text:
Resources to learn computational chemistry?
I am interested in learning computational chemistry to explore the possibility of postgraduate work in quantum/theoretical chemistry and would like to attempt self-studying.  I know the basics of quantum mechanics on par with a p.chem course and have an understanding of linear algebra and differential equations.  Are there any other resources that would help me in learning?
As far as I know, there is no software suited for learning computational chemistry on a desktop PC.  But would there be anything similar that would assist in my learning?
Also I am lacking in the programming side of things, would that be a significant barrier to my learning, especially when using computational software in the future?

Here is a good place to go, http://molecularmodelingbasics.blogspot.co.uk/ and the software can be obtained here http://www.msg.ameslab.gov/gamess/ then there are youtube tutorials


text:
Determine products of redox-reaction (KMnO4) by Standard electrode potential
I want to figure out a way how to determine the products of a redox-reaction by looking at the documented Standard electrode potential (for example here). But with my "way" I have some trouble/contradictions with common redox reaction, in my case the reduction of $\ce{MnO4-}.
Here is my procedure for the reduction of $\ce{MnO4-}$:

Collect all available reactions with their Standard Potential:*
\begin{align}
\ce{MnO4- + 2H2O + 4e- &-> MnO2 + 4OH-}&
E(\text{Red})&= \pu{0.83 V}\tag{a}\label{a}\\
\ce{MnO4- + H+ + e- &-> HMnO4^-}
& E(\text{Red}) &= \pu{0.72 V}\tag{b}\label{b}\\
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+}&
E(\text{Red}) &= \pu{1.23 V}\tag{c}\label{c}\\
\ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}&
E(\text{Red}) &=\pu{1.6 V}\tag{d}\label{d}
\end{align}
Then figure out the reaction with the highest reduction potential, because according to $E = E(\text{Red}) - E(\text{Ox})$, $E$ is maximal with the biggest $E(\text{Red})$ and the reaction is going to happen with the highest $E$.
In my example, $\eqref{d}$ has the highest potential, so according to my theory $\ce{MnO2}$ is formed in the reduction of $\ce{MnO4-}$ (in combination with for example the oxidation of $\ce{I-}$. But in all textbooks, internet sites, the reduction $\eqref{c}$ is named - not reaction $\eqref{d}$ which I had determined.

I am aware that these potentials are pH- and concentration dependent, I fit the $E(\text{Red})$-data with the Nernst-Equation to $\mathrm{pH} = 3$. Nevertheless, $E(\text{Red})$ of $\eqref{d}$ is always higher in $E(\text{Red})$ than $\eqref{c}$. According to my calculations (graph of $\eqref{d}$) which is dependent on the $\mathrm{pH}$ is always higher than $\eqref{c}$).
I have drawn a diagram which shows the $E(\text{Red})$ potential of the two reaction $\eqref{c}$ and $\eqref{d}$ (see legend). Reaction $\eqref{d}$ has always the higher standard potential - is the better oxidation reaction. Why does nevertheless reaction $\eqref{d}$ happen in acid solutions?

Functions calculated with Nernst-Equation:
\begin{align}
E(\text{pH}) &= 1.507 - \frac{0.059}{5} \cdot 8 \text{pH}\tag{c'}\\
E(\text{pH}) &= 1.7 - \frac{0.059}{3} \cdot 4 \text{pH}\tag{d'}
\end{align}
(other concentrations are assumed as equal)

Please check some unbalance equations and corresponding $\mathrm{E^o}$ (red) values in equations (a)-(d).

What you have listed are standard redox potentials. You need two of them to determine a reaction. As I'm not sure what reaction you intend, I'll give an example below, which you can use to work out your own reaction. The diagram helps to sort out what can happen.

Note that a reducing agent is an electron donor, and an oxidising agent an electron acceptor and $$\ce{oxidised form + electrons <=> reduced form}.$$
For example:
The oxidiser $\ce{Sn^{2+}}$ reacts completely with reducer $\ce{Cr^{2+}}$ (at standard redox potentials).
\begin{align}
E^\circ &= \pu{-0.136 V}:&
\ce{Sn^2+ + 2e- &<=> Sn}\\
E^\circ &= \pu{-0.41 V}:&
\ce{Cr^3+ + e^- &<=> Cr^2+}\\
&&
\ce{2Cr^2+ + Sn^2+  &<=>> 2Cr^3+ + Sn v}
\end{align}
With energy
$$E= (\pu{-0.136 V}) - (\pu{-0.41 V}) = \pu{0.274 V}$$
 which corresponds to a negative free energy and hence is spontaneous:
\begin{align}
\Delta G &= -nFE= (-2\times  96485\times 0.274) \pu{J};&
(\pu{1J} &= \pu{1C*V}).
\end{align}
Although the reverse reaction, $\ce{Cr^{3+}}$ to oxidise $\ce{Sn}$ metal could occur, this happens to a negligible extent.

First thank you for your answer,
I understand your point, but if I look at following reaction:
2I- -> I2 + 2e- E(Red) = -0.536 V
with c) $\Delta E =$ E(Red) - E(Ox) = 1.23 V + 0.536 V = 1,766 V
with d) $\Delta E =$ E(Red) - E(Ox) = 1.6V + 0.536 V = 2,136 V
=> reaction d) (in redox with the I- oxidation) will happen according to the "theory" - but in reality, there is no brown, dirty solution - clear, weak pink (indicates c)!)

OK, bad example. Its brown from the I2. But all reactions in the internet and books claim the formation of Mn2+...

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Counterions in adsorption of ions
When an ion is adsorbed from an aqueous solution onto a solid, what happens to its counterion? Obviously, the solution does not become charged - does this mean that it too is adsorbed?
(Example: $
\ce{NaF}$ with activated alumina as the adsorbent)
I suppose an alternate explanation would be that the remaining cation hydrolyses , but this does not seem likely in the example above as $\ce{Na^+}$, being the salt of a strong base, will not hydrolyse.

Want to elaborate?

@Maecel Yes, sure.

Since this got no answers, I'm posting an answer from Reddit user midnight-cheeseater, which can be found here.

There are two main possibilities here, depending on how the ions are adsorbed:

If the solid only adsorbs a type of ion, either positive or negative. Such as the example quoted of alumina adsorbing fluoride ions. The alumina will become negatively charged at the specific points where the fluoride ions stick to it. These negative charges will attract anything nearby which is positively charged. So the $\ce{Na+}$ ions will associate themselves with the negatively charged areas on the alumina.
As a solid adsorbs one ion, it may release another. This process is commonly seen in ion-exchange resins. Such as those which are used to produce "de-ionized" water. These adsorb positively charged metal ions and release $\ce{H+}$ ions at the same time. They can be regenerated by running fairly concentrated nitric acid through them.

The counter-ions are typically chloride, sulfate or nitrate, as well as some other halides, carbonate and phosphate. So this essentially converts a dilute solution of metal salts into a dilute solution of hydrochloric, sulfuric and nitric acids. Which explains why "de-ionized" water is rather acidic, as opposed to distilled water which is neutral.
Another common example found in dishwashers and water softening systems is an ion-exchange resin which adsorbs $\ce{Ca^2+}$ or $\ce{Mg^2+}$ ions while releasing $\ce{Na^+}$ ions at the same time. This converts dissolved calcium or magnesium salts (usually bicarbonates or sulfates) into sodium salts. These are regenerated by running saturated $\ce{NaCl}$ solution through them, which is why dishwashers and water softeners require a supply of salt.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Structure of aminoglycosides

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

With biomolecules like these, I've often seen the numbering being chosen by convention rather than with IUPAC rules. I don't know anything about these systems, so I can't say for sure.

DOS is not an abbreviation I commonly come across. Can you include what it means, please?

deoxystreptamine

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

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

The numbering is logical - if we would start from the other amine then the glucosamine substituent would be at atom 6 (against the usual rules of numbering where we prefer lower numbers for substituents). Most aminoglycosides can be divided into two groups: 4,5-disubstituted (neomycins, paromomycins) and 4,6-disubstituted (kanamycins among others). Therefore the red numbering in the original question is the right one, if we assume the synthetic pathway described in this article:
Park J.W. et al, "Discovery of parallel pathways of kanamycin biosynthesis allows antibiotic manipulation" Nat. Chem. Biol. 2011,  7(11):843-852 (DOI: 10.1038/nchembio.671).

If I were to have name it with a traditional way, which start from either one amine, I will be having 2 alternatives, one in blue and another in red (I have attached new photo on my original post). So if I were to follow the blue ones, I will have the left sugar at position 4 of 2-DOS I would name it with prime and the right sugar with double prime. However, if I were to use the red route, the left sugar would be numbered with double prime as it is now at position 6. BTW I got this numbering technique from here: https://www.bioscience.org/1999/v4/d/wright/fulltext.htm (2.0 Introduction)


text:
Which of the following samples contains the largest number of atoms?

Which of the following samples contains the largest number of atoms?
a.    2.0 moles of $\ce{H3PO4}$
b.    3.0 moles of $\ce{H2SO3}$
c.    4.0 moles of $\ce{HNO3}$
d.    6.0 moles of $\ce{HClO}$
e.    8.0 moles of $\ce{HBr}$

My logic would tell me that every mole has $6.022\times10^{23}$ atoms in it. Then you divide the amount of grams of each compound by the number of moles?
so for A I'd  use the following:
(Avg # *2) / 98 grams/per mole of $\ce{H3PO4}$   = $1.22\times10^{22}$ atoms
and so on, but I don't get the right answer :/
$(2\times(6.022\times10^{23})$ = number of atoms = $1.20\times10^{23}$
$\ce{H3PO4}$ = 98 grams/mole
The answer is C btw, I just don't know to set up the equation right to get to this answer.

Geeze, NaCl has 1 mole of Na and 1 mole of Cl per mole of NaCl. So that is 2 moles of atoms per mole of NaCl.
$\ce{K2SO4}$ has 2 moles of potassium, 1 mole of S and 4 moles of O per mole of $\ce{K2SO4}$. So that is 7 moles of atoms per mole $\ce{K2SO4}$.
You don't need to convert to atoms using Avogadro's number then back to moles. That complicates the problem needlessly. Just use the stoichiometry and you can do the arithmetic in your head.


text:
Could one synthesise unsubstituted "cyclohexa-1,3,5-triene"?
Is the cyclohexa-1,3,5-triene (non-aromatic benzene) possible to make or will it destabilize (or should I say stabilize?) into benzene?
Cyclohexa-1,3,5-triene is alternative name of benzene better describing its hypothetical, non-aromatic version. However they are similar and I was wondering if you made cyclohexa-1,3,5-triene would the electrons delocalise to form benzene or not?

but i thought benzene and cyclohexa-1,3,5-triene are the same!
how are they different?

@PrakharSankrityayan Edited it to be more clear.

You cannot make (= synthesise) isolated, nonaromatic cyclohexa-1,3,5-triene[1] much like you cannot make (= synthesise) pseudoaromatic, biradicallic cyclobuta-1,3-diene. Hypothetical unsubstituted cyclohexa-1,3,5-triene will immediately give benzene, as this answer demonstrates.
The only way to actually generate nonaromatic cyclohexa-1,3,5-triene is to apply an intramolecular constraint that disallows the aromatization. Since cyclobuta-1,3-diene is the most obvious antiaromatic, planar system, that will attempt to always adopt a localised double bond structure, structure 1 below immediately suggests itself as a model system for the localised cyclotriene.

