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Role of heat in combustion
What role does heat play in the combustion process.Specifically, does the amount of heat  determine whether complete or incomplete combustion occurs, and is the amount of heat impacted by the amount of oxygen

Welcome to chemistry S.E?Why did you tag your question with inorganic-chemistry? Which combustion do you refer?

Yes, temperature and heat do affect the spontaneity and the equilibrium of reactions. The term heat can be referred to enthalpy, which can be quite confusing in context, as I believe you are asking about raising the temperature.
Let's first take a look at the equation for the Gibb's Free Energy and raising temperature:
$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
The Gibb's Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibb's Free energy describes how much "free energy" is available, or how much work is possible.
Since, entropy, enthalpy are state functions, we can calculate them by using Hess's Law and using tabulated values for standard thermodynamic data:
$$\Delta H^\circ_{rxn} = \Sigma \Delta H^\circ_{products} - \Sigma \Delta H^\circ_{reactants} $$
$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$
For simplicity's sake, I will use the combustion of methane that mannaia has so kindly provided and run the reaction at STP using heats of formation:
$\Delta H^\circ_{\text{comb of} \ \ce{CH_4}} = -890 \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H^\circ_{\text{f of} \ \ce{H_2O} \ \text{(l)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H^\circ_{\text{f of} \ \ce{CO_2} \ \text{(g)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$
$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$
$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
$$ \Delta G^\circ  = -604.53 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -633.7 \text{kJ} \cdot \text{mol}^{-1} $$
If we use the enthalpy of combustion for methane:
$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
$$ \Delta G^\circ  = -1569.34 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -1598.38 \text{kJ} \cdot \text{mol}^{-1} $$
What would occur if we raise the T term? Obviously, the Gibbs Free Energy would become larger in magnitude and more negative, thus being more spontaneous. Inversely, if we lower the temperature to such a degree that $T\Delta S^\circ = \Delta H^\circ$, the reaction would not occur at all, and if $T\Delta S^\circ > \Delta H^\circ$, we would need to input energy in order for the reaction to proceed forward.
In general:
If $\Delta H<0$ and (-) and $\Delta S>0$ (+), the reaction is spontaneous at any temperature.
If $\Delta H>0$ and (+) and $\Delta S<0$ (-), the reaction is NOT spontaneous at any temperature.
If $\Delta H<0$ and (-) and $\Delta S<0$ (-), the reaction is spontaneous at low enough temperatures.
If $\Delta H>0$ and (+) and $\Delta S>0$ (+), the reaction is spontaneous at high enough temperatures.
It is possible to think of heat as a product or reactant. Combustion reactions are highly exothermic. Le Chatlier's principle includes raising the temperature as a stress. Alternatively, we can think about the equilibrium of this reaction to see how much product will be formed by looking at the magnitude of the equilibrium constant, $K$.
Accordingly the van't Hoff isochore equation is written as:
$$ln \frac{K_2}{K_1} = \frac{\Delta H}{R}(\frac{1}{T_1} - \frac{1}{T_2}) $$
In summary, an increase in temperature favors the formation of products if the reaction is endothermic. Inversely, a decrease in temperature favors the formation of products if the reaction is exothermic.
As, Uncle Al has stated:

Ambient oxygen makes no difference

The atmosphere is composed of about 70% nitrogen and 20% oxygen. If this reaction was to be observed in nature, there would be almost a limitless amount of oxygen for the reaction to rpoceed. Temperature does play a role in the stoichiometry of the reaction and thus whether or not the combustion is complete or incomplete. However, it is possible to limit one of the reactants as mannaia has done, specifically oxygen in a laboratory setting.
As related to the kinetic molecular theory of matter, increased temperature will increase the frequency of collisions of particles with sufficient energy to cleave bonds and reform them, driving the reaction further to the right. This is commonly done with the commercial production of ammonia using Van Haber process. Although this is thermodynamically unfavorable, when run at high temperatures, the rate of formation of ammonia is much faster.
In regards to say, an open environment, the temperature would increase the kinetic energy of the oxygen molecules, and so the particles would be spaced further apart. Describing this quantitatively... well, an experiment should be carried out. I would assume if one was so inclined to find if that would increase the equilibrium, rate, or completeness of combustion I'm sure they would find differences.

could you please elaborate your sentence: "While you can limit one of the reactants ..., relatively speaking, it does not have an effect." Relatively to what ? Effect on what ? Thanks !

check also the value of $\Delta H$ for methane combustion. thanks again !

hopefully one last point ... Uncle All's statement ("Ambient oxygen makes no difference") referred to a specific case, but it can't be generalized. What happens if oxygen is not enough, for methane combustion?(http://chemistry.stackexchange.com/questions/9552/ch4-o2-co2-2h2-whats-wrong-with-it/9564#9564).

