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Does caffeine isolated from coffee contain harmane?
I'm not sure whether the caffeine contained in otc caffeine pills is made synthetically or isolated from coffee and other plants, but if the latter is the case, is it likely that it contains the chemical harmane?
It's claimed that isolated caffeine is 99% pure and I'm wondering whether the remaining 1% is likely to contain any appreciable amount of harmane.

Given that caffeine can be readily extracted from a variety of abundant and exceedingly cheap natural sources, I doubt it would be economical to synthesize. Various decaffeinated products are popular, and the caffeine removed during the decaffeination process is normally recovered, purified, and re-sold for use in various products, which makes the synthetic route even less economically attractive. As far as I'm aware, the vast majority of caffeine sold commercially is extracted from plant sources. You can do your own research to determine the exact statistics, but since your concern relates to caffeine extracted from natural sources that might also contain harmala alkaloids, I'll address that case specifically.
There are a number of different methods of caffeine extraction (see the Wikipedia page on decaffeination), but all use solvents and/or adsorbents at some stage in the separation and purification processes. From the perspective of chemistry, then, the question is how similar are the solubilities of caffeine and harmala alkaloids, and what percentage by mass do they constitute of the natural caffeine sources used? I was able to find one article that quantified the chemical constituents of coffee beans. If you examine the data table therein, you'll see that the main constituents of those beans are carbohydrates, lipids, amino acids, and chlorogenic acids, not one of which is a category that applies to the harmala alkaloids. Lipids, carbohydrates, and amino acids are generally the main constituents of virtually all plants, so I think it's probably safe to extend those statistics to other natural sources besides coffee beans. Purely as a matter of probability, it's reasonable to assume the impurities are likely to consist mostly of substances from those categories. With respect to solubility, the harmala alkaloids are mostly somewhat heavier nitrogenous bases which are less polar than the caffeine molecule (at least in their neutral, "free base" form). Given that, I'd expect that methods using water extraction, which appear to be popular, would separate caffeine from harmala alkaloids fairly efficiently (I found one paper indicating that harmala alkaloids are water insoluble, while caffeine is anywhere from moderately to highly soluble depending on temperature).
An impurity of 1% is already quite small, and that 1% comprises a variety of compounds, both organic and inorganic (the overwhelming bulk of which are likely not harmala alkaloids, given the natural composition of caffeine-containing plant sources I described above). Based on the above information, I'd suspect that only a tiny fraction of those impurities, if any, would be likely to consist of harmala alkaloids at a detectable level.

Thanks for the detailed answer. Here is a study about harmane in brewed coffee. Maybe it isn't present in raw coffee beans? I'm also not sure whether caffeine is isolated from raw or roasted beans, but I will check out your Wikipedia link to decaffeination. http://www.ncbi.nlm.nih.gov/m/pubmed/15764332/

@MorseBott, yes, I did some more searching and found another article [http://www.hindawi.com/journals/jt/2011/628151/#B13] with some primary literature references confirming the presence of harmane in coffee, so I'll delete that first paragraph.

@MorseBott, see also this article [http://www.tandfonline.com/doi/abs/10.1080/02652030210145892#.Ufcuyqz5Dis]. It suggests the alkaloids formed during roasting, while decaffeination is performed on green coffee beans. The maximum content of harman they found in instant coffee was 1.64mcg/g, which is 0.000164% by mass. If it is indeed produced during the roasting process, and given that tiny percentage, then I suspect the content in pure extracted caffeine is truly miniscule, possibly even below the limits of detection.


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Would concentration of a compound in equilibrium increase even if it's not favored by a volume decrease?
So say I have a reaction where there are x moles of gas on the reactant side, and y moles of gas on the product side, where x>y. The volume decreases, favoring the products side. The textbook I am using says that the equilibrium concentration of the reactants would decrease, but wouldn't it end up increasing due to the lowered volume?

Suppose you have the following generic reaction:
$\ce{xA_{(g)} <=> yB_{(g)}}$
With the condition, $x > y$. The reaction quotient, written in terms of concentration would be:
$Q = \frac{(\frac{n_B}{V})^y}{(\frac{n_A}{V})^x}$
Where $n_A$ and $n_B$ are moles of the respective species and $V$ is volume. With some algebraic manipulation, this can be rewritten as follows (note here I'm explicitly treating $Q$ as dimensionless, q.v., the relevant Wikipedia page on equilibrium constant units, which applies to the reaction quotient as well):
$Q = \frac{n_B^y}{n_A^x} \cdot V^{x-y}$
Written in that form, it should be apparent that, if $x > y$, then $x - y > 0$, and therefore $Q$, the ratio of products to reactants, varies directly with $V$. In other words, as volume grows, $Q$ grows, and as volume decreases, $Q$ decreases. Suppose that, prior to some volume change, the reaction is at equilibrium, i.e., $Q = K$. If volume instantaneously begins to increase, $Q$ becomes instantaneously greater than $K$, and therefore the rate of the reverse reaction becomes greater than the rate of the forward reaction, favoring generation of $A$ until equilibrium is reestablished. Conversely, if volume instantaneously begins to decrease, the opposite happens and production of $B$ is favored. This is precisely what one would expect based on an intuitive understanding of Le Chatelier's principle: if volume decreases, equilibrium shifts to favor production of fewer moles of gas to counteract the increasing pressure; if volume increases, equilibrium shifts to favor production of more moles of gas to counteract the decrease in pressure.
Edit: I misread part of your question initially. In short, if the volume decreases, generation of product $B$ becomes favored, so the concentration of reactant $A$ must decrease relative to $B$ (since $x$ moles of $A$ are consumed in the production of $y$ moles of $B$, and the reaction quotient approaches the equilibrium constant over time).
Just to clarify this further, it's true that, in an absolute sense, if volume decreases in an instantaneous fashion, then the concentrations of the reactant and product, considered separately, would both have to increase. However, in relative terms, the concentration of $B$ must increase relative to $A$ as the reaction proceeds. The ratio between the two must grow to compensate for the diminished volume, otherwise $Q$ will remain perpetually smaller than $K$, when in fact $Q$ always converges to $K$, for a given set of conditions, over time.

Yeah, I knew that relatively [A] would have to decrease with respect to [B], but I wasn't sure if absolutely they would both increase or not. My book only asks "Which of the following is increased by decreasing the volume of the reaction system", and then lists "Equilibrium concentration of reactants" as one of the options, so it reads as if it means absolutely, but the answer says that it meant relatively. I guess it's just a case of bad book writing (seeing as it's an SAT review book, I wouldn't expect the greatest quality though). Thank you for your response.

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Consequences of using reagent preserved with sodium azide on gold film
I am running experiments using reagents preserved with 0.1% sodium azide, which are applied to a gold film.  Unfortunately, there are very few options for using reagents without sodium azide.
I understand that sodium azide reacts with gold to form heavy metal azides, which are explosive.
I would like to know:

How long will it take for sodium azide to react with gold?
Generally, what is going to happen to the gold film?
Any other info you can think of

just out of interest, what are the reagents? Is it possible to deactivate the azide and/or move the reagents into a different solvent?

Sodium azide can be neutralised using nitric acid. Is this an option for you?

Even if it form explosive azide, its going to be so trivial as to not affect your process. Best to ask your chemical supplier if you are unsure.

http://www.ehs.neu.edu/hazardous_waste/fact_sheets/sodium_azide/
"Sodium azide reacts with heavy metals (such as silver, gold, lead, copper, brass, or solder in plumbing systems); and metal salts to form an accumulation of the highly explosive compounds such as lead azide and copper azide."
Assume reaction is fast.  How many micromoles of gold/cm^2 constitute your film?  How many micromoles of azide/volume are in your application?  Will the gold be corroded through?
Is the gold there as a convenience or as a necessary surface?  A brief soak in dilute decanethiol solution plus rinse will cover the surface with a monomolecular layer of hydrocarbon, each chain being anchored by its sulfur.  The surface will be passivated and probably hydrophobic. 1-hydroxy-10-decanethiol will give a hydrophilic surface.  Is it worth the bother and expense to diddle with the gold surface?
Gold is selling for $(USD)42.15/g today.  How much gold is in your annual discard pile? When my school dumped a load of ancient undergrad combustion bombs I went through them.  They originally shipped each with a platinum crucible.  One remained.  Half a Troy ounce of platinum was a good hour's tax-free wage.

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Which gems can be prepared on a small scale basis?
I am interested to know the following:

Which gems can be prepared at home (without sophisticated instruments)?
What will be the quality (appearance) of the product for a certain method?

I want these information just for hobby-scale purposes(no academic/commercial production). In addition to common lab equipment, I have $\pu{1000 K}$ of temperature and few atmospheres of pressure at disposal.

It's not off-topic per se, so I think you're okay there.  I think this gets at issues of crystallization, so it could be interesting.

I wonder if rock candy counts as a gem.

@jonsca : Thanks. Updating the question, please have a look.

Most likely the only gems you will be able to make at home are plastic gems.
Wikipedia sheds some light on the production of some synthetic gems, Corundum and Cubic Zirconia, both of them not possible at home (or at least my home).

