.g. $\ce{MgCl2}$ or $\ce{CuSO4}$. There's also the issue that this battery is liquid and wouldn't do so well if you tipped it over, the fact that the voltage (which is really the difference between the charge on the metal bars) is only about 1.1 volts, yada yada. That's not central to the explanation.

The fact that certain compounds are more likely to gain electrons and some are more likely to lose them is the central reason why things like batteries can exist. If you'd like to read more about it, I pulled this explanation from chemguide.co.uk, run by Jim Clark. I am currently ready to graduate with a B.Sc in chemistry, and his website is still one of the best resources for conceptual learning that I've come across---as a matter of fact, I learned about things on that website (eutetic mixtures and azeotropes) that I still have yet to come across in four years of university-level study. If you'd like to read more into redox reactions, the redox equilibria page is at this link right here.


How would you synthesise tetrahydrohyperforin and octahydrohyperforin from hyperforin? The chemical structure of the compounds in question are:

Figure 1: Hyperforin

Figure 2: Tetrahydrohyperforin

Figure 3: Octahydrohyperforin

Would heating hyperforin in the presence of water manage to saturate the required bonds and thus allow one to synthesise these compounds?

That route won't work.

You need a hard base reduction that does not attack the enone, or you need to protect the enone to make Tetrahydrohyperforin.

Catalytic hydrogenation that is selective against carbonyl reduction is necessary to make Octahydrohyperforin.


What is the purpose of the electrolyte in the half-cell where oxidation is taking place? In an electrochemical cell that consists of two half-cells, there is an electrolyte and an electrode in each half-cell.

Example:
$$
\begin{align}
\ce{CuSO4(aq) + 2e- &-> Cu(s) + SO4^2-(aq)} \\
\ce{Zn(s) + SO4^2-(aq) &-> ZnSO4(aq) + 2e-}
\end{align}
$$

In the half-cell where the reduction takes place, $\ce{Cu^2+}$ gains two electrons. Here the electrode and the electrolyte serve a purpose, as it is the $\ce{Cu^2+}$ ion from $\ce{CuSO4}$ that goes through a reduction reaction.

However, in the half-cell where the oxidation takes place, it is $\ce{Zn(s)}$ that loses two electrons and forms $\ce{Zn^2+(aq)}$. Thus, the $\ce{ZnSO4}$ does not seem to play any role in the oxidation. Why is it included in the electrolyte in the zinc oxidation half-cell

The primary role of the anode chamber electrolyte (the anolyte) in this case is to provide sufficient conductivity in that chamber.

If deionized water were used in the anode chamber, the conductivity of that liquid would be extremely low. Thus, it's important for there to be something dissolved there in order for any appreciable amount of current to pass through the overall electrochemical cell. The overall electrochemical circuit involves both electronic conduction (in the wires and electrodes) and ionic conduction (in the two electrolytes), and it will not function efficiently/effectively if any element of that circuit is too poorly conductive.

That said: It's not crucial that it be specifically $\ce{ZnSO4}$, however.  $\ce{NaCl}$, $\ce{H2SO4}$, $\ce{NH4NO3}$, or a wide variety of other chemicals might work just as well (or even better!).  However, a sulfate salt makes sense here because (at least):

It provides relatively high conductivity
Its salts with metal cations, including $\ce{Zn^{2+}}$, are relatively water-soluble.
In this specific cell configuration, using sulfate makes sense because it's the same anion as in the $\ce{CuSO4}$ used for the catholyte, and thus you don't have to worry about counter-diffusion of different anions across the salt bridge.

It further makes sense to use $\ce{ZnSO4}$ specifically as the anolyte salt because it provides the same cation as you'll be forming from your anode reaction. This will result in a more stable electrochemical cell, because having a different, potentially electrochemically-active cation in the anolyte would most likely change both the thermodynamic and the practical operating characteristics of the cell.


Can aged acetylcholinesterase be reactivated? it is a well-known fact that certain phosphorus compounds can inhibit the enzyme acetylcholinesterase, and that the treatment of poisoning with such compounds is administration of atropine and some kind of oxime-containing compound. This needs to be done quickly, since another reaction, called "Aging" usually takes place, whereby a part of the phosphorus compound that does not act as the initial leaving group is hydrolyzed off, leaving a negatively-charged phosphonate or phosphate, which forms a salt bridge with the protonated, positively-charged histidine residue at the active site of the enzyme. Aging renders AChE resistant to reactivation by oximes. Is it possible to somehow reactivate AChE that has aged?

Until the late 2010s, it was considered impossible to reactivate acetylcholinesterase that has undergone aging. However, in recent years, new advancements have been made that show promise in this area.
A group of researchers investigated several sulfonium-containing alkylating agents that were supposed to alkylate the phosphonate moiety bound to the serine of aged AChE, in hopes that this would allow the enzyme to be reactivated by traditional methods.

Computational simulations suggested that all three of these compounds would be able to alkylate the aged enzyme.
Another group of researchers studied two aminoalcohols that would act by transferring a proton to the aged adduct. One of the compounds showed promise in reactivating aged AChE, as it was able to do so according to computer simulations. This is the molecule:

What is notable about this aminoalcohol is that it is predicted to be capable of reactivating the enzyme without alkylating the adduct.
Yet another series of compounds was shown to be successful in reversing the aging of acetylcholinesterase. These compounds reverse the aging reaction by alkylating the phosphonate/phosphate-AChE adduct, and also exhibit the potential to reactivate the enzyme entirely.

Unfortunately, they were quite slow to reverse the aging of acetylcholinesterase, with the most active compound, C8, resurrecting 12% of methylphosphonate-aged AChE in one day at a concentration of 4 mM.
In conclusion, while several compounds have shown promise in reactivating acetylcholinesterase that has undergone aging by traditional nerve agents or pesticides, therapeutic methods of administering those compounds have yet to be developed. Furthermore, these compounds have not yet been tested in vivo, under physiological conditions, or against other anticholinesterase agents such as Tabun or the Novichoks.
Sources:
Franjesevic, A. J., Sillart, S. B., Beck, J. M., Vyas, S., Callam, C. S., & Hadad, C. M. (2018). Resurrection and Reactivation of Acetylcholinesterase and Butyrylcholinesterase. Chemistry - A European Journal. doi:10.1002/chem.201805075
Khavrutskii, I. V., & Wallqvist, A. (2017). β-Aminoalcohols as Potential Reactivators of Aged Sarin-/Soman-Inhibited Acetylcholinesterase. ChemistrySelect, 2(5), 1885–1890. doi:10.1002/slct.201601828
Zhuang, Q., Franjesevic, A. J., Corrigan, T. S., Coldren, W. H., Dicken, R., Sillart, S., … Hadad, C. M. (2018). Demonstration of In Vitro Resurrection of Aged Acetylcholinesterase after Exposure to Organophosphorus Chemical Nerve Agents. Journal of Medicinal Chemistry, 61(16), 7034–7042. doi:10.1021/acs.jmedchem.7b01620