text:
What is total concentration of buffer used in derivation of buffer capacity?
ChemBuddy — The buffer capacity provides the following:
$$\ce{HA <=> H+ + A-}$$

Total concentration of the buffer $c_\mathrm{buf}$ is given by
$$c_\mathrm{buf} = [\ce{HA}] + [\ce{A-}]\tag{19.3}$$
From dissociation constant definition we have
$$[\ce{HA}] = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm a}\tag{19.4}$$

But shouldn't
$$c_\mathrm{buf} = [\ce{HA}] + [\ce{A-}] + [\ce{H+}]?\tag{1}$$
The total concentration at equilibrium is
$$c(1 - \alpha) + 2c\alpha = c(1 + \alpha)\neq c,\tag{2}$$
hence the initial concentration of buffer is different from that at equilibrium.

you asked a question, then you replied correctly. The quantity of matter is conserved.

I don't understand. Does that mean they have given wrong equation? I have proved that its different in the last line.

At the equilibrium,  [HA]=c(1−α) and [A-]=cα. The quantity of matter of the specie A is conserved. So, the total amount of all species having A is c. So, c=[HA]+[A-]=c(1−α)+cα

Using the given definition of total concentration of buffer makes sense because that value does not change when the acid dissociates. Your proposed definition would make everything very complicated.

Initially in the solution the following equilibria exist
$$\ce{HA <=> H+ +A-}$$
and
$$\ce{H2O <=> H+ +OH-}$$
Let initial concentration of $\ce{HA}$ be $a_1$ and that of $\ce{A-}$ be $a_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K_a$ and the ionic product of water, $K_w$. By adding acid, you have disturbed the equilibrium, and from Le Chatelier's principle, both the reactions go backwards.
$$\ce{HA <<=> H+ + A- }\tag{1} \label{eqn:1}$$
$K_{eq}=K_a$
$$\ce{H2O <<=> H+ +OH-} \tag{2} \label{eqn:2}$$
$K_{eq}=K_w$
Let the new equilibrium concentrations of $\ce{HA},~\ce{A-},~\ce{H+},~\ce{OH-}$ be $a_1+\mathrm dx,~a_2-\mathrm dx,~\frac{K_a\cdot a_1}{a_2}+\mathrm dn-\mathrm dx-\mathrm dy,~\frac{K_w\cdot a_2}{K_a.a_1} - \mathrm dy$ respectively.
At equilibrium, for $\ref{eqn:1}$,
$$K_a = \frac{(\frac{K_a\cdot a_1}{a_2} +\mathrm dn-\mathrm dx-\mathrm dy)\cdot (a_2 - \mathrm dx)}{a_1+\mathrm dx} \tag{3} \label{eqn:3}$$
and for $\ref{eqn:2}$,
$$K_w = (\frac{K_a\cdot a_1}{a_2}+\mathrm dn-\mathrm dx-\mathrm dy)\cdot (\frac{K_w\cdot a_2}{K_a.a_1}-\mathrm dy) \tag{4} \label{eqn:4}$$
Dividing $\ref{eqn:3}$ by $\ref{eqn:4}$, we get
$$(K_a)\cdot (\frac{K_w\cdot a_2}{K_a\cdot a_1}-\mathrm dy)=(K_w)\cdot (\frac{a_2-\mathrm dx}{a_1+\mathrm dx})\\
K_w\cdot a_2+\frac{K_w\cdot a_2}{a_1}\cdot \mathrm dx-K_aa_1\mathrm dy-K_a\mathrm dx\mathrm dy=K_wa_2-K_w\mathrm dx$$
Neglecting the term $K_a\mathrm dx\mathrm dy$, we get
$$dx= \frac{K_a\cdot {a_1}^2}{K_w(a_1+a_2)}\cdot \mathrm dy$$
and also from $\ref{eqn:3}$
$$K_a\cdot (a_1+\mathrm dx)=(\frac{K_a\cdot a_1}{a_2}+\mathrm dn-\mathrm dx-\mathrm dy)\cdot (a_2-\mathrm dx)$$
Which on simplifying and neglecting terms of the type $\mathrm da\cdot \mathrm db$ will give,
$$\mathrm dn= (\frac{{K_a}^2\cdot {a_1}^2}{K_w\cdot {a_2}^2}+\frac{K_a\cdot {a_1}^2}{K_w\cdot (a_1+a_2)}+1)\cdot \mathrm dy \tag{5} \label{eqn:5}$$
By definition,
$\beta = \frac{\mathrm dn}{\mathrm{dpH}}$ and $\mathrm{pH}=-\log_{10} {[\ce{H+}]}$.
On differentiating this, we get
$$\mathrm{dpH} = \frac{-1}{[\ce{H+}]}\cdot \frac{1}{\log_e10}\cdot \mathrm d[\ce{H+}]$$
Substituting this in the definition of buffer capacity,
$$\beta = \frac{-\log_e10\cdot [\ce{H+}]\cdot \mathrm dn}{\mathrm d[\ce{H+}]} \tag{6} \label{eqn:6}$$
Here,
$$\mathrm d[\ce{H+}] = \mathrm dx+\mathrm dy-\mathrm dn = -\frac{{K_a}^2\cdot {a_1}^2}{{a_2^2}\cdot K_w}\cdot \mathrm dy$$
Substituting this expression of $\mathrm d[\ce{H+}]$ and $\mathrm dn$ from $\ref{eqn:5}$ in $\ref{eqn:6}$, we get,
$$\beta = \log_e10\cdot [\ce{H+}]\cdot \frac{(\frac{{K_a}^2\cdot {a_1}^2}{{a_2}^2\cdot K_w}+\frac{K_a\cdot {a_1^2}}{K_w\cdot (a_1 + a_2)} +1)}{\frac{{K_a}^2\cdot {a_1}^2}{{a_2}^2\cdot K_w}}\\
\beta = \log_e(10)[\ce{H+}](1 + \frac{K_w}{{[\ce{H+}]}^2} + \frac{{a_2}^2}{(a_1+a_2)\cdot K_a})\\
\beta = \log_e10( [\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{{a_2}^2\cdot [\ce{H+}]}{(a_1+a_2)\cdot K_a})$$
Substituting $[\ce{H+}]= \frac{K_a\cdot a_1}{a_2}$ in the third term, we get
$$\beta = \log_e10( [\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{a_1\cdot a_2}{(a_1+a_2)})$$
Multiplying numerator and denominator of the third term with $K_a$,
$$\beta = \log_e10( [\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{K_a\cdot a_1\cdot a_2}{K_a\cdot (a_1+a_2)})$$
Taking $a_2$ common from the denominator and cancelling the $a_2$ in the numerator and substitute $\frac{k_a\cdot a_1}{a_2}=[\ce{H+}]$ and multiplying numerator and denominator with $[\ce{H+}]+K_a$, we get
$$\beta = \log_e10([\ce{H+}] + \frac{K_w}{[\ce{H+}]} + \frac{(\frac{K_a\cdot a_1}{a_2})\cdot (K_aa_1+K_aa_2)}{{([\ce{H+}]+{K_a})}^2}$$
Taking $K_a$ out from the numerator and substituting $[\ce{H+}]=\frac{K_a\cdot a_1}{a_2}$, we get
$$\beta= \log_e10([\ce{H+}]+[\ce{OH-}]+ \frac{[\ce{H+}]K_a(a_1+a_2)}{([\ce{H+}]+K_a)^2})$$
So, in your expression, of course, $C_{\text{buf}}=a_1+a_2=[\ce{HA}]+[\ce{A-}]$ and not $[\ce{HA}]+[\ce{A-}]+[\ce{H+}]$
$$\beta= \log_e10([\ce{H+}]+[\ce{OH-}]+ \frac{[\ce{H+}]K_aC_{\text{buf}}}{([\ce{H+}]+K_a)^2})$$
You will get the same result if a small amount of base is added.

You should read Which symbols are written in roman (upright) font and which are italicized?. I have corrected it for now. And it is better to use $$...$$ (block mode) over $... $ (single line mode) when writing long block of equations.

@NisargBhavsar Thanks for the edit and I'll definitely look into it

Time flies. I asked this question like 5 years ago and I don't even understand what I even asked right now.


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Storage life of 3% hydrogen peroxide
I am using 3% hydrogen peroxide for a series of operations, and the common practice is to prepare it on-site from 30% hydrogen peroxide and saline. This is time consuming and somewhat more prone to error than preparing a 3% stock. I was told this is not done because 3% hydrogen peroxide has a short shelf-life (~a week).

Is this true?
If so, why?
Would the shelf-life be extended if the solution were prepared with phosphate-buffered saline or just with deionized water (instead of saline)?

Would it be a problem if the hydrogen peroxide used was 2.8% instead of 3%? 3% seems pretty low for hydrogen peroxide, it decomposes but not that fast, at this concentration. Unless you need exactly 3% I don't think storing refrigerated would be a problem. It's worth saying that unless your stock solution is always new or the 3% concentration is somehow measured, it's probably not exactly 3% anyway.

any value from 2 to 5 should be ok. Would you recommend I prepare a 4% solution instead? what would be the decomposition time course? why is it more stable at higher concentrations?

These links say that 3% and 30% hydrogen peroxide both decompose at 0.5% per year when stored at room temperature.

I would use pre-made (probably, medical grade) 3% purely for convenience. 30% staff requires gloves (you REALLY do not want this thing on your skin), while 3% does not mess with skin.

It may pretty well be true, and here's why. Not only hydrogen peroxide is highly unstable by itself, but its decay can be catalyzed by a wide variety of compounds. And by wide, I mean really wide; at times it seems to me that those compounds that don't catalyze it are a minority. Now, I don't know what you have in your saline, but chances are that it works on the peroxide, too.
With deionized water, this should not be a problem... or should it? See, that link by @ron says that the 30% thing is stabilized. Guess what? It means that they've put some additive into it, which somewhat slows down the decomposition. When you dilute your peroxide, the stabilizer gets diluted too, and hence can't work as well as before. This is a possible reason why you may get shorter shelf-life even when diluting with pure water.
I'm not sure what is the stabilizer they use. In any case, if you find that out, then add it during the preparation, and don't add anything that can catalyze the decay, your 3% solution should be fine (in terms of shelf-life) - but I'm afraid these additions and subtractions may stand in the way of your primary objectives.

