text:
Which solution is more conductive
Which of the following solutions is more conductive:
1) 0.1 M sodium hydroxide and 0.1 M hydrochloric acid mixture (effectively NaCl?)
2) 0.1 M sodium hydroxide and 0.1 M acetic acid mixture (effectively NaCH3COO?)
I know the answer is the first one but I can't explain why.  I have been taught (high school chemistry) that conductivity of ions is simply relative to [ions].  If that were true, both solutions would be equally conductive.  I know that strictly speaking, based on the initial acids and bases, 1) would be more conductive, but I can't understand this because there should be none of the original reactants present anyways.  This came up on a test, and I can't justify 1) being correct based on what I have learned.

What you have been taught is wrong. Conductivities of different ions at the same concentration vary a great deal.

Right, but that's why I am trying to figure out how to explain this.

Conductivity of different ions carrying the same amount of charge can vary as per the comment of Ivan. In this question, acetic acid is a weak acid so the final acetic anion will combine with the hydrogen ion from the autodissociation of water molecule to form acetic acid molecules. This is known as the hydrolysis of $\ce{NaAc}$. Hydrochloric acid, on the other hand, is a strong acid so the hydrolysis won't happen. Hydrolysis is the main reason that $\ce{NaAc}$ solution is basic while $\ce{NaCl}$ is neutral.

There should be none of the original reactants present anyways.

This is one fallacy in your reasoning. There are some $\ce{CH3COOH} \rm(aq)$ formed due to hydrolysis in the final solution since it is weak acid.
To summarize, in the first case the conducting particles are equal amount of $\ce{Na+}$ and $\ce{Cl-}$, while in the latter they are $\ce{Na+}$, $\ce{Ac-}$ and $\ce{OH-}$, with the concentration of the first equal the sum of the other two. This is what I believe you need to know. Beyond that without looking up the conductivity table or analyzing the structures of the ions we cannot deduce whose conductance is higher.
In paricular, given a strong acid $\ce{HA}$ and a weak acid $\ce{HB}$, saying that for the solution of the same concentration, $\ce{NaA}$ will be more conductive than $\ce{NaB}$ is simply wrong and there are counterexamples.

This is exactly what I was looking for.  You cannot make that assumption.  I know I did a bad job at wording it, but the bottom line is that it would be wrong to assume the weak acid/weak base soln. is more conductive.  And I did in fact find the counter examples that you speak of.  I will just have a tough job convincing my teacher given that this appeared on a government issued examination...

@Jordan I believe it will be a hard job. Showing that you have done some research and giving the counter example will be more helpful than providing a link to my answer. Good luck!

Thanks a lot, I will try my best.  Its just frustrating to see a question like this issued on an exam that thousands of students have written.  If you aren't reading into it, you just see two strong acids/bases vs weak acid/strong base and chose the answer accordingly.  When you think about it though, it really counters the logic of all ions conduct equal amounts of electricity (I know this is wrong).  Just... disappointing that I have spent days unable to explain one question.

Yes, and it's frustrating as well to see the question being voted down for reasons I don't know... As long as you know that (in my example above) $\ce{HA}$ will be more conductive than $\ce{HB}$ it's fine I guess.

Well, good news.  I am going to go with the hydrogen oxalate anion as the basis for my argument.  It is far more conductive than chloride, yet is the conjugate base of a weak acid. :)

@Jordan Yes, that is one of the counterexamples. Glad you found it through your own efforts.


text:
Methods for dissolving Scotch/Kaptop/etc. Tape for Electron Microscopy?
The famous method for the Nobel prize in studying Graphene is the so-called "scotch tape" method. Here, one takes a weakly-bonded van der Waals-type material and peel a layer off. From that layer, another piece of tape is taped onto the original piece of tape with the layer to split it into two pieces. This is repeated until nanometer thick layers are produced.
At this point, you have a single monolayer on a piece of tape, and need to transfer it onto a grid to look at it with a TEM (transmission electron microscope). However, the monolayer is much too thin to peel off the tape, so the tape must be completely dissolved.
The question I have is: what are the best options for dissolving tape?
Obviously there are many, many variables at play here. But here are some relevant constraints:

Completely dissolves tape, leaving little sticky residue
Solvent is relatively non-reactive to inorganics, metals
Safe and inexpensive

For example, I found a question on Research Gate that suggests using an alkaline solution to completely dissolve Kapton. I wonder if there are other methods for scotch tape
Since this method for TEM preparation is so common there must be some great literature resources on this topic, so resource recommendations would be great!
I don't fully understand the mechanics of "stickyness" so a description from the atomic perspective would be useful, especially in understanding what good solvents should do.

Acetone? Dobatex?

How thoroughly do those solvents work? The plastic part of the tape would dissolve easily in acetone,  but thoroughly does the adhesive dissolve? I wouldn't expect it to go to the monomer level, but l may be wrong.

No idea. Trial and error? Ask the manufacturer of the tape?

I see... the reason I am asking this question on chemistry.se is to avoid reinventing the wheel, surely this problem has been thoroughly studied in some research article.

@ user157879 Generally is used Acetone. If you are  interested also in details for scotch tape methodology I can post them

@Jojostack, yes please! I would be interested in knowing as much about the method as there is out there.

I've added an answer on the methods I think it is one of the clearer reference around. I have also more references if needed

Kapton and scotch tape are completely different and are dissolved by different solvents, so this is really two separate questions. (Just a warning for people who end up on this page; half the answers are for the wrong type of tape.)

Generally you use Acetone. The following is taken from
Enoki, Toshiaki, and Tsuneya Andō. 2018. Physics and Chemistry of Graphene : Graphene to Nanographene p. 94
The Scotch tape method is very common in the fabrication of
graphene devices, which are mainly used in research into the
fundamental properties of graphene. It can produce high-quality
graphene with lateral sizes ≤ 0.1 mm; however, controlling the size
and position of the graphene is almost impossible. The following is
the typical procedure of the Scotch tape method:
(i) Place a few flakes of graphite (≈ 1 mm) on the adhesive side
of a plastic sticky tape with tweezers (Fig. 3.1(a)). Scotch tape
(3M) and Nitto tape are commonly used. Natural graphite,
Kish graphite, or HOPG (highly oriented pyrolytic graphite) is
usually used for the starting material.
(ii) Fold the adhesive side to sandwich the graphite flakes and
press the tape firmly (Fig. 3.1(b)).
(iii) Peel the tape apart slowly, so that the graphite flakes are
cleaved and attached on both sides of the tape (Fig. 3.1(c)).
(iv) Repeat the second and the third steps with slightly shifting
the fold line, so that the graphite flakes do not overlap (Fig.
3.1(d)).
(v) Stop repeating the cleavage when graphite flakes spread over
the sticky tape (Fig. 3.1(e)).
(vi) Prepare a silicon substrate with a silicon dioxide layer on
the top surface (Fig. 3.1(f)). Place address markers on the
surface in advance using photolithography so that one can
easily locate the position of a graphene flake in an image.
These markers are indispensable for further microfabrication
on graphene. An example of the address markers is shown in
Fig. 3.2
(vii) Attach the adhesive side of the tape with the graphite flakes to
a silicon substrate, and gently rub the surface to remove any
air between the substrate and the tape (Fig. 3.1(g)).
(viii) Slowly peel the tape off the substrate (for example, more than
2 min for a 1 cm substrate) (Fig. 3.1(h)). Not only graphite
flakes but also some adhesive remains on the substrate. The
latter can be removed by submerging the substrate in acetone
for a few seconds.

It is also reported that in the scotch tape method, not only graphene flakes but also a large amount of adhesive is attached to the substrate. You get rid of it immersing the substrate in Acetone. Other methods are reported in the reference too.
About adhesion
There are lots of aspect about adhesion. About scotch tape the most interesting answer you can find here "https://www.scientificamerican.com/article/what-exactly-is-the-physi/" is the following
"The simplest answer that I can give to the question is that pressure-sensitive adhesives (which are polymers) are 'tacky' or 'sticky' because they are essentially very high viscosity liquids that also have some elastic characteristics--in technical terms, they are 'viscoelastic.' This property means that they exhibit some of the characteristics of liquids, and so they will 'wet' a surface to which they are pressed. But then, because of their elasticity, they will resist separation when stressed. Thus, 'stickiness' is strictly a physical (viscoelastic) phenomenon, not a chemical one."

Great! If you have a few more references is love to read them, essentially if they discuss this in the context of TEM.

Kapton is a polyimide:

It contains an $\ce{N}$ substituted phtalimide as a substructure.
This kind of compound is used in the Gabriel synthesis to make primary amines.
The synthesis is very ingenious because no secondary or tertiary amines can be formed, which is normally the problem with direct alkylation.
In the final step the phtalimide is hydrolyzed with hydrazine, a reaction that works very well.
A 35-50% solution of hydrazine in water is used.

Hydrazine is not the most innocuous of products, although in solution it is not so dangerous, and should be used in a fume cupboard.
If you do not have a minimal Chemistry lab at your disposal, then you can better try to use a potassium hydroxide solution.
I have never tried it because the hydrolysis with hydrazine gives such good results.

This is great. How does the solution handle the adhesive commonly used in kapton tape? Could you also add on some discussion about ordinary scotchpe?

source: Stack Exchange

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

sort: votes


text:
Angular momentum - Question on Zare's use of the Vector Model to Estimate Probability
My question regards a specific example, although I get the feeling that the answer is simpler than I imagine. In Zare's Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics pgs 53-54, He discusses the vector model for coupled and uncoupled representations for the addition of two angular momenta.
I've scanned and included the picture I am going to refer to for clarity:

The image above is a modified representation of the uncoupled representation, although it appears very similar to the coupled diagram, the difference is that j1 has been arbitrarily fixed, and j2 is precessing in the dashed circle, around an axis parallel to z.
Zare expands the dot product of $\mathbf{j_1}$ and $\mathbf{j_2}$
(Eq 2.34):
$\mathbf{j_1 . j_2} = |\mathbf{j_1}||\mathbf{j_2}| cos(\gamma)$
$=|\mathbf{j_1}||\mathbf{j_2}| [ cos(\theta_1) cos(\theta_2)+  sin(\theta_1) sin(\theta_2) cos(\theta_1) cos(\phi_1 -\phi_2)]$
So far, so good, no problem. Then he instructs:
"Let us differentiate Eq (2.34) with respect to time to obtain the rate of change of the length of $\mathbf{j}$. From the law of cosines
$|\mathbf{j}|^2 = |\mathbf{j_1}|^2 + |\mathbf{j_2}|^2 - 2 |\mathbf{j_1}| |\mathbf{j_2}| cos(\gamma)$
$=|\mathbf{j_1}|^2 + |\mathbf{j_2}|^2 - 2 \mathbf{j_1} . \mathbf{j_2}$
So that
$\frac d{dt} (\mathbf{j_1} . \mathbf{j_2}) = -\frac 12 \frac d{dt} [ |\mathbf{j}|^2 + |\mathbf{j_1}|^2 - |\mathbf{j_2}|^2 ]$"
And THIS next step is where I get lost:
He IMMEDIATELY equates that expression with the following:
$$ = -|\mathbf{j}| \frac {d|\mathbf{j}|}{dt} $$
So that there's no confusion, what I do not understand is the statement:
$$\frac d{dt} (\mathbf{j_1} . \mathbf{j_2}) = -\frac 12 \frac d{dt} [ |\mathbf{j}|^2 + |\mathbf{j_1}|^2 - |\mathbf{j_2}|^2 ] = -|\mathbf{j}| \frac {d|\mathbf{j}|}{dt}$$

