How to calculate the temperature at which a reaction becomes spontaneous? At $25~^\circ\mathrm{C}$, $298~\mathrm{K}$, the reduction of copper(I) oxide, $ \Delta H = 58.1~\mathrm{kJ}$, $\Delta S = 165~\mathrm{J/K}$, is nonspontaneous, $ \Delta G = 8.9~\mathrm{kJ}$.
  Calculate the temperature at which the reaction becomes spontaneous.

So for this question I'm thinking I use the equation

$$ \Delta G = \Delta H - T\Delta S$$

Then plug in the values and solve for $T$? $$8.9~\mathrm{kJ} = 58.1~\mathrm{kJ} -T(0.165~\mathrm{kJ}) $$

You have the right equation.  But the problem already gives you the temperature at which $\Delta G$ is 8.9 kJ.

Assume that $\Delta H$ and $\Delta S$ are constant and do not vary with temperature.  $\Delta G$ still will though because of the $T$ in the $\Delta G = \Delta H - T\Delta S$ equation.

So plug in the values for $\Delta H$ and $\Delta S$, and also the $\Delta G$ value that you think represents the transition from spontaneous to non-spontaneous.  You should be able to take it from there.


Ideal gas constant applied to water, lead, and other non-gases I've asked a number of chemists why the ideal gas constant is in the equation for Gibbs free energy, and all say the same thing: "The name is a relic of the argon gas chamber used to determine energy and temperature at different pressures and temperatures, but it applies to every thermodynamic system." This sounds great and I understand names of constants can be relics, but why can a conversion factor that only appears to pertain to ideal gases be used for other substances? R assumes elastic collisions, where energy is not lost during molecular collisions, but only ideal gases have this property. Wouldn't you use a different conversion factor for something much denser and fluid like water or even more dense like lead to convert from temperature to energy? I've heard that R is used for free electrons as well, which are pretty different from atoms much less gases. Would "R-specific" be best to use everywhere, as I've heard they do in engineering, so maybe every substance has its own temperature-energy conversion factor?

It is a very good question: Why does R appear everywhere in equations even though it is apparently related to or derived from the ideal gas law? Yes historically R was derived from the limiting gas law in the 1830s.

The beauty of R is that it is related to two physical constants called the Boltzmann constant and Avogadro's constant. Read here

https://en.wikipedia.org/wiki/Gas_constant#Relationship_with_the_Boltzmann_constant

The Boltzmann constant appears everywhere in thermodynamics, semiconductors and statistical mechanics: https://en.wikipedia.org/wiki/Boltzmann_constant

  "The name is a relic of the argon gas chamber used to determine energy
  and temperature at different pressures and temperatures, but it
  applies to every thermodynamic system."

I feel this is nothing but fiction. Not sure why people make up stories. William B Jensen traces the history of R in the Journal of Chemical Education.


Regenerating a Cobalt Acetate Catalyst I have an eluent from an ion-exchange column containing Cobalt (II) ions and Chloride ions. In order for the cobalt catalyst to be recycled to the main process, it must be in its acetate form.
Is there a route from cobalt chloride to acetate that is feasible at an industrial scale?

Thanks

Quick Laboratory method:

$$\ce {CoCl2(aq) + 2NaOH(aq)->Co(OH)2(s) + 2NaCl(aq)}$$
$$\ce {Co(OH)2(s) + 2CH3CO2H(aq)->Co(CH3CO2)2(aq) + 2H2O(l)}$$
$$\ce {Co(CH3CO2)2(aq) + \Delta->Co(CH3CO2)2\cdot 4H2O (s)}$$

You should separate $\ce{Co(OH)2}$ by filteration and washed with water ($\pu {K_{sp} = 1.0 \times 10^{-15}}$) to remove all $\ce{NaCl}$ before subjected to react with $\ce{CH3CO2H}$.

This might work in industrial scale, if be careful. $\ce{Co(OH)2}$ is unstable to heat but no need to dry before react with $\ce{CH3CO2H}$.


Platinum group metals: Why are platinum and palladium great catalysts and not the others? Platinum and palladium are great catalysts. At the same time, other metals of the so-called platinum group metals are not. What are the atomic level reasons for this?

I think some of the other transition elements have also catalytic properties. For example, rhodium is generally used in addition to platinum or palladium in catalytic converters to oxidize carbon monoxide and hydrocarbons on one hand, and reduce $\ce{NO_x}$ to nitrogen on the other hand.

So, the choice is often made on the basis of relative cost.

But of course, not all transition elements are adequate for these catalytic properties:

Metals such as silver and copper have a high affinity for sulfur and form metal sulfides and sulfates. As this happens, there will be progressively less metal available to function as a catalyst. While platinum tends not to become totally or irreversibly poisoned, i.e It can be efficiently recycled.
Another point is linked to "Tammann temperatures" of these metallic elements:  Actually, in a catalytic converter, metal is used in the form of nanoparticles dispersed on the surface of a porous support material. When  temperature increases the particles starts to becomes mobile and coalesce (i.e. the catalytic activity is lost). This process is well-known under the name "sintering" and it becomes tangible as the metal is close to its Tammann temperature. This temperature is often defined as the half of the metal's melting point on the absolute temperature scale. In fact, metals such as gold, silver and copper have a Tammann temperature that is well below the average temperature of exhaust-gas ($600-700^0\mathrm{C}$).
Tammann Temperature is $750^0\mathrm{C}$ for platinum, and $640^0\mathrm{C}$ for palladium. While it is only $405^0\mathrm{C}$ for copper, $345^0\mathrm{C}$ for silver and $395^0\mathrm{C}$ for gold.

To sum up, the candidate transition element for use as a catalyst in a catalytic converter:
1) has a low affinity to poisons  such as sulfur compounds;
2) can be efficiently recycled;
3)  has a relatively high melting point.


In nuclear chemistry, how does a neutron split to form a proton and an electron? I'm studying radioisotopes at the moment and balancing nuclear reactions isn't making sense in that more matter is coming out of the equation in negative β⁻ decay equations:

$$\ce{_6^{14}C -> _7^{14}N + e- + \bar{\nu_e}}$$

Notice how the original element has decayed into a new element with an unchanged mass number but an atomic number that has increased by one.

In positive β⁺ decay equations, it makes sense:

$$\ce{_{12}^{23}Mg -> _{11}^{23}Na + e+ + \nu_e}$$

How can you create something with the same mass but with another proton?

One neutron has changed into a proton, that's what has happened. Our convention in chemistry is to identify nuclear species by the proton number (or the "atomic number") and that is why a new nuclide with an increased atomic number is formed. Both the $\beta^+$ and $\beta^-$ decay follow identical paths: in $\beta^-$ decay, a neutron changes into a proton (thus giving no change in mass number and increase in atomic number), whereas in $\beta^+$ decay, a proton decays into a neutron giving no change in the mass number and a decrease in the atomic number.


Why is polymer B more dense than polymer A? The official solution is that A has branched / side chains, weak Van der Waals forces between the polymer chains, less compact packing than B, and large inter-chain distances.
B has no branched chains, hydrogen bonds between the polymer chains, compact packing, and small inter-chain distances.

What does a branched chain mean? How does this influence the density? Doesn't B also have side chains?

What does 'compact packing' mean in this case? Does it mean that separate chains are packed closely together, or that in a single chain, the molecules are closer together?

First polymerisation proceeds through free radical mechanism, which may cause intermediate radicals to abstract H from other parts of growing chain, thus creating new branches in the process.
Refer Wikipedia for detailed mechanism.

On the other hand, second polymerisation proceeds through condensation reaction, where no such thing is possible (condensation can occur only at terminals).

Furthermore, the chains of B interact through H-bonding, which is much stronger than van der Waal interaction of polymer A. This causes chains to come closer and increases the overall density of polymer.


Why Acetone does not behave like its computational values? I am trying to simulate the excitation state of acetone. I ran TDDFT for it both in gas phase and solvated state in water (both implicit and explicit water).

The experimental data say that acetone undergoes n->π* transition, which means it has a higher wavelength in gas phase ($\mathrm{\lambda_{max}}=$~$276~\mathrm{nm}$) and a shorter wavelength in the solvated state ($\mathrm{\lambda_{max}}=265~\mathrm{nm}$).  I started with gas phase and expected something a wavelength of about $276~\mathrm{nm}$ ($4.49~\mathrm{eV}$ ), but surprisingly I got $\mathrm{\lambda_{max}}=136.172~\mathrm{nm}$ ($9.326~\mathrm{eV}$)! I really cannot understand why there is such a big discrepancy!

Here is my GAMESS input file. What is wrong with this molecule ?

!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NGAUSS=6 $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END
 $CONTRL ICHARG=0 MULT=1 $END
 $TDDFT NSTATE=9 $END
 $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END
 $SYSTEM MWORDS=1000 PARALL=.TRUE. $END
 $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END

 $DATA
Title
C1
O     8.0     0.00000    -1.27900     0.00300
C     6.0    -0.00000    -0.05800     0.00100
C     6.0     1.29700     0.69100    -0.00000
C     6.0    -1.29800     0.69000    -0.00000
H     1.0     1.35900     1.32900    -0.90600
H     1.0     1.35900     1.33200     0.90300
H     1.0     2.15700    -0.01300     0.00100
H     1.0    -2.15700    -0.01400     0.00100
H     1.0    -1.35900     1.32900    -0.90600
H     1.0    -1.35900     1.33200     0.90300

 $END

It seems that the results of the calculations are more or less fine and the OP just misinterpreted the NIST data. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper[1] referenced on the NIST page it is said (emphasis mine):

  There are two diffuse ultraviolet bands; the first at 2800 Å. is very
  weak with its oscillator strength $f \sim 0.0004$ and the second at
  about 1900 Å  is moderately intense with a maximum extinction
  coefficient $\epsilon_{\mathrm{m}} \sim 1000$. McMurry has identified
  the 2800 Å. band as a forbidden $\pi^* \leftarrow n$ transition
  involving excitation of a non-bonding $\ce{O}$ electron to an
  anti-bonding $\pi$ orbital between the $\ce{C}$ and $\ce{O}$ of the
  carbonyl group.

A high-resolution photoabsorption spectrum in the energy range 3.7-10.8 eV can be found, for instance, in this recent study[2] and is more or less consistent with the result of the OP's calculations.

Noel S. Bayliss, Eion G. McRae, J. Phys. Chem., 1954, 58 (11), 1006–1011.
M. Nobre, A. Fernandes, F. Ferreira da Silva, R. Antunes, D. Almeida, V. Kokhan, S. V. Hoffmann, N. J. Mason, S. Eden, P. Limão-Vieira, Phys. Chem. Chem. Phys., 2008, 10, 550-560. (available at researchgate.net)


Synthetic ruby - damaged by heat less than its melting point? A bit of a jewellery question but chemistry related. I want to encase a synthetic ruby (corundum, Al2O3) in silver by melting the silver around it. Either in a furnace or by casting. The MP of silver is 960 °C.

The ruby will be facteded and polished. Now, in theory the ruby should be unharmed given that its MP is double that of the silver and is much harder. However, would there be any damage to its optical properties?

However, would there be any damage to its optical properties?

