text:
Chemical equilibrium and Le Chatelier Braun for ammonia production
In the book of Chemical, biochemical and engineering thermodynamics 4th edition from Stanley I. Sandler (Wiley, 2006), there is a problem about the ammonia producing reaction at page 770.
13.17 The simple statement of the LeChatelier-Braun principle
given in Sec. 13.1 leads one to expect that if
the concentration of a reactant were increased, the
reaction would proceed so as to consume the added
reactant. This, however, is not always true. Consider
the gas-phase reaction
Show that if the mole fraction of nitrogen is less than
0.5, the addition of a small amount of nitrogen to the
system at constant temperature and pressure results
in the reaction of nitrogen and hydrogen to form ammonia
whereas if the mole fraction of nitrogen is
greater than 0.5, the addition of a small amount of
nitrogen leads to the dissociation of some ammonia
to form more nitrogen and hydrogen. Why does this
occur?
So I approached this problem the following way:
\begin{align}
  \ce{N_2 + 3H_2<=>2NH_3}
\end{align}
The equilibrium constant:
$$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$$
Now when we write this in terms of reactant consumption X:
$$K_c=\frac{(2x)^2}{([N_2]-x)([H_2]-3x)^3}$$
Now, when I use some values to try to prove this. I never get a negative X value, so I am not getting ammonia dissociation? Mind that I am not an expert on chemical thermodynamics and this question was also proposed to us in an industrial chemical engineering class where we discussed the Haber-Bosch process.
Can someone help me out here? Why is this not working? Or what is the right strategy to approach this problem.
Thank you in advance

In effect, your $x$ is ${1\over2}\ce{[NH3]}$. Well, of course it can never be negative, because the negative concentration of ammonia (or any other substance, for that matter) is physically impossible.

I think you should write a mass balance for the initial gas so that you can express $[\ce{N2}]$ and  $[\ce{H2}]$ as mole fractions.  It is also seems as if the question is asking you to start with an assumed equilibrium mixture of gas and then calculate the change in this equilibrium composition as $y_{\ce{N2}}$ is changed.  In other words, you need to find $\frac{d x}{d y_{\ce{N2}}}$.

But that is rather easy, you just fill in some values, you get the equilibrium values. Then you take Kc constant. Iteration 1 --> you find the equilibrium, then you adjust the N2 value and look how the equilibrium shifts. Then with the new values, you can adjust N2 again and look how the equilibrium shifts.


text:
At what pKa value do we consider an acid strong?
In between the traditional strong mineral acid that dissociate completely in aqueous solution, and obvious weak organic acid like acetic acid, things get murky.
For example, oxalic acid has a $\mathrm{p}K_\mathrm{a} = 1.46$ which is  about 1000 times stronger than acetic acid, but no one seems to qualify it as a "strong" acid. How about trichloroacetic acid, $\mathrm{p}K_\mathrm{a}\approx 0.5$? Well, some people start calling it "strong" acid.
So, what $\mathrm{p}K_\mathrm{a}$ value would qualify an acid as a strong acid?
I would like to know a concrete numbers, not merely “smaller $\mathrm{p}K_\mathrm{a}$ means stronger acid” as found in other questions. Just like a specific body mass of $90\ \mathrm{kg}$ would qualify a boxer to be a heavyweight boxer.

Your question is not as easy answered as you might want it to be. The reasons for that are mainly given in the answer by Zen to What is a "Strong" Acid?
The main problem is, that there hardly can be an absolute acid strength and you will always have to take the reference system into account. Another problem arises, that there is no consistent (or official by IUPAC) definition of the acid constants. Usually, when you are talking about $\mathrm{p}K_\mathrm{a}$ values, then you are most likely already implying aqueous solution.
It is a quite accepted classification of strong acids (in water), to state that they completely dissociate. Hence in the equilibrium is entirely shifted to the right side:
$$\ce{HA (aq) + H2O (aq) <=> A- (aq) + H3O+ (aq)}.$$
The equilibrium constants for this reaction is
\begin{align}
K  &=       \frac{a(\ce{A-})\,a(\ce{H3O+})}{a(\ce{HA})\,a(\ce{H2O})}
    \approx \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})\,c(\ce{H2O})}
    \approx \frac{K_\mathrm{a}}{c(\ce{H2O})}\\
K_\mathrm{a} &=
    \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})}\\
\mathrm{p}K_\mathrm{a} &= -\lg K_\mathrm{a}.
\end{align}
For a strong acid we can see that the concentration of undissociated acid is (nearly) zero. Therefore we write
\begin{align}
\lim_{c(\ce{HA})\to0} \left(K_\mathrm{a}\right) &= \infty&
&\therefore&
\lim_{c(\ce{HA})\to0}\left(\mathrm{p}K_\mathrm{a}\right) &= -\infty.
\end{align}
Strictly speaking, in aqueous solutions the acidity constants cannot be measured, since the hydronium ion is the strongest acid in this medium.
So in priniciple, the value that separates strong from weak acids would be the $\mathrm{p}K_\mathrm{a}$ of the hydronium ion. With that we arrive at a completely new problem, which is quite efficiently discussed here: What is the pKa of the hydronium, or oxonium, ion (H3O+)?
From this question we can take home the suggestion the value of $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = -0.7$ for the purpose of comparing it to other acids. In other words $\mathrm{p}K_\mathrm{a}(\text{strong acid}) < -0.7$ (in water). Taking this literal, however, might lead to many misconceptions, and using a relative approach of “this acid is stronger than this acid” should  be preferred in any case.

Good answer, but for measuring the Ka of strong acids in aqueous solution see: "Ionization of Strong Electrolytes. III. Proton Magnetic Resonance in Nitric, Perchloric, and Hydrochloric Acids" http://scitation.aip.org/content/aip/journal/jcp/22/12/10.1063/1.1739994 and "The Dissociation Constant of Hydrochloric Acid" http://pubs.rsc.org/en/content/articlepdf/1936/tf/tf9363200743

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text:
Is H+ (in an aqueous solution) = H3O+?
I was learning about autoionization in science and my science teacher wrote $\ce{H3O+}$ as $\ce{H+}$ — now, my intuition would lead me to assume that:
$$\ce{H3O+} = \ce{H+}$$
But, $\ce{H+}$ is just an atom with a positive charge because its lack of electrons, while $\ce{H3O+}$ is the product of $\ce{H2O}$ nabbing that $\ce{H+}$, right?
So, is saying that $\ce{H3O+} = \ce{H+}$ just for the purpose of simplicity in the sense that we are assuming that in the solution (aqueous, is the context I am dealing with), the positive hydrogen atom (is it more specifically called an ion?) will bond with a water molecule?
P.S. I know that you would call $\ce{OH-}$ hydroxide, and $\ce{H3O+}$ hydronium, so  what would you call $\ce{H+}$ (if it even has a name)?

@DavePhD, no - I am not asking whether or not the hydrogen ion exists, although that is a very intriguing question... that I will check out.

well after you read my answer there, explain what more you want to know

When Hydrogen ion is solvated in water $\ce{H+_{\mathrm{aq}}}$, it gives$ \ce{H3O^{+}}$

@DavePhD - Ok, I will see if any further clarification is needed.

@Jan - No, my question is regarding more formality rather than the actual chemistry behind it.

When dealing with an aqueous solution, you are correct that the $\ce{H+}$ ion is equivalent to $\ce{H3O+}$ for all intents and purposes. Due to the abundance of water in solution, molecules of $\ce{H2O}$ will readily pick up the hydrogen ions, meaning that most of the $\ce{H+}$ in an aqueous solution is actually of the form $\ce{H3O+}$. While this is the case, using $\ce{H+}$ over $\ce{H3O+}$ is more correct in general contexts. If you are using something like ethanol ($\ce{CH3CH2OH}$) as your solvent, then instead of having hydronium ions, you would have $\ce{CH3CH2OH2+}$ as your measure of concentration of acid.
As for your other questions, the $\ce{H+}$ would be considered an ion rather than an atom. This is because it is a charged species. Either $\ce{H+}$ can be referred to as a proton or as a hydrogen ion, since $\ce{H-}$ is too unstable a species to be formed. While some of your $\ce{H+}$ in solution might be deuterium or tritium, most people will understand what you are talking about if you just call the hydrogen ion a proton.

Great info! Thanks a lot -
Just one quick clarification:
Are you saying that we literally just call a hydrogen ion a proton in the specific context?
e.g. Context: autoionization of water
"Yo, dude, guess what? If a water molecule takes the proton from another molecule it makes hydronium!"

Yeah, that works. You can either refer to it as a proton or a hydrogen ion.

In water, $\ce{H+}$ never exists, the proton is always part of some other chemical (or ion). So, it would be a bit incorrect to say that $\ce{H+}$ bonds with water, at least in the sense that a reaction has occurred between a species best described as $\ce{H+}$ and something else (water).
I am tempted to claim that the answer by Bz is wrong, but it's reasonably close to being correct so will refrain. $\ce{H+}$ in water exists as both the $\ce{H3O+}$ ion and other $\ce{H + $n$\ H2O}$ species such as $n=2$ $\ce{H5O2+}$ and $n=3$ $\ce{H7O3+}$. These latter can be thought of as polymeric ions. (Because of charge repulsion, doubly negative species are a negligible component of liquid water.) The solvated (or aqueous) proton is generally just called the hydrogen ion, but other names are certainly possible. $\ce{H+}$ is the most common species in the Universe, but it doesn't exist in water. Since, for reactions in solution, the solvent is quite important, but beginning students are usually 'protected' from this level of detail, $\ce{H3O+}$ is a good way to express the concept how the $\ce{H+}$ (reactive) species is being 'carried' (or where it has gone to). Keep in mind that the hydrogen bonding that goes on in water is important for its physical and chemical properties. Any real reaction involving the transfer of $\ce{H+}$ in water will be similar to $\ce{XH + Y -> HY + X}$  (where these species may be charged or neutral (of course charge will balance) and definitely not involve $\ce{XH + Y -> XY- + H+}$, but the last could be $\ce{-> XY- + H3O+}$ (with the understanding that $\ce{H3O+}$ is shorthand for $n=1$ mostly with a bit of $n=2$ and less $n=3$, and less and less $n=4,5,6,7$).

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text:
Is sulfide ion a stronger base than hydroxide ion?
I received these three responses. They are all, except the last response, incorrect to varying degrees. I am, however, unsure about how to grade these responses because the question itself, in my opinion, misleads the students by saying that both have the same pH and to use this information to determine the strength of the bases. Their responses also raised a few questions of my own. What are your opinions?

Question
It is an experimental fact that the pH of 1 M $\ce{Na2S}$ is essentially the same as the pH of 1 M $\ce{NaOH}$. Based on this information, is $\ce{S^{2-}}$ or $\ce{HO-}$ the stronger Bronsted-Lowry base? How can you tell?