Now, if I explicitly expand all components of the vector norms: $\mathbf{j_1}$,$\mathbf{j_2}$,and $\mathbf{j}$, and try to solve for what's inside the brackets on the LHS, I just regenerate the expression:
$|\mathbf{j}|^2 =|\mathbf{j_1}|^2 + |\mathbf{j_2}|^2 - 2 \mathbf{j_1} . \mathbf{j_2}$, which is of no use in aiding me as far as I can see.
Furthermore, it is explicitly stated that the only time-dependent component of this system is the dihedral angle $\phi_1 - \phi_2$, and in any expansion of the vector $|\mathbf{j}|$ into its components, a time derivative would render the any other independent parts mute.
I am thoroughly confused. I have consulted the following textbooks and not found any sign of this exact derivation:
Cohen-Tannoudji (volumes I and II), Feynman's lectures on Physics Vol III, Shankar, Atkin's physical chemistry, Engel and Reid's physical chemistry, Gaziorowicz, and Eisberg and Resnick.
What I suspect is that I'm simply lacking in some fundamental piece of calculus or arithmetic. Whether there's some simple trick of Vector Calculus that I haven't employed to give me the identity placed here? I would love a direct answer, but will also accept a nudge in the right direction. Right now it just feels like I'm leafing through books without a compass.
Best,
and THANKS!

+1 for citing Dick Zare, a truly significant contributor to the field of chemical physics.

Nothing special here, just that the moduli of  $\mathbf j_1$ and $\mathbf j_2$ are fixed, by definition of the problem.

As mentioned in the comment by Weijun Zhou, the length of the vectors $\mathbf{j}_1$ and $\mathbf{j}_2$ does not change as $\mathbf{j}_2$ precesses about $\mathbf{j}_1$, only the total vector $\mathbf{j}$ changes its length. Knowing this just requires application of the sum and chain rule of derivations that is
$
-\frac{1}{2}\frac{d}{dt}|\mathbf{j}|^2=-2|\mathbf{j}|\frac{1}{2}\frac{d}{dt}|\mathbf{j}|
$
and the time derivatives of $\mathbf{j}_1$ and $\mathbf{j}_2$ are zero.

Ah! Thank you. I expected it to be obvious. It's been a while since I've had to apply calculus to vectors.

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


text:
How are such ionic equilibrium equations derived? (H3O+ concentration)
I missed my classes on ionic equilibrium & I came to know that certain equations were derived for cases such as 'Monobasic acid', 'Two monobasic acids', 'Dibasic acid'etc.
These equations relate H3O+ concentration with the acid's initial concentration, its dissociation constant & water's dissociation constant.
These equations were told to me that, they provide highly accurate answers, instead of the commonly used Henderson–Hasselbalch equation.
These equations are of either cubic or bi-quadratic form, taking the H3O+ concentration as the variable.
Can someone explain this? Unfortunately; I couldn't find any sources on the internet (or perhaps I didn't use the proper keywords?)
If anyone is able to drop by a link to a webpage that explains these derivations and the use of these equations, it would be greatly helpful. Thank-you.

Possible duplicate of Determine the pH of mixture of two weak acids

Before anything else I must clarify that there is nothing wrong with Henderson–Hasselbalch equation. It is formally exact because it comes from the definition of the $K_a$ as long as you ignore ionic strength which I believe is not what you are interested in at least at this stage. If you say there's anything inaccurate about it it is because people are making assumptions when applying the equation.
Now for the main matter. If you take a look at this note (not what you are looking for I am afraid, but we'll come to that in a moment), near the end of page 4 we have the quadratic form of the equation
$$K_a=\frac{[\ce{H3O+}]^2}{c(\ce{HA})-\ce{H3O+}}$$
based on approximation that $[\ce{H3O+}]\approx[\ce{A-}]$. Now, what you are looking for is the formula without making this assumption. So first examine what's wrong about this assumption.
Mass balance for $\ce A$ states,
$$c(\ce{HA})=[\ce{HA}]+[\ce{A-}],$$
charge balance states,
$$[\ce{H3O+}]=[\ce{A-}]+[\ce{OH-}],$$
you see that the approximation ignores the second term in RHS,
and we do know,
$$[\ce{H3O+}][\ce{OH-}]=K_{\rm w},$$
Now it is not difficult to see what we need to put into the the definition of $K_a$.
$$K_a=\frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{[\ce{H3O+}]\left([\ce{H3O+}]-\frac{K_{\rm w}}{[\ce{H3O+}]}\right)}{c(\ce{HA})-[\ce{H3O+}]+\frac{K_{\rm w}}{[\ce{H3O+}]}}$$
This is the cubic equation you are looking for, it is not that difficult to derive but quite a pain to solve numerically if you don't have a calculator that solves equation for you (there are various numerical techniques but that's beyond the scope of this answer). The reason that it is not explicitly included in many books is that you don't need to go for such complex form under most occasions and, more importantly, these forms do vary a lot if, say I am talking about $[\ce{H2A}]$ instead of $[\ce{HA}]$.
So the bottom line: instead of pursuing these complex forms (which may seem interesting, but you just can't remember all of them), stick to the charge balance equation and mass balance equation and optionally, if you have been taught, the proton balance equation, and also the definitions of $K_a$'s. With these in mind you can derive the most accurate form of high-order equations for $[\ce{H3O+}]$ whenever you want to, and these balance are not that hard to grisp compared to these exotic high-order equations.


text:
Single Crystalline Thin Films is detect by GIXRD?
I understand epitaxal thin film, have lattice match with substrate.
But what about those films that doesnt match but have only one XRD peak? Is this proof of single crystale thin film?

What is characterized by one peak? Whether the lattices match or not?

single crystal thin film @andselisk on GIXRD you will observe only one pick, is this correct?

I'm still not following. What do you mean by "pick"? To select the certain lattice from an array? Or was it a typo and you meant "peak"? This is still the same x-ray diffraction, so there is going to be as many peaks as Bragg's law and systematic extinctions allow. Please refine your question as now it's quite hard to read and to understand what' you are trying to say.

@andselisk XRD pick, in xrd analysis results

@andselisk Please sorry, I edited question,  I now realize that it was not clear

I know what "XRD peak" is, but "XRD pick" makes no sense whatsoever (the question says "peak" though, but in the comments you insist on "pick").

yes, you are right @andselisk english is not my mother tongue, I was referring to how one identify single crystal thin film- is it by occurance of single peak

English isn't my mother language too, I understand the struggle. I would't rely on a single peak only; this can serve as a lousy proof of crystallinity, but in general I'd do some supplementary tests such as XRF, AFM/SEM, maybe SAXS.


text:
How long does it take to saturate H2O with O3
Let’s assume I have one liter of distilled $\ce{H2O}$ in an stainless steel container – from the bottom side of the container I do bubble ozone trough the $\ce{H2O}$.
Of cause it is not pure O3, so let’s assume that the gas consists of:
7% $\ce{O3}$
83% $\ce{ O2}$
10%. $\ce{ N2 / NO2 / NO}$
The gas flow is about 5 liters per minute and the corona discharge consumes about 500W.
How long will it approximately take, till full saturation of the
  $\ce{H2O}$   with   $\ce{O3}$  will take place?

The solubility of ozone in water as per one website is:

First thing is to establish which curve you need. In 1 m$^3$ gas you have 70 L ozone:
$\frac{70\ L\ \ce{O3}}{1\ m^3}\frac{1\ mol}{0.082\cdot 298\ L}\frac{16\cdot 3\  g\ \ce{O3}}{1\ mol}= 138\ g\ \ce{O3}/m^3$
Let us take the curve of 150 g/m$^3$ which has an ozone solubility of 30 mg/L water.
Then:
$\frac{30\cdot 10^{-3}\ g\ \ce{O3}}{1\ L\ \ce{H2O}}\frac{1\ mol}{16\cdot 3\ g\ \ce{O3}}\frac{0.082\cdot 298\ L}{1\ mol}\frac{1\ min}{5\cdot 0.07\ L\ \ce{O3}}= 0.044\ min$
Which is about 3 seconds.

please explain why time enters your last expression, and there is no mention of the diffusion coefficient of ozone in water. Just because the gas flows in does not mean that it is all absorbed or if so how long this takes.

Because the gas flow is 5 L/min with a content of 7% ozone. Of course it is supposing that the mixing is perfect, etc. I have not complicated things because I think that the OP just wants to know if it will take long before the output will stabilize. On basis of the other questions posted by the same OP.

I understand what you write. The distance diffused in time $t$ is $L=\sqrt{6Dt}$ and for, say, oxygen in water ,$D\approx 10^{-5}$ cm$^2$/s (which is a typical value) meaning that in 1 sec the gas has diffused only $\approx 0.01$ cm. So I think it will take far longer to dissolve that  than you estimate. Of course it will depend on the gas pressure also.


text:
Formatting a chemical equation
I recently came across this chemical equation :
$$\ce{2HNO3 + Na2Co3 -> 2NaNO3 + CO2 + H2O}$$
The equation describes the reaction between Nitric Acid and Sodium Carbonate, the equation states that the resulting compounds $\ce{2NaNO3}$ carbon dioxide and water would be created. However my question is how the formula should be written out. I have always written a chemical equation like this :
$$\ce{(HNO3)2 + Na2Co3 -> (NaNO3)2 + CO2 + H2O}$$
Do both ways of writing the equation state the same thing? If so which one should I use. Which form of writing the equation is considered "more correct", used more commonly or considered "more standardized".

You have always written it wrong.

Not to mention that Co3 is tricobalt, whereas CO3 is carbonate. Mind capitalization too, it's very important. Voting to close as homework (seriously, read any introductory book on chemistry).

@andselisk 'A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself.' Imho this doesn't apply here.

This question is about a fundamental misunderstanding of how the rules  work. Perhaps the OP should have done a little research first, and this question is essentially too simple, but to shut it down this way doesn't seem to fit. To take a whole species and add parents and subscript kinda says it is 2 or 3 of them.....

According to the IUPAC guidelines arabic numerals are used:

(a) As right subscripts, to indicate the number of individual constituents (atoms or groups of atoms)
(c) To indicate the composition of (formal) addition compounds or non-stoichiometric compounds. The numeral is written on the line before the formula of each constituent.

Source: IUPAC recommendations found
here
This essentially means that the first variant is correct.

@andselisk Point taken. Is it better now?

Thank you, that's much better IMO, -1 -> +1 :)


text:
HOW STRONG IS ACETIC ACID?
Will acetic acid cause irritation on our skin?
 Does it have any specific smell? what is it's pH range? Is it too strong to cause irritation in our nasal passage?

Your second question is a proof that you've never encountered acetic acid and hence probably should not worry about your remaining questions.

Ever smell vinegar?

