Two possible chains - isopropyl or ethyl and methyl? I see two possibilities for choosing the longest branch, and they yield pretty different names:

3-isopropyl-3-hexene
3-ethyl-2-methyl-3-hexene

Is there a rule to determine which to choose? Does one of these prefixes take priority over the other, for example? Or is the second option preferable because of the lower number? I suspect this is a very straightforward rule...

The first important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

Since the compound that is given in the question does not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-44.1.1 is not relevant in this case.

The second criterion given in P-44.3.2 is the longest chain. However, both possible names, (3Z)-3-ethyl-2-methylhex-3-ene and (3Z)-3-(propan-2-yl)hex-3-ene, indicate a principal chain with six skeletal atoms.

Also, both possible principal chains include the double bond. Therefore, the criteria given in P-44.4.1 (a) (greater number of multiple bonds) and (b) (greater number of double bonds) cannot be used to reach a decision.

The criterion given in P-44.4.1 (h) (lower locant for the suffix) is not relevant since the compound that is given in the question does not have any suffix.

Finally, the criterion given in P-44.4.1 (j) stipulates that the principal chain shall have the lower locant for the double bond, i.e., for the ‘ene’ ending. However, both names, (3Z)-3-ethyl-2-methylhex-3-ene and (3Z)-3-(propan-2-yl)hex-3-ene, have the locant 3 for the ‘ene’ ending.

Further criteria can be found in Section P-45:

  P-45 SELECTION OF PREFERRED IUPAC NAMES

  P-45.0 INTRODUCTION

  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of
  the criteria in this section will generate the preferred IUPAC name.

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  (…)

In accordance with P-45.2.1, the preferred principal chain corresponds to (3Z)-3-ethyl-2-methylhex-3-ene, which has a principal chain with two substituents (3-ethyl, and 2-methyl), whereas the principal chain of (3Z)-3-(propan-2-yl)hex-3-ene has only one substituent (3-(propan-2-yl)).


Empirical formula of Berthollides Can the empirical formula be defined for berthollides? Is it equal to the molecular formula, or it can't be defined?

"Berthollide" seems to be a word for "non-stoichiometric compound" that has fallen out of use.

You have to look at the structure, as a rule things happen for a reason.  The formula can be derived from the structure.

Consider the series of alums, which is  KAl(SO4)2·12H2O, or more generic:  X+Y3+(SO4)2·12H2O.  The generic structure can accomodate a monovalent cation, provided it's sufficiently large (lithium is out of the question, sodium is an edge case), and a tervalent cation (Al, Cr, Fe, pretty much anything), and the proportion can be freely chosen within the constraints of the structure.


How to balance the disproportionation reaction of manganate anion in basic conditions? $$\ce{MnO4^2- (aq) -> MnO2(s) + MnO4^-(aq)}$$

How would I balance this in basic conditions? I know we need to separate this into two half reactions but how do we do this if there is only one reactant?

You are referring to the disproportionation reaction of $\ce{Mn(VI)}$ to $\ce{Mn(IV)}$ and $\ce{Mn(VII)}$, which is:

$\ce{MnO4^{2-}(aq) -> MnO2 (s) + MnO4^{-}(aq)}$

Reduction: $\ce{MnO4^{2-} + 2e^{-} + 2H2O -> MnO2 + 4OH-}$

Oxidation: $\ce{2(MnO4^{2-} -> MnO4^{-} + 1e^{-})}$ = $\ce{2MnO4^{2-} -> 2MnO4^{-} + 2e^{-} }$

Combine both, and the final balanced equation is:

$\ce{3MnO4^{2-} + 2H2O -> MnO2 + 2MnO4^{-} + 4OH-}$


Sulfuric acid recovery I know sulfuric acid is an important catalyzer and is used in various processes. My question is, how do I recover the remaining sulfuric acid? It will be impure, and I don't know how to do the "standard" procedure (is there one?)

In laboratory, nobody cares. It's too cheap to care about.

In industry, other catalysts are usually used (heterogeneous ones).


Partial Pressure and Partial Molar Volume When we define partial pressure and partial molar volume of a gas in a mixture, suppose $P$ is partial pressure of a gas in a mixture and $V$ is partial molar volume
then why the following equation doesn't hold
$PV=nRT$, (where $n$ is number of moles of gas) ?

It doesn't work because partial molar volume of a species in a gas mixture is defined as the partial derivative of the total volume of the system with respect to the number of moles of that species at constant total pressure and temperature of the system.


Why are atom spherical in shape ? I am learning about structure of atom, in which i saw right from J.J. Thomsom atomic model to modern nuclear atomic model all are spherical in shape.

I have seen how different discoveries help to develop the atomic model but i have't came through a single discovery which proves that atoms are spherical in shape.

So, Why atoms are spherical in shape?

I think you should think this question a little through:

The easy answer is that atoms are not spherical in shape. Sphere is a solid object, atoms to not have well defined boundaries with "solid cutoffs". Atoms are more like clouds consisting distributions of electrons, neutrons and protons.
Atoms, per definition, are electrons + a single nucleus, ie. the huge positive charge of a point-like object (nucleus) pulls together the whole thing. That result is that electrons approximately move in a central field. Such central field generates spherical symmetry of the electronic states, and that is why you have e.g. spherical harmonics when you describe the atomic orbitals. Check the solution of the Schrodinger equation how it leads to states, and how wave-functions with only radial dependence (i.e. sphere-like) appears.

+++++ Incorrect part of the answer was deleted based on the comments of DavePhD.


What lab equipment did Marie Curie use to isolate radium? My experience in growing crystals for condensed matter physics has been sealing grams of material in ampoules which get heated in laboratory furnaces, so I don't have the experience to understand how Marie Curie managed to go through tons of pitchblende to isolate radium, especially given her limited resources. How were the samples heated, and how would she have kept her samples from getting oxidized? It seems like making lots and lots of ampoules would be impractical.

After searching Google Books, I found autobiographical notes from Marie Curie. Apparently, it was radium chloride that was isolated first, not elemental radium. Salts can often be precipitated from solution at much lower temperatures than those needed to melt many inorganic materials, and you typically wouldn't need to shield a solution from air, so ampoules wouldn't be needed. This would be the case for fractional crystallization, which she mentions in her autobiographical notes. (She also mentions spending whole days "mixing a boiling mass with a heavy iron rod". I'm not sure if she's referring to mixing molten material or a solution in water or acid.) She also says they eventually used a factory; as time passed and their work became "more and more known", she was able to do the "crude processes" in a factory, allowing herself "more time to the delicate finishing treatment".
source: the book Pierre Curie: with Autobiographical Notes by Marie Curie


Which electrodes do not corrode at all? I had used spare pieces of metal to perform electrolysis. They all had the disadvantage that they corrode when used as anode - and some of the oxides are toxic (copper, chromium).

I've found that carbon, which can be harvested from pencils and old batteries does not corrode contrary to my expectations (I'd expect it to produce $\ce{CO2}$). But soldering on carbon is completely impossible. Also, I need some bigger electrodes.

So are there any metals that do not corrode during electrolysis?

I'm using sodium carbonate as electrolyte.

This highly depends on what you are electrolyzing. When using solutions that are not very acidic, oxide-coated metals sometimes work, like lead-oxide electrodes, that can be harvested from old lead accumulators (attention, lead is toxic, wear lab coat and gloves when working with them. Lead accumulators contain ~20% sulfuric acid, so wear gloves as well and unassemble them in an acid-proof working space)

If you search the web for recipes of perchlorate preparation by electrolysis at home, you will find extensive discussions on the topic of anode materials and making oxidation-resistant electrodes.

Platinum is really hard to corrode, but even it corrodes in some cases.


Subscript on Orbitals What do the subscripts on the orbitals (such as px, py, pzetc) mean and how can one derive these or at least gain a fundamental understanding of each? Do they describe the geometry of the shape? And if so- how do derive them?

Please describe in an advanced high school level.

As a high school student, here’s how I considered it:

The wavefunction of an orbital describes the region of space that an electron residing in that orbital would be in about 90% of the time.

The wavefunctions would have a radial component (depending on period), which is why the 3p orbitals Arena larger than the 2p etc.

The wavefunction would also have an angular component, which would describe its shape.

In the case of the p orbitals, the x, y and z labels simply means that their angular components are x/r, y/r and z/r respectively (hence the subscript). Do note that we usually use the radial forms of the wavefunctions though.

Hope this helps.


Shape of the P1/2 Orbital I always thought that p-orbitals had a dumbbell shape as pictured below.

(image source)

However, I was reading an article (see Table 1, item 2) that says, "...the spherical $\mathrm{p_{1/2}}$ subshell..." (my emphasis). The implication being that the $\mathrm{p_{1/2}}$ orbital (subshell?) has a greater electron density near the nucleus than the $\mathrm{p_{3/2}}$ orbital (subshell?). Why is this and why do they call the $\mathrm{p_{1/2}}$ orbital spherical?

As orthocresol mentioned, this is all about relativity, so let's talk about it. I am hardly an expert myself, but I'll try to give an answer to the best of my limited knowledge. For an interesting and accessible overview of the incorporation of relativistic effects into chemistry, I recommend the 2012 review article "Relativistic Effects in Chemistry: More Common Than You Thought" by Pekka Pyykkö (free access!), one of the great names in the field.

The theoretical basis for most of our work as chemists is "old" quantum mechanics, developed up to around 1925, and one of the hallmarks of this era is the Schrödinger equation. The whole concept of orbitals comes as the solution of the Schrödinger equation for the hydrogen atom, (i.e. a two-body system composed of an electron in a spherically symmetric electric field originating from the nucleus). The stationary states in this system are spherical harmonics, described by the quantum numbers $n$, $l$ and $m$ (or $m_l$). The first two numbers determine the shape of the orbital, but we've gotten used to calling the second number by a letter (s, p, d, f, g, ... for 0, 1, 2, 3, 4, ...). So that's what orbital names, such as $\textbf{4s}$, $\textbf{3d}$ and $\textbf{5f}$ mean.

That said, old quantum mechanics is fundamentally incomplete, in a way which very much matters to chemists. In particular, in 1922 the Stern-Gerlach experiment discovered the property of spin. While its existence could be accommodated within the prior framework by "tacking it on" and adding an extra quantum number $s$, there was no theoretical background to explain it. This changed a few years later, when Paul Dirac incorporated special relativity into quantum mechanics. Old quantum mechanics implicitly assumes the speed of light is infinite, and by adequately introducing its finitude, many new experimentally-verified phenomena appear, such as antiparticles and, indeed, spin. In other words, spin is a purely relativistic quantum mechanical property!