Student 1's response:

A base's strength is measured by how much it can increase the hydroxide ion concentration in a solvent relative to its own initial concentration.
Given that both the 1 M $\ce{Na2S}$ and 1M $\ce{NaOH}$ solutions have the same pH, they both have the same $\ce{[H3O+]}$ concentration and therefore the same $\ce{[HO^{-}]}$ concentration. Therefore, the two bases are the same strength as each creates a $\ce{[HO^{-}]}$ concentration equal to their initial molarities in water.

I think this response could be improved if the student mentioned the leveling effect and recognized how all bases stronger than hydroxide ion are "leveled." It's just like the high striker game at the fair; you can hit it hard enough to make the puck go all the way to the top, and so can someone else, but that doesn't necessarily mean you guys are equal in strength; it just means that the device quantifying strength doesn't differentiate between "strong" and "super strong."
Student 2's response

$\ce{S^{2-} + H2O ->HS- + HO-}$
From this equation and the problem statement, it is clear that $\ce{S^{2-}}$ hydrolyzes water to create a hydroxide ion concentration in water equal to its own initial molarity. Hydroxide ion isn't hydrolyzing water to create hydroxide; the below equation is non-sensical. Therefore, sulfide ion is the stronger base.
$\ce{HO^{-} + H2O ->H2O + HO-}$

Is the second equation really non-sensical?
Student 3's response:

The stronger base has the weaker conjugate acid.
Sulfide ion's conjugate acid is $\ce{HS-}$.
Hydroxide ion's conjugate acid is $\ce{H2O}$.
From the $\ce{K_{a}}$ table in the beginning of the lab manual, one knows that $\ce{HS-}$ is a weaker acid than $\ce{H2O}$. Therefore, sulfide is the stronger base.

@Jan - my dilemma too; 1 starts off okay but makes the wrong conclusion. 3, however, I think is succinct and to the point.

3 is technically correct but it's hardly based on the information given in the question. The point made in 1 is complicated by the fact that the resulting species HS- also displays acid-base properties, in order to do it properly consult an analytical chem textbook, they will show you how to set up a system of equations to solve for the pKa.

@orthocresol is the information in the question helpful? I think it's irrelevant at best and misleading at worst.

How about "Is $\ce{Na^17OH}$ or $\ce{Na^16OH}$ as stronger base, given that a 1 M solution of each have essentially the same pH?". Same setup, different answer?

Equilibrium Expressions Related to the Solubility of the Sour Corrosion Product
Mackinawite Ind. Eng. Chem. Res. 2008, 47, 1738-1742 compiles 18 different studies concerning the second Ka of H2S and shows that they vary over 7 orders of magnitude from $10^{-19}$ to $10^{-12}$
So firstly it isn't accepted whether or not Na2S is stronger than NaOH.
Putting that aside, the only way that 1M sulfide could yield 1 M hydroxide (and hence pH be equal to pH of 1 M hydroxide) is if sulfide was infinitely strong.  If they were equally strong, it would only yield 0.5M hydroxide.
Student 1: clearly wrong.
Student 2: correct (based upon the information given)
Student 3: can't say without seeing the table, but if the table actually says that SH- is weaker than water than the answer is true, but not correct in the sense that the instructions say "Based on this information".

what do you mean sulfide has to be infinitely strong?

@Dissenter I mean sulfide would need to completely react with water to create hydroxide, with no sulfide remaining, for 1 mole of sulfide to yield 1 mole of hydroxide.

oh I see now. The question accounted for that by writing that the two (1 M Na2S and 1 M NaOH) are "essentially" the same in terms of pH.

What is the second Ka of water?

@Dissenter http://chemistry.stackexchange.com/questions/24342/what-is-the-ka-of-oh-and-kb-of-h3o

It does depend a lot on your marking scheme, but looking at it zealously I read two partial questions:

Is the information given enough to discern the strength of either base? If yes, which parts led to the conclusion? If no, why can we not tell?
Which base actually is stronger according to the information given.

The question attempts to ask within its own boundaries, so no previous knowledge of $\mathrm{p}K_\mathrm{a}$ values seems assumed.

A perfect answer should thus include the points:

levelling effect of water, i.e. sulphide cannot be the weaker base or its pH value would be lower;
It is impossible to tell whether $\ce{S^2-}$ is stronger or whether the two are equally strong.

Additional bonus points could then be given to answers like student 3’s which further add additional information such as the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HS-}$ and $\ce{H2O}$.

Student 1’s explanation of base strength is incorrect, no questions asked. (Base strength is measured by $K_\mathrm{b}$ values.) The levelling effect is not present and apparantly not known. The conclusion is possible but incomplete.
Strictly speaking, this should probably be awarded zero.

Student 2 somehow starts talking about the levelling effect by mentioning that sulphide abstracts a proton from water to generate hydrogen sulphide and hydroxide. However, the water equation is far from nonsensical; it is an equilibrium that actually occurs on a molecular scale. The conclusion drawn may be chemically correct, but it is only semi-correct within the boundary of the question.
Maybe give him half of what you allocated for the levelling effect and nothing for the rest.

Student 3 gives chemically correct facts, but fails to address the problem in the question. Students should be taught to read the question, accept its premises, and then answer according to what is inside unless the question clearly asks for external knowledge.
We can give him a bonus mark for the knowledge to answer the question ‘extra-universally’, but the informatio I would be expecting is simply not there.

Also remember that you should generally give marks towards the lower end of the possible spectrum when correcting a test. Students are likely to come to their TAs to ask for better grading but very unlikely to say ‘but I got this completely wrong, please deduct the points.’

The second student's comment about nonsensical is very similar to what the following journal article says: http://pubs.rsc.org/en/Content/ArticleLanding/1998/AN/a705491b#!divAbstract

@DavePhD I reject that articles assumptions and conclusions. (Note that I never learnt a $K_\mathrm{a}$ of water being defined by the equation $\ce{H3O+ + H2O <=> H2O + H3O+}$; it was defined with the help of the autoprotolysis that the article later write to ‘correspond to a physical process.’)

the article is saying that equation is for the Ka of H3O+, not the Ka of H2O

Thinking that if they were equally strong, 1M $\ce{S^2-}$ should give only 0.5M OH- isn't really correct.
Another thing is that the $\ce{S^2-}$ is in fact, stronger base than $\ce{OH-}$, but weaker base than $\ce{O^2-}$.
$\ce{S^2-}$ removes 1 H+ from $\ce{H2O}$ and forms $\ce{HS-}$ and OH- (same quantity for each), then an equilibrium state is reached ($\ce{HS-}$ and OH- can't deprotonate each other, even OH- is a little bit stronger than $\ce{HS-}$).
1M $\ce{S^2-}$ will give exactly 1M OH-, so Na2S will have the same pH than NaOH in water, even the first is stronger base than NaOH. It's also true for other much stronger bases like NaNH2 or NaH which will give the same pH than NaOH, but are much stronger than OH- as a base.

Question
It is an experimental fact that the pH of 1 M Na2S
is essentially the same as the pH of 1 M NaOH. Based on this information, is S2− or HO− the stronger Bronsted-Lowry base? How can you tell?

Assumptions
Questions asked as part of a course usually lack context for people not taking the course. The question does not state which solvent was used and at what temperature and pressure measurements were done. It also does not say what level of theory to use in the answer. Lastly, it uses the terms "essentially the same" and "stronger base" without defining them. Here are the assumptions I will make in this answer:

The solvent is water.
The measurement was taken at room temperature and normal pressure where pKw is 14.
Ionic strength differences are ignored. Activity coefficients are set to 1.
"essentially the same" means a difference of 2% in concentration is fine.
The criterion for "stronger base" is the pKa of the conjugate acid.

Answering a different question first
I will first compare a 1 M solution of NaOH and a 1 M solution of 17-oxygen NaOH which I will abbreviate as NaQH. To make things easier, I will assume that the NaOH and the water are made with isotopically pure 16-oxygen.
For the NaOH solution, the final concentrations are 1 M $\ce{OH-}$ and $\pu{1e-14 M}$ $\ce{H+}$, and the pH is 14.
For the NaQH solution, I will use 49 M for the concentration of water to simplify the math. So there is 1 M of total 17-oxygen atoms in the system. For simplicity, I am assuming no isotope effects on pKa values. The 17-oxygen will distribute equally among the hydroxide and the water, with 2% of oxygen atoms being 17-oxygen. The final concentrations are 0.02 M $\ce{QH-}$, 0.98 M $\ce{H2Q}$ and $\pu{1e-14 M}$ $\ce{H+}$. From this, we can calculate the Ka as:
$$\ce{H2Q <=> QH- + H+}$$
$$K_a = \frac{[\ce{QH-}][\ce{H+}]}{[\ce{H2Q}]} = \frac{1}{49} \cdot \pu{e-14}$$
So the value of the Ka of oxygen-17 hydroxide is that of the Kw divided by 49, and the equilibrium concentrations are given above. The Ka value of hydroxide is the same, of course, because we excluded isotope effects.
What happens if the base has the same strength as hydroxide?
Now we turn the problem around and start with a base $\ce{B-}$ that has the same strength as hydroxide. Because it has the same strength, the equilibrium concentrations will be essentially the same, i.e. 0.02 M $\ce{B-}$, 0.98 M $\ce{HB}$, and $\pu{1e-14 M}$ $\ce{H+}$. The only difference is in the hydroxide concentration, which is 0.98 M instead of 1 M (well, the hydrogen ion concentration will be slightly different, too, 2% higher, and the pH will be 13.99 instead of 14.00).
Why are hydroxide and $\ce{B-}$ not at the same concentration, even though they have the same strength? It is because the conjugate acid (HB or water) are present at different concentrations (about 50-fold difference).
What is the answer to the question the OP cited in the posted question
From the given information (together with the assumptions I had to make), sulfide could be a stronger base than hydroxide, or of the same strength.
What is the answer to the OPs question?
According to a 2018 paper by May et al title "Goodbye to $\ce{S^2-}$ in aqueous solution" (https://doi.org/10.1039/C8CC00187A), the sulfide ion has never been detected in aqueous solution. The hydroxide ion has, so sulfide is the stronger base by that argument. As for the OP's question in the body "What are your opinions?", that is mainly opinion-based and should be closed.

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text:
Can we safely consume ions?
When we drink salt water, we are consuming the sodium and chloride ions, right? We can't consume elemental sodium as it's highly reactive but the consumption of its ions seems to be fine. Does that mean that we can consume ions of all elements?

No it does not. I strongly recommend against consuming Hg(2+) ions

Everything is toxic in the right dosage.

It's just like berries. You should only eat the edible ones. All the others may harm you.

Nope:

Lead
Arsenic
Cadmium
Mercury (H/T @orthocresol)

And multiple others...