It totally depends on the concentration of acetic acid you use. We have dilute acetic acid in our kitchens (vinegar) so low concentration (Approx 3-9% v/v) is okay. The smell is acidic but is bearable. The pH of vinegar is around 2.4.  On the opposite scale glacial acetic acid (99%), causes severe skin burns and is flammable. It has a very strong acidic smell and will most probably burn your nasal passage or severely irritate it. Hope this helps.


text:
Dilution calculations

$\pu{5 mL}$ of $\pu{0.012 M}\,\ce{Pb(NO3)2}$ is mixed with $\pu{5 mL}$ of $\pu{0.030 M}\,\ce{KI}$. $\ce{PbI2}$ forms. Calculate the diluted concentrations of $\ce{Pb^{2+}}$ and $\ce{I-}$ ions.

I couldn't figure out whether it is an equilibrium question, or something else entirely.
I have currently tried using the moles of the limiting reactant (Pb(NO3)2) to obtain the moles of PbI2 formed and using the molar ratio to get concentration of each ion. However I have no idea whether I have done the correct calculation for ion concentration. The values I am getting are 0.003M Pb2+ and 0.012M I-
update: I have looked at the questions that follow and am able to say for certain this is not an equilibrium question.

I have currently tried using the moles of the limiting reactant (Pb(NO3)2) to obtain the moles of PbI2 formed and using the molar ratio to get concentration of each ion. However I have no idea whether I have done the correct calculation for ion concentration. The values I am getting are 0.003M Pb2+ and 0.012M I-

I would edit your question to include that comment in your original post.

You're dealing with a solubility product. This is absolutely an equilibrium question.

I haven't checked your numbers but you're assuming the $\ce{PbI2}$ to be perfectly insoluble. That's a reasonable assumption but in reality its solubility product is $K_{sp}=4.41\times 10^{-9}$ (Wikipedia), so it does have some solubility.
To take this into account it becomes an equilibrium problem. Assume the initial concentrations (just after mixing)  to be $\ce{C(Pb^2+)}$ and $\ce{C(I^-)}$.
After precipitation the lead concentration has been reduced by$X$ and the iodide concentration by $2X$.
With the solubility product after precipation:
$$K_{sp}=[\ce{Pb^2+}]\times [\ce{I^-}]^2$$
So we have:
$$K_{sp}=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$
Note that this is a third degree polynomial equation in $X$. Solve for $X$ to find the real equilibrium concentrations $[\ce{Pb^2+}]$ and $[\ce{I^-}]$.
We can also use the equation for the case where $K_{sp}=0$. Then:
$$0=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$
First case: $\ce{C(Pb^2+)}-X=0 \implies \ce{C(Pb^2+)}=X \implies [\ce{Pb^2+}]=0$
Also: $[\ce{I^-}]=\ce{C(I^-)}-2\times\ce{C(Pb^2+)}$
In the second case we assume $\Big(\ce{C(I^-)}-2X\Big)^2=0$ and proceed as above.
Which case applies depends on which reagent is the limiting one.


text:
How does the mathematical definition of entropy (ΔS=Qrev/T) give us the degree of disorder?
I think degree of disorder in thermodynamics means the number of possible locations that a molecule can take
How does the mathematical equation signify this?

Entropy isn't "degree of disorder".

@Mithoron In most of my text book it is defined that way.

It's not "defined" like that, it's just a lame description. Also misleading apparently...

While 'degree of disorder' is ok to use (but not 'disorder' on its own ) and  more or less correct it has little real meaning in the sense that we cannot easily imagine some  quantitative a degree of disorder. It is better to think of the number of different ways  that molecules can be arranged into their  various energy levels and/or the number of different ways $\Omega$ they can be arranged in a given volume. (These are often called the number of complexions in older literature). Boltzmann found that $S=k_B\ln(\Omega)$ where $k_B$ is a constant.

At about the same time this poster asked this question he also asked whether entropy was related to unavailable energy. That question was up - voted. This question seems no less worthy of consideration than that one but it has been down voted and put on hold. I'm new here so I'd love to know why.  In answering this question I have to say that I gained insight into what an old boss used to call "the ghost of a quantity that never existed" beyond any I have ever had before so this question was extremely valuable to me. Thanks.

$-\sum\limits_{i}^{N}p_ilog p_i$ is a quantitative measure of disorder

Disorder is perhaps not the best term to use in describing what entropy measures. It is perhaps better described as a measure of what we don't know about a system. If the system is likely to be in one of a handful of states then the entropy is low and we have a pretty good idea as to what state, or which of the few states, the system is likely to be. If, OTOH, it can be in any of a multitude of states with roughly equal probability the entropy is high and we have less likelihood of being able to deduce in which one of those states it might be. Consider a stick of dynamite. If we ask, before detonation, "Where are the carbon atoms in the nitroglycerine?" we can confidently answer "They are all in that little cardboard tube". After the explosion our answer has to be "They are all over the place - maybe as far away as the next county by now." Entropy is related to the dispersion of available states. Distance of a carbon atom from the original location of the dynamite stick is a state variable but clearly not the only one. Nonetheless, it helps me, at least, to understand the concept. Originally the distances of carbon atoms from the center of the stick are in a tight distribution. For purposes of illustration lets assume that this is Gaussian with standard deviation 5 cm. After the explosion it is reasonable to suppose, continuing to assume a Gaussian distribution, that the standard deviation is now 10's of meters. The reason for picking the Gaussian for this ilustration is that the entropy of a Gaussian distribution varies directly as the standard deviation. This falls out of the definition of entropy: $S = -k_BE{log p_i} =-k_B\sum\limits_{i}^{N}p_ilog p_i$ where there are $N$ possible states numbered with subscript $i$. The probability the system is in state $i$ is $p_i$. $k_B$ is the Bolzman constant. $E$ is the expectation operator. If we have, for example, 4 possible states and one of them is much more likely to be occupied than the others then we might have $$ -(.01*log(.01)  + .01*log(.01) + .01*log(.01) + .97*log(.97)  ) = 0.0728314$$ If, on the other hand, each of the 4 was equally likely to be occupied then we'd have $$  - (.25*log(.25)  + .25*log(.25) + .25*log(.25) + .25*log(.25)  ) = 0.60206$$
We can no longer be assured that that one state is the one we would expect the system to be in. We know less about the system. Uncertainty has gone up. Entropy has increased.
I'm editing this to add the important fact that if all the possible states are equally likely then $$-\sum\limits_{i}^{N}p_ilog p_i = -\sum\limits_{i}^{N}(1/N)log (1/N) = -log(1/N) = log(N)$$
Thia gives the definition often found in the textbooks $S= k_BlogN$ usually written as $$S = k_Blog\Omega$$
in which thhey use $\Omega$ instead of $N$ to represent the number of available states. The important thing to realize when confronted with this formula is that the states have to be equally probable. In the Shannon based definition they don't.


text:
Entropy is a measure of unavailable energy?
I came across a statement in my text book which said that entropy is a measure of unavailable energy.
What does this statement signify?

When the first law was first formulated it was assumed that the most efficient utilisation of a chemical reaction to produce work was to use all the heat  to produce work. The second law was not fully understood and thus at maximum efficiency $-\Delta H$ was expected to represent the maximum amount of work. Many experiments failed to confirm this. We now know that it is the free energy which measures the maximum capacity to do work, $\Delta G=\Delta H-T\Delta S$. Clearly $\Delta G$ and $\Delta H$ are equal only if there is no entropy change between the beginning and end of an isothermal reaction. If this is not the case the work obtainable in a reversible process can be either greater than or less than the heat of reaction.
From the first law the external work performed must be equal to the loss in energy of the system, unless some heat is taken from or given to the surroundings. This is exactly the point first clearly seen by Gibbs. In a reversible isothermal reaction $T\Delta S$ is the heat absorbed from the surroundings and if this is positive the work done will be greater than the heat of reaction.
So 'available/unavailable energy' seems to depend on the particular case. Hopefully some thermo experts will give you a more detailed argument.
edit: It occurs to me that perhaps the textbook was referring to the fact that one can 'convert all the work into heat but not all the heat into work'. In other words not all the random thermal motions of molecules can be made to do work.

Another way of seeing this issue is to write the energy conservation law two ways:
$$dU = \delta Q - \delta W= TdS - \sum_{k}Y_kdX_k $$
and this is combined with Clausius's inequality
$$\delta Q \le TdS \\ \delta W \le \sum_{k}Y_kdX_k$$
Here $T$ is the temperature at which $\delta Q$ heat transferred to and $\delta W$ is the work done by the system whose internal energy and entropy have thereby changed by $dU$ and $dS$. The $Y_k$ and $X_k$ are the internal intensive and extensive system parameters.
Evidently less work is done by the system for the same amount of change $dU$, $dS$ and $dX_k$ when the process is irreversible (strict inequality) than when it is reversible (equality).
Because of the Clausius inequality $0 \le \delta \mathcal{N} = TdS- \delta Q$ and this, of course, is the same as $\sum_{k}Y_kdX_k - \delta \mathcal{N} =  \delta W$. The quantity $\delta \mathcal{N} = T\delta \sigma $ with $\delta \sigma \ge 0$ represent the amount of internally generated "heat" and entropy, resp., by the irreversible process. Because of the internally produced entropy $\delta \sigma $ the total entropy change $dS$ will be greater by this amount than in a reversible process connecting the same states, and the resulting $\delta \mathcal {N}$ is then the amount of deliverable work that is "lost" by the irreversibility.


text:
Why Mn atom show maximum valency of 4 with F atom?
Mn ($3d^5$, $4s^2$) can have maximum 7 unpaired electrons in excited state so it should have formed MnF7 molecule but it can form only MnF4 molecule.
Is it because:
$d_{xy}, d_{yz}, d_{zx}$ orbitals are not axially oriented and therefore not suitable for head on sigma overlap but can form pi bonds as in Mn2O7 molecule.

Actually even MnF4 is unstable and easily decomposes to MnF3 and F2.

For some reason $\ce{MnO4-}$ with $\ce{Mn}$ at valence $+7$ is stable but there is no $\ce{MnF7}$ or $\ce{Mn^{7+}}$.

Answer could be similar as in https://chemistry.stackexchange.com/questions/74528/formation-of-mno2-vs-mnh2o64-in-aqueous-solution MnF4 is already about as strong oxidant as fluorine, MnF5 would be even stronger.

@Mithoron This is surprising that MnO2 even oxidize HF to F2. But my doubt is that MnF5 is even possible? considering _ "dxy, dyz and dzx  orbitals are not axially oriented and therefore not suitable for head on sigma overlap but can form pi bonds as in Mn2O7 molecule."_

There's no problem with any orbitals, there can be even $\ce{MnF7^3-}$ anion.


text:
What does pO2 of blood mean and why do we use it?
I understand the basic Dalton's law of partial pressures in gases. Also, Henry's law of diffusion, says, the concentration of gas dissolved in a fluid is proportional to the partial pressure above it.
So if we say that the $p(\ce{O2})$ of oxygenated blood is $\pu{100 mmHg}$, where is the free gas existing in equilibrium with dissolved gas? Does it mean that the blood has a concentration of oxygen equal to that when placed in a surrounding of $p(\ce{O2}) = \pu{100 mmHg}$? If yes, why don't we directly report in concentrations instead? Is it easier to measure?
Wikipedia also says that the Henry's law doesn't stand if the gas is reacting. But isn't oxygen reacting with the Haemoglobin?

https://en.wikipedia.org/w/index.php?title=Oxygen#partial_pressure

No, it's in equilibrium. The reaction is reversible, and so it's still proportional though the constant of proportionality is not the same as in just pure water.