The equivalent of the Schrödinger equation in relativistic quantum mechanics is the Dirac equation. Much like the Schrödinger equation, the Dirac equation can be solved for a two-body system composed of an electron in a spherically-symmetric electric field created by the nucleus. The resulting solutions ("Dirac orbitals"), however, are somewhat different. In particular, the quantum numbers $l$ and $s$ are no longer "good" quantum numbers (that is, they no longer describe stationary states, due to spin-orbit coupling), but their sum ($\overrightarrow{j}=\overrightarrow{l}+\overrightarrow{s}$) is. For the electron, $s=\frac{1}{2}$, and thus for each value of $l$, there are now two possible values of $j$, given by $|l+\frac{1}{2}|$ and $|l-\frac{1}{2}|$. This means that the s, p, d, f... subshells from the Schrödinger solutions are now split according to their $j$ values. The s subshells are unique in that they don't split ($|0+\frac{1}{2}| = |0-\frac{1}{2}| = \frac{1}{2}$), but are now termed $\textbf{s}_{1/2}$ subshells. The p subshells, which are triply degenerate via the Schrödinger equation, now split into a single $\textbf{p}_{1/2}$ orbital and two degenerate $\textbf{p}_{3/2}$ orbitals. The d subshells, previously quintuply degenerate, split into two degenerate $\textbf{d}_{3/2}$ orbitals and three degenerate $\textbf{d}_{5/2}$ orbitals. The trend continues similarly for the further subshells. Here is a figure from "Relativistic quantum chemistry: The electrons and the nodes", which will hopefully make it clearer:

So what are the physical consequences of these different mathematical solutions? Firstly, the orbitals change in energy. Orbitals with a lower $j$ for a given $n$ and $l$ have lower energy (e.g., the single $\textbf{4p}_{1/2}$ orbital is lower in energy than the two degenerate $\textbf{4p}_{3/2}$ orbitals). The shape of the orbitals also change; all orbitals with $j=\frac{1}{2}$ are spherical, regardless of $l$.

According to "Contour diagrams for relativistic orbitals":

  Contour plots for $^2P_{1/2}$ and $^2S_{1/2}$ are spherically symmetrical,  while those for $n = 2$ and $3$, $l = 1$, $j= 3/2$, and
  $m = \pm 3/2$ look very similar to those for p orbitals already published in this journal.

Similarly, all orbitals with $j=\frac{3}{2}$ are dumbbell-shaped, including $\textbf{d}_{3/2}$ orbitals. The article brings a helpful schematic:

As additional confirmation, "Pictorial Representations of the Dirac Electron Cloud for Hydrogen-Like Atoms" states:

  As an example of these two theorems the angular charge distributions for the magnetic states $m = \pm 3/2$ for a $^2P_{3/2}$ term are not only the same but are also identical with the two magnetic states $m = \pm 3/2$ of the $^2D_{3/2}$ term. It should be pointed out that the radial charge distributions of the $^2P$ and $^2D$ terms are different. [...]
  In this connection one of the consequences of the Dirac theory is that
  a single p electron or two similar p electrons, depending on whether $j=\frac{1}{2}$ or
  $j=\frac{3}{2}$ respectively, present spherical symmetry. Not only are all S states
  spherically symmetrical, as on the Schrodinger theory, but also all one electron systems having but one valence electron and that electron in a $^2P_{1/2}$ state,
  eg., the normal states of B, Al, Ga, In, Tl.

The article also shows a table of the angular distribution functions for the wavefunction. Note how the function is equal to $1$ for $\textbf{s}_{1/2}$ and $\textbf{p}_{1/2}$ orbitals. That is, there is no angular dependence, implying spherical symmetry.

The article also brings some diagrams and pictures of simulated orbital shapes.

I wouldn't be able to tell you why this is the case any more than I would be able to explain the shapes for the solutions to the Schrödinger equation - that's just how it pans out.

Another difference is that the Dirac orbitals do not possess nodal points or planes. They are now replaced with "pseudo-nodes", where the wavefunction assumes very small values, but never reaches zero anywhere. This is nicely shown in this figure from "Relativistic quantum chemistry: The electrons and the nodes":


Impact of detergent on internal living tissue? There was some spirited debate between the members of my chemistry lab today about the theoretical consequences of ingesting a large amount of the detergent Triton X-100 at a 100% concentration. The MSDS provided by Sigma Aldrich indicates that it is "harmful if swallowed," with a LD50 of 1800mg/kg in rats, but does not provide further detail.

Given that it is capable of decellularizing animal-derived tissues, would it literally "wash" one's extracellular matrix of its cells after degrading the lipid bilayers?

If Triton X-100 is incapable of doing so, are there more powerful detergents that would pose a genuine threat if ingested?

Caustic Ingestion Injury
Triton X-100 has a pH of 9.7 at 100% concentration (according to Wikipedia) which can cause alkaline caustic injury, however it is less likely to cause severe full thickness injury, as those injuries tend to happen with ingested products of pH <2 or >12.5 or with other corrosive properties that Triton X-100 does not have. Ingestions at pH 6.0-8.0 are unlikely to cause injury unless there is long term exposure (as in gastro-oesophageal reflux disease or bile reflux); note that saliva itself often has a pH of 7.8!
Alkaline caustic injury has a large spectrum. The alkali can cause liquefactive necrosis via saponification and protein dissolution. In the best case scenario, there is only superficial injury to mucous membranes, causing pain and exudates. In the worst case scenario there is full thickness involvement, usually involving the oesophagus, where you get a perforation. One of the worst cases I have had any involvement in was a young lady who drank a large quantity of Drano (concentrated NaOH with pH 11.5-13.5) and had oesophageal, gastric and duodenal perforation with extensive chemical damage to her pancreas, liver and fortunately to a much lesser extent, mediastinum.
For reference, Zargar's grading system for caustic ingestion (usually based on findings on gastro-oesophagoscopy +/ pharyngolaryngoscopy) is as follows:

Grade 0: Normal examination
Grade 1: Oedema and hypermia of the mucosa
Grade 2a: Superficial ulceration, erosions, friability, blisters, exudates, hemorrhages, whitish membranes
Grade 2b: Grade 2a plus deep discrete or circumferential ulcerations
Grade 3a: Small scattered areas of multiple ulceration and areas of necrosis with brown-black or greyish discoloration
Grade 3b: Extensive necrosis

Triton X-100 would most likely cause Grade 1-2b injury. Grade 3 injury is unlikely but I suppose it depends on dose as well.
Ingestion at lower concentrations and effects not related to caustic ingestion:
This link contains information on the general toxicity of non-ionic surfactants such as Triton X-100.
In summary:

They cause mild irritation to mucous membranes
This is primarily due to foaming
Foaming and irritation can cause vomiting and diarrhoea
Foam can also be aspirated causing chemical pneumonitis
Alcohol ethoxylates (as Triton X-100 appears to be) are readily absorbed into the bloodstream via mucous membranes and is likely to have toxic effects in this way

The rest of the evidence I could find also states:

They can cause haemolysis- makes sense, as surfactants in the circulation would come into contact with red cell membranes
Triton X-100 causes direct cytotoxicity due to cell membrane disruption
It also inhibits NAD(P)H, which is a coenzyme involved in cellular respiration
There is dose-dependent destruction of the cytoskeleton
There is also some associated double-stranded DNA breakage
It is possible for cells to recover from this poisoning

Thus, the likely scenario from swallowing dilute Triton X-100 is this:

First you get foaming, oral & oesophageal pain and abdominal pain
Next you get vomiting and diarrhoea, at which point you may inhale some of the foam
If you've inhaled any foam, you develop respiratory distress
You develop dose-dependent haemolysis which presents as tiredness (and/or shortness of breath, chest pain, palpitations), red or dark brown urine, jaundice
In severe cases potential multi-organ failure as a result of direct cytotoxicity, haemolysis, anaemia and/or chemical pneumonitis

It seems it'd be a huge spectrum all the way from general foaminess and some pain all the way to multi-organ failure and death. I'm not entirely sure what doses would be toxic as it appears no-one has forced animals or humans to drink the stuff!
Regarding whether there are more toxic and corrosive but chemically similar compounds out there
yes.


If electrolysis splits water, why does only either hydrogen or oxygen (but not both) bubble up on one pole? If electrolysis splits water, then that means that $\ce{H2O}$ is split into $\ce{H}$ and $\ce{OH}$ or $\ce{O}$.

How come that if a water molecule is split at e.g. the negative pole (anode), only the $\ce{O}$ bubbles up. What happens to the $\ce{H}$ and/or $\ce{OH}$? Do they react with the pole and stay there?

Water, as you may know, has a dualist nature between covalent and ionic bonding; oxygen is the second most electronegative element in the periodic table, while hydrogen's simplistic construction makes it very zen about how it forms bonds (it defines the center point of most electronegativity scales). While the bond between the hydrogens and oxygen of a water molecule are termed covalent, the electrophilic nature of the oxygen makes the bonds border on the ionic, and so water is just as easily thought of as $\ce{2H^+ + O^2-}$, or in more stable terms as $\ce{H^+ + OH^-}$.

In electrolysis, this ionic character of water is exploited in a series of "half-reactions"; at each node, a gas is formed by adding or removing electrons from water to liberate one of its two components, and when that gas precipitates out, what's left remains dissolved in the water as charged ions, which are repelled by the node that created it and attracted to the other node.

When electrons are added to water at the cathode, they are attracted to the partial positive charge of the "hydrogen side" of water, and can then form an orbit around a hydrogen. That liberates the hydrogen; it no longer needs the electron it's sharing with the oxygen to balance its charge, but it still needs a second electron to stabilize its 1s shell, and the best bet is to pair with another hydrogen to form hydrogen gas, which precipitates out. That leaves soluble hydroxide ions in the water, which are repelled by the cathode (and each other, so these ions wouldn't form any compounds) and move toward the anode. It's highly unlikely that both hydrogen atoms would be liberated from a single oxygen, and even if it did, the oxygen, with extra electrons, couldn't bond with another negative-charge oxygen to form gas; it would be much more likely to deprotonate another water molecule forming two hydroxides.

$$\ce{4H2O + 4e^- -> 2H2 (g) (^) + 4OH^- (aq)}$$

At the anode, the positive potential causes water to give up electrons, liberating the hydrogens again, this time as naked protons, which are attracted to water's partial negative pole to form hydronium ($\ce{H_3O^+}$ or simply $\ce{H^+ (aq)}$). The now chargeless but bond-seeking oxygen is left to pair up as oxygen gas, which precipitates out. Hydrogen peroxide is a possible intermediate, but very short-lived, as this peroxide will either find another peroxide and pair up the extra oxygens, or it will give up its hydrogens as protons under the positive voltage potential the same way water would, and become oxygen gas.

$$\ce{2H_2O - 4e^- \to O_2(g)\uparrow + 4H^+ (aq)}$$

... Then, somewhere in the middle, the hydronium and hydroxide meet, and as these proton donors and proton acceptors will do, they reform water:

$$\ce{4H^+(aq) + 4OH^- (aq) \to 4H_2O }$$

You'll notice that only one gas precipitates out at each node; the other atoms are trapped as soluble ions, either protons/hydronium or hydroxide, which cannot form other compounds because their like charges repel them from each other. Instead, they remain dissolved as local acidic and basic solutions. If the potential charge in the nodes is high enough, dipping a litmus strip at each end of the water bath (or dropping in a few mL of pH indicator dye) will yield an acidic reading at the anode and a basic reading at the cathode.


Energy per atom in a parallel radioactivity equation I was reading an example in my textbook and the question goes like this:

In a parallel radioactive decay, find energy liberated per atom of 'A' in MeV.