To clarify a bit further, the hazards from sodium metal and from, say, ionic mercury are different:
Sodium metal is so "eager" to cast away an electron that it will do so quite aggressively when it contacts any aqueous solution, including saliva.  It thus represents a "reactivity" or "chemical" hazard in the mouth.  Once each $\ce{Na^0}$ has forcibly donated its electron, though, it becomes the non-hazardous $\ce{Na+}$.
Ionic mercury, on the other hand, represents a relatively minor "direct" hazard to the body.  Instead, the various forms of mercury interfere with biological/biochemical processes in the body, leading to its severe toxicity.  One particular example highlighted in the Wikipedia article on mercury poisoning is the irreversible inhibition of selenoenzymes:

One mechanism of mercury toxicity involves its irreversible inhibition of selenoenzymes, such as thioredoxin reductase .... Although it has many functions, thioredoxin reductase restores vitamins C and E, as well as a number of other important antioxidant molecules, back into their reduced forms, enabling them to counteract oxidative damage. Since the rate of oxygen consumption is particularly high in brain tissues, production of reactive oxygen species is accentuated in these vital cells, making them particularly vulnerable to oxidative damage and especially dependent upon the antioxidant protection provided by selenoenzymes. High mercury exposures deplete the amount of cellular selenium available for the biosynthesis of thioredoxin reductase and other selenoenzymes that prevent and reverse oxidative damage, which, if the depletion is severe and long lasting, results in brain cell dysfunctions that can ultimately cause death.

Thanks Brian! Your explanation was amazing and really cleared my doubt. However, another question that came to mind after reading your answer was whether or not the position of an element on the periodic table(metals, transition metals and non-metals) has anything to do with the danger its ions poses to humans. For example, as you and @orthocresol have said, sodium ions pose no danger to humans but ionic mercury does. Is this due to sodium being a metal and mercury being a transition metal or is there no way to generalize?

@Goodyear Yes, there are general positional trends. This warrants more explanation that just a comment, though -- I recommend asking it as a new question.


text:
What intrinsic property do metals have that causes their delocalisation?
So I understand that electrons in metals delocalise completely and are free to move around. But my question is, why would something like Nitrogen not delocalise, whereas say Iron or Bismuth does? What is special about their orbital configuration that causes this?


text:
Are these chemicals suitable for disinfecting my water tank?
This is the same question that I asked before, only for another product this time. I have more detailed information about this product, including a data sheet.
There is an 8000 liter tank containing water for household use like showers and washing dishes and clothes (but not drinking or cooking). For more details, read here or here.
I need to combat harmful organisms and microorganisms in the tank, but I live in a remote rural area and can't just buy any chemicals I want. Thus, I am asking whether this product will work.
Here are the ingredients of the product:

Water
  Sodium Hypochlorite
  Sodium Chloride
  Sodium Carbonate
  Sodium Chlorate
  Sodium Hydroxide
  Sodium Polyacrylate

Here is a full data sheet.
If this product is suitable, what is a good amount and frequency?
If this product is not suitable, which ingredient is the culprit?

The last time I checked it, microorganisms were pretty small (I think even so small so that you can't seem them with the bare eye). While the chemicals are all more or less harmless I wouldn't underestimate the potential danger of the microorganisms -- even if you can't see them after cleaning. So I would definitely consider checking the water quality afterwards. There should be lots of options available to let the water quality get checked by a professional institution (probably depends on the country), so I think this is very commonly done/cheap.

When you start asking questions about chemistry you are only left with more questions... in general terms of safety, using household bleach to create a concentration of 1 - 3 ppm sodium hypochlorite is relatively safe (especially compared with sewage). In specific terms, the answer is two-fold; you have to consider the amount of exposure and the component's toxicity (the difference between medicine and poison is the dose).
It is important to note that the SDS for household bleach only lists sodium hypochlorite in the hazardous ingredients section. This directly implies that the only chemical of significant hazard in the bottle is sodium hypochlorite, and the rest of the ingredients are not present in sufficient amounts to cause issues.
First let's look at what thise things are:

Water : you know
Sodium Hypochlorite : bleach- household bleach is made from chlorine, caustic soda, and water and DC current (electrolysis).
Sodium Chloride : table salt, leftover from production or evaporation of water and chlorine.
Sodium Carbonate : washing soda- kind of like baking soda but about 40% stronger. This may be considered a buffer or stabilizer or byproduct.
Sodium Chlorate : when bleach is cooked/boiled (and dried), the leftover crusty stuff is salt and sodium chlorate... so this would be a byproduct, probably only in a very tiny amount.
Sodium Hydroxide : lye (caustic soda, NaOH) - household bleach solutions are typically stabilized by lye as part of the manufacturing process. Lye is used for raising the pH of drinking water and for making many things including soap.
Sodium Polyacrylate : thickening agent - It may be used in food, toothpaste, and shampoo.

APPROXIMATE Household Bleach Amounts:

Water : 93%
Sodium Hypochlorite : 5.5%
Sodium Chloride : 0 - 2%
Sodium Carbonate : guestimating 0 - 0.2%
Sodium Chlorate : guestimating ~ 0 - 0.2%
Sodium Hydroxide : since the pH of bleach is about 11, the amount of sodium hydroxide must be less than 0.001 molarity, so less than 0.004%
Sodium Polyacrylate :  based on the viscosity of bleach and my personal experience with this thickening agent, about 0.5%.

Now lets look at the LD50 (a general measure of toxicity). The LD50 is the amount of material necessary to kill 50% of the test subject (usually rats). For these chemicals I listed acute (hard and heavy) oral dosage. Chronic exposure may be more relevant because you will not be drinking this, but chronic exposure data is also far more difficult to measure accurately, and with respect to all of these chemicals, chronic effects are generally nil, so I wouldn't have anything to say.

Water : greater than 90,000mg/kg, rat (yet drowning causes more deaths than all of the following chemicals combined)
Sodium Hypochlorite : 8910 mg/kg, rat
Sodium Chloride : 12357 mg/kg, human
Sodium Carbonate : 4090 mg/kg, rat
Sodium Chlorate : 1,200- 7,000 mg/kg, rat
Sodium Hydroxide : 100mg/kg, rat
Sodium Polyacrylate :  greater than 2000 mg/kg, rat. When something is above 2000 mg/kg the exact lethal dose is often not pursued.

Great answer. Much appreciated. I have applied this for the first time. It smelled and felt pretty much like a swimming pool, sort of like expected. I will have to find an ideal amount considering both effects on skin/smell and disinfection properties.

@Fiksdal Don't forget about the pump/filter. It will help prevent settling and/or uneven concentrations of chlorine, and keep debris out of the pipes.

Such a product isn't available in this remote area. I'd have to try and get it imported from elsewhere. The inlets are covered by filters, so it's not a huge problem. Some dead bugs do accumulate in the tank, but I could use a net to scoop them out. I'll think about it.

@Fiksdal I'm sorry to hear that you have difficulty obtaining stuff. Any water pump that can pump 2500 gallons per hour (9500 liters per hour) should work... it only needs to run a few hours per day. Circulating the water is kind of important.

Alright, I'll look into it.

The biggest concern that I see is that this product causes skin and eye irritation, and thus I would not advise using a water supply treated with it for taking a shower. Also, the safety recommendations say to take off and wash clothing contaminated with this compound. As such, I would be careful using this product in a water supply for laundry, showering, and dish washing. While you should be careful with this product, if it is your only option, you should be able to use it to clean the water tank if it is rinsed out. I don't know if this was your plan, but just adding this to your water would probably be a bad idea. On another note, I am a student of chemistry, and you should take what I have to say about this with a grain of salt. I hope this helps!

I think the key here would be the concentration. Normally, one would put 100 ml of this product into a bucket of 10 liters. That's the dilution ration they are warning about. At that ratio the solution will bleach clothes. I'm thinking of putting 100 ml into a tank of 8000 liters. I know that many of the ingredients in the product are used for water treatment. Are there any of the ingredients you are concerned about?

I would be very careful about using it. I don't know if this is an option, but you will probably be better off emptying the tank, using the disinfectant to rinse it clean, and rinse it out before refilling. IDK if this is possible for you, but you don't want to damage your eyes.

You should probably have some one else verify that this will work too, I would not just take my suggestion.

Sure! I'm waiting for more answers :)


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What are these fluffy white crystals growing out of my floor tiles?
Does anyone know what these white crystals are?
As you an see they are growing out of some floor tiling in my house. They have been appearing for some years now out of the floor, which is cleaned fairly regularly (at least, I assume so, since my cleaner does it!). They may appear more frequently when the weather is humid.
For scale, the tile is 20cm each side. The crystals form light fluffy heaps, rather like snow and would be a few millimeters in height.
My guess is that they could be common salt, or possibly something connected with the bleach or cleaning products used on the floor. The latter case prevents me from trying the taste test to prove the former.

To me, this looks more like the salts washed out from concrete (or brickwork, or whatever is beneath those tiles), akin to the thing known as "wall saltpetre".

Does is give bubbles with vinegar? https://en.wikipedia.org/wiki/Efflorescence

No bubbles in vinegar.

Having Googled it, it seems probably it would be wall saltpetre. I had a soil pipe break under the house a few years back and the ground would likely have become saturated with a good source of water, potassium and nitrogen, not to mention the ammonia in cleaning agents. I guess the tiles must be porous, especially where the surface is damaged, as in the picture.

This is probably efflorescence, i.e. salts that are first dissolved from the mortar or other components of the floor and then recrystallize on the surface of the floor when water evaporates, see e.g. https://www.ceramictilefoundation.org/blog/understanding-efflorescence-ugly-white-powder-on-tile.
According to the source above, if there are soluble components in the floor and water moving to the surface, these crystals are able to form.
Wikipedia has the following definition:

In chemistry, efflorescence (which means "to flower out" in French) is the migration of a salt to the surface of a porous material, where it forms a coating. The essential process involves the dissolving of an internally held salt in water, or occasionally in another solvent.

According to the wikipedia article, there is primary and secondary efflorescence, and a variety of substances that crystallize.

My guess is that they could be common salt, or possibly something connected with the bleach or cleaning products used on the floor. The latter case prevents me from trying the taste test to prove the former.

In this context, the term "salt" does not typically refer to table salt, but could include a wide range of highly toxic substances. Getting rid of the crystals will require different methods depending on the nature of the crystals. It seems more likely that the source of the salt is in the floor rather than a cleaning product.


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Why does the Grignard reagent of CH3I can only be prepared using diethyl ether as solvent but there is no reaction when using THF?
Does anyone know why Grignard reagent of CH3I can only be prepared using diethyl ether as a solvent and yields no reaction when developed in THF? Theoretically, I is the most reactive halogen for Grignard reactions, so using THF or diethyl ether should make no difference.

Can you provide a reference for this? I have never heard of this problem.

It was something my lab teacher said. He tried several times doing the reaction in THF but it didn´t work.

MeMgI is usually prepared in ether, but it can also be prepared in THF. However, magnesium iodide is sparingly soluble in THF, so RMgI reagent in THF should not be very concentrated, so that no MgI2 precipitate forms.

It seems like $\ce{MgI2}$ should be equally sparingly soluble in diethyl ether. Do you have a reference for your solubility claim?

@BenNorris It is repeatedly pointed in Gringnard-related literature that MgI2 is soluble in DEE, but not THF (and similarly for MgCl2 ). I was unable to find trustworthy values on quick googling, though.