Scheme 1: Diercks’ and Vollhardt’s synthesis of a cyclohexa-1,3,5-triene system 1.
Indeed, that was the approach that Diercks and Vollhardt used to synthesise a model of cyclohexa-1,3,5-triene.[2] They report a crystal structure of 4 ($\ce{R}= \ce{TMS}$) wherein the alternating single and double bonds are clearly seen. The cyclobutyl sub-species adopts a cyclobutene-structure, meaning that the central cyclohexatriene ring’s double bonds are localised between the cyclobutane cores as indicated by the structure drawn. The outer benzyl rings are relatively aromatic (but not perfect hexagons).

Note and reference:
[1]: IUPAC nomenclature would allow systematically naming benzene as cyclohexa-1,3,5-triene. This is because IUPAC nomenclature assumes no knowledge of the chemistry of a given structure and requires names to be applicable ab initio and a priori. However, henceforth in this answer I shall make a distinction of (nonaromatic, localised, $D_\mathrm{3h}$-symmetric) cyclohexa-1,3,5-triene and (aromatic, $D_\mathrm{6h}$-symmetric) benzene.
[2]: R. Diercks, K. P. C. Vollhardt, J. Am. Chem. Soc. 1986, 108, 3150. DOI: 10.1021/ja00271a080.


text:
Is CO2 produced when KHCO3 reacts with H2O?
We have been doing experiments on the rate of photosynthesis to find out the importance of $\ce{CO2}$ and we used $\ce{KHCO3}$ in one beaker and found that the plant kept in that beaker produced greater amount of oxygen that those without it. Sources say that $\ce{KHCO3}$ acts as a source of $\ce{CO2}$ but I'm not sure of it. What is the reaction taking place to produce $\ce{CO2}$?

$\ce{HCO3-(aq) + H2O(l) <=> H2CO3(aq) + OH-(aq) <=> H2O(l) + CO2(aq) + OH-(aq)}$ Now, that does not mean that if you dissolve $\ce{KHCO3}$ in water, it will immediately decompose and you get bubbles of $\ce{CO2}$. It is just that the plant is using the $\ce{CO2}$, so that pulls the equilibrium over, Le Chatelier's principle, etc.

or as I like to think of it: $$\ce{HCO3^-(aq) + H2O(l) <=> H2CO3(aq) + OH^-}$$ $$\ce{H2CO3(aq)  <=> H2O(l) + CO2(aq)}$$

I think one of the crucial steps is also $\ce{CO2 (aq) <=>> CO2 ^ }$ since plants are absorbing carbon dioxide from the air (mostly).

@Martin-マーチン The transfer of atmospheric CO2 was cut off before the readings were taken.

I was never implying that, if you read correctly, I am saying that the carbon dioxide needs to diffuse out of the solution before the plants can absorb it.

Compiling the comments by Max and orthocresol:

What is the reaction taking place to produce CO2?

$$\ce{HCO3− (aq) + H2O (l)
<=> H2CO3 (aq) + OH− (aq)
<=> H2O (l) + CO2 (aq) + OH−(aq)}$$
or
\begin{align}
\ce{HCO3− (aq) + H2O (l) &<=> H2CO3 (aq) + OH−}\\
\ce{H2CO3 (aq) &<=> H2O (l) + CO2 (aq)}
\end{align}

$\ce{KHCO3}$ does not readily decompose into $\ce{CO2}$ when dissolved in water.
  It is just that the plant is using the $\ce{CO2}$, so that pulls the equilibrium over, Le Chatelier's principle.

I am not sure about the previous answers but when KHCO3 reacts with water it forms a complex [K(H20)6]+ and bicarbonate ion. And hence no Carbon dioxide is formed.
But if Potassium carbonate decomposes between 373K and 393K it forms CO2.


text:
In general, are carbonyl compounds (ketones/aldehydes) more susceptible to a nucleophilic attack then alkynes?
Are carbonyl compounds, in general, better electrophiles than alkynes?
For example, would an amine prefer to attack a carbonyl like benzaldehyde (to form an imine) or an alkyne like phenylacetylene (also to form an imine)?
Is there a way to measure their relative electrophilicity's?

Why was this downvoted?

The double bonds in alkenes are usually electron-rich. So firstly they aren't good or even average electrophiles. Thus there are only few cases of nucleophilic addition. So I think definitely, carbonyl compounds are way better than alkynes. Also relative electrophilicity can be easily measured by having a nucleophilic attack on them and analysing the rate of reaction or product ratio which can be easily done with modern chemical methods

Right I learned that alkynes are much more susceptible to nucleophilic attack than alkenes, especially in phenylacetylene where the resulting carbanion would be stabilized by resonance with the aromatic ring.

According to your  question,first there's way lot of difference among alkenes andcarbonyls.and the reason to your previous comment is alkynes can show minor electrophilicity character but alkenes don't show. They are purely nucleophilic due to the electron cloud around them and accept on electrophilic attacks


text:
What are the practices to naming organic chemicals by abbreviations?
Is there a preferred method to writing organic chemicals by abbreviation?
I am finding it hard to decipher the methods I see in writing.
For example $\ce{EtOH}$ is ethanol, which makes sense, but $\ce{PhAc}$ is acetophenone, where the constituents appear to be backwards.
What about more complicated chemicals like 1,1-diphenylethanol, would $\ce{EtPh2OH}$ be considered appropriate, or maybe $\ce{Ph2EtOH}$?

The best practice is to avoid abbreviations as much as possible, but when unavoidable to define the meaning of an abbreviation and then stick to it throughout the whole text.

Say, if in the beginning of a paper you write: "We studied the reactivity of 1,1-Diphenylethanol (EtPh2OH) bla-bla-bla..." and then use the EtPh2OH abbreviation to mean 1,1-Diphenylethanol to the very end, everything will be fine.

The only accepted practice without any further constraints is to only use element symbols and sum or constitution formulae. Taking your and a few other examples, that would result in:

$\ce{C2H5OH}$ or $\ce{H3CCH2OH}$
$\ce{C6H5COCH3}$
$\ce{H3C(C6H5)2COH}$
$\ce{H3CCOOC2H5}$
$\ce{(CH3)3COCH3}$

The more complicated these get the less clear everything is. Thus, many people make use of a set of common abbreviations. However, these must be defined in a list of abbreviations, footnotes or wherever.

$\ce{EtOH}$
$\ce{PhAc}$[1]
$\ce{MePh2COH}$
$\ce{EtOAc}$
$\ce{$t$BuOMe}$

Where the symbols $\ce{Et, Ph, Ac, Me, $t$Bu}$ would need to be defined to mean ethyl, phenyl, acetyl, methyl and tert-butyl, respectively.
Finally, of course, you can suggest other common or your own abbreviations and use them (consistently!). For example, in my group it is rather common to abbreviate $\ce{EtOAc}$ as $\ce{EE}$ (because of its retained German name Essigester). Or a common way to denote the last compound in that list is $\ce{MTBE}$ or $\ce{TBME}$ (methyl-tert-butylether or tert-butylmethylether). Again, these abbreviations must be defined in footnotes or a list of abbreviations in best practice.
Finally note that an abbrevation typically includes all the hydrogens. Thus, methyl is always $\ce{CH3}$; $\ce{MeCl}$ is permissable for chloromethane but $\ce{MeCl2}$ does not mean dichloromethane. Consequently, neither $\ce{EtPh2OH}$ nor $\ce{Ph2EtOH}$ are permissable for 1,1-diphenylethanol, since carbon $\ce{C{1}}$ has no remaining hydrogens (not 2 as $\ce{CH3CH2}$ — ethyl — would imply).

[1]: Note that $\ce{Ac}$ can also mean actinium. If your paper uses actinium anywhere, do not use $\ce{Ac}$ as an abbreviation for acetyl or it shall cause confusion. The same is true for $\ce{Pr}$ and praseodymium and a few even less common others.

In addition to Ac (actinium or acetyl) and Pr (praseodymium or propyl), there may also be Cm (curium or carboxymethyl), Nb (niobium or p-nitrobenzyl), Np (neptunium or p-nitrophenyl), and new Ts (tennessine or 4-toluenesulfonyl).

Didn’t know about tennessine, never encountered the nitrobenzyl and nitrophenyl abbreviations afaicr. But thanks @Loong!

Generally, you should avoid jargon. Abbreviated terms should be used with care, and their use should be limited to those cases where it is not likely to cause confusion. Therefore, it is often preferable to avoid abbreviations. However, if a long term is repeated many times throughout a text, an abbreviation may be warranted. You may also use abbreviations when it is necessary to prevent awkward construction.
If a list of abbreviated terms is not given in your document, then the first time that an abbreviated term is used, the full term shall be given with the abbreviated term following in parentheses. After defining the abbreviation in the text, you may use it throughout the document. However, if the abbreviation is used in the abstract of a publication, define it at first use in the abstract and again at first use in the text.

2,2′-bis(diphenylphosphino)-1,1′-binaphthyl (binap)

An exemplary list of recommended abbreviations is given in Appendix 10-2 of The ACS Style Guide. This list includes various structural abbreviations that are used in organic chemistry, for example

Ac – acetyl
  Me – methyl
  Et – ethyl
  Ph – phenyl

Note that this list also includes some symbols that are not abbreviations (such as unit symbols or element symbols). The ACS Style Guide also includes a short list of common abbreviations that do not need to be defined (see page 158).
According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), structural abbreviations may also be used as labels in structural diagrams.  These recommendations also include rules for the preferred ordering of structural abbreviations and other symbols in contracted labels (e.g. $\ce{-CHPh2}$ and not $\ce{-CPh2H}$).
According to the 2008 IUPAC recommendations (see Table II on page 406), the structural abbreviations Me, Et, Pr, iPr, Bu, iBu, s-Bu, t-Bu, Ac, Ph, Ms, Ts, and Cp may be used freely without further explanation. Other structural abbreviations may be used if they are accompanied by a clear explanation (e.g. a graphical representation) of the represented structural fragment. The use of structural abbreviations that are identical to an element symbol (e.g. Ac, Cm, Nb, Np, Pr, and Ts) should be restricted to situations where they are unlikely to be mistaken for the element symbols.

That's a wonderful resource! I've read through some of it and realized that I make a lot of common mistakes, whoops!

@CayetanoGonçalves Note that I included links to two different documents, and both are mainly recommendations. If you deviate from the recommendations, it is not necessarily a mistake.

I was referring to the IUPAC recommendation for structure diagrams, it's very thorough.


text:
Why is a buffer solution best when pH = pKa i.e. when A-/HA=1
Buffers work best when  $\mathrm{pH}$=$\mathrm{p}K_\mathrm{a}$
From the Henderson-Hasselbalch equation,
$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log_{10}\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$
If $\mathrm{pH}$=$\mathrm{p}K_\mathrm{a} \implies\log_{10} \ce{\frac{[A-]}{[HA]}} = 0 \implies \frac{[\ce{A-}]}{[\ce{HA}]}= 1$
But why does having $\frac{[\ce{A-}]}{[\ce{HA}]}= 1$  make it a good buffer solution?
When I look at a $\log_{10}$  graph, it looks like the further to the right $x$ i.e. $\ce{\frac{[A-]}{[HA]}}$ is, the less effect any change in $\ce{\frac{[A-]}{[HA]}}$ has on $\log_{10}\frac{[\ce{A-}]}{[\ce{HA}]}$ (and hence effect on $\mathrm{pH}$ as well).
So wouldn't this indicate that a good buffer would have a large $\log_{10} \ce{\frac{[A-]}{[HA]}}$ (as e.g. going from $\ce{\frac{[A-]}{[HA]}} = 10$   to $\ce{\frac{[A-]}{[HA]}}= 11$  will have a smaller change (0.04 in this example) in $\log_{10}\ce{\frac{[A-]}{[HA]}}$ than going from $v= 1/1$   to $\ce{\frac{[A-]}{[HA]}}= 2/1$) (0.3 in this example). Where both of these changes show addition of 1 M of base.