Given as promised. But I am not after plastic gems.

Taking into account that gems often are doped oxides with rather high melting points, your project is quite a challenge.
A historical example for synthetic emerald is Igmerald, produced by the I.G. Farbenindustrie (hence the name) from 1929 until 1942 at the Bitterfeld facility.
The synthesis was performed at 800 °C in platinum crucibles, reacting $\ce{BeO}$ and $\ce{Al2O3}$ and $\ce{Li2CrO4}$ (as the colour-giving dopant). Lithium molybdate was used as a solvent and natural emeralds were used as seeds.
The crucibles were covered with platinum lids having platinum tubes attached, through which more material was fed in during the crystallisation process.
After 18 growth periods over a year, prismatic synthetic emeralds with a length of 2-3 cm could be obtained.
The information giving above is taken from the article The colour of Igmerald: I.G. Farbenindustrie flux-grown synthetic emerald by Karl Schmetzer and Lore Kiefert, hosted on the site of the SSEF Schweizerisches Gemmologisches Institut.

We made emeralds and also rubies from scratch (no seeds) in our advanced inorganic lab course, they were close to 1mm-sized in a week.

You can purchase bulk raw or cut stones much cheaper than you can synthesize bulk raw,
http://www.syntheticgems.org/
Otherwise, Dennis Elwell, Man-Made Gemstones, 1979 or following. It includes Appendix 3, "How to grow your own rubies," mentioning dopant that suppresses the otherwise platey crystal habit.


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How can silver work as a disinfectant in a bathtub if precipitates in the presence of chloride?
I use a silver based disinfectant in my hot tub. It is in the form of silver nitrate beads. There is chlorine in the tub that I presume becomes chloride ion. Why doesn't the silver form a precipitate of silver chloride in the water?

Ag+ is bioactive, it kills bacteria. There is chlorine in the tub in what form? In the paint? Please be more specific.

Oh, I assume you are tossing the AgNO3 breads in the hot tub full of water with "chlorine", is that it? Cl is mostly in the form of HOCl+ in the tub, which is an disinfectant. If that's the case, I guess you should use either "chlorine" or silver as disinfectant. You're probably just not seeing the precipitate, it's white and hard to see in small amounts (compared to all the water on a hot tub I doubt you'd see anything, and there's probably not much Ag+ on that).

If you're referring to chlorination of public water supplies, keep in mind that this is likely not the main source of chloride ion. Elemental chlorine disproportionates via the following reaction:
$\ce{Cl2 + H2O <=> HCl + HOCl}$
Most chloride ion probably derives from various natural and industrial sources, as well as miscellaneous human activity. One source (somewhat outdated) I found indicates an average chloride ion concentration in U.S. public water of $11.5$ mg/L, which equates to a molarity of $3.24 \cdot 10^{-4}$. Using the $K_{sp}$ of $AgCl$ you can calculate the concentration of silver ion necessary to produce a precipitate, but given that it has a value of $1.77 \cdot 10^{-10}$ under standard conditions, you can assume that a fairly minimal amount of silver would be required, even with the large volume of a tub and increasing solubility in hot water. It's likely the case that any precipitate is just finely dispersed in solution and not clearly visible.

The chloride ion I am referring to is from the addition of dichloro-s-triazenetrione, which generates HOCL and HCL when added to the water, and then, I assumed, further dissociates into Cl-.

Since most of the Ag is in the form of insoluble AgCl, would it still have anti-microbial properties?

@Richardbernstein, yes, HCl would dissociate to produce chloride ion, but the quantity of chloride introduced in that manner constitutes a small percentage of the total. As for whether the bactericidal action is inactivated, I frankly don't know; it's not my area of expertise.


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How can one test if a ruby is real?
My friend recently bought a ruby gem from a store and asked me how to test whether its an original one or a duplicate.
Are there any ways to test it using only domestic materials?

If the concern is that the ruby might be synthetic, then I'm fairly certain there are no chemical tests that can distinguish them, since synthetic rubies have the same basic elemental composition as natural ones, aside perhaps from trace impurities in natural ones. If that's the issue, only a professional will be able to reliably determine its authenticity. If you just want to identify it as a ruby and not some other type of stone, my suspicion is that testing it for hardness on the Mohs scale is the only unambiguous test you could do without professional expertise and equipment.

Ruby glows red under a UV lamp.

To test if it's ruby at home check it hardness. If it's harder than 7 even if it's not ruby it still something precious. :-) You can also try to find refraction index however for this you need some equipment. If you really want 100% prove, ask on a local chemistry faculty if they could "as a teaching example" make single crystal diffraction experiment. Simple run with 30 frames would be enough to find a cell what is enough to prove it's ruby.
Fitting your ruby on diffractometer would be challenge as a new diffractometers are working on crystals with size of 10--300 micrometers. To find if it natural or synthetic you need to go to specialize jeweler. They look for imperfections (natural stone are usually less perfect than synthetic).

Thanks for the answer,Well i have another query.is it possible to test for the hardness using domestic materials? I would be very much grateful to you if you please suggest me some ways to do the same.

@javaprogrammer: as you have another question, raise it as another proper question

@javaprogrammer Yes, it's possible.


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How much does butter shrink when frozen?
I just bought a butter dish which is a millimeter or two too small (the problems are that the corners are rounded which they forgot to mention before I ordered it and asked for the dimensions) to fit a piece of butter as it is sold in the supermarket. Since I don't want to return it my idea was to freeze the butter before I put it in this butter dish but for different reasons I can't make a "field test" just now (will update the question when I get the opportunity to do that). Instead I have to fall back to on theoretical answers.
How much does butter shrink when it's cooled from approximately 4°C (39°F) to -18°C (-0.4°F)?
Typically butter has about 1% salt, 82% fat and the rest is water.

There is one clearly technically superior solution to freezing the butter: cutting the extra bit at the end, spreading it on fresh bread, and eating it!

@F'x, you read my mind, but I wanted to do the math to prove it.

Well, such a practical solution wouldn't be a good ground for a fun question at StackExchange, would it? :-)
Besides, to be serious, I want to avoid greasing around with the butter before it is in the box.
However, the butter is in fact 150 mm including the paper cover in the direction that is the bottleneck so 7‰ combined with removing the paper might be what does the trick. Otherwise, if it still does not work, Alessi should be ashamed of their industrial design of butter dishes.

The answer here would solve half of the problem. But the harder part is: how much does the butter dish shrink in the freezer?

Given that fat (consisting mainly of assorted triglycerides of varying saturation) is the major constituent by a wide margin, it's probably reasonable to compare the change in density w/r/t temperature of butter to the change measured in various vegetable oils, since they have roughly comparable chemical composition (and because extensive data on them actually exists). I found a study in which the densities of various vegetable oils were measured over a range of temperatures. On average, the increase in density resulting from cooling from 110.0°C down to 23.9°C was roughly 6-7%. The data points graphing density against temperature arrange very linearly, and the regression lines generated fit them exceedingly well. I see no reason to suppose that the trend wouldn't continue as temperature decreases further. Assuming the comparison is reasonably valid, you could extrapolate that the increase in density you'd observe over the temperature range you specified should be on the order of approximately 2% at best.
As for the amount of shrinkage, strictly considering volume, the ratio of densities (assuming the 2% approximation is reasonable), would be:
$\frac{D_2}{D_1} = 1.02$
That is, the density after cooling, $D_2$, is 2% greater than $D_1$. That implies, holding mass constant:
$\frac{V_1}{1.02} = V_2$
Hence, given that volume is in units of cubic length, you can get a rough idea of how much the length of each side/edge should shrink by taking the cube root of both sides of the equation (obviously, this is only completely accurate for a perfect cube, but it emphasizes that the shrinkage in each dimension is smaller than the shrinkage in volume):
$\sqrt[3]{V_2} \approx 0.993\sqrt[3]{V_1}$
Unless your stick of butter is really gargantuan, a decrease of 0.7% on an edge is not going to account for 1-2mm.
Edit: Additionally, I've read that, in general, the magnitude of relative thermal expansion/contraction that liquids experience is usually greater than that experienced by solids, meaning that liquid density varies more over a given temperature range than solid density. Of course, I'll add the caveat: that's a very broad statement that admits of various exceptions. Nevertheless, this suggests that the estimate of 2% I made above by comparison to data on liquid vegetable oils is, if anything, more likely to be too large than too small.

Would it be more efficient to heat up the stainless steel dish? Either with boiling water, in the oven or using my gas stove? How much does stainless steel expand when heated?

@user23122, I don't know off hand the degree to which it would expand, and I'm sure it varies with the exact composition. Unless you physicallly deform it, though, it will simply contract to its previous volume upon cooling.

@user23122, in general, I know that metals and alloys typically thermally expand much less than amorphous solids (e.g., plastics), and very much less than organic liquids. I think only inorganic salts and minerals are less apt to expand, but I could be mistaken.


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Why don't metals form covalent bonds in bulk metal?
I really don't understand why metals form metallic bonds. I mean, it makes no sense. It would make much more sense for them to form covalent bonds with themselves and have a 'pseudo-full' outer shell. How does freeing off electrons make them any more stable?