@Zhe equilibrium with what? Alveolar oxygen?

Your interpretation is correct with regard to 100 mm being the Henry's law value.  It is reported in pO2 because that is historically the way it has been done.   If the reaction of oxygen in blood is slow, the Henry's law value is a good approximation.  In any event, even if the reaction is not slow, the Henry's law value is a surrogate for the concentration in molar units.

@Chester Miller I don't think that's the case cause 98%of oxygen exists in combined form with Hb. Then how are we still using pO2?

There is a good explanation in Relating oxygen partial pressure,
saturation and content: the
haemoglobin–oxygen dissociation
curve Breathe 2015; 11: 194–201

The partial pressure of oxygen (also known
  as the oxygen tension) is a concept which often
  causes confusion. In a mixture of gases, the
  total pressure is the sum of the contributions
  of each constituent, with the partial pressure of
  each individual gas representing the pressure
  which that gas would exert if it alone occupied
  the volume. In a liquid (such as blood), the partial
  pressure of a gas is equivalent to the partial
  pressure which would prevail in a gas phase in
  equilibrium with the liquid at the same temperature.
  With a mixture
  of gases in either the gas or
  liquid phase, the rate of diffusion of an individual
  gas is determined by the relevant gradient of its
  partial pressure, rather than by its concentration.
  While in a gas mixture, the partial pressure
  and concentration of each gas are directly proportional,
  with oxygen in blood the relationship
  is more complex because of its chemical combination
  with haemoglobin. This allows blood
  to carry an enormously greater concentration
  (content) of oxygen than, for example, water
  (or blood plasma). Measurement of $p_\ce{O_2}$, therefore,
  does not give direct information about the
  amount of oxygen carried by blood.

So blood $p_\ce{O_2}$ does not correspond to a particular concentration of oxygen, because the concentration of haemoglobin can vary, and most of the oxygen is bound to the heme iron.
$P_\ce{O_2}$ is the partial pressure of oxygen in a hypothetical gas phase which would make the blood oxygen and gas phase oxygen be in equilibrium.

I believe medical oximeters read % saturation relative to the partial pressure of oxygen in the atmosphere. Thus a reading of 94 - 95% (normal) implies that the amount of oxygen in the blood is 94 - 95% of that which would be found were the blood in equilibrium with air. I think the reason it's done that way is for convenience. If the oximeter reads 50% the gas passer knows right away to increase the partial pressure of oxygen he is feeding the patient by 45% to get saturation back to around 95%.

SpO2 is a different thing. It is the percentage of hemoglobin saturated with oxygen. A completely different parameter.

Yes, I believe that's right.


text:
Cell notation for the lead-acid battery
Reactions for the lead acid battery are:
$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + HSO4^-(l) &-> PbSO4(s) + H+(l) + 2e-}\\
\text{Reduction}&\ce{PbO2 + HSO4^-(l) + 3H+(l) + 2e- &-> PbSO4(s) + 2H2O}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2HSO4^- +2H+ &-> PbSO4 + 2H2O}\\
\end{array}
$$
What will be the cell notation for this battery? My attempt:$$\ce{Pb(s), Pb^2+(s)| HSO4^-(l)| PbO2(s), Pb^2+(s),Pb(s)}$$
The things that I have in mind,

In both sides the electrode material is $\ce{Pb(s)}$
Separate every element that is in the same phase with a comma
Separate every element with different phases with a single bar
Separate the two half reactions with a single bar (as I find no salt bridge here)

I am having a tough time learning cell notation. Anyways this was a challenging one for me and I tried my best shot. I am really not sure whether everything is right or not. So please identify my mistakes and show me the right way to do it.

There are a couple of things wrong here. First off, your final reaction is unbalanced. Once you've fixed the balancing, read the other mistakes:

The ions do not exist in the liquid state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the aqueous (aq) phase instead.
While there is certainly no salt bridge here, there is still an electrolyte - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.
The $\ce{PbSO4}$ formed at the anode is in solid state. Hence, writing it as $\ce{Pb^2+(s)}$ is incorrect, as it is not dissociated into the ions $\ce{Pb^2+}$ and $\ce{SO4^2-}$.

With all these corrections, your final, correct cell representation should be:

 $$\small{\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$

Some websites (like KhanAcademy) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:
$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + SO4^2-(aq) &-> PbSO4(s) + 2e-}\\
\text{Reduction}&\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &-> PbSO4(s) + 2H2O(l)}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &-> 2PbSO4(s) + 2H2O(l)}\\
\end{array}
$$
assuming the bisulphite ion to be further ionized. The cell notation in this case would then be:

 $$\small{\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$


text:
Are artificial ingredients better than natural ones because they are made in laboratory conditions?
If we have something like an artificial vanilla flavouring, is it any worse than a natural one?
Will both the natural and artificial ones have the same chemical formula?
Won't the artificial one be purer since its made in lab conditions?

https://en.wikipedia.org/wiki/Vanillin

The question is by no means restricted to vanilla. It refers to any natural product that can be reproduced artificially. Ditto for the answer.
Sure, the artificial product might have the right chemical formula. Also, it can be made as pure as you want. The problem is that the natural product is not pure. It is a bunch of various compounds, and they all contribute to our perception of the flavour. You identify the compound that makes up 90%, investigate its structure, develop the synthetic route, and finally you hold it in your hands, and it feels just about right, but something is amiss. So you look for the next component, which constitutes 9% and hence is a lot harder to isolate, and after a long and tedious procedure you identify it, and repeat everything, and finally you end up with the product that feels totally right. Well, almost so. Is it the remaining 1% that is responsible for the difference? You never know until you try.
So it goes.


text:
UV-Vis and Polylactid Acid (PLA)
I am trying to use UV-VIS for a 3D printed ~4mm PLA design. However I keep running into a problem because the polymer is not showing detectable under the UV-VIS. Does anyone have any suggestions that would allow for visibility of PLA? Any recommended papers would be of great use as well.
I also thought of possibly placing spider silk inside but this would implement an array of complexity in the product.
Thanks guys

Welcome to Chemistry.SE! Find out how to beautify your posts in the help center and take the [tour] to learn more about this site. Enjoy!

Based on your question, I think you're trying to estimate the UV/Vis absorption of the PLA sample? What wavelength range are you examining?

The wavelength was between 200 and 800. We could not see any data form between 400 and 800 nm


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

Buffer capacity depends on the $K_\mathrm{a}$ of an acid. I will use the explanation that can be found in David Sheehan: Physical biochemistry: principles and applications. John Wiley & Sons, 2013.
Any aqueous solution containing both $\ce{A-}$ and $\ce{AH}$ is, in principle, capable of resisting change in $\mathrm{pH}$ according to
$$\ce {A^- + H^+ <=>> AH}.$$
If alkali is generated in the solution (that would tend to remove protons) we have
$$\ce{AH + OH- <=> A- + H_2O}.$$
Most buffers consist of mixtures either of a weak acid and its salt or of a weak base and its salt. The ability to resist change in $\mathrm{pH}$ is finite, especially if the number of protons involve is especially large. This limit is represented by the buffering capacity of the buffer $\beta$.
This is defined as the amount of substance in moles of $\ce{[H^+]}$ which must be added to a liter of the buffer to decrease the $\mathrm{pH}$ by one unit. It can be mathematically calculated by
$$\beta = \frac{2.3 K_\mathrm{a}[\ce{H+}][C]}{(K_\mathrm{a}+[\ce{H+}])^2},$$
where $[C]$ is the sum of conentrations of $\ce{A-}$ amd $\ce{AH}$. This relationship means that buffering capacity increases with the buffer concentration. Buffers work best at $\mathrm{pH}$ values around their $\mathrm{p}K_\mathrm{a}$  most pH are effective one $\mathrm{pH}$ unit above an one below their $\mathrm{p}K_\mathrm{a}$.
I hope that the formula previously reported help in clarifying how the buffer capacity is related to $\mathrm{p}K_\mathrm{a}$ (since they are indeed related).
Example (based on ZUMDAHL, Steven. World of chemistry. Cengage Learning, 2012.)
A chemist needs a solution buffered at $\mathrm{pH}~4.30$. choosing from ethanoic,  propanoic, butanoic acid. We can calculate the ratio  $\ce{[HA]/[A^-]}$ required for each system. Considering that a $\mathrm{pH}~4.30$ corresponds to $\ce 5\times 10^{-5} M$ using the equation
$$[\ce{H+}]=K_a \frac{[\ce{HA}]}{[\ce{A^-}]}$$
we substitute the required $[\ce{H+}]$ and $\ce{K_a}$ for each acid to calculate the ration $[\ce{H+}]/[\ce{A-}]$ needed in each case is:
\begin{array}{llr}
\text{Acid} &
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &
5.0 \times 10^{-5} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &
2.88 \\[1ex]
\text{Propanoic} &
5.0 \times 10^{-5} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
3.749\\[1ex]
\text{Butanoic} &
5.0 \times 10^{-5} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
3.98  \\ \hline
\end{array}
In this case the choice would be ethanoic acid since it has the ratio closes to 1.
If we need a buffer near $\mathrm{pH}~5.5$ the best choice is butanoic acid:
\begin{array}{llr}
\text{Acid} &
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &
3.2 \times 10^{-6} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &
0.18 \\[1ex]
\text{Propanoic} &
3.2 \times 10^{-6} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
0.23 \\[1ex]
\text{Butanoic} &
3.2 \times 10^{-6} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
0.25 \\ \hline
\end{array}

Thank you for your effort in continuously improving the quality of the answer.

I tried using this formula for an ethanoic, sodium ethanoate buffer, made by reacting 200 mL of 1M acetic acid and 100 mL of 1M NaOH solution. Am I correct in making the following assumption: $[H^+] = K_a \implies \beta = \frac{2.3 K_\mathrm{a}[\ce{H+}][C]}{(K_\mathrm{a}+[\ce{H+}])^2} = 2.3[C]/4$? If so, wouldn't this mean that this buffer, and a buffer prepared identically but with propanoic acid, would have the same buffering capacity?

@Jojostack My question isn't about which is the best buffer for a specific pH. I am asking about how to calculate the buffering capacity of a buffer with equal concentrations of acid and conjugate base.

I hope I've understood well. So if you use same amount of acid and conjugated salt you would obtain three solution with different pH. Each of them would be a buffer that will work better in proximity of the pKa of the acid of each buffer solution

That formula, which is demonstrably not correct when pH = pK because the constant is wrong but otherwise is OK (at pH = pK, I haven't checked other values) just shows what should be obvious: for buffers  which are strong enough that the charge is dominated by the acid (i.e. not water) and in which if there are multiple protons these have pKs separated by a few pH units buffering capacity does not depend on pK. So no surprise that you found all the acids when used in equal strength produced buffers of the same capacity.