Note: 2 similar questions where asked before: Why is buffer capacity at a maximum when the ratio of its components is 1? and Why does a buffer work best at the pH closest to its pKa?
The questions didn't show what exactly they didn't understand which I am attempting to do here. Also the answers were not useful as one said said the buffer is best when X (which I know), but not didn't explain why this is the case. The other questions answer provided a link which I read but was beyond me.

Preliminaries
There are many ways in which you can go about defining a "Buffer Capacity". A good intuitive way to start would be to say, it is tha maximum amount of acid (or base) the solution can tolerate before it starts showing a significant change in pH. While this is not bad, it obviously begs the question what is deemed "significant"?
However, let us start at the very beginning by examining (in very basic terms) how a buffer works.
Say, you have some amount of $\ce{A^-}$ in the solution and some amount of $\ce{HA}$. You add a strong acid to the solution, and this leads to an influx of $\ce{H^+}$ ions; the buffer responds by forming more $\ce{HA}$ through the combination of $\ce{A^-}$ and $\ce{H^+}$. Similarly, you add a strong base to the solution, and you have a lot of $\ce{OH^-}$ floating around, and the soultion responds appropriately, by favouring the disassociation of $\ce{HA}$, the $\ce{H^+}$ so released combines with the $\ce{OH^-}$ and neutralises it.
The equilibrium we are interested in is $$\ce{HA <=> A^- + H^+}$$
an increase in acidity causes a shift to the left, and an increase in basicity causes a shift to the right.
So, in this way a buffer acts a proton sink, absorbing/releasing protons as appropriate to maintain the pH of the solution.
The buffer is said to be broken when we add to much acid/base, and this consumes all of the $\ce{A-}$ / $\ce{HA}$ respectively.
A solution with more weak base, $\ce{A-}$, has a higher buffer capacity for addition of strong acid. A solution with more weak acid, $\ce{HA}$, has a higher buffer capacity for addition of strong base.So although the pH of a buffer is determined by only the ratio conjugate base and acid, the ability of the buffer to absorb strong acid or base is determined by the individual concentrations of the conjugate base and acid.
So it might seem intuitive to have the optimum at equal concentrations of conjugate base and acid.

Intuitive definitions
$$\ce{HA + OH^- <=> A^- + H2O }$$
$$\ce{A^- + H3O^+ <=> HA + H2O}$$
$$pH + pOH = 14$$
The following discussion is largely inspired by this paper (free to read)
Namely, if you you have $$\mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{A^-}{HA}\right)$$
Since, $\mathrm{pK_a}$ is a fixed number, an changes that can take place to the pH value are caused by the $\log \left( \frac{A^-}{HA}\right)$ term.
so, we can say $$ \mathrm{pH} = \mathrm{pK_a} + \delta \\ \text{where}\ \delta = \log \left( \frac{A^-}{HA}\right)$$
make $\delta = 0$ and you are done. You can call that your ideal case, an ideal buffer.
Now, you can consider a buffer's resistance to base and acid separately, and you can define an acid and base buffer capacity ($\beta_a$ and $\beta_b$), for very simple cases (read caveats).
The stoichiometric ratio of of conjugate base and acid is 1:1. $\Delta$ denotes an arbitrary change in $\delta$ after $\alpha$ mol of $\ce{HA}$ has reacted (after addition of a strong base)
$$\delta + \Delta = \log \left( \frac{[A^-] + \alpha}{[HA] - \alpha}\right)$$
One then notes that bases attempt to increase the pH of a solution so $\Delta = 1$, and also let us denote $\beta_b$ as the buffer base capacity. Also, note that $\alpha = \beta_b$. We are interested in the maximum amount of base a 1:1 buffer solution can tolerate before the pH increases by one unit. We are assuming a 1:1 stoichiometry between the the weak acid in our buffer and the strong base that is being added to it.
$$10^{\delta + 1 } = \left( \frac{[A^-] + \beta_b}{[HA] - \beta_b}\right) $$
We solve for beta and substitute for $\delta$ to ultimately get,
$$ \beta_b = \frac{9[\ce{HA}][\ce{A^-}]}{10[\ce{HA}] + [\ce{A^-}]}$$
Now, let us consider that we are adding a strong monoprotic acid. The approach is similar to the one described above. A few key differences is one lets $\Delta = -1$ and let $\alpha = -\beta_a$
I am omitting the algebra involved. You can attempt as an exercise it if you'd like, and ask for clarifications in the comments if you run into trouble. We obtain a similar relation
$$ \beta_a = \frac{9[\ce{HA}][\ce{A^-}]}{10[\ce{HA}] + [\ce{A^-}]}$$
$\beta_a = \beta_b$ if and only if $\delta = 0$ Otherwise, the buffer either displays stronger resistance towards acids or bases.

Caveats
This is a good exercise to develop a good basic understanding, however doesn't have a lot of practical uses. Here, only a simple buffer was considered, i.e one without multiple weak bases and acids. Additionally, polyprotic acids haven't been accounted for.

Formal defintion of buffer capacity
The formal definition of buffer is a little bit intimidating, but I will throw it in here anyway. It obviously makes little sense to have two different kinds of capacities, one for acids and one for bases in everyday life. So we need something more general, which still incorporates the intuition we have developed so far (although it might not seem obvious)
Let $n$ be number of equivalents of added strong base (per 1 L of the solution). Note that addition of $dn$ moles of acid will change pH by exactly the same value but in opposite direction. I will try to derive a formula connecting buffer capacity with pH, pKa and buffer concentration- all numbers we can obtain easily.
For simplicity, I assume that the strong base is monoprotic, and we have volume of 1 (which will allows me to use concentration and number of moles interchangeably)
$$\beta = \frac{\mathrm{d}n}{\mathrm{d}pH} \tag{1}$$
Charge balance of the solution is given by the equation
$$[\ce{A^-}]+[\ce{OH^-}] = [\ce{B^+}] + [\ce{H^+}] \tag{2}$$
$[\ce{B+}]$ is nothing but concentration of the strong base present, or the number of moles present since we assume a volume of 1 L(:D ), so it is simply $n$ in the solution.
$$C_{b} = [\ce{HA}]+[\ce{A-}] \tag{3}$$
and $$[\ce{HA}] = \frac{[\ce{A-}][\ce{H+}]}{K_a} \tag{4}$$
from 3 and 4,
$$ [\ce{A-}] = \frac{C_bK_a}{K_a+ [\ce{H+}]} \tag{5} $$
Using the definition of $K_w$, equation 2 and 5
$$n = \frac{K_w}{[\ce{H+}]} - [\ce{H+}]+ \frac{C_bK_a}{K_a+ [\ce{H+}]} \tag{6}$$
$$\beta = \frac{\mathrm{d}n}{\mathrm{d}pH} = \frac{\mathrm{d}n \mathrm{d}[\ce{H+}]}{\mathrm{d}[\ce{H+}]\mathrm{dpH}} \tag{7}$$
Now, you can evaluate the derivative given above, and after some alebraic manipulations you get
$$\beta = (2.303) \left( \frac{K_w}{[\ce{H+}]} + [\ce{H+}]+ \frac{C_bK_a}{(K_a+ [\ce{H+}])^2} \right )$$
Generalising this for solutions containing several buffers
$$\beta = (2.303) \left( \frac{K_w}{[\ce{H+}]} + [\ce{H+}]+ \sum_{}^{} \frac{C_bK_a}{(K_a+ [\ce{H+}])^2} \right )$$
I'd like to point out that the first two terms in the buffer capacity formula are not dependent on the buffer presence in the solution. Why? They just tell us solutions of high (or low) pH are resistant to pH changes. Simple.

Above plot shows how the buffer capacity changes for the 0.1M solution of acetic buffer. As expected buffer exhibits the highest resistance to acid and base addition for the equimolar solution (when pH=pKa). From the plot it is also obvious that buffer capacity has reasonably high values only for pH close to pKa value. The further from the optimal value, the lower buffer capacity of the solution. Solution containing only conjugate base (pH 8-10) has buffer capacity of zero, for the higher pH presence of the strong base starts to play an important role. In the case of pure acetic acid solution (pH below 3) pH is already low enough to be resistant to changes due to the high concentration of $\ce{H+}$.

What will make a good buffer solution, a buffer of weak acid or when a buffer of strong acid is used

@Pole_Star Strong acids don't make buffers as the dissociate completely and can't be forced to turn back into pure acid in solution, atleast not for reasonable pH

You need to correct your equation.
When you add A- or HA to your solution, the concentration of one raises while the concentration of the other falls, so the equation should look more like this:
pH = pKa + lg[(A- + x) / (HA - x)
In my example pKa = 5

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

why did you assume their concentrations were equal initially?

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text:
How to calculate the concentration of a monoprotic organic acid that was extracted into an ethyl ether layer?

Compound A is a monoprotic organic acid. If $\pu{25.0 mL}$ of a $\pu{0.456 M}$ solution of A in water was extracted with $\pu{25.0 mL}$ of ethyl ether and it was determined experimentally that $\pu{5.45 mL}$ of $\pu{0.523 M}$ $\ce{NaOH}$ was required to neutralize the bottom layer from the extraction, what is the concentration of compound A in the ether layer?

My attempt:

$\pu{25.0 mL}$ of a $\pu{0.456M}$ solution of A in water

This gives the total millimoles of your unknown acid that is split between the two layers.

$\pu{5.45 mL}$ of $\pu{0.523 M}$ $\ce{NaOH}$ was required to neutralize the bottom layer

That gives the number of millimoles in the bottom layer.
So the number of millimoles in the ether layer must be the difference, and there are $\pu{25 mL}$ of ether.
I think I figured the concentration out: $0.523\times\frac{0.00545}{0.25} = \pu{0.114 M}$
Or do I need to determine the amount in the aqueous layer by subtracting this number? I.e. $0.456 - 0.114 = \pu{0.34 M}$
I am left with either the answer being $\pu{0.34 M}$ or $\pu{0.114 M}$.
I think the answer is $\pu{0.34 M}$, because the $\pu{0.114 M}$ applies to the bottom layer which is water and needs to be neutralized by the $\ce{NaOH}$ solution, therefore the top layer (ether) has the remaining $\pu{0.34 M}$ of compound A.