This is due to the low ionization energies of the metals. It's easier for them to release few electrons from the outer shell to obtain a noble gas configuration rather than consuming several ones. However, the difference between an ionic and a polar covalent bond is always fringe.
Again, it's not true that metals don't form covalent bonds at all. I guess you never heard of quadruple bond or δ-bond. There are several examples such as $\ce{K2[Re2Cl8]·2H2O}$ and Chromium(II) acetate hydrate.

This is due to the low ionization energies of the metals || This is BS.  silicon: ionisation energy 8 eV (nonmetallic bond); mercury: ionisation energy 10 eV(metallic bond).

Silicon is not entirely nonmetallic, it becomes a metal when melted.  The point is, what makes things metallic is not as simple and brute force as ionization energy.

Atoms in molecules $\ce{Mo2}$ and $\ce{W2}$  in a gaseous phase form even a sextuple bond

Bulk d-metals and especially intermetallic compounds often do have significant covalent bonding.
Strictly speaking, metal bonding is a kind of covalent bonding in a sense. It is a common knowledge, that 3- or more atoms may be bound by one pair of electrons, like $\ce{H3+}$ ion. In bulk metals, similar bonding bond is working for valence electron.
Even more strictly speaking, to fully understand the matter you have to consider the matter from position of theory of molecular orbitals.
Let's assume we have a nanocrystal of 100 3d row element atoms. From 400 orbitals of said atoms 400 molecular orbitals are formed, with strict distribution of their energies depending on shape and size of the crystal and crystal cell. Then, the only difference between metals and non-metals is that metals have semi-full group of orbitals of same energy (so called conductivity zone, while non-metals have only full and vacant orbitals.

3 is rather misleading, and in this form does not explain why metals would have such orbitals or what is special about them

The difference between covalent and metallic bonding looks much clearer when written on paper than it does in the real world.
If you take the elements from group 15 from phosphorus downwards and analyse them, you have ‘true’ covalent bonding for one allotrope of phosphorus (the white one, $\ce{P4}$) and ‘true’ metallic bonding for bismuth. The entire transition downwards from phosphorus — technically already including phosphorus’ red and black allotropes — is a transition from covalent to metallic bonding.
If you look at an orbital picture of metallic bonding, you see the clear similarity to covalent bonding, except that you aren’t dealing with discreet molecules but a huge network.

Arguably, the bonding in bulk metals is covalent bonding
The bonding in metals is often treated as a special type of bonding because it has unusual properties where many electrons are delocalised enough to make the bulk material conducting. So we don't describe the bonding the same ways we describe the bonding in, for example, diamond.
But, from a molecular orbital point of view the major difference between a metal and diamond is not which orbitals exist, but which bonds are occupied. To simplify a lot, in molecules when atoms join up we get molecular orbitals from combinations of the atomic orbitals. These fill up with the spare electrons. In bulk solids (as opposed to small molecules) the same process happens but there are far more orbitals to fill and there is a lot of duplication of their energy levels. In diamond the filled orbitals are very localised and the bonding orbitals confine most electrons between adjacent carbons. In metals, the highest level filled orbitals are delocalised over many atoms and the electrons can wander fairly freely over the bulk. WE could still describe them as "covalent" bonds but many of their properties are better understood by completely different tools than those used to describe small molecules.
That this is not a completely silly way to look at metallic bonding is illustrated by the bonding in graphite (or graphene). Hardly anyone would describe that as a solid held together by anything other than covalent bonds. But the pi-orbitals in the plane of the hexagonal rings are delocalised over the whole plane. Electrons can wander about fairly freely in those orbitals and the conductivity in the plane can reach metallic levels. Despite the 2D metal-like behaviour in the plane we still call these bonds "covalent". Bonding in actual metals isn't qualitatively different to this: it is just 3D not 2D in how electrons can move.
So from a chemistry point of view metallic bonds are just a special case of covalent bonds where the electron delocalisation is in 3D not 2D as it is in graphite. But we treat that special case very differently as we need a whole different set of tools to analyse and think about the properties of metallic bonds that matter and the tools for thinking about the bonding in small molecules are of little to no help.

Certain metals under certain conditions form highly covalently bonded structures.
Certain allotropes (phases) of metals have no metal qualities such as alpha-tin. Formed at about 13.2 degrees celcius with pure tin- it has no metallic qualities and pure covalent bonds.

Alpha-tin is a semi-metal, it certainly has metallic qualities (it still is conductive for one thing). The issue is that while the bottom of the conduction band overlaps with the top of the valance band in energy, those occur at different $k$ values.

First of all! Metal do form covalent bond. It is very common in transition metal like platinum, palladium.
However, it is not the way you are talking about. Typically, when pure metal atoms bond together, they prefer metallic bond.

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Is a negative pH level physically possible?
A friend of mine was looking over the definition of pH and was wondering if it is possible to have a negative pH. From the equation below, it certainly seems mathematically possible—if you have a $1.1$ (or something $\gt 1$) molar solution of $\ce{H+}$ ions:
$$\text{pH} = -\log([\ce{H+}])$$
(Where $[\ce{X}]$ denotes the concentration of $\ce{X}$ in $\frac{\text{mol}}{\text{L}}$.)
If $[\ce{H+}] = 1.1\ \frac{\text{mol}}{\text{L}}$, then $\mathrm{pH} = -\log(1.1) \approx -0.095 $
So, it is theoretically possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)?

The $0$ on the pH scale is an artefact of our system of units.  Physically, there is nothing special about the concentration $1\textrm{ mol dm}^{-3}$ (pH $0$) any more than there is about the concentration $1 \textrm{ lb-mol ft}^{-3}$.

I won't go into technicalities as it has been abundantly discussed above but highest recorded $\mathrm{pH}$ is of fluoroantimonic acid with $\mathrm{pH}\ {-25}$, so yeah it's possible.

Adding a little more background and maybe some boring technicalities may actually be the way to a good answer. In it's current state, this answer does not provide any more novel insight into the topic.

It's not a pH but pKa value

This was featured in a Stack Overflow blog post (near "Chemistry isn’t collecting any rust").

One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006, 83, 1465. Quoting the abstract in full:

The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books.

The article text provides some counterexamples:

For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$.

It's certainly possible theoretically. Solve for $\ce{pH < 0}$:
$\ce{-log[H+] < 0\\
log[H+] > 0\\
[H+] > 1}$
So, as you said, a solution in which the hydrogen ion concentration exceeds one should theoretically have a negative $\ce{pH}$. That said, at those extremes of concentration, the utility and accuracy of the $\ce{pH}$ scale breaks down for various reasons.
Even acids conventionally categorized as "strong" do not in fact dissociate 100%. In reality, their dissociation is also essentially an equilibrium process, though this only becomes apparent at surpassingly high concentrations. As the solution becomes more concentrated, any additional acid cannot be as thoroughly solvated, and the chemical equilibrium begins to favor dissociation progressively less and less. Hence, as the solution becomes increasingly saturated, the extent of dissociation begins to plateau and the hydrogen ion concentration approaches some practical upper limit. Furthermore, $\ce{pH}$ measured via molar concentration as a proxy for thermodynamic activity is inherently inaccurate at the extremes of concentration. Other phenomena, such as the formation of distinct chemical species by self-ionization in a concentration-dependent manner further complicate things (e.g., generation of $\ce{H3SO4+}$ in concentrated sulfuric acid, $\ce{H2F+}$ in concentrated hydrofluoric acid, etc.).
For highly concentrated solutions of strong acids, alternatives/extensions to $\ce{pH}$ exist that are functional beyond the limits of $\ce{pH}$ (see, for example, the Hammett acidity function).
As for whether solutions of negative $\ce{pH}$ have actually been experimentally prepared or observed, the answer is yes. Here's a link to one article describing the measurement of $\ce{pH}$ in acidic mine waters, which cites a figure of $-3.6$.

I completely agree with your answer. As I recall the pH is derived from the mass action law in aqueous solution (at $25^\circ{}C$), hence $\ce{pH = -\log_{10} c(H3O+)}$, with $\ce{K_{w} = [H3O+]\cdot[{}^{-}OH]\approx14}$. This is also one reason why those acidities are hard to measure. It is also worth mentioning that it is highly temperature depending. ($K=\exp{-\frac{\Delta G}{RT}}$).

@GregE.  It's somewhat contradictory to discuss pH as -log[H+], and then say that pH = -3.6 is possible.  There is no way you can cram over 1000 moles of hydronium ions into a liter! The -3.6 value only makes sense if you explain that pH = - log (H+ activity), and that it is the deviation of activity from concentration that makes the value -3.6 possible.   -log[H+] instead would be about -1 for the mine water.

@Dave I am not so sure of your comment.  Thought the brackets technically meant activity.  We just assume it's proportional to concentration in typical usage.

Any strong acid solution with concentration more than 1 mol/L has the negative pH. Think about any concentrate commonly used strong acid solution such as 3M $\ce{HCl}$, 6M $\ce{HNO3}$. Negative pH is actually very common.