The solution is proceeded like this:
$$
 \lambda _{1}/\lambda _{2}= \frac{(T_{1/2})_2}{(T_{1/2})_1}=\frac{20}{60}=\frac{1}{3}
$$
Energy per atom of A is $$= 20\times\frac{1}{4}
+40\times\frac{3}{4}=35 MeV
$$
I have understood how we have got the first step, but I'm unable to understand the second step. Why are 1/4 and 3/4 specifically being multiplied to 20 and 40, respectively? Also, where are we using the ratio we got in the first step, in the second step? It has got me all confused. In short, how is energy per atom of A calculated?

You may state that, in $60$ s, one A atom has liberated $20$ MeV by the first reaction. In the same time, three A atoms have liberated $3·40$ MeV by the $2$nd reaction. The total energy liberated in $60$ seconds, is $120 + 20 = 140 $ MeV.- As a consequence, the average energy liberated by one A atom is $140$ Mev$  /4 = 35$ MeV


Why benzenediazonium bromide and iodide are not important? When we think of aryldiazonium compounds, the first compound that comes to mind is benzenediazonium chloride. Sure thing, the compound is versatile and has been used to make a vast array of organic compounds and thus serves its purpose as a useful reagent. But what about other halides like benzendiazonium fluoride, bromide and iodide? Chemists and scientist do not take them noteworthy as there is little to no literature evidence of their synthesis. Are the compounds so unstable that chemists neglect them? Nobody has attempted to make the compounds in the way benzenediazonium chloride is made. Why is it so? Is it because of unfavourable reaction conditions? Can't we use hydrogen bromide or hydrogen iodide to make benzenediazonium bromide/iodide respectively? Like the following reaction:

$$\ce{C6H5NH2 + HNO2 + HX -> C6H5N2+X- + 2H2O  ~~~~~~~~ X = F, Br, I}$$

What are the reasons for the non-existance of/not preparing other benzenediazonium halides? Is it just because of their instability or feasibility or any other reasons? Or is there is any (other) way to (theoretically) make these compounds (if they exist)?

There is indeed very little to be found in the literature.

A recent Russian paper (Ref.1) mentions benzenediazonium bromide as a reactant. This thesis from 1975 (Ref.2) contains a preparation of benzenediazonium bromide and an attempted preparation of benzenediazonium iodide; it notes that the iodide is unstable as does this reference here.  The discussion section of this Organic Syntheses' preparation of fluorobenzene (Ref.3) mentions an alternative preparation by formation of benzenediazonium fluoride and its decomposition referencing work from the 1900s.

My overall thought is that they are probably less stable than the well-known chlorides and undergo reaction in lower yields so they are of little interest.

References:

Yu. V. Ostapiuk, V. S. Matiichuk, N. I. Pidlypnyi, N. D. Obushak, "Convenient synthesis of $\alpha$-bromo ketones by the Meerwein reaction," Russian Journal of Organic Chemistry 2012, 48, 519–522 (https://doi.org/10.1134/S1070428012040094).
Geoffrey D. Picard, "Sulfonium Salts, Aliphatic and Aromatic Diazonium Salts, Formation and Isolation," Honors Theses 1975, 2119 (https://digitalworks.union.edu/theses/2119)(PDF).
D. T. Flood (Checked by W. W. Hartman and J. R. Byers), "Fluorobenzene," Org. Synth. 1933, 13, 46 (DOI: 10.15227/orgsyn.013.0046) and Organic Syntheses, Coll. Vol. 1943, 2, 295 (PDF).


What is the difference between IPA and Ethanol in terms of solubility? I have read an article online that says that ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why?

From some preliminary internet searches, I have discovered that:

"Isopropyl alcohol is a slightly bigger molecule than ethanol (Methanol is $\ce{CH4O}$, Ethanol is $\ce{C2H6O}$ and Isopropyl alcohol is $\ce{C3H8O}$) and is often cheaper to make. It is not as poisonous as methanol (an ounce of methanol can kill you and a third of an ounce causes blindness). So it is useful as a solvent in the same way acetone is useful as a solvent."

So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?

One final question. It would be best to use 99% pure solvents (mostly free of water), however for an ethanol replacement, I have seen a product used for ethanol fireplaces where the MSDS mentions 95% ethanol and 5% 2-propanol. Would this be an acceptable substitute for 99% ethanol? Thanks a lot.

[...] ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why? [...]

They hopefully didn't write about the "concentration of isopropanol" but mentioned that a larger amount of the solvent has to be used.

It is not uncommon that organic compounds with polar substituents are less soluble in higher alkanols.

Vanillin is a typical example, the solubilities are in the range of 4.15 mol/L in methanol, 2.5 mol/L in ethanol and 1.8 mol/L in 1-propanol. (Data are taken from a freely accessible collection of experiments performed by students: OpenNotebook Science Challenge). I couldn't find any data for 2-propanol.

  So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?

No, not really. A ketone and a secondary alkanol, although looking similar on first sight, may have behave completely different as solvents. Polystyrene (foam), as an example is pretty much insoluble in isopropanol but highly soluble in acetone!


Si(OH)4 vs. H4SiO4 vs. SiO2.2H2O Can someone please explain to me what is the differences between $\ce{Si(OH)4}$, $\ce{H4SiO4}$, and $\ce{SiO2.2H2O}$? I am currently studying sandstone acidizing geochemical models and this question came to me. It seems that $\ce{SiO2.2H2O}$ and $\ce{Si(OH)4}$ are basically the same thing and are considered minerals that may precipitate during the process, but $\ce{H4SiO4}$ is the aqueous form.
References

Quinn, M. A., Lake, L. W., & Schechter, R. S. (2000, February 1). Designing Effective Sandstone Acidizing Treatments Through Geochemical Modeling. Society of Petroleum Engineers. doi:10.2118/60846-PA
Walsh, M. P., Lake, L. W., & Schechter, R. S. (1982, September 1). A Description of Chemical Precipitation Mechanisms and Their Role in Formation Damage During Stimulation by Hydrofluoric Acid. Society of Petroleum Engineers. doi:10.2118/10625-PA
Sevougian, S. D., Lake, L. W., & Schechter, R. S. (1995, February 1). A New Geochemical Simulator To Design More Effective Sandstone Acidizing Treatments. Society of Petroleum Engineers. doi:10.2118/24780-PA

First, my answer here is relevant: What is chemical formula of Bauxite?, and also follow links inside.
Regarding your question specifically, there is not much difference. However, there are some conventions that come into play here.
The three formulae describe compounds that stoichiometrically have four atoms of oxygen and hydrogen for each atom of silicon.
$\ce{SiO2.2H2O}$
This notation simply says that for each component of "silica" there is a single component of "water". This notation is often used for solids, and never for dissolved species in aqueous solution. In some cases, this can denote that this entire thing has a crystalline structure, in which there are actual molecules of $\ce{H2O}$ held together with "molecules" of $\ce{SiO2}$. This is often called water of crystallisation, but this is not the case here because hydrated silica (which is what this thing is) is not crystalline, but is amorphous.
$\ce{Si(OH)4}$
This is a slightly more simplified way of writing the formula for hydrated silica, withour referring to specific chemical components (such as water or silica). In some cases this indicates that the crystal structure contains hydroxide groups ($\ce{OH-}$), but again, as hydrated silica is amorphous this is not the case here.
$\ce{H4SiO4}$
This is the formula for "silicic acid", which when in solid form, is exactly the same as "hydrated silica". However, this notation is often used when discussing hydrated silica in aqueous solution chemistry, as you correctly inferred. This does not necessarily mean that it is an actual acid, at least in the sense we think of when discussing things like sulfuric or phosphoric acid. We have a bottle of commercially acquired "silicic acid" in our lab. It's not an acid - it's just some hydrated silica, but written in the "acid"-type notation.

Something to remember here, is that some of these types of notation imply that the hydrogen occupies a specific structural site in the crystal lattice. The first implies molecular water, the second implies hydroxide groups, the third implies hydrogen protons.
This is misleading, because in fact hydrated silica can contain all three structural types of hydrogen simultaneously, and figuring out exactly which and their relative abundance is often of little importance in most studies. Adding to the confusion is that the compound is not really crystalline (like opal). And finally to confuse you even more, a ratio of 2-to-1 of water to silica isn't required, and it can vary. In fact, sometimes the formula is written as $\ce{SiO2.nH2O}$ because we simply don't know (and don't care) exactly how much water is in there. Finally, to add even more confusion, some groups use one notation while others use another for exactly the same thing. The same author can even use two different notations in two different papers!


Recrystallization Process Questions Should a crude compound should be dissolved completely in cold solvent and then heated up?

I think non-polar solvents are best choice for recrystallization of polar compounds, no?

Recrystallized product is then isolated using vacuum filtration

Can scratching the inside of the flask can help initiate crystallization?

If you chose to recrystallise by variation of temperature, you have to find a liquid that dissolves your product much better when heated (or even heated to reflux), than at room temperature.  A qualitative test may be performed with a small quantity of substance, for example a tip of a spatula, that you fill in a test tube, followed by a small amount of liquid.

Does your compound dissolve well in the liquid already at room temperature? This is then not a good choice of sole solvent for a recrystallisation, as much of the product will be retained by the liquid.  Intentionally I use liquid, because while you likely find a commonly used solvent suitable for recrystallisation, some other chemicals not typical perceived as "solvents" may be useful, too.

Depending on your equipment available, you may extend the accessible temperature range towards lower temperatures by use of an ice bath, or a freezer.  Be aware, however, some of the solvents that are liquid at room temperature may solidify then, too (glacial acetic acid, for example).

Often, you face a situation like "ethanol (as one solvent) has a too high solvent capacity for X, and water (as an other solvent) has a too low solvent capacity -- regardless if warmed, or heated to reflux".  Then it is time to use a mixture of mutually miscible liquids as solvent of recrystallisation, to tune the solvent properties.  This is why, for example, aqueous ethanol (ethanol with an intentionally added proportion of water) is used as solvent of recrystallisation, too.

There are some seemingly obvious choices of solvents of recrystallisation, like methanol and ethanol for not so polar organic compounds, and other somewhat surprisingly well working examples (acetic acid for less polar compounds); hence, recrystallisation offers a spectrum of science, craftsmanship, art, and obviously some luck, too.  Note that recrystallisation aiming for purification may be well a different one as aiming for nice crystals (for the purpose of single crystal diffraction analysis, for example).

If you made your choice, retain a small quantity of non-recrystallised compound which may be helpful later.  Mix your solid (or tar-like) product with the solvent / solvent mixture at room temperature and heat it up; if you want to heat up to reflux, consider a reflux condenser.  At high temperature, all visible solid should be dissolved.  If this is not the case, there is to little of solvent present and carefully add more.  After each addition of some new solvent, wait a little bit until the temperature of your mixture is the same as prior to this addition. (Caveat: If your product to recrystallise is a mixture, one component may dissolve better, than the other.  If so, consider a fractional recrystallisation or other means of purification first.)  Then allow the solution a slow cooling to room temperature, for example switch out the heater (and stirrer, too) of the oil bath yet keep the round flask in the oil bath.