Were this  posed "Make a 5 buffer from these 3 acids. Which has the greatest buffering capacity?" Then the answer is "The one with the pK closest to 5 (butanoic)". The capacity of a buffer is always greatest at one of its acid's pK's and that's why we always try to get as close to a pK as possible. There is nothing inherent in the nature of acetic acid that makes it a better buffer than butanoic and that's why we answer that the capacity is pK independent. If you specify a design pH then capacity does depend on acid choice (and hence pK). But you didn't.

Taking "buffering capacity" to mean the amount of acid or base required to change $\mathrm{pH}$ by a fixed small amount at $\mathrm{pH}$ values near one of the $\mathrm pK$'s of the acid on which the buffer is based buffering capacity is not $\mathrm pK$ dependent. To see this write Henderson–Hasselbalch for each proton in the form $\mathrm pr_1=\mathrm pK_1-\mathrm{pH}$, $\mathrm pr_2=\mathrm pK_2-\mathrm{pH}\ldots$ where $\mathrm p(\cdot)=-\log(\cdot)$. Taking the antilog we see $r_j = 10^{(\mathrm{pH}-\mathrm pK_j)}$ Note that $r$ is dependent on the difference between $\mathrm{pH}$ and $\mathrm pK$. Thus we find that $r$ is the same one $\mathrm{pH}$ unit either side of $\mathrm pK$ irrespective of whether $\mathrm pK_\mathrm a$ be 1 or 7.3. This is your first clue that buffering is $\mathrm pK$-independent.
Assuming you have started with $C$ moles of acid the fraction of those that remain undissociated and uncharged at $\mathrm{pH}$ is $f_0(\mathrm{pH})=1/(1+r_1+r_1r_2+r_1r_2r_3\ldots)$, the fraction that has lost one proton and thus carries a single negative charge is $f_1=f_0r_1$, the number that is doubly charged is $f_2=f_1r_2$ and so on so that the total charge on acid anions is $$Q(\mathrm{pH})=C(-f_1-2f_2-3f_3...)$$ (moles of charges or Eq if you prefer). This is clearly a function of $\mathrm{pH}$ (shown) as well as each of the $\mathrm pK$'s (not shown). The number of protons required to shift the buffer's $\mathrm{pH}$ from $\mathrm{pH}$ to $\mathrm{pH}+\delta$ is $\Delta Q=Q(\mathrm{pH}+\delta)-Q(\mathrm{pH})$. This is the buffering (alkalinity, acidity) between $pH$ and $pH+\delta$. The buffering capacity is the slope of the buffering curve. At $pH$ it is:
$$\partial Q(\mathrm{pH})/\partial\mathrm{pH}=\lim_{\delta\to0} (Q(\mathrm{pH}+\delta))-Q(\mathrm{pH})/\delta\approx(Q(\mathrm{pH}+\delta)-Q(\mathrm{pH}))/\delta$$
in units of $\mathrm{mol}\cdot\mathrm{pH}^{-1}$. It is the number of moles of protons (1 M strong acid delivers 1 mole of protons per liter) which must be added or absorbed per unit $\mathrm{pH}$ shift.
Now these formulas aren't very complex and if you will take the trouble to put them into a spreadsheet or visualization program and make some plots you will see that the buffering capacity of a buffer near a $\mathrm pK$ depends only on $C$ as long as the $\mathrm pK$'s are separated.
I have been, for simplicity, ignoring the buffering capacity of water here. When $3\lt\mathrm{pH}\lt10$ water does exhibit appreciable buffering and needs to be considered. The net charge on a liter of water ions is $Q_\mathrm w=10^{-\mathrm{pH}}-10^{\mathrm{pH}-\mathrm pK_\mathrm w}$ and the buffering of water, between $\mathrm{pH}$ and $\mathrm{pH}+\delta$, is $$\Delta Q_\mathrm w=10^{-(\mathrm{pH}+\delta)}-10^{-\mathrm{pH}}+(10^{\mathrm{pH}-\mathrm pK_\mathrm w}-10^{\mathrm{pH}+\delta-\mathrm pK_\mathrm w})$$ $Q_\mathrm w$ is added to $Q$ as given above when differentiating to find buffer capacity. Keep in mind that $Q_\mathrm w$ is per liter and $Q$ per mole.

So if pKa is unrelated to the differing buffering capacities of these acids (assuming that they are all made with the same concentration of acid, and same of concentration of NaOH, at the same temperature), then could you point me to some resources that account for why the difference arises?

Based on what we have discussed here there is no difference so I can hardly point you to literature explaining why there is. Perhaps it is in the way you are defining buffering. If you had 0.1 mole of each of these acids and added 0.05 mole of strong base you would wind up at pH 4.76 for acetic acid, 4.88 for propionic and 4.82 for butyric. If you wanted to get acetic acid to pH 4.88 it would take more than 0.05 mole. You might conclude from this that acetic acid has more buffering than butyric acid and it does - to pH 4.88. But the slope of the buffering near the pKs is the same for all 3.

This should really be the accepted answer. Don't see why it is downvoted.

You'll note that I have added water to the answer. In cases where the pK of the acid is small ( < 3 ) or large (> 10) then you are not titrating just your acid but a mix of acids with the other being water. In such cases the buffering capacity will depend on the pK. This would not the case with any of the three acids mentioned in the question unless they were extremely dilute..

I don't understand that either. If someone sees a problem please let me know what it is so I can fix it.

Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms". Since buffer capacity is definitely $\mathrm{pH}$ dependent, and the $\mathrm{pH}$ of the solution depends on the $\mathrm{p}K_\mathrm{a}$ of the weak acid, I fear this is incomplete.

Proper according to whom? I have no problem with referring to mass as the number of killograms. Or the size truckload as the number of potatoes. This is pedantry.  It's very clear from the formulas that I gave that buffering is pH dependent. How could you miss that?  But note that buffering capacity near a pK depends only on pH - pK. It should also be quite clear that in finding the pH of a solution of mixed weak acids using the technique my post implies requires including the pKs but the question wasn't about finding the pH, it was about the buffering capacity. Suggest a re-read.

Apparently an offensive statememt is :" It is the number of moles of protons(1 M strong acid delivers 1 mole of protons per liter) which must be added or absorbed per unit $pH$ shift." Does anyone here think its meaning would be enhanced if it read: "It is the the amount of substance  (1 M strong acid delivers 1 mole of protons per liter) which must be added or absorbed per unit $pH$ shift." ??

I am not pedantic, I find that very offensive. I even have linked to the recommendations of the IUPAC; I don't expect you to follow these, but I wanted you to be aware of it. This is also not the reason why I down-voted. Every time I read your answer, I'm more and more sure that in the way it is written it is at least incomplete. The caveat and your principle assumption is that you are looking only when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$; and you fail to mention that explicitly.

The reason there is no explicit statement of the sort you expect is because you have conjured up a "caveat" and a "principle assumption" which do not exist. The shape of a titration curve depends only on |pH - pK|, not the absolute values of either. This is, of course, untrue at extreme pH or pK neither of which pertain in this question. You have decided that my statement that the questioner is probably interested in buffering near pK (why would he be interested far from it?) implies that the formulae I use to show that it is pK independent is so restricted. Not so. Valid at any pH.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Does Acetate Cause Major E2 On Tertiary Haloalkanes?
This question really confuses me. I've narrowed it down to E2 being favored in step one (giving 2-methylpropene), but I don't see how a relatively weak base like acetate could cause E2 to be the major product. The only other explanation I could think of is that E2 is a substantial minority product, so the reaction isn't synthetically useful. Am I on the right track? Thanks!

SN1 then ester hydrolysis.


text:
Why is tautomers not possible in benzoquinone?
1,4-Benzoquinone, commonly known as para-quinone, is a chemical compound with the formula $\ce{C6H4O2}$
When there are alpha Hydrogen's why can't it show tautomerism?

Look at the supposed tautomer.

You could sort of devise a tautomer by transferring an $\alpha$ hydrogen from carbon to oxygen.  But if you draw the resulting structure you find a pair of cumulated double bonds.  That tends to be unstable, and since the cumulated double-bonded carbons favor a 180° bond angle putting them into a relatively small ring makes them a lot more unstable.  Your proposed tautomer is way out there in energy.
Tautomerization of a ketone works better when you have an $\alpha$ hydrogen on a saturated carbon atom.


text:
Is there a way to quantify the 'surface area' or simply the shape of organic compounds?
I wish to write an investigation into the effects of the shape of organic compounds and the location of the functional groups on the enthalpy of combustion of the organic compounds. Thus, I wonder if there is a way of quantifying these variables, as databases I have looked at do not provide much of an answer.

It depends upon what the surface stops. There are different ways to map the surface as for a representation of a single atom. In your case it might be the shape /surface as sensed by a dioxygen.

It is possible to compute the surface area of a molecules, and to compute the surface area of each individual atom in a molecule. These areas depend on the conformation of the molecule. An example of a software that does this is MSMS (http://mgltools.scripps.edu/packages/MSMS).
With respect to the shape, it is possible to categorize molecules in a triangular diagram describing similarity with "rods", "disks" and "spheres" (e.g. http://pubs.acs.org/doi/10.1021/ci300415d). Details about this analysis were published here: http://pubs.acs.org/doi/10.1021/ci025599w.


text:
Bond order of H2O and NH3 using group theory to construct MOs
The bond order of H2O is 2, and that of NH3 is 3, which makes sense, when considering the number of bonds they have.
However, if using basic group theory, you get the MO diagrams as presented in this paper:
http://www.scielo.br/pdf/qn/v35n7/v35n7a32.pdf
Upon examination, we see that in the case of H2O, there are two orbitals that are strongly bonding, one weakly bonding orbital, and one non-bonding orbital. If we then tried to apply the bonding order equation to this MO diagram, would we not end up with a bond order of 3 - despite the a1 orbital being only weakly bonding, its occupation still stabilizes the molecule relative to its constituent atoms. Is it the case that the concept of bonding order breaks down using these kinds of MO diagrams?
A further example is given using the NH3 MO diagram. Here, they calculate the bond order as 3, ignoring the fact that the NH3 a1 orbital is weakly bonding. If we were to in theory calculate the MO diagram for NH3 in a trigonal planar geometry (same as for BH3), we would also get the bond order as 3. Of course, the occupied a1 orbital is weakly bonding in the trigonal pyrimidal NH3 geometry, leading to a preference over the trigonal planar geometry, but why is this not reflected in the bond order?

It depends on your definition of bond order. If you are using the simple formula
$$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$
then, yes, it won't be able to capture the subtleties that you describe. After all, when you use that, you're categorising electrons into bonding vs antibonding in a black and white manner. So it's really no surprise that it doesn't reflect "weakly bonding" MOs.
You gave the example of water. Well, for that weakly bonding MO, you need to make a choice: is that considered a bonding MO, leading to a bond order of 1.5 per O–H bond in water (doesn't make much sense)? Or is it considered a nonbonding MO, leading to a bond order of 1 (makes more sense but then you have this problem you brought up)? Or is it somewhere in between, leading to a bond order between 1 and 1.5 – and crucially, if this is the route you want to go down, how do you quantify the bonding character of the MO?
I don't have any experience using different definitions of bond order, but if you used one of these, I would not be surprised to see an O–H bond order that deviates ever so slightly from 1.


text:
What is the iupac name for ch3ch2cooag?
Like ch3coo is acetate group and is named accordingly,how to name this one.