It is very much possible.
Let’s say you put 3 moles of $\ce{HCl}$ into 1 mole of water.
$\ce{HCl}$, being a strong acid dissociates completely into $\ce{H+}$ and $\ce{Cl-}$ ions as:
$$\ce{HCl -> H+ + Cl-}$$
so after complete dissociation, $[\ce{H+}]=3~\mathrm{mol/L}$   (ignoring the very tiny contribution from water itself)
By definition,
$$\mathrm{pH} = -\log[\ce{H+}]$$
therefore, $\mathrm{pH}= -\log 3= -0.48$
So it is very much possible to have solutions of  strong acids whose $\ce{[H+]}$ is 1 molar or more, and thus whose pH is negative.

While I agree that a concentrated solution of a strong acid is an excellent counterexample, it should be mentioned that this treatment is bordering on being a bit simplistic. The pH is strictly defined as the negative logarithm of the activity of $\ce{H+}$, and the value of the activity deviates from the value of the concentration, especially at higher concentrations. Just as an example, the standard hydrogen electrode actually uses 1.18 M HCl to ensure that $a_{\ce{H+}} = 1$.

While negative pH values are most certainly possible (see all of the other answers), there is a practical element to the idea that the pH scale goes from 0-14.
Most commercial pH meters, the ones used in educational settings (and many commercial settings) have a significant amount of error as you move to very high and very low pH.  In addition, measuring pH values below 1 can also damage many pH meters, as the acid attacks the glass and other components of the probe.

What is the actual measured result? How much does it deviate?

It is possible to have $\mathrm{pH}<0$ and you don't need to create any substance. Take a concentrated solution of one of the strong inorganic acids (i.e. one with dissociation constant above 1000 like sulfuric acid) and here you are.

The pH scale is taken on our reference as 0 to 14 for concentration values from $1~\mathrm{M}$ to $\mathrm{10}^{-14} \, \mathrm{M}$. This range is such that our normal calculations in lab can be worked out easily. It is to be noted that this scale is at room temperature. If you increase the temperature then the limits changes. For example pH of pure water at $100 \, \mathrm{^\circ C}$ is $6.14$ and not $7$. Hence we can see that the scale has shifted with temperature.

This is downright incorrect. Yes the pH changes at different temperatures but that does not in any way mean that there are "limits" on the pH scale. "This range is such that our normal calculations in lab can be worked out easily" - no such thing.

If pH is about concentrations, then probably we have endless values for them. Depending them on temperature will increase those number of values just manifold.

$\mathrm{pH}$ is essentially a convention. It is defined as $$-\log_{10} [\ce{H+}]$$ since the concentrations of the solutions commonly used lie in the interval $$[10^{-14}\ \mathrm{mol/L},1\ \mathrm{mol/L}]$$ and thus the $\mathrm{pH}$ lies in $$[0,14]$$ But nothing constrains an aqueous solution from having a $\mathrm{pH}$ that does not lie in this interval. The only constraints are:
$$[\ce{H+}]\lt[\ce{H2O}]_\text{liquid}$$ and $$[\ce{OH-}]\lt[\ce{H2O}]_\text{liquid}$$
The first limiting case is when you suppose all water has turned into $\ce{H+}$, which is not quite true, because there must be some water that has turned into $\ce{OH-}$ in order to $$K_\mathrm w=[\ce{H+}][\ce{OH-}]$$
But $$[\ce{H2O}]_\text{liquid}=\frac{1\ \mathrm{mol}}{18\ \mathrm g}\frac{1000\ \mathrm g}{1\ \mathrm L}=55.6\ \mathrm{mol/L}$$
And then we have $$[\ce{H+}]\lt55.6\ \mathrm{mol/L}$$
The last case implies
$$[\ce{OH-}]\lt[\ce{H2O}]_\text{liquid}$$
which means (considering $K_\mathrm w=10^{-14}$) $$[\ce{H+}]\gt\frac{10^{-14}}{55.6}\ \mathrm{mol/L}$$
Then $$\frac{10^{-14}}{55.6}\ \mathrm{mol/L}\lt[\ce{H+}]\lt55.6\ \mathrm{mol/L}$$
$$-\log_{10}(55.6)\lt-\log_{10} [\ce{H+}]\lt-\log_{10}\left(\frac{10^{-14}}{55.6}\right)$$
$$-1.74\lt\mathrm{pH}\lt15.74$$

This is valid for aqueous solutions at room temperature, in which $K_w$ assumes that value and the density of water is approximately 1g/mL

I'm afraid you started off with an approximate definition of pH, which makes the remainder of the derivation incorrect. pH is defined by the chemical activity of protons, not their concentration. The activity coefficient can differ enormously from 1 in concentrated solutions, allowing your proposed bounds to be exceeded.

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


text:
Is Mn(C₂H₃O₂)₂ soluble in water?
Also, can you explain why, as in which solubility rules apply in this case?

Welcome to Chemistry.SE. This looks like a homework level question of the type. Help us help you by letting us know what you have tried or where you are stuck.

Yes it is. But have no idea on its $\ce{Ksp}$ or $\ce{g/L}$ value! If I'm not wrong, it's Manganese(II) acetate.
'Which rule applies' is the following :
All nitrates, acetates and perchlorates are soluble in water.
Possibility of H-bonding adds to acetates' solubility in water.  Have a look at the first few lines of this abstract, it's quite explanatory.  However, few acetates like those of $\ce{Ag}$, $\ce{Hg}$ or  $\ce{Pb}$ are slightly soluble.


text:
Suitability of liquid plastic resin for plumbing application
Which synthetic resins are suitable for fabrication of plumbing components e.g pipe sections?
I'm trying to find an easy way to mold/house/encase a custom designed data logger.
The water temperature ranges between 10 to 50 degrees C.
EDIT: I'm planning this for both indoor & out door use. Under direct sunlight.

Epoxy should work OK and you have different options that are easy to get.

Different options that are easier than resin casting?

You have puttys that you can shape by hand. I'm supposing a prototype so you might not need a mold.

How well do puttys handle UV/sunlight? I've always known putty to be a Terminal emulator link until I Googled.

I meant a dough you can give a shape to. Other forms are used for car paint and antiacid lining. I don't think you will have a problem with outdoor use.


text:
What is the effect of temperature during cryotrapping?
I'm working on a lab report that has some specific questions that need to be included. The lab work itself dealt with GC–MS.
What is the effect of the temperature selected during cryofocusing/cryotrapping? What is the effect of making this temperature warmer or colder?
My understanding was that the gas based samples would adsorb to the sample loop and then rapidly desorb when the temperature is increased. However, I am not sure what the effect of changing the cold temperature is.

It depends on the temperature regime you of your cryo-trap.  If you are trapping in the temperature range from dry ice to water ice then you must be sure any significant water vapor in you sample has been removed.  If you are using liquid nitrogen, there is some danger of condensing oxygen.  This can be mitigated by using the enough-warmer-that-this-doesn't-happen-with-air liquid argon.


text:
Why does ice water get colder when salt is added?
It is well known that when you add salt to ice, the ice not only melts but will actually get colder. From chemistry books, I've learned that salt will lower the freezing point of water. But I’m a little confused as to why it results in a drop in temperature instead of just ending up with water at 0 °C.
What is occurring when salt melts the ice to make the temperature lower?

I think it's related to the Raoult's law

There are some assumptions here, I think: The salt is NaCl, and the system is at thermal equilibrium (ice and water at 0 °C throughout). Also, the system is isolated (i.e the surrounding is not heating it up or cooling it down).

there is also an explanation here https://chemistry.stackexchange.com/questions/116302/where-does-the-energy-come-from-to-lower-the-temperature-of-a-brine-solution/116316#116316

When you add salt to an ice cube, you end up with an ice cube whose temperature is above its melting point.
This ice cube will do what any ice cube above its melting point will do: it will melt. As it melts, it cools down, since energy is being used to break bonds in the solid state.
(Note that the above point can be confusing if you're new to thinking about phase transitions. An ice cube melting will take up energy, while an ice cube freezing will give off energy. I like to think of it in terms of Le Chatelier's principle: if you need to lower the temperature to freeze an ice cube, this means that the water gives off heat as it freezes.)
The cooling you get, therefore, comes from the fact that some of the bonds in the ice are broken to form water, taking energy with them. The loss of energy from the ice cube is what causes it to cool.

…so not all of the water is getting colder.  Some of it is getting warmer?

...maybe you could give your opinion on the question, whether an ice cube that is heated will invest all the absorbed energy into phase transition and thereby not heat its core up until it melts, or an temp. gradient will form like in any other body.

Correct answer, but terminology can be confusing! If water "gives off heat as it freezes", it sounds like the water is heating something up - well, it is heating up the cooling device, but it is clearer to say that heat is withdrawn from the water to make it freeze. IMHO

The OP said it more clearly: the freezing point of salt water is lower than the freezing point of pure water. Saying that the melting point of ice cubes changes when adding salt is a bit strange because we still have pure water in the ice cube. It is the melted water that is no longer pure.