AFAIK, it is stabilized by bounding cations of 3d transition metals, mostly Mn and Fe as most common in natural water.

Strangely, the link to the 3% solution specifies the same shelf-life, but says nothing about stabilization. An omission?

It's difficult to tell what would happen in your particular situation, because as Ivan Neretin mentions, hydrogen peroxide decomposition is catalyzed by a huge number of substances.  The easiest way to find out whether your solution is actually decomposing or not would be to test it.
Make a solution of known concentration.
If you're using an accordion bottle, cap it tightly and let it sit for a week.  One liter of 3% hydrogen peroxide will contain roughly one mole of solute, and when it all has decomposed it will produce about ten liters of oxygen gas.  If the bottle doesn't accordion substantially in a week's time, you know that it isn't decomposing at a rate that should concern you.
If you're not using an accordion bottle then you can cap the bottle with a balloon (or maybe even a rubber glove + rubber band) instead, assuming you've got a reasonably safe place to put the thing for a week.  Again, it's going to make roughly three liters of oxygen for each 1% loss of peroxide per liter of solution, so you can just watch the balloon and wait a week.  A balloon will leak somewhat, but if the rate of decomposition is significant, it will still inflate.
In general you can drastically slow the rate of decomposition by storing peroxide solutions in brown bottles and keeping them refrigerated.  Unless you're using something other than sodium chloride solution or storing it in bright light in a warm room in a clear container, I'd be surprised if your solution is decomposing at a rate of more than 1% a year.  However, if the concentration is important, and it sounds like it is, then a test is really the only way to know for sure.

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text:
Do polymerized cooking oils release free radicals
I am seasoning a carbon steel wok and from what I understand it basically involves heating cooking oil on it until it gets to the smoke point and then forms a polymerized layer that makes it like a non stick pan.  I have also heard that as oil gets heated close to and past the smoke point it breaks down releasing free radicals and other unhealthy compounds.  My question is that when the heated oil polymerizes do these compounds get permanently locked up in the polymer or will they eventually get into the food cooked in the pan?

I read several articles in the topic in the past 5 hours, and I have a different conclusion than the other answerer.
My experience that the TG polymer layer on the surface of your wok is far from being stable. Every time you use it, smaller or bigger particles of the polymer can end up in your food. Some of them are even visible, that's why I stopped using a wok a long time ago. If you want to use a seasoned wok no matter what, then do the seasoning with an oil high in polyunsaturated fatty acids (e.g. flax seed oil), so the polymer will be somewhat more stable, because of the additional bonds, but there is no guarantee that it won't break down in time. Don't use these heat sensitive oils for cooking, they are good only for seasoning or for cold salads.
High polymers can be absorbed depending on whether the enzymes of the digestive tract can handle them. For example starch is a high polymer of glucose and amylases can easily hydrolize it, while cellulose is a different high polymer of glucose, which we don't have the enzymes to digest. As of TG polymers, they contain mostly fatty acid monomers, dimers and trimers. So the polimerization is far from being complete, you won't get long fatty acid polymers by cutting down the glycerols with a lipase. According to the articles I read, lipases can hydrolize around 10-50% of the polymers (they are inhibited by them at some degree), so the resulting dimers, trimers, oxidized monomers, cyclic monomers, etc. created in the seasoning process, can be absorbed. It is hard to find studies about their health effect, but they are not considered healthy.

Any cook can tell you, that seasoning a pan is not a permanent, but ongoing affair.  High heat, acidic food and mechanical damage can and does occur, which is why you need to season your pan on a regular basis. So it is clear, that some of this stuff ends up in your food, no way around that.

If you use fresh cooking oil containing unsaturated fats (like olive or canola oil) to season your pan, it will thermally polymerize to form a stable layer that gives the cooking surface its nonstick properties. Note that a saturated fat like coconut oil, lard, or ghee will not work very well or at all for this, since it does not have any double bonds to undergo polymerization. The radical formation you are concerned about results over time from air oxidation of unsaturated fats to form peroxides--these in turn break down to give aldehydes and acids that cause the bad flavor of rancid fat, and have adverse health effects. (Not a problem for coconut oil.) Heating fresh oil in a pan to create a nonstick coating may result in some air oxidation in addition to polymer formation, but I think it is not a big cause for concern. High polymers are generally nontoxic because the body does not absorb them.

I am not sure I understand your answer. I am pretty sure, that heating oil above its smoke point is pretty bad for your health and as far as I know this is exactly what produces trans fats, please correct me if I am wrong.

@user1721135 No this is not how trans fats are created. And, since the point of seasoning a pan is to create a semi-permanent coating not to cook food, the concerns are irrelevant if the treated pan is washed before you cook with it.

Based on inf3rno's answer, this is from "Deep Frying Chemistry, Nutrition, and Practical Applications" by Michael Erickson

Digestibility of oxidized fatty acid monomers was considerably high,
averaging 76.6%. Among polymeric fatty acids, nonpolar dimers had the
lowest digestibility (10.9% on average),whereas oxidized dimers and
polymers possessed higher apparent absorbability than expected,
ranging from 22.7-49.6% (96).

Basically, the question around all safe cookware questions is, "is a coating of X on the surface" considered safe. In stainless steel it would be chromium oxide. Nonstick coatings are manmade thermoplastic polymers. Glass is silicon dioxide. There seems to be a blind consideration of "cast iron" as safe yet it is quite difficult to figure out what the coating actually is, so this question of asking if it releases free radicals makes a lot of sense.

While this is interesting, I don't see how it actually answers the question.

I don't have enough points to comment on the above post.

My train of thought was not explained well. Since inf3rno stated that "dimers, trimers, oxidized monomers, cyclic monomers" are created, I wanted to quantify how well they are absorbed. I agree that they are not healthy.

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Why does zinc plate copper in silver penny lab
In class, we performed a lab that involved heating a solution of zinc sulfate with zinc strips at the bottom of the beaker. A penny was placed on top of the zinc strips and soon after the penny became covered in a layer of zinc. What process causes the zinc layer to develop?
The explanation was that the zinc metal transferred electrons to copper which in turn reduced the zinc ions in solution, but this does not seem to make any sense as the zinc is being oxidized to just be reduced again. Why does this happen.  Is a thin zinc layer more stable?
Here are the instructions for the experiment

Was any electricity involved in this plating?

No the metals simply contacted each other

I made a video on the process of electroplating copper (a penny): https://www.youtube.com/watch?v=JOeJpoVqBMQ
Using Ni here, not Zn. Same principles though

None of the answers to this question account for the electrochemical potential induced by the strain defects present from milling present in the zinc powder.

The system of $Zn$ metal in a solution of $\ce{Zn^{2+}(aq)}$ is an equilibrium. Which means that the rates of the reactions $\ce{Zn -> Zn^{2+} + 2e-}$ and $\ce{Zn^{2+} + 2e- -> Zn}$ are happening in exactly the same rate and this rate is not zero. It is critical no notice that this rate is non-zero.
By heating up, you are accelerating both reactions, probably at different rates so there will be a new equilibrium point.
The precipitation is happening on any surface that it can nucleate on and this surface might be the surface of the penny or some other zinc. This is immaterial.
The counter reaction doesn't really matter here since it just happens on whichever metal the other reaction is happening and it is just the reduction or oxidation of water.

Electricity is involved: the copper (well, copper plated zinc in the US) penny is in direct contact with the zinc, forming a short-circuited connection. Forcing the plating action "backwards" (copper being more electronegative than zinc) requires current.
Try the same experiment but with a separator, such as a piece of filter paper, between zinc and copper; there should be no plating.

But i still dont understand why electrons would go from the zinc metal to the copper metal just to go back into the zinc

That would make reaction have no net change as you start and end with zinc solution, copper metal, and zinc metal.

The explanation was that the zinc metal transferred electrons to copper which in turn reduced the zinc ions in solution, but this does not seem to make any sense as the zinc is being oxidized to just be reduced again. Why does this happen. Is a thin zinc layer more stable?

Your intuition is correct and both of the given answers are wrong. The zinc plates the penny because there is a driving force. but the question then is "what is the driving force?" it is an electrical reaction but it is not as simple as one rate just happens to out pace another or is out accelerated by heating.  First lets look at the half reaction of the powder:
$$\ce{Zn (s) -> Zn^{2+} (aq) + 2e-}\tag{E$^\circ$  = - 0.76}$$
and that of the penny:
$$\ce{Zn^{2+} (aq) + 2e- -> Zn (s)}\tag{E$^\circ$  = + 0.76}$$
as you have already noted the standard electrode potential are equivalent, so what now? Well the operative word here is standard. standard electrode potentials assume that the materials is at 298K, 1 bar of pressure, 1 M electrolyte concentration and also that the surface is flat and well annealed.  Zinc powder produced by milling is neither flat nor well annealed, it is quite sharp with a lot of residual stress. These conditions increase the energy of the zinc material which lowers the standard electrode potential for the powder such that it is now lower in magnitude than the electrode half reaction on the penny.  i.e.:
$$\ce{Zn (s) -> Zn^{2+} (aq) + 2e-}\tag{$\color\red {\mathrm E  <}$ - 0.76}$$
This reduced half potential creates a non net zero electrical potential which in turn produces a plating of zinc on the penny which is flatter with less internal stress. Thus your net reaction is:
$$\ce{Zn (s, powder) -> Zn (s, plate)} \tag{$\color\red {\mathrm E  < 0.0}$}$$
As for why being sharp and internally stressed cause higher energy, That is beyond the scope of the question, I have have given a brief and hand-wavy explanation below but I strongly encourage you to do your own research for a better explaination.

 First for flatness, remember that chemical reactions are kinetic reactions relying on one rate happening faster than another. Knowing this we can infer that for a dissolution reaction, the more directions an atom has to dissociate, the more likely it will be to dissociate (See:Kelvin Equation) since energy directed towards the solid  will not result in dissociation. As for the internal stress, stress is caused by atoms not being in a regular low energy state. Some perturbation of the microstructure causes them to be in a non-equilibrium state.  Since equilibrium states are the lowest energy states, any configuration not at equilibrium must have higher energy and thus when given the chance will try to return to a lower energy state.

I'm a little confused by your answer because the question mentions zinc strips not zinc powder. Otherwise your argument makes good sense.