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Why is EDTA most soluble at a pH 8? (addition of NaOH)
Although Wikipedia says that EDTA is a very soluble salt, preparing a solution (.5M) is a pain in the neck. It takes a lot of time and still doesn't dissolve. Although addition of a few drops of NaOH makes the process a bit easier (still takes a lot of time)
Now, it did not make sense to me of why NaOH made EDTA more soluble. The salt I'm using is a disodium salt of EDTA, so wouldn't Na+ turn out to be the common ion only to decrease the solubility? Then, it might be the effect of pH(?) But, being a salt of a weak acid and a strong base, theoretically shouldn't such a salt be more soluble in lower pH? The EDTA divalent ion thus formed would be a stron conjugate base and accept H+ increasing the solubility?
I know both my assumptions are wrong since adding NaOH actually helped. What might be the actual reason which is counterbalancing the above mentioned 2 effects (are they even valid in this case?)

Adding also an acid would encounter the common ion effect since it is only the disodim salt and 2H+ are still out there? Or is it that the base pull out these 2 H+ and makes it more polar??

sigh If you neutralise it further it gets more soluble and tetrasodium EDTA is much better soluble, what's your problem with that?

Then why pH 8? If it neutralises completely shouldn't it be pH 7. But in manuals, it clearly says that the end product should be pH 8 @Mithoron

A disodium salt of EDTA also has two remaining carboxylic-acid groups, which can be deprotonated by addition of base, like $\ce{NaOH}$. By the way, higher pH is generally better especially if you plan to use it for chelation, i.e. cation-capture. The labs I've taught in the past use buffers of pH ~10 with EDTA for complexometric titration experiments.
Deprotonating an organic acid in general improves solubility in water because the electrostatic interactions of $\ce{-O^-}$ and the solvent (the hydrogens on water) are stronger than interactions of $\ce{-OH^0}$ and water.
A similar example would be 4-t-butylphenol, which is nearly impossible to dissolve in water without addition of a base, e.g. $\ce{NaOH}$. In experiments I had to reach pH of about 12 to strip the compound of the alcohol-proton, yielding a negative charge which interacts more strongly with water.
Nonetheless you can still make a decently concentrated (~0.1M) solution of EDTA in water by stirring alone, if you allow a stir-plate to do the work for ~30 minutes.
In short, it's not the addition of $\ce{Na+}$ that matters, but rather the addition of $\ce{OH-}$, to increase solubility.

thanks! I'm actually doing a western blot which has a buffer of pH of 7.4 So, why don't they directly sell the tetrasodium salt? Or why is the dis odium salt more feasible?

welcome! Pure (then fully deprotonated, e.g. with pH ~10) EDTA is ideal for chelation. That is, a non-sodium salt.
The sodium/other salts have their uses. But in general it's the base that changes solubility, much more than the sodium. A tetrasodium EDTA salt would be more soluble, I imagine. Never used/seen it though.

Also be aware that the so called disodium salt is almost never sold as declared (it's not that easy to get the stoichiometry right on industrial skale), and you usually end up with a mixture of compounds that have 2/3 carboxylic acids protonated per molecule. That is the reason why EDTA is not a primary analytical standard, and needs to be standardized after dissolving it. Adding NaOH also increases the ionic strength of the soln. which shifts equilibrium towards dissolution. Common ion effect doesn't play a big role, as Ksp value is quite large.


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Difference between freezing point and melting point
I do not understand why for mixtures freezing point is lower than melting point. Are not the two representing the same equilibrium between solid and liquid which we are approaching from two different sides, so why should the temperature at which this happens be different for the two cases?

related http://chemistry.stackexchange.com/questions/42300/why-does-freezing-point-decrease-on-adding-impurities

No, they are not. They represent two different situations with different compositions of solid and liquid phases (this could be understood in two ways, and both are correct: the composition is different between solid and liquid, and also different between situations). So no, it is not the same equilibrium.

@Ivan Neretin will you please explain in detail why these do not represent the same equilibrium.Thanks in advance.

The Bodner research group at Purdue University has posted this excellent description of the difference between melting point and freezing point:

Pure, crystalline solids have a characteristic melting point, the temperature at which the solid melts to become a liquid. The transition between the solid and the liquid is so sharp for small samples of a pure substance that melting points can be measured to $\pu{0.1^\circ C}$. The melting point of solid oxygen, for example, is $\pu{-218.4^\circ C}$.
Liquids have a characteristic temperature at which they turn into solids, known as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small differences between these quantities can be observed.
It is difficult, if not impossible, to heat a solid above its melting point because the heat that enters the solid at its melting point is used to convert the solid into a liquid. It is possible, however, to cool some liquids to temperatures below their freezing points without forming a solid. When this is done, the liquid is said to be supercooled.
An example of a supercooled liquid can be made by heating solid sodium acetate trihydrate ($\ce{NaCH3CO2 . 3H2O}$). When this solid melts, the sodium acetate dissolves in the water that was trapped in the crystal to form a solution. When the solution cools to room temperature, it should solidify. But it often doesn't. If a small crystal of sodium acetate trihydrate is added to the liquid, however, the contents of the flask solidify within seconds.
A liquid can become supercooled because the particles in a solid are packed in a regular structure that is characteristic of that particular substance. Some of these solids form very easily; others do not. Some need a particle of dust, or a seed crystal, to act as a site on which the crystal can grow. In order to form crystals of sodium acetate trihydrate, $\ce{Na+}$ ions, $\ce{CH3CO2-}$ ions, and water molecules must come together in the proper orientation. It is difficult for these particles to organize themselves, but a seed crystal can provide the framework on which the proper arrangement of ions and water molecules can grow.
Because it is difficult to heat solids to temperatures above their melting points, and because pure solids tend to melt over a very small temperature range, melting points are often used to help identify compounds. We can distinguish between the three sugars known as glucose ($\vartheta_\mathrm m = \pu{150^\circ C}$), fructose ($\vartheta_\mathrm m = 103{-}\pu{105^\circ C}$), and sucrose ($\vartheta_\mathrm m = 185{-}\pu{186^\circ C}$), for example, by determining the melting point of a small sample.
Measurements of the melting point of a solid can also provide information about the purity of the substance. Pure, crystalline solids melt over a very narrow range of temperatures, whereas mixtures melt over a broad temperature range. Mixtures also tend to melt at temperatures below the melting points of the pure solids.

In my opinion (but not in Martin’s, for example) this is a plagiarising answer because all it does is copy from a source and state that source.

plagiarism is passing someone else's work off as your own. if you cite the source, it's not plagiarism


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Building a Voltaic Cell Without Metal Ions
For an assignment in my Chemistry class we have to build a voltaic cell using materials that are provided to us and that outputs a voltage as close to 0.8v as possible. The list of materials is as follows:

Glass plate
4 filter paper strips

1 ml of the following 1.0M solutions

HCl
NaOH
NaCl
Water

Small metal strips:

Cu
Fe
Pb
Mg
Zn

I originally thought of doing something like this:

This is essentially a small scale version of a 2 beaker voltaic cell with a salt bridge that you see in textbooks. I would attach a multimeter between the Zn and Cu strips and read a voltage.
What I'm confused by however, is that I'm not provided with any solutions of aqueous metal ions, which are required for the half reactions powering this cell to go forward. I know that this has something to do with the HCl and NaOH that I'm provided with, but can't figure out exactly how they factor in.
I've done quite a bit of googling but I'm completely stuck. How would one make a voltaic cell out of the materials I've listed? Thanks for any help in advance.

Try something along the lines of a lead acid battery https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery.  It doesn't need any metal ions.

You can dissolve some of the metals to have aqueous metal ions..

If you just put a copper strip and a zinc strip into some NaCl solution and try to write the Nernst equation for the whole system, you will notice that the electrochemical potential will be undefined. Since the electrochemical potentials involve the cation concentrations which are all zero by preparation.
What this means is that there is infinite driving force for some metal to get oxidized into the solution until there is some equilibrium. Therefore, as one of the comments to your question indicate, you will be dissolving some of the metals into the solution by the very act of putting them into such an arrangement.

Also think in terms of metal/air battery - just be careful about the expected potentials. Mg would not be good in this case (potential is somewhat high - about 3100 mV vs. SHE as stated in - see DOI: 10.1039/C3MH00059A).


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Is ΔS of a system related to temperature and change in enthalpy?
I often see the equation $\Delta S_\text{universe}$ = $\Delta S_\text{surroundings}$ + $\Delta S_\text{system}$, where $\Delta S_\text{surroundings}$ can be re-expressed in terms of the enthalpy change of the system as
$$\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$$.
This leaves me wondering why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$ (Maybe as $$\Delta S_\text{system} \overset{?}{=} \frac{- \Delta H_\text{surroundings}}{T}$$ although this would make $\Delta S_\text{universe}$ equal $0$).
Instead, $\Delta S_\text{system}$ is usually given as a standard value, seemingly independent of temperature. Is there a reason why we can only express the change in entropy of the surroundings, but not the change in entropy of the system, as related to the change in enthalpy and the temperature of a reaction?

If the system experiences a change at constant pressure, the change in enthalpy of the system $\Delta H_{system}$ is equal to the heat Q transferred from the surroundings to the system.  So the heat received by the surroundings is $-\Delta H_{system}$.  Typically, the surroundings is regarded as an ideal constant temperature reservoir, and the reservoir is considered as being "internally reversible."  So, if T is the temperature of the reservoir, the change in entropy of the surroundings is $\Delta S = -\frac{\Delta H_{system}}{T}$.  The system is typically not regarded as experiencing an internally reversible change, so the same kind of approach can not be applied to the system.

Could you specify what you mean by "internally reversible?"

It's a term used in Fundamentals of Engineering Thermodynamics by Moran et al.   I particularly like their approach in this book.  "A reversible process is one for which no irreversibilities are present within the system or its surroundings.  An internally reversible process is one for which there are no irreversibilities within the system.  Irreversibiities may be located within the surroundings, however.  Using the internally reversible process concept we refine the definition of the thermal reservoir ...we assume no internal irreversibilites present within a thermal reservoir."

What would be a case where irreversibilities only occur in the surroundings (but not the system)?

Covered in section 5.3.4 of the book.  "If there were a spatial variation in temperature, say, there would be a tendency for a spontaneous energy transfer by conduction to occur within the system in the direction of decreasing temperature." Another example would be if a spontaneous chemical reaction is occurring.  Other examples are mixing and diffusion. "For reversibility (within the system), however, no spontaneous processes can be present.  From these considerations it can be concluded that the internally reversible process consists of a (continuous) series of equilibrium states.

Oh.  I didn't answer your question.  Suppose you have a process for expanding an ideal gas reversibly, but, instead of gradually removing tiny weights from a piston, you control the motion of the piston by hand.  In that case, your body is part of the surroundings, and there are a huge number of irreversibilities occurring  within your body as you gradually decrease the force on the piston.  So the gas receives the same reversible expansion in both cases, but, in the first case, the surroundings are handled reversibly and in the second case, they are not.