It is almost guaranteed. Corundum is incredibly sensitive to thermal shock and unless you can do this in a furnace that heats very slowly, it will fracture/crack.

As far as any other properties, your ruby should not see any change in other optical properties for such a short duration, but for longer durations (hours at high temperature) metals can diffuse into the ruby and cause color changes, but I doubt this is an issue for what you describe.


How can you tell if a metal ion will hydrolyze to form an acidic solution? I saw somewhere that all metal ions except for the alkali metals and alkaline earth metals will hydrolyze to increase acidity. If I'm interpreting things correctly, sodium ions, for example, will not hydrolyze because sodium hydroxide is strong, thus sodium ions will not react with water to form sodium hydroxide and hydrogen ions. But I was told that magnesium will not hydrolyze, but magnesium isn't on the strong bases list that was given. By the way, I'm a grade 12 student, so I would appreciate if you kept things down to earth. Can someone clarify please?

So the ability of a metal cation to hydrolyze water molecules to form acidic solutions is based on the charge of the ion and the size of the atom. The larger atoms will stabilize charge better (i.e. it spreads it out over a larger space). Unstable cations form hydroxides or oxides with water molecules and release hydronium ions.
Estimations of the acidity for metal cations can be found with the equation below,
$pka = 15.14 - 88.16(\frac{Z^2}{r})$
Z being the charge of the ion and r being the atomic radius. I bold the word estimation cause this isn’t a perfect system and I want to make that clear. However, better equations do exist.
Electronegativity does play a role too. The more electronegative a metal cation is, the more acidic it will be.
I would recommend taking a look at this link for more information


Which of the following statements is true about the trend down group 7? Which of the following statements is true about the trend down group 7 (halogens) with increase in atomic number?

  The reactivity increases.
  The atoms get smaller
  The boiling points decrease
  The melting points increase
  The reactivity decreases

For me D and E must be right but only one can be right.

The answer is D, the melting points increase. This is absolutely true (source for values):
\begin{array}{lrr}
\text{Halogen} & \text{Melting point}/^\circ\mathrm{C}& \text{Boiling point}/^\circ\mathrm{C}\\\hline
\text{fluorine}   & -220  & -188  \\
\text{chlorine}  & -101  & -35   \\
\text{bromine}  & -7.2  & 58.8  \\
\text{iodine}   & 114   & 184   \\
\text{astatine}  & 302   & 337   \\\hline
\end{array}

And the boiling point increases, too, so answer C is definitely wrong.

Another obviously wrong answer is B, the atoms get smaller. With increasing main quantum number the atoms obviously have to become bigger since the main electron density moves further away from the nucleus. In numbers:
\begin{array}{lrr}
\text{Halogen} & \text{Covalent radius}/\mathrm{pm}& \text{Ionic radius }\ce{(X^{-})}/\mathrm{pm}\\\hline
\text{fluorine} &  71  & 133  \\
\text{chlorine}   &  99  & 181  \\
\text{bromine}  & 114  & 196  \\
\text{iodine}   & 133  & 220  \\
\text{astatine}  & 150  &      \\\hline
\end{array}

For answers A and E it cannot be unambiguously answered, because reactivity is not a well defined concept, see "What is reactivity really, and can it be quantified?"  for example. The IUPAC goldbook states, that reactivity is a kinetic property. It goes on to describe, that it can only be absolutely used in a given context or in reference to another system. However, it is also often used in expressing general trends, which is the case when looking at this question.
Given this context, you can find the following portion on UC Davis ChemWiki:

  Reactivity of Elements: decreases down the group
  The reactivities of the halogens decrease down the group ( At < I < Br < Cl < F). This is due to the fact that atomic radius increases in size with an increase of electronic energy levels. This lessens the attraction for valence electrons of other atoms, decreasing reactivity. This decrease also occurs because electronegativity decreases down a group; therefore, there is less electron "pulling." In addition, there is a decrease in oxidizing ability down the group.

In this context, answer A is wrong. However, I would not consider E to be strictly true either. Have a look at the electron affinities, and you will find, that fluorine behaves anomalous. This was also discussed in "Why does chlorine have a higher electron affinity than fluorine?"
However the general statement will usually be given as: "Electron affinity decreases down the group."
\begin{array}{lc}
\text{Halogen} & \text{Electron Affinity}/\mathrm{kJ\cdot mol^{-1}}\\\hline
\text{fluorine}  &  -328.0  \\
\text{chlorine} &  -349.0  \\
\text{bromine}   &  -324.6  \\
\text{iodine}    &  -295.2  \\
\text{astatine}  &  -270.1  \\\hline
\end{array}
The same trend can be observed with bond enthalpies of the $\ce{X2}$ series (source):

This trend is broken for the hydrogen halides $\ce{HX {(g)}}$.

Your assumption is therefore correct, maybe you have already done the same analysis. However, we can only definitely state and proof, that answer D must be correct, while answer E might be correct depending on the reference system.


Acid Catalysed Ring Expansion – Mechanism? Ring expansion of vinyl cyclobutane -

Could someone please help me with a detailed mechanism for the above conversion?

It is the acid-catalyzed rearrangement of 4,7,7-trimethylbicyclo[3.2.0]hept-3-en-6-one to 3,3-dimethyl-6-methylidenebicyclo[2.2.1]heptan-2-one.

I couldn't find the mechanism anywhere, and couldn't propose a good one either.
Please help.

P.S.
As one of the users in the comments is interested in seeing my best effort, I'll explain my thought process-
As the carbonyl oxygen is the most nucleophilic, it is the first to be protonated in acidic medium. Furthermore, this species has a resonating structure (less contributing, but the only one that can lead to a ring expansion, I believe) – one obtained by shifting the π bond towards oxygen – a carbocation.
After this, I'm clueless. I think, that the expansion could have something to do with the hydrogen at alpha position to the sp² hybridised carbon atom in the five membered ring – as the product has an exocyclic double bond.

You are correct about protonating the carbonyl group. [The first of your products is the same as the reactant.] The cyclobutanone 2 was prepared from a mixture of geranic acids. (J. J. Beereboom, J. Org. Chem., 1965, 30, 4230). The structures of 2, 4 and 5 were determined by Beereboom. I am providing the most reasonable mechanism. Cyclobutanone 2 must have a more stable cis ring juncture owing to its method of formation or exposure to p-TsOH. The trans isomer of 2 cannot give the trans isomer of 3 because of strain. Tertiary carbocation 3 arises by addition of the double bond to the protonated carbonyl. Collapse of the carbocation produces bicyclic ketone 4, which in the presence of p-TsOH isomerizes to the exocyclic isomer 5. In general, 1,1-disubstituted double bonds are less stable than tri-substituted ones. In this case, ring strain may be present in the endocyclic alkene 4.


How to find pH by mixing two solutions of different concentrations? A mixture is made by combining 110 mL of 0.15 M $\ce{HCl}$ and 215 mL of 0.055 M $\ce{HI}$.  What is the pH of the solution?

$$\ce{HCl -> H+ + Cl-}$$
$$\ce{HI -> H+ + I-}$$
According to what I thought, the molarity of $\ce{H+}$ is the same as $\ce{HCl}$, because it is a strong acid and the mole ratio.  So, I added the molarity of both acids: $\pu{0.15M} + \pu{0.055M} = \pu{0.205M}$ for $\ce{H+}$ in solution.

$$\mathrm{pH} = -\log[\ce{H+}]$$
$$\mathrm{pH} = -\log[0.205]$$
$$\mathrm{pH} = 0.689$$

You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. here.
Therefore you can assume, that the concentration of protons is equal to the concentration of the acid.
\begin{align}
c_{\ce{HI}}(\ce{H+}) &=  c(\ce{HI}) &
c_{\ce{HCl}}(\ce{H+}) &=  c(\ce{HCl}) &
\end{align}

Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$

You have to add the amount of protons in each solution and determine the combined volume.
\begin{align}
n_\mathrm{tot}(\ce{H+}) &= n_{\ce{HI}}(\ce{H+}) + n_{\ce{HCl}}(\ce{H+})&
V_\mathrm{tot} &= V(\ce{HI}) + V(\ce{HCl})
\end{align}

To calculate the pH of the resulting concentration, just take the negative logarithm.
$$\mathrm{pH} = -\log c_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n_\mathrm{tot}(\ce{H+})}{V_\mathrm{tot}}\right]$$

   Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.

The approach you chose can only work, if you have equal volumes of the two solutions.


Why doesn't sodium carbonate decompose? I was asked to find the hypothetical $\Delta H$ and $\Delta S$ of the decomposition of sodium carbonate and then asked to explain why it doesn't decompose.

I calculated the enthalpy and entropy change and from that I calculated the Gibbs free energy change. I was hoping to get a positive answer, but actually I got a negative answer, indicating that the process is actually spontaneous.

So is the reason that sodium carbonate doesn't decompose due to its kinetic properties, i.e. the reaction is too slow?

From the values on the Wikipedia page for sodium carbonate, the Gibbs free energy of formation is negative and thus, formation is thermodynamically preferred over decomposition at $\pu{298K}$. Your given values seem to be wrong after all.

By the way, $\Delta H$ and $\Delta S$ should be the negative of those given on that page for your case, making $\Delta G$ a positive number.


What's the mechanism of water catalyzing a reaction of hydrogen sulfide gas and sulfur(IV) oxide to yield sulfur? The reaction

$$\ce{2 H2S(g) + SO2(g) -> 2 H2O(l) + 3 S(s)}$$

goes in presence of a catalyst, usually $\ce{Al2O3}$ or $\ce{TiO2}$, but also just in presence of moist or water (YouTube video)

I don't understand the catalytic mechanism - why doesn't the reaction go without water and how exactly does presence of water help?

Yes, hydrogen sulfide and sulfur dioxide reacts with each other in the presence of moisture. The reaction has been studied previously and was noted that the reaction did not proceed if there was no water vapor(dry gases). The mechanism of  reaction was determined and was adsorption. The gases adsorped on the surface of water and collided with each other to deposit sulfur and water. The formation of water further catalysed the reaction. This is a excerpt from the paper which extensively studies the mechanism of the reaction.$\ce{^{[1]}}$

  The results of experiment 3 are best explained by assuming that both
  the  $\ce{H2S}$ and $\ce{SO2}$ are adsorbed and that reaction takes
  place only between these  adsorbed molecules.