You arrive at the correct answer, but the way you get there is a bit messy. (And it only works because both phases have the same volume.) You should be a lot clearer in defining what properties you have been given in the question and which ones you need to calculate. Setting up the equation is a lot more important then calculating it in the end, because this does not only show that you know what you are doing, but it also is a way to condense a difficult sentence into a calculator understandable way.
\begin{align}
V(\mathrm{water})       &= \pu{25.0 mL}\\
c_\mathrm{tot}(\ce{AH}) &= \pu{0.456 mmol/mL}\\
V(\mathrm{ether})       &= \pu{25.0 mL}\\
V(\ce{NaOH})            &= \pu{5.45 mL}\\
c(\ce{NaOH})            &= \pu{0.523 mmol/mL}\\
\color{red}{c_\mathrm{ether}(\ce{AH})} &= ?
\end{align}
Then you must always carry the units with the quantity; they belong together. This way you will always have a way to tell if you have made a mistake setting up the equation. If the units don't work out, it cannot be correct.
Since it is a monoprotic acid the following will hold true:
\begin{align}
\ce{AH &<=>[H2O] A- + H+} &
&& n_\mathrm{tot}(\ce{AH}) &= c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water})
\end{align}
Now you extracted your solution and split the amount of substance between the water and the ether phase. You can calculate the remaining substance in the water phase from the neutralisation experiment.
\begin{align}
\ce{AH + NaOH &-> A- + Na+ +H2O} & n_\mathrm{water}(\ce{AH}) &= n(\ce{NaOH})\\
&& n_\mathrm{water}(\ce{AH}) &= c(\ce{NaOH}) \cdot V(\ce{NaOH})
\end{align}
The extracted amount of substance of the acid is simply the difference, from that you can easily calculate the concentration in the ether phase.
\begin{align}
c_\mathrm{ether}(\ce{AH}) &=
  \frac{n_\mathrm{ether}(\ce{AH})}{V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{n_\mathrm{tot}(\ce{AH}) - n_\mathrm{water}(\ce{AH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water})
        - c(\ce{NaOH}) \cdot V(\ce{NaOH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{\pu{0.456 mmol/mL} \cdot \pu{25.0 mL}
        -  \pu{0.523 mmol/mL} \cdot \pu{5.45 mL}}
       {\pu{25.0 mL}}\\
c_\mathrm{ether}(\ce{AH}) &\approx \pu{0.342 mmol/mL}
\end{align}

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text:
Dish soap reaction on braided stainless steel hose
Just wondering if leaking dish soap can have a reaction on a braided stainless steel hose used for hot water intake of dishwasher over time (several months)? The hose which was new at the time perforated without explanation and caused a flood.
Thanks.

Highly unlikely in my opinion (as a chemist and dishwasher owner). If the water coming in was/is hard, there might have been some buildup on the intake screen (if there is/was one), but we'd need to know your water chemistry and more about the plumbing. Over-tightening the hose and/or having a bad gasket (usually old, thus easily cracked) can also be the cause. Dish soap would be way down the list.


text:
Why is the boiling point of water and ammonia so different?
Boiling point of water is 100 degrees Celsius, while boiling point of ammonia is minus 33 degrees Celsius, which makes 133 degrees difference. Now when we discuss value of boiling point, we also say that it depends on inter-molecular forces, and in case of both water and ammonia I can see two such forces: London dispersion forces and hydrogen bonds. Relative molecular mass of water is 18 amu, and relative molecular mass of ammonia is 17 amu, so unlikely we can explain the difference in boiling points by difference in London dispersion forces: as they depend on mass, the LDF should be about the same for two compounds. Speaking about hydrogen bonds, it looks as each molecule (water or ammonia) is able to form 4 hydrogen bonds. And in this way amount of hydrogen bonds again should be about the same. So what is an explanation for the difference in boiling point between water and ammonia?
UPDATE: I found some sort of explanation, which partially seems OK, but I see at least one problem with it though:

Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of δ+ hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. That means that on average each ammonia molecule can form one hydrogen bond using its lone pair and one involving one of its δ+ hydrogens. The other hydrogens are wasted. In hydrogen fluoride, the problem is a shortage of hydrogens. On average, then, each molecule can only form one hydrogen bond using its δ+ hydrogen and one involving one of its lone pairs. The other lone pairs are essentially wasted. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system.

Source: http://www.chemguide.co.uk/atoms/bonding/hbond.html
OK, so each ammonia molecule forms 2 hydrogen bonds and so does each hydrogen fluoride molecule. Their masses are 17 amu and 20 amu respectively (difference is very small). Boiling point of fluoride is 19.5 degrees Celsius, while boiling point of ammonia is minus 33 degrees Celsius, which makes 53.5 degrees difference. So what is an explanation for the difference in boiling point between hydrogen fluoride and ammonia?

Man, you could really use some formatting. Now, you've pretty much answered your titular question yourself. As for the difference between $\ce{HF}$ and $\ce{NH3}$, the former is much more polar, hence its hydrogen bonds are stronger.

It’s not related to you actual question, but please note that the quantity “relative molecular mass” $M_\mathrm r$ is a dimensionless quantity (to be precise: a quantity of dimension one). Thus, the atomic mass unit $(\mathrm{amu})$ is not a unit of relative molecular mass. Please also note that the unit $\mathrm{amu}$ is obsolete (in Chemistry since 1961).

I would like to add another relevant question with very good answer here: http://chemistry.stackexchange.com/q/64191/25519

If I interpret your edited question correctly, it now boils down to why is the difference between the boiling points of $\ce{HF}$ and $\ce{NH3}$ so large, even though they have almost the same molecular masses ($M_\mathrm{r}(\ce{NH3}) = 17, M_\mathrm{r}(\ce{HF}) = 20$).
Ammonia can self-ionise much like water can according to the following equation $(1)$. And much like water, the process is important to determine a $\mathrm{pH}$ value, but rather neglegible when it comes to determining attractive forces between ammonia molecules where ‘traditional’ hydrogen bonding predominates.
$$\ce{2 NH3 <<=> NH4+ + NH2-}\tag{1}$$
On the other hand, self-ionisation in hydrogen fluoride, especially in the liquid and solid states, is much more important and leads to the two fragments shown in equation $(2)$.
$$\ce{3 HF <=> H-F-H+ + F-H-F-}\tag{2}$$
These two linear fragments both feature de facto four-electron-three-centre bonds and are thus much more prevalent than the corresponding ammonia-derived ones. (Indeed, something analogous to $\ce{FHF-}$ doesn’t exist for ammonia.) Since these also introduce ionic interactions into the equation, the intermolecular forces between $\ce{HF}$ molecules are much stronger than those between ammonia molecules contributing to $\ce{HF}$’s higher boiling point.


text:
Brackets in chemical formulas
When you have brackets such as $\ce{Zn(NO3)2}$, if there is a coefficient around the formula such as $\ce{2Zn(NO3)2}$, how would you read this? Would there be 2 zinc atoms, 4 nitrogen and 12 oxygen atoms if I am multiplying that out correctly?

It means $\ce{Zn(NO3)2}$ contains a $\ce{Zn^{2+}}$ ion and 2 $\ce{NO3-}$ ions

@Atticus283blink the coefficient indicates that there are two units of that molecule.

@JM97 right i know that but would how does the 2 affect the (NO3)2 ? Would this means there are 4 N and 12 O atoms?

@Atticus283blink yes, you are right.

$\ce{2Zn(NO3)2}$ can be interpreted differently, depending on the context.
The '2' outside of the parenthesis could either refer to two molecules or two moles of substance. In either case, the chemical formula tells you the same thing: that there are two units of $\ce{NO3-}$ for every unit $\ce{Zn^2+}$.
I would write it as zinc (II) nitrate, (read as 'zinc-two-nitrate) The roman numeral (II) refers to the oxidation state of the zinc ion, which is $+2$.


text:
Why don't the products of combustion stop combustion?
We know that combustion of fuels give water and carbon dioxide (during combustion of hydrocarbon) as a byproduct, but why don't these byproducts stop combustion?

Normal flames are open to atmosphere, so the byproducts, being in gaseous state (due to the flame's temperature), escape off by themselves. An engine, being a closed environment, would need to expel them before the next combustion event occured (the byproducts would not carry away much of the heat generated). That is why an exhaust system is present in automobiles.

You are exactly correct that in a closed system, the buildup of products would inhibit the reaction. As the concentrations of $\ce{CO2}$ and water vapor increase, the concentration of the oxygen will effectively decreases until the reaction slows and eventually stops.
Fortunately we rarely have to worry about this because we rarely try to do such a combustion in a closed system. Rather, as in the case of a car engine for example, there is a system in place to actively vent out the exhaust waste products ($\ce{CO2}$ and water) while pumping in more oxidant (oxygen) and fuel (hydrocarbon gasoline).


text:
Is concentration polarisation an observable phenomenon over time in a galvanic cell?
If polarisation means: "any mechanical side-effect by which isolating barriers develop at the interface between electrode and electrolyte."
And if concentration polarisation is "changes in the electrolyte concentration due to the passage of current through the electrode/solution interface. Here polarization is understood as the shift of the Electrochemical potential difference across the cell from its equilibrium value. The cause of the changes in concentration (emergence of concentration gradients in the solution adjacent to the electrode surface) is the difference in the rate of electrochemical reaction at the electrode and the rate of ion migration in the solution from/to the surface."
Thus, should concentration polarisation cause a decrease in the performance/efficiency of a galvanic cell over time and would this be shown with a decrease of voltage/current through the cell? Or does voltage and current remain the same throughout a galvanic cell reaction?

There are two different factors that cause a battery to have a lower energy output.
First regardless of the "state" of the battery, there is the open cell voltage and the obtained voltage under load. State meaning here how much of the battery is used up. For the open cell voltage the voltage is measured with a high impedance voltmeter so that essentially no current (meaning a very very small current) is being drawn. Next the battery is put under some load typical for its use and the voltage is measured again while under the load. This second load voltage measurement will always be lower that the first measurement for the open cell voltage. This voltage drop is caused by the diffusion limits of the chemicals in the battery to and from the electrodes.
This behavior can be essentially modeled by assuming that the battery has an internal resistance. So as the current draw increases the voltage drop internal to the battery increases and the output voltage drops.
Now the second problem is related to the Nernst equation. Let's assume that the battery is used up in 10% increments in a series of tests. At the start of each test the voltage and current is measured. Of course during using each 10% of the battery the reactants in the vicinity of the electrodes are used up and both the voltage and current out for a given load drops.
Now if the battery is left to equilateral for some time then the reactants redistribute and the battery regains some power. However the battery will not recovery to the level at the start of the previous load test. Essentially the bulk concentration of the reactants is dropping by about 10% after each test. So even after a recovery period both the voltage and current that the battery can supply for a constant load drops.
Now with all of this being said there are many types of batteries. Some types produce a more constant power output than others.

So in an experiment with a simple Cu/Zn electrochemical cell, if you attached a load (i.e. 3 ohms) into the circuit and left the cell to operate over a period of time, would the cell decrease in voltage and current (as the electrodes are being used up)? And would a larger load drain the "bulk concentration of the reactants" quicker and thus also reduce the voltage and current quicker?

Yes and yes. A larger load of course means less resistance in the load so that more current flows from the battery.

What do you mean by "less resistance in the load"?

The load has a resistance. Since V=IR, in order to increase I you have to decrease R.

Ah okay that makes sense, thanks for the help.

Sorry, this post was made a while ago, but upon returning to finish some investigations, I have another question. What is the reason for the initial instantaneous voltage drop when a resistor is added in series to the cell? (Change in cell potential due to diffusion limits only occurs over time?!)


text:
van der Waals coefficients for helium and neon
The values of the coefficient $b$ for helium and neon are
$$\begin{array}{cc}
\hline
\text{Element} & b\text{ / }\mathrm{dm^3\ mol^{-1}} \\
\hline
\ce{He} & 2.38 \times 10^{-2} \\
\ce{Ne} & 1.67 \times 10^{-2} \\
\hline
\end{array}$$
(source: Atkins & de Paula, Physical Chemistry, 9th ed., p 916)
This doesn't make sense, however, considering that neon should have a larger atomic radius than helium. Why is $b_\ce{He} > b_\ce{Ne}$?