OK, I think I understand what internally reversible means now, but why does the system not typically experience an internally reversible process?

Let us continue this discussion in chat.

One of the purposes of defining a system and its surrounding in thermodynamics is to separate the interesting stuff from the boring stuff.

$\Delta S_\text{surroundings}$ can be re-expressed as $\dfrac{-\Delta H_\text{system}}{T}$

The best boring surrounding is a large homogeneous phase with the same temperature as the system, for example a stirred water bath. (If the temperature of the system changes, you have to quickly swap the bath to match the temperature.) In this specific scenario, the entropy change of the surrounding will be equal to the heat $-q$ transferred into the surrounding divided by the temperature. If the process in the system is happening at constant pressure, the system is closed and there is no non-PV work (no electrical, mechanical, etc. work, just expansion work), then the heat $q$ transferred out of the system will be equal to the change in enthalpy of the system, $\Delta H$.
So there are a lot of disclaimers to make before you can set $\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$.

[...] why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$

Remember we want the interesting stuff to happen inside the system. In general, there is no way of predicting the entropy change in the system from other thermodynamic values unless the system is as boring as the surrounding (for example when the system is a block of hot metal and the surrounding a block of cold metal). Of course, if you put the interesting stuff in the surrounding and keep the system boring, $$\Delta S_\text{system} = \dfrac{-\Delta H_\text{surroundings}}{T}$$ would be appropriate.

Is ΔS of a system related to temperature and change in enthalpy?

In general, it is not. Entropy and enthalpy change are not coupled. They can both be negative, both positive, one negative and the other positive and vice versa. This shows you that there is no fixed relationship between entropy and enthalpy.
There is one special case where entropy change and enthalpy change of the system are related - when the change in Gibbs energy is zero. The change in Gibbs energy is expressed in terms of entropy and enthalpy changes as (in the following, the subscript "system" is omitted because all the quantities refer to the system):
$$\Delta G = \Delta H - T \Delta S$$
When $\Delta G$ is zero (for example when a substance is at its melting temperature and both solid and liquid phase are present), the following relationship holds:
$$ \Delta S = \frac{\Delta H}{T}$$
Notice that there is no negative sign. This relationship can be used to estimate the melting point of a substance. Notice also that for chemical reactions at equilibrium, $\Delta_r G^\circ$ is not equal to zero, but $\Delta_r G$ is.
So while at equilibrium,
$$\Delta_r S = \Delta_r S^\circ + R \ln K = \frac{\Delta_r H}{T}$$
is correct, you can't calculate $\Delta_r S^\circ$ from $\Delta_r H^\circ$, i.e.
$$ \Delta_r S^\circ \neq \frac{\Delta_r H^\circ}{T}.$$
The only exception is when $K$ happens to be one, as it is by definition for melting when the substance is at the melting point.

$ΔS_\text{system}$ is usually given as a standard value, seemingly independent of temperature

Both $ΔS_\text{system}$ and $ΔH_\text{system}$ are dependent on temperature, and the heat capacity is used to describe the temperature-dependency. If you are evaluating the temperature-dependency of $ΔG_\text{system}$, however, the largest dependency comes from having the temperature directly in the defining equation:
$$\Delta G = \Delta H - T \Delta S$$
So even if entropy and enthalpy were temperature-independent (again, they are not), $\Delta G$ would still depend significantly on temperature. Oddly enough, a rigorous treatment of the temperature-dependence of $\Delta G$ gives the result that you would get from differentiating the defining equation with respect to $T$:
$$\frac{d \Delta G}{dT} = - \Delta S$$

I don't know if you have done some statistical mechanics, but maybe this is related to counting of states. First of all, we express the entropy change of the surroundings in terms of enthalpy change and temperature of the system because we simply can define our system in most cases and so can circumvent and measures quantities that defines a system (enthalpy change, for example, can be measured in a calorimetry expt.) then the entropy change of the surrounding can be inferred because the that of the universe is zero (only for reversible processes).
Second, for the counting I mentioned earlier, Boltzmann defines entropy as
$$S = k\ln W$$
$W$ represents the count of possible states of the system/surrounding. But the  number of states of the system multiplied by those of the surroundings equal the states of the universe. So if you know two quantities (here, system and universe) through knowing their entropies, you can get the remaining. Sure, this doesn't prevent you from doing what you suggested, which is expressing system entropy change as surroundings' enthalpy change and temperature, but you can't define them specifically for the surroundings and don't have a readily available tool to measure them.

We don't know the entropy change of the universe (nothing the OP said indicates a process that is at equilibrium). Even if it were at equilibrium, we still could not express system entropy via the surrounding's enthalpy change unless your system is very simple (no reactions, and enthalpy equal to heat transfer).

I probably don't have the full answer to your question, but I may have some insight. I like this topic, because it's the central idea in Chemistry, but textbooks care a lot about the math of it (which is easy), but not as much about explaining the ideas behind it in detail.
Since the reaction is assumed to take place at constant temperature, the kinetic-rotational energy of the particles of the reacting system does NOT increase of decrease following the reaction. If it were increased or decreased, the temperature would have changed.
Where does the heat released in the reaction come from (in an exothermic reaction)? The heat that is released in the reaction does NOT come from a decrease in the speed, rotation, or vibration of the particles, but from changes in electrical potential. So that's why heat can get released - bonds form with lower electrical potential energy and the difference in electrical potential is released - while the speed, vibration, rotation of the system stays the same - which is why temperature is constant.
So you see, the amount of heat released cannot be used to easily predict how the entropy of the system changed. The entropy change in the system is the result of the fact that when the system reacts, while it has the same amount of kinetic-vibrational energy, that energy is distributed in a way in which either more of it is in a form that cannot be used to do work (increase in entropy) or in a way that more of it is in a form that can be used to do work (more concentrated kinetic-vibrational energy and lower entropy).
By the way, notice that the entropy of the system is not about the electrical potential stored in the system (the source of enthalpy changes in a reaction), but is a fraction of the kinetic-vibrational energy of the particles (the higher percentage of that energy, whose amount in the system does not change when the reaction occurs in constant temperature, is distributed in a way that can NOT be used for work, the higher the entropy).

Is it possible to use the fact that since ΔHsurroundings = ΔHsystem and Tsurroundings = Tsystem, if ΔSsystem were to equal -ΔHsurroundings/T, then from the equation ΔSuniverse = ΔSsurroundings + ΔSsystem, ΔSuniverse would always be 0, in order to show that -ΔHsurroundings/T CANNOT equal ΔSsystem?

"If ΔSsystem were to equal -ΔHsurroundings/T" - I believe that you can't relate these in a straightforward way as in principle you could, for example, have a reaction which releases heat (changes in electrical potential), while experience no opposite change in the entropy of the system (no more concentrated kinetic-vibrational motion in the products compared to the reactants).

Entropy and energy have different dimensions, so the entropy can't be "a fraction of the kinetic-vibrational energy".

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


text:
Different masses for the same element and isotope?
I am watching this Khan Academy video on mass defect and binding energy, it uses helium-4 as an example to go through the concept. The person finds the predicted mass (protons + neutrons) to be 4.0318274 amu. He also finds that the experimentally measured mass for He-4 to be 4.00150608 amu.
What's baffling me is, where did he find the experimentally measured mass for He-4? I have been searching for the atomic mass of He-4 all over, and every source I found seems to agree that its mass is, in fact, 4.002602 u ± 0.000002 u, that's a 0.001 difference!

The difference looks mighty like the weight of two electrons. That is, you are comparing $\ce{^4He}$ to $\ce{^4He^2+}$. Also, welcome to Chem.SE.

Also note that the video shows the obsolete unit $\mathrm{amu}$ (which is deprecated since 1961), although the given numerical values rather correspond to the unit $\mathrm u$ ($1\ \mathrm{u}=1.660539040(20)\times10^{-27}\ \mathrm{kg}$). The proton mass actually is $m_\mathrm p=1.007276466879(91)\ \mathrm u$ and the neutron mass is $m_\mathrm n=1.00866491588(49)\ \mathrm u$. The difference is small but significant in this case. For example, the atomic mass of C-12 is about $12.0005\ \mathrm{amu}$ (as used in chemistry until 1961) or $12.0000\ \mathrm{u}$ (by definition, since 1961).

Do protons and neutrons have exactly the same mass in an atom regardless of the combination of quarks. For example, you could conceivably have a proton consisting of Up/Red, Up/Blue, Down/Green quarks and another one U/B, U/G, D/R. In fact, given that a neutron has slightly more mass than a Proton, it seems to me that the charge/color charge combination does result in different combined masses.

As noted by Ivan in his comment, the difference is likely due to using the helium nucleus $\ce{^4He^2+}$ and not a helium atom. Shame on the Khan Academy video for not making this clear.
However, there is a good reason to only consider the nucleus. The mass defect and binding energy are consequences of the strong nuclear fore, and thus properties of the nucleus.
The choice of helium-4 is convenient: the helium-4 nucleus is the alpha particle - a common product of radioactive decay and thus well-characterized. Thanks to mass spectrometry, we can determine the mass of an alpha particle very precisely to be $4.00150608\ \text{u}$. Here is the source of that mysterious number. Again, shame on Khan Academy for not being explicit.
The electron has a mass of $5.48579909\times 10^{-4}\ \text{u}$, so a helium-4 atom would be expected to have a mass of one alpha particle and two electrons:
$$4.00150608\ \text{u} + 2\left(5.4858\times 10^{-4}\ \text{u}\right) = 4.00260324\ \text{u}$$

While it appears there might be a small mass defect from adding the electrons $\left(4.002603 > 4.002602\right)$, however the value I calculated is within the $\pm 0.000002$ error on the experimental value.


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Estimation of calcium EDTA
How can i estimate calcium EDTA in a given sample?
We tried estimation by using mercuric nitrate but in vain we cant get a sharp end point.

@JinnySara Could you provide more information about the sample?

Something crossed my mind - you could try titrating with Fe(III) standard using SCN- as indicator. First excess of Fe3+ would produce a sharp end-point.