There is another paper which extensively discuss the mechanism of reaction. It says that both the gases react with a reaction to form a solution. This is called Wackenroeder solution which is a mixture of sulfur based acid(polythionic acids). These acids dissociate in aqueous solution and form and water. Have a good read at the paper:$\ce{^{[2]}}$

  Volynskii proposed a mechanism for the reaction of $\ce{H2S}$ and
  $\ce{SO2}$ in aqueous solutions. The formation of sulfur follows the
  steps below.

  $$\ce{H2S + SO2 -> HSSO2H }$$ $$\ce{HSSO2H -> S(OH)2 + S }$$
  $$\ce{S(OH)2 + H2S -> 2H2O + 2S (or~S2) }$$

  The overall equation for the three steps is reaction 2. The formation
  of polythionic acids results from the interaction of ionized $\ce{SO2}$ and
  monatomic or polyatomic sulfur, i.e.,

  $$\ce{SO2 + H2O -> HSO3- + H+ }$$
  $$\ce{HSO3- + S -> HS2O3- }$$
  $$\ce{S_{x} + 2HSO3- -> S_{(x +1)}O6- + H2S }$$ $$\ce{S_{x} + 2HS2O3- ->S_{(x +2)}O6- + H2S }$$
  $$\ce{S_{x} + HSO3- + HS2O3- -> S_{(x +3)}O6- + H2S }$$

  where, x can range from 1 to 7. Tiwari believed that the reaction of
  $\ce{H2S}$ and $\ce{SO2}$ in dilute sulfuric acid (below 10 wt %)
  follows the scheme from (7) to (9). His kinetic data show that the
  reaction is first order with respect to both $\ce{H2S}$ and
  $\ce{SO2}$, consistent with the scheme because he assumed reaction 7
  to be the rate-controlling

Reference

$\ce{^{[1]}}$ http://pubs.acs.org/doi/abs/10.1021/j150272a005

$\ce{^{[2]}}$ http://pubs.acs.org/doi/abs/10.1021/ie020275i


Why are different colors, according to sugar concentration observed in Benedict's test? Colors range from green yellow orange to red. How does the sugar concentration and Cu2O concentration cause this?
I know Cu2O is red. Why is there a green or yellow color instead of just red or some kind of red?

Benedict's Reagent is just simply $\ce{Cu^2+}$ ions in the form of copper citrate. When Benedict's Reagent finds an aldose (a sugar with an aldehyde group), it can oxidize the aldose to a carboxylic acid. For example:

Since D-Glucose is oxidized, Cu is reduced to a red precipitate ($\ce{Cu2O}$). The colors range form green to red because the original Copper Citrate ($\ce{C6H8Cu2O7^4+}$) is blue in color.

Source: Quora


Could the difference between absorption time of Diclofenac Sodium and Diclofenac Potassium be related to the cation? Diclofenac is a common non-steroidal anti-inflammatory drugs (NSAID) that comes in a variety of formulations. Two of the most common forms are diclofenac sodium and diclofenac potassium (the links refer to the FDA prescribing information of both forms). I have heard several times that the Potassium salt starts to act faster than the sodium salt.

In the second page of both leaflets above there is a table with pharmacokinetic parameters:

For diclofenac potassium, Tmax (the time it takes to reach maximum concentration after one dose) is 1.0 hour, and the half-life (the time it takes the drug concentration to go down by 50%) is 1.9 hours.
For diclofenac sodium, Tmax is 2.3 hours, and the half-life (the time it takes the drug concentration to go down by 50%) is 2.3 hours too. This means that for the sodium salt is being "processed" (absorbed, metabolized, eliminated) by the body more slowly than the potassium salt.

One more important point of difference to note is that the sodium salt comes in enteric-coated tablets, while the potassium salt comes in immediate-release tablets.

My question is: Can the difference in "processing time" of both salts by the body stem from the different salts, or is it due to the pharmaceutical formulation (EC tablets compared to IR tablets)?

It might seem at first glance that the difference in formulation should be the answer, but if it is so, why is the cation different between formulations?

I hope anyone can shed light on the mechanism, I was unable to find any useful information that actually addressed this issue.

Under physiological conditions, the counter-ion is almost entirely dissociated from the drug, meaning that the identity of the counter-ion isn't directly responsible for how well a drug performs. This has been looked at extensively by medicinal/formulation chemists, who often screen many salt forms looking for desirable properties.

From the graph below, you can see that the plasma concentration curves are essentially identical for sodium, potassium, and calcium salts (slight variation in Cmax and Tmax but within experimental error):

Source: Drug Like Properties: concepts, structure, design and methods, Chapter 7

One case in which the identify of the cation does matter is where there is a common ion in the body. This 'common ion' effect is most common when amine drugs are presented as their hydrochloride salts. The stomach is acidic (hydrochloric acid), and therefore the presence of a high concentration of chloride ions can prevent the salt from turning into the free drug.

What is likely to be different between different cations the form that the salt takes. Different salt forms have different packings (some might be amorphous, some might be highly crystalline and ordered). The formulation may also differ (powder coated tables vs gelatine capsules), which will change the time it takes for the drug to be released/absorbed.


PLP-Dependent Cyclization and Decarboxylation I am looking at Compound A (see below) and am trying to figure out a decarboxylation mechanism that results in Compound B (see below). I have come up with an internal mechanism that might work that I have attached to this question. However, I am trying to connect it to enzyme catalysis, and I am thinking that a PLP-dependent enzyme is involved I just cannot quite figure out the mechanism.

A PLP-dependent enzyme could certainly accomplish a reaction like this. An example of something similar is the PLP-dependent enzyme KAPA synthase in the biotin biosynthesis pathway. In that reaction, alanine is decarboxylated and a new bond is formed between the alpha carbon and an electrophilic thioester. Interestingly, the proposed mechanism involves deprotonation of the alpha carbon rather than decarboxylation as the initial step. Only after the new C-C bond is formed is the carboxylate lost:

1: Source: Webster, S.P, et al. (1998) Characterisation of S-amino-7-oxononanoate synthase: A bacterial PLP-dependent, acyl CoA condensing enzyme. Biochemical Society Transactions 26(3):S268. DOI: 10.1042/bst026s268
Although this is an intermolecular reaction, your proposed intramolecular reaction should be even easier to accomplish. An example of a PLP-dependent enzyme catalyzed reaction that results in intramolecular cyclization (but not decarboxylation) is that of aminocyclopropane carboxylate synthase.


How do I calculate the amount of preservatives one would need for a cleaning solution? I plan on making a very simple cleaning solution to clean metal contacts on PCB boards. It cannot contain anything that is flammable or hazardous so I've come up with the solution below.

Distilled Water ~92%
Ethoxylated Alcohol (CAS 68439‐46‐3) ~5% "surfactant"
Tetrasodium Glutamate Diacetate (CAS 51981-21-6) ~1% "chelating agent"

I am not making this solution in a perfectly sterile environment so I would like to add a preservative to increase shelf life. I've been researching for weeks and I cannot find anything definitive. I just want something that will easily dissolve into the solution above and not be harmful to metal. The solution will be applied to the metal via a cotton swab. Even if I find a suitable preservative how would I go about calculating how much to use?
My ideas for preservatives:
I think Citric Acid (CAS 77-92-9) would be okay. I think a combination of Methylchloroisothiazolinone (CAS 26172-55-4) and Methylisothiazolinone (CAS 2682-20-4) would be okay as well.
Other concerns are: Do I need an emulsifier? Is there an additional non-hazardous chemical I can add to decrease drying times?

Most producers/suppliers have recommendations for concentration ranges of their products. If you can´t get it directly, it is often possible to find the information on their websites (e.g. Schülke or BASF).
The ranges are typically around 0,1% - 0,5%, but different products are often mixtures of different preservatives and with different concentrations, so it can be difficult to compare products directly.
Citric acid has some biocidal effect but is not very efficient, so I would not recommend it.
Why do you need a chealant? You use demineralized water and a nonionic surfactant, so ions in the product should not be a big issue.
The CAS-number you refer for the surfactant covers a variety of chemicals, with different C-chains and degrees of ethoxylation. They are generally classified as mostly soluble in water, but it may be necessary to add a hydrotrope (e.g. Cumenesulfonate) or co-solubilizer (e.g. Butylglycol).


Is it correct to say that ethenol is an alcohol? IUPAC Gold Book defines alcohols as:

  Compounds in which a hydroxy group, $\ce{–OH}$, is attached to a
  saturated carbon atom $\ce{R3COH}$.

However, Wikipedia says that:

  Vinyl alcohol, also called ethenol (IUPAC name), is an alcohol.

Is Wikipedia wrong about this?

And if ethenol is not an alcohol, what would be the correct name for compounds in which an $\ce{–OH}$ group is attached to an unsaturated carbon atom?

The golden book states that ethenol is an alcohol:

  Enols

  Alkenols; the term refers specifically to vinylic alcohols, which have the structure $\ce{HOCR'=CR2}$.

However, the modifier "vinylic" is rather important, so I would never omit it.


Mechanism of sulfur trioxide reaction with water to make sulfuric acid This question touches on whether sulfur is hypervalent (whether it expands its octet when bonding).

The first picture shows both sulfurs with an expanded octet.

The problem is that in the top half of the below picture, we are supposed to end up with sulfuric acid, but we actually end up with sulfuric acid plus two lone pairs (if I drew the mechanism correctly). Did I? (I actually separated out the top part of the picture into two parts - first, hydrogen attacking oxygen; second, oxygen attacking sulfur - so the arrow pushing wouldn't get confusing).

The second picture follows updated chemical fact. According to Wikipedia, sulfur trioxide has one double bond and two dative single bonds. Sulfuric acid is also a charge separated species with four single bonds, not two single bonds and two double bonds. I seem to end up with the proper product here without any hand-waving.

I've drawn a more correct mechanism for the reaction of dilute $\ce{SO3}$ with water in the liquid phase:

$$\ce{SO3 (aq) + 3H2O (l) -> SO4^2- (aq) + 2H3O+ (aq)}$$

$\ce{SO3}$ is a strong electrophile, enough to react quickly with water, which is a relatively weak nucleophile. A water molecule is added to the structure, facilitated by the dislocation of a $\ce{S=O}$ bonding electron pair in $\ce{SO3}$ to the oxygen atom, forming the intermediate structure shown. Additional water molecules then quickly abstract hydrogen atoms from the intermediate (actually, the first abstraction is far easier than the second, so it might have been more proper to show each proton abstraction separately in sequence). The final product of the reaction is sulfate anions and hydronium cations ($\ce{H2SO4}$ is a strong acid in water, so it doesn't really exist as a molecule. The second proton isn't nearly as acidic as the first, but in dilute solution both are ionized). The formal global reaction is shown below the mechanism. Notice that at every point the formal charge on the central sulfur atom is exactly the same, zero.

Now, with regards to our little discussion about the actual structure of $\ce{SO3}$. Classical valence bond theory, enhanced with the concept of hypervalency and VSEPR theory, attributes the structure shown, with three double bonds. The Wikipedia article for $\ce{SO3}$ mentions another structure, based on modern valence bond theory computational calculations, with one double and two single bonds. This structure is quite interesting as it does not invoke hypervalency. Indeed, hypervalency is thought to be an old concept, and not really necessary in more modern theoretical treatments of chemical bonding.

However, if one opts to display the structure according to modern valence theory, they would do well to make it explicit. Almost all structural formulae today are drawn in the context of classical valence bond theory. Certain assumptions from the classical theory do not carry over to the modern theory. For example, the authors of the article "Modern valence bond theory", Chem. Soc. Rev., 1997,26, 87-100 (free access mirror) find that the bonding in $\ce{N2O}$, described classically as a mixture of the canonical forms $\ce{^{-}N= N^{+}=O}$ and $\ce{N≡N^{+}-O^{-}}$, is best described as $\ce{N≡N=O}$. Even with five bonds, the central nitrogen atom has a calculated occupancy close to 7, i.e. it is not hypervalent. Since the structural construction rules of classical and modern valence bond theory aren't the same, the structures generated from both do not mesh and cannot be compared directly. The simplest thing to do, then, is to draw $\ce{SO3}$ according to classical valence bond theory, with three double bonds.