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Freezing all the water on earth with a seed crystal
I was recently reading the satirical, fictional book Cat's Cradle by Kurt Vonnegut. In this book, one of the major subplots is the invention of a mythical polymorph of ice called ice-9 in the book. In the book, it is described that if this polymorph of ice were to touch actual water, all the water, inevitably touching through oceans, rivers, lakes, etc. would freeze into this polymorph of ice which is purported to melt at a temperature over 100 degrees Fahrenheit.
This raises two questions for me:
First, would something like this actually happen? I know, for instance, a method people use for purifying crystals is to create a supersaturated solution and disturb the substance and it will crystallize quite dramatically. Could something similar happen here in the sense that the mythical ice-9 structure would be introduced into liquid water (which to ice-9 is a supersaturated solution of water) and create a massive crystal out of all water?
Second, and more in reality, is there any solid that is like this? A solid which has two stable crystal structure that exist at temperature reachable in everyday life at about atmospheric pressure? And if so, can you actually convert one crystal structure to another easily? And would one of the liquids behave as a supersaturated solution of the crystal?
Sorry if that is a little unclear. I am sincerely interested in this and can clarify if there is any confusion.
Anything interesting along those lines would be considered relevant in my mind.

This is probably more suited at Worldbuilding.SE

There is no creative or world-building element to this since it's a question inspired by a book that's already written.... The question of forming a crystal using a seed-crystal is very much a chemistry question for which I am only interested in a chemistry answer. I'll edit the second part to make it more appropriate to Chemistry.SE though.

The only I way I see for water to solidify like in Vonnegut's novel is if all liquid water on Earth is in a false ground state (a metastable state), blocked from the true ice-9 ground state by a kinetic barrier, so that once a small portion crystallises, the released energy prompts the conversion of the rest of the Earth's water into ice-9 (strong parallels can be made with a vacuum metastability event).

Should my previous comment be the case, then there's an argument which is very interesting for its simplicity, which I have previously heard used as a quick dismissal of a quack theory in chemistry (hydrino theory). Basically, the Earth has had large liquid bodies of water for over 4 billion years, subject to all sorts of conditions. That's more than enough time for water to have found its true ground state one way or another, so we can only infer that liquid water is stable under the present conditions.

Interesting. More generally, do you know of any crystal which commonly exists in a metastable state just out of curiosity? Or is this quite rare?

A precise physical argument against the existence of an Ice-9-like state of water is probably not trivial (ab initio methods required?). However, there is a simple but powerful argument against it; we haven't found anything like it yet. Water is a very common substance, and humans have been applying the scientific method to it for hundreds of years, humans have manipulated it for hundreds of thousands of years, but most importantly, the entire Earth has been churning massive amounts of water around for billions of years. If an Ice-9-like state of water did exist, we should have seen it by now.
To be clear, superficially this argument may seem to be a fallacious argument from ignorance. However, we can qualitatively back it up with a bit more knowledge. Assuming that Ice-9 is the true ground state of water, and that all liquid water on Earth is currently in a false ground state, separated only by a kinetic barrier, and realising that the conversion of liquid water to Ice-9 is autocatalytic, we must ask ourselves, if it only had to happen once, why hasn't it happened yet?. The only reasonable conclusion is that, if it can happen, then the kinetic barrier to conversion must be enormous. The interconversion of diamond to graphite is a chemical process with a famously large kinetic barrier (around $\mathrm{540\ kJ\ mol^{-1}}$), yet we have plentiful evidence of it happening many times in geological time scales (and we can do it in the lab too!). Thus the kinetic barrier for conversion between liquid water and Ice-9 would have to be even larger, which is unlikely. There's also a limit to how high we can expect chemical kinetic barriers to be, which we can put very generously at $\mathrm{100\ eV \approx 10\ MJ\ mol^{-1}}$. Water molecules on Earth have had many, many, many interactions at energies way higher than this, as they have been incessantly exposed to cosmic radiation, nuclear decays, lightning strikes, high temperatures, etc. In short, the collective history of all water molecules on Earth has explored a vast region of the space of allowed states, and not once has an Ice-9-like state been found. It is thus fair to assume it simply does not exist.
Regarding your second question, I just remembered a great example: pure tin. At ambient pressure, solid tin can exist in two different polymorphs, and the transition between the most stable form is conveniently close to room temperature. Above 13.2 ºC, tin is most stable as a ductile metallic material, called white tin, whereas below 13.2 ºC, the most stable form is a brittle non-metal called grey tin. It is possible to sustain white tin below 13.2 ºC for a period of time, creating a metastable system, much like a supersaturated solution. However, an adequate disturbance will cause the white tin to fall to its ground state. Like Vonnegut's Ice-9, this transition is autocatalytic; once conversion begins, there's no stopping it below 13.2 ºC, creating the so-called tin pest. There are actually several great videos showing the process, which you can easily find on YouTube, such as this one.

I forgot to mention, but not long after Kurt Vonnegut published Cat's Cradle, there was a lot of commotion in the scientific community over the possibility of water being able to polymerise, forming a substance called polywater, which could conceivably have had the ability to convert liquid water as we know it into a new form with different properties. It turned out to be a messy mistake, though, and there are several interesting accounts on the web about how polywater developed from its first reports to the final nails in its coffin.

Polywater is a great example. Considering all the work (both theoretical and experimental) into the structure of water and various ice polymorphs, it's extremely unlikely that there is a more stable ground state.

As an aside:  Vonnegut's ice-9 was never meant to be represented by the real Ice IX, which was discovered five years after Cat's Cradle.  Vonnegut simply used a number high enough to represent an unknown material, which of course made it all the more difficult to contend with in the story.  Today one would use ice-19 (or higher).

Just to offer a counter to arguments based on energetics...

However, there is a simple but powerful argument against it; we
  haven't found anything like it yet

But the same could be said for replicating, living organisms outside of the one known case here on Earth.
Even though the necessary components were present and the kinetic barrier to the formation of the key ingredients was relatively moderate, the probability of it happening is (apparently) extremely remote.  Catalysts, such as the chiral face of a crystal and special magnetic and electrical events may have been simultaneously required to bring about the first chiral organism (or its inanimate, chiral precursor). The odds of all of this being in the same place and happening at the same time may be incredibly small.
Perhaps the same could be true for ice-9 formation.  Maybe a crystal face with a very specific morphology is required to catalyze the transformation of water into ice-9.  Even if the mineral is known on Earth, perhaps the specific crystal face\morphology does not presently exist on earth.  However, it's conceivable that it could be formed on a planet with extreme temperatures and pressures not found in Earth geology.  Maybe this planet, in some remote corner of the galaxy, has since been destroyed through a collision and a resulting planetary fragment containing crystals with the alien morphology is now headed towards Earth.

You are, of course, entirely correct. The problem with my qualitative argument is that, though it suggests water in Earth has explored a large state space, there is no information provided about the full size of the permissible state space (is there even a way to know this?). It could be that there are many more configurations for water that have not yet been reached (and indeed, that may never be reached in the entire future of the Universe!). In the end, all I can do is put some weak constraints. Sorry I can't do any better!

No, on the contrary, your arguments are very persuasive.  I'm just offering an equally unprovable point of view.


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Why does ice freeze the way it does?
I was just recently looking into some of the details of the different crystal structures that ice can freeze in. I was aware that many of the unusual properties of ice $I_h$ (the structure of “everyday ice”) are a result of the proton disorder necessary for the hydrogen-bonding arrangement it adopts.
What I also found is there are other arrangements of ice which are proton ordered and have nearly identical density to ice $I_h$. Now, I know density has no predictive value on what crystal structure will be taken (rather crystal structure can be used to predict density), but it seems unusual to me that there is such massive divide between ice $I_h$ and the rest of the ice srystal structures.
Is the entropic gain of proton disorder an explanation why ice freezes the way it does? Also, is there a large energy minimization associated with the hexagonal lattice that ice $I_h$ takes on? Doesn't it seem like a proton-ordered arrangement would be preferable over a proton-disordered arrangement? Or is it all much more complicated than that? It just seems like there should be a crystal structure of ice which is comparable to ice $I_h$ in energy, stability, etc. and yet there isn't.
I suppose a true answer to this question, because I admit it's rather broad, would deal with why the hexagonal lattice is so much more preferable for water than some other lattice arrangement. By dealing strictly with water, rather than crystals in general, I believe an answer to this question is more within reach.

Related, but not a duplicate http://chemistry.stackexchange.com/q/35251/15489

Some of what you are asking are still open questions… and the few data published on the topic are often controversial. But let's try to get some of the facts that appear to be accepted rather broadly:

There is a proton-ordered ice structure, called ice XI that is structurally equivalent to our everyday hexagonal proton-disordered ice (aka ice Ih). In the commonly accepted version of the ice phases diagram:

it can be seen that it is the most stable phase of ice at ambient pressure and low temperature. (Although obtaining this phase in practice can be difficult, and is done by doping the ice crystals with dilute KOH.)

Thus, the proton ordering is energetically favored over proton disorder, while proton disorder is entropically favored. Following simple combinatorial arguments [G. P. Johari, J. Chem. Phys., 109, 9543 (1998)], one can estimate the entropy difference at $R \log(3/2) \approx 3.37$ J/mol/K. From the transition temperature of $T_0 = 72$ K, one can then work out the enthalpy difference at $\Delta H = -T \Delta S \approx  -0.243$ kJ/mol. We can note that both terms are very small.
The hexagonal arrangement is not the only one accessible: there is a metastable cubic ice ($I_c$) phase accessible at ambient pressure. The enthalpy differences between cubic and hexagonal ices are even smaller than with ice XI, of the order of 40 J/mol.
There are, finally, numerous other phases of ice — some experimental, some hypothetical — at positive or negative pressures. Since their boundaries are almost horizontal on the phase diagram, it implies that these transitions are dominantly density-driven.
There are also amorphous ices!

To your final question: why ice prefers the hexagonal net? Well, in order to form low-energy crystalline structures, water must adopt a tetrahedral / four-fold coordination, with each molecule surrounded by four nearest neighbors: two on the donating end of hydrogen bonds, two on the receiving end. So the question is: how many periodic structures (crystals) are mathematically possible with this arrangement? They are called four-connected nets, and there are an infinite number of those. However, it happens that among those with relatively low density and “simplicity” (i.e. relatively few inequivalent vertices), most are of high symmetry, including hexagonal.
There is actually an interesting parallel between ice structures and another category of solids having four-connected nets: silicas.


text:
Does dissolving salt in water change the pH?
I am particularly interested in the effect or lack of effect of dissolved NaCl in regards to pH in water. It would be interesting also if anyone has any insight for other salts and their effects. thanks.