Thanks a lot for the information you gave me. Actually it helped mei a lot and v did this using ferric ammonium sulphate using potassium thiocyanate as indicator which will give you a blood red colour. We recently manufactured calcium disodium edatate hard gelatin capsule so for that we were trying to develop a method


text:
Decomposition of calcium carbonate in a sealed vessel
For the equilibrium
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$
$\Delta H^\ominus = 178\ \mathrm{kJ\ mol^{-1}}$ and $\Delta S^\ominus = 160\ \mathrm{J\ mol^{-1}}$.
I'm asked to suppose that $1\ \mathrm{mol}$ of $\ce{CaCO3}$ and $1\ \mathrm{mol}$ of $\ce{CO2}$ are heated reversibly in a closed vessel under a pressure of one bar.
How much $\ce{CaCO3}$ is present in the container when the temperature reaches $1100\ \mathrm K$?
I know that at this temperature $\Delta G^\ominus = 2\ \mathrm{kJ\ mol^{-1}}$ and $K_p = 0.80$. I'm thinking that since the equilibrium constant $K = p(\ce{CO2})/p^\ominus$ is less than $1$, then the $\ce{CO2}$ would combine with any $\ce{CaO}$ around to make $\ce{CaCO3}$, but since there isn't any the composition must remain unchanged at $1\ \mathrm{mol}$ $\ce{CaCO3}$ and $1\ \mathrm{mol}$ $\ce{CO2}$. Is this right?
Now suppose the temperature reaches $1120\ \mathrm K$, at which $K_p = 1.14$. Some of the $\ce{CaCO3}$ will start to decompose, right? Suppose $n$ mol do, so that I have $1-n$ mol of CaCO3, $n$ mol of $\ce{CaO}$ and $1+n$ mol of $\ce{CO2}$.
How do I find $n$?

Is your system Isobaric or Isochoric?

Isochoric, I assume, since the vessel is described as "closed".

In that case Gibb's free energy equations are not appropriate.  You should use Hemholtz energy for an isochoric system.

Hmm, you're right. If the pressure is constant, then the volume must change, though?

Well a closed system can expand, it just cannot exchange matter

The system is isobaric, the question specifies "under a pressure of 1 bar".

At $\pu{1100 K}$, $\Delta G>0$, favouring the backwards reaction. The backwards rate is 0, as there is no $\ce{CaO}$ present. Therefore, at $\pu{1100 K}$ there is 1 mole of $\ce{CaCO3}$ present. At $\pu{1120K}$, $\Delta G<0$ and the decomposition is favoured. See my working below to show why the decomposition goes to completion:
[![showing decomposition goes to completion][1]][1]
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$
Let's say $\ce{A <=> B + C}$. We start with 1 mol of $\ce{CaCO3}$ and 1 mol of $\ce{CO2}$. What is the extent of the reaction, given the equation composition for $\Delta G < 0$ in a closed vessel at $\pu{1 bar}$
$$ G_{tot} = \sum_i\left(n_i\mu_i\right) \implies G_{tot} = n_\ce{A}\mu_\ce{A} + n_\ce{B}\mu_\ce{B} + n_\ce{C}\mu_\ce{C}$$
Let $\alpha$ be the degree of dissociation, Recall $n_\ce{A} = 1$, $n_\ce{C} = 1$ initially
So,
$$G_{tot} = (1-\alpha)\mu_\ce{A} + \alpha\mu_\ce{B} + (1+\alpha)\mu_\ce{C}$$
As $\ce{A}$ and $\ce{B}$ are solids, $\mu_{A} = \mu_\ce{A}^\circ$ and $\mu_{B} = \mu_\ce{B}^\circ$
$\ce{C}$ is gaseous so we use, $\mu_{A} = \mu_\ce{C}^\circ + RT\ln \frac{P_\ce{C}}{P^\circ}$ (since C is assumed to be a perfect gas)
We now have,
$$G_{tot} = (1-\alpha)\mu_\ce{A}^\circ + \alpha\mu_\ce{B}^\circ + (1+\alpha)\left(\mu_\ce{C}^\circ + RT\ln\frac{P_\ce{C}}{P^\circ}\right)$$
$P_\ce{C} = P^\circ$ due to the pressure being 1 bar within the container.
$\therefore$ Rearranging we get,
$$\begin{align}G_{tot}  &= \mu_\ce{A}^\circ + \mu_\ce{C}^\circ + \alpha (\mu_\ce{B}^\circ + \mu_\ce{C}^\circ - \mu_\ce{A}^\circ) \\  &= \mu_\ce{A}^\circ + \mu_\ce{C}^\circ + \alpha \Delta G^\circ (< 0) \end{align}$$
Therefore $G_{tot}$ is minimized at maximum $\alpha$, i.e. $\alpha = 2$
Thus at 1120 K, we have 0 moles $\ce{CaCO3}$. Physically, this is rationalised as there being no entropy of mixing between the reactants and products. The same rationale can be used to explain why there is no ice present in liquid water above 273.15 K at atmospheric pressure.

You are right that at $\pu{T_o}$ = $1100~\mathrm{K}$ there will not be any decomposition of $\ce{CaCO3}$ in the presence of $\ce{CO2}$.
But, as the temperature increases further to $\pu{T_{1}}$ = $1120~\mathrm{K}$, $\ce{CaCO3}$ starts decomposing because at this temperature $Q_\mathrm p < K_\mathrm p$.
As the reaction is taking place in a closed container of volume $V$, by viewing the reaction mixture as an ideal gas mixture, we can calculate the amount of $\ce{CaCO3}$ decomposed at $1120~\mathrm{K}$:
$$p_o(\ce{CO2}) \times V = 1 \times R \times 1100 \tag1$$
$$p_o(\ce{CO2}) = 0.80 \mathrm{bar}$$
$$p_1(\ce{CO2}) \times V = (1+n) \times R \times 1120 \tag2$$
$$p_1(\ce{CO2}) = 1.14 \mathrm {bar}$$
On solving, $n= 0.399$.

The system is isobaric, the question specifies "under a pressure of 1 bar". – orthocresol♦ Jan 17 at 12:33


text:
Immersing copper and zinc in hydrochloric acid
I am wondering what would happen if a metal strip of copper and a metal strip of zinc were immersed in solution of $\ce{HCl}$ in standard electrochemical conditions. I have calculated the standard change in Gibb's free energy $\Delta_\mathrm rG^0$ using the relation $\Delta_\mathrm rG^0=-zF\Delta E^0$ where $z$ is the number of electrons exchanged, $F$ is faraday's constant, and $\Delta E^0$ is the standard change in electrode potential.  I have found the following values:
$$\Delta_\mathrm rG^0_{\ce{Zn}}=-147\ \mathrm{kJ/mol}, \Delta_\mathrm rG^0_{\ce{Cu}}=-66\ \mathrm{kJ/mol}$$
It appears that the reaction between $\ce{Zn}$ and $\ce{H+}$ is more spontaneous. Does this mean that the reaction with $\ce{Cu}$ will not take place at all?

Enthalpy or free energy? These are two different things. Make up your mind.

Thermodynamics spontaneity and kinetics of reactions are two different things. Just because something has more driving force, doesn't mean it is going to go faster. In fact, there are cases where higher driving force actually causes the reaction to go slower. For more, I would recommend the reading the Nobel lecture of Rudolph A. Marcus who developed theories explaining what is now known as the "Marcus Inverted Region".
Kinetics of reactions depend on more complex issues that are harder to control and predict.


text:
Kinetics energy expression of photon
I need to calculate the kinetics energy of a photon and I'm trying to deduce a correct expression for calculating it.
So we have that
$K= \frac{1}{2} m_fv_f^2$
So $v_f= c$
For De Broglie equation we have that $m_f = \frac{h}{\lambda c}$
And then
$K = \frac{hc}{2 \lambda}$
But apparently the correct expression is:
$K = \frac{hc}{\lambda}$
Can someone help me figure out what I'm doing wrong?
Thanks!

Well, photons have no mass, so using the de Broglie equation is inappropriate.

The whole $mv^2\over2$ thing stops working as we approach the speed of light.

A photon has no mass and moves at relativistic speeds, therefore you can't use the usual expression for the kinetic energy.
The energy of a photon is given by (as you already stated)
$$
E = \frac{hc}{\lambda}
$$
where $h$ is Planck's constant, $c$ the speed of light and $\lambda$ the wavelength of the photon.
You can derive it from the relativistic energy
$$
E^2 = p^2c^2 + m^2c^4
$$
with $m =0$ and using the fact that for a photon
$$
p = \hbar k = \frac{h\nu}{c} = \frac{h}{\lambda}.
$$


text:
LC-MS: Can a "bad" peak still be interpreted
I have done an untargeted metabolomics study using LC-MS. There is one peak that matches to a biologically relevant compound but analyte eluted at the beginning of the LC run and the peak shape is far from ideal.
Do I assume the peak is an artifact, or can I say that there is a compound with that m/z in my sample? Ideally I would run the experiment again with standards, but I cannot do that due to time constraints. Someone else will follow up on the interesting compounds that I've identified and I wouldn't want to send them in the wrong direction.
Below is the peak in question, and following it is an example of a "good" peak from the same dataset.

This is low-quality data.  It's very difficult to say whether this signal is meaningful for several reasons.

It appears as soon as the LC-MS starts recording data at very low retention times.  That means it could be an artifact of column regeneration.  There is no sign that the data contains a "void volume" peak, meaning it is unclear whether the low-RT signal is from sample injection, from the previous sample, or some other artifact.

You don't say whether you randomized sample injection order.  Therefore the differences between the red and black sets of samples could arise because of changing conditions in the LC-MS instrument over time (e.g. column fouling, sample aging in the autosampler) or from legitimate differences in the samples that are biologically meaningful.

You don't show the full mass spectrum for your proposed analyte, so it is possible that the single m/z you showed here has an isotopic pattern consistent with your analyte, or inconsistent with it.

Thus, to answer your questions:

It basically wrong on the basis of info presented to claim that you detected a "biologically interesting" metabolite on the basis of the information presented.

It is also bordering on misleading to say you detected "a molecule" with the given m/z at all.  Maybe you could say you detected an unknown signal at m/z of xyz and are unsure of its origin.  That's about it.


text:
What are better acids for removing rust from an iron nail than phosphoric, or oxalic acid?
I'm planning to conduct an experiment in which I test how fast rusting occurs with changes to the concentration of NaCl in the solution.
I want to measure that by finding the change in mass of an iron nail over time, but in order to do that I need to clean off all the rust on the nail before weighing. I have a relatively short time to conduct the experiments.
I've heard acids like phosphoric acid, oxalic acid, or citric acid will work. I'm looking for the best acid to separate the rust from the nail preferably without using steel wool or any abrasive method. I can, however, spray down the nail with distilled water if necessary. Any suggestions?

In industry, rust is cleaned using hydrochloric acid, because besides being an acid, it is also a reducing acid.

In the automotive industry, pickling of steel plates is done with a diluted solution of sulphuric acid. You might use hydrochloric acid, too, but be sure to use pure (for lab purposes) $\ce{HCl}$. Commercial $\ce{HCl}$ (also known as muriatic acid) quite often contains traces of other metals (iron included, and $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions give muriatic acid its yellowish colour, that may turn your nail inert to some extent.


text:
Why are spherical compounds with a half filled HOMO concidered aromatic?
In planar annulenes, huckel's rule effectively states that compounds with a full HOMO for their conjugated $\pi$ systems are considered aromatic while those with a half filled outer shell are antiaromatic. Aromatic compounds are stabilized both by delocalization across their $\pi$ complex and possessing a full HOMO spread across the ring. On the other hand, antiaromatic compounds break degeneracy among their orbitals, preventing resonance, as it would only result in a half filled outer shell. Despite this, according to this source, a half filled HOMO in a spherical compound will result in aromaticity on a sphere. Is this because there is so much potential for delocalization in spherical compounds, since they contain so many conjugated pi bonds, or can this be explained with something else?