Thermochromism in PbI2 I noticed while heating some PbI2 powder that it reversibly changes from a yellow to a dark red. However, I was unable to find anything on this thermochromism.

There are related cases in TlI and AgI but there it is mostly about occupying the free sites in the crystal structure or if Ag is present there is also the possibility of Ag coming close to Ag (so-called argentophilia). This, however, doesn't seem to be the case for Pb(II).

It is also no thermal charge transfer, for Hg(II) + I(-) a thermal charge transfer to Hg(I) + I(0) was often mentioned and it makes sense there but Pb(I) is not really useful here.

The structure is also different forming layers. I read they have weak VdW-bonds so perhaps those break in the process of heating it? If impurities are introduced there is the possibility of forming a perovskite structure but I doubt this would be reversible. So this has to be a smaller shift in the structure.

PbI2 has quite many different room temperature and high-temperature polytypes, so perhaps it's really a change in the crystal structure? I couldn't really find much on this because I don't know how the colors change for the different polytypes.

Does anyone have an idea what might be the dominating effect for a reformation in the structure, like an actual effect that might take place here?

You see often it is the case in these solid-state thermochromic compounds, that the high-temperature form seems to be the best situation, but cooling it down allows weaker forces like this argentophilia, for example, to distort the structure a bit. Like TlI is more or less an orthorhombically distorted NaCl in its low-temperature form.

Given the fact however that Pb(II) still has a lone-pair that can be positioned quite well in the gap between the layers I don't see a real demand for any change at all.

Just found something:

So the most dominant thermal change, and we are talking here about annealing single crystals for several days to around 150°C seems to be a polytypic change from the 2H-Type to the 12R-Type. Or, as there are several hexagonal and rhomboedric phases let's just assume a change from H to R. You may know polymorphs, like allotropes for example, diamond vs graphite or different phosphorus allotropes. And polytypes are structures that usually appear in layers where the overall composition and connections are pretty much the same but the stacking order of the layer is different. I found a sketch how the layers have to change:

1 E.Salje, B.Palosz, B.Wruck, Structural and Magnetic Phase Transitions in Minerals, Advances in Physical Geochemistry, vol 7, Springer, New York, p.231, 1988.

It was often reported that the change from H to R seems to be thermally while the change from R to H takes a long time. As the crystals were grown from sodium based gels traces of sodium could be found and the higher the sodium content the easier and faster they turned back to H. I didn't grow any crystals but precipitated it using NaI, so I probably have a high sodium content which acts as an impurity to fastly change the structure back, which appeared to me as being reversibly thermochromic.

I don't know whether this also causes the color change but many other color changes, like in TlI depend on minor distortions in the crystal structure as well so it could be plausible.


Solubility: Is the common ion effect proportional to individual solubility values? I'm trying to grasp how solubility works in real applications. Is it so that in a solution with only one salt, it's solubility will change upon the addition of another salt with a common ion in a manner strictly proportional to the latter salt's solubility in water alone? (i.e. that highly soluble salts will decrease the solubility of common ion salts the most and vice versa) If so, is the solubility of each salt in a combined solution a ratio of their respective solubility values when alone in the solution?

There are basically 3 effects.

The solubility is controlled by the solubility product, the constant being the multiplication of ion activities

$$K_\mathrm{sp}={a_\mathrm{M}}^m\cdot  {a_\mathrm{X}}^n$$
for a salt $\ce{M_{m}X_{n}}$

For low concentration of ions, it can be approximated by concentrations, related to activities via activity coefficients $\gamma$, that are supposed to be equal to 1.

$$K_\mathrm{sp}={c_\mathrm{M}}^m\cdot  {c_\mathrm{X}}^n$$

where $a_\mathrm{M}=\gamma_\mathrm{M} \cdot c_\mathrm{M}$, the same for $\text{X}$

Therefore the presence of other salt with the common ion decreases the salt solubility, as the common ion contributes to the rate of precipitation.

The second effect is general effect of dissolved non common ions, given by ionic strength

$$I=\sum_i {c_\mathrm{i}\cdot {z_\mathrm{i}}^2}$$

and its effect on the ion activity, given for diluted solutions approximately by Debye-Hueckel equation

$$\log \gamma=-0.5 \cdot z_+ \cdot z_- \cdot \sqrt I$$

where $\gamma$ is the mean activity coefficient for given salt and $z$ are respective ion charges. The coefficient $0.5$ is rounded up.

For a particular ion, it is often written as

$$\log \gamma=-0.5 \cdot {z}^2 \cdot \sqrt I$$

For less diluted solutions, it is often modified as

$$\log \gamma=-0.5 \cdot {z}^2 \cdot \frac {\sqrt I}{c + \sqrt I}$$

where $c$ is ion dependent constant close to 1.

Generally, the higher is overall ion concentration, the higher is salt solubility. But the effect is smaller than effect of common ions.

For high concentration of general salts, ion activity coefficients revert their decreasing trend and start raising. For very high concentration to values much higher than 1.

As result, the solubility decreases.


How to understand Nelson-Obert charts? The compressibility factor is given by:
$$
z=\frac{P\overline{V}}{RT}
$$
The Nelson-Obert chart is:

This chart brings the following information:

At very low pressure (Pr << 1), gases behave as an ideal gas regardless of temperature.
At high temperatures (Tr > 2), ideal gas behavior can be assumed with good accuracy.
The deviation of a gas from ideal gas behavior is greatest in the area of the critical point.

Where:

$$
T_R=\frac{T}{T_{(At\ critical\ point)}}\\
P_R=\frac{P}{P_{(At\ critical\ point)}}
$$

What I don't understand is:

Does this chart describe the ideal behavior of a supercritical fluid or a superheated gas or plasma?

Having in mind any phase diagram:

It describes the entire region to the right of the vertical dashed line through the critical point.


Does Magnesium metal react with carbonic acid? I use rainwater in my house for my water supply.  I collect the water in plastic lined tanks.  Some of my piping at home is copper, and I have blue stains in my shower, because I think the copper from my pipes is being dissolved by the carbonic acid in the rainwater.  I'm planning to hang magnesium rods into my rainwater tanks, in the hope that these rods will react with the carbonic acid instead of my copper pipes.  Does that make sense?

First, thanks for reporting some obvious unexpected chemistry occurring. Others in your community may also have similar metal leaching issues, which especially for children, may be a health concern (see, for example, this article).

My speculation, on what is occurring, is that the Iron pipe in your shower is displacing (upon standing over time) some dissolved copper (possibly from created bicarbonate). The likely acidic rainwater may also contain chloride or nitrate salts, which may further complex with the fresh active copper precipitate. Here is a related illustrative source, per an article 'Electrochemical detection of nitrate and nitrite at a copper modified electrode' to quote:

  The development of a reagentless electrode system for the determination of nitrate is presented. The approach is based upon the deposition of a macroporous copper deposit which shows marked selectivity for nitrate ion under mildly acidic conditions (pH 3) with a linear range extending from 10 to 200 mM nitrate.

So, do not employ Magnesium, address the iron presence. The coloration means there is some, but in actual amounts, most likely limited copper ion presence along with nitrate.


Could a strong acid become a strong base? Is it technically possible for a strong acid to change into a strong base, through a chemical reaction?

Example: if an acid with pH 1 became a base with pH 12.

Take nitric acid for instance. Reduce the nitrate until ammonia forms. Then using a stronger base than amide ion, you can generate amide base.

Therefore, it is required to make a number of reactions to get a strong base from a strong acid. If you meant that could a strong acid be changed to a strong base by one step reaction, it might be possible with complex big organic or organoboron molecules by replacement of acid functional group with a basic one, but I am unsure about this part.


Primary and secondary standard Why are secondary standard solutions used in titrations? Why don't we always use primary standard solutions?

A primary standard is critical for an analytical result. If we dissolve a known mass of a material into a known volume, we must be able to know its concentration with confidence. Such a standard must be easily made very pure (> 99.95%), it should not decompose, and it must be fairly robust to say heating or vacuum. We use such standard solutions as, for example, titrants. This is surprisingly hard to come by, and so we say that secondary standards are materials that do not meet these strict criteria.

Say you want to titrate a sulfuric acid solution for quantitative work, you could use a known concentration of NaOH or KOH and volumetrically determine the concentration of the sulfuric acid solution. In fact, dissolving pellets of these particular bases in a known volume of water is not sufficient to calculate concentration with any confidence in analytical work because the solids are contaminated with carbonate from atmospheric CO$_2$. In general these carbonates contaminate alkaline solutions quite readily through this reaction: $\ce{OH- + CO2 -> HCO3-}.$ This is why we would have to titrate the concentration of $\ce{NaOH} $ solution with a primary standard before using it as a secondary standard.

Now there are acid and base primary standards available, so why would one ever need to use a secondary standard? Usually it’s a matter of specific needs and convenience. Whenever possible, use a primary standard if your results actually matter.

Let us look at a specific case where a secondary standard is useful: potassium permanganate is exceptionally useful as a titrant because it serves as its own indicator where the presence of a purple color signifies the end point, but it is unstable in light or heat due to this reaction: $\ce{4 MnO4^- +2H2O -> 4MnO2 + 3O2 + 4OH-}$. This is why solutions of it must be stored in a dark cool place; even still, we must standardize the concentration of such a solution and we can only trust its measured concentration for a short time thereafter.

A secondary standard solution of $\ce{KMnO4}$ is usually titrated by oxalic acid, a primary standard, and then used to determine the concentration of a number of other analytes.

The reaction with oxalic acid can be summarized as $\ce{5 H2C2O4 + 2 MnO4- + 6H+ -> 10CO2 + 2 Mn^{2+} + 8H2O}$ where everything except $\ce{MnO4-}$ is colorless. The end point of this titration is a marked by a purple color due to unconsumed $\ce{MnO4-}$. This allows you to make a standardized solution of potassium permanganate that can be used for a short while.

Now, how would we use this? Solutions containing $\ce{Ca^2+, Pb^2+, Ag+, Mg^2+, Co^2+, Zn^2+}$ and etc. can be precipitated as a metal oxalate. You would then digest this ppt in an acidic solution and titrate the released $\ce{H2C2O4}$, where the end point is marked by a purple color, allowing you to infer the original concentration of the metal ion in the original solution.

$\ce{KMnO4}$ was used in mineralogical surveys of uranium prior to the Manhattan project, where better techniques were developed to quantitate.


Assigning Re and Si face for Carboxylic group Suppose I have a compound $\ce{CH3COOH}$ and I have to assign Re and Si faces. Which of the two oxygens get a higher priority and why?

The carbonyl oxygen gets higher priority because it has a real oxygen bonded via $\sigma$ bond and also the double bond, but we count that as the oxygen bonding back to the carbon. We don't treat it like a real carbon; it's a "ghost" or "phantom" carbon with nothing attached to it. This carbon counts as a carbon but is lower priority than any other real carbon. That means that -O-(C) outranks -O-H.