To a first approximation the answer is no, but I'd be curious to see if someone can pull up a more precise answer, taking into account the finite values of the acidity/basicity constants and ionic strength of the medium.

When you dissolve a salt in water, you have to distinguish 4 cases:

The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$  do hydrolyze and give ions  $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$  do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:

$K_a= K_b$, the aqueous solution is neutral.
This is the case, for example, of $\ce{NH4CH3COO}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{CH3COOH}$. The ions  $\ce{NH4^+}$ and $\ce{CH3COO^-}$ do hydrolyze. As the hydrolysis constant of ion acetate equals the hydrolysis constant of ion ammonium, the same concentarion of ion hydronium and ion hydroxide is produced, and the pH of the solution is neutral.
$K_a> K_b$
This is the case, for example, of $\ce{NH4F}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HF}$. The ions  $\ce{NH4^+}$ and $\ce{F^-}$ do hydrolyze  and as  ammonium ions hydrolyze to a greater extent than  fluoride ions, i.e more hydronium ion is produced, and the pH of the solution is acidic.
$K_a< K_b$
This is the case, for example, of $\ce{NH4CN}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HCN}$. The ions  $\ce{NH4^+}$ and $\ce{CN^-}$ do hydrolyze  and as  cyanide ions hydrolyze to a greater extent than  fluoride ions, i.e less hydronium ions are produced, and the pH of the solution is basic.

$\ce{NaCl}$ is a salt of a strong acid and strong base. As you know, strong acids give weak conjugate bases and strong bases give weak conjugate acids. Hence, the $\ce{Cl-}$ ions are too weak to accept protons from water and $\ce{Na+}$ ions are to weak to react with $\ce{OH-}$. So there is no change in the concentration of $\ce{H+}$ or $\ce{OH-}$, so no change in pH.
But of a salt like $\ce{CH3COONa}$ is added, $\ce{CH3COO-}$ accepts protons and changes pH. This is because, weak acid $\ce{CH3COOH}$ gives strong conjugate base $\ce{CH3COO-}$.


text:
Ether nomenclature: Oxygens on alkyls and alkyls on alkyls

Does the ether hold priority over longest chain? If so, my bet would be on:
4-(2-ethylheptanyl)oxy decane
If not, I would say:
4-(octan-2-yl methoxy) decane

Again, if ether takes precedence and not longer chain (which, if you ask me, would be ideal), I'd say:
4-[(1-methyl,2-propylheptan-1-yl)oxy]decane
If not:
4-(ethan-1-yl oxy nonan-4-yl)decane
What is correct?

Names for ethers can be formed by five different methods. Furthermore, various traditional names are retained for use in general nomenclature.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), preferred IUPAC names of ethers $\ce{R-O-R'}$, when $\ce{R}$ and $\ce{R'}$ are both aliphatic groups, are formed by substitutive nomenclature. In substitutive nomenclature, when $\ce{R}$ is different from $\ce{R'}$, $\ce{RH}$ is chosen as parent hydride and $\ce{R'-O-{}}$ is cited as a substituent to it.

P-63.2.2.1.1 Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. (…)

For the first example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.
The substituent prefix name for the group $\ce{R'-{}}$ is ‘2-ethylheptyl’.
Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(2-ethylheptyl)oxy’.
Therefore, the preferred IUPAC name of the compound is ‘4-[(2-ethylheptyl)oxy]decane’.

For the second example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.
The substituent prefix name for the group $\ce{R'-{}}$ is ‘3-propyloctan-2-yl’.
Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(3-propyloctan-2-yl)oxy’.
Therefore, the preferred IUPAC name of the compound is ‘4-[(3-propyloctan-2-yl)oxy]decane’.

Take a look at this page This Wikipedia article contrasts systematic and alternate notations of butyls. Applying the same to larger alkyls, don't you think that "(1-methyl,2-propylheptyl)oxy" is also a correct name for the alkoxy substituent in the second case?


text:
What is the difference between an element and an atom?
First, I would like to quote sentences from a book introducing elements and atoms:

An element is a fundamental (pure) form of matter that cannot be broken down to a simpler form.
Elements are made up of particles called atoms. An atom is the smallest unit of any element [...]

The first quote states that element cannot be broken down to a simpler form, but the second quote says elements are formed by atoms, which means element can be split into a simpler form (atoms)?
The difference between atom and element confuses me.

The problem in the definitions you give is that they fail to distinguish between the bulk properties of a substance and the atomic composition of a substance. Bulk properties depends on more than the composition: they also depend on how the atoms are connected. A lump of pure carbon is made from only one type of atom (that makes it an element). But diamonds and graphite a both pure carbon. We can't separate a different kind of atom from either but their form is not the same.

The wording in your quote is a bit unclear, I agree. A very nice definition is given by the Jefferson Lab page What is the simplest way of explaining what atoms, elements, compounds and mixtures are?
Atoms

are the smallest bits of ordinary matter

Whereas an element is

a substance that is made entirely from one type of atom

Atoms themselves can be 'broken down' into smaller sub-atomic particles (protons, neutrons and electrons) - the amounts of these (particularly the number of protons) define each element.
For example, in their un-ionised form, the element carbon is made entirely from carbon atoms, which have 6 protons and 6 electrons each, and cadmium atoms, which have 48 protons and 48 electrons, make up the element cadmium.
The amount of neutrons can vary with atoms of a particular element, forming isotopes, which (from the link):

For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6, which means that every carbon atom has 6 protons, so that the neutron numbers of these isotopes are 6, 7 and 8 respectively.

Different arrangements of atoms of single elements can form allotropes, such as for carbon, with diamond and graphite (among others - shown below). They both are composed of atoms of the element carbon, but in different configurations.
Below are diagrams of the allotropes of carbon featured in the article Work Function Engineering of Graphene (Garg et al. 2014)

Probably worth clarifying that atoms can be broken down into smaller particles.

@bon Okay, I have added a bit more

Since "an element is a substance that is made entirely from one type of atom" and since isotopes involve different atoms, would a block of lead containing a natural abundance of isotopes be considered a mixture of elements?

@ron It is a mixture, but it is not considered a mixture of different elements, since all of the isotopes are isotopes of lead.  (The category boundaries are not perfectly crisp.  Deuterium, for instance, is different enough from ¹H that in some contexts it makes sense to treat it as a separate element.)

Does water (H₂O) contain elements? Not in the sense of elemental substance, but elements in general.

@IncnisMrsi I don't know what you mean by "not in the sense of elemental substance, but elements in general", but water definitely does contain both hydrogen and oxygen, which are elements. There are things in nature that are not composed of elements; the most familiar example is light.  But you have to go pretty far afield to find something material that is not composed of elements, such as a neutron star.

@zwol: Stay on topic, please (I won’t discuss neutronium here, although I perfectly can). Could you post your definition of an element? Santiago seemingly ignores me here, so there is little merit in further discussion on his definition.

@ron that would be a mixture of isotopes (using the broad definitions from the Jefferson Labs link at the least) - that could make for an interesting question in itself.

@IncnisMrsi I would define "element" as a type of atom, classified by the number of protons it contains.  Thus, things which are composed of atoms are by definition composed of elements (since all atoms have some number of protons), and things which are not composed of atoms (light, neutrinos, etc) are not composed of elements. You're right that I shouldn't have brought up neutron stars, since we don't fully understand their composition.

If we define elements as "a substance that is made entirely from one type of atom" then wouldn't $\ce{O2}$ be an element as it composed entirely from one type of atom?

Given that you seem to be a beginner in Chemistry, I thought you might benefit from a description from somebody who only studied up to high school chemistry.
We'll use water as an example because you've probably heard it referred to as H2O.

A single "piece" of water, H2O is a molecule.
It contains 2 elements: Hydrogen, and Oxygen.
But it contains 3 atoms, 2 Hydrogen, and 1 Oxygen.

Atoms contain even smaller pieces called subatomic particles. There are three kinds: Protons, Neutrons, and Electrons. the number of protons defines what the element is.
1 proton - Hydrogen
2 protons - Helium
12 protons - Carbon, etc.
The total number of protons + neutrons is the atom's atomic weight
Elements can contain different numbers of neutrons. These are called isotopes.
Hydrogen-1 has 1 proton
Hydrogen-2 has 1 proton 1 neutron
Hydrogen-3 has 1 proton 2 neutrons
These different forms of hydrogen are isotopes of one another.

Obviously it gets more complicated that this, but for you at the moment, it doesn't need to :)

Different isotopes are considered as same element but can we call them different atoms?

@Apurvium Yes, they are different atoms since they exist independent of each other.

The word element has two slightly different meanings in chemistry:

an element as an abstraction, an “atomic species” (ex.: $\ce{{}_8O}$ oxygen),   and
an elemental substance (ex.: $\ce{O2}$ dioxygen gas), that represents an element in practice.

Textbooks on chemistry, possibly, downplay the distinction since they are not textbooks on philosophy.
Different elements are distinguished by their atomic number (ex.: 8 for oxygen). Atomic number is a characteristic of an atom, whether is it free or bound, neutral or ionized. Atomic number is the number or protons in the nucleus (that is chemically immutable) and, the same, electric charge of the nucleus (in appropriate unit). It is an invariant preserved in all chemical processes. Elements are a classification of atoms. For each natural Z, all atoms with atomic number Z form a chemical element: A generalized atomic species (although different atoms of the same element are not always identical). Any atom represents (pertains to) some element, and to only one element.
An elemental substance consists of atoms of the same element, and doesn't contain atoms of other elements. It may be formed of free atoms (as for noble gases), molecules (as for $\ce{O2}$), or form a crystal. When a chemist tries to isolate an element, i. e. to gather its atoms in one place, an elemental substance results. Hence chemists use one word “element” with two slightly different denotations.
Again, the difference between abstract elements (atomic species) and elemental substances. When we ask: “of which elements does water consist?”, we assume species (hydrogen and oxygen, of whose atoms does the water consist), and don’t assume the water contains $\ce{H2}$ or $\ce{O2}$ (although it might, a slight quantity).
And how are atoms different from all this? An atom is a physical body. We can count them as things: There is one atom, two atoms, three atoms, etc.. How many atoms are in a water $\ce{H2O}$ molecule? There are three: Two atoms of hydrogen and one atom of oxygen. How many elements are in water (molecule)? Two elements: Hydrogen and oxygen.