It's a good question, but please, always include the units whenever you are discussing science! It may sound pedantic, but science is not science without units.

I'm guessing charge-to-volume ratio.

So you mean if the charge density is higher than the atom would be "tougher" and any particle would therefore bounce back immediately.

What does that even mean, there is no charge on the atoms.

I should be careful about the terminology. It should had been "electron density" here

I suspect this depends very heavily neon's full $2p$ orbital and how this affects the volume actually excluded by the atom. I'd also be skeptical of values you find for van der waal's radii unless you trust the method by which they're found because there are quite a few different ways to get a guess experimentally.

related: http://chemistry.stackexchange.com/a/37514/16683

The question intregues me so I looked up the $a$ and $b$ values in Wikipedia on the web page Van der Waals constants. This doesn't answer the question, but adds more data to focus the question.
$$\begin{array}{cc}
\hline
\text{Element} & a\text{ / }\mathrm{L^2 \ bar\ mol^{-2}} & b\text{ / }\mathrm{L\ mol^{-1}}\\
\hline
\ce{He} & 0.0346 & 0.0238 \\
\ce{Ne} & 0.2135 & 0.01709 \\
\ce{Ar} & 1.355  & 0.03201 \\
\ce{Kr} & 2.349  & 0.03978 \\
\ce{Xe} & 4.250  & 0.05105\\
\ce{Rn} & 6.601  & 0.06239 \\
\hline
\end{array}$$
A plot of the Van der Waals $a$ and $b$ values.

So is the b value for He high, or is the b value for Ne low ?!?
I also looked up for each element what was listed for the covalent radius and the Van der Waals radius on each elements individual Wikipedia webpage.
$$\begin{array}{cc}
\hline
\text{Element} & Covalent\text{ (pm) } & Van~der~Waals \text{ (pm)}\\
\hline
\ce{He} & 28 & 140 \\
\ce{Ne} & 58 & 154 \\
\ce{Ar} & 106±10  & 188 \\
\ce{Kr} & 116±4  & 202 \\
\ce{Xe} & 140±9  & 216 \\
\ce{Rn} & 150  & 220 \\
\hline
\end{array}$$

So Is the b value for He high, or is the b value for Ne low ?

What is the plot for ?

@MathTheNovice - As the text indicates "A plot of the Van der Waals a and b values [for the six noble gases]." The gist is to look at the curve and decide if the He Or Ne value is out of whack for the trend.

A coefficient stands for the intereaction between molecules while b for molecule volume (In the interpretation from physics ) and I don't really get how it is related though the plot show a significant correlation between them

I mean is the straight line appeared by coincidence

@MathTheNovice - I don't know if helium or neon should be on the curve. The other point is that the "a" and "b" coefficients are not "pure" in the sense that both coefficients depend on both the interaction between particles and the volume of the particles.

@MaxW Also the value of $a$ is very low for $\ce{He}$ is very low compare to even $\ce{H2}$. The only explanation I can think of is '$imperfection$ of van der Waals equation'.


text:
How to determine the reactivity of a set of compounds that may undergo nucleophilic substitution?

Which of the following compounds is most reactive under $\mathrm{S_N1}$ conditions?

Arrange the following compounds in order of decreasing rate of $\mathrm{S_N1}$ hydrolysis:

A. II > III > IV > I
  B. IV > III > II > I
  C. III > IV > II > I
  D. I > II > III > IV

Both these questions are based upon the rate of reactivity of $\mathrm{S_N1}$ reaction.
1) In this I thought the answer should be B because the rate of $\mathrm{S_N1}$ reaction depends on the ease of formation of carbocation. Being more electronegative, the electron density between carbon and chlorine bond is quite shifted towards chlorine, so it should be easier for chlorine to leave, hence forming the carbocation. Options A, B, and C allow the carbocation to be resonance stabilized so D cannot be the answer.
But the answer given is A.
Where am I wrong with this logic? I guess it might be because of more polarizing power of iodine, if it is this, then how are we supposed to figure out which one will be the dominating factor?
2) I think we check this on the basis of carbocation stability as well. Carbocation will be formed as $\ce{Br}$ leaves. Alkyl chain is electron releasing, so it must increase the electron density on the benzene ring which should be better for the stability of carbocation (as there is more electron density on the benzene ring to stabilize the positive charge). Hence the answer should be I > II > III > IV
But the answer is A.
Where am I wrong with my logic in both?

Think about leaving group ability. 2) draw resonance forms, see where the charge appears, and especially consider how that charge is going to to be stabilised by the differing alkyl substituents when it appears at their position.

@getafix We know poor bases are good leaving groups. Hence iodide ion being the worse base amongst other halide ions here must be the best leaving group. But what if I think in terms of electronegativity as in chlorine is able to hold the electron pairs better than others( higher electronegativity)

I was gonna write an answer, but Jan's answer addresses your problem perfectly.

@getafix Sorry I sniped you … but you’ve had your fair share of sniping what I wanted to answer, so … ;)

@Jan Oh no it is perfectly alright, your answer is probably better than anything that I could've come up with. (up voted it myself). Excellent job, like always!

In question 1), you are wrong about the tendency of leaving groups. While you are correct that the carbon-chlorine bond is more polarised towards chlorine, iodide is still a better leaving group. This is because it is much larger and can thus stabilise the resulting negative charge much better.
You can observe this in $\mathrm{S_N2}$ reactions, where iodides are much more reactive than bromides which are more reactive than chlorides (and fluorides are basically unreactive) even though the same polarisation argument could have applied.

In question 2), the stabilisation that alkyl groups give carbocations, or the electron density they donate into a phenylic ring actually comes from $\ce{C-H}$ bonds. Thus, the para-xylene derivative can stabilise a cation better since there are three $\ce{C-H}$ bonds it can use for stabilisation. No matter which way the methyl is oriented, it can stabilise. para-ethylbenzyl bromide can only be stabilised in two out of three orientations (when a $\ce{C-H}$ bond is orthogonal to the phenyl ring) and para-isopropylbenzyl bromide only in one. Thus, the rate of hydrolysis is accordingly.

I do understand that iodide ion is a better leaving group because of its large size but I don't understand why we cannot extend the electronegativity reasoning to predict the better leaving group

In question 2, I was supposed to account for hyperconjugation since it is more dominant than inductive effect of alkyl chain. Right?

Is it that the net effect of size and electronegativity factors make iodide ion the best leaving group?

@Blue Re 2nd comment: Hyperconjugation is how methyl groups exercise their $+I$ effect.

@Blue Re 1st and 3rd comments: It’s size and thus charge distribution. Electronegativity doesn’t play too much of a role when determining nucleofug…acity?


text:
On Electrochemical Cells and Salt Bridges
We're studying the topic Electrochemistry at school now, and I've got a few doubts regarding electrochemical cells and salt bridges.
What I am going to do is listing out (point-wise) what I understood from class, and then pose my questions (in bold). The logic behind listing out what I understood about the topic beforehand? I've a very strong feeling I'm harboring some serious misconceptions regarding the topic, so if I list out what I do know, then hopefully someone can alert me to an incorrect point, and also correct it.

LHC - Left half cell (contains the anode, which is negatively charged in electrochemical cells)
RHC - Right half cell (Contains the cathode, which is positively charged in electrochemical cells)

During the course of the reaction,

In the LHC, metal atoms of the electrode release their electrons to the anode and enter the solution as positive ions.
The positive metal ions, under the influence of the negatively charged anode, accumulate around the anode, resulting in a Solid-Liquid Junction. The Solid-Liquid Junction is also known as the Helmholtz Double Layer.
Similarly in the RHC, negatively charged ions (formed from the electrolyte solution) accumulate around the cathode, resulting in another Solid-Liquid Junction.
It is the potential difference between the LHC and RHC Solid-Liquid Junctions potentials that gives rise to the emf which results in the flow of electrons (electric current) from the LHC to the RHC via the metal wire that connects their corresponding electrodes.
However, as the reaction proceeds, the quantity of positive and negative ions builds up (in the LHC and RHC respectively). So much, that these ions under their mutual attraction, override the force of attraction that binds them to the electrodes and accumulate at the porous asbestos partition that separates the two half cells. This is called the Liquid-Liquid Junction. The formation of the Liquid-Liquid Junction comes at the cost of the Solid-Liquid Junctions, which disintegrate.

As the Liquid-Liquid Junction potential doesn't involve the electrodes, it cannot set up a potential difference across them, therefore no current flows along the wire. This is why an electrochemical cell stops generating current after some time has elapsed.
This problem can be overcome by use of a salt bridge, which in effect, neutralizes the net charges on either side of the asbestos partition, thus releasing the ions back into the solution. These ions are once again attracted by the electrodes, and the two Solid-Liquid Junctions are re-established. So current once again begins flowing through the wire.

I'd appreciate it if anyone could proof-read those points. Also I have doubts whether the Solid-Liquid Junction can also be called the Helmholtz Double Layer.
I made these points in class, and I let my teacher go through them, and he's in agreement with them, so I don't think this is a case of talking down the points wrong, at least not with respect to what was mentioned in class.
Now I asked my teacher, Q1: Screw the partition. Why not simply have the two half cells in separate compartments kept at a distance from each other, instead of using a 'partial' barrier like the porous asbestos partition? Wouldn't that eliminate the possibility of a Liquid-Liquid Junction being formed?

His answer: Well they're ions buddy, so they'll attract each other anyway and accumulate at the sides of the compartment, after all....they're charged. So even though that wouldn't result in a Liquid-Liquid Junction, there still wouldn't be any potential difference across the electrodes, so nil current. Plus, if you separate the compartments like that, it'll just take up space;  compact electrochemical cells are more convenient to use.

Q2: Is my teacher entirely right here?
And yet again, Q3: Are all the points I've mentioned correct?

Related: Difficulties understanding how salt bridge works

It would have no current because the circuit is open.

@DHMO I beg your pardon?

Print this post on paper and burn it. This should help you to forget the idea of Liquid-Liquid Junction.

@IvanNeretin What?!

The what about the Solid-Liquid Junction?