Like acetic then is propionic. IUPAC name as in answer @Oscar Lanzi. By the way simply typing ch3ch2cooh in Google you get it as first or second result depending on language setting. https://pubchem.ncbi.nlm.nih.gov/compound/propionic_acid

Start with the acid $\ce{CH_3CH_2COOH}$.  This is clearly the one and only simple carboxylic acid derived from an unsubstituted propane chain, so it's
"Propan-" from the propane chain
plus "-oic acid" from the carboxyl group (we don't need a locant because there is only one place to fit that group)
= propanoic acid
Then the silver salt is silver propanoate, the "-ic acid" part becomes "-ate" just like inorganic salts.


text:
Orientation in benzene rings with more than one substituent
Why is it that the chlorination of 1-chloro-3-nitrobenzene (18) gives 1,4-dichloro-2-nitrobenzene (19) as the major product, 1,2-dichloro-3-nitrobenzene (20) as a minor product and 1,2-dichloro-4-nitrobenzene (21) is not formed at all?

I know $\ce{-NO2}$ is a deactivating, meta-directing group and hence it decreases the electron density at positions 2, 4, 6 while position 5 is unaffected. On the other hand, $\ce{-Cl}$ is a deactivating, o-p directing group and hence it directs the incoming group to positions 2,4 or 6. Taking into account the fact that $\ce{-NO2}$ is a more powerful deactivating group than $\ce{-Cl}$, the incoming $\ce{Cl+}$ should be attached at position 5 as it has more electron-density than 6,2 or 4 position. But the product with $\ce{Cl}$ at position 5 is not formed. Why so? I am unable to rationalise this.
Source: page 583 of "March's Advanced Organic Chemistry, 7th ed." under the heading "Orientation in benzene rings with more than one substituent".

I quote March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition, page 666.

Some substituents have a pair of electrons (usually unshared) that may be contributed
  toward the ring. The three arenium ions would then look like this:
The stability of these two ions is increased by the extra form not only because it is another canonical form, but because it is more stable than the others and makes a greater contribution to the hybrid. Every atom (except of course hydrogen) in these forms (C and D) has a complete octet, while all the other forms have one carbon atom with a sextet. No corresponding form can be drawn for the meta isomer.The inclusion of this form in the hybrid lowers the energy , but also because it spreads the positive charge over a larger area—out onto the group Z. Groups with a pair of electrons (e.g., as the halogens) to contribute would be expected, then, in the absence of field effects, not only to direct ortho and para, but also to activate these positions for electrophilic attack. (my emphasis)

In your example, if chlorine electrophile attacks at C-4 or C-6, the resulting intermediate  is stabilized by an extra resonating structure, lowering the energy of this intermediate.Hence electrophillic aromatic substitution takes place.

If chlorine electrophile attacks at C-5 , the resulting intermediate is not stabilized compared to attack at C-6 .Hence the resulting electrophillic aromatic substitution is not seen.

This is the possible reason  that  the product with Cl at position 5 is not formed.

Quoting from chemistry.msu.edu:

The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:
The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring.
The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance).

The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs.

The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites.

Therefore chlorine is an ortho para directing group with minor meta product.
References:

Advanced Organic Chemistry 5th edition Part A: Structure and Mechanisms, Francis A. Carey and Richard J. Sundberg
March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition

According to your answer, the attack at "your"(as I noticed that you numbered the carbons in a different way than I did) position 6 is favorable but my question itself says that pdt is not formed at all. And you used Cl to stabilize the +ve charge through $+M$ effect but I suppose $-I$ of halogens is greater than their $+M$, so Cl should destabilize it.

@Carrick first numbering was done as per IUPAC numbering system.Secondly you are correct in saying  -I effect of Cl is evident.How ever  +M effect of Cl cannot be discounted.Going by your arguments chlorbenzene  should not give ortho and para products as major products.Note one more point, -I effect is through a sigma bond while +M effect is through a Pi bond .

btw I took that question from jerry march, and it said $21$ is not formed at all which I think is indeed true because if you delocalize the +ve charge further in the benzene ring, you shall notice that the +ve charge lands on the carbon right beneath the $NO_2$ group. So, don't you think that it should really destabilize that??

@Carrick i have added NOTES regarding chlorine's ortho para directing nature.But you also get an extra resonance structure in which +ve charge is stabilized by lone pair on chlorine. 3 to be exat.

@Carrick i have since added electrophillic attack at C4.I have given plausible mechanisms .Among which attack at C5 does not give carbocation stabilized via +M  of chlorine and there for this product is not formed.This answers your question.How ever 21 may not be formed via ortho effect  as mentioned in JERRY MARCH.

why haven't you taken the deactivating effect of $NO_2$ gp into account in any of your proposed explanations.


text:
Why is Cr+ unstable?
We know that half-filled and fully filled orbitals are highly stable. In ground state $\ce{Cr}$ has a $\ce{3d^5\! 4s^1}$ configuration. Therefore, the elctronic configuration of $\ce{Cr+}$ should have $\ce{3d^5\! 4s^0}$ state. The half-filled $\ce{3d}$ orbital should make the ion stable.
However, we don't see many $\ce{Cr(I)}$ compounds. There are one or two rare examples, but other than them, most compounds of $\ce{Cr}$ are $\ce{Cr(II)}$, $\ce{Cr(III)}$ and so on. Why does this happen? Why $\ce{Cr+}$ is unstable?
[If you find a reference, add it to your answer. As I am a student, it will help me immensely. But if you don't have a reference, don't hesitate to post an answer. Getting an answer is the priority.]

Cr(II) and Mn(III) - their oxidizing and reducing properties? conceivably a duplicate.

I'm afraid it might be one of these cases where there is no intuitive explanation and we're left with "because when we go through all the quantum mechanic computation we get such and such ionization and hydration energies".

@JanRzymkowski , Yes, that might be true. But in most cases, there is often a simple and intuitive explanation. I put the question on the assumption that there is an explanation. I still think there can be a good explanation without the need to go into quantum mechanics.


text:
Why does pyrolysis activate carbon?
I am doing a project that involves activating carbon. The method my team is using is simply to heat it within the range of 400-500°C, and then testing for activity by using a dye.
Why does simply heating it (without any additional reactants like zinc chloride and so on) activate it?

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Comparison of stability of cyclic imines

I am looking for the reason for the lowest stability of cyclohexane-1,3,5-triimine (D).
I was not able to find any explanation using resonance or aromaticity.
All help is appreciated.

The cyclohexane-1,3,5-triimine tautomer is benzene-1,3,5-triamine, which is aromatic. This makes this trienamine form much more stable than any of the other possible tautomeric forms.

Not really, this question is pretty much nonsense, as all of them are "unstable", whatever this is supposed to even mean...

I agree this is clearly a textbook unrealistic question. But my best interpretation was that the condition are such the imines don't solvolyse or anything as such and that the question is about tautomerism (which, I agree, assumes presence of some Lewis bases).

Your answer is not very clear. Why does that make trienamine the least stable?

It makes triimine the least stable, because it readily tautomerizes to trienamine, which is aromatic therefore stable. (I'm guessing that in the context of the original question, stable means not transforming into an enamine)


text:
Can a molecule be neither gerade nor ungerade?
Or can any molecular orbit always be written a linear combination of gerade and ungerade basis states?

Of course, if you have a linear combination where both gerade and ungerade components are nonzero, the sum is neither.

g/u is not a property of a molecule; I assume you meant molecular orbital.
All functions can be written as a linear combination of even + odd functions: there is a short explanation on Wikipedia. Basically, you have a function $f(x)$; now define
$$\begin{align}
f_\mathrm e(x) &= \frac{1}{2}[f(x) + f(-x)] \\
f_\mathrm o(x) &= \frac{1}{2}[f(x) - f(-x)] \\
\end{align}$$
It is clear from the definition of parity that $f_\mathrm e$ is even (since $f_\mathrm e(x) = f_\mathrm e(-x)$) and likewise that $f_\mathrm o$ is odd. Now since $f(x) = f_\mathrm e(x) + f_\mathrm o(x)$, we have shown that any function $f(x)$ can be expressed as a linear combination of even and odd functions.
In three dimensions where you have $\psi(x,y,z)$ you simply need to define
$$\begin{align}
\psi_\mathrm g(x) &= \frac{1}{2}[\psi(x,y,z) + \psi(-x,-y,-z)] \\
\psi_\mathrm u(x) &= \frac{1}{2}[\psi(x,y,z) - \psi(-x,-y,-z)] \\
\end{align}$$
and analogously to above one can see that every function $\psi(x,y,z)$ can be expressed as the sum of a gerade component $\psi_\mathrm g$ and an ungerade component $\psi_\mathrm u$. These components don't have any physical meaning (unless $\psi$ itself is either g or u, in which case one component is simply $\psi$ and the other is zero), but that's somewhat beside the point.
If a molecule does not possess a centre of inversion then its (canonical) molecular orbitals will not possess g/u symmetry.
So the answer is yes, and yes. There is no "or".


text:
Can H– act as a strong Brønsted base like CH3–?
In the reduction of carboxylic acids with $\ce {LiAlH_4}$ we get an alcohol. But isn't there a possibility of the hydride donor abstracting $\ce {H^{+}}$ from the acid to form $\ce {H_2}$?

You are right. The hydride would first act as a base deprotonating the acid before performing the hydride attack and subsequently reducing it.

Related: https://chemistry.stackexchange.com/questions/4940/reduction-of-carboxylic-acids-by-lialh4?rq=1


text:
Conversion of Z-matrix to Cartesian coordinates
To begin with, I wrote a script that gets Cartesian coordinates of molecule as input in the below. These are $x,y,z$ coordinates of H2O2 molecule.
1 O          -1.7529   -0.5188    1.3324
2 O          -0.4737   -0.1091    0.7774
3 H          -2.2902    0.1678    0.8933
4 H           0.0636   -0.7957    1.2164

Then, the script constructs a Z-matrix with them, like this:
Z-mat :
O
O   1   1.45335189476
H   1   0.976176039452  2   96.5694760083
H   2   0.976131061897  1   96.5720363573   3   -179.995395182

Now I need to perform the reverse operation and use this Z-matrix as input and define $x,y,z$ coordinates for each atom. This is converting a Z-matrix to Cartesian coordinates.
My question is after setting first atom as 0,0,0
1 O          0   0   0

and the second one as 0,0,(distance from first) to put it on the z-axis
2 O          0   0   1.45335189476

How should I treat the 3rd and 4th atoms?
The 3rd atom must have coordinates that are something like this if read correctly:
3 H          0   distance*sin(angle)  z2+distance*cos(angle)

Taking z2 as the z-coordinate of atom 2. I am not sure if I should calculate this as z2 + distance.cos(angle) or z2 - distance.cos(angle) and what it depends on, if both are possible.
For the 4th atom, I use spherical coordinates
r, theta, phi = (0.976, 96.572, -179.995)

to calculate $x,y,z$ values from the formulas
x = r * sin(theta) * cos(phi)
y = r * sin(theta) * sin(phi)
z = r * cos(theta)

If there aren't any mistakes up to this point, how I will calculate Cartesian coordinates of the 4th atom using these $x,y,z$ values?