Here is the IUPAC definition of an element:

https://doi.org/10.1351/goldbook.C01022

A species of atoms; all atoms with the same number of protons in the atomic nucleus.
A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.

Now that we have had evidence of atoms as nucleus plus electrons for over 100 years, there is no more need for the classic definition ("smallest particle, can't separate") that is now known not to be true if you allow for nuclear reactions.
Atom is a term at the particular scale, it refers to one particle. Element traditionally is a term at the macroscopic scale, it refers to the nature of a substance (either element or compound).
However, it is also used to refer to an atom-type, such as in the statement that "acetic acid contains oxygen", meaning that it contains oxygen atoms. Of course, acetic acid is not an element, and it is not a mixture of elements. In order to make it from the elements, you need a chemical reaction, so then the elements are no longer elements (but the oxygen is still an oxygen).
The confusion comes from the first part of the definition: An element is a species of atoms. We call those atoms that have 8 protons in their nucleus "oxygen atoms". It would be nice to have separate words for the particular and the macroscopic scale, but we have not.

An elements is a substance which is made out of atoms that all have the same proton number. So an element can consist of multiple atoms (like oxygen which goes around in pairs). An atom consists of a nucleus and some electrons.
Source: IGCSE Chemistry

Does water (H₂O) contain elements? Your source, apparently, defines an elemental substance, not an element in general.

The elements that make up water are hydrogen and oxygen. Hydrogen and oxygen are elements on the periodic table. The oxygen element goes around in pairs in air but can be broken to form other molecules and compounds.

An atom is a thing; an element is a type of thing.
An atom is a collection of protons, neutrons and electrons.  A single, isolated atom in its neutral state has some number of protons, the same number of electrons and some number neutrons (about as many as protons for the lighter elements, up to about 50% more, for heavier elements).  The number of neutrons or protons in an atom only changes as a result of radioactive processes or very high-energy interactions such as you get in particle accelerators.  And I mean really high energy: even if you think about blowing up sticks of dynamite, that isn't nearly enough energy to start messing with the protons and neutrons. Chemistry happens when atoms come together and share electrons or give electrons to each other.  Chemical reactions happen all the time and many of them don't need a whole lot of energy: moving electrons from atom to atom is often very easy.
So, the chemistry of an atom depends on the number of electrons and the number of electrons in an isolated atom depends directly on the number of protons. Electrons are so easy to add and remove from atoms (just rub a balloon against your hair: the static electricity is because you transferred electrons between your hair and the balloon) so we classify atoms according to the number of protons in them.  Neutrons aren't so relevant: I'll put a note about them at the end.
So, the element of an atom is determined by the number of protons.  All atoms of hydrogen have one proton, and all atoms with one proton are hydrogen. Two protons is helium, three is lithium, seventeen is chlorine, 79 is gold, etc.  A pure sample of an element contains only atoms of that type: for example, a pure sample of iron contains only atoms with 26 protons.  Water, on the other hand, is not an element: a molecule of water consists of two hydrogen atoms (one proton each) sharing electrons with an oxygen atom (eight protons).
Now, what does it mean to claim that an element "cannot be broken down into a simpler form," and why aren't atoms "a simpler form"?  Well, they're not a simpler form because an iron atom is iron: it's the same form, not simpler.  Think of it this way.  If I give you a lump of pure iron, all you can do is break it into smaller lumps of iron, or make it into a more complex substance, for example by letting it rust – rust is formed of iron and oxygen.  The smallest possible lump of iron you could make is a single iron atom, but that's still just an unbelievably tiny lump of iron. If you wanted to break down a lump of iron beyond single iron atoms, you'd need to use a nuclear reactor or a particle accelerator or something like that and then, finally, you'd be able to get something that wasn't iron, because you'd have changed the number of protons in the atoms.
Let's contrast that with water.  If I give you a bucket of pure water then, just like the lump of iron, you can divide it into smaller and smaller samples, ultimately ending up with a single water molecule.  But you can do something else: if you pass electricity through the water, it splits into pure hydrogen and pure oxygen.  These are "simpler" substances because each consists of atoms of just one element, whereas water has atoms of two elements.

What about neutrons? Well, in terms of chemistry, they don't do a whole lot and atoms that have the same number of protons but different numbers of neutrons are much more alike (they have essentially the same chemistry, for example) than atoms that have the same number of neutrons but different numbers of protons.  It makes much more sense to classify by the number of protons, since that determines the number of electrons and that determines the chemistry.
Suppose you tried instead to classify atoms according to the number of neutrons. Well, most argon atoms (18 protons) have 22 neutrons, but some chlorine atoms (17 protons) and a decent fraction of potassium atoms (19 protons) also have 22 neutrons. As you probably know, argon, chlorine and potassium are absolutely nothing like each other. On the other hand, potassium atoms with 22 neutrons behave almost identically to the most common kind of potassium atoms, which have 21 neutrons.

Did not downvote, but I think this answer is inaccurate, in the sense that it ignores molecular elements/allotropes, which can be broken down, changing the substance but not the element.

@MarchHo That's a good point, though I'd describe it as incomplete, rather than inaccurate.

I didn't downvote either, but in your first paragraph, "approximately the same number of neutrons" is flat-out wrong for the heavier elements.  I'd say "some number of protons, some (usually a little larger) number of neutrons, and the same number of electrons as protons".

Is element “a form of matter”? If it is, then which namely, and is it a substance of something else? You made a lengthy lecture on neutrons and radioactivity, but didn’t address these questions.

After going through all the answers, I think things are not very well defined. May be this will add some clarity to distinction between atom and element, and between molecule and compound:
$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$
In the above reaction, two atoms of element copper combine with one molecule of compound oxygen to give two molecules of compound cupric oxide.
We cannot write it in other way round as: … two elements of atom copper combine with…


text:
Why is sodium sulfate a neutral salt
When I have a solution of $\ce{Na2SO4}$, the sodium sulfate should ionize:
$$\ce{Na2SO4(aq) -> 2Na+(aq) + SO4^{2-}(aq)}$$
In my mind, the sodium ions should just float around, but because $\ce{HSO4-}$ is a weak acid:
$$\ce{HSO4- (aq) + H2O <-> H3O+ (aq) + SO4^2- (aq)}$$
The sulfate ion should be a weak conjugate base, and hence the sodium sulfate salt should be a weak base. This was my line of reasoning, until today my friends told me it is actually a neutral salt, which was confirmed by searching online. So why is it a neutral salt? (i.e. what's wrong with my reasoning?)

$\mathrm{HSO}_4^-$ can hardly be called weak. It is an acid of medium strength, stronger than carbonic or acetic.

Weak ions conjugate bases do not accept protons.

Sulfate anions are weakly basic according to Bronsted theory - you're right, but it's called neutral salt according to Arrhenius theory in which it would need to contain OH- ions to be called basic.

The sulfate ion is indeed a weak base, so weak that it hardly participates in the equilibrium reaction you have written.

For a solution of $\ce{Na2SO4}$, the sodium sulfate will ionize as you indicated:

$\ce{Na2SO4(s) ->[aq] 2Na+(aq) + SO4^{2-}(aq)}$

This reaction is essentially irreversible in that the ions in an unsaturated solution won't suddenly form solid $\ce{Na2SO4}$ again.
$\ce{HSO4-}$ is a not a weak acid, but has a $\text{pK}_\text{a2} = 1.99$ which makes a moderately strong acid. So for  pH > 3 essentially all the $\ce{HSO4-}$ will ionize:

$\ce{HSO4-(aq) + H2O ->[pH > 3] H3O+ (aq) + SO4^2- (aq)}$

Thus $\ce{SO4^{2-}}$ is such a weak base that reverse reaction doesn't happen when you dissolve $\ce{Na2SO4}$, and a solution of $\ce{Na2SO4}$ in distilled water stays neutral rather than becoming basic.

$\ce{\require{cancel}H2O + SO4^2- \cancel{->} HSO4- + OH-}$

To think about a different example acetic acid is a weaker acid, $\text{pK}_\text{a} = 4.76$, so the acetate anion is a stronger base than the sulfate anion. A solution of sodium acetate would become weakly basic.
One last point - the statement should be "sodium sulfate a neutral salt in aqueous solution." In other solvents sodium sulfate could act as a base.

Here is the somewhat weak explanation given by  Wikipedia:

Sodium sulfate is a neutral salt: its aqueous solutions exhibit a pH
of 7. The neutrality of such solutions reflects the fact that sulfate
is derived, formally, from the strong acid sulfuric acid. Furthermore,
the Na+ ion, with only a single positive charge, only weakly polarizes
its water ligands provided there are metal ions in solution. Sodium sulfate reacts with sulfuric acid to give the acid salt sodium bisulfate.

It definitely leaves some unanswered questions like what they mean by "metal ions in solutions": what metal ions (sodium ion?), what concentrations?
But the gist of it makes sense and makes a reasonable answer to me.


text:
How to find the number of stereoisomers of a tetra-substituted allenes, exhibiting both axial and central chirality?
I need a reliable method for determining the number of diastereomers and enantiomeric pairs of the following series of compounds bearing in mind that the substituent groups A are chiral on their own (exhibiting central chirality).

There are two planes of interest bisecting the allene sistem, the one constains the left A groups, the other is orthogonal to it. If two A groups are to be reflected through one plane (so it becomes a symmetry plane) their configuration must be different. That is all regarding the central chirality, but I don't know how to include axial chirality in this description (so I don't count one stereoisomer twice).
I think that these are the four diastereomers. Instead of A, I wrote the configuration of A (R/S). Now, as R and S are connected with a mirror plane, diastereomers 2 and 4 shouldn't exhibit axial chirality, as R and S can be regarded as same substituents on one side of the allene system (because of the mirror plane). That's why the third diastereomer exhibits axial chirality (P/M). I am not sure whether my reasoning is correct. The rest beats me. The answer should be 4 diastereomers and 3 enantiomeric pairs (last page), but how?

The groups A may differ in, and only in, stereo configuration ‘R’ or ‘S’?

The only method that I am aware of is to draw and compare the various possibilities.