Scratching the inner walls of the flask with a glass rod may be dangerous as you accidentally may pierce the wall (and consequently loose product to the oil bath).  A long Pasteur pipette is a better choice, too much pressure applied will break it, but not the flask.  But in general, it is one good method to generate crystal seeds for nice crystals.  (The optimal temperature for crystal seeding need not be the same as the optimal temperature for growing larger crystals.)

If possible (vide supra), complete the crystallisation by cooling below room temperature (ice bath) as this often lowers even more the solvent capacity.  If accessible, collect the crystals at this / near this low temperature, too.  If you want to rinse your crystals with a "bad solvent" (i.e. with low solvent capacity), equally cool it in advance by an ice bath.  If you crop your crystals by vacuum suction, prior cooling the equipment (Buchner filter, suction flask) by an ice bath / in a freezer may be helpful, too.

As a final note (here), do not throw away the mother liquor / filtrate of your recrystallisation too early.  It may contain a considerable amount of product.  Concentrate it (for example, at the rotary evaporator), and repeat the recrystallisation for a second crop.  Do not merge crops of subsequent crystallisation until an analysis reveals them as of sufficient similarity, like comparable purity.


Titration of sodium carbonate and nitric acid I am trying to find the pH of the following problem (Answer key is $\mathrm{pH}=10.77$)

  Calculate the $\mathrm{pH}$ of a $\pu{100 mL}$ solution containing $\pu{40.0 g}$ of $\ce{Na2CO3}$ after   $\pu{400 mL}$ of $\pu{1.00 M}$ $\ce{HNO3}$ has been added. Ignore  the volume  of water    produced    by  the reaction

My approach:
$$\ce{Na_2CO_3~(aq)~+~2HNO_3~(aq)\rightarrow CO_2~(g)~+~H_2O~+~2NaNO_3~(aq)}$$

I start with $\pu{0.377 mol}$ $\ce{Na2CO3}$ and $\pu{0.4 mol}$ $\ce{HNO3}$.  Because of the 2:1 ratio, I use up all of the $\ce{HNO3}$, and create $\pu{0.4 mol}$ $\ce{NaNO3}$ and have $\pu{0.177 mol}$ of $\ce{Na2CO3}$ left over.

I am unsure how to go about setting up the ice table.  Should I use that
$$\ce{NaNO3 + H2O -> HNO3 + Na+ + OH-}$$
where I have the initial amount of $\ce{NaNO3}$?

To solve this problem, you need to know the pKas of carbonic acid.

The pKas are 6.35 and 10.33.

If you add 1 equivalent of acid to $\ce{Na_2CO_{3}}$, the product will be $\ce{NaHCO_{3}}$ (sodium bicarbonate).

In this case, you are going 0.4-.377 = 0.023 moles past the point of 1 equivalent.

I would model the final solution as 0.354 moles of $\ce{HCO_{3}^-}$ and 0.023 moles $\ce{H_2CO_{3}}$ , and use the pKa of 6.35 and the Henderson-Hasselbalch equation to calculate pH.

The proposed answer 10.77 is incorrect.


Can the guanidinium ion give stable N-nitrosamines under mild metalworking conditions? Can the guanidinium ion react with inorganic nitrite under mild basic metalworking conditions ($\mathrm{pH\ 8 - 10}$ and temperature $\lt \pu{70^\circ C}$) to give stable N-nitrosamines?

Usually, the carcinogenic nitrosamines are generated from secondary amines. But guanidinium ion is not a secondary amine. Guanidinium ion is a mesomerism-stabilized system, and the nitroso group could be also stabilized by mesomerism therefore. So I'm afraid that it can build nitrosamines readily. But is this possible under basic conditions as mentioned above?

By definition, N-nitrosamines are aliphatic or aromatic derivatives of secondary amines, which have a nitroso group ($\ce{–NO}$) attached to nitrogen. In general, N-nitroso compounds are usually formed by the interaction of a nitrosating agent ($\ce{NOX}$) derived from either nitrite salts or nitrogen oxides with an amino substance such as secondary and tertiary amines, secondary and tertiary amides, peptides, substituted ureas, guanidines, and urethanes. Some important N-nitroso compounds are shown below:

As structures indicated, resonance in the framework has nothing to do with $\ce{-N-N=O}$ function (e.g., 1-Methyl-3-nitro-1-nitrosoguanidine is stable at room temperature and commercially available). Nonetheless, guanidine derivatives interact with $\ce{NOX}$ to give stable nitroso-guanidine derivatives (which are type of N-nitrosamides). With $\ce{NOX}$ derived from nitrite salts, the formation of N-nitroso derivatives of guanidine rarely occurs above $\mathrm{pH \ 4}$, but increases steadily with increasing acidity i.e. decreasing $\mathrm{pH}$, a maximum at  $\mathrm{pH \ 2.5-3.5}$ where the yield is inversely proportional to the basicity of the amino substance  (Ref.1).

On the other hand, guanidines do not react with $\ce{NOX}$ derived from nitrogen oxides in aqueous media (Ref.2,3). Note that $\ce{NOX}$ derived from nitrogen oxides worked best with strongly basic amino substances (e.g., aliphatic amines and heterocyclic amines) under non-aqueous or neutral and alkaline aqueous conditions (Ref.2,3).

References:

Sidney S. Mirvish, "Formation of N-nitroso compounds: Chemistry, kinetics, and in vivo occurrence," Toxicology and Applied Pharmacology 1975, 31(3), 325-351 (DOI: https://doi.org/10.1016/0041-008X(75)90255-0).
B. C. Challis, "Nutrition and nitrosamine formation," Proceedings of the Nutrition Society 1985, 44(1), 95-100 (DOI: https://doi.org/10.1079/PNS19850015) (PDF).
Brian C. Challis,  Soterios A. Kyrtopoulos, "The chemistry of nitroso-compounds. Part 11. Nitrosation of amines by the two-phase interaction of amines in solution with gaseous oxides of nitrogen," J. Chem. Soc., Perkin Trans. 1 1979, 299-304 (DOI: https://doi.org/10.1039/P19790000299).


Do electrons in covalent bonds always obey the octet rule? Do electrons in covalent bonds always necessarily obey the octet rule? Always?

The elements in borons row do not always follow the octet rule as in the case of boron fluoride as boron is in the center of the periodic table and will usually only bond to fill up its outer shell with 6 electrons. Additionally, elements with can form expanded octets which can share more than 8 electrons such as phosphorous pentachloride.  Of course hydrogen only has 2 electrons to bond with so it is also an exception to the octet rule.


Using charcoal to clean water I have read about charcoal being used to clean water before it is then boiled or mixed with a UV light stick. I have some questions about this:

  1) Does it have to be activated charcoal?

  2) Can I use normal charcoal from any old shop (in the UK) as I assume
  these days everything is synthetic or have added harmful chemicals?

  3) Can it purify irradiated water? Yes I am a huge Fallout 3 fan!

If there is anything else that I might not have thought of asking, or might be interesting, feel free to let me know :-).

Activated charcoal cleans water mainly by adsorbing the impurities.

You can also couple normal charcoal with sand and form alternate layers of charcoal and sand and pass water through it. It's a very effective Indian style of filtering water. Nearly all suspended particles gets blocked by the above mentioned filtering style.

You can also convert normal charcoal into activated charcoal to serve your purpose.

Now irradiated water cannot be treated by these methods alone. Normal carbon filtering is used along with several other methods such as ion exchange methods and other such high level water processing methods to treat irradiated water. High levels of radiation can't be treated fully though.

You can refer to this link to learn about effective radioactive water purification techniques.


Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable? Predict major product:

In this question first I've protonated the OH group and then water is removed to form a carbocation. Now I have to expand the ring, but I'm confused which ring to choose: 3-membered ring or 4-membered ring?

Note:  I can make an alkene (which would be the final product of the reaction) by removing an alpha hydrogen if it is known which ring to expand.

It is clear to me that no matter which ring you open the first, you'd get the same product as A. Michael Lautman gave the correct product , first opening the cyclopropyl ring. Michael's mechanism is fair, but I'm reluctant to say it is the most preferable path because it gives you bicyclobutyl $2^\circ$ carbocation as an intermediate, which needed extra 1,2-hydride shift to get stabilized as $3^\circ$ carbocation (total of 5 steps). However, if you have opened the cyclobutyl ring first, you'd get the same final product A, through a relatively low energy cyclopropylcyclopentane carbocation (total of 5 steps) even though it is also needed a 1,2-hydride shift (in relatively more stable cyclopentyl $2^\circ$ carbocation when compared to $2^\circ$ carbocation in cyclobytyl ring in Michael's mechanism) to get stabilized as $3^\circ$ carbocation. Complete mechanism is depicted below:

This mechanism is supported by following reference:

G. K. Surya Prakash, V. Prakash Reddy, G. Rasul, J. Casanova, G. A. Olah, “The Search for Persistent Cyclobutylmethyl Cations in Superacidic Media and Observation of the Cyclobutyldicyclopropylmethyl Cation,” J. Am. Chem. Soc. 1998, 120(51), 13362–13365 (https://doi.org/10.1021/ja9828962).


Carbon Dioxide Solubility in Water I did a lab where I tested the solubility of carbon dioxide in water at different temperatures. I took 50 mL of carbonated water at different temperatures and then I titrated that with sodium hydroxide solution. The indicator I used was phenolphthalein with an endpoint of 8.3 pH. I'm unsure how to calculate the amount of carbon dioxide that was dissolved. I'm stuck about deciding which equation to use.

This equation would mean that there is a 1:1 ratio.
$$\ce{CO2(aq) + NaOH(aq) → NaHCO3(aq)}$$
And this equation would mean that there is a 1:2 ratio.

$$\ce{CO2(aq) + H2O(l)->H2CO3(aq)}$$
$$\ce{H2CO3(aq) + 2NaOH(aq)->2H2O(l) + Na2CO3(aq)}$$

Which one would be the correct equation to use?

Also, there are literature values for the solubility of carbon dioxide in water at different temperatures. Do those solubility values apply to my lab? I ask this because I measured the amount of carbon dioxide present in a given amount of carbonated water, I didn't add carbon dioxide to pure water to measure how much can actually dissolve.

Sodium carbonate ($\ce{NaCO3}$) is basic and gives solutions with a pH of around 10.  Sodium bicarbonate ($\ce{NaHCO3}$) creates solutions with a pH of approximately 7.3.  This is in fact what the human body uses to regulate pH.  You want to used the sodium bicarbonate since it will yield a nearly neutral solution.


What exactly is entropy? I have looked at other questions, videos and articles but  I kept reading disorganization, messiness or energy distribution (probability) or a thermodynamical equilibrium and stuff. What do we mean by disorganization? And some say that it is a wrong definition. Also, there is this spontaneity and what is its relation to entropy? I am very confused and I would be very pleased if anyone explained it.

Entropy has two general precise definitions.  That which is most useful here is from Boltzmann, S(E) = k ln W(E) where S is the entropy, E is the energy of an isolated system (which is the easiest system to consider), k is Boltzmann's constant, and W(E) is the number of states of the system that have energy E.

For example, if the system has exactly one state with energy E, then S = 0.  If there are 2 states with energy E, then S = k ln 2.  And so on.