You have it all backwards. Liquid-liquid junction does not exist at all in your supposed meaning. Solid-liquid junction does, but it is very much not what you think it is, to the point that you might just as well assume it does not exist either.
The complete story would hardly fit in an answer (and, moreover, it is explained in any decent textbook on the subject), so I'll focus on the details that are especially wrong. First of all, the double electric layer, being electrically charged, creates an electric field of its own, and thus shields the rest of the solution from the electrodes and their field. Second, the electric forces are really, really strong (like a thousand times stronger than you seem to think, and even stronger than that), and therefore the said layer at any point contains but a very tiny portion of metal ions. That portion suffices for creating an electric field which would completely negate the field of electrodes. The vast majority of metal ions float freely inside the solution, having no idea at all that they are being electrolyzed.
As the ions build up, the shielding becomes ever more perfect. Dissolve some $\ce{NaCl}$ in a glass of water without any electrodes or anything; the ions would float freely in the solution. Do they feel mutual attraction? Well, in a way, they do. Do they assemble into strange walls across the glass? I've never seen anything like that. Now, the ions in you cell (the majority that does not participate in the double layer on either side) feel exactly the same way as ions in that glass.
Electrochemical cell, if set up properly, will give current until one of the electrodes is completely eaten up. The need for a salt bridge has absolutely nothing to do with electric effects. Rather, it serves to prevent the mixing of solutions, which are presumably different and better left that way, for fear that otherwise they might react with the electrodes directly.
As for the cell with separate spaces and no bridge, it will stop working right away, much like the light bulb connected with a single wire to one pole of a battery.


text:
Removing Rust with Coke
I've seen a few videos on cola being used to remove rust, I've tried it myself too so I know that it works.
Now the thing that's bothering me is that this apparently contradicts what I've learnt about corrosion in class. If I'm not wrong, cola contains phosphoric acid, and therefore would most probably have free H+ ions floating around in there. I've been taught that H+ promotes (catalysis) the rusting of iron.
Then how come coke removes rust instead of promoting its formation? Also, why doesn't coke form rust on iron?

http://www.redandblack.com/uganews/science_health/she-blinded-me-with-science-why-can-coca-cola-be/article_a3385d64-a325-11e3-9b52-001a4bcf6878.html

Rust is a hydrated iron oxide that has a volume greater than the iron it contains. When iron rusts, the oxide grows in layers and curls away from the iron, exposing it to more corrosion.
Phosphoric acid is a moderately strong acid, but it forms an iron phosphate that is insoluble but adherent to iron (and steel). In that sense, phosphoric acid deactivates corrosion on iron (not perfectly, but pretty good), and there are commercial products (like Naval Jelly) which are essentially gelled phosphoric acid, designed to rub on iron and steel to give some corrosion protection. It is a sort of passivation.
Phosphate ions adhere to rust, making rust particles negative and promoting dispersion in the liquid and disadhesion from the iron metal. Some elbow grease is generally needed.
Overall, the extra corrosion from Coke or other phosphoric acid is minimal, and the dispersive effect is helpful, and the iron is better protected after the treatment than before.

Well, yes, $\ce{H+}$ ions accelerate rusting. The coke has phosphoric acid and $\ce{CO2}$.
The phosphoric acid is used for rust removal industrially.
The rate of acceleration of rusting of iron (with $\ce{H+}$ ions) is very slow and
all the rust will have already reacted with phosphoric acid to form iron phosphate.
All the rust is already destroyed and the $\ce{H+}$ ions just increase the rate catalytically, but now there is  no rust left for this effect to work.


text:
Does the term 'Cation' always refer to a positively charged particle?
From what I was taught in middle school, cations are those ions that move towards the cathode, likewise anions are those ions which move towards the anode.
I didn't have issues with this back then, since all we studied were electrolytic cells. But now that we've crossed over to electrochemical cells, I'm having doubts.
Every piece of chemistry literature I've come across so far, always deals with positive ions as cations and negative ions as anions; even my school textbook does it! That makes sense with respect to electrolytic cells, since the cathode's negative and anode's positive therefore positive ions would move to the cathode (hence 'cations') and negative ions would move to the anode (hence 'anions').
But in an electrochemical cell, the cathode's positive and the anode's negative. So if ions are to be classed according to the electrodes they move over to, then positive ions would be anions and negative ions would be cations, which is exactly opposite to the first case (on the basis of electrolytic cells).
This is really confusing....
So should I refer to positive and negative ions as cations and anions respectively or as anions and cations respectively? Or are both acceptable, depending on the scenario?

Yes, cations always have a positive charge and anions always have a negative one.
The difficulty is that the term cathode and anode do not always correspond to the same pole. The cathode is that pole of an electrolytic/electrochemical cell where reduction takes place (cathodic reduction) while the anode is where oxidation takes place (anodic oxidation). Since it depends on the type of cell which will be oxidised and which will be reduced, the terms cathode and anode mean different things in different cells.

You’re German, right? My mnemonic was anodische Oxidation ist das A und O.

@AaronAbraham Yeah, you are correct, it means anodic oxidation is the alpha and omega; in German, alpha and omega is typically written as A und O to the point that many people don’t know what it actually means.

To potentially add to the "anode/cathode" confusion, in the battery community, it is conventional to refer to an electrode as the "anode" or "cathode" based on whether oxidation or reduction occurs during discharge. (Otherwise, it would flip depending on whether you are charging or discharging, and would get confusing in a paper)

From what I was taught in Middle-school, cations are those ions that
move towards the cathode, likewise anions are those ions which move
towards the anode.

Nope, the definitions are as follows (from the IUPAC Goldbook):

cation
A monoatomic or polyatomic species having one or more elementary charges of the proton.
anion
A monoatomic or polyatomic species having one or more elementary charges of the electron.

So, yes, cations are always positively charged, while anions are always negatively charged.


text:
Relative acidity of aqueous LiF, SnCl2, and YCl3
Which one of the following salts make the highest and lowest pH when dissolved in water?
$$\ce{LiF, SnCl2, YCl3}$$
My approach was :
considering the corresponding aqua acids, the order of acidity should be $\ce{Li(H2O)_n+ < Sn(H2O)_n^2+ < Y(H2O)_n^3+}$ and $\ce{F-}$ is less stable than $\ce{Cl-}$. So $\ce{LiF}$ is the weakest acid and $\ce{YCl3}$ the strongest.
But considering aqua acid looks weird (do they exist?) and I cannot be sure this of solution. I don't even know the right answer. Can anyone give some help?

Hi, that's absolutely beautiful formatting, but please don't use MathJax in titles on chem.SE. And, regarding the question, I would expect the same trend that you have described.

Yes, acidity of aquocomplexes is important here but it's more tricky than just charge.

Salt solutions are acidic because of hydrolysis.
LiF is obviously neutral, because it's a salt of very strong acid and very strong base.
Ironically, Wikipedia is wrong about Sn2+: it's much more acidic in water than Ca2+ or Zn2+, and more acidic than Y3+,
because Sn2+ forms very stable and non-soluble hydroxo compounds and hence readily undergoes hydrolysis.
The answer is: LiF < YCl3 < SnCl2
Table data:
YCl3 has pH 4.5-5.8 (source),
SnCl2 has pH 1-2 (source 1, 2).

I can’t start to explain how wrong it is to define deprotonation of aquacomplexes as hydrolysis (even if Wikipedia does so). Also, $\ce{LiF}$ is not the salt of ‘a very strong acid and a very strong base’.

Well, if you can't start to explain your thoughts about hydrolysis, then at least explain why LiF is not a salt of strong acid and strong base.

Because neither is $\ce{HF}$ a strong acid nor is $\ce{Li-}$ — if that would even make sense — a strong base (You could argue about $\ce{LiOH}$, but that only semi made up the salt). On hydrolysis: It is the lysis (i.e. splitting) of a compound via the action of water. However, the acidic action in solution is not due to any lysis of any kind except for protolysis (displacement of a proton). The aqua ligands remain coordinated to the Lewis acid with one aqua ligand transforming itself (autolyticly!) into a hydroxido ligand.

A basic salt is formed when a weak acid reacts with a strong base. $\ce{LiF}$ is formed from strong base $\ce{LiOH}$ and weak acid $\ce{HF}$ to give a basic salt.
$\ce{HF}$ is not a strong acid.

I attempted to restore intended formatting; please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also it seems like the answer is only partial; feel free to elaborate on two remaining compounds to make it complete.

Welcome to Chemistry.SE! To help you become acquainted with the site, I would just like to suggest if you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and question policies. You will find StackExchange (SE) to be a different kind of Q&A website from the mainstream. To learn more visit Meta.Chemistry, Meta.SE, or chat


text:
What is the composition of the oil-like fluid that seeps out of dry-cells over time?
An experience a lot of us might've had; a dry cell (not completely depleted) when left in an electrical device for long periods of time, has this viscous fluid coming out that usually corrodes the terminals of the battery.
I almost always find this fluid, present as a sorta thin film over the terminals and the casing of the battery.
I'm told that the Ammonium Chloride in the cell, being slightly acidic reacts with the zinc casing of the cell, which is why dry cells don't last long (My teacher mentioned this during our Electrochem. class, he didn't mention anything about that fluid I was talking about though). Could this fluid possibly the result of that reaction? What are its chemical constituents?

Exactly which type of dry cell do you want to know about?All of them have different electrolytes.For example in alkaline cells it is potassium carbonate.

@ZOZ the Le Clanché kind......

......now I'm not sure if it's Leclanché or Le Clanché.....

I doubt that's what you want.Because those cells used to be used in the 1800s.I wrote an answer regarding more recent cells. Lechlance is quite similar.

Not really :-) https://en.wikipedia.org/wiki/Leclanch%C3%A9_cell

There are various kinds of dry cells.I will give some examples as to what is that oil-like fluid:
In zinc-carbon cell:

Zinc carbon cells have a short shelf life as the zinc is attacked by
  ammonium chloride. The zinc container becomes thinner as the cell is
  used, because zinc metal is oxidized to zinc ions. When the zinc case
  thins enough, zinc chloride begins to leak out of the battery. The old
  dry cell is not leak proof and becomes very sticky as the paste leaks
  through the holes in the zinc case. The zinc casing in the dry cell
  gets thinner even when the cell is not being used, because the
  ammonium chloride inside the battery reacts with the zinc. An
  "inside-out" form with a carbon cup and zinc vanes on the interior,
  while more leak resistant, has not been made since the 1960s.

In alkaline cells:

Technically, that white fluffy corrosion that develops at the ends of
  the battery (most often the negative end) is called potassium
  carbonate.
All batteries will slowly gradually self-discharge over time. This
  will occur whether they are setting on the shelf (a much slower
  process) or installed in a device (which often occurs much quicker) –
  and dead batteries will eventually leak.
Very high temperatures can also cause batteries to rupture and leak
  (e.g. in hot car during the summer).
The “alkaline” of the battery is potassium hydroxide (it’s the alkali
  equivalent of acid’s hydrochloric acid), and this will leak out,
  forming a white “fluff” of potassium carbonate, typically on the
  negative end of the battery cell – because apparently the positive end
  is vented better.

So the reason for the cell-corrosion varies with the type of the cell.
Read the Wikipedia page to know about the composition of electrolytes of different dry cells.https://en.wikipedia.org/wiki/Dry_cell

Would you also happen to know the product of the reaction between the ammonium chloride and the Zinc canister? I vaguely recall some source telling me it includes the complex  [Zn(NH3)4]2+, but I can't find sources backing that up....great answer by the way....

Yes you are correct.See https://en.wikipedia.org/wiki/Metal_ammine_complex that's the most probable product as far as I know!


text:
Derivation of the Heisenberg uncertainty principle
The Heisenberg uncertainty principle states that
$$\Delta x \Delta p \geq \frac{\hbar}{2}$$
where
$\Delta x$ is the uncertainty in the position,
$\Delta p$ is the uncertainty in linear momentum, and
$\hbar = 1.054571800(13) \times 10^{-34}\ \mathrm{J\ s}$[source] is the reduced Planck constant
This means that, regardless of what quantum mechanical state the particle is in, we cannot simultaneously measure its position and momentum with perfect precision. I read that this is intrinsically linked to the fact that the position and momentum operators do not commute: $[\hat{x},\hat{p}] = \mathrm{i}\hbar$.
How can I derive the uncertainty principle, as given above?