Axial and planar are just adjectives we can use to describe what type of chirality a molecule possesses.  Usually, just saying the molecule is chiral is good enough.  Being planar or axially chiral does not create any new enantiomers or diastereomers compared to if we just said the molecule is chiral.
Let's analyze your problem using standard methods - no special techniques just because the molecule happens to be axially chiral.
I get the same 4 diastereomers that you've drawn above (good job!).  When analyzing allene stereochemistry I've always found it convenient to analyze them using Newman projections.  I've redrawn the 4 diastereomers below using Newman projections.

In the top diastereomer, all 4 chiral substituents are the same; the molecule has a non-superimposable mirror image (enantiomer).  Therefore this diastereomer is chiral.
In the second diastereomer, 3 chiral substituents are the same, one is different; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.
In the third diastereomer, two chiral substituents with different chirality are attached to the same carbon; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.
In the fourth diastereomer, two chiral substituents with the same chirality are attached to the same carbon.  This diastereomer belongs to the interesting point group $\ce{S_4}$.  I say interesting because this point group does not contain a $\ce{\sigma}$ plane, but is still nondissymmetric (achiral).  Look at my Newman projection for this diastereomer and picture it lying behind the plane of the screen.  Reflect it in the plane of the screen and then rotate it 90° (n=4, 360/4 = 90°) and you have recreated the original molecule. This molecule is achiral (all molecules belong to point group $\mathrm{S}_n$ [n is an even integer] are achiral).
So what do we wind up with?  4 diastereomers, 3 of which are chiral (enantiomeric pairs) and one is achiral.

Now when I think, P or M couldn't even be assigned to thee molecules.

Maybe it could be assigned for the third diastereomer only (R has higher priority than S). So the third diastereomer is the only axially chiral one, but it is complement to the central chirality (the same pair of enantiomers arises from both)?

Definition: "An axis about which a set of ligands is held so that it results in a spatial arrangement which is not superposable on its mirror image" - therefore each of the 3 chiral allenes is axially chiral.

Now I got confused. Firstly, there is no way to assign P or M notation for the other diastereomers. Secondly, The allene in my second question gave rise to a diastereomer by axial chirallity, while here it is not the case.

See the example here  The front is always ranked 1,2, the back is always ranked 3,4.  Note in the example Cl > OH, but still Cl = 3 because it is in back.

But if on the same side of the allene system are two identical groups with identical configurations, none has higher priority.

Maybe you are correct, or maybe (?) there is a rule that "top" or "right" R takes priority over bottom of left R, I dunno.

That would be reasonable. It seems unreasonable trying to divide central and axial chirality in this case, but there is the other allene in my second question where two of the diastereomers are connected by axial chirality only.

I am wrapping my head around this for quite a while now and I am certainly not sure about anything I am going to state now. I don't think that the (R,R,R,R) and (R,R,R,S) compounds and their enantiomers exhibit axial chirality. The reason why they cannot be superimposed to their mirror image comes from the fact, that you cannot superimpose R on S. According to the goldbook definition they are chiral, but I think this was written with the background of having achiral moieties attached to the allene backbone.

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


text:
Titration of H3PO4 with NaOH
I came across an ionic equilibrium problem stating:

Find the pH when 150 ml 1 M $\ce{NaOH}$ has been added to 100 ml 1 M $\ce{H3PO4}$.

I'm stuck with this question. What I know:
$$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$$
$$\ce{NaH2PO4 + NaOH -> Na2HPO4 + H2O}$$
Finally 50 millimoles $\ce{Na2HPO4}$ and 50 millimoles $\ce{NaH2PO4}$ remains.
Should I apply acid buffer equation?

The answer to your question does not need a single calculation.
You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.
Now what is between those two volumes, at your desired volume of 150 mL?
A simple buffer system that can be described using the Henderson–Hasselbalch equation:
$$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$
As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).
$\hskip3.5cm$

This is a simple question  involving a mixture of strong acid (not that strong) and a strong base. Since $k_a$ values are not given the book might have wanted you to solve it like this. They want you to assume complete dissociation of both compounds. That is, 5 moles of $\ce{NaOH}$ gives 5 moles of $\ce{OH^-}$ ions after complete dissociation.
Find the equivalents of $\ce{H^+}$ from phosphoric acid (a) and the equivalents of $\ce{OH^-}$ from $\ce{NaOH}$ (b). Also calculate the total volume. Number of equivalents=$NV$. But be careful since concentration is given in molarity.
If $b>a$, then the resulting solution is basic. Calculate the equivalents of $b$ that remain. From this you will get pOH.
$$[\ce{OH^-}]=\frac{b-a}{V_1+V_2}$$
If $a>b$, then resulting solution is acidic. Calculate the amount of $a$ that remains. From this, you will get pH.
$$[\ce{H^+}]=\frac{{a-b}}{V_1+V_2}$$
Where N is normality and equivalents of acid is greater than equivalents of base.


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Are bisulfates and bicarbonates soluble in water?
Generally, are bisulfates and bicarbonates soluble in water?
Specifically, are calcium hydrogen carbonate and magnesium hydrogen carbonate soluble in water?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Other guys: This isn't necessarily a homework question.

According to it's Wikipedia page, calcium bicarbonate is not a known solid compound, but it does exist in solution, presumably prepared by dissolving calcium carbonate and bubbling carbon dioxide through the solution until bicarbonate dominates the following pH-dependent disproportionation:
$$\ce{2HCO3- <=> H2O + CO2 + CO3^2-}$$
It apparently has a solubility of 16.6 g per 100 mL of water at room temperature. which is about a 1 M solution.
Magnesium bicarbonate similarly exists only in aqueous solution, though it is much less soluble at 0.077 g / 100 mL.
As solids, these salts probably fall victim to the following equilibrium( $\ce{M}$ is a generic divalent metal:
$$\ce{M(HCO3)2 <=> H2O + MCO3 + CO2(g)}$$
The carbon dioxide escapes, preventing the system from ever staying at equilibrium.


text:
Calculating change in enthalpy of combustion of 1.12g of hexane
I will calculate the change in enthalpy of combustion of 1.12g of hexane.
1) Calculate the energy transferred to 200g of water using the equation: J = mass of water in grams (200g) x specific heat capacity of water (4.18 J/gK) x temperature increase in Kelvin (24C = 297.15K). This results in the amount of energy transferred to the water of a total of 248,417.4J or 248.4kJ.
2) Calculate number of moles of hexane burnt using n=m/M. Weight of hexane before experiment = 222.07g. Weight of hexane after experiment = 220.92g. Therefore 1.15g of hexane was burnt during the experiment. 1.15g of hexane = 0.013mol of hexane.
3) Find enthalpy change of reaction. Combustion of 1mol of hexane produces (by my calculations) -4194 kJ/mol.  Therefore, 0.013mol will produce (0.013/-4194) -3.1x10-6 kJ/mol.
So this is the procedure that I used to find out the answer! Was it correct?
Because, apparently, you can use the values of steps 1 and 2 to find out the enthalpy change?

When you convert °C to K then only add 273.15 when you have absolute values but not a difference (like in this case), because they have the same scale.

I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:
1.) 0.2 kg Water was icreased by 24 °C:
U = 0.2 kg * 4187 J/(kgK) * 24 K = 20,100 J = 20.100 kJ
2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol
3.) 1.15 g caused the temperature increase of 24 °C:
20.100 kJ / 0.013 mol = 1,510 kJ/mol
So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane.

Ah so you find the amount of energy transferred to the water (20.1 kJ) and divide this by the number of moles of fuel burnt (0.013 mol) to obtain the change in enthalpy of the exothermic reaction (1,510 kJ)!
Thanks a lot!
One thing that is throwing me, however, is that the paper tells me that the published enthalpy changes of combustion for hexane is -4163 kJ/mol, so how could we get -1,510 kJ from burning 0.013 mol? @AstronAUT

That is indeed a good question. If this was a real experiment i'd say it could be because of some energy being lost into the environment. That would mean your efficiency factor is around: 1,510 kJ / 4163 kJ = 36 %. Other factors that could have lead to an error are incomplete burning of the haxane or the fact that you are extrapolating your results from a rather small test amount of hexane (1.15 g), therefore also small errors have an bigger impact.

Ahh okay, I reckon it was simply down to not measuring the mass of the hexane immediately after burning and the small amount burnt. Thanks a lot for your help!


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Is there a generic graphical language to represent chemical reactions?
I want to show a schematic representation of chemical reactions from start to finish in a way that will include more detailed preparation instructions than a plain chemical formula: volumes, concentrations, waiting time, temperature, etc.
Is there a graphical language for this kind of thing? I'm looking for something more sophisticated than a typical flow chart, and more targeted towards chemical reactions.

Are you looking for something like a chemistry-UML for diagramming laboratory reactions? If so, I had dug up this in response to a query a few months ago.

@ToddMinehardt yes, kinda. That software is more Visio-like and focuses too much on graphics I don't really need. More UML-like is better.

I poked around. You might find something of use relating to process flow diagramming; or from the open source diagramming tool Dia, which apparently has support for PFD; or look at Chemical Markup Language.


text:
Titration of alpha-amino acid with strong base
Question:
$\mathrm{50~ml}$ of the conjugate acid $\ce{H3N+CH(R)COOH}$ of an $\alpha$-amino acid required $\mathrm{30~ml}$ of $\mathrm{0.1~M}~\ce{NaOH}$ to completely convert into the dipolar ion $\ce{H3N+CH(R)COO-}$. The pH of the solution was 6.4 during the titration with $\mathrm{15~ml}$ of $\mathrm{0.1~M}~\ce{NaOH}$ and 7.8 at the isoelectric point. Calculate the dissociation constants for the amino acid.
Attempt:
Reactions involved:
$$\ce{H3N+CH(R)COOH <=> H3N+CH(R)COO- + H+}$$
$$\ce{H3N+CH(R)COOH + OH- -> H3N+CH(R)COO- + H2O}$$
At the isoelectric point, $\mathrm{pH}=\frac{\mathrm pK_{\mathrm a1}+\mathrm pK_{\mathrm a2}}{2}$.
Initially, number of milli moles of conjugate acid = number of milli moles of NaOH = $\mathrm{30\times 0.1=3~mmol}$ of salt during titration (neglecting the amount produced by acid) = 1.5
A buffer forms during titration. My doubts is, does the concentration of acid which was present initially, change while adding NaOH? Using Henderson equation will give pH of Buffer.
I know the concentration of salt and pH. But I am stuck with the concentration of acid. 1.5 milli moles of acid will react to form salt. If I use 1.5,I get wrong answer. If I use 3, I get right answer.