@KarstenTheis I have removed my poor comment. A better explanation  is here https://chemistry.stackexchange.com/questions/116302/where-does-the-energy-come-from-to-lower-the-temperature-of-a-brine-solution/116316#116316

We know that melting or freezing is an equilibrium process. The energy that is required to melt an ice cube will not contribute in elevating its temperature until all the solid water is molten.
If we take two ice cubes and add salt to one of them, then put each of them at room temperature, both of the ice cubes will absorb energy from the surroundings, and this energy as we said will contribute in breaking down the bonds between water molecules.
The cube that salt has not been added to, has a melting point $0~\mathrm{^\circ C}$ and so if we measure its temperature during melting it will remain zero until all ice is molten. That ice cube to which we have added salt, the salt that is added lowers the melting and freezing points of water because it lowers the vapor pressure of water. This ice cube will absorb energy from the environment to help break bonds between water molecules. We know that the salt added will dissolve in the melted portion of the ice. This formed solution of salt will have a lowered freezing point, so the equilibrium between the solid phase and the aqueous phase will be shifted towards the liquid phase since such a solution will freeze at say $-2~\mathrm{^\circ C}$. Since both phases are close together, the ice will absorb energy from the salt solution and will reduce its temperature to the $-2~\mathrm{^\circ C}$ to maintain the equilibrium. When all ice is molten we end up with a salt solution that has got a temperature of say $-1.5~\mathrm{^\circ C}$. This is due to the solution being diluted now. After that, it will start absorbing heat from the room and reach zero and above. So, in conclusion that is how salt melts ice.

The question was about why the temp would drop, not how salt melts ice, so the concluding "that is how salt melts ice" is misleading. but the text does add some explanation. However, it might be the reason why the other answer got more upvotes, as it more directly focussed on the temp decrease.

A mixture of water and ice stabilizes at the freezing point of water.
If the ice were any colder, it would absorb heat from the water, in the process raising its own temperature while freezing some part of the water.
If the water is any hotter, it will cool down by melting some of the ice.
This works because ice thawing is endothermic; energy (heat) is used up to turn solid into liquid even though the temperature is staying the same.
The freezing point of water is $0 \pu{°C}$, so water-ice slush stays at $0 \pu{°C}$. If it was lower, it would stabilize at the lower temperature. By adding salt, you are lowering the freezing temperature. The mixture stabilizes there and is colder.

Not as technical as the higher voted answers, but done in a very easy (non-chemist) to understand way. Combined with the others, this answers all my questions, thanks.

When you dissolve $\ce{NaCl}$ in water, it will have to take energy from the system to break its structure so it can dissolve in water. This is the reason the water gets colder because the salt uses the energy from the water to solve it.
Now let's look at why ice melts when salt is added. This is based on a so-called colligative attribute. These attributes are only dependent on the amount of substance. When you add particles to a solvent, its vapor pressure lowers. This will result in a higher boiling point(using salt for cooking) and a lower freezing temperature for the solution.
I hope this gives a starting point for further reading consult books on physical chemistry(for e.g. Atkins).

Asking why
When you ask why, you want to know about causality. If I ask "why does the cold pack show a decrease in temperature" and the answer is "because the reaction is endothermic", this might be considered a tautology. After all, endothermic means that energy is needed, and this energy can come from the surrounding, lowering the temperature.
Why does ice water get colder when salt is added?
As the OP states, this lowers the freezing point of the liquid. The system is no longer at equilibrium, and some ice will melt in an endothermic process. As a consequence, the temperature drops and the salt water gets diluted. The melting process stops when salt concentration and temperature are matched again, i.e. the freezing point of the liquid is equal to the temperature of the system.

It is well known that when you add salt to ice, the ice not only melts but will actually get colder.

The melting process is at the interface of liquid and solid, so both the solution and the ice will get colder.

From chemistry books, I've learned that salt will lower the freezing point of water. But I’m a little confused as to why it results in a drop in temperature instead of just ending up with water at 0 °C.

So the question is given that some ice melts, why does the temperature drop. Saying that ice melting is an endothermic process maybe does not fully answer the question (explain the causality).

What is occurring when salt melts the ice to make the temperature lower?

In terms of kinetics, the salt does not melt the ice. Instead, it lowers the rate of water freezing. The net effect is that ice melts. At the molecular level, according to https://www.nyu.edu/pages/mathmol/textbook/info_water.html, "In liquid water each molecule is hydrogen bonded to approximately 3.4 other water molecules. In ice each each molecule is hydrogen bonded to 4 other molecules." So upon melting, water loses about half a hydrogen bond. Also, the remaining hydrogen bonds might have less ideal distances and angles. So that's what makes the process endothermic. The NaCl has little role in the energetics, as any other solute has pretty much the same effect (colligative property).

When you add salt to the ice it melts, I won't go into why since you didn't ask that; all you need to do is that it does if you don't believe me ->
http://science.howstuffworks.com/nature/climate-weather/atmospheric/road-salt.htm
Moving on, whenever a substance undergoes a phase change it's temperature does not rise and usually stays relatively constant, if you looked at a graph of most substances when undergoing different phase changes (i.e. solid to liquid to gas) you will observe regions that are 'flat' or horizontal this is because the energy is no longer causing a rise in temperature but a change in state. Since you have dissolved salt in the ice it will lower the freezing point (note that freezing and melting point of any substance is the same they can be seen as mirrors for one another) this means that water can now exist at lower temperatures and not turn into ice or in other words it will begin to melt at lower temperatures this could attribute as to why the temperature would LOWER as it no longer needs to reach as high a temperature to begin to melt. I haven't explained it very clearly but I hope you understand it consulting a physics and chemistry text book as the person above has suggested is a good idea.

Melting is endothermic and freezing is exothermic. We never observe water warm up when it freezes because more energy has to be lost from the system before more water freezes. When water freezes from being in cold air, the release of heat actually slows down the freezing. When you add salt to a mixture of water and ice, it causes more ice to melt by depressing the freezing point and not by adding internal energy so it actually gets colder.

Let’s do a thought experiment to see what is going on. Combine 500 mL of pure H2O and 500 grams of ice, each at 0 C, in a perfectly insulated container. Then add 300 grams of NaCl (also at 0 C). At first, not all the NaCl will dissolve, because NaCl is only soluble 26% in water at 0 C. (The temperature of the water will decrease slightly because of the energy required to dissolve the salt: about 1 kcal/mole of salt.)
The melting point of the ice cube is still 0 C. However, the liquid surrounding it, although at approximately 0 C, is a near-saturated salt solution, not pure water.
Water molecules from the ice cube will diffuse into the salt solution, diluting it. Energy is required for the transition from solid ice to water: the amount of heat required to melt ice is 68.3 kcal/mole (heat of fusion). This energy comes from the salt solution, reducing its temperature. The ice cube will continue to melt (turns from solid to liquid) as long as it is not in equilibrium with the solution. The ice cube does not melt because the surrounding liquid has a lower freezing point, it melts because the surrounding solution is less than 100% water. The surrounding solution is already colder than 0C and the ice cube is still pure solid water at 0C. In other words, the driving force is a non-equilibrium of concentration. At the beginning of the experiment, there was thermal equilibrium (everything at 0C), but a concentration non-equilibrium forced the melting of ice by heat energy from the solution. Freezing point depression characterizes a solution, but is not a driving force.

No. Unfortunately we seem to all be forgetting a fundamental property of ice - I kicked myself when I realized it. Think of a large chunk of ice taken out of your home freezer, sitting in a bucket, floating in its own melt-water. All the liquid water surrounding that ice is at $\pu{0 ^\circ C}$. The extreme surface of that ice, that is exposed to the meltwater, (the layer that is about to transition to liquid) is also at $\pu{0 ^\circ C}$.  BUT the solid (non-transitioning) interior of the ice chunk is still at about the temperature of the freezer that it came from, about $\pu{-18 ^\circ C}$! So we shouldn’t think of all the water in the bucket (liquid & solid) as being at $\pu{0 ^\circ C}$. Instead, most of the frozen water is much colder than $\pu{0 ^\circ C}$. So now we see that within the chunk of ice there is a significant temperature gradient. The exact center is coldest (at perhaps $\pu{-18 ^\circ C}$). Moving out from there the temperature increases until you reach the surface where the ice has warmed to a temperature of $\pu{0 ^\circ C}$ and has begun melting. Now we can readily see why adding salt to the liquid surrounding the ice chunk would cause a lowering of the temperature of that salty melt-water. By adding salt you have lowered the melting point. The surface layer of the ice that was $\pu{0 ^\circ C}$ would rapidly melt because the salt is in close contact with it and acts on it such that the ice surface is now much too warm to maintain the solid state. Underneath that quickly melting ice layer is another layer at say $\pu{-1 ^\circ C}$. The salt would be able to act on that colder ice in turn and cause it also to melt while maintaining that negative temperature creating a small amount of $\pu{-1 ^\circ C}$ briny water. This below-zero melting action caused by the briny liquid would continue toward a new temperature equilibrium (between the solid surface and the liquid) somewhere below $\pu{0 ^\circ C}$ in accordance with the salinity level of the melt-water.