Related: Significance of 4π in the uncertainty principle; Justify the value of the product of the uncertainty in position and momentum

The proof I will use is taken from Griffiths, Introduction to Quantum Mechanics, 2nd ed., pp 110-111.
Defining "uncertainty"
Let's assume that the normalised state $|\psi\rangle$ of a particle can be expanded as a linear combination of energy eigenstates $|n\rangle$, with $\hat{H}|n\rangle = E_n |n\rangle$.
$$| \psi \rangle = \sum_n c_n |n\rangle \tag{1}$$
The expectation value (the "mean") of a quantity, such as energy, is given by
$$\begin{align}
\langle E\rangle &= \langle \psi | H | \psi \rangle \tag{2}
\end{align}$$
and the variance of the energy can be defined analogously to that used in statistics, which for a continuous variable $x$ is simply the expectation value of $(x - \bar{x})^2$:
$$\sigma_E^2 = \left\langle (E - \langle E\rangle)^2 \right\rangle \tag{3}$$
The standard deviation is the square root of the variance, and the "uncertainty" refers to the standard deviation. It's more proper to use $\sigma$ as the symbol, instead of $\Delta$, and this is what you will see in most "proper" texts.
$$\sigma_E = \sqrt{\left\langle (E - \langle E\rangle)^2 \right\rangle} \tag{4}$$
However, it's much easier to stick to the variance in the proof. Let's generalise this now to any generic observable, $A$, which is necessarily represented by a hermitian operator, $\hat{A}$. The expectation value of $A$ is merely a number, so let's use the small letter $a$ to refer to it. With that, we have
$$\begin{align}
\sigma_A^2 &= \left\langle (A - a)^2 \right\rangle  \tag{5} \\
&= \left\langle \psi \middle| (\hat{A} - a)^2 \middle| \psi \right\rangle \tag{6} \\
&= \left\langle \psi \middle| (\hat{A} - a) \middle| (\hat{A} - a)\psi \right\rangle \tag{7} \\
&= \left\langle (\hat{A} - a)\psi \middle| (\hat{A} - a) \middle| \psi \right\rangle \tag{8} \\
&= \left\langle (\hat{A} - a)\psi \middle| (\hat{A} - a)\psi \right\rangle \tag{9}
\end{align}$$
where, in going from $(7)$ to $(8)$, I have invoked the hermiticity of $(\hat{A} - a)$ (since $\hat{A}$ is hermitian and $a$ is only a constant). Likewise, for a second observable $B$ with $\langle B \rangle = b$,
$$\sigma_B^2 = \left\langle (\hat{B} - b)\psi \middle| (\hat{B} - b)\psi \right\rangle \tag{10}$$
The Cauchy-Schwarz inequality
...states that, for all vectors $f$ and $g$ belonging to an inner product space (suffice it to say that functions in quantum mechanics satisfy this condition),
$$\langle f | f \rangle \langle g | g \rangle \geq |\langle f | g \rangle|^2 \tag{11}$$
In general, $\langle f | g \rangle$ is a complex number, which is why we need to take the modulus. By the definition of the inner product,
$$\langle f | g \rangle = \langle g | f \rangle^* \tag{12}$$
For a generic complex number $z = x + \mathrm{i}y$, we have
$$|z|^2 = x^2 + y^2 \geq y^2 \qquad \qquad \text{(since }x^2 \geq 0\text{)} \tag{13}$$
But $z^* = x - \mathrm{i}y$ means that
$$\begin{align}
y &= \frac{z - z^*}{2\mathrm{i}} \tag{14} \\
|z|^2 &\geq \left(\frac{z - z^*}{2\mathrm{i}}\right)^2 \tag{15}
\end{align}$$
and plugging $z = \langle f | g \rangle$ into equation $(15)$, we get
$$|\langle f | g \rangle|^2 \geq \left[\frac{1}{2\mathrm{i}}(\langle f | g \rangle - \langle g | f \rangle) \right]^2 \tag{16}$$
Now, if we let $| f \rangle = | (\hat{A} - a)\psi \rangle$ and $| g \rangle = | (\hat{B} - B)\psi \rangle$, we can combine equations $(9)$, $(10)$, $(11)$, and $(16)$ to get:
$$\begin{align}
\sigma_A^2 \sigma_B^2 &= \langle f | f \rangle \langle g | g \rangle \tag{17} \\
&\geq |\langle f | g \rangle|^2 \tag{18} \\
&\geq \left[\frac{1}{2\mathrm{i}}(\langle f | g \rangle - \langle g | f \rangle) \right]^2 \tag{19}
\end{align}$$
Expanding the brackets
If you've made it this far - great job - take a breather before you continue, because there's more maths coming.
We have1
$$\begin{align}
\langle f | g \rangle &= \left\langle (\hat{A} - a)\psi \middle| (\hat{B} - b)\psi \right\rangle \tag{20} \\
&= \langle \hat{A}\psi |\hat{B}\psi \rangle - \langle a\psi |\hat{B}\psi \rangle - \langle \hat{A}\psi | b\psi \rangle + \langle a\psi |b\psi \rangle \tag{21} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - a\langle \psi |\hat{B}\psi \rangle - b\langle \hat{A}\psi | \psi \rangle + ab\langle \psi |\psi \rangle \tag{22} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - ab - ab + ab \tag{23} \\
&= \langle \psi |\hat{A}\hat{B}|\psi \rangle - ab \tag{24}
\end{align}$$
Likewise,
$$\langle g | f \rangle = \langle \psi |\hat{B}\hat{A}|\psi \rangle - ab \tag{25}$$
So, substituting $(24)$ and $(25)$ into $(19)$,
$$\begin{align}
\sigma_A^2 \sigma_B^2 &\geq \left[\frac{1}{2\mathrm{i}}(\langle\psi |\hat{A}\hat{B}|\psi \rangle - \langle \psi |\hat{B}\hat{A}|\psi\rangle) \right]^2 \tag{26} \\
&= \left[\frac{1}{2\mathrm{i}}(\langle\psi |\hat{A}\hat{B} - \hat{B}\hat{A}|\psi \rangle ) \right]^2 \tag{27}
\end{align}$$
The commutator of two operators is defined as
$$[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} \tag{28}$$
So, the term in parentheses in equation $(27)$ is simply the expectation value of the commutator, and we have reached the Robertson uncertainty relation:
$$\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{A},\hat{B} ]\rangle \right)^2 \tag{29}$$
This inequality can be applied to any pair of observables $A$ and $B$.2
The Heisenberg uncertainty principle
Simply substituting in $A = x$ and $B = p$ gives us
$$\sigma_x^2 \sigma_p^2 \geq \left(\frac{1}{2\mathrm{i}}\langle[\hat{x},\hat{p} ]\rangle \right)^2 \tag{30}$$
The commutator of $\hat{x}$ and $\hat{p}$ is famously $\mathrm{i}\hbar$,3 and the expectation value of $\mathrm{i}\hbar$ is of course none other than $\mathrm{i}\hbar$. This completes the proof:
$$\begin{align}
\sigma_x^2 \sigma_p^2 &\geq \left(\frac{1}{2\mathrm{i}}\cdot\mathrm{i}\hbar \right)^2 \tag{31} \\
&= \left(\frac{\hbar}{2}\right)^2 \tag{32} \\
\sigma_x \sigma_p &\geq \frac{\hbar}{2} \tag{33}
\end{align}$$
where we have simply "removed the square" on both sides because as standard deviations, $\sigma_x$ and $\sigma_p$ are always positive.

Notes
1 I have skipped some stuff. Namely, $\langle \hat{A}\psi |\hat{B}\psi \rangle = \langle \psi |\hat{A}\hat{B}|\psi \rangle$ which is quite straightforward to prove using the hermiticity of both operators; $\langle \psi |\hat{A}|\psi \rangle = a$; $\langle \psi |\hat{B}|\psi \rangle = b$; and $a = a^*$ since it is the expectation value of a physical observable, which must be real.
2 This does not apply to, and cannot be used to derive, the energy-time uncertainty principle. There is no time operator in quantum mechanics, and time is not a measurable observable, it is only a parameter.
3 Technically, it is a postulate of quantum mechanics. (If I am not wrong, it derives from the Schrodinger equation, which is itself a postulate.)

Concerning your footnote 3: In the book by Ballentine (chapter 3) the author derives the commutation relation(s) just by assuming invariance under Galilei transformations and then identifying the generators for the Galilei group with the dynamical variables. He even derives from this the Schroedinger equation for the free particle. It is a very interesting view on the foundations of QM.