There is no such thing as spherical aromaticity.

related http://chemistry.stackexchange.com/questions/37641/is-buckminsterfullerene-aromatic-or-not?s=9|0.1343

That being said, I kindly ask everyone to leave this question open, at least for a while.

I reworded the question. Spherical aromaticity exists and buckminsterfullerene is antiaromatic.

OK, let's get this straight. There is no such thing as spherical aromaticity. The so-called Hirsch's rule is sheer nonsense and never works.
Now don't get me wrong, the premises were sound. When we have dozens of atoms arranged kind of uniformly on the surface of a sphere, the whole thing starts to look mighty like continuous - reminiscent of a huge "atom" consisting of surface only, without any depth. Naturally, the eigenfunctions (AKA molecular orbitals) of such a system would be the pure angular functions without the radial part, and their energy will be determined by the angular quantum number $\ell$. The bottommost orbital would be like $s$ (a sphere of one sign), then would go three $p$-like orbitals (each with one change of sign), then five $d$-like orbitals, and so on. Look at the lowest energy levels of $C_{60}$ (pictured), or any other fullerene, for that matter; they would go just like that - one, three, five... (With any other fullerene, the correspondence would be approximate; with $C_{60}$, because of its high symmetry, it is exact). And of course, when we count the electrons up to the $n$'th level, we get $2n^2$.

Sounds good, does it?
Well, it just does not work. There are too many crude approximations on top of each other. Most importantly, when we get all the way up to the HOMO, our eigenfunctions start changing sign too often, so we can't any longer pretend that the surface of a fullerene is continuous. Consequently, the pattern breaks and all conclusions fall apart.
One might point out that pretty much any rule in chemistry is like that, including the Hückel's rule itself. It is also built of crude approximations, isn't it?
That's right, it is, but there is a huge difference, too. Hückel's rule works. Hirsch's rule doesn't. The former predicts the outstanding stability of benzene, and not-so-great stability of cyclooctatetraene, and that cyclopentadienyl would prefer to be an anion, and so on - and they all obey. The latter predicts... well, see for yourself.
What are the magic numbers of electrons for the presumed spherical aromaticity? $2,\,8,\,18,\,32,\,50,\,72,\,98$ and so on. The first five are too small for a fullerene, so ignore them. 60 is not magic, so one would expect it to be unstable. WRONG. It is pretty stable. The paper you mentioned says it is strained. Now what was that about? Of course it is strained, as are all fullerenes and nanotubes; if you don't strain that $sp^2$ carbon, it would just form a flat sheet known as graphene. But what in the world does this have to do with [anti]aromaticity? Think again of the old good Hückel's aromaticity. Think of the cyclopropenyl cation; is it aromatic? Positively yes (pun intended). And yet it is like ten times more strained than fullerene. Aromaticity and strain are totally unrelated things; if people bring up one of them to judge the other, they are either clueless or deliberately misleading, even if they managed to get a paper published in a peer-reviewed journal.
Antiaromaticity, if anything, is about having a half-filled HOMO and thus being prone to Jahn-Teller distortion, radical reactions, and other unpleasant things. Is this the case with $C_{60}$? Look again at those energy levels. Look at the completely filled HOMO, and the completely empty LUMO, and that huge gap between them. Look and think twice before calling this guy antiaromatic.
Now comes the "magic" number 72. That would be $C_{72}$. Does it exist? Well, yeah, it kinda does, like any $C_n$ where $n$ is even and $\ge70$. Is it particularly stable? No, it does not seem so; it is way less abundant than the smaller $C_{70}$ and the bigger $C_{76}$, which themselves are way less abundant than $C_{60}$. Also, starting at $C_{76}$, the fullerenes would have multiple isomers, and as soon as $C_{80}$, we find an example where one isomer is unstable for electroniс reasons (antiaromatic indeed, if you'd like) and the other is stable. So the stability is not determined by the number of electrons alone, after all.
Now what would you call a seemingly reasonable rule which works $\bf{0\%}$ of the time?

Thanks for the information. Hirsch's rule was causing a lot of confusion for my understanding of aromaticity in general.

Another aspect is that you cannot render a spherically symmetric shell with a finite number of atoms.  The best you can do is icosahedral symmetry.  Under icosahedral symmetry the "s", "p" and "d" orbital combinations retain their relative energy ordering and the full degeneracies of the "p" and "d" levels.  In that sense $2, 8, 18$ could be magic numbers.  But when you get to the "f" level the icosahedral symmetry allows splitting which destroys the pattern presumed by the Hirsch rule, so the true Hirsch sequence stops well short of where it would become relevant for fullerenes.


text:
Relation between strength and proticity of an acid
The strength of an acid is related to the number of molecules which have dissociated into hydronium ions in aq solution of that acid while proticity is the number of hydronium ions furnished by 1 molecule of that acid in water. But what is the relation between the two.
Does a strong acid have a high proticity or vice-versa?

There's no real connection. Higher proticity can make a compound  slightly more acidic but it depends on what you compare with what. Also that kind of usage of "basicity" is misleading.

I'm a chemist. I've never heard the term "proticity". If your definition of it is correct, then consider phosphoric acid, H3PO4. It has a proticity of 3. Is it a stronger acid than HCl or H2SO4? There's really NO useful relationship I can see between strength and the number of donate-able protons. The term "acid" has three of four different meanings. The three most common are Arrhenius, Brønsted, and Lewis acids, but note that there are other definitions (usually used in very specialized contexts). The difference between Ar. and Br. is that Arr. applies only to water as the solvent while Br. applies to any compound which will donate an H(+) ion (solvent unspecified and possibly without any solvent present), the most general definition, L., has to do with compounds (or ions) which accept electron density, so it is often used in descriptions of covalent (not (necessarily) ionic) reactions. You can look up the dissociation constants for each of the three HnPO4 species which can give up an H(+) ion. None of them are "strong" acids, on the other hand H2SO4 is a very strong acid while H2CO3 is a very weak acid. (If phosphoric acid were strong, those of us who drink carbonated soda wouldn't have any teeth left since they add it in order to increase the solubility of the carbonate ion (CO2 +H2O → HCO3(+) + OH(-)...you do the mass balance! (it's incomplete the way I've written it). Both sulfuric and hydrochloric acids can easily have pHs well below 0. (anhydrous HCl is a gas, and H2SO4 is often sold as 98% H2SO4 in water (what's the molarity of that?!)). You can also have NaOH solutions in water with a pH well above 14, but don't tell your teacher, s/he may argue with you (but I know better!).(although to be truthful, at some concentration NaOH(aq) is no longer usefully thought of as a "water solution",(its more a solution of water IN NaOH (or in H2SO4)).

You should really consider breaking this up into paragraphs. It's hard to look at.

I tend to think of acids as proton donators and bases as proton acceptors. You are correct that an acid is anything that increases the concentration of $\ce{H+}$ in solution. A general acid will have the form $\ce{HA}$, which after losing an $\ce{H+}$ becomes $\ce{A-}$. While there are a couple of different ways to calculate the strength of an acid, one of the most common methods is to see how much the acid tends to dissociate. The $pK_a$ of an acid is equal to $-log(K_a)$, where $K_a$ is the acid dissociation constant for the particular acid. The lower an acid's $pK_a$, the larger its $K_a$, and thus the stronger the acid.
While the $pK_a$ is one way to think about the strength of an acid, another thing to think about is the strength of the conjugate base. A stronger acid is one that dissociates more readily, meaning that its conjugate base is more stable. A more stable conjugate base in turn means that after losing a proton, the acid is unlikely to pick up one from solution. This means that a strong acid will create a weak base while a weak acid will create a strong base after dissociating.

OP's asking about multibasic acids...

http://chemistry.stackexchange.com/questions/16804/what-is-the-word-for-how-polyprotic-an-acid-is also  @Pearu

This is a general explanation about acid strength. At the time of writing, it was unclear what the question wanted.

For Brønsted acids and bases, a strong acid (with high acidity) means it will easily give up a hydrogen ion. A strong base (with high basicity) will easily accept a hydrogen ion. For Lewis acids and bases, a strong acid will accept electrons more easily than a weak acid, and strong bases will donate electrons more easily than a weak base.

I'm afraid you didn't get OP's point.

Then answer instead of writing useless condescending comments. Basicity is related to the strength of the acid. And acidity is related to the basicity of the acid. In aqueous solution at 25 degrees Celsius there is an equilibrium constant, determined by the concentration of the acid and base products, which means that acidity and basicity are always related. Compounds with weak basicity will have strong acidity, and vice versa.

OP is asking about so called "basicity" with the meaning equal to proticity i.e. about mono and multibasic acids - dissociating one or more protons. I also at first thought he was asking about "normal" basicity. Sorry for being overly laconic.


text:
Preparing a 0.030 M solution of potassium ferrocyanide trihydrate?
I am a little bit confused on what I am supposed to do. The instructions say to prepare 250 mL of this solution by measuring out in grams the amount of PFT using a 250 mL volumetric flask. Then it says I have to add dI water until te meniscus hits the full mark on the flask, then it says to calculate the actual molarity of the solution. Am I supposed to multiply the molarity by the liters given to me to find the grams of PFT I am supposed to use? How would you go about this?

It is quite trivial. Just calculate how many moles of potassium ferrocyanide trihydrate you need to measure out (n = C x V), then multiply with molar mass to see how many grams you need to measure out. After measuring the sample on analytical balance, quantitavely transfer it to the volumetric flask. Add some water and dissolve the sample. After homogenizing the solution, add water carefully to the full mark. You can't actually calculate the exact molarity of the prepared solution: you need to determine it experimentally, by performing a standardization with silver-nitrate standard using chromate salt as an indicator.


text:
Are titration, recrystallisation, functional group analysis, etc. outdated?
In modern chemistry,
- pH meter, ion electrode, AES, HPLC, etc. replacing titration
- HPLC replacing recrystallisation
- IR and NMR replacing melting point analysis, derivative analysis, functional group analysis
In other words, are the techniques that being replacing only being used inside school and teaching labs?

It is true that modern techniques make it faster to elucidate the identity of compounds and isolate them from the reaction mixture, however to call the traditional methods like titration, recrystallization, or column chromatography (basis of HPLC) outdated is far too broad. You could probably get away with calling them outdated if all you wanted to do was to characterize compounds/final product. But when you're in synthesis, you still need the 'traditional' methods to prepare larger amounts of intermediate.
For example, HPLC is only able to separate small amounts of compounds (possibly enough for characterization) and if you wanted enough to go to the next step of your reaction (or if you wanted to make a lot), HPLC would be a highly inefficient way of doing this. Recrystallization, Distillation, and/or Column Chromatography (depending on the compound) would be much better for this purpose.