To warm up one gram of ice from -18 to zero degree celsius requires about 36 joules. To melt the same amount takes about 334 joules. So even if the ice were colder than zero degrees celsius, the effect would be small compared to the effect of melting some ice. The question, admittedly, is a bit vague about the starting conditions, but a water/ice slush is pretty much almost zero degrees celsius throughout.

So what is the point of using salt on ice in an ice cream making machine?

It would be quite the feat to debunk all those ice cream makers with nothing but a thermometer and bowl of brine slush - heck, an elementary school child could do it. Challenge taken!

My comment might have been unclear: Pure water/ice slush is zero degrees celsius. Ice/brine slush is much colder (about -21 degree celsius, which is why our ice boxes are still set to that temperature to reproduce the conditions in an old fashioned ice box).

So points not in dispute: 1) Ice / water slush = 0 degrees. 2) ice / water / salt = below 0. Question in dispute: How does the ice slush temperature go FROM 0 degrees TO sub-0 when salt is added? My answer: The input into the system of sub-0 temps is the sub-0 ice which the salt melts while it is still below 0. Instead, everyone here seems to say that the sub-0 ice has no role / is irrelevant to the drop in temp below 0. You all say that the entire source of the loss in temperature of the water slush upon adding salt is a chemical reaction between the salt and the water.


text:
How is the acid HSO₄⁻ formed?
How is $\ce{HSO4-}$ formed? If Hydrogen is 1+ and $\ce{SO4}$ has 2- charge? I don't understand this. Could someone please clarify?

This doesn't usually exist by itself but is rather found as a polyatomic ion in solution - $\ce{H2SO4}$ is a strong polyprotic acid which dissociates in aqueous solution into $\ce{H+}$ and $\ce{HSO4-}$. ($\ce{HSO4-}$ then behaves as a weak acid.)

$\ce{HSO4-}$ look at the valencies of the hydrogen and sulphate ion, because of the numbers don't match i.e. hydrogen has a +1 charge that means it is deficient by 1 electron while sulphate has a -2 charge which means it has two extra electrons (it can also be viewed that it wants to 'give away' these two electrons) these valence numbers basically show the amount of bonds the molecules can form (since at this level of chemistry bonds can be seen as the sharing of electrons) thus the sulphate can form two while the hydrogen can only form one and in this case since it forms $\ce{HSO4-}$ that means there is one 'unhappy electron' or one that is unbonded since the bond between the sulphate and hydrogen makes hydrogen's charge neutral and takes an electron from the sulphate ion leaving the unbonded electron thus giving the molecule an overall charge of -1. Note that here I am just explaining the bonding that is occurring between the molecules but $\ce{H2SO4}$ DOES form I am just explaining the charges and values involved and also note that bonding and chemical reactions usually involve the VALENCE electrons!.
However this molecule ($\ce{H2SO4}$) will disassociate in water or solvent of some sort, and form ions as the person before me pointed out. into $\ce{H+}$ ions and $\ce{HSO4^-}$ thus forms the conjugate base of the acid ($\ce{H+}$ is the basis of most acidic reactions)
If you're confused as to why $\ce{H2SO4}$ doesn't form, it does but $\ce{HSO4-1}$ is what happens after it disassociates in solution.
I don't know which part of the question or problem is where your confusion lies, hope I answered or helped!
(I am only a first year university student if my answer is wrong please notify me but it correlates with my notes, textbooks and what I was taught back in high school)


text:
What is the reaction order of a reaction?
Here's is what I understand so far and the following is referenced directly from UNSW CHEM1061 lecture notes by Professor Pall Thordarson:

order of the reaction SHOULD be equal to the number of molecules of that reactant involved in each collision ( however this is not always the case)
it is only possible to predict the order of a reaction is it occurs in a single step (elementary reactions)
reaction order is experimentally derived from rate equation

so I understand all of the above but what I fail to fully comprehend is that actual DEFINITION of 'reaction order' what is it in it's entirety, what does it's value represent if it only sometimes correlates with molecularity (no. of molecules which react in an elementary reaction) ? If someone can answer this question could they please provide a qualitative answer that explains the value conceptually? Or is it just a value to relate concentration to reaction rate? that helps us work out more information?

Very Shortly, Rate of reaction is directly proportional to product of the active masses of reactants raised to some power which is Order of the Reaction.
Consider an elementry chemical reaction
$$\ce{2A(g) + B(s) -> C }$$
As reaction is elementry, stoichiometric coefficient and order will be same. The order of the reaction will be 2 (as the other component is solid, therefore will not affect concentrations). The rate law for reaction will be given as
$${R = K[A]^2}$$
where K is reaction constant.
[A] is concentration of A.
If the order of the reaction would be 0, then Rate would be equal to reaction constant itself as the rate law would be given as
$${R = K[A]^0}$$
This explains the maximum use of Order of the reaction and also what order of reaction is.
Edit: for more wide explaination on this, see wikipedia. (Same explaination is provided but a widely expanded.)


text:
Negative charge of methoxide anion
Why is the methoxide ($\ce{CH_{3}O^{-}}$) anion negatively charged?
It has 13 valence electrons and has 13 free electrons. In spite of that the methoxide anion is negatively charged.

I couldn't figure out the count of 13 free electrons. What you did actually mean by free electron.

Probably, you are having problem with formal charge calculation.

The negative charge is due to the unpaired electron in the oxygen atom, remained after all bonds(molecular orbitals) are filled. Oxygen has 2 unpaired electron in its valence shell and is singly bonded to carbon in this molecule. Hence being un-bonded, the other electron exhibits the charge.
Check this link, hover the cursor on the $\ce{O}$-atom, you will get a graphical representation of the same.

Simply put: there are more electrons than protons.
Protons:

Hydrogen x 3 = 1 x 3 = 3
Oxygen x 1 = 8 x 1 = 8
Carbon x 1 = 6 x 1 = 6

3 + 8 + 6 = 17 protons.
Electrons:
Looking at the electron shells and covalent bonds in the Lewis diagram provided by blackSmith, you have the following:

4 single covalent bonds = 4 x 2 = 8
3 lone pairs on oxygen = 3 x 2 = 6
Unseen $1s^{2}$ pair on both carbon and oxygen = 2 x 2 = 4

8 + 6 + 4 = 18 electrons.
As there is 1 more electron than proton, it has a charge of -1.
As to where that negative charge tends to be, one must create and look at the Lewis diagram where you see that the oxygen atom has an extra valence electron.

From what you have given, this question can not be answered in terms of electrons.
It is meaningless to do any electron calculation, because you have to know it has one negative charge before you can do the electron counting. How can you get a correct Lewis structure without knowing that it has one extra electron? The negative charge is part of the given condition, not your result you get from calculation.
Since I saw you are talking about 13 valence electrons, I guess you want to count valence electrons. If you add valence electron from all atoms, you will get 13 valence electron: 4 (C) + 3 * 1 (H) + 6 (O) = 13. Then because you have one negative charge, you have to add one more electron and then you get 13 + 1 = 14 valence electrons. Now everything agree with each other.
If you want to find out which atom the charge is on, you have to do the formal charge calculation. Find it here


text:
energy efficiency, transformation from chemical to thermal energy, for district heating plants
If I have a system in which I introduce $\mathrm{100~e.u.F}$ (energy units) derived from chemical energy (i.e. fuel) and generate thermal energy to heat water what are e.u. absorbed by water? I.e. the efficiency of the transformation?
Basically I would like to know the "average/general" efficiency of energy transformation for a district heating plant. Considered that the only energy input is fuel and the only energy output is heat (no electricity)?
What are the energy loss?

This is an interesting question. Usually we are concerned with the efficiency of converting chemical energy to work, with the loss being heat.


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What are simple applications of Monte Carlo simulations in chemistry, that can be done in Matlab?
I am trying to learn simulation myself and I have installed Matlab on my machine for that purpose. In web I have gone through some of the basics of Monte Carlo simulation, but I am unable to find a simple Monte Carlo simulation experiment for chemistry. I am trying to simulate growth of a nano particle during a wet synthesis. I have tried searching research articles, but as I am new to simulations most of them I didn't understand. If possible can you explain a simple Monte Carlo experiment in chemistry, which can be done in Matlab. Is it possible to simulate the growth of nano particles using the Monte Carlo method?

I've done a little simulation as a senior year elective, though not in Matlab. Can you be a little more specific about how/what you're trying to model with nanoparticle growth? Perhaps direct us to an abstract or two.
The first Monte Carlo simulation you should attempt is the drunken sailor problem. The basic idea is that you have a drunken sailor who has equal probability of taking a step forwards or backwards, and you let him stagger around on a one dimensional line for some number of steps (1000) and repeat the experiment a large number of times. You then get some summary statistics and plots around your experiment.
After you're comfortable with that concept, you can look at increasing complexity into 2 or 3 dimensional random walks, changing the weighting function (not 50-50 for each direction), or some specific algorithm such as Metropolis-Hastings Monte Carlo.
I hope this helps. If you're looking for specific programming help, posting on stack exchange would be a good idea.