Finding is the concentration of H⁺ in a benzoate buffer solution after adding HCl Q: A buffer with a $\mathrm{pH}$ of 4.44 contains $\pu{0.21 M}$ of sodium benzoate and $\pu{0.12 M}$ of benzoic acid. What is the concentration of $\ce{[H+]}$ in the solution after the addition of $\pu{0.050 mol}$ of $\ce{HCl}$ to a final volume of $\pu{1.3 L}$?

A: So these are the steps that I have taken for this:
The reaction is:
$$\ce{C6H5COOH +H2O -> C6H5COO- + H+}$$
The $K_ \mathrm{a}$ of benzoic acid is $6.3 × 10^{-5}$
But, $K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}$
We don't know the concentrations of $\ce{[H+]}$ so we solve for that. We then assume at equilibrium $\ce{[C6H5COO-]}$ is equal to it.
Then: $6.3×10^{-5}=\ce{\frac{[C6H5COO- ][H+]}{[C6H5COOH]}}$ and
$\ce{[H+]}=0.00272$
But this is wrong. I think I'm doing something wrong in my problem, but really can't figure out what.

Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid:
$$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+_\mathrm{(aq)}}$$
When you add  $\ce{HCl}$ to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)}  <=>  C6H5COOH + H2O }$$
Now, the equilibrium constant of this reaction is:
$$K=\frac{1}{K_ a}=0.16 \times 10^5>10^3$$
The reaction is quantitative (total). So, we can rewrite it:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)} ->  C6H5COOH + H2O }$$
Now, we have to construct an ICE table:
Assuming that we didn't dilute the solution when we added $\ce{HCl}$ (this point is not clear in your question): $$\ce{[C6H5COOH]_0} = \pu{0.12 M}$$
$$\ce{[C6H5COO^-]_0} =  \pu{0.21 M}$$
$$\ce{[H^+]_0} = (\frac{0.050}{1.3} + 10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$
As you can clearly seen $\ce{[H^+]_0}$ is the lowest, and so ion hydronium is the limiting reactant.
At the end of the reaction, we have:
$$\ce{[C6H5COO^-]_\mathrm{(eq)}} = (0.21-0.0385) = \pu{0.1715 M}$$
$$\ce{[C6H5COOH]_\mathrm{(eq)}} = (0.12+0.0385) = \pu{0.1585 M}$$
$$\ce{[H^+]_\mathrm{(eq)}} = \epsilon$$
Substituting all these numbers in the expression of $K$
$$K = \frac{ 0.1585}{0.1715\epsilon}=0.16 \times 10^5$$
By solving the above equation, we find:
$$\ce{[H^+]} = \epsilon = \pu{5.82 \times 10^{-5} M}$$
$$\mathrm{pH} = 4.23$$


What does this lambda–sigma notation mean? I am currently reading the CRC Handbook of Phosphorus-31 Nuclear Magnetic Resonance Data by John C. Tebby (CRC Press, 1991) and on several figures (ex. pages 9 to 14) there is a caption that reads along these lines

  $\ce{^{31}P}$ NMR chemical shifts of three coordinate (λ3 σ3) phosphorus compounds. (Tables B to E.)

What does the (λ3 σ3) signify? I've not seen that notation before, and it is on the captions of some of the charts I need.

The notation refers to the valency (number of valence electrons involved in bonds, i.e., either 3 or 5) and coordination number (number of substituents attached) in organophosphorus compounds. For example, it could refer to the tautomeric forms of phosphonate esters:1

So the λ3σ3 refers to a P with 3 bonding valence electrons (the lone pair doesn't count) and 3 substituents attached while λ5σ4 refers to a P with all five of its valence electrons involved in bonds and 4 substituents attached.

Reference

Kraszewski, A.; Stawinski, J. H-Phosphonates: Versatile synthetic precursors to biologically active phosphorus compounds. Pure Appl. Chem. 2009, 79 (12), 2217–2227. DOI: 10.1351/pac200779122217.


How do hydrogen bonds break the ionic bonds of a crystal lattice? In the case of water dissolving sodium chloride. I was under the impression that hydrogen bonds were much weaker than the ionic bonds in sodium chloride, so how do they overcome the ionic bond strength? Do the ions simply get swarmed with enough water molecules so that the combined dipole interaction forces break the ionic bond? Or is it possible that because the sodium chloride dissolution reaction is endothermic the hydrogen bonds on their own are not enough to break the ionic bonds so the reaction must draw energy from the surroundings to do so (the reaction must be entropically favorable to do so, right?)?

Your presumption that it is hydrogen bonding that breaks the ionic bonds is wrong. Forget H-bonds. It is true that they exist in both pure water and in a sodium chloride solution. Water molecules surround both the Na+ and Cl- ions. The interaction there is dipole-ion. Na-OH2 and Cl-HOH. As is obvious, only one of those involve an H atom. (note that the "-" I used is not intended to suggest either a covalent or an ionic bond, rather it is intended to show the electronic interaction between the two oppositely charged species (the Cl & Na ions have full integer charge, the H & O atoms have only partial opposite charge, mostly induced.) This question can be answered just by hand-waving about the thermodynamics: the energy of the solvated Na and Cl ions is less than that of those two ions in the crystal. So, on average, the ions will leave the crystal and enter the liquid (until their concentration is so high that equilibrium is established between the two phases). So, while it is true that HOH...Cl bond can be called a H-bond, thinking about it this way has no obvious advantage as far as I can see. The dipole-ion interactions that are occurring are indiscriminate. Each ion is surrounded by lots of (oriented) water molecules. To learn more see https://en.wikipedia.org/wiki/Solvation_shell and especially https://en.wikipedia.org/wiki/Solvation


Why is a spin change favourable in intersystem crossing? All phosphorescent molecules go through the following transitions:
 $$\text{excited singlet state}$$
$$\Bigg\downarrow$$
$$\text{[intersystem crossing]}$$
$$\Bigg\downarrow$$
$$\text{ excited triplet state}$$
Why does that spin change always occur, given it’s a “forbidden” transition and, thus, unfavourable and much less likely to take place?

There are two things going on here. You have conflated with "forbidden" with unfavorable. In many systems, the first excited triplet is lower in energy than the first excited singlet. Hence, the order of energies excited singlet > triplet > ground state singlet is correct and sensible. If your question is "why is this the order?" that should be a separate question.

The forbidden part of the transition has to do with the fact that, because of conservation of (electron spin) angular momentum, you can't excite directly from a singlet ground state to a triplet excited state with a photon. The forbiddenness is a statement about not being able to get there, not unfavorability. The triplet is still lower in energy and the molecule has to do some other shenanigans (often involving spin orbit coupling) to do the intersystem crossing into the lower energy triplet state. If your question is "what is the mechanism of intersystem crossing?", that should be a separate question.


What is the total number of isomers, considering both structural and stereoisomers of cyclic ethers with the molecular formula C4H8O? According to me, it should be 11 and this is how I have worked out as shown in the attached diagram.

However, I am not sure whether all these 11 isomers can be considered as those belonging to cyclic ether.

I had erroneously uploaded incorrect diagram earlier. Now I have uploaded the correct diagram. My specific query is whether the 11th Isomer shown in this diagram, wherein Oxygen is not part of the ring, can be considered as Cyclic Ether.

I would not include the last structure, even if it was clever of you to think about it.

I think that the term "cyclic ether" restricts the scope to "cyclic compounds which have an ether moiety inside the ring". Your last compound would be considered as a linear ether with an alkyl and a cycloalkyl substituents.

Oh and your 6th structure does not have 3 isomers but just 2 because the (R,R) an the (S,S) isomers are actually equivalent by rotation of the molecule. This compound is said to be "meso".


SN1 reaction of ethanol at tertiary alkyl halide I know this undergoes an $\ce{S_{N}1}$ reaction:

Ethanol is the solvent. Weak nucleophile. Neutral nucleophile. This points toward $\ce{S_{N}1/E1}$ (competing pathways).
The leaving group is on a tertiary carbon. This precludes backside attack, $\ce{S_{N}2}$.
The professor labeled this reaction as $\ce{S_{N}1}$.

My question: where's the substitution part of this reaction? I don't see any substitution. Is there supposed to be no reaction here? From what he wrote it looks like a bromine left as bromide ion and a hydrogen appeared out of nowhere.

My expected transition state is a tertiary carbocation, which will not rearrange, and then the nucleophilic oxygen part of the ethanol molecule attacks the carbocation. Bromide ion deprotonates the oxonium ion, and I get an ether.

Who's right?

Just to leave this question not unanswered. As ron already pointed out, you reasoning is correct. Based on the information given, the ethanol may perform a nucleophilic attack at the carbocation. To a very minor extent you might find the elimination product, but I guess you will have to cook it quite well. The elimination pathway is more likely to occur, when there is a stronger base than ethanol.

Just to get your terminology right. The carbocation is a reactive intermediate, not a transition state. The $\ce{S_{N}1/E1}$ reaction have always (at least) two transition states. The first one corresponds to the carbon leaving group bond break and the second one corresponds to the nucleophile carbon bond formation, proton abstraction, respectively.


Electrochemical cell polarity and electrodes In an electrochemical cell, the anode undergoes oxidation and the cathode undergoes reduction. This means that the anode loses electrons while the cathode gains electrons.

However, the anode is negatively-charged while the cathode is positively-charged. Why is this so?

The electrons which are "given up" by the oxidized species at the anode flow from the anode through the electric circuit to the cathode, where they are "needed" for the reduction reaction. The oxidation creates an excess of electrons (negative charge) at the anode, which is compensated by the consumption of electrons at the electron-deficient cathode (positive charge) for the reduction reaction. In summary, the flow of electrons in the electrochemical cell is from anode (electron source) to cathode (electron drain).


Why does only the cis-isomer of potassium dioxalatodiaquochromate(III) form? A friend of mine performed a standard synthesis reaction of potassium dioxalatodiaquochromate(III) by adding potassium dichromate to ethandioic acid dihydrate and a little water. The synthesis instructions suggested that only the cis-form is created. When looking online about this, I found numerous dodgy documents corroborating this idea, such as this or this.

Why is this? I can't find an explanation as to why only one stereoisomer forms nor any mechanisms anywhere.

I'm not an inorganic chemist, but to my understanding, both cis- and trans-dioxalatodiaquochromate(III) ions exist (Note that some also called it as Diaquabis(Oxalato)Chromate (III)). According to what I found online is that solid state reaction gives cis-form while aqueous reaction predominantly forms trans-isomer (it is mentioned in one of your attachments). However, it's still up to debate.

Hamm in 1952 had shown that trans—cis Isomerization using UV-vis spectroscopy ($trans \rightarrow cis$; Ref.1). According to Hamm, all previous workers have prepared trans-isomer because its lack of solubility. Yet, the existence of cis- and trans-isomers has been first recognized by Werner and has presented the preparation methods of two isomers individually (Ref.2), thus it is not true that the your concept of only the cis-stereoisomer forms.

Werner had mentioned that aqueous solution of trans-isomer is stable only in the cold temperatures. That may be the reason why only cis-isomer forms in solid state preparations because the preparation needs heating at high temperature.