Now "organization" has no precise definition, but is rather an descriptive label we might apply to categorize states of the system.  Some of those states look "organized" to us, some look "disorganized."  For example, consider a model of atoms adsorbed on a crystal surface.  Let's model the surface as a square 3x3 array of sites and put 5 atoms on it.  A priori, there are 9!/(9-5)!4! = 126 ways of putting 5 atoms on 9 sites.  Now which of these states are "organized?"  That's to some extent a matter of opinion, but a common definition is "even" or "symmetrical."  So, for example, the state where there's an atom on every other site, in a checkerboard pattern, looks "organized" -- as if the sites formed a 2D crystal.  There's only one such state.

So now let us supposed we observe  this system, and we have some way of knowing, either from measurement or method of preparation, that it is in an "organized" (symmetric, checkerboard-like) or "unorganized" state.  There is only one "organized" state, so if we know the system is in an organized state then its entropy is k ln 1 = 0.  On the other hand, if we know the system is in an "unorganized" state there are 125 possible states it could be in, so its entropy is k ln 125, higher.

This is generally the case.  States we are willing to call "organized" generally have severe restrictions on the degrees of freedom: their values are distributed symmetrically in some way, e.g. the atoms are distributed uniformly, or in a regular array, the velocities are all the same, and so forth.  Because of these restrictions, it is almost always the case that there are fewer states of the system that satisfy the restrictions than that don't.  So the entropy of the system when it's known to be "organized" is less than when it is known to be not.

In short, it might help if when you hear "organized" you mentally translate that to "has some symmetry restrictions on how the degrees of freedom (e.g. atoms) are arranged."  Hopefully it is intuitive that there are far fewer ways of arranging atoms symmetrically than not, and therefore that the entropy of systems known to be symmetric is lower than the entropy of systems that are not.


Stereochemistry of hydrogenation with Raney nickel If I react the following with Raney nickel and $\ce{H2}$,

$\hspace{65 mm}$

then I am getting a meso compound, as the $\ce{H-}$ anion will have anti addition, right? One $\ce{H}$ will attack on right $\ce{C}$ from bottom and the other on the left $\ce{C}$ from the top. That makes it a meso. But in the book it says, there is racemization. How?

The hydrogenation of alkenes using heterogenous catalysts such as Raney nickel is syn-stereospecific; i.e. both hydrogen atoms are added to the same face of the double bond.

Therefore, the hydrogenation of (2E)-(2,3-2H2)but-2-ene yields
(2R,3R)-(2,3-2H2)butane and
(2S,3S)-(2,3-2H2)butane.


Calculating pH when weak base is added to an strong acid $\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is $\pu{25.0 cm3}$ and $\pu{26.0 cm3}$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is ${9.25}$.

So I started with the equation:

$$\ce{HCl + NH3 <=> NH4Cl}\tag{1}$$

EDIT #1:

As the initial concentration of both base and acid are same, I calculated how much of base is left in the solution:

  $V_\text{miscon} = \pu{26.0 cm3 - 25.0 cm3 = 1.0 cm3}$

That's how I tried to calculate pH of the solution when $\pu{26.0 cm3}$ of $\ce{NH3}$ is added:

  $\ce{H+}$ concentration:
  $$[\ce{H+}]_\text{miscon} = [\ce{H+}]_\text{initial} \frac{V_\text{initial}}{V_\text{miscon}}\tag{2} $$ $$=  \pu{(0.1 mol dm-3} \times \pu{0.025 dm3)/0.001 dm3} = \pu{2.5 mol dm-3}$$

But, in this way, the results I am getting are completely different from marking, and even negative pH values.

But, when I tried to find $\ce{H+}$ concentration (to calculate pH) of the solution when $\pu{25.0 cm3}$ of $\ce{NH3}$ is added, I found out that initial $\ce{HCl}$ volume is completely reacted. Only ammonium chloride is left in the solution.

The marking only gives the answers, $\ce{pH = 5.28}$ and $\ce{pH = 7.85}$.

How can I calculate pH of this solution in above two conditions separately?

I can't figure it out, I am stuck in the above equation. Please give me some hints to start solving. Thanks in advance.

EDIT #2:

I finally figured it out (Adding $\pu{25.0 cm3}$ of $\ce{NH3}$).

$$\ce{[NH4+(aq)]}=n(\ce{NH4+})/V_{\mathrm{\ce{total}}}$$
$$=(\pu{0.1 mol dm-3}\cdot\pu{0.025 dm3})/\pu{0.05 dm3}$$
$$=\pu{0.0025 mol}/\pu{0.05 dm3}$$
$$\ \ \ =5.0\times 10^{-2}\  \pu{mol dm-3}$$

$\ce{NH4+}$ reacts with water to produce $\ce{H3O+}$ ions.

$$\ce{NH4+(aq) + H2O(l)  <=> H3O+(aq) + NH3(aq)}$$

Using acid dissociation constant for $\ce{NH4+}$:

$$\ce{K_\mathrm{a} = [H3O+(aq)] [NH3(aq)] / [NH4+(aq)]}$$
$$\ce{[H3O+(aq)] = \pu{( K_\mathrm{a} \cdot [\ce{NH4+(aq)}] )0.5}}$$
$$\ce{= \pu{( 5.62\times 10^{-10}\pu{mol dm-3} \cdot 5.0\times 10^{-2}\pu{mol dm-3} )0.5}}$$
$$\ce{= \pu{( 2.81\times 10^{-11}\pu{mol2 dm-6} )0.5}}$$
$$\ \ \ =5.301\times 10^{-6}\  \pu{mol dm-3}$$

  $$ \text{pH} = -(\log 5.301 - 6\log 10) = -(0.7244 - 6) = 5.28 $$

The question
This is the question, corrected according to Poutnik's comment, and adding (i) and (ii) to distinguish the two scenarios:

$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is
(i) $\pu{25.0 cm3}$ and
(ii) $\pu{26.0 cm3}$.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH4+}$ is ${9.25}$.

The chemical reactions
The chemical equation (1) in the question is incorrect and missing something. It should reflect that in aqueous solution, hydrochloric acid is completely dissociated (strong acid), and ammonia is partially dissociated (weak base). One way of writing this is:
$$\ce{HCl(aq) + NH3(aq) -> NH4+(aq) + Cl-(aq)}\tag{A1}$$
$$\ce{NH4+(aq) <=> NH3(aq) + H+(aq)}\tag{A2}$$
Equation A1 shows that the reaction of a strong acid with a weak base goes to completion (no HCl will be present, chloride does not get protonated). Equation A2 shows that ammonium is a weak acid.
If you find it confusing that ammonia first picks up a proton, and then partially loses it again, you could also write it like this:
$$\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\tag{B1}$$
$$\ce{NH3 + H+ <=> NH4+(aq)}\tag{B2}$$
If you learned not to write $\ce{H+}$ but instead  $\ce{H3O+}$, you could write the B-equations slightly differently:
$$\ce{HCl(aq) + H2O(l)-> H3O+(aq) + Cl-(aq)}\tag{B1'}$$
$$\ce{NH3 + H3O+(aq)  <=> NH4+(aq) + H2O(l)}\tag{B2'}$$
In the end, each pair of equations says the same thing: Hydrochloric acid is as a strong acid, ammonia is a weak base. If ammonia is a weak base, we also know that the ammonium cation is a weak acid (with ammonia its conjugate base).
The calculation
There are two steps. You missed one (the one that involves the equilibrium). Let's work with equations A1 and A2 to work through the calculations.
You had the correct intuition about what happens with the hydrochloric acid: When adding 25 mL ammonia solution to the hydrochloric acid solution, the reactants are present at stoichiometric ratio, and the only species that is left to influence the pH is ammonium cation. When adding 26 mL ammonia solution, however, the ammonia is in excess, so some of it turns into the ammonium cation, and some remains. Scenario (i) is called a weak acid in solution, scenario (ii) is called a buffer solution. We can ignore the chloride ion, it is a spectator ion.
The amount of hydrochloric acid present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{HCl},initial}} = c_{\mathrm{\ce{HCl},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\  \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\  \mathrm{mol}$$

Scenario (i): weak acid

The amount of ammonia present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\  \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\  \mathrm{mol}$$
You calculate the concentration of species by first figuring out the amount, and then dividing by the volume of the solution, which is 50 mL. In scenario (i), all of the ammonia was turned into ammonium, so you know:
$$n_{\mathrm{\ce{NH4+}}}=2.5\times 10^{-3}\  \mathrm{mol}$$
$$c_{\mathrm{\ce{NH4+}}}= \frac{n_{\mathrm{\ce{NH4+}}}}{V_\text{solution}} = \frac{2.5\times 10^{-3}\  \mathrm{mol}}{\pu{50 mL}} = \pu{0.05 mol/L}$$
This concentration is lower by a factor two compared to the initial concentration of ammonia because the volume doubled when we added hydrochloric acid.
Now you can calculate the pH (pH of a weak acid with a pKa of 9.25 and a concentration of 0.05 mol/L:
$$ \text{pH} = \frac{1}{2} 9.25 - \frac{1}{2} \log 0.05 = 5.28 $$

Scenario (ii): buffer

$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.026\  \mathrm{L}$$
$$\ \ \ =2.6\times 10^{-3}\  \mathrm{mol}$$
HCl is limiting in reaction A1, so that determines the amount of product:
$$ n_{\mathrm{\ce{NH4+}}} = n_{\mathrm{\ce{HCl},initial}} = \pu{2.5e-3 mol}$$
Some of the ammonia is still present:
$$ n_{\mathrm{\ce{NH3}}} = n_{\mathrm{\ce{NH3},initial}} -  n_{\mathrm{\ce{NH4+}}} = \pu{2.6e-3 mol} - \pu{2.5e-3 mol} = \pu{0.1e-3 mol} $$
You can plug these concentrations into the Henderson-Hasselbalch equation (buffer equation) to get the pH:
$$ \text{pH} = 9.25 + \log \frac{0.1}{2.5} = 7.85$$


Neutralization between Ca(OH)2 + H2SO4 in a 30% hydrogen peroxide solution I have performed an experiment where I added an excess of $\ce{Ca(OH)2}$ base to a solution consisting of 5 mL of 30 % hydrogen peroxide (buffered at pH 5) and a very small amount of sulfuric acid (such that the pH of the original solution of 30 % $\ce{H2O2}$ and acid was 1.7).

I am having trouble understanding what reactions maybe occurring between these three substances to give a final pH of 10.5.

I know that $\ce{Ca(OH)2}$ and $\ce{H2O2}$ form $\ce{CaO2}$ when reacted, but shouldn't the final pH be equal to that of $\ce{Ca(OH)2}$ considering it is in excess? What reactions could be occurring here and why are they not allowing the pH to reach that of $\ce{Ca(OH)2}$?

Any help at all would be very much appreciated. Thank you in advance.

in your question formulation, you have forgotten to take into account $\ce{H2O2}$ is a weak acid.