@mannaia Thanks! I used the heat of formation for methane instead of combustion. I corrected that as well as making the last two paragraphs more clear and concise. And as you stated in the comment above and in your answer below, limiting one of the reactions will also limit completeness of combustion.

could you still have a look at the numbers. I think there's something wrong in $\Delta G$ calculation. Then, why do you calculate $\Delta G$ twice and get two different values for the same reaction at STP ?

@mannaia Apologies, I used enthalpy of formation from its constituent elements instead of enthalpy of combustion.

Indirectly yes, the amount of oxygen will have an impact on the amount of heat. This because, if there is enough oxygen, combustion will tend to be complete.
Consider methane combustion, complete and then incomplete:
$$
\ce{CH4(g) + 2O2(g) ->CO2(g) + 2H2O(l)} \qquad \Delta H^{o}_{comb}=-891\,kJ/mol
$$
$$
\ce{CH4(g) + 3/2O2(g) ->CO(g) + 2H2O(l)} \qquad \Delta H^{o}_{comb}=-608\,kJ/mol
$$
The value of enthalpy, $\Delta H^{o}_{comb}$ ($25^{o}$) is larger for a complete combustion, as reported above.
That means that more heat (= enthalpy in conditions of constant pressure) is released if the combustion is complete.

I think the question was a bit different: how the heat itself affect on combustion process.

@pmod Well, I've tried to address the point "amount of heat impacted by the amount of oxygen" and differentiate between incomplete and complete combustion. But, anyway, I haven't probably interpreted the question in the good way (the title and body of the question look different to me).

In contrast, the explosion of triacetone triperoxide (83% the pop of TNT) has hard by zero exotherm.  Ambient oxygen makes no difference.  Much of the acetone can be recovered and reused -  a near-zero carbon footprint, highly recyclable terrorist explosive.  Why do people complain?  J. Am. Chem. Soc. 127, 1146 (2005)
TATP, $\mathrm{Δ_{f}H°}$ = +151.4 kJ/mol
DOI: 10.1002/prep.201100100

Short answer is, heat is a byproduct of the chemical reaction. It is released by the reaction. It is such that we now use combustion for (most of the time) the sole purpose of getting the heat out of it. We don't want the CO2, but it comes with it (if you burn hydrocarbons).

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$.


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Testing a sodium aluminate solution
I prepared some sodium aluminate by reacting a strong $NaOH$ solution with aluminium foil and filtering the result. I now have a clear solution and I'm not sure what it is. I stopped adding aluminium when the reaction rate showed down dramatically in a hot water bath. Basically what I'm asking is what are the properties of sodium aluminate since Wikipedia is severely lacking in information on sodium aluminate

Sodium aluminate is a caustic hazard, http://nj.gov/health/eoh/rtkweb/documents/fs/1675.pdf http://www.usalco.com/wp-content/uploads/2013/05/Sodium_Aluminate_SDS_USALCO.pdf http://www.chemone.com/default/msds/Sodium%20Aluminate.pdf

There are multiple reactions going on simultaneously. Aluminum metal reacts with the oxygen in surrounding air in order to create a plating of aluminum (III) oxide.
$$\ce{4Al(s) + 3O2(l) -> 2Al2O3(s)}$$
Upon addition of sodium hydroxide, the oxide layer reacts with the base, leaving the aluminum free to attack by water for instance.
$$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$
$$\ce{Al(OH)3 + NaOH -> Na+ + [Al(OH)4]-}$$
In summary, we can write the sum of the reactions as:
$$\ce{2Al(s) + 2NaOH(aq) + 6H2O(l) -> 2Na+(aq) + 2[Al(OH)4]- + 3H2(g)}$$
The reason why the reaction proceeded slowly at first was due to the aluminum (III) oxide layer covering the aluminum metal.
From this link, http://www2.uni-siegen.de/~pci/versuche/english/v44-10.html:

This reaction is used in drain cleaners. They are mostly made out of strong alkalis, to which alumunim or zinc has been added. The alkalis break down organic residues chemically. In addition, the formation of hydrogen leads to a bubbling effect which adds an additional mechanical cleaning mechanism.

Considering the properties of aluminates, one can go to PubChem hosted by the NIH which has a lot of information for almost every single compound necessary as well as literature linked to the specific compound in question:
http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=14766&loc=ec_rcs

source: Stack Exchange

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Preference of grignard between aldehyde and double bond?
Quick question, in this reaction:
2-Butenal + CH3MgBr  -->  ??
Will the grignard attack the double-bond or the aldehyde carbon?
Please explain this preference, thanks!

The Grignard will first attack the aldehyde carbon, giving you a secondary alcohol. Since the alcohol is strongly polarized, it would be difficult for the double bond to be attacked.

Thanks! So just to clarify, the major product will be the secondary alcohol and that is what prevents a second addition.