While searching I found TMPChem's work on GitHub and it does exactly what I want. However, in his work, there is a mathematical part that I don't understand:
# get local axis system from 3 coordinates
def get_local_axes(coords1, coords2, coords3):
    u21 = get_u12(coords1, coords2) #calculating vector between that points 1-2
    u23 = get_u12(coords2, coords3) #calculating vector between that points 2-3
    if (abs(get_udp(u21, u23)) >= 1.0):
        print('\nError: Co-linear atoms in an internal coordinate definition')
        sys.exit()
    u23c21 = get_ucp(u23, u21) # unit cross product
    u21c23c21 = get_ucp(u21, u23c21) # unit cross product
    z = u21
    y = u21c23c21
    x = get_ucp(y, z)
    local_axes = [x, y, z]
    return local_axes

What is "getting local axis system from 3 coordinates"?
Here is some context of how this function is used:
bond_vector = get_bond_vector(atom.rval, atom.aval, atom.tval)
disp_vector = np.array(np.dot(bond_vector, self.atoms[i].local_axes))
for p in range(3):
    atom.coords[p] = self.atoms[atom.rnum].coords[p] + disp_vector[p]

The bond vector definition is here:
def get_bond_vector(r, a, t):
    x = r * math.sin(a) * math.sin(t)
    y = r * math.sin(a) * math.cos(t)
    z = r * math.cos(a)
    bond_vector = [x, y, z]
    return bond_vector

Again, the only part I don't understand is # get local axis system from 3 coordinates. What is that function doing?

Not really an answer, but MOLDEN converts xyz to z-matrix, IIRC, and the source should be available to you. You may have to provide a certain command line parameter to avoid a reordering of the molecule.

https://chemistry.stackexchange.com/q/55702/16683 is this relevant?

@TAR86 Thank you for your answer. However I'm trying to understand mathematical concept behind it and implement it to this script.

@orthocresol I checked that question. OP asked somewhat related question trying to do a similar thing. But his question is not clear, at least I dont understand what exactly is his problem. This might be helpfull maybe. I'll try to use that source.

I will try to answer the question in bold from a mathematical perspective.
Basically, the first two lines in get_local_axes are to build the vectors $\vec r_{12}$ and $\vec r_{23}$. The next line builds the cross product $$\frac{\vec r_{23}\times\vec r_{12}}{|\vec r_{23}\times\vec r_{12}|},$$
The next vector is
$$\frac{\vec r_{12}\times(\vec r_{23}\times\vec r_{12})}{|\vec r_{12}\times(\vec r_{23}\times\vec r_{12})|},$$
This is a vector perpendicular to $\vec r_{12}$, and lies in the plane where $\vec r_{12}$ and $\vec r_{23}$ lie.
The next lines are defining the three axes of the system. $\vec e_z$ is just $\vec r_{12}$ normalized, $\vec e_y$ is chosen so that $yz$ is the plane where the three points (atoms) locate. and $\vec e_x$ is chosen so that $(\vec e_x, \vec e_y, \vec e_z)$ forms a orthonormal set and are the versors of a right-handed coordinate system. This system is the local axis system.
So this is the meaning of "getting local axis system from 3 coordinates".

Thank you for answer. If I understand correct, dot product of this vector with bond vector(the one comes from angles) should give a scalar as displacement vector. How it has 3 component ?  that allows this
"for p in range(3):
atom.coords[p] = self.atoms[atom.rnum].coords[p] + disp_vector[p]"
to work

@OnurOzcan Local axes has three components ($\vec e_x,\vec e_y,\vec e_z$), each of which is a vector, so it is in fact 3-by-3. Dot product of it with the bond vector gives the bond vector's representation (the coordinates) in the local axis system, which is composed of three elements.

Ah I missed it. I think this might be enough, thank you again.


text:
what is the product of this reaction and mechanism?
![picture from national chemistry Olympiad]

I cant get the mechanism.
Answer shows there will be formation of keto group.
But how?
As I can understand the $\ce{H^+}$ is going to attack the $\ce{O}$ and can remove the $\ce{CH3}$ groups forming two adjacent alcohols which may join to form keton.
I'm not sure.

The mechanism to convert acetal to ketone in acid based hydrolysis is in following steps-

Protonation of (1) to make OCH3 a good leaving group forming  (2)
O lone pair allows LG to leave forming oxonium ion and alcohol(3)
Water nucleophile attacks C of C=O (3) to give (4)
Base removes proton to give (5).
Reprotonates (5) to make OCH3  a better leaving group(6).
CH3OH leaves to give (7)
Base deproronates to give ketone (8)

Note :In this scheme, the base, B:, could be R-O-R, R-OH, H2O or the conjugate base of the acid catalyst

References:

http://www.chem.ucalgary.ca/courses/353/exams/3513/353w05/353fin05me.html
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/aldket1.htm

Base is confusing. It is actually water.

that was awesome! in the 5 step it will be h2o Not base.

@RoulKessels  In this scheme, the base, B:, could be R-O-R, R-OH, H2O or the conjugate base of the acid catalyst

@SharifulIslam In this scheme, the base, B:, could be R-O-R, R-OH, H2O or the conjugate base of the acid catalyst

Yes, the oxigen is protonated, water enters, methanol leaves and another $\ce{H^+}$ is released.
Then the oxigen of the other $\ce{CH3O}$ group is protonated, the keto group is formed while the methanol leaves and $\ce{H^+}$ is released.
Edit:
The first part results in:

and $\ce{CH3OH}$.

what do you mean by water enters?

I have edited. I do not have a proper drawing program with me to draw the whole mechanism

in the first step the lower O will be protonated,forming OH+(CH3).Then the CH3- Will be removed and OH+ WILl become alcohol group.From where does CH3-gets Oh-?And why the same will not occur for the upper och3 group?

No. Water enters and methanol is removed. Then the other -OMe is protonated and methanol is removed and the ketone is formed.

I have found an image

yes the image is from http://www.chem.wisc.edu/areas/reich/Handouts/ElecPush/ep-mechanisms.htm

thatz so great!


text:
pH value of distilled (pure) water
What is exactly the pH value of distilled water?
which is correct?

Neutral(not acidic not alkaline)pH 7?
pH 6?

This is a different question. You just edited your last question  and made it into something else. It would have been better if you had created a new question...............

You really should not do so ... The comments (which now have been deleted) seemed so confusing after you changed the entire question.

Distilled water usually has a pH of 7. However, pH might change with temperature. See this answer for more detail.


text:
Ionization energy of beryllium
Suppose I have $\ce{Be^3+}$. What would be its 4th ionization energy?
By trying to solve the issue I saw that its a "hydrogen-like" atom – means that beryllium left with only $1$ electron in his valance shell. Hence, I can use the Bohr atom model to solve the problem.
Is my guess right? If so, what values I need to plug in the equation and why so?

You are right on target.  Go ahead and use the Bohr model.  You should be able to get the energy levels of the electron in the Bohr model from your text and you would use the most stable (ground) state.

In terms of Bohr model ionization potential $E_\mathrm{i}$ is the work $A_\mathrm{i}$ on eliminating an electron in vacuum from its current non-excited orbital level to infinity:
$$E_\mathrm{i} = \frac{A_\mathrm{i}}{e}$$
$$A_\mathrm{i} = h\nu = \frac{hc}{\lambda}$$
Unknown wavelength $\lambda$ can be determined from the Rydberg formula:
$$\frac{1}{\lambda} = R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
so that final equation for ionization energy looks like this:
$$E_\mathrm{i} = \frac{hc}{e}R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
Since we are determining 4th ionization energy of beryllium ($Z = 4$), $n_1 = 1$ and $n_2 = \infty$:
$$E_\mathrm{i}^\mathrm{IV} = \frac{\pu{6.63e-34 m^2 kg s-1}\cdot\pu{3e8 m s-1}}{\pu{1.602e-19 C}}\cdot\pu{10973732 m-1}\cdot 4^2\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = \pu{217.86 eV}$$
This value is in a good agreement with the one listed in CRC Handbook of Chemistry and Physics [1, p. 10-204]: $E_\mathrm{i}^\mathrm{IV}(\ce{Be}) = \pu{217.71865 eV}$.
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; CRC Press, 2016. ISBN 978-1-4987-5429-3.

I was also taught the straight forward formula for energy of the $n$-th orbit in atom $Z$, as $E_n^Z=-13.6\cdot\frac{Z^2}{n^2}$, using the Bohr model. Posting just in case anyone else finds it useful.


text:
Which is more stable out of the given carbocation?

Which is more stable?

$\ce{^+C(CH3)3}$
$\ce{^+C(CD3)3}$

I have read in Solomon's and Fryhle that the +I effect of D is more than that of H but here in this problem it is given that $\ce{(CH3)3C+}$ is more stable.
How is this possible? Since, deuterium enriches the electron density over the central carbon resulting in the diminishing of a positive charge eventually making the second structure more stable.
Second explanation can be on the basis of hyperconjugation which says that $\ce{-CH3}$ is better in showing hyperconjugation than $\ce{-CD3}$, but why is this so?

$\ce{C-D}$ bonds are stronger. Hyperconjugation is thus more easily seen  with hydrogen rather than deuterium. Hyperconjugation> Inductive effects at stabilization.I believe you can reason the rest out by yourself.  To see why $\ce{C-D}$ bonds are stronger.

See this

While true, this answer is extremely unsatisfying. The link from bond strength to hyperconjugation is not obvious to me at all. (And I'm not saying I have any better explanation. This is something which I have never seen a good explanation for.)

@orthocresol Maybe this contains the required answer.


text:
NIOSH 7300 - ICP OES calculation
need some help with NIOSH 7300 method https://www.cdc.gov/niosh/docs/2003-154/pdfs/7300.pdf
I have two 37mm cellulose filter paper, one is my filter blank while other is sample. The sample has 80L of natural gas passed through and I plan to test for Mg , final concentration in mg/kg.
Both are digested and top up to 10 ml with dilution acid.
ICP OES Result:
Filter Blank : Mg = 0.022 mg/L , Pb = 0.002 mg/L
Sample       : Mg = 0.038 mg/L , Pb = 0.006 mg/L

(From Method)
Given the formula is
Solution concentrations for the sample, Cs(µg/mL),
Filter blank , Cb(µg/mL)
Volumes of sample, Vs(mL), and Filter blank , Vb(mL),
calculate the concentration,C (mg/m3), of each element in the air volume sampled, V (L):
C= (CsVs - CbVb)/V = mg/m3
NOTE: µg/L / mg/m3

My calculation,
[(0.038 mg/L x 0.01 L) - (0.022 mg/L X 0.01 L)]/ 80L
= 0.00016 mg/ 80L
Edit:
According to the composition result, density is = 17 gmol-1 ,
and I calculate this under STP.
= 0.00016 mg / [(80/22.4)mol * (0.017kg/mol)]
= 0.00263 mg/Kg
is this correct?