The title should rather be:
Neutralisation between calcium hydroxide and 30% hydrogen peroxide"

Unless $\ce{Ca(OH)2}$ was in excess over $\ce{H2O2}$ - and it was said it was not - $\mathrm{pH}$ would be always significantly lower than pH of the hydroxide.

$$\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$

where $\mathrm{p}K_  \mathrm{a,\ce{H2O2}}=11.75$$ by Wikipedia, but see the links below.

If we consider reaction

$$\ce{Ca(OH)2 +  H2O2 ->  H2O + Ca(OH)(HO2)}$$

we need to neutralize 50% of $\ce{H2O2}$ to reach $\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}}$

The hydroxide forms from $\ce{H2O2}$  the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.

$$\begin{align}
\ce{Ca(OH)2  &<=>> CaOH+  + OH-}\\
\ce{CaOH+ &<=>> Ca^2+  + OH- }\\
\ce{H2O2 &<<=> H+ + HO2-}\\
\ce{H+ + OH- &<=>> H2O}\\
\end{align}$$

$\ce{Ca(OH)2}$: $\mathrm{p}K_\mathrm{b1} =1.37$, $\mathrm{p}K_\mathrm{b2} =2.43$
( Wikipedia )

Additionally, $\ce{HO2-}$ is partially eliminated by precipitation, therefore ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.

$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O v}$$

Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid and that it's $ \mathrm{p}K_  \mathrm{a}$ depends on $\ce{H2O2}$ concentration.

H2O2 pH-and-Ionization-Constant

The solubility constant of calcium peroxide octahydrate in relation to temperature; its influence on radiolysis in cement-based materials


Does hydration show kharasch effect? In the conversion of propene to 1-propanol, can we just use water in presence of peroxide instead of first converting it into 1-bromopropane and then reacting with $\ce{NaOH}$?

No, you can't.

If you observe the mechanism of Kharash effect, an $\ce{HBr}$ molecule first undergoes homolytic bond cleavage to produce a $\ce{Br}$ free radical. To undergo that homolytic bond fission, the electronegativities of the two elements, attached by that bond, have to be very close, which is possible for $\ce{H}$ and $\ce{Br}$ ($\chi _ {\ce{H}} = 2.1; \chi_{\ce{Br}} = 2.96$). The benzoyl peroxide is just a free radical initiator.

Now if you come to water, electronegativity difference between $\ce{H}$ and $\ce{O}$ is huge $(\chi _ {\ce{H}} = 2.1; \chi_{\ce{O}} = 3.44)$. So, it is not possible at all to undergo homolytic bond cleavage of $\ce{H-O}$ bond even in presence of peroxides.

To do hydration according to anti-Markownikoff's rule (without any rearrangement), the most widely used method is Hydroboration-Oxidation. This is an alternative and better than doing first bromination in presence of peroxides and then hydrolysis, because in many other cases chances of rearrangement also remains there.


Why does the cathode in an OLED require a low work function and the anode require a high work function? I am currently doing some research on organic light emitting diodes and am just trying to get my head around some of the terminology and concepts because I am more from the organic chemistry side don't have a strong material sciences/physics background.
My reading indicates that the work function is the amount of energy required to extract an electron from a material. Electrons are injected via the cathode and electrons are extracted from the anode (i.e hole injection)
I keep reading that for OLEDs, the cathode should have a low work function to match the energy of the LUMO of the electroluminescent layer, and the anode should have a high work function to match the HOMO. I've read this in many journals and review papers but none of them explain why.
Any help would be appreciated.

The operation of light emitting diodes involves the electrochemical creation of radical - cations and -anions that upon recombination lead to molecules in their emitting excited states.
The formation of the above charged species happen when the applied forward potential is positive enough to inject (reduction) electrons from the cathode to the LUMO of the molecule and extract electrons (oxidation) from the HOMO to the anode.
Therefore, the electrodes must have work functions about the levels you want to operate onto.
This means that the cathode must have oxidation potential / work function near the LUMO energy level. Opposite, to work on the extraction of electrons from the HOMO, a metal lining down at higher oxidation potential is needed.
Note that this design not only reduces the potential needed for the functioning of the LED, but minimise energy barriers (if any) at the emitting material / electrode interfaces.
Also note that a reverse order can still lead to light emission, but the ON potential to be applied to this "unnatural" configuration is inevitably higher.


What is the difference between ∆G and ∆G°? In Brady's Molecular Nature of Matter, I read that $\Delta_\mathrm{r} G^{\circ}$ is $\Delta_\mathrm{r} G$ at $25~^\circ\mathrm{C}$. But later, it gives a value for $\Delta_\mathrm{r} G^\circ$ at another temperature.

What is the meaning of $\Delta_\mathrm{r} G^\circ$?  Specifically:

Does it only refer to $25~^\circ\mathrm{C}$?
Does the $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T \Delta_\mathrm{r} S^\circ$ equation only refer to $T=298~\mathrm{K}$?
If at $25~^\circ\mathrm{C}$, a system that has reached equilibrium ($\Delta_\mathrm{r} G=0$), and thus

$Q=K$
$\Delta_\mathrm{r} G^\circ=0$
$\Delta_\mathrm{r} G^\circ=0=-RT\ln K$
$RT\ln K=0$
$\ln K=0$
$K=1$

But doesn't $K$ vary from reaction to reaction? It shouldn't always equal 1, should it?

Short answer

  Does it need to be at $25~^\circ\mathrm{C}$?

No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change.

  Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ$ always use $T = 298~\mathrm{K}$?

No. You use whatever temperature you are running your reaction at.

  (...maths...)

Yes, at equilibrium, $\Delta_\mathrm{r} G = 0$ and $Q = K$.

However, everything after the first bullet point is wrong. You cannot conclude that $\Delta_\mathrm{r} G^\circ = 0$, nor can you conclude that $K = 1$. The equation $\Delta_\mathrm{r} G^\circ = -RT \ln K$ does not analogously translate into $\Delta_\mathrm{r} G = -RT \ln Q$! The accurate relation is:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$

Setting $Q = K$ and $\Delta_\mathrm{r} G = 0$ in this equation does not tell you anything about the value of $K$. In fact, if you try doing it, all you will find out is that $\Delta_\mathrm{r} G^\circ = -RT \ln K$ - no surprises there!

Long answer

Any book that writes that $\Delta_\mathrm{r} G^\circ$ is the "special case" of $\Delta_\mathrm{r} G$ at $T = 298~\mathrm{K}$ is wrong.

The Gibbs free energy of a system is defined as follows:

$$G = H - TS$$

Under constant temperature and pressure (from now on, I will just assume constant $T$ and $p$ without stating it), all systems will seek to minimise their Gibbs free energy. Equilibrium is reached when $G$ is minimised. When $G$ is at a minimum, any infinitesimal change in $G$, i.e. $\mathrm{d}G$, will be $0$. Therefore, this is equivalent to saying that the condition for chemical equilibrium is $\mathrm{d}G = 0$.

Clearly, we need a way to relate this quantity $\mathrm{d}G$ to the actual reactants and products that are in the system. This can be done by using the Maxwell relation (see any physical chemistry text for details):

$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$

Under constant $T$ and $p$, $\mathrm{d}p = \mathrm{d}T = 0$ and therefore

$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i$$

where $\mu_i$ is the chemical potential of species $i$, defined as a partial derivative:

$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$

So, we now have a refined condition for equilibrium:

$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i = 0$$

We can go further by noting that the values of $\mathrm{d}n_i$ for different components $i$, $j$, etc. are not unrelated. For example, if we have a reaction $i + j \longrightarrow k$, then for each mole of $i$ that is consumed, we must also use up one mole of $j$; this means that $\mathrm{d}n_i = \mathrm{d}n_j$.

This can be formalised using the idea of a stoichiometric coefficient $\nu_i$, which is defined to be positive for products and negative for reactants. For example, in the reaction

$$\ce{3H2 + N2 -> 2NH3}$$

we have $\nu_{\ce{H2}} = -3$, $\nu_{\ce{N2}} = -1$, and $\nu_{\ce{NH3}} = 2$.

By stoichiometry, if $1.5~\mathrm{mol}$ of $\ce{H2}$ is consumed, then $1~\mathrm{mol}$ of $\ce{NH3}$ has to be produced. We could write $\Delta n_{\ce{H2}} = -1.5~\mathrm{mol}$ and $\Delta n_{\ce{NH3}} = 1~\mathrm{mol}$. These quantities are proportional to their stoichiometric coefficients:

$$\frac{\Delta n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{-1.5~\mathrm{mol}}{-3} = 0.5~\mathrm{mol} = \frac{1~\mathrm{mol}}{2} = \frac{\Delta n_{\ce{NH3}}}{\nu_{\ce{NH3}}}$$

The quantity $0.5~\mathrm{mol}$ is a constant for all chemical species $\ce{J}$ that participate in the reaction, and it is called the "extent of reaction" and denoted $\Delta \xi$ (that is the Greek letter xi). If the reaction is going forward, then $\Delta \xi$ is positive, and if the reaction is going backwards, then $\Delta \xi$ is negative. If we generalise the above result, we can write

$$\Delta \xi = \frac{\Delta n_i}{\nu_i}$$

and if we make $\Delta n_i$ smaller and smaller until it becomes an infinitesimal, then:

$$\begin{align}
\mathrm{d}\xi &= \frac{\mathrm{d}n_i}{\nu_i} \\
\mathrm{d}n_i &= \nu_i\,\mathrm{d}\xi
\end{align}$$

If we go back to our condition for equilibrium, we can substitute in the above to get:

$$\mathrm{d}G = \sum_i \mu_i\nu_i\,\mathrm{d}\xi = 0$$

Now, $\mathrm{d}\xi$ is no longer dependent on $i$, since we have established already that $\Delta \xi$ (and by extension $\mathrm{d}\xi$) is a constant for all chemical species. So, we can "divide through" by it to get:

$$\Delta_\mathrm{r} G \equiv \frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$

where $\Delta_\mathrm{r} G$ is defined to be $\mathrm{d}G/\mathrm{d}\xi$.

Note that $\Delta_\mathrm{r} G$ is an intensive property and has units of $\mathrm{kJ~mol^{-1}}$, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$. This ensures that the units we use are consistent: since we know that $\Delta_\mathrm{r}G = \Delta_\mathrm{r} G^\circ + RT\ln Q$, $\Delta_\mathrm{r}G$ must have the same units as $RT$.

How do we interpret the physical significance of $\Delta_\mathrm{r} G$, or in other words, what does it even mean? There are two ways, each based on a different mathematical expression.

We have $\Delta_\mathrm{r}G = \sum \nu_i \mu_i$. This means that $\Delta_\mathrm{r}G$ is simply the difference between the chemical potentials of the products and the reactants, weighted by their stoichiometric coefficients. For the reaction $\ce{3H2 + N2 -> 2NH3}$, we have:

$$\Delta_\mathrm{r} G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$

We have $\Delta_\mathrm{r}G = \mathrm{d}G/\mathrm{d}\xi$. This means that it is the slope of a curve of $G$ against $\xi$:

Note that up to this point, we have not stipulated any particular temperature, pressure, amounts of species present, or any conditions whatsoever. We have only said that the temperature and pressure must be constant.