This does however not explain why Grignard reagents react different than Gilman cuprates ;)

Actually, under the reaction conditions, the magnesium is an ion for the oxygen. This creates a strong negative characteristic to the molecule so the Grignard reagent cannot do a second attack. However, if the alcohol is purified and then reacted with more Grignard reagent, then addition at the double bond is likely. (I'm not familiar with Gilman cuprates.)

@ Klaus, I do remember my professor saying something about how Gilman reagants were "soft nucleophiles" and would preserve the carbonyl group. I don't think he explained why this happened though; it would definitely make for another good question.

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Incomplete Combustion products
In class we performed an experiment which I am unable to understand the results for
Why does the the amount of soot produced vary? What is the amount of soot  produced  dependent on?
Experiment:

We had to have four test tubes upside down in a beaker of water. The test tubes were filled with water in them as well. We had to insert pieces of calcium carbide under the test tubes and have them as follows: one full of acetylene, one 3/4 full, one 1/2 full, one 1/10 full. Then remove the rubber stopper and bring a burning splint to the mouth of the test tube to ignite the acetylene gas.

Results:

The full one: burned slow and produced a little bit of soot
3/4: burned fast and lots of soot
1/2: burned fast and lots of soot
1/10: burned fast and no soot

Can you provide some picture of the experiment setup?

The reaction for complete combustion of acetylene is
$$\rm 2C_2H_2 + 5O_2 \rightarrow 4 CO_2 + 2H_2O$$
...you'll get complete combustion when acetylene and oxygen are mixed in a 2:5 mole (or volume) ratio. If you have a higher ratio than that, combustion will be incomplete. The incomplete combustion products will appear as soot and carbon monoxide.
Air is 1/5th O$_2$, so you need 5 times as much air as you'd need pure O$_2$. The optimal acetylene/air ratio should be about 2:25.
Look at the acetylene/air ratios for each tube. The gases are both at the same temperature and pressure, so the volume ratio is equal to the mole ratio. Any ratio that's greater than 2/25 will probably produce soot, and the higher the ratio, the more soot you'd expect. Any tube that contains an appreciable amount of oxygen mixed in with the acetylene will burn quickly.
The full tube burned slow because oxygen needed to mix with acetylene at the mouth of the tube. The little bit of acetylene that burned there produced soot, because the acetylene/air ratio was probably much higher than 2/25.
The tubes that were 3/4 and 1/2 full had 3:1 and 1:1 acetylene/air ratios. Both ratios are fuel-rich, so soot was produced (you'd expect more soot with the 3/4 full tube than the 1/2 full tube). Both tubes burned quickly because there was a fair amount of O$_2$ mixed with the acetylene.
The tube that was 1/10 full had a 1:9 acetylene/air ratio. 1/9 is just a bit higher than 2/25, so this tube was fuel-rich, too, but much less so than any of the other tubes. Very little soot would be expected; maybe none, if more air was drawn into the tube as the acetylene burned.


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Why is NH4OH used in Polarimetric analysis?
Why is a sugar solution pre-treated with ammonium hydroxide before analyzing? And why is it specifically ammonium hydroxide instead of sodium hydroxide or calcium hydroxide?

If I'm not wrong, this is about polarimetry, not polarography. (No dripping mercury, but optical rotation).
If your sugar is a hexose, the following species will exist in aqueous solution:

open form (a bit)
furanose  (traces)
pyranoses (major amount)

The pyranoses will adopt the $\alpha$- and the $\beta$-form. Both rotate polarized light by a different angle.
The interconversion between these forms (mutarotation) is facilitated by addition of acid or base. Thus, an equilibrium, resulting in a constant angle in polarimetry, is reached faster.

Do you see a potential corresponding to the sugars, or to their Schiff bases, or to said imines' trimers?  Perhaps ammonia gives you a nice pH buffer, as opposed to alkali hydroxides that will wildly swing pH and allow Cannizzaro disproportionation of reducing sugars' aldehyde groups at high pH. Ammonia does not reduce at the cathode.


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CH4 + O2 => CO2 + 2H2, what's wrong with it?
It says in my textbook that:
$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$$
which seems like a displacement reaction to me. Shouldn't the reaction be:
$$\ce{CH4 + O2 -> CO2 + 2 H2}$$
since $\ce{4H}$ are displaced by $\ce{O2}$ (becoming $\ce{2 H2}$), which makes the reaction balanced?
Why is the reaction $\ce{CH4 + O2 -> CO2 + H2O}$, which is further balanced as my textbook says?

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen.
So in "first" step:
$\ce{CH_4 + O_2 -> CO_2 + 2H_2}$
but then the hydrogen will be also combusted in reaction :
$\ce{2H_2 + O_2 -> 2H_2O}$
so the overall reaction will be:
$\ce{CH_4 + 2O_2 -> CO_2 + 2H_2O}$
As was pointed out in the others answers the real mechanism of a methane combustion at a low oxygen pressure is more complicated:

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida, http://enu.kz/repository/2011/AIAA-2011-94.pdf
another:

Intermediate Species Proﬁles in Low-Pressure Methane/Oxygen Flames Inhibited by 2-H Heptaﬂuoropropane: Comparison of Experimental Data with Kinetic Modeling, http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf

Thanks, you are the first guy who actually got it. I had another question: This is a displacement reaction or an oxidation reaction? I think it is displacement but I am not sure.