There is for example preparative chromatographie: http://www.buchi.com/us-en/products/preparative-chromatography/prepchrom-c-700 up t0 1 g for preparative HPLC and there are bigger versions. So separation with hplc is quite possible.

As someone working in the analytical industrie, I can tell you that these techniques are far from outdate. This has several reasons:
If you're working in the pharmaceutical analysis, and therefor working with the so called Pharmacopoeias (USP, JP, BP, PhEur) many methods are still based on the classical working techniques. Heavy metals are still often tested as Sulfites, a lot of titration methods are used (Karl-Fischer for water, hydroxyl value...), dry matter, ash testing...
This is also due to the limitations of pharmacies, who have to test themselfes and don't own the big and expensive instruments. On another side, this is due to the enormous efforts that neds to be done to update this compendias.
This leads to the next points: costs.
A lot of these modern and trustworthy machines are expensive and require a lot of know-how. Not every company can or wants to invest this much money in the analytical equipment, if a technician can do a lot of this cheeper by himself.
Please don't think that you can buy a GC/HPLC/NMR..., just put it in place and it works reliable the next 20 years. You have to maintain it, qualify and calibrate it quite often... This also costs a lot of money and time and if you don`t use it 24/7, this will raise the costs per analysis quite a lot.
I will update this further, after work.


text:
Homemade insecticide, is it safe or will a harmful chemical reaction occur?
I live just outside of Tucson Arizona and there are a lot of venomous spiders, scorpions and other potentially harmful critters.  I have 2 retrievers, 2 cats and 4 turtles and would like to avoid using harmful, toxic pesticides.  I reviewed a few articles online and various ingredients were suggested.  I took a few from each website with the intentions of creating a fail proof mixture.  Can I mix the below ingredients without having to worry about a chemical reaction?  Are any of these harmful to turtles, cats, dogs or humans?  I only intend on spraying the outside perimeter of the house, windows and door ways.

Organic Dish Soap
Lavender oil
Citronella oil
Eucalyptus oil
Cinnamon bark oil
Peppermint oil
Organic Tobacco
Fresh Garlic
Fresh Lemon and Lime

Sounds like a recipe for gumbo.

Why do padawans always think that such "natural products" aren't c-h-e-m-i-c-a-l-s ?!? I mean if you seep "Organic Tobacco" in water you don't get a nicotine solution but some "natural" concoction?

You have a lot of ingredients and a lot of questions. I can't see anything here that raises a concern in terms of chemical reactivity. Most of the oils are terpene based mixtures. Cinnamon bark will contain aldehyes. Nearly all of these ingredients are used in essential oil therapies, and are mixed on a regular basis. Presumably you will mix this up and dilute it with water before spraying. Probably what you are making is in fact a creepy crawly deterrent, rather than an insecticide. You would probably need to spray directly onto the critter of choice to kill it. I have in the past used an aqueous mix of orange oil, citronella and pure soap as a spray to deter web-building spiders on brickwork around the house. Works wonderfully well, and lasts several weeks between applications. Our dogs, cats, chickens, geese and ducks all lived through it well enough. We just get the pest man in to do the job now, and the animals are still alive.
I'd suggest speaking to your local vet about possible toxicology effects of these ingredients with your animals. Most of the ingredients listed can be toxic to animals if ingested. I know that citronella and citrus based extracts are hazardous to cats and dogs. Cats are especially susceptible and using citrus dog shampoos etc on cats can prove fatal. Garlic is also toxic to dogs and cats; again cats are more sensitive, although the toxicity comes through ingestion. Some types of detergents are dermal irritants for dogs, although pure soaps are ok. Eucalyptus oil, although commonly used as a disinfectant, is toxic to dogs and cats (and horses!) if ingested. Even tobacco can cause fatal nicotine poisoning if ingested. Has that put you off yet? Unless your pets are going to lick the areas you have sprayed, then it is unlikely to cause many problems. But speak to your vet - they will know.
I can't say I know anything about turtles, other than they are a wonderful vector for salmonella. But you probably already know that.


text:
Hybrid functional calculations using different approach
In hybrid functional such as B3LYP, PBE0, HSE etc, some part of HF exchange and/or correlation is added to some part of DFT X/C part in order to get better results. My question is if I am adding 1/4 th of HF exchange part that means I already calculated HF exchange part and HF is undoubtedly better than DFT, then why don't we just use the full HF solution that we already have in our hand?

I guess the theoretical reason is, the artificial system of KS is not the mean field. Hence, the exchange effect is not exactly the same as wave-function based schemes.

HF is usually less accurate than DFT because it lacks electron correlation

In hybrid functional such as B3LYP, PBE0, HSE etc, some part of HF
  exchange and/or correlation is added to some part of DFT X/C part in
  order to get better results.

First, there is no correlation in HF, only exchange, so you can not add part of the HF correlation into DFT, as you said. In principle, of course, if you want to add not only exchange from the wave function theory (WFT) methods but also some correlation, you could use other that the HF sources, but the conventional hybrid DFT corresponds to mixing in just (a part of) the HF exchange.
Now, you very question of

why don't we just use the full HF solution that we already have in our
  hand?

is addressed in almost any textbook on DFT. I suggest that you read, for instance, Section 6.6 Hybrid Functionals in Koch & Holthausen, specifically, pages 78-80. The authors provide a relatively simple argument for for the well-known failure of the exact exchange/density functional
correlation combination approach.
In short, the source of the problem is that the separation of an exchange-correlation entity (would it be energy, or hole, or something else) into exchange and correlation contributions is artificial. Only the whole exchange-correlation entity has a physical meaning, while the above mentioned individual contribution are introduced just for convenience. Usually, it means that exchange and correlation contributions has to be chosen consistently so that their sum, an exchange-correlation entity, has some desired properties.
For instance, the exchange-correlation hole in molecular systems is usually relatively localized, so that the exchange hole and the correlation hole must both be either localized or delocalized (in a very special way so that combined they result in a localized hole). In LDA- and GGA-DFT both the exchange and the correlation hole are localized, but when we mix HF exchange and LDA/GGA-DFT correlation in the most straightforward way, i.e.,
$$
E_{\mathrm{XC}} = E_{\mathrm{X}}^{\mathrm{HF}} + E_{\mathrm{C}}^{\mathrm{DFT}} \, ,
$$
then, I quote:

We combine the exact, delocalized exchange hole with a localized model
  hole for correlation. Because the cancellation between the two
  individual holes cannot take place, the resulting total hole has the wrong characteristics.

1) Koch, W., & Holthausen, M. C. (2001). A Chemist's Guide to Density Functional Theory, Second Edition, Wiley-VCH Verlag GmbH.

As already noticed by Jan, Hartree-Fock is a poor theory, since it completely neglects correlation effects. DFT is in principle an exact theory but in practice its accuracy depends on the approximations and the self-interaction is not exactly cancelled as it is in Hartree-Fock theory. Both approximations have strengths and shortcomings and it is by combining them to some extent that we can obtain better results.
A striking example of why we don't use the whole Fock exact exchange in hybrid functionals is found in solid-state physics. Semilocal approximations (LDA and GGA) largely underestimate the band gap of semiconductors and insulators: this is known as the band gap problem of DFT. Hybrid functionals can be used to reproduce the experimental band gap by tuning the portion of exact exchange taken into account (1/4 is absolutely not a magic number and higher or lower mixing coefficients can and should be used). The portion of exact exchange needed to reproduce the experimental band gap depends on the system, but it is never 100%: this is because Hartree-Fock theory tends to overestimate the band gap.

[1] A. Alkauskas, P. Broqvist and A. Pasquarello, Phys. Status Solidi B, 248, No. 4, 775–789 (2011).


text:
Reverse single replacement reaction
Is it possible using electrochemistry or some other method to reverse a typical single replacement reaction to go in the opposite direction.  For example, is it possible to take silver metal and copper nitrate to get copper metal and silver nitrate?  I would like to do this, because nitric acid is very expensive to ship, and I can easily buy both silver metal and copper nitrate.  I don't know how to isolate the silver the silver nitrate as it is formed because it will simply react with copper metal in a electrochemical cell.

Have you heard of an electrolytic cell?

Yes, but I don't know what electrolyte I could use for the silver half of the cell without polluting the product or causing side reactions.

Use copper sulfate as the electrolyte and do it in a solvent where copper sulfate is soluble but silver sulfate is not.


text:
How to convert 2-propanol to 2-butanol?
Can someone confirm that this is a correct way to make 2-butanol starting from 2-propanol?

First add Cl2 and UV-light to it to become 1-chloro-2-propanol
Make from CH3Cl a Grignard reagent by adding Mg and dry ether
Let a nucleophile substitution happen between CH3MgCl and 1-chloro-2-propanol
You get 2-butanol?

I would not do that.

Radical chlorination is inherently unselective because of the high reactivity of the chlorine radical, if you have to propose this as part of a synthetic route there's a 99% chance you're not doing it right
Your Grignard will get immediately destroyed by the hydroxyl group present
Grignard SN2 reactions on alkyl halides don't go very well either

Off the top of my head, I would think a much better way would be to oxidise the starting material to acetone, make an enolate with LDA, do a SN2 with MeI, and reduce the compound back with NaBH4.

Methyl iodide would probably be attacked by the oxygen and lead to a methyl-vinyl-ether.


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why does silver chloride darken when exposed to electromagnetic radiation
All I know about this is that it does but I have no idea as to why and when I ask the only answer I get is "Because it does" which isn't helpful in the least. If I could get a good, well-said response, that would be great.

And lead to the development of photography! Silver halide crystals suspended in a gelatine film on exposure to light darkens!

The energy in the electromagnetic radiation decomposes AgCl into its components, silver and chlorine. This produces finely divided silver particles, which look dark because yet solid silver (in the form of an ingot, for example) has a typical metallic 'colour', silver powder is dark.


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Order of ionization enthalpy
The order of successive ionization enthalpies of group 15 are as follows :
$$\Delta H_1<\Delta H_2<\Delta H_3$$
But as all the fifteen group elements have a very stable electronic configuration with half filled p orbital so it would require more energy to remove an electron from an extraordinarily stable configuration so the order according to me should be $\Delta H_1>\Delta H_2>\Delta H_3$.
please guide me if I am wrong.

@Jan: the answer states, "Second ionization energies are almost always higher". In which cases for the second, third, or any of the further ionisations does the "successive ionisations cost increasing amounts of energy" rule not work?   The answerer hasn't been online for a while, which is why I pinged you.

@harry The ping isn’t really appreciated. I am not aware of any case. The answerer may have chosen a phrasing that allows wiggle room in case there is indeed one case somewhere so that the answer remains correct. I don’t know, I can’t look into their brain.

While you are correct that the first ionization energy should have a larger value for all the group 15 elements, this has a minuscule effect on the ionization energy compared to the energy that is required to separate charges. Second ionization energies are almost always higher than first ionization energies because the nucleus already has a positive charge, creating a larger coulomb force that needs to be overcome as part of the ionization process. In addition, additional ionization increases the charge on the nucleus of the ion, which progressively makes the ion less and less stable.