It is important to realise that $\Delta_\mathrm{r}G$ is a well-defined quantity at all $T$, all $p$, and all possible values of $n_i, n_j, \cdots$! The shape of the curve will change when you vary the conditions. However, no matter what the curve looks like, it is always possible to find its gradient ($= \Delta_\mathrm{r}G$) at a particular point.

What exactly, then, is $\Delta G^\circ$? It is just a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:

For a gas: pure ideal gas when the pressure $p$ is equal to the standard pressure $p^\circ$.
For a liquid or solid: pure liquid or solid at $p = p^\circ$
For a solution: ideal solution when the concentration $c$ is equal to the standard concentration $c^\circ$.

$p^\circ$ is most commonly taken to be $\pu{1 bar}$, although older texts may use the value $\pu{1 atm} = \pu{1.01325 bar}$. Since 1982, IUPAC has recommended the value $\pu{1 bar}$ for the standard pressure (Pure Appl. Chem. 1982, 54 (6), 1239–1250; DOI: 10.1351/pac198254061239). However, depending on the context, a different value of $p^\circ$ may prove to be more convenient. Likewise, $c^\circ$ is most commonly – but not necessarily – taken to be $\pu{1 mol dm-3}$.

Note that in the above definitions, no temperature is specified. Therefore, by defining the standard Gibbs free energy, we are fixing a particular value of $p$, as well as particular values of $n_i, n_j, \cdots$. However, the value of $T$ is not fixed. Therefore, when stating a value of $\Delta_\mathrm rG^\circ$, it is also necessary to state the temperature which that value applies to.

When a reaction vessel is prepared with all its substances in the standard state, all the components of the system will have an activity of exactly $1$ by definition. Therefore, the reaction quotient $Q$ (which is a ratio of activities) will also be exactly equal to $1$. So, we could also say that $\Delta_\mathrm{r}G^\circ$ is the value of $\Delta_\mathrm{r}G$ when $Q = 1$.

Returning to the graph of $G$ against $\xi$ above, we note that at the left-most point, $Q = 0$ since there are only reactants; at the right-most point, $Q \to \infty$ as there are only products. As we move from left to right, $Q$ increases continuously, so there must be a point where $Q = 1$. (In general, the point where $Q = 1$ will not be the same as the equilibrium point.) Since $\Delta_\mathrm{r}G$ is the gradient of the graph, $\Delta_\mathrm{r}G^\circ$ is simply the gradient of the graph at that particular point where $Q = 1$:

The gradient of the graph, i.e. $\Delta_\mathrm{r}G$, will vary as you traverse the graph from left to right. At equilibrium, the gradient is zero, i.e. $\Delta_\mathrm{r}G = 0$. However, $\Delta_\mathrm{r}G^\circ$ refers to the gradient at that one specific point where $Q = 1$. In the example illustrated above, that specific gradient is negative, i.e. $\Delta_\mathrm{r}G^\circ < 0$.

Again, I reiterate that the temperature has nothing to do with this. If you were to change the temperature, you would get an entirely different graph of $G$ versus $\xi$. You can still find the point on that graph where $Q = 1$, and the gradient of that graph at the point where $Q = 1$ is simply $\Delta_\mathrm{r} G^\circ$ at that temperature.

We have established the qualitative relationship between $\Delta_\mathrm{r} G$ and $\Delta_\mathrm{r} G^\circ$, but it is often useful to have an exact mathematical relation.

$\Delta_\mathrm{r} G^\circ$ is exactly the same as $\Delta_\mathrm{r} G$ except for the imposition of the standard state. It follows that if we take the equation

$$\Delta_\mathrm{r} G = \sum_i \mu_i \nu_i$$

and impose the standard state, we get

$$\Delta_\mathrm{r} G^\circ = \sum_i \mu_i^\circ \nu_i$$

Thermodynamics tells us that

$$\mu_i = \mu_i^\circ + RT\ln{a_i}$$

where $a_i$ is the thermodynamic activity of species $i$. Substituting this into the expressions for $\Delta G$ and $\Delta G^\circ$ above, we obtain the result:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$

where the reaction quotient $Q$ is defined as

$$Q = \prod_i a_i^{\nu_i}$$

When equilibrium is reached, we necessarily have $\Delta_\mathrm{r} G = 0$ (see the discussion above). The equilibrium constant $K$ is defined to be the value of $Q$ at equilibrium. Therefore, at equilibrium, $Q = K$. Plugging this into the equation above gives us the famous equation:

$$\Delta_\mathrm{r} G^\circ = -RT\ln K$$

Again, no temperature is specified! In general, $K$ depends on the temperature as well; the relationship is given by the van 't Hoff equation.


What reaction occurs when soaking a piece of meat in apple cider vinegar? I would like to know what reaction occurs when  soaking a piece of meat in apple cider vinegar. Additionally, how does it happen?
I read a topic which was about the bad reputation of apple cider vinegar, which can actually "cook" your skin texture.

This "cooking" phenomenon happens because of the basic building blocks of a meat (protein) and how they can be broken down (which is called "denaturation"): heat (traditional cooking), pH changes (your vinegar question) or enzymes (bromelin or papain, which are typical ingredients of meat tenderizer)

When exposed to these factors, the proteins that make the bulk of the meat "break", and the result you can observe is very similar, which is the "cooking". Since it varies with the exposure level to the denaturating agent, this is why you only see the "cooking" happening on the surface of marinaded meats.

pH changes is also an alternative "cooking" method used in some dishes, e.g. ceviche, which is raw fish finely chopped that is soaked in lemon juice to "cook"

Hope it helped :)


Why is chloride measurment 0 In class we did an experiment to measure chloride in different water sources, including a lake.  We used Ion-selective electrode of Ag/AgCl technique.
We did the calibration curve and all steps and measurments in different sources of water are consistent, for example drinking water was a bit below 250ppm. But when measuring lake water we get a measurment of 0, why could this be happening?
I can't find a reason

Most likely this was an artificial "lake" sample and someone gave you distilled water instead. Making artificial water samples is not an unusual practice in teaching laboratories.
Natural water cannot have zero concentration of chloride ion. Sodium and chloride ions are very common contaminants even in pure reagents. There is no "zero" concentration concept in analytical chemistry at all. The proper practice is to write "Not detected" instead of saying 0 ppm. All it means that it was below the detection limit of the ion-selective electrode. Ion selective electrode usually measure samples in parts per million (mg/L) range. Typical limits of detections are in single digit ppm.


Why does pasta really boil over? I was making pasta, and I noticed the pasta boiling over. I thought about it some more, and I realized I had no idea why this was happening. When the lid is on, the foam rises. When the lid is off, the foam dies back. Clearly there's some surfactant at work allowing bubbles to form, and the bubbles are mostly full of steam. If you remove the lid, they cool and condense down to a tiny size. This article is honestly just not specific enough. Why is the starch important?(It also happens with potatoes) How does a polysaccharide, a hydrophilic molecule, become a surfactant(if indeed it does)? Is it phospholipids from the cells that became pasta? Do starch molecules form a polymer and trap steam underneath like a balloon? Is it some kind of hybrid where bubbles are stabilized by viscosity? What's going on here?
This question on seasoned advice tries to tackle it but gets surface tension wrong(I think) so that makes me a little wary of it. Surfactants like soap reduce the surface tension, allowing bubbles to form by letting water molecules spread out into thin films. They also keep your alveoli in your lungs from collapsing. If starches increased the surface tension, wouldn't that reduce the likelihood of forming bubbles by increasing the energetic penalty? High surface tension materials act like mercury.
This answer is all over the internet but doesn't make any sense to me. You need less surface tension for bubbles, surely?
If it's just phospholipids acting as detergent, could I 'boil over' pasta in cold water with a whisk? (I tried this, you can't.)
I considered putting this on seasoned advice, but I'm not interested in how to prevent boil-over(or really any practical results), and the supplied answer there lacks scientific rigor.

The starch forms a loosely bonded network that traps water vapor and air into a foamy mass, which expands rapidly as it heats up.

Starch is made of glucose polymers (amylopectin is one of them, shown here):

Some of the chains are branched, some are linear, but they all have $\ce{-OH}$ groups which can form hydrogen bonds with each other.

Let's follow some starch molecules through the process and see what happens. In the beginning, the starch is dehydrated and tightly compacted - the chains are lined up in nice orderly structures with no water or air between them, maximizing the hydrogen bonds between starch polymers:

As the water heats up (or as you let the pasta soak), water molecules begin to "invade" the tightly packed polymer chains, forming their own hydrogen bonds with the starch:

Soon, the polymer chains are completely surrounded by water, and are free to move in solution (they have dissolved):

However, the water/starch solution is not completely uniform. In the middle of the pot of water, the concentration of starch is low compared to water. There are lots and lots of water molecules available to surround the starch chains and to keep them apart. Near the surface, when the water is boiling, the water molecules escape as vapor. This means that near the surface, the local concentration of starch increases. It increases so much as the water continues to boil, that the starch can collapse back in on itself and hydrogen bond to other starch molecules again. However, this time the orderly structure is broken and there is too much thermal motion to line up. Instead, they form a loosely packed network of molecules connected by hydrogen bonds and surrounding little pockets of water and air (bubbles):

This network is very weak, but it is strong enough to temporarily trap the air as it expands due to heating - thus, the bubbles puff up and a rapidly growing foam forms. Since they are very weak, it doesn't take much to disrupt them. Some oil in the water will inhibit the bubbles from breaking the surface as easily, and a wooden spoon across the top will break the network mechanically as soon as it touches it.

Many biomolecules will form these types of networks under different conditions. For example, gelatin is a protein (amino acid polymer) that will form elastic hydrogen-bonded networks in hot water. As the gelatin-water mixture cools, the gel solidifies, trapping the water inside to form what is called a sol-gel, or more specifically, a hydrogel.

Gluten in wheat is another example, although in this case the bonds are disulfide bonds. Gluten networks are stronger than hydrogen-bonded polysaccharide networks, and are responsible for the elasticity of bread (and of pasta).

DISCLAIMER:

pictures are not remotely to scale, starch is usually several hundred glucose monomers long, and the relative size of the molecules and atoms isn't shown.
there aren't nearly enough water molecules - in reality there would be too many to be able to see the polymer (1,000's).
the starch molecules aren't "twisty" enough or showing things like branching - the real network structure and conformations in solution would be much more complicated.

But, hopefully you get the idea!


Can I anodize an aluminum coating on steel to have better corrosion inhibition for the steel? I am planning to coat aluminum on steel and I want to give the steel its maximum corrosion protection. Is anodizing the coating necessary?

The aluminum oxide layer formed by anodizing will have substantially higher corrosion and abrasion resistance than non-anodized aluminum.  For most applications, simple Type II sulfuric acid anodizing is sufficient.  For maximum protection, Type III hard anodizing will yield a much thicker coating with substantially higher corrosion and abrasion resistance. If you require a thinner film, the older Type I chromic acid anodizing process may be used.

If you plan to paint or dye the part, you will most likely need to stick to Type II sulfuric acid anodizing or Type I chromic acid anodizing.  Type III hard anodizing, depending upon the alloy, will result in a dark grey to an almost black finish, so dyes will not show. Paint adhesion is also an issue with Type III.

See MIL-A-8625 for more complete information about the various types of aluminum anodizing.