@ Klaus Warzecha - thanks I will
@ user2468338 - It is an oxidation reaction:
An oxidation/ reduction reaction is one where there is a change in the oxidation states of two or more species

Yeah, but displacement reaction is also a type of oxidation reaction right.

for example reaction of natrium hydroxide with hydrochloric acid is displacement reaction and not oxidation/reduction reaction.

@JaroslavKotowski Looks strange to me: with little oxygen, you fully oxidize carbon ($\ce{CO2}$). I've never thought that methane combustion could occur via two steps. Could you  please provide reference for that ?

@mannaia The world "first" is quoted because it is not real 1st step. The reaction in low oxygen pressure is a bit more complicated:
http://enu.kz/repository/2011/AIAA-2011-94.pdf
http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf

@JaroslavKotowski Looks indeed very complex. Thanks for pointing it out in your last edit.

I kinda potentially disagree with the first sentence, because I’m not sure if $\ce{CO2}$ and $\ce{H2}$ will actually form given the lacking of oxygen, but other than that, this answer deserves the badge it is getting now.

Look at the enthalpies of formation of $\ce{CO}$, $\ce{CO2}$, and $\ce{H2O}$, then the net enthalpies of reaction.  In a deficiency of oxygen, and trimmed by the temperature and kinetics versus equilibrium conditions, are you more likely to obtain $\ce{CO}$ plus $\ce{H2O}$ than free hydrogen?  Given less oxygen, soot and water?  Note reaction energetics below,
http://en.wikipedia.org/wiki/Water_gas
http://en.wikipedia.org/wiki/Syngas
http://en.wikipedia.org/wiki/Gasification
http://www.precision-combustion.com/High_Pressure_CPOX_methane.pdf

There's nothing wrong. It's incomplete combustion.  It is one of a number of oxidizing reactions that can and will be occurring if you are burning this hydrocarbon in an oxygen-restricted environment. But the dominant reaction will be the one where the lowest energy state is reached. So you will likely wind up with one CO molecule, one water molecule and one hydrogen molecule instead.


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Best solvent for crystallization + mass recovered from crystallization + purity of the product

I have a chemistry final tomorrow and this question on my sample exam through me for a loop. I'm not entirely confident I can work out problems B and C, but those questions rely on the answer to A.
If you could please especially explain how I'm supposed to read the solubility chart to understand which solvent is the best, that would be awesome.

The best solvent would be the one with the greatest ratio of solubility for 3-phenylacrylaldehyde and a high solubility for benzaldehyde. You should be able to work out why from what I said.

Hi, thanks for answering, but I still dont quite understand.

Expanding on LDC3's comment a little.
The information in the table (irrelevant of how good or bad the method proposed is) tells you what the saturation points are for solvent mixtures at given temperatures for each compound.
Fundamentally, you want to separate your desired product from your undesired product. In this case, you are trying to recrystallise the 3-phenylacrylaldehyde, leaving the benzaldehyde in solution. You know this because for all temperatures and solvent mixtures it has lower solubility (this cannot be assumed for a more complex solvent array).
You need the minimum of hot solvent, here 100 degrees, to dissolve all of your crude compound, and you want to maximise the return of pure compound on cooling. This explains why you want the biggest ratio of hot to cold solubility as LDC3 indicated.
You also want to minimise the amount of your starting material that crashes out with your product. So, calculate the same ratios for the starting material, but here you want to minimise the return on cooling, so you want the smallest ratio.
You now have a table of solubility ratios, columns being 'solvent mix', 'rP' (ratio of product solubilities) and 'rSM' (ratio of starting material solubilities). The largest value of rP/rSM gives you the best solvent mixture.

Is the double bond cis ${Z}$ or trans ${E}$?  (Terrible nomenclature.) What is the scent of the product?  What is it commonly called? Don't taste it. A tiny bit on your tongue will excavate a deep ulcer.  The ${E}$-isomer melts at -7.5 C.  How did you get the putative  ${Z}$-isomer, and how can it melt at least 40 C degrees higher than the ${E}$-isomer?
Two solvents are used, not five.  Different mixtures are used.  The technique is terrible!  You would dissolve contaminated product in warm ethanol (below product mp) with stirring, then titrate with water until the cloud point.  Add another drop of ethanol to clear.  Very slowly cool without stirring (cheesecloth wrap) to obtain large clean crystals.  Refrigerate for more crystals. Repeat a second time to maximize product yield.  Do not combine to two crops!  The second is not as pure.
As the product is most likely a liquid, you would fractionally distill it under reduced pressure (aspirator).
http://chemistry.st-andrews.ac.uk/labs/misc/nomo2.gif

Although an interesting piece of information, this has hardly any connection with the question the OP is asking. OP wanted to know how to select the best solvent for the reaction from the solubility chart.