Using the method given by Werner, trans-dioxalatodiaquochromate(III) isomer has been prepared and its crystal structure has been studied (Ref.3 & 4).

Later studies have showed this trans-isomer converted to cis-isomer by long standing in aqueous solutions (Ref.5). Based on their results, the authors have concluded that these color changes (dark red trans-isomer to blue cis-isomer) are due to trans–cis isomerization, uncomplicated by olation phenomena and that the transformation to the cis configuration proceeds to practical completion. They also stated that:

  Kinetically, the reaction (trans–cis isomerization) obeys a first-order law, and the dependence of the observed velocity constant upon temperature is found to be adequately described by the Arrhenius equation, where $A = 3 \times 10^9$ and $E = \pu{−17.9 kcal/mol}$.

Finally, in recent studies have been done on computational analysis of trans- and cis-Isomers of Potassium dioxalatodiaquochromate(III). The studies revealed that  cis-Isomer is slightly stable than trans-Isomer (Ref.6):

  The trans- and cis-isomers of potassium diaquabis(oxalato)chromate (III) were studied computationally and experimentally. The structures of trans- and cis-configurations of $\ce{[Cr(H2O)2(C2O4)2]-}$ were optimized by DFT methodology with various functionals namely: B3LYP, CAM-B3LYP, TPSS, PBE, M06-L and ωB97X-D along with the more sophisticated MP2 method. The calculations show that the most stable forms for both isomers are in quartet states. The results from all DFT methods reveal that the cis-isomer is literally more stable than the trans-isomer with the lower average relative energy of $\pu{2.1 kcal/mol}$. These are consistent with the results from MP2 calculation and experimental observation. The absorption wavelengths for the excited states of trans- and cis-structures were calculated by the time-dependent density functional theory (TDDFT) method. For the experiments, the trans- and cis-isomers of potassium diaquabis(oxalato)chromate (III) were synthesized and characterized by UV-Vis spectrophotometry. Both isomers have two maximum absorption wavelengths at $415$ and $\pu{560 nm}$.

References:

Randall E. Hamm, “Complex Ions of Chromium. II. trans—cis Isomerization of Potassium Dioxalatodiaquochromate(III),” J. Am. Chem. Soc. 1953, 75(3), 609-611 (https://doi.org/10.1021/ja01099a026).
A. Werner, W. J. Bowis, A. Hoblik, H. Schwarz, H. Surber, “Über Metallverbindungen mit komplex gebundener Oxalsäure. Zweite Mitteilung: Über Dioxaloverbindungen),” Justus Liebigs Annalen der Chemie 1914, 406(3), 261-331 (https://doi.org/10.1021/ja01099a026).
J. N. van Niekerk, F. R. L. Schoening, “Structure of Trans-Potassium Dioxalato-Diaquo-Chromiate,” Nature 1950, 166, 108 (https://doi.org/10.1038/166108a0).
J. N. van Niekerk, F. R. L. Schoening, “The crystal structure of trans potassium dioxalatodiaquochromiate, $\ce{K[Cr(C2O4)2(H2O)2].3H2O}$,” Acta Cryst. 1951, 4(1), 35-41 (https://doi.org/10.1107/S0365110X51000064).
G. E. Cunningham, R. W. Burley, M. T. Friend, “Potassium Dioxalato Diaquo Chromium (III) Complexes: Trans–Cis Isomerization,” Nature 1952, 169, 1103 (https://doi.org/10.1038/1691103a0).
Songtham Ruangchaithaweesuk, Juthathip Chorkate, Thana Maihom, Potjaman Poolmee, Piti Treesukol, Tipawan Rungsawang, Phornphimon Maitarad, Bundet Boekfa, "Combined Computational and Experimental Studies of Trans- and Cis-Isomers of Potassium Diaquabis(Oxalato)Chromate (III)," Key Engineering Materials 2017, 757, 103-107 (https://doi.org/10.4028/www.scientific.net/KEM.757.103).


Which graph shows Chemical Equilibrium? So my teacher started the explanation of Chemical Equilibrium saying..
The concentration does not change with time.
And, gave us certain graphs asking which ones show chemical equilibrium.
I cannot understand what's written in my copy since I took the lesson two years ago. I need to be explained what I wrote in my copy.

I know that the concentration does not change with time.
If concentration is same or not, does not matter. That only the rate of forward reaction and the rate of backward reaction needs to be same.

These are the graphs: (I am sorry I could not draw better.)
:
Now, I know 1,4,5 and 6 show chemical equilibrium.Reasons:

The concentration of the reactant is lost but same amount is achieved by the product.So here, the rate of forward reaction and rate of backward reaction is same.
For 5 and 6- The concentration of reactant and product may not be the same but it does not change with time.

I need help regarding :

1.Explanation as to how the other graphs do not show chemical equilibrium(if possible in terms of how the answer for the other graphs are explained).

2.Explanation for graph 5 and 6. I cannot understand the reason I had written.

3.An overall explanation and answer of the question.

Thank you very much for your help and understanding!!

Firstly, you need to understand what it means when a system is in equilibrium. Equilibrium is the state of a system where the rate of the forward and reverse reaction are equal.

Note: this doesn't mean that the reaction completely stops, all this means that the concentration of the reactants and products remain constant.

A simple example of this can be shown in an experiment where a cube of sugar is placed in a saturated solution of sugar. If you leave it for a few days and come back, you will find that the shape of the sugar cube has changed. Instead of being a smooth, perfect cube, it is a irregular shape. However when if weigh the solid sugar, you find that it will weight exactly the same as the original sugar cube. Now this can be explained by the fact that the solid sugar is in equilibrium with the dissolved sugar in the saturated solution. Despite the forward and reverse reaction being equal (as no more sugar will dissolve) the reaction is still occurring. At a microscopic level, glucose molecules in the sugar cube are dissolving into the solution while dissolved sugar molecules precipitate out onto the sugar cube. However where they precipitate on the cube is random, hence they don't form a perfect cube. This is why the sugar cube changes shape but has the same original mass.

To do this question, it is important that you understand the concept that the reaction still occurs when a system reaches equilibrium. Now, lets look at each graph separately (remember, when we are looking at these graphs, we are only interested at what is happening at the end, what happens at the start is irrelevant):

1: Here the forward and reverse reaction are equal, so from our definition of equilibrium, this graphs shows a system in equilibrium. You should note that despite the system being at equilibrium, the reaction is still going

2,3: Here in both graphs the forward and reverse reaction are not equal. Therefore the system can't be in equilibrium. This is because in graph 2, the rate of the backward reaction is greater than the rate of the forward reaction. Hence the concentration of the products will be decreasing while the concentration of the reactants will increase. Vice verse for graph 3

4,5,6: Here, since the concentration of the reactants and the products are not changing over time, this means that the forward and reverse reaction is the same. Therefore, the system is in equilibrium

7: In this graph, the concentration of the products is increasing while the concentration of the reactants is decreasing. Hence the forward reaction must be greater than the backward reaction. Therefore, the system isn't in equilibrium.

So to recap the ideas that we heave just learnt, here is the  2 main things to check to work out if a system is in equilibrium or not:

Is the forward and backward reaction equal? if they are equal, this means that the system is in equilibrium. If they aren't equal, the system isn't in equilibrium
Is the concentration of the products and the reactants constant over time? if they are constant, the system is equilibrium. Note, the concentration of the products and reactants don't have to be equal. If concentration of the product and reactants are aren't constant, the system isn't in equilibrium.


R,S in Cyclic Compounds - Path of Highest Priority? Consider (R)-3-bromo-1,1-dimethylcyclohexane.

I wish to clarify how I assign R,S priorities in a cyclic compound. My book says to compare the atoms along the "path of higher priority." The problem is in a cyclic compound, if you go far along this path of higher priority without finding a point of difference, you may collide (picture below):

I'm assuming that because the path taken initially results in no difference, we can still assign the top path a higher priority than the lower path because the top path has two methyl groups. My book says that if the paths are identical then choose the one with more of the same atoms.

If you "collide" without finding some difference, then the molecule is not chiral.  In the example at hand, it appears that you have analyzed it correctly.  First, you look at the 2 carbons adjacent to the chiral center.  They are both $\ce{CH_{2}{'}s}$; no difference, therefore move out to the next carbon on each side of the ring.  Here we find a $\ce{C-CH_{2}-C}$ on one side and a $\ce{C-CMe_{2}-C}$ on the other side.  The latter has higher priority than the former.  Stop at this point, no need to go further, and complete the "1-2-3-4" prioritization.


Why a low conductance ion such as lithium is used in batteries? Lithium ions have a relatively low conductance (Wikipedia) so why are they used in batteries, and not sodium, potassium, etc.?

The conductivity values you cite are for ions in water. there is no water in a lithium battery otherwise, it would explode and burn before leaving the factory. Beside low conductivity can be compensated by increasing concentration of the ions or a variety of other architectures.

Further, there's a few reasons lithium is optimal in batteries.  It's not environmentally toxic (like beryllium), has a higher oxidation potential than the other alkali metals, but most importantly is its high energy density in both Joule per weight and per volume, as well as power per weight, which no other metal can beat.  The only disadvantages for lithium are its flammability and cost.

(source)


Quantum mechanical (or QFT) description of mesomeric effect This is rather a question about finding the right literature, since the answer would probably exceed a typical forum post.

In organic reactions, mesomeric and inductive effects are of tremendous importance for the prediction of reactivity. I can work with Lewis structures and electronegativity, but I still wonder if there are more accurate descriptions arguing in terms of symmetries of the potential that is generated by the nuclei; how it allows for different coexisting standing waves; and that also consider 3D geometry dependences of the inductive effect.

Most texts argue in terms of atomic orbitals that are somehow bent and combined and hybridized, but electrons don't have a memory so they don't mourn after the times when they belonged to single atoms. Is there any readable description that looks at the whole molecule from the start, possibly written by physicists? Or is this just a thing for computers?

I searched the local university library and google scholar, to no avail. I have Clayden and Anslyn/Dougherty, but those don't really answer this question.

As far as I understand you search a text which provides you with the information about how to quantify the mesomeric and inductive properties of parts of an three dimensional organic molecule. The quantification should account for symmetries in the underlying potential.

From my point of view, any modern quantum chemical program provides you exactly this. So a good way to go are the usual books from Szabo and Ostlund or (a bit more modern and complete) from Helgaker, Jorgensen and Olsen. Both start from a quantum description of fermionic systems and later on introduce the potentials arising due to the physical nature of the system. However, geometrical symmetries are not covered in these books (maybe apart from solving the Schrödinger equation for a particle in spherically symmetric field - $\ce{H}$-like systems). If you're interested in general in molecular symmetries, then the book of Bunker and Jensen might be of interest.

However, those texts usually use the linear combination of atomic orbitals for molecular orbitals (LCAO-MO) approach, i.e., a certain number of atomic orbitals is combined to form molecular orbitals (MOs). Effects like "mesomerization" are obtained directly as a result because MOs are "spread over the whole molecule", i.e., the probability density to find an electron at position $\mathbf{r}$ is roughly constant along the bonds of the mesomeric system.

Your question states that

  [m]ost texts argue in terms of atomic orbitals that are somehow bent and
  combined and hybridized [...]