Stability of transition and post-transition metal alkoxides Many transition metal alkoxides are seen in chemistry, for instance those of $\ce{Ti, Sn}$ and $\ce{Al}$, and in particular one of those of aluminium has its own role in the MPV reduction reaction.

I was wondering: how stable these $\ce{M(OR)_n}$ structures are?

To simplify the question:

Are they susceptible to acid/basic attacks (which could possibly hydrolyze them, maybe)?
Being them metal-alkoxide compounds, do they present similar reactivity as alkali metal oxides (eg: use in Williamson's synthesis of ethers)?
Would it be possible to obtain an alkali metal alkoxide from a transition metal alkoxide, hence synthesizing those compounds without using alkali metals?
When used in a reaction as a catalyst (eg: MPV or Oppenauer reactions), a metal alkoxide can be recovered and reused?

Alkali metal, main group, and transition metal (TM) alkoxides all have significantly different chemistry. Early- and Late-TM alkoxides also have different behaviour, so we should split this family into at least four categories. All of them are basic and will decompose via hydrolysis if dissolved in water; however, this type of chemistry is not usually carried out in presence of water.

In TM chemistry, alkoxide or phenoxide are mostly used as ancillary ligands, meaning they do not usually participate in the chemistry. Although they are liable to be hydrolyzed under protic conditions, alkoxides are actually known for being especially stable TM ligands because they do not undergo β-hydride elimination (BHE) (1), which is one of the most common decomposition pathways for TM alkyls. For instance; you often cannot make a stable t-butyl TM complex because the β-hydride will be transferred to the transition metal, liberating 2-methylpropene unless the metal is totally coordinatively saturated. An important example of a BHE-stable TM t-butoxide is the following Mo(IV) Schrock catalyst, which is stable:

$\hskip2.4in$

It should be noted, though, that the Schrock catalyst doesn't actually have any β-hydrides. Later, I'll show an example of a TM-alkoxide that does.

An important feature of alkoxides as ligands for TM chemistry is that they are electronically malleable. Since an alkoxide has 5 electrons it can behave either as an σ-donating X ligand or as a σ- and π-donating LX ligand (and formally as L2X). Presumably, this electronic malleability has something to do with their stability to BHE. An alkoxide's π-donor character might stop the metal center from liberating a coordination site to accept the β-hydride. For more information on BHE-stability, see (2), as cited by Schrock in (1).

Since early-TMs are more electropositive, the bonding character of their alkoxides is more ionic, although not to the same degree as that of alkali metal alkoxides. Alkoxides are very good bridging ligands, and if they are not sufficiently bulky, alkoxide complexes often polymerize. With a sufficiently bulky monomeric early-TM alkoxide, the main decomposition pathway is formation of the oxo complex. For example; the decomposition of a metal diolate shown in the next image was observed by Gable et al. (3) Their Hartree-Frock calculations implicated an empty metal MO at rhenium in the M=O bond formation, as shown in the MO picture in the next image.

$\hskip0.2in$

Late-TM alkoxides are expected to have weaker M-O bonds; and not as many have been isolated, but a new study (4) has examined these M-X (where X = O, N) bonds.

A really nice example illustrating a few important TM-alkoxide concepts is tantalum(V) ethoxide, Ta2OEt10. See wikipedia: https://en.wikipedia.org/wiki/Tantalum(V)_ethoxide

$\hskip1.5in$

Ta2OEt10 is a stable TM-alkoxide with true β-hydrides, bridging oxygens, and it's readily hydrolyzed when exposed to protons. It can be synthesized via salt metathesis with sodium ethoxide, a reaction that has equilibirium lying in the opposite direction of the alkali metal alkoxide formation you proposed.

In MPV reduction (and OPP oxidation), Al(O-i-Pr)3 is usually used as the catalyst. Aluminium is a metal, but it is a main group metal, and its chemistry is somewhere between that of boron, a semimetal, and that of early transition metals, except, of course, that it does not have a partially filled d shell. Al(O-i-Pr)3 is an electron deficient Lewis acid, comparable in chemistry to boronate esters, but much more reactive. The proposed mechanism for MPV/OPP [R]/[O] goes something like this:

$\hskip0.2in$

It is a true catalytic reaction, so the Al(O-i-Pr)3 is indeed regenerated, but I don't know if it's a routine part of the procedure to recover it. Al(OR)3's are not, however, soluble in IPA, but alcohols and ketones usually are, so you could probably precipitate Al(O-i-Pr)3 in IPA providing you have a way to seperate your ketone and your secondary alcohol afterwards.

Interestingly, Ti(O-i-Pr)4, is analogous to Al(OR)3, but its reactivity totally different, since titanium is a true TM and not just a Lewis acid main group metal like aluminium. Ti(O-i-Pr)4 is the catalyst for sharpless epoxidation, so this might be a good reaction to look at as an example of early-TM alkoxide reactivity.

To answer your questions in list form:

Pretty much all alkoxides are susceptible to acidic attacks, but, being fairly strong bases themselves, they are probably stable to weaker bases
Although they will react with acids, they will not necessarily react with electrophiles unless they are stronger electrophiles than the metal center; however, early-TMs are fairly strong electrophiles. I don't think people use TM alkoxides for ether synthesis though, and you always have to keep in mind d orbitals and the unique rules of TM chemistry when you're discussing TM complexes.
As mentioned above, and in my comment; "You can't have MX + M'(OR) -> M(OR) + M'X, where M=alkali metal and M'=TM, because the equilibrium would lie in the opposite direction, since the highly nucleophilic alkali metal alkoxide M(OR) will attack the electrophilic transition metal M'X (especially if it's an early-TM). As Schrock says in (1), "the relatively high electrophilicity of an early transition metal makes dissociation of alkoxide ion, at least in a neutral species, unlikely".
If you can find a suitable means of seperation like a solvent that will precipiate your metal alkoxide, you should be able to seperate it, although some catalysts will be deactivated after a certain number of cycles.

(1) R. R. Schrock, Polyhedron, Vol. 14., No. 22, pp. 3177-3195, 1995

(2) D. C. Bradley, R. C. Mehrotra and D. P. Gaur, Metal Alkoxides. Academic Press, New York, 1978

(3) K. P. Gable, J. J. J. Juliette. J. Am. Chem. soc, 1996, 118, 2625-2633

(4) J. R. Fulton, A. W. Holland, D. J. Fox, R. G. Bergman. Acc. Chem. Res., 2002, Jan; 35(1): 44-56


Can we have an adiabatic process in a sealed metal container? Let us consider a non-isothermal process inside a metal container. If we somehow are able to make the temperatures on the inside and outside the system remain same throughout the process, can it be called an adiabatic process?

Adiabatic means that energy is transferred between the system and the surroundings only as work and not as heat.

Let's look at your process. You have a non-isothermal process, so $\Delta T_{sys} \ne 0$ for the process. An adiabatic process does not need to be isothermal, so this condition is okay. What makes your process non-adiabatic is the condition that constant temperature is somehow maintained (likely through a constant temperature bath which serves as the surroundings). This requires heat exchange between the system inside the metal container and the surroundings outside the container.

If you remove the constraint that the overall temperature is held constant, then you can have an adiabatic process.


Why aren't the R-side-chains of acidic and basic amino acids the site of polypeptide synthesis? If polypeptides are synthesized by acid-base reactions between the carboxyl group of one amino acid and the amine group of another amino acid, what stops the formation of a peptide bond between the alpha amine group of one amino acid and the carboxyl group (that's in the side chain) of another amino acid?

For example, why doesn't the side chain of glutamic acid bond to the alpha amine of another amino acid (essentially creating a new branch at which polypeptide synthesis can occur)?

Chemically

If you were making a small polypeptide you'd have a protecting group on the side-chain of glutamic acid, and preferably also an activating group on the carboxyl where you want to attach it. If you're just mixing glutamate with another amino acid you'll get a mixture as you expect (including all kinds of possible oligomers).

Biologically

Protein synthesis in the ribosome actually works the other way around (the amine-end is in the ribosome, tRNAs are attached to residues at the carboxyl side). So your expected error would be the continuation of protein synthesis at another alpha amine: the lysine sidechain. Apparently the ribosome is quite good at distinguishing this 4-carbon difference.

My 2 minute hypothesis is that this is one of the reasons why there are residues with shorter acid sidechains (glutamate and aspartate) while a shorter version of lysine is not used in biological protein synthesis.

How the tRNA synthases are so specific that they don't load the amino acid with the wrong carboxyl group is a different question, but tRNAs are extremely specific enzymes in many ways (they have no problem recognizing the difference between a methyl and a hydroxyl group).


How does the ionic strength formula work? (simple version) The ionic strength formula provides a way to estimate the Debye screening length. I’m having some trouble understanding the structure;

$$I = \begin{matrix}\frac{1}{2}\end{matrix}\sum_{{\rm i}=1}^{n} c_{\rm i}z_{\rm i}^{2}$$

It seem a bit tricky to find the derivation for this, but all I’m really looking for is an intuitive answer to a couple of questions;

Firstly; why is there a factor of a half out the front?
Secondly; why do we square the charge number of the ion?

I am a physicist, so please explain any chemistry concepts like I’m 5. Thank you for your patience.

Firstly; why is there a factor of a half out the front?

To avoid double-counting. For a salt such as $\ce{NaCl}$ comprising a monovalent cation and a monovalent anion, the ionic strength is defined such that it should be equal to the molar concentration of the salt, as opposed to the sum of both the anion and cation concentrations.  Therefore, since such a salt has two constituents, each constituent contributes half of the total ionic strength.

  Secondly; why do we square the charge number of the ion?

For this there's not much choice but to turn to the Poisson-Boltzman equation, which provides a differential expression for the electric field inside the electric double-layer. (Sorry, 5-year-old version of Jekowl!) Below is the version I have in my graduate transport book, Eq. (15.4-6):

$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\exp{\left(-{z_iF\over RT}\Phi\right)}}
$$

$\Phi$ is the electric potential, $\mathrm{J} \over \mathrm{C}$ or $\mathrm{V}$
$F$ is Faraday's constant, $96,485\ {C \over \mathrm{mol}\ q}$
$\epsilon$ is the electrical permittivity, $\mathrm{C}\over \mathrm{V\ m}$
$z_i$ is the valence of ion $i$, $\mathrm{mol}\ q \over \mathrm{mol\ ion}$
$C_{i\infty}$ is the concentration of ion $i$ far from the surface of interest, $\mathrm{mol} \over \mathrm{m^3}$
$R$ is the ideal gas constant, $8.314\ {\mathrm{J} \over \mathrm{mol\ K}}$
$T$ is temperature, $\mathrm{K}$
$q$ is the elementary charge

If the magnitude of the electric field is small such that $\left|z_iF\Phi/RT\right|\ll 1$, the exponential can be expanded as a Taylor series:

$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\left(1-{z_iF\over RT}\Phi+\dots\right)}
$$

If you then pull everything inside the parentheses, ignoring signs the second term becomes:

$$
C_{i\infty}z_i^2{F^2\over \epsilon RT}\Phi
$$

The second-order dependence of $z_i$ is readily visible in the above.