The knowledge of trivial names can be helpful :) As for the isomerization, the answer lies in the 300 nm range ;)

And yes, the options given in the question are terrible, the nomogram is helpful and distillation under reduced pressure an obvious choice. Storing the product under nitrogen or argon in the freezer in a flask wrapped with aluninium foil isn't a bad idea either.


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Correlation Diagram Doesn't Match Selection Rules?
From Miessler's Inorganic Chemistry, 5th edition, there is this correlation diagram for $d^{2}$ in a octahedral field.

On the left hand side, if it's not clear, it goes (from lowest to highest):
$^{3}F, ^{1}D, ^{3}P, ^{1}G, ^{1}S$.
But by Hund's rules mentioned even in the same chapter, one would expect the $^{3}P$ to show up right after the $^{3}F$ because of their higher spin multiplicities. And even after that, G should come before D. (the ordering for that being s,p,d,f,g in reverse for same spin multiplicities). I don't know of any reason for why the $^1D$ term is just above the $^3F$ term in this graph though. Is this an error or does anyone have an explanation?

No, it's not an error. Take a look at the Tanabe Sugano diagram for $d^2$ (from Wikipedia):

You can see that again, the $^1D$ state is below the $^3P$ state. Why? If we were considering a $p^2$ configuration, the $^3P$ is indeed lower in energy than $^1D$. And again anyone might predict that $^3P$ is lower for $d^2$ too, for the exact reasons you give.
We know this result from spectroscopy. The difference is because of the Racah parameters for the different terms (i.e., the electrostatic repulsion between electrons in the same orbital. The Wikipedia article had a nice link to how you'd compute the difference in this case.
Now of course a $^3F \rightarrow ^1D$ transition is not an allowed transition, which is shown in the textbook on the next page.
The moral of the story is that Hunds' rules and the basics will always get you the lowest energy term symbols for multi-electron atoms, but the details can be tricky for higher-level states.

This is why I only teach how to find the lowest energy term symbol. I gave up trying to get through all the details of higher-energy states - there are too many subtle effects.


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Spectrophotometry for Colourant
I have been looking for a paper that by using spectrophotometry method, we can determine rate of reaction of colourant. I could not find any paper. I was wondering if any body could help me.

What do you mean by "colorant"? Please clarify this a little more.

@tschoppi I mean dye or pigment...

So basically you're looking for a paper about spectrophotometry per se...? I think that is a very broad question...

What do you mean for a "reaction of colorant"? A reaction with what?

@GM For instance, degradation or formation colourant

I'm not sure I have answer to your question maybe, if you consider natural colorant too, these are the articles that go more close to what you want:
Prediction of phenolic compounds in red wine fermentations by visible and near infrared spectroscopy
Spectroscopic, stability and radical-scavenging properties of a novel pigment from gardenia
Try to have a look a these too:
Mimani, T., and Samrat Ghosh. "Combustion synthesis of cobalt pigments: Blue and pink." Current Science 78.7 (2000): 892-896.
Bernardi, M. I. B., et al. "Comparison of blue pigments prepared by two different methods." Journal of the European Ceramic Society 22.16 (2002): 2911-2919.
Visinescu, Diana, et al. "The environmentally benign synthesis of nanosized Co< sub> x Zn< sub> 1− x Al< sub> 2 O< sub> 4 blue pigments." Dyes and Pigments 87.2 (2010): 125-131.


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Are atoms really round?
I'm not sure if this is a silly question, but I was sitting here with a cup full of cheezey poof balls thinking, "My goodness, it's like an amazing cheesey delicious liquid - huge water molecules!"
Of course my next thought was, "Wait a minute - water has two hydrogen atoms bonded to an oxygen, so that's not quite right. They wouldn't be round like this."
Then I started thinking about the diagrams we see in chemistry textbooks, etc., and how the atoms are always pictured as round balls. How do we know this is accurate? Is it possible for the atoms to be configured in more complicated shapes (e.g. not solid, crystalline, some type of lattice)?

...delicious?