I improved the formatting of your post using the \ce command. For more information on how to do this yourself please see here and here.


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Water Droplet permeability through Graphene Oxide

TL;DR: The above image is an over simplified model to my question begging to know whether the permeation behavior exhibited by water droplets through graphene oxide is the top image or the bottom.

I couldn't find information about this anywhere:
It has been confirmed that water vapor and pools of liquid water can pass through graphene oxide films (prepared for stability in water on aluminium substrates) assuming the interlayer distances and quantity of layers are appropriate for optimal passage.
However, I do not know whether or not a droplet of water (placed on one side of graphene oxide) will permeate through the graphene oxide film and emerge intact on the other side of the film. I could simply confirm this by spraying water onto graphene oxide but I cannot make a legitimate film on my own just yet is why I ask.
Basically, stated differently, I don't want to end up spraying graphene oxide with water droplets on one side and end up with a stream of water on the other side; they are to ideally remain as droplets.
The question is very simple but also very important in determining to what extent G.O. can be utilized. There are also different studies that conclude different permeation mechanisms for water through G.O. so I can't even reason the answer out on my own. Quite maddening.
Thank you all and sorry for the pure info. extraction but it's for the greater good!
EDIT1: Oh, the preferred droplet max diameter would be 100 micrometers.
EDIT2: Thanks for the reads; I'll try and find the best method for making graphene alumina oxide to try and find the answer on my own. If anyone can point me to the best method meanwhile, that would also be great. Thanks again

You may want to clarify the "emerge intact on the other side of the film" part. The situation seems to be VERY dependent on experimental conditions, type of graphene oxide, actual amount of water, etc. GO seems to be an umbrella terms for a diverse set of materials, with different edge structure, oxygen content, holes on the sheet etc.  Also, if a droplet is stable on the surface, why would lay out as a water film? Or the other way, if the surface is wetable, and water spreads out as film, why would you have separate droplets at the first place?

"Emerge intact on the other side of the film"
---- Hi! By this I'll use an example of water sprayed downward and centered onto a napkin versus onto graphene oxide:

For the napkin, continued spraying makes the whole napkin wet until eventually water gathers and drips at the center. Individual droplet paths cannot be traced because of the merging.
For graphene oxide, will the same occur, or will each droplet 'emerge intact on the other side of the [graphene oxide] film'?

"...type of graphene oxide"
----Stability of Graphene Oxide in water has been attributed to aluminium filters:

"During filtration, the pH value of the graphene oxide solution is typically around 3 to 5. In this range of pH values, AAO corrodes to release Al3+ cations, which were incorporated into the graphene oxide membranes. These trivalent aluminum cations act as crosslinkers, holding the negatively charged graphene oxide sheets together in water."
-http://bit.ly/1HExnui

"if the surface is wetable, and water spreads out as film, why would you have separate droplets at the first place?"

---- Well I don't know whether or not the water spreads out. The primary goal for me is just to know the behavior of water as it passes through graphene oxide. Everyone knows vapor passes as well as pools of water, but what is not explained is the behavior of water droplets passing through G.O.
Thanks for reading this essay~

(you may want include relevant parts in the question to improve it and help others, instead of answering me the questions)

My answers to you are just me repeating myself but I've included an image for clarification. Thanks for being the first to actually respond!!!


text:
Bond Dissociation vs Hess' Law Enthalpy
I am trying to determine two of the theoretical values for the enthalpy of the combustion of methanol.
I have calculated (using Hess' law) that the $ \Delta H_{\mathrm {reaction}}\; = -1453.418~\mathrm{kJ~mol^{-1}} $, which is correct according to several sources. However, when I tried to calculate the $ \Delta H_{\mathrm{reaction}} $ using the bond dissociation and formation method: $ \Delta H_{\mathrm{reaction}}\; =\; \sum{\Delta H_{\mathrm{bonds\; broken}}}\;-\; \sum{\Delta H_{\mathrm{bonds\; formed}}} $
I got a value of $\mathrm{-464.9~kJ~mol^{-1}}$. I am just wondering if this is correct, or if I have made a mistake somewhere?
Ps: My balanced equation looks like this: $$\ce{2CH3OH + 3O2 -> 4H2O + 2CO2}$$
I used this table to get the known values for the various bond energies.
Thanks

Could you show us your actual calculations? Otherwise it is difficult to see where the mistake is.

Could the discrepancy be partly due to the fact that the bond energies in your table are given in kcal per mole?


text:
Dissolve/alter cellulose properties for printed circuit boards
I'm trying to find any readily available chemical/substance (i.e. household, hardware store etc.) or combination of such that breaks down paper/cellulose.
I'm a home electronics hobbyist and am using my laser printer to print circuit board layouts.  I place the paper on the blank copper board and run it through a laminate machine (high temp & pressure) to 'melt' the laser toner to the copper.  I then soak the board and paper in water to try and slush the paper free of the toner, after which I can put the board in acid and remove the exposed copper and leave my circuit pattern.
This actually works pretty well except that releasing the paper from the toner is maybe only 95% efficient/reliable.  Sometimes the toner gets pulled off the copper and/or fibers/slag from the paper remain on the board.
From doing a little research on the internet, it would seem that dissolving/breaking down cellulose is not an easy process and is quite involved for commercial applications.  However, for my purposes I don't think I need some 'perfect' solution.  If I can just get the paper to soften, or partially break down or just change its properties a bit so that it will let go of the toner more efficiently, then that would probably be sufficient for my needs.
Any ideas?

Schweizer's reagent is known to dissolve cellulose. Not sure whether it will be good for the toner and board, though.

@IvanNeretin It  will dissolve copper on prolonged contact.

Timber is hydrolyzed industrially in 0.5% sulfuric acid at 180-200 Celsium under pressure. My guess, boiling with diluted  (say, 10%) sulfuric, or better hydrochloric* acid for 10-12 hours should work. BUT! Be aware that prolonged contact with aerated acidic solution may dissolve some copper as well. || * common paper often contains a bit of calcium carbonate, which will not readily dissolve in sulfuric acid, but will do so in hydrochloric acid.

Many thanks Ivan & Permeakra!  Schweizer's reagent sounds like a possibility.  The sulfuric and/or hydrochloric acid for 10-20 hours would be impractical for my purposes.  I seem to be frustratingly close to being able to make fairly intricate, clean, neat and professional looking circuit boards simply and cheaply to satisfy my inner electronics inventor!  Just getting the toner to release from the paper is proving to be the spoiler.

Oops!  sent above reply prematurely.  Where might I be able to acquire Schweizer's regent?  I'm not a chemist but the Wikipedia description leads me to believe that this may not be all that stable and has little shelf life and that it needs to be created and used all in one go?  Also, how toxic and/or what special handling precautions must be considered?


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Why is a bridgehead atom bearing a negative or a positive charge unstable?
Why is a bridgehead atom bearing a negative or a positive charge unstable? Is it something with the structure of bridge bonds? I am unable to understand why.

Since the answer by Aditya Dev has already dealt with the carbocation part, I'll write about the carbanion one. The OP states that:

Why is a bridgehead atom bearing a negative or a positive charge unstable?

This is partially incorrect. Carbanions in bridgehead position are infact stable. From March Advanced Organic Chemistry, 6th ed, Chapter 5, Carbanions:

Carbons at bridgeheads, although extremely reluctant to undergo reactions in which they must be converted to carbocations, undergo with ease reactions in which they must be carbanions and stable bridgehead carbanions are known.

(my emphasis)
It also says that:

For other evidence that carbanions are pyramidal, see Streitwieser, Jr., A.; Young, W.R. J. Am. Chem. Soc. 1969, 91, 529; Peoples, P.R.; Grutzner, J.B. J. Am. Chem. Soc. 1980, 102, 4709.

This makes the picture completely clear. Bridgehead carbanions are not planar, but in fact, pyramidal. So, they are definitely  not sp2 and hence, the Bredt rule isn't violated by their being stable.

The hybridisation of a bridgehead carbon when its not charged is sp3. Now, if you remove a hydrogen atom, its hybridisation becomes sp2.
Normally, in sp2 hybridisation, the 3 hybridised orbitals lie in a single plane for minimised repulsions. But the bridgehead carbon with positive charge would not have hybridised orbitals in a single plane. This increases repulsion and makes the compound unstable.
Here is a link - Bredt's Rule - that explains the same thing, but instead of a positively charged bridgehead, they discuss a double bond on the bridgehead atom.


text:
How to use this salt analysis chart to determine the cations present in a given solution?

With the only possibilities being $\ce{Ba^2+, Ca^2+, Cu^2+, Fe^2+, Fe^3+, Pb^2+}$, what cations are present in a solution which:
a) Forms a white precipitate with sulfate and with fluoride, no precipitate with chloride, and a blue precipitate with hydroxide which dissolves in ammonia solution

I have the answer with me, but I am not too sure how to use the above chart for determining cations in a solution, to answer the question. I can sort of make it out, but I am not completely sure.

What is your thinking?  Please include that detail into your question

@santiago yeah i'm not really sure..I can see the precipitates that form as mentioned in the question but I'm having trouble linking them all together...I think I'm over complicating this in a way or another

For this question rather than reading the question and trying to understand what is going on all in one go, I suggest just work your way through the flow chart and refer to the question as you go along so that you don't overload your brain with excess information and confuse yourself.
So starting from the top, it asks if a precipitate will form if $\ce{HCl}$ is added. Looking from the information in the question, there is no precipitate formed. Therefore we can rule out $\ce{Pb^2+}$ being in the solution. Now following the steps in the flow chart, it asks if a precipitate forms if sulfuric acid is added to it. Looking from the data in the question, it says that a white precipitate is formed with sulfate. Therefore, we follow the next step which asks if a precipitate is formed when $\ce{NaF}$ is added. In the question it states that a white precipitate forms with fluoride. Therefore we can conclude that $\ce{Ca^2+}$ ions are in the solution.
However, if we look again into the question, it states that a blue precipitate also forms when mixed with hydroxide ions. Therefore, if we look back to the flow chart regarding the section on testing with $\ce{NaOH}$, it can be seen that traces of $\ce{Cu^2+}$ ions also exist in the solution. Therefore, both $\ce{Ca^2+}$ and $\ce{Cu^2+}$ cations are in the solution.
As you can see, once you know the general method, it is really quite simple and self-explanatory.