Using the Cauchy-Schwarz inequality, as @orthocresol did in his answer, is certainly the most straightforward way to derive the Heisenberg uncertainty principle. There is however another very important inequality (that can, by the way, be used to derive the Cauchy-Schwarz inequality) that such a derivation can be based on, namely
\begin{align}
  \langle \phi | \phi \rangle \geq 0 \ , \tag{1}
\end{align}
where $| \phi \rangle$ can be any vector in Hilbert space. This is a fundamental property of the inner product. In order to arrive at the desired equation we just have to plug the right vector into this inequality. For this I will adopt @orthocresol's notation and lean a bit on the first paragraph of his excellent answer, i.e. I will introduce two hermitian operators, $\hat{A}$ and $\hat{B}$, whose expectation values (which are real numbers) shall be denoted by $a$ and $b$, respectively. Furthermore the variance of the observable $A$, which is represented by $\hat{A}$, is $\sigma_A^2 = \left\langle (A - a)^2 \right\rangle = \langle \psi | (\hat{A} - a)^2 | \psi \rangle$ and analogously $\sigma_B^2 = \left\langle (B - b)^2 \right\rangle = \langle \psi | (\hat{B} - b)^2 | \psi \rangle$.
With that out of the way let's get started: First we use $\hat{A}$ and $\hat{B}$ to define another operator, $(\hat{A} - a) + \mathrm{i}\lambda (\hat{B} - b)$, where $\mathrm{i} = \sqrt{-1}$ and lambda is some arbitrary real constant. Since an operators map vectors onto vectors, letting our new operator act on some wave vector $| \psi \rangle$ will just give another wave vector, say $| \tilde{\psi} \rangle$,
\begin{align}
\left( (\hat{A} - a) + \mathrm{i}\lambda (\hat{B} - b) \right) | \psi \rangle = | \tilde{\psi} \rangle \ .
\end{align}
This new wave vector will have to satisfy inequality $(1)$:
\begin{align}
\langle \tilde{\psi} | \tilde{\psi} \rangle \geq 0 \ .
\end{align}
Using the hermiticity of $\hat{A}$ and $\hat{B}$, i.e. $\hat{A}^{\dagger} = \hat{A}$ and $\hat{B}^{\dagger} = \hat{B}$, the bra vector $\langle \tilde{\psi} |$ corresponding to the ket vector $| \tilde{\psi} \rangle$ is
\begin{align}
\langle \tilde{\psi} | = \langle \psi | \left( (\hat{A} - a) + \mathrm{i}\lambda (\hat{B} - b) \right)^{\dagger} = \langle \psi | \left( (\hat{A} - a) - \mathrm{i}\lambda (\hat{B} - b) \right)
\end{align}
we get
\begin{align}
\left\langle \psi \middle| \left( (\hat{A} - a) - \mathrm{i}\lambda (\hat{B} - b) \right)\left( (\hat{A} - a) + \mathrm{i}\lambda (\hat{B} - b) \right) \middle| \psi \right\rangle &\geq 0 \\
\biggl\langle \psi \Bigg| (\hat{A} - a)^{2} + \lambda^{2} (\hat{B} - b)^{2} + \mathrm{i} \lambda \underbrace{\left( (\hat{A} - a) (\hat{B} - b) - (\hat{B} - b) (\hat{A} - a)\right) }_{= \left[ (\hat{A} - a) , (\hat{B} - b)\right] } \Bigg| \psi \biggr\rangle &\geq 0 \\
\left\langle \psi \middle| (\hat{A} - a)^{2} \middle| \psi \right\rangle + \lambda^{2} \left\langle \psi \middle| (\hat{B} - b)^{2} \middle| \psi \right\rangle + \mathrm{i} \lambda \left\langle \psi \middle| \left[ (\hat{A} - a) , (\hat{B} - b)\right] \middle| \psi \right\rangle &\geq 0 \tag{2}
\end{align}
The commutator $\left[ (\hat{A} - a) , (\hat{B} - b)\right]$ evaluates just to the commutator of $\hat{A}$ and $\hat{B}$,
\begin{align}
\left[ (\hat{A} - a) , (\hat{B} - b)\right] &= (\hat{A} - a) (\hat{B} - b) - (\hat{B} - b) (\hat{A} - a) \\
&= \hat{A}\hat{B} - \hat{B} \hat{A} - b\hat{A} + b\hat{A} - a \hat{B} + a\hat{B} \\
&= \hat{A}\hat{B} - \hat{B} \hat{A} = \left[ \hat{A} , \hat{B} \right] \ ,
\end{align}
so that, together with the definitions of the variances of $A$ and $B$ from above and the expectation value of the commutator $\langle [ \hat{A} , \hat{B} ] \rangle = \langle \psi | [ \hat{A} , \hat{B} ] | \psi \rangle$ , inequality $(2)$ becomes
\begin{align}
\left\langle \psi \middle| (\hat{A} - a)^{2} \middle| \psi \right\rangle + \lambda^{2} \left\langle \psi \middle| (\hat{B} - b)^{2} \middle| \psi \right\rangle + \mathrm{i} \lambda \left\langle \psi \middle| \left[ \hat{A} , \hat{B} \right] \middle| \psi \right\rangle &\geq 0 \\
\sigma_A^2 + \lambda^{2} \sigma_B^2 + \mathrm{i} \lambda \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle  &\geq 0 \ . \tag{3}
\end{align}
Inequality $(3)$ already contains all the ingredients we need to arrive at the Robertson uncertainty relation. We just need to extract it out of it. To achieve this we have to recognize that the left-hand side of inequality $(3)$ depends on the arbitrary real constant $\lambda$ and that the strongest form of this inequality will be obtained when we choose a value for $\lambda$ that minimizes the left-hand side. This minimization is done in the usual way of taking the first derivative of the left-hand side with respect to $\lambda$ and setting it equal to $0$ (I leave out the check of the second derivative to ascertain that we really have a minimum):
\begin{align}
\frac{\mathrm{d}
\left( \sigma_A^2 + \lambda^{2} \sigma_B^2 + \mathrm{i}\lambda \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle \right)}{\mathrm{d} \lambda}  &= 0 \\
2 \lambda \sigma_B^2 + \mathrm{i} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle &= 0 \\
\Rightarrow \lambda = \frac{1}{2\mathrm{i}} \frac{\big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle }{\sigma_B^2} \ .
\end{align}
Plugging this value for $\lambda$ into inequality $(3)$ yields the Robertson uncertainty relation
\begin{align}
\sigma_A^2 - \frac{1}{4} \left(\frac{\big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle }{\sigma_B^2} \right)^{2} \sigma_B^2 + \frac{1}{2} \frac{\big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle }{\sigma_B^2} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle  &\geq 0 \\
\sigma_A^2 \sigma_B^2 \underbrace{- \frac{1}{4} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle^{2} + \frac{1}{2} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle^{2}}_{= \frac{1}{4} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle^{2}}  &\geq 0 \\
\Rightarrow \sigma_A^2 \sigma_B^2 \geq - \frac{1}{4} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle^{2} = \left(\frac{1}{2i} \big\langle \big[ \hat{A} , \hat{B} \big] \big\rangle \right)^{2} \ . \tag{4}
\end{align}
And how we get from here to Heisenberg's uncertainty relation has already been nicely demonstrated in @orthocresol's answer. Just replace $\hat{A}$ with the position operator $\hat{x}$ and $\hat{B}$ with the momentum operator $\hat{p}$, use the known expectation value of their commutator, $\big\langle \big[ \hat{x} , \hat{p} \big] \big\rangle = \mathrm{i} \hbar$, then take the positive square root of the resulting inequality and you get
\begin{align}
\sigma_x \sigma_p &\geq \frac{\hbar}{2} \tag{5} \ .
\end{align}


text:
What exactly is hydrolysis reaction in organic chemistry?What mechanism does it follow?
According to my study material:

Hydrolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) where water acts as both nucleophile and a solvent molecule.

Is this definition correct? If so how to solve the following problem (picked up from my textbook exercise which has no solution)?

When alkyl bromides (listed here) were subjected to hydrolysis in a
  mixture of ethanol and water ($80\ \%\ \ce{C2H5OH} / 20~\%\ \ce{H2O}$)
  at $55~\mathrm{^\circ C}$ the rates of reaction showed the following order:
  $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$.
Provide an explaination for this order of reactivity.

But my doubt is how is this order $\ce{(CH3)3CBr} > \ce{CH3Br} > \ce{CH3CH2Br} > \ce{(CH3)2CHBr}$ possible?
I mean I know the first compound will form a tertiary carbocation. That is okay for $\mathrm{S_N1}$ but how will the second compound undergo $\mathrm{S_N1}$ (because methyl carbocation is very unstable)? Moreover if really $\mathrm{S_N1}$ is taking place the fourth compound should form a more stable carbocation compared to the third compound. These things are confusing me. Is hydrolysis really $\mathrm{S_N1}$ or something else?
According to me the order should have been $(1)>(4)>(3)>(2)$.

Do you call it a hydrolysis reaction? ... Because I just call it a substitution reaction...

@DHMO I stated what my textbook and study material wrote.I did'nt make up the definitions myself.See https://en.wikipedia.org/wiki/Solvolysis

Hydrolysis is not necessarily SN1, it can also refer to a SN2 process (especially if there is OH-). Or it can be any process really. There are many different ways of hydrolysing esters, for example. However, in this particular case, it is weird. I would expect only SN1 reactivity under the stated conditions, so I don't know why MeBr is so high up.

Your textbook is almost correct. Here is the corrected version of its statement:

Hydrolysis is a special type of nucleophilic substitution where water acts as both nucleophile and a solvent molecule.

Note me omitting the $\mathrm{S_N1}$ part. This is because a hydrolysis mechanism can be any nucleophilic substitution, whether $\mathrm{S_N1, S_N2, S_N}$ for heavy atoms or $\mathrm{S_NAr}$.
That also explains the given order of reactivity. tert-Butyl bromide is very fast at $\mathrm{S_N1}$ reactions. Bromomethane very rapidly undergoes $\mathrm{S_N2}$ reactions. The other two are in-between with bromoethane being just a tad better at $\mathrm{S_N2}$ reactions than 2-bromopropane is at $\mathrm{S_N1}$ reactions.
($\mathrm{S_N1}$ is preferred in 2-bromopropane as the solvent is polar protic and it can stabilize the intermediate carbocation well)

By the extension of your definition: In the solvolysis definition in Wikipedia too should the $\mathrm{S_N1}$ be removed? See https://en.wikipedia.org/wiki/Solvolysis

@ZOZ Yes.$%MathJax comments hurray!$

Morever,I cannot understand how can water in this case act as nucleophile in 20 percent ethanol.Water is considered to be a weak nuceophile.Can it really initiate a $\mathrm{S_N2}$ ?

@ZOZ Water isn’t that weak a nucleophile, tbh.

Well by that logic if it were 80 % ethanol and 20 % water would ethanol have acted as the nucleophile? I mean the ethyl group exerts a $+I$ effect on the O atom.So would ethanol be nucleophile in that case? @Jan

@ZOZ But ethanol is bulkier, allowing water to be a kinetically better nucleophile.

Thank you for clearing the doubt.I was really confused with that Wikipedia definition.Maybe I should edit that Wiki page :-P.


text:
Why isn't CASPT2 used in surface hopping simulations?
I will write what I think the answer is. I'd like someone to confirm and maybe explain in detail.
I think that since surface hopping dynamics simulates movements on PES, it has to calculate gradients in each step. Since gradient for CASPT2 method can be evaluated only numericaly, CASPT2 quantum mechanical base for surface hopping dynamics would make calculations imspossibly slow.
Is this correct?

CASPT2 analytic energy gradients are available in some codes (e.g. in MOLPRO). 2) A quick look at surface hopping dynamics reveals that  nonadiabatic coupling terms has to be computed, which looks like an actual issue, since they seem to be not available in codes as of now.

I'm pretty sure nonadiabatic couplings are available in Molcas 8, since you can optimize conical intersections with Molcas.

It probably comes a little late, but you might want to check this out!
https://arxiv.org/abs/1701.02259


text:
Change of potential difference during electrolysis while adding salt
I was conducting a simple experiment at home on electrolysis. Obviously, when you build a circuit with a 9V battery, a lamp, and put two electrodes in water you measure a potential difference of ~9V on the two electrodes while the potential difference on the lamp is 0 (really close to 0). Then as I kept adding salt in water, I first measured that the potential difference across the lamp got higher and higher (reaching 6V), then suddenly, as I got from 40g /250ml of salt in water to 50g /250ml, the potential difference across the lamp fell to 4V.
Why is that? Maybe is it because of the ion association occurring in water as the concentration of salt gets higher?
4.5 V lamp. 8.6 e.m.f battery.

What was the voltage across the electrodes at 40g and 60g of solute?

what kind of lamp?

It sounds like you might be witnessing corrosion on the electrodes, which I would expect if you allowed electrolysis to continue. You should try different concentrations with new (or at least vigorously cleaned) electrodes and see if the results are similar.

It would be good to clarify the exact experimental set up. Lamp as commented before but more importantly what were the electrodes made from? By salt, I am assuming culinary table salt, which is mostly sodium chloride, but often contain sodium iodide or potassium iodide and an anti-caking agent such as sodium aluminosilicate. Were your solutes fully dissolved? Were the solutions at the same temperature? What else did you observe?

Tungsten lamps change resistance as they warm up. At room temperature, the resistance of a small flashlight lamp (e.g. #222) might be 10 ohms when brightly lit, but only 1.5 ohms at room temperature.
At first, the tungsten filament acts like a short circuit, so most of the voltage of the battery is dropped across the series resistance of the electrolytic cell. Once enough current goes through the lamp, though, its resistance abruptly rises and more of the voltage is dropped across the filament.
If you use an LED instead of a tungsten-filament bulb, the change in voltage would be more linear as you add salt. Note that an LED will be effectively an open circuit until a minimum voltage is reached (~1.6 V for a red LED, ~3 V for blue or white LED), but then current increases linearly until ~20-50 mA, at which point some may burn out.

This does not explain why the voltage dropped at higher solute concentrations.

Yes it would The tungsten element got warm, and the resistance increased. At the same time, more solute was added. The observed drop in voltage was correlated with higher salt concentration. Correlation <> causuality.