It depends how you define the surface of an atom. Atoms maintain no surface in the normal sense; only regions of space where you have a better chance of finding electrons. So in fact it is not correct to say they have a true shape at all.
Shapes of Atomic Orbitals
However if you plot the region of higher probability of finding electrons in an atom you can obtain something like this:

These are the shapes of the first five atomic orbitals: $1\mathrm{s}$, $2\mathrm{s}$, $2\mathrm{p}_x$, $2\mathrm{p}_y$, and $2\mathrm{p}_z$ from Wikipedia. $1\mathrm{s}$ orbital is sphere-shaped but other orbitals have more complex shapes so atoms with many electrons have orbitals very different from a sphere.
Shapes of Molecular Orbitals
Molecules have more electrons and so even more orbitals. They can have very strange "shapes". I've calculated for you with GAMESS and Avogadro the LUMO 2$b_2$ water's orbital that is one of the molecular orbitals of water. This is the result:

Shapes of Atomic Constituents
For answering your question in the comment: In fact even protons, neutrons and electrons don't have a real shape due to wave-particle duality. However we can assume in many cases that neutrons are particles (so we suppose a spherical symmetry) but the  de Broglie hypothesis states that they have also a wavelength!

Same thing for neutrons?

Man... particles are weird. Would it be safe to say that really we don't actually know what makes up "stuff" (i.e. matter) - we just have really good evidence of how it behaves in different conditions?

@WayneWerner absolutely! We can describe very well some quantic phenomena and do some accurate prediction, but we are really far from a true understand of matter! Sometimes science give us the illusion of true Knowledge but in fact it give us only instruments for controls matter!

Everything has a wave-particle duality, including atoms and even large molecules (only, normally thermal motion causes decoherence on scales smaller than their own size, so we don't really perceive quantum effects much except in very complicated cryogenic experiments). — BTW, spherical symmetry has little to do with particle vs. wave, it's a property fully describable in the exact Hilbert-space formulation. Fermions actually don't have a spherical symmetry but a strange two-rotations symmetry that has no classical correspondence.

(Though you're right in a sense: in the absence of strong magnetic field the spin states are degenerate, so in the classical / statistical limit they're averaged out to spherical symmetry)

All full shells of isolated atoms are spherically symmetric. And, yes, this is not obvious from the visualization images you have selected, but it is clear in the math.

@dmckee Indeed that is true, because the electron density outside of the selected cutoff for drawing the orbitals also need to be taken into account. But just to be sure, I understand that half-filled subshells are also spherically symmetric? Thus any atom with a $s^1/s^2/p^3/p^6/d^5/d^{10}/f^7/f^{14}/...$ valence configuration (assuming all other non-valence subshells are fully populated) would be spherically symmetric.

@NicolauSakerNeto I think that is correct, at least for atoms whose nuclei do not exhibit a strong magnetic moment. Not sure if it can be guaranteed if the nuclear moment is significant.

@NicolauSakerNeto why is it limited to half filled or filled?  For a free atom, say hydrogen in a 2p state, is the electron in a particular p orbital or a superposition of all three p orbitals?

@DavePhD Interesting remark. After a little thinking and with the great help of a previous answer, it would seem my original thoughts were incorrect. Without any disturbance, all atoms have spherically symmetric electron shells a priori regardless of how filled their subshells are. Thank you for the heads up.

If you can find a single atom in vacuum with a net dipole moment, its electron cloud is obviously not spherically symmetric.  Go across the periodic table's second row.  They are all $\ce{1s^2}$ $\ce{2s^2}$, so their atomic cores absent chemical combination and hybridization are first order spherically symmetric, filled s-orbitals.
B $\ce{2p^1}$, C $\ce{2p^2}$, N $\ce{2p^3}$, O $\ce{2p^4}$, F $\ce{2p^5}$, Ne $\ce{2p^6}$
There are three orthogonal 2p oribtals shaped like dumbbells: $\ce{2p_{x}}$, $\ce{2p_{y}}$, $\ce{2p_{z}}$.  Maximum multiplicity says they each fill before electron paring occurs,  One then suspects B, C, O, and F single atoms in vacuum would have dipole moments.  If they superpose orbital hybridization absent chemical combination, symmetry says they do not have a dipole moment. Do they?  Google/Google Scholar are your friends.
http://arxiv.org/ftp/arxiv/papers/1010/1010.2425.pdf
For alkali metals.  Is the vacuum phase spontaneously dimeric?
http://home.physics.wisc.edu/~tgwalker/056.singlet_PRA.pdf
Complicated.

Firstly, the probablity function for a 2px has the symmetry of a cylinder (a plane of symmetry perpendicular to an infinite-fold rotational axis of symmetry like O=C=O), so a dipole moment would not result.  Secondly, for B, C, O, and F single atoms in vacuum the three p orbitals would be degenerate and the electrons would not occupy a particular subset, but an linear combination of all.  Finally, there is much interest in looking for permament electric dipoles in atoms due to time reversal violation implications; I would be skeptical of any article that is not peer reviewed.


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What reaction takes place when potassium iodide is added to a tetraammine copper(II) complex?
I am given a blue solid W and it contains a tetraamine copper(II) complex and a $\ce{SO_4^{2-}}$. I added dilute sulfuric acid until the solution is pale blue; then added potassium iodide solution. A white precipitate forms ($\ce{CuI}$) in a brown solution ($\ce{I_2}$). However, I am confused about what reaction is actually going on.