I think this is a good idea. The 2D random walk can lead into diffusion-limited aggregation (DLA) with "sticky" particles, which is related to some models of nanoparticle growth.


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chemically inert molecule with piezoelectric properties
For a research project, I'm looking into existing  chemical molecules that have the following properties:

It is mostly chemically inert, safe for animals and children
It exhibits piezoelectricity -  production of high electric oscillating potentials under pressure
paramagnetic moment a plus, but not required

Are there known candidates for this?

What's wrong about quartz !

The ur-piezoelectric crystal is quartz. Being composed of $\ce{SiO2}$, it more or less epitomizes chemical/biological inertness. It has the additional advantage of being comparatively cheap and common. There are a number of other natural minerals which are piezoelectric (e.g., various perovskites). Most of those minerals are, as far as I know, relatively unreactive and non-toxic. Naturally, if toxicity is a concern, avoid the common lead-based compounds.
I've seen mention of materials that exhibit both ferrimagnetic and piezoelectric properties. If ferrimagnetism is acceptable to you instead of paramagentism, then you may wish to consider that as an avenue for research. If you need paramagnetic materials specifically, then it's probably not worth your time (unless your project is compatible with subjecting those materials to their respective Curie temperatures).

Tourmaline equally exhibits piezoelectric (as well as pyroelectric) properties.


text:
Reaction of an electrode
Why is the cathodic reaction of $I^-/AgI/Ag$ is $$\ce{AgI + e-  -> Ag + I-}$$ but not,
$$\ce{Ag+ + e- -> Ag }$$
whereas, in $Ag/AgCl(s)$, $KCl$(0.2M); it is
$$\ce{Ag+ + e- -> Ag}$$

The real reaction is $\ce{Ag+ + e- -> Ag}$ for both cases. Equation written in this way, ${E^0}$ will be equal to the reduction potential of Ag(I) in its 1 M solution.
If it is written as $\ce{AgI + e- -> Ag + I-}$, then ${E^0}$ will be equal to the reduction potential of Ag(I) in 1 M $\ce{I-}$ solution -or in other terms in ${K_{sp}(AgI)}$ Molar Ag(I). solution
Hence, technically, $\ce{Ag+ + e- -> Ag}$ is same with $\ce{AgI + e- -> Ag + I-}$. Latter is more descriptive and practically more relevant, however, because the presence of $\ce{I- /Cl-}$ ions in the medium limits the amount of $\ce{Ag+}$ cations via ${K_{sp}}$ and the notation readily implies this.


text:
Best place to take off vacuum for distillation
I have seen vacuum take-off adapters that are essentially plugs, which could be placed at the top of a three-way above the fractionating column, and also straight and bent vacuum take-offs with an inner stem that seem meant to be placed right before the collection vessel.
Would there be any difference in which place is better to vacuum from? Would it be best to maybe use a T-connector on the vacuum line and vacuum the top of the fractionating column as well as the collecting flask?

Assuming that you are using a typical distillation set up (boiling container, fractionating column, distillation head, condensation tube, collecting adapter, and collecting vessel), and the the "three-way" you are talking about is the distillation head. It is much better to put the vacuum take off adapter right before your collecting vessel. The reason is this set up creates a vapor flow that goes through your condensation tube, and most of your fractions will be condensed and collected.
If you attach the vacuum take off adapter on top of your distillation head, most of your vapor will take shortcut to your pump through your adapter instead of going to the condensation tube and the collecting vessel (this direction is a dead end).
Actually the top of the distillation head is the place to put your thermometer to measure the temperature of vapor, not for anything else.


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Dead-body Ashes: biodegradable or not?
In many cultures, human dead bodies are burnt to ashes, which is then thrown in rivers and some time buried in earth.
Are human ashes after burning dead bodies biodegradable?
Note: Earlier I posted this question here
But I was suggested to post it here.

According to one source, the major elemental constituents of cremation remains are calcium and phosphorus, presumably from the hydroxylapatite in bone (which consists primarily of calcium phosphate). The rest is mainly assorted trace metals, which exist in relatively minute concentrations in the human body and don't vaporize at the temperatures typical of a crematorium furnace.
Virtually all organic matter, comprising mainly oxygen, carbon, hydrogen, nitrogen, and sulfur, is oxidized into a variety of simple compounds and is removed as a mixture of gases in the furnace fumes (and perhaps some finely dispersed, light solid particulates).
Consequently, what remains is largely inorganic and non-calorific in nature. While the mineral and metal content of the ash can be dissolved and subsequently absorbed by various organisms in trace amounts, there's probably little or no organic material left that can be metabolically decomposed by bacteria, fungi, or plants as a direct energy source. Of course, there are chemical processes that can further break down the ash and isolate specific components via redox reactions (dissolution in strong acids to give salts, electrolysis, etc.), but obviously that's distinct from biodegradation.

It could be used to correct P and Ca deficiency in the earth for plants.


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Oxidation number and Pourbaix diagrams
In a Pourbaix diagram (E-pH diagram) where the regions are given , but the assignment of these to the elements we have is not done yet , how can the oxidation numbers helps me out assigning each element in its appropriate region?

If you do not have a list of the species that belong in the Pourbaix diagram, then it is not possible to accurately predict what species occur where, unless you have extensive experience with the system in question.  If you have a list of species and need to place them in the appropriate regions of a Pourbaix diagram (fill in the blank style), then the higher oxidation state species occur at the top of the diagram and the more basic (e.g. OH containing) species occur towards the right.
If you have Latimer diagrams of all the species in the Pourbaix diagram (this will require the Latimer diagrams for both acidic and basic conditions, typically) then you can more accurately predict the structure of a Pourbaix diagram by (a) writing balanced half reactions and (b) adjusting them for the appropriate pH using the Nernst equation.


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Why is H₂O V shaped?
We know that the molecule of H₂O is V-shaped. This is what makes it a dipole.

But why is that? I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other, until they are diametrically opposite to the oxygen molecule. But that doesn't happen. Why is this?

If you're interested at why molecules have a given shape, read up on VSEPR theory (Valence shell electron pair repulsion), which is the typical undergraduate-level model for understanding/explaining these things.

The oxygen atom fills its octet rule by forming two bonds. It shares 1 electron in a covalent bond with each hydrogen and has 4 remaining valence electrons. It is  sp$^3$ hybridized and has the 4 non-bonding electrons in two lone pairs. An sp$^3$ hybridized atom has four attachment points spaced approximately 109$^\circ$ apart and has the shape of a pyramid with a triangular base. See this image for an example. You don't typically see the water molecule drawn in 3 dimensional space with the lone pairs of electrons, all you see is a planar molecule with H-O-H bond angle of ~109$^\circ$.

It is not the + charges on hydrogens, but the negative charges on the O-H bonds, and on the O it self that is dominant in determining the molecule's shape. There are four non-bonding electrons on O -two lone pairs- on top of the two O-H bonds.
So you should think of four groups around O, not two.

I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other.

Well, this is not a complete answer to your question, but this could be a factor I guess.
Bond pair – lone pair repulsion is always greater than bond pair – bond pair repulsion
So what I want to say is that the repulsion between the O---H bond and the lone pair is so heavy such that it is really hard for two hydrogen atoms to get away each other as you are expecting to happen.

In the bonding of Oxygen with Hydrogen, Oxygen ends up with 2 extra electrons, giving it what is called the "bent" shape. This is because the attachments and the electrons all repel each other but the electrons are not drawn except in a Lewis Dot structure such as the one here: http://en.wikipedia.org/wiki/File:Water-2D-flat.png

Oxygen has four extra electrons: two lone pairs.


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Why is the carbon dioxide we exhale transparent?
Why is carbon dioxide that comes from dry ice opaque and smoky while the carbon dioxide that we exhale transparent?

Essentially, because the carbon dioxide sublimates from solid (dry ice) to gas at a very low temperature (roughly −78 °C at 1 atm), it causes water vapour in the air to condense, causing a visible fog. Thus what you are seeing is not carbon dioxide, but rather water.
When we exhale and it is reasonably warm, the carbon dioxide expelled is roughly body temperature (36–37.5 °C) and so we do not see any fog forming.
What happens when we exhale in cold weather is a bit different. Our breath contains a high percentage of water, higher than the percentage of water that can stay in gaseous form at a low temperature. Thus the ‘extra’ water condenses into droplets.
For reference, here is the phase diagram of $\ce{CO2}$:

And here is the phase diagram of $\ce{H2O}$:

And here is the saturation fraction of water at various temperatures at 1 atm (sea level):

How does saturation fraction relate to saturation vapour pressure?

@gerrit saturation fraction is the concentration while vapour pressure is a pressure. They're related by the formula $p=nRT/V$ where p is pressure, n is amount of the gas in mol, V is volume, T is temperature and R is the gas constant.