The blue precipitate from the hydroxide suggests the presence of Cu2+ as well

Yep, you right, I didn't read the question properly. I didn't read 'cations'.

No problems, edit it in to your answer if you'd like.

There could also be some $\ce{Ba^2+}$ mixed in; the chart cannot distinguish $\ce{Ba^2+ + Ca^2+}$ from just $\ce{Ca^2+}$.


text:
How to determine the number of stereoisomers of a di-substituted allene, exhibiting both axial and central chirality?
What are the diastereomers and enantiomeric pairs of the compound below where A are chiral groups (central chirality)? There should be 3 diastereomers and 3 enantiomeric pairs (last page).

I previously asked a question about tetra-substituted allenes.

Have you tried to apply the Newman projection method I described in your previous question? If so, what problem do you run into?  Can you draw the the diastereomers you came up with?  Which ones are chiral?

Can't find three of them. I got the R,R (S,S), R,S (S,R), those are just two.

There are 2 R,R diastereomers, both chiral and 1 chiral R,S diastereomer - does that help?

Are the two R,R isomers are connected through axial chirality?

yes, sighting down the allene (Newman) one R,R diastereomer has front R at 0° and back R at 270°; other R,R diastereomer has front R at 0° and back R at 90°

Okay, I've got it, but that returns me to my previous question. Could you check my comment there?

These are the three diastereomers with their enantiomers on the right side. The top and middle diastereomers are connected via axial chirality (R and S are used to mark the configuration of A groups).

Nice work, but all 3 are axially chiral.

I'm not that good at this but here is something to work on. Each row is a pair of mirror images. Grey structures in 3rd row are identical to the respective 2nd row ones.


text:
Resources where I can find approachable applied literature to understand general chemistry topics?
I am trying to find resources where I can learn about general chemistry topics outside the scope of a textbook, preferably in primary literature. Note, I have read through a general chemistry text and finished studying topics in the scope of a general chemistry course, so the informational foundation is there. My ideal "resource" is a place where I can find something akin to a 1-5 page experimental setup or design, or maybe a discussion, in which these concepts are applied and I'm forced to think about relationships in general chemistry, or what happens when things are changed. For example, I'd love to see some paper where I can see the ideal gas law applied as an approximation in the lungs or in deep sea divers, or some maybe some experiment where ion solubility is discussed in the scope of "hard water" in pipes. I know these applications exist, but I want to read about them on the level that makes me think about general chemistry concepts.
I can use VPN connections to login as a institution from my school and access some primary literature. However, sometimes I can't get full access becaue my institution is small and does not a subscription to a journal. Moreover, sometimes this primary literature is too complex in explanation, too longwinded, and goes beyond the scope of a general chemistry student. Now, I still want to think critically, but most of these papers include applications in engineering and use quantum physical assumptions beyond the scope of wave function solutions and electron configuration in general chemistry courses. My background includes a year of cell and molecular biology, an understanding of physics I and II, and math a little past integral calculus. I'd like the mathematics to be minimized, however, as I want concept.

You mean books on general chemistry? If that's the case, Atkins chemical principles is a really good book.

No, I don't mean general chemistry books. I've already read a general chemistry book. I want to read content that assumes you know general chemistry, but forces you to think critically about its applications as it pertains to the real world, experiments, etc. Something like a paper about gas laws in breathing, or ion solubility and treatment of hard water pipes.

While not perfect for what you looking for, the ACS Journal of Chemical Education often has articles directed towards applying basic chemical principles to real world situations. The articles are more geared towards educators, but I think it could be useful to you.


text:
Calculating the volume of 2 M solution needed to make a 0.1 M solution
I need help with a basic concentration problem.
If I have a $\mathrm{2~M}$ solution, and want to make a $\mathrm{0.1~M}$ solution of $\mathrm{100~ml}$, how would I go about that?
My work:
$$M_1V_1 = M_2V_2$$
Where $M_1 = \text{initial concentration}$, $M_2 = \text{final concentration}$, $V_1 = \text{volume needed from initial}$, $V_2 = \text{volume needed from final}$
$$2~M \times V_1 = \pu{0.1 M \times 100 ml}$$
$$V_1 = \frac{\pu{0.01 M\cdot L}}{\pu{2~M}} = \pu{5 ml}?$$
Therefore I need $\pu{5 ml}$ of my $\pu{2 M}$ solution, increased to $\pu{100 ml}$ to make a $\pu{0.1 M}$ solution of$\pu{100 ml}$?

Please note that I have amended your post in one area where the units were incorrect; and I have explicitly used liters, because molarity is in units of moles per liter.

If you have a simple solution of say $\pu{0.1 M}$ sodium chloride, you have $\pu{0.1M}\, \ce{NaCl}$ in $\pu{100 ml}$.
Which is equivalent to "$\pu{100 mL}$ of $\pu{0.1 mol}\,\ce{NaCl}$ per litre", or,
$$n_\ce{NaCl} =\frac{100}{1000} \times 0.1 = \pu{0.01 mol}$$
$\pu{5 mL}$ of $\pu{2 M}\ \ce{NaCl}$ is equivalent to "$\pu{5 mL}$ of $\pu{2 mol}$ $\ce{NaCl}$ per litre", or,
$$n_\ce{NaCl} = \frac{5}{1000} \times 2 = \pu{0.01 mol}$$
So congratulations, you did it right. If ever you aren't sure, just write out all the steps and follow it through.
You could also look at it like this - $\pu{2 M}$ is $20\times$ more concentrated than $0.1\times\frac{2}{0.1}$, so you expect to need $20\times$ less of the $\pu{2 M}$ than the $\pu{0.1 M}$.
$\pu{5 ml}$ is $20\times$ less than $\pu{100 mL}$.


text:
Propose a reaction mechanism that explains the rate law for the formation of dinitrogen pentoxide
The reaction is:
$$\ce{NO(g) + NO2(g) + O2(g) -> N2O5(g)}$$
From the experimental data, I have found the rate law to be: $$\ce{Rate~=~k[NO][NO2]}$$
This means the overall reaction does not depend on $\ce{O2(g)}$.
Can someone suggest how I go about building the elementary steps to form the reaction mechanism?

First step - the rate determining step or the slow step (activation energy is high).
$$\ce{NO(g) + NO2(g) -> N2O3(g)}$$
2nd step - the fast step.
$$\ce{N2O3(g) + O2(g) -> N2O5(g)}$$
Since the rate determining step does not include $\ce{O2(g)}$, the rate is independent of the initial $\ce{O2(g)}$ concentration.


text:
Charles's Law confusion
Charles's Law states that for every 1 degree Celsius increase in temperature of a gas, the volume increases by 1/273 of the original volume (at constant pressure).
What I am wondering is, doesn't gas have an indefinite volume, therefore it is not possible for it to have volume at all? Am I missing something here?

No it doesn’t. A certain amount of gas (counting molecules/atoms) will always occupy a specific volume at a certain pressure. Otherwise Earth’s atmosphere would extend into infinity, too, wouldn’t it?

A gas occupies the volume of whatever the container it is in. Take a balloon filled with Helium for example. Now the volume of Helium gas is the volume of the ballon.
If you increase the temperature by 1 °C at constant pressure the volume of the balloon increase by 1/273th of the original volume of the balloon. Hope this might clear things up.

Yes, but what if it was in a hard container, such as a glass jar? It wouldn't be able to expand 1/273 of the volume. (What I'm meaning to say is, if a balloon is filled up with gas that matches the volume of a glass jar, and the same glass jar is filled with the same amount of gas (in mass) than when the temperature rises by 1 degree Celsius, the glass jar couldn't expand to fit the gas.)

If it is a rigid container then the pressure rises by a factor of 1/273. According to my knowledge it is impossible to rise the temperature by keeping both pressure and volume constant.

That makes sense. But in my textbook it speaks as if the volume must be increased

The glass jar your are referring to, may blast due to the increase in volume. But I assume that is not the case.

Yes, so is this a limitation to Charles's law? The pressure may increase but does the Volume HAVE to increase? It is the same case in balloons, surely the balloon limits the increase of volume at least somewhat?

The 3 physical quantities are related by the equation PV/T = k where k is a constant. And yes , the balloon may blast upon heating under constant pressure due to the increase in volume.

Charles law is for gases at constant presure which in the real world requires a vessel for containment and thus the gas would occupy the volume of its conatiner.
$$ \text{Charles Law:}\qquad \frac{V_1}{T_1} = \frac{V_2}{T_2} \implies \frac{T_2}{T_1} = \frac{V_2}{V_1 } \qquad{P_1 = P_2}$$
For Example: for $T_1 = \pu{273K}$ and $T_2 = \pu{274K}$:
$$\frac{T_2}{T_1} = \frac{\pu{274K}}{\pu{273K}} = \frac{V_2}{V_1} = \frac{274\times  V_1}{273\times V_1} = \left(1 + \frac1{273}\right)V_1$$
Your statement that a 1 degree Celcius increase in temperature results in a $\frac1{273}$ increase in volume is only true at $\pu{0^\circ C}$ or rather $\pu{273K}$. A $\pu{1^\circ C}$ increase at $\pu{1^\circ C}$ or $\pu{274 K}$ results in a $\frac1{274}$ expansion of the gas. More generally:
$$\color{blue}{1 + \frac{\Delta T}{T_1}} = \frac{T_1 +\Delta T}{T_1} = \frac{V_1+\Delta V} {V_1} = \color{blue}{1 + \frac{\Delta V}{V_1}}\\
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\color{blue}{1 + \frac{\Delta T}{T_1}  = 1 + \frac{\Delta V}{V_1}} \implies \frac{\Delta T}{T_1}  = \frac{\Delta V}{V_1}\implies \Delta V = \left(\frac{\Delta T}{T_1}\right)V_1$$
Thus $\Delta V$ is dependent on $T_1$ for a given temperature change:
$$\Delta V = \left(\frac{\Delta T}{T_1}\right)V_1$$

Simply Charle's Law(change in volume is directly proportional to change in temperature) is true with the assumption of constant pressure.