Copper(II) is a hard Lewis acid that would be octahedrally coordinated but for Jahn-Teller distortion.  Instead, it is square-planar coordinated with two distant weakly bound apical ligands.  The lovely deep blue color, hard Lewis base ammonia versus hard Lewis acid water coordination, illustrates the nephelauxetic effect.  What would obtain with triethylamine coordination of copper?  Cu(II) is substitution labile.

Jahn-Teller effect
Nephelauxetic effect
Kinetics (and thermodynamics) of substitution reactions

Adding acid protonates the ammonia, its lone pair now in a dative bond with a proton and unavailable for coordination.  Tetraamminecopper(II) (note the spelling of "ammine" when coordinated) becomes tetraaquocopper(II) with fast kinetics.
Aqueous Cu(II) is a modest oxidizing agent; iodide is a modest reducing agent.  Cu(I) is a soft Lewis acid, iodide is a soft Lewis base, $\ce{CuI}$ is very insoluble (look up its $\ce{K_{sp}}$ -- see here.)
Iodine reacts with excess iodide to give triiodide anion.  However, isolate the "$\ce{CuI}$," wash, and titrate with thiosulfate.  $\ce{CuI}$ strongly adsorbs elemental iodine.  It is also modesty light sensitive.  Cyanide, being a pseudohalide, follows the same path to afford exceptionally insoluble CuCN plus cyanogen.
Remember who earned the "A," or do your chair parade for better references, then understanding for the next question.

This was really helpful. Just one thing, when you said "Adding acid protonates the ammonia," Are you referring to this equilibrium: $$[Cu(H_2O)_6] + 4NH_3<=> [Cu(NH_3)_4(H_2O)_2] + 4H_2O $$. So, by protonating the $NH_3$, The $NH_3$ is used up and the equilibrium shifts to the left forming more hexaaquacopper(II) ions?

Ammonia as such, plus acid, is removed as ammonium. There is nothing remaining but water to coordinate Cu(II).  Co(NH3)6 is instantly disassembled by acid.  Co(NH3)6 will boil in concentrated HCl for weeks with no change.  Co(en)6 plus formaldehyde and nitromethane in base is exceptionally elegant re Sargeson, DOI: 10.1021/ja00451a062  Chemistry is great stuff.

The formation of the tetraammine (II) copper sulfate ion complex is made through the addition of excess ammonia to copper (II) sulfate solution:
$$\ce{CuSO4(aq) + 4NH3(aq) -> [Cu(NH3)4]^2+(aq) + SO4^{2-}(aq)}$$
If we add sulfuric acid in solution, we push the equilibrium to the left, as there will be an increase in the concentration of sulfate ions as sulfuric acid ionizes. As the equilibrium shifts to the left, there will be an increase in concentration of ammonia and copper (II) sulfate in solution according to Le Chatlier's Principle:
$$\ce{[Cu(NH3)4]2+(aq) + SO4^{2-}(aq) + H2SO4(aq) -> CuSO4(aq) + 4NH3(aq)}$$
If we add potassium iodide, an oxidation and reduction reaction takes place between the copper (II) sulfate solution and potassium iodide. Since the reaction is taking place in basic solution, we can balance the electrochemical reaction.
$$\ce{2KI (aq) + CuSO4(aq) -> CuI2(s) + K2SO4(aq)}$$
Here we can see that solid white copper(II) iodide precipitates and potassium sulfate dissociates into its ions.
Since copper (II) iodide is very unstable, it decomposes into copper(I) iodide and iodine:
$$\ce{2CuI2 (s) -> 2CuI (s) + I2(s)}$$

"CuI2" does not exist.

@UncleAl Thanks, I fixed that accordingly. I forgot that it is unstable and immediately decomposes after formation.

@Jun-Goo Kwak: so the acid was added just to tip the equilibrium to the left? Just asking, If I didn't add the sulphuric acid, would the tetraammine complex react with KI?

@Eliza Hi, yes. Otherwise, you would have very little copper (II) sulfate solution. By shifting the equilibrium left, you produce more copper (II) sulfate which can then react with the potassium iodide.

@Jun-Goo Kwak: That makes so much sense. Just one thing, instead of thinking of the equilibrium like as above, can I think about it like this: $$[Cu(H_2O)_6]^{2+} + 4NH_3<=> [Cu(NH_3)_6(H_2O)_2]^{2+} + 4H_2O $$ The $H^+$ added causes the $NH_3$ to be protonated - this uses it up. This in turn causes equilibrium to shift to the left producing more hexaaquacopper(II) ions?