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Why are only sigma bonds used in determining the geometry and bond angle in VSEPR theory? Why aren't pi bonds used?
I have read that in VSEPR theory, multiple bonds are considered or treated as single bonds when predicting the geometry of a molecule? I've read in Yahoo! Answers that it is because only sigma bonds are used in determining their shape. What's the reasoning behind this?

VSEPR is used to predict highly idealized geometries corresponding to Platonic solids or combinations thereof (e.g., the trigonal bipyramidal geometry being essentially the result of fusing two faces of a tetrahedron). The basic rationale behind its use, and the reason for its success in many situations, is the recognition that electrons, being like-charged, repel one another electrostatically. Hence, as a first approximation, it's reasonable to assume that the most stable spatial configuration of any molecule would have the atoms spaced as uniformly and as distant from one another as possible. VSEPR achieves this by placing the central atom of a simple molecule (or some molecular sub-structure) at the center of a geometrically perfect polyhedron having as many vertices as non-central atoms and lone pairs, and then placing the non-central atoms and lone pairs at those vertices.
The reason that $\sigma$ bonds (and lone pairs) determine the geometry is that they form the basic skeleton of the molecule. Recall that $\sigma$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented along the imaginary axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the two nuclei. $\pi$ bonds, on the other hand, are essentially orthogonal to the $\sigma$ bond skeleton, and are substantially weaker. Moreover, $\pi$ bonds do not exist in isolation, meaning any $\pi$ bond between two given atoms is always formed secondarily to the $\sigma$ bond between said atoms. As such, $\pi$ bonds do not alter the basic idealized geometry of a molecule as dictated by $\sigma$ bonding, although in practice, because they do actually introduce additional electron density and require closer orbital overlap, bond multiplicity affects bond length and bond angles (as does, for that matter, the size of the atoms involved). The magnitude of those deviations from the idealized VSEPR geometry could, in principle, be quite large, although most often those deviations are slight enough that VSEPR remains useful as a first approximation.

Perhaps it's worth noting that pi bonds do contribute to the shape of a molecule and VSEPR takes this into consideration as a deviation from the first approximation.  Examples would be $\ce{CF2O}$, $\ce{NO2}$ and $\ce{SO3^2-}$ where the change in bond order leads to repulsive effects and deviations from the 120 degree ideal angle.  See Miessler & Tarr 5th edition pg 56.

@bobthechemist, thanks for the reference. As I noted, the change in electron density with increasing bond multiplicity does distort bond angles and lengths, causing deviations from the ideal geometry. Would you agree that it's correct to say that the basic root geometry is solely determined by steric number? My understanding of VSEPR has always been that it doesn't distinguish between multiple and single bonds, and that all corrections to the ideal geometry predicted by VSEPR are the result of empirical observations of the real molecular structures (e.g., by X-ray diffraction).

@bobthechemist, I've read various guidelines that have suggested adjustments based on, e.g., the presence of lone pairs in place of bonded atoms (in those cases, some sources suggest that, as a correction to the first approximation, the bond angles between remaining atoms should be collapsed on the order of ~4 degrees, while others maintain angles equal to the angles of the "root" parent structures even for the "derived" shapes). Is there a standard practice for making those types of predictive corrections?

Yes, I agree that the heart of VSEPR is steric number.  The presence of multiple bonds and lone pairs distorts angles from their ideal values.  Personally, I've never seen any (semi)quantitative estimates of these changes and I only use  these deviations to explain periodic trends (e.g. the decreasing angles upon going from methane, to ammonia to water).

Actually, if there were a type of pi bonding that is the first bond between the binding atoms -without any sigma bonds- molecule shape wouldn't deviate from what VSEPR predicts more than it deviates now.

@sencer, if there's an actual example of pi-bonding without accompanying sigma-bonding, I'd be fascinated to hear about it. I can't imagine why that would ever be energetically favorable, since that implies zero internuclear electron density, which should result in repulsion between the two nuclei, and less stabilization of the pi electrons than if they were in a sigma bond. Obviously, I could be mistaken. As for the relationship that would have to the VSEPR-predicted shape, it seems to me a purely hypothetical exercise, but I think you'd probably be right.

It was hypothetical. I don't think such a molecule exists, not at least in its ground state. Though, one might be able to excite a molecule to get something like that, for a short period of time.
Cyclopropane, however, isn't really sigma bonded. Perhaps some where between pi and sigma.

@sencer, yeah, cyclopropane is an oddity. From what I've read, there's only one sigma-type bonding MO that encompasses the three carbon atoms, but of course symmetry requirements preclude head-on overlap all the way around the ring, so the electron density of that MO is directed toward the center of the ring. All other C-C bonding is apparently accounted for by pi-type MOs. That probably goes some way towards explaining its peculiar reactivity.

I think the trigonal bipyramidal geometry is better described as the fusing of two trigonal pyramids ... tetrahedrons have 109.5 degree bond angles. Trigonal bipyramidal geometries have 120 degree bond angles.

@Dissenter, I was strictly discussing the idealized geometry of these shapes as mathematical constructs. I didn't imply any congruence w/r/t bond angles between the tetrahedral and trigonal bipyramidal molecular geometries. A triangular bipyramid is not a platonic solid, but it can be constructed from them by fusing two tetrahedrons. Besides, what exactly do you mean by a "trigonal pyramid" that distinguishes it from a tetrahedron? A tetrahedron is just a special case of a triangular pyramid when all the faces are congruent.

I mean a triangle with an equilateral base.

@Dissenter, well, that's a tetrahedron if the other three faces are congruent equilateral triangles.

But trigonal bipyramidal molecules don't have equilateral triangles for their other faces.

@Dissenter, if you have five identical ligands around the central atom, then I'm almost certain all of the faces of the bipyramid would be congruent equilateral triangles. See the wikipedia page.

@Dissenter, also, just to clarify, I am most emphatically not stating that this is actually the case. Of course, in reality, axial bonds are longer, multiple bonds are typically shorter, steric demands differ between disparate ligands, etc. My point is only that the most basic VSEPR analysis would predict that idealized geometry.

Question actually implies a misunderstanding. VSEPR has nothing to do with number of sigma bonds.
VSEPR is a primitive theory, relying only on simple electrostatic effects. In this view, nature of the bond is not relevant but only the number of groups surrounding the central atom is. This is the number of bound atoms, an number of unbound electron pairs.
All VSEPR says is that, groups surrounding the central atom should be separated as much as they can. This yields linear shape for two groups, triangular for three, tetrahedral for four etc...

You're obviously correct to note that the VSEPR geometry is determined by steric number. However, the OP's question is tantamount to the question of why multiple bonds and single bonds are treated identically in calculating said steric number. Clearly, the specific nature of sigma bonds is relevant to that. Moreover, any bonding between two given atoms will always include exactly one (and only one) sigma bond, so as a surrogate for counting bonded atoms, counting sigma bonds directly will always yield the same result.

Thank you for your kind note, Greg. You are right, with a few exceptional cases, when you are saying steric number is same with #sigma bonds. This is a neat -let me say- "coincidence". Though it is absurd to count sigma bonds when you are trying to figure out steric number, since only way to count sigma bonds is actually counting steric number. Also, relating VSEPR to MO theory is neither historically, nor logically reasonable. I understand what OP is asking, and I think the fair answer to this question is: What you read is not correct, bond symmetry is irrelevant.

I take your point, and I voted you up. As I think more about it, I think the fundamental point of the nature of bonding being irrelevant is correct. The gist, if I've understood you correctly: as it happens, sigma bonds are the primary form of covalent bonding, but even if some other form of covalent bonding were holding two atoms in close proximity, then the combination of internuclear and electron-electron repulsion would still demand maximum spatial separation, within the limits allowed by the bonding. That strikes me as an entirely accurate point.

Because any $\pi$ bond is always affiliated to a $\sigma$ bond, which participates the determination of the geometry. It is no way that a $\pi$ bond can be different orientation to its corresponding $\sigma$ bond.


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Finding molar mass of an unknown acid

"In the titration of a solid acid, an endpoint is reached after 22.0mL
  of 0.120 M $\ce{NaOH}$ has been added. Assuming that each acid particle
  contains two acidic hydrogens and both are neutralized, what is the
  molar mass of the unknown acid?"

I really don't know how to solve this. I've figured out how to calculate the moles of acid were in the sample but don't see how to find the molar mass from that...any ideas?

From the data above, all you'll get is the moles. For molar mass calculation you need more data, like the weight of acid added, or the density of the solution.

If the acid is solid, then the final solution has the same density as pure water due to neutralization, so you basically know the volume and the total mass contained, and you know the mass of $\ce{NaOH}$ added; I think the book may be right...

Your data is not enough to calculate molecular weight. You may have missed the mass of the solid acid you have weighed.
Let's assume you have weighed $a$ gram of acid. Then:
$Molecular Weight = \frac{Weight}{Moles_{acid}} = \frac{Weight}{2 \times Moles_{NaOH}} =  \frac{Weight}{2 \times Molarity_{NaOH} \times Volume_{NaOH}} = \frac{a}{2 \times 0.120 \times \frac{22.0}{1000}}$

That's what I thought, must be an error in the book.

authors: الالماسة السوداء