You need density of the gas at the " pumping through filter" conditions.


text:
Stability of carbanion
I am confused about how one can determine the stability of carbanions using pKa-values. I know that the stability of carbanions can be determined by the inductive effect, hybridization of the charged-bearing atom and resonance, but when searching about how pKa can do so I don't find anything. The lower the pKa-value, the stronger the acid and hence, the weaker the conjugate base, does this have anything to do with the stability since weak bases are unreactive and thus more stable?

Except for the mistake SteffX has pointed out, I think your argument is valid but you need to pay attention to the context of "stability".

As commented by SteffX, less the $\mathrm{p}K_\mathrm{a}$, more stronger the acid. More stronger acid means that the conjugate base or the anion formed after donating the proton is more stable compared to the carbanion of an acid of a  higher $\mathrm{p}K_\mathrm{a}$ value. So, if the anion formed is a carbanion, then lower the $\mathrm{p}K_\mathrm{a}$, the stable the carbanion is.

Why the downvote? Have I written anything wrong?

The stability of every molecule is related to the spreading of charges: the more you can share a charge among atoms, the more stable the molecule is.
For a carbanion, every group WITHDRAWING electrons to the carbanion center will stabilize it (i.e. the negative charge will be shared by several atoms)
For a carbocation, every group GIVING electrons to the carbocation will stabilize it (i.e. the localized positive charge will be compensated for... at least partially)


text:
D3d point group matrices
I am trying to construct all the matrices for the point group D3d operations.   I find that the matrices of E, i, S6, inverse of S6, C3, and inverse of C3 are easy,  but I do not know how to construct the matrices of the three C2s and the three dihedral reflections.   When I construct these matrices,  they do not look like those of usual reflections or rotations.   I cannot find any resources about finding these more complicated matrices.   Any help would be appreciated.

One of the usual setup for the coordination system would be that $x$ axis coincident with a two-fold axis. If we use this setup, we can easily write out the matrix for rotation around this 2-fold axis:
$$\begin{bmatrix}
1&0&0\\
0&-1&0\\
0&0&-1
\end{bmatrix}.$$
Then you can combine this with the transform matrix of the 3-fold axis to get the transform matrices of the other two 2-fold axes. That's just matrix multiplication.
In this setup, $yz$ is one of the reflection plane and the corresponding matrix is
$$\begin{bmatrix}
-1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}.$$
Note that this is just the combination of the previous matrix with inversion. You can apply the 3-fold axis rotation to it (once again, matrix multiplication) to obtain the matrix with regard to other two reflection planes.


text:
Assignment of the IR spectrum of 2,4-pentanedione (acetylacetone)
The IR spectrum is shown here:

My assignments so far:

The 1728 and 1708 stretches correspond to $\ce{C=O}$ stretches in the keto form - the symmetric/asymmetric stretches respectively.
The 3003 stretches are typical $\ce{C-H}$ stretches.
The 1606 stretch corresponds to the $\ce{C=C}$ in the enol form. However, hidden under this peak there is also the $\ce{C=O}$ stretch in the enol form, which is lowered by conjugation to the $\ce{C=C}$ and the $\ce{O}$ of the $\ce{-OH}$ respectively.

I have two questions about the spectrum, however:

Why is there no $\ce{-OH}$ peak? Even if the enol is hydrogen bonded to itself, there should still be an $\ce{-OH}$ peak in the $\pu{3400 cm^{-1}}$ range?
Does the fact that the carbonyl peak in the enol form is hydrogen bonded change its shift? Assuming the only contribution to the lowering of the $\ce{C=O}$ stretching frequency is conjugation to the one $\ce{C=C}$ bond, this would lower the peak to ~$\pu{1655 cm^{-1}}$: Surely this would be seen as a peak in its own right? It seems to be shifted even lower to be fully hidden by the 1606 peak.

You're looking for the OH stretch, but what does that look like in this molecule? Because of internal hydrogen bonding, you might expect the bond strength to be very different, so the stretch frequency is not going to be where you would expect...

OH str for beta diketones is tabulated from 3200 to 2400 1/cm. The shift is explained by resonance of the hydrogen bonded structure.
Its absorption features is small or unseen as for

commonly the ketonic form is highly dominant; i.e. we won't expect to see OH str in the acetaldehyde spectrum, even
owing to the above mentioned resonance,  the dipole change associated to symmetric OH str is null, and it is minimal for the AS one.

The symmetry of 2,4-pentanedione makes the last point even more important.
Concerning the second part. The fact that the O of the carbonyl of the enoform is H bonded does further shift down the related CO str. Moreover, in addition to resonance with the C=C, note that again there will be resonance with the CO in beta as above, and this further shift the frequency to lower wn.


text:
How can chlorine have 10 electrons around it in chlorine trifluoride?
I'm relatively new to chemistry, and while doing some exercises about $\ce{ClF_3}$ I tried to draw where the electrons are. I checked Google and got this:

How is this structure possible? From what I've learned from high school every element wants 8 valence electrons. From the figure, $\ce{F}$ has 8, but $\ce{Cl}$ has 10.

"Every element wants 8 valence electrons" is just a simplified description for people to get started with chemistry. Octet rule has its limits and many exceptions, $\ce{ClF3}$ is a typical example of hypervalent molecules. I think you can find pretty much from the two links above and if there's anything unclear, come back and ask more specific questions regarding your doubts.

@WeijunZhou Ok, thank you! I'll have a look! Should I delete this question?

No need to do so. It can help others who may ask questions similar to yours, and that's how the site works.

There's a total of 4*7 = 28. One of the flourine's shares its lone electron with chlorine's lone electron. The other two flourines get  shares of two of the chlorine's remaining  six thus allowing covalent bonds to be formed )and leaving two unshared pairs).

possible duplicate of What is the favoured geometry of chlorine trifluoride?

Interhalogens consist of one $\ce{X-F}$ pair whith a regular two electron bond, and pairs of flourine atoms taking over one free electron pair of X (chlorine, bromine, iodine).
What forms are three-centre four-electron bonds (one pair of electrons from the chlorine, two single electrons from two fluorine atoms). That's why interhalogens always have an uneven number of fluorine atoms (except for the dimer $\ce{Cl_2F_6}$).
https://en.wikipedia.org/wiki/Three-center_four-electron_bond
If this is still octet rule is debatable, or at least not directly obvious (see the wp article). Let's just say that fluorine is a bad, bad guy who has things his way.


text:
Relative reactivities of alkyl halides in an E2 reaction
What is the exact order of reactivity of alkyl halides in an E2 reaction?
Organic Chemistry by Paula Yurikanis Bruice says that the order is $3^\circ>2^\circ>1^\circ$. And I think this order is correct because the transition state of the tertiary alkyl halide will be better stabilized by more hyperconjugating structures as compared to the secondary and primary halides.
But my teacher made me write that the order is $3^\circ<2^\circ<1^\circ$ and I even saw the this order in a book.
So I'm kind of confused about the correct order in E2  mechanism.
Any help will be appreciated.

I believe you might be mixing up E1 and E2 elimination reactions. In the first case the reaction first proceeds by formation of a carbocation from the alkyl halide and then a proton extractio. In this case, the order of reactivity is indeed $3^∘>2^∘>1^∘$ as such is the order of carbocation stability - in order the E1 reaction to proceed the carbocation must be formed.
But the situation is completely different in an E2 elimination. In this case the mechanism is the abstraction of the proton that is oriented anti to the halide with the simultaneous formation of the double bond and halide leaving the molecule as anion. That means that in this case the stability of carbocation plays no role and the significant factor are the sterical considerations. If the carbon atoms around the halide make it hard for the base to get to the anti proton then the reaction is less likely to proceed via E2 mechanism.
But bear in mind that usually both E1 and E2 elimination can take place and because of that tertiary halides would usually undergo elimination more readily than primary halides - just not via E2 mechanism.
More details on the E2 mechanism, with pictures: Master Organic Chemistry

Nobody is talking about carbocation. We are talking about transition state.

@Jan Rzymkowski I am aware of E1, E2 & E1cb and i'm really not mixing them up. According to you, the order should be $3^\circ<2^\circ<1^\circ$ right? but what about the stable transition state that can be attained when a tertiary halide undergoes elimination?

If you have a stable transition state with a carbocation then you have an E1 reaction, not E2 reaction. As I said, tertiary halide is generally more reactive than primary, because of that possible carbocation - but it's not more reactive in E2 mechanism.

Here you have a transition state for E2 mechanism: http://iverson.cm.utexas.edu/courses/310M/ReactMoviesFl05%20/E2scheme.gif
If your transition state is a carbocation, then the reaction proceeds via E1 mechanism, not E2 mechanism.

Not a transition state with a carbocation. In fact a carbocation cannot be called a transition state. It is a reaction intermediate, and please stop talking about E1 here. We are not confused about what is E1 and what is E2.  For your reference here is one pdf: http://home.iitk.ac.in/~madhavr/CHM102/Lec13.pdf. Check page 6. Here is another: http://ocw.uci.edu/upload/files/51b_chapter8_f2014.pdf. Maybe the problem is that we have different understanding of the word "reactivity".

Perhaps it would be more helpful to think of the transition state as alkene-like instead; also stabilised by hyperconjugation, but not a carbocation

In E1 and E2 both are same reactivity order of alkyl halide because both of reaction 3^• alkyl halide give easily product because after elimination reaction on 3^•alkyle halide. alkene is attached more alpha hydrogen which is more stable.so reactivity order in E1and E2 both are same 3^•>2^•>1^and is also same  in SN1 and but different in SN2


text:
Reactions of differents halohydrins with tert-butoxide
I have two compounds (1R,2S)-3,3-dimethyl-2-bromocyclohexanol and (1R,2R)-3,3-dimethyl-2-bromocyclohexanol, both of them are treated with $\ce{KO^tBu}$ (a strong but bulky base) to form two different isomers of $\ce{C8H14O}$ which is not an alcohol.
I though that (1R,2R)-3,3-dimethyl-2-bromocyclohexanol is going to form an epoxide due the configuration of the halogen and the hydroxyl (although I don't know if it is required an specific configuration) and (1R,2S)-3,3-dimethyl-2-bromocyclohexanol is going to have a E2 reaction due steric hindrance but that will end with an alcohol anyway.

Would be real nice if you upgraded to good English ;)

Chair conformation RR-1 (shown as its alkoxides) is in equilibrium with conformation RR-2, which is the unfavorable partner owing to the strong 1,3-diaxial Me/alkoxide interaction. It is this conformation that leads to the epoxide 1. This route is possible if the rate of formation is much greater than the rate constant for RR-1 --> aldehyde 2. RR-1 has concentration on its side and anti-periplanar alignment of bonds for ring contraction. I prefer ring contraction but this does not preclude epoxide formation.
The RS isomer has two chair conformations, RS-1 and RS-2.  Again, RS-2 is an unfavorable conformation (vide supra) but has proper bond alignment for ring contraction. Conformation RS-1 has the necessary bond alignment for E2 elimination to form cyclohexanone 4 via the enol 3.
A priori, it is not easy to predict with certainty what the product distribution will be. The epoxide can only arise from the RR isomer while aldehyde 2 can arise from either bromohydrin. Organic chemistry is an experimental science.