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Why does NaCl dissolve in H2O despite its strong ionic bond?
If $\ce{NaCl}$ is such a strongly bonded ionic compound, why it does it disassociate so readily in $\ce{H2O}$? One would think that because of the very strong ionic bond it would not disassociate it all. Further, why do other compounds like $\ce{AgCl}$ or $\ce{HF}$ not disassociate as readily even though they also have strong bonds?
Although others have marked this as duplicate, I prefer the wording of this question as it explicitly refers to NaCl has having a strong bond (which includes the search term I used to find this website). I have been writing a manuscript in which I have been attempting to answer this question (among others). The real answer becomes a bit complicated, but I will try to give a short answer.
It is my opinion that thermodynamic data is not always properly understood. The first thing we should understand is that the same thermodynamic change in energy should result whether the reaction were to take place in the gas phase (Born-Haber cycle) or in solution. This is a principle of conservation of energy, Hess's Law. In the Born-Haber Cycle, the sum of the energies for vaporization, ionization, and lattice energies are determined. If a hypothetical solution phase Born-Haber calculation were to be performed, the same amount of energy would exchange, except for the solvation energy, which is small. That is what the poster was asking.
If you were to think about the conditions of a solution phase Born-Haber cycle, then you would take the elements and convert them directly to solvent separated ions. What this should tell you is that you have just calculated the solution phase energy of a redox reaction of sodium and chlorine to give sodium and chloride ions. This bypasses the implication of sodium chloride having a strong affinity for its ions. Yes, it would take a lot of energy to separate sodium and chloride ions in the gas phase. That is why the lattice energy is so large, but if we did the reaction in solution, we will not have a gas phase lattice energy component in our calculations. Of course this seems plausible as the solution phases of HCl and NaOH are ionic. This is consistent with our expected properties of HCl to act as an acid (chloride having a low affinity for cations) and NaOH to act as a base (sodium having a low affinity for electrons). The sum is we should not expect NaCl to have a strong affinity (not none) for the ions of the opposite charge.
By the way, I think you will also find a good correlation of heterolytic solution phase bond strength with bond length (actually, ion gap). This is different from the homolytic bond energies you may find reported in many tables, including the poster's question, "Why does NaCl dissolve in H2O despite its strong ionic bond?"
PW

Dr OChem's edit of this question is an answer and should have been posted as such, not as an edit to the question. :-(

The electrostatic forces between the positive cations and anions are strong. Coulomb's Law expresses the potential energy of two charges separated by a distance.
$$E_p = \frac{Q_1Q_2}{4\pi \epsilon_0 r} $$
As you may already know, water molecules stick to each other through dipole-dipole forces due to the permanent dipole that water faces due to the high electronegativity of the oxygen atom.
As ionic solids are added to water, water molecules proceed to surround each ion on the surface of the solid, forming a sphere of hydration. In the process, ions are separated from each other.
The $\delta^-$ charge on the oxygen atoms of water are attracted to cations and inversely, repels the $\delta^+$ hydrogen atoms. Thus, for cations, the oxygens of water point inward, and for anions, the hydrogens face inward respectively. The most important thing is that the ion-dipole interactions and separation of ions with little change in energy.
We can relate the potential energy of the ions to the two partial charges of a polar molecule like water:
$$ E_p \propto - \frac{|z|\mu}{r^2}$$
Z is the charge number of the ion and $\mu$ is the dipole moment of the polar molecule. Potential energy is lowered by the interaction between the solvent molecules and the ion. The $r^2$ term indicates that the interaction between ions and dipoles depends more on distance than the charges between two ions.
Thus, for hydration to occur, ion-dipole interactions must occur at the surface of the ion, and thus, ion-dipole interactions are strong for small, highly charged ions such as $\ce{Mg^{2+}}$, $\ce{Li^{2+}}$ etc.
$\ce{AgCl}$ is very slightly soluble in water and will not dissociate into it's ions. $\ce{HF}$ is a weak acid thus it does not deprotonate easily.

Ok, this is what I understood so far based on your answer and the link in the comment to my question. When NaCl is added to water, the electronegative oxygen is attracted to the Na+ and the electropositive hydrogen is attracted to the Cl-. However, this is still not strong enough to break the ionic bond so external energy is needed and this is thus an endothermic reaction. However, the reaction is still spontaneous because there is an increase in entropy.

@user2612743 Yes, looking at spontaneity is one way of looking at the reaction. Another way is too look at potential energy as matter will always try to achieve the lowest possible energy state.

How does being disassociated have a lower energy state than remaining as a solid? Because doesn't becoming disassociated requires an addition of energy, making it have a higher energy state?

@user2612743 Yes, in general that is correct. However, the partial charges of the water molecules replace the charges of the ionic neighbors and thus there is a minimal change of energy. In looking at the potential neergy between the full charge of an ion and the two partial charges of a polar molecule, we can relate it through the proportion above. As the solvent molecules and the ions interact, the potential energy is indeed lowered.

@Jun-Goo Kwak, CaCl2 gives off heat when it dissolves. I would say that there