I agree that the usual chemistry starter books do this. And as you state, the bendedness of such an orbital (like the $sp^3$) hybridization arises from the symmetry of the underlying potential (in the $sp^3$-case the tetrahedral [$T_d$] potential of the four $\ce{H}$ atoms around $\ce{C}$ in methane). In LCAO-MO theory, however, symmetry is not necessary, i.e., you can place your atomic orbitals with their angular momenta at the atoms, and you would still get the correct answer. But by symmetry adaption of your basis (basis, because the atomic orbitals form the basis within which the MOs are spanned), you can obtain basis functions that already have a symmetry corresponding to the molecular geometry. This step is not entirely complicated but requires a bit of knowledge about group theory. If you are interested particularly in this part, I would recommend a book about Hückel MO (HMO) theory, like, e.g., Elektronenstruktur organischer Moleküle by Martin Klessinger or the book of Heilbronner and Bock. HMO theory is a simplified LCAO-MO theory in order to treat organic molecules that consist mainly of $\pi$-conjugated systems. In these books they explain how to symmetry-adapt a basis of $p_z$ orbitals e.g. for the calculation of the MOs of $D_{6h}$ symmetric molecules like benzene. While for the non-symmetry adapted basis you need to solve a $6\times6$ eigensystem, it simplifies very much in a symmetry adapted basis.

Lastly, I would like to mention that inductive as well as mesomeric a mere qualitative models in order to quickly come up with an idea how a system changes its properties under substitution of specific parts. None of those quantities is are actual calculable properties (there is no operator that returns the expectation value for this property when applied on a wave function). In this way a quantum chemical theory cannot teach you directly about

  3d dependences of the inductive effect[.]

But you can calculate your system and its properties (absorption spectra, electron distribution, electronic/charge density, electric dipole moents, ...) and do the same for a substituted system. From both data you can come up with a hypothesis about the source and change of mesomeric and inductive effects. The same arguments hold, of course, for aromaticity. There, it suits to cite a professor:

  Aromaticity is like a unicorn - something everybody knows, but nobody has ever seen.


Minimizing Gibbs free energy with Helmholtz free energy expressions If I have an expression for the Helmholtz free energy (from statistical associating fluid theory for a polymer system), can I still find the Gibbs free energy minimum under constant $T,P$ conditions?

Seems that since

$\mu_i = \left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j} = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_j}$

where $A$ is the Helmholtz free energy and $G$ is the Gibbs free energy and $\mu_i$ is the chemical potential of component $i$.

If I can just find the expressions for $\mu_i=\left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j}$ I should be able to use this in the expression for $G = \sum_i \mu_i n_i$ and find its minimum, without having to worry about an equation of state to relate change in volume ($dV$) with respect to constant pressure ($P$)?

Edit I should clarify that I wonder about the Gibbs free energy because I'm interested in the constant $T,P$ (isothermal, isobaric case), and in the case especially for liquids. If you needed to introduce an equation of state to parameterize the change in V as a function of $T,P$ this would not be too difficult for an ideal gas, but for liquids it would be a challenge. So in this case, I wonder how to handle the $PV$ term (for constant $P$) in the Helmholtz free energy which does not appear in the Gibbs.

I don't think you can use the approach you are outlining without an equation of state.

$G = \sum_i \mu_i n_i$ is valid only at constant $T$ and $P$.

$A = \sum_i \mu_i n_i$ is valid only at constant $T$ and $V$.

If $P$ and $V$ are both constant, then you can equate the two and use $G=A= \sum_i \mu_i n_i$.  Actually, that assumption is probably not horrible for most liquids at "normal" pressures: I think it amounts to assuming the liquid is incompressible.


why is delocalization less effective when there is a separation of charge? I've been reading Peter Syke's book "A guidebook to mechanisms in organic chemistry" wherein he has frequently mentioned that delocatlization is less effective when there is a seperation of charge i.e instead of a double bond formation and a negative charge, there is a positive and a negative charge occurance eg. Formate ion and Formic acid

It takes energy to separate the charges, even when the difference in electronegativity is in the same direction as the proposed charge separation.  The energy that must he put in becomes part of the energy in the charge-separated contributing structure, making that contribution less stable and less likely to contribute to the molecular ground state.


Is this an exception to the rule that strong acids/bases always replace weak acids/bases? According to the $\mathrm{p}K_\mathrm{a}$ table: hydrochloric acid is stronger than sulphuric acid. It has a $\mathrm{p}K_\mathrm{a}$ around $-6.3$ while the $\mathrm{p}K_\mathrm{a}$ of sulfuric acid is around $-3$. The second dissociation in the case of sulfuric acid is not very extensive and doesn't contribute much to its acidity.

\begin{aligned}
\ce{H2SO4 + H2O ~&<=>~ H3O+ + HSO4^{−}&&pK_{a}~=~-3}\\
\ce{HSO4- + H2O ~&<=>~ H3O+ + SO4^{−2}&&pK_{a}~=~2}\\
\ce{HCl + H2O ~&<=>~ H3O^+ + Cl^{-}&&pK_{a}~=~$-6.3$}\\
\end{aligned}

Now $\ce{BaSO4}$ reacts with  $\ce{HCl}$ to produce very less of $\ce{BaCl2}$ because of the insolubility of $\ce{BaSO4}$.

My teacher told me that in a reaction "Strong acids or bases replace weak acids or bases". This holds good for this reaction as $\ce{HCl}$ is stronger than $\ce{H2SO4}$.

However $\ce{NaCl}$ reacts with $\ce{H2SO4}$ to produce $\ce{Na2SO4}$ and $\ce{HCl}$. In this case $\ce{H2SO4}$  replaced a stronger acid $\ce{HCl}$. How do we account for this?

Every reaction is considered reversible to some extent and feasibility of a reaction is mainly dealt in the terms of favourability of either products or reactants. When we say that creating a weaker acid from a stronger base is feasible, we actually mean that the equilibrium favours the formation of the products in such a case. However some other factors can be at play and can well influence an equilibrium. One such governing principle is Le Chatalier's.

$$ \ce{H2SO4 + 2NaCl <=> 2HCl + Na2SO4 }$$

Here's a youtube video showing the production of hydrochloric acid under laboratory conditions. Notice the $\ce{HCl}$ produced is a gaseous product, and hence there's an imbalance in our equilibrium; $\ce{HCl}$ escapes out as vapors from the solution, lowering $Q_c$ of the reaction thus making the reaction go forward.


ph & pOH - Autoionization of Water So I was learning about pH and pOH and it made sense. I looked at strong, weak acids & bases and learned that they dissociate completely in water. So then I moved on to water and learned about the autoionization of $\ce{H2O}$.

Here is the process:

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

So then I learned about the equilibrium constant and then I figured this:
$$\ce{{[H3O+][OH-]} = 1E -14}$$

  What does this mean?

So after that I was doing some basic pH and pOH calculations withe strong acids and bases.

Here is one of them:

Find the pOH of a $\ce{1E-6M} \ce{HCL}$
So here are the steps:

$$\ce{-log[1E-6M] = 6}$$
Ok so the solution had the pH of 6

At first I was like that's nice and I proceeded.

$$\ce{6+pOH=14}$$
$$\ce{pOH=8}$$

So then I was like wait a minute where did the $\ce{OH-}$ come from if it dissociates like this:

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

So then I thought about it and this is what I came up with:

  Since there is some $\ce{OH-}$ then there must be some water more water that autoionized into OH and that is the only place where $\ce{OH-}$ could've come from. So that means that even though $\ce{HCl}$ completely dissociated there must be more water than $\ce{HCl}$, because referring to the original equation there is $\ce{1M}$ quantities of $\ce{H2O}$ and $\ce{HCl}$ and so in the $molarity$ of the $\ce{HCl}$ given that means there was more water and less $\ce{HCl}$ and the $\ce{OH-}$ could've only come from the autoionization of that excess water.

So that is what I thought could be the only possible explanation of the $\ce{OH-}$ in a strong acid. And that could only come from $\ce{H2O}$ .

So am I right, and If I am wrong then please tell me where the $\ce{OH-}$ came from.

  Why is the $\ce{pH}$ scale only to $14$? Because if I took a $\ce{2M}$ $\ce{HCl}$ and then I get a super acid in which the $\ce{H+}$ ions is a lot more than the $\ce{H2O}$ available to absorb them. So why is it only to $14$? Does it have to relate to the original autoionization of $\ce{H2O}$?

And If I totally grasped this wrong then can you please accurately tell me what $\ce{pH}$ & $\ce{pOH}$ and what autoionization of $\ce{H2O}$ is for a beginner and still make sense.

I think you have it all right. The OH- present must come from the autolysis of the water present, which is far in excess of the HCl. If you figure the pOH of pure water would be 7, then it makes sense to think that adding 1E-06 M (a small concentration) of HCl would lower the basicity (raise the pOH from 7 to 8).

The reason a scale of 0-14 is used for pH is that when you get to either extreme end of the scale (close to 1M in either acid or base), the sheer concentration of ions present, relative to water, is a game changer and the system no longer behaves in a nice linear fashion. So the scale does not work well outside the 0-14 range. Conceptually, you can extend the pH scale down below zero using concentrated mineral acids or strong acids like fluorosulfonic acid, and empirically measure their effectiveness in protonating a very weak base as a proxy for pH.


Distinguishing Test for Organic Compounds Given the following pairs of samples, how do I distinguish between them?

benzamide and ethanamide
ethanoyl chloride and ethanoic anhydride

My Approach:

We could easily convert the amides to amines by reducing them with $\ce{LiAlH4}$. Now, the question reduces to identifying the methyl group against the benzene group.

I am really lost here but my guess would be that we are looking for some characteristic reaction for anhydrides.

Since this sounds like an assignment, I am not going to give it all away.

Does you have to use qualitative chemical tests? I don't even teach those anymore. They were supplanted by spectroscopic techniques in most professional labs decades ago.

IR spectroscopy would answer both questions quickly.

Benzamide would have peaks characterizing an aromatic ring, for example:  $\ce{C}_{sp^2}\ce{-H}$ stretches around $3050-3100\ \mathrm{cm^{-1}}$. Ethanamide will not have these peaks, but will instead have $\ce{C}_{sp^3}\ce{-H}$ peaks $2800-3000\ \mathrm{cm^{-1}}$. There are other differences as well.
There are differences in the position (and sometimes number!) of carbonyl peaks between an anhydride and an acid halide. What are they?

So would mass spectrometry. Benzamide and ethanamide have different molar masses and different fragmentation patterns. Ethanoyl chloride... chlorine has two major isotopes. How can that help?

NMR spectroscopy would also be useful. However, IR is cheap, fast, and easy.


When 3,4-dimethyl-2,4-hexadiene reacts with HBr what is the major product? I found out from drawing it out that there are 2 products: 4-bromo-3,4-dimethyl-2-hexene and 2-bromo-3,4-dimethyl-3-hexene. I want to know which forms most of the time.

I used the fact in allyl carbons, the double bonds will travel to the carbon-carbon single bonds to form a double bond. I also used the fact that tertiary carbocations are more stable due to sigma bonds that can participate in hyperconjugation - that is delocalization of electrons by the overlap of sigma bond orbital with an empty p-orbital on an adjacent carbon.

I would like to see some of these reactions in a lab setting, but for now I'm learning organic chemistry reactions like as if it were math.

Conjugated dienes can undergo two types of additions with HX, they can undergo 1,2-additions and 1,4-additions.  A 1,2-addition occurs when HX adds to only one of the double bonds, not interacting with the other.  a 1,4-addition occurs when HX adds X to one of the double bonds, and because the system is conjugated, the hydrogen will be added three carbons away from X.  The 1,2-adduct and 1,4-adduct are either thermodynamically or kinetically stable.  The kinetic product will form quicker, but is less stable, it will dominate at low temperatures and short reaction times.  The thermodynamic product is more stable but takes more time to form.  It will be the major product at high temperature and long reaction times.  In this case, the kinetic product would be 2-bromo-3,4-dimethyl-3-hexene, the thermodynamic product would be 4-bromo-3,4-dimethyl-2-hexene, because the thermodynamic product has the bromine on the most substituted carbon.  Depending on the reaction conditions, either could form.  At low temps. and short reaction times, 2-bromo-3,4-dimethyl-3-hexene will form, at high temps. and long reaction times, 4-bromo-3,4-dimethyl-2-hexene will form.


How do molecules know how to behave when a disturbance occurs in an equilibrium? I am well aware of Le Chatelier's principle, but what causes molecules to behave in such a way, my guess is that it all has to do with rates of reactions and frequency of collisions.

When at a constant volume, the addition of an inert gas causes no change in the equilibrium, since the concentration of each species will remain constant, I understand that there will be the same amount of molecules per unit volume as before the addition, however wouldn't the rate or probability of collision increase since now each unit of volume has less space in which the molecules can move about due to the addition of the inert gas in that unit of volume, hence if a reaction had more moles in the forward then the reverse the probability of the forward occurring would be greater, causing a forward shift?

The molecules don't know, they aren't alive in any way for that matter!

Here's an easy way to see it (note that I'm avoiding a mathematical approach to give an intuitive reason):

Consider a hypothetical reaction where $\ce{A}$ and $\ce{B}$ combine and gives $\ce{C}$:

$$\ce{A + B <=> C}$$

Let's say initially, there's an equilibrium. The rate of combination of $\ce{A}$ with $\ce{B}$ is the same as the rate of dissociation of $\ce{C}$. Let's say we added an excess of $\ce{B}$. If there's lots of $\ce{B}$ around, molecules of $\ce{A}$ have more chances of bumping into molecules of $\ce{B}$. This pushes the equilibrium forward, until there's enough $\ce{C}$ to counteract the amount of $\ce{A}$ and $\ce{B}$ present. A similar argument can be given if $\ce{A}$ was in excess than in $\ce{B}$.

You could now think about what happens if we removed some amount of $\ce{C}$. In such a case, the rate dissociation of the limited amount of $\ce{C}$ in the vessel would not balance with the rate of combination of $\ce{A}$ and $\ce{B}$, and thus equilibrium would again flow towards the products.

Now, lets see what happens if you add an inert gas $\ce{D}$ to that mixture. In such a case, $\ce{C}$ will continue to dissociate in it's usual rate, but $\ce{A}$ and $\ce{B}$ cannot combine with each other as fast because they may bump into $\ce{D}$ instead of each other. Thus equilibrium shifts towards the reactants.

I hope this makes sense. Leave a comment if you're not understanding any part of this.


Rounding of atomic weight isotopes When we look at atomic mass, for example, $\ce{^134Ba}$ (barium) has a mass of $133.90 \,\mathrm{Da}$. Why is this rounded to $134 \,\mathrm{Da}$? This is shown in my book.

The way my teacher first explained this is due to the definition of atomic mass. According to Google:

  The mass of an atom of a chemical element expressed in atomic mass units. It is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.

Now, the key words are

  ...the average number allowing for the relative abundances of different isotopes.

What does that mean? We go back to the definition of an isotope. An isotope of an element is a species that has the same number of protons in the nucleus, yet a different number of neutrons. For example, $\ce{C^12}$ and $\ce{C^14}$ are isotopes of each other. Sidenote, isotopes are not to be confused with allotropes, which are different forms of the same element. For example, graphite and diamond are allotropes of carbon, since both, either, or neither might be made up of $\ce{C^12}$ ($6p$, $6n$)  or $\ce{C^14}$ ($6p$, $8n$). Now how do we represent this with the periodic table? Simply, we take the weighted average. Now, I do not have enough information or data to calculate the relative abundances of each isotope of carbon, and the periodic table does not either. We have an average. The atomic mass of $\ce{C}$ is given as $12.011$. That means a few different things. Firstly, there are more isotopes that are closer to having a sum of $p+n=12$, and less that have $p+n=14$ or $15$. Secondly, this means that although this number is not exact, it represents the most common $p+n$. Averaging all of those different $p+n$'s up, from each isotope, multiplying they by their relative abundances, and averaging should give our periodic table value of $12.011$.

Now it is important to remember that this number is not rounded. You might be asking, "Then why, in our calculations, do we sometimes round this number?" Well it depends on the problem, because for many problems, you do not round, but for the ones that you do, it is because it is more convenient to assume that the number that is given rounded is the most common isotope. Was that sentence confusing? Essentially, it is easy and generally correct to say that $12.011$ for carbon can be rounded to $12$, and $12$, meaning $\{p,n\}\in\{6,6\}$. We can do this by assuming that carbon in its most common form has $6p$, and $6n$.

In orthocresol's comment above, he is also correct. He has a perfectly valid point regarding common word usage in speaking. However, I do not think that this is the main reason.

Regarding Jon Custer's comment above, yes, that would be true, but only if we had a single isotope, which we do not.

Now, make sure you can differentiate between atomic mass and atomic weight! :)

Hope this helps!


DNA pairs (adenine-thymine, guanine-cytosine) Well I was wondering why adenine pairs with thymine and cytosine pairs with guanine.

From those pictures:

Thymine has the lowest acidity and adenine has the biggest acidity.

So it is logical that the intermolecular bond will be strong.

Adenine and Cytosine have pretty much the same acidity so why is it energetically favorable for guanine to make an intermolecular bond with cytosine?

Is it due to the three combinations this (adenine-thymine, cytosine, guanine) has the lowest energy?

In Principles of Nucleic Acid Structure, W. Saenger argues that hydrogen-bonded bases contain at least two hydrogen bonds (forming a "cyclic" pattern). Often, there is a tautomeric form possible that also makes two hydrogen bonds, with two covalent bonds turning into hydrogen bonds, two hydrogen bonds turning into covalent bonds, and double bonds moving around so that each atom still has an octet. In the OPs first example (A:T pair), for example, the ketone would turn into an enol, and the protonation of the amino and imino groups would change.

Saenger goes on to say that 28 pairs are possible.

Many of these are observed (the ones that are labeled are so common that they were given names), in structures of DNA, RNA, and in crystal structures of dinucleotides.

  I was wondering why adenine pairs with thymine and cytosine pairs with guanine.

They pair because they make strong hydrogen bonds with a geometry of the base pairs that allows base stacking and is compatible with base pairing of the other nucleotides in a DNA strand. As the discussion above shows, many other base pairs are theoretically possible, and some of them are actually observed in nature.

  Thymine has the lowest acidity and adenine has the biggest acidity. So it is logical that the intermolecular bond will be strong.

I'm not sure about the premise, and the conclusion is incorrect, as other base pairs also form. As long as a group has the right protonation state, the pKa does not directly determine hydrogen bond strength.


Why does ice crack when a drink is poured on it? I think most people have experienced this, but while pouring a beverage over ice I noticed the ice cracked.

Before

After

My initial thought was that because the beverage has an antifreeze property to it, that some sort of thermal shock might occur due to the rapid local cooling from dissolving the ice, but I know that this also occurs when other beverages such as various flavored carbonated sugar water products are poured on ice which do not depress freezing as much. So then What would cause the ice to crack?

I believe the ice cracked due to residual strains from freezing. Since ice freezes from the outside inward and it expands as it freezes, that as the inner water freezes, it imparts a tensile force on the surrounding ice (like the opposite effect of tempering glass).  As the warm liquid removes ice, the cross sectional area under tension decreases while the tensile force remains the same, causing an increase in stress. Once the stress reaches a certain point, the ice fails to counter the tension mechanically, causing the cracks.


What does it mean for chemical reaction network to be in equilibrium? What does being in "equilibrium" mean for a set of chemical reactions? If we have some number $n$ of $\ce{A}$ molecules and $m$ of $\ce{B},$ and the following reaction where $\ce{A}$ and $\ce{B}$ can reversibly bind:

$$
\begin{align}
\ce{A + B &->[$k_\mathrm{f}$] AB} \\
\ce{AB &->[$k_\mathrm{r}$] A + B}
\end{align}
$$

Suppose $k_\mathrm{f} = 2,$ $k_\mathrm{r} = 1.$ Can this system be "in equilibrium" and what would it mean? Does it mean all reactions have to be reversible and that as $t \to \infty$ go on at the same rate? In that case it seems that this system cannot be in equilibrium, since $k_\mathrm{f} \neq k_\mathrm{r}.$

Ignoring the math, "in equilibrium" means the forward and reverse reactions are proceeding at the same rate, so there is no gross change in the amount of reactants. In a closed system, if initially one reaction was faster, then it's product will start to dominate, and, by the law of mass action, the reverse action will "catch up" because there is now a larger supply of those reactants.

A simple physical analog:

Balance a ruler on two fingers, one finger near the end of the ruler, the other finger inset from the end towards the middle.
Gradually move your fingers together.

No matter where you start, the fingers will come to rest near the middle of the ruler.


Methylcyclohexane chair conformers I am trying to create the molecule of the methylcyclohexane in the equatorial and axial conformers. I read in this site about the chair conformation. My molecule looks like the following picture (created in Molden)

Can someone point out if this is actually good? I am quite doubtful about the "mid" carbons, should they be push to the center of the molecule like in the drawings in the website?

It's a little hard to tell from this angle, but that looks like a pretty sensible representation of the axial methyl geometry given on the website you linked to.

I'm not entirely sure what you mean by "push to the center of the molecule", though.

ETA: These might help a little more: this isn't exactly the most accurate geometry, but it's a pretty good representation.


Different coloured chromium salts Evaporating cold aqueous chromium(III) chloride produces violet crystals. However when hot aqueous chromium(III) chloride crystallizes, green crystals result.

Could someone please explain the chemistry behind this... I mean both the crystals contain $Cr^{3+}$, so the difference in colour cannot be due to two different oxidation states. Also, they contain the same ligand $Cl^-$. Can someone please tell me where I am going wrong?

Well, actually when we evaporate aqueous chromium(III) chloride, the hydrate of the crystal will be gone, so we get anhydrous chromium(III) chloride ($\ce {CrCl3}$). This type of anhydrous produce violet color.

But if we crystallizes aqueous of the chromium(III) chloride, the hydrate won't go, so we get hydrous chromium(III) chloride (usually hexahydrate, like $\ce {CrCl3.6H2O}$). This crystal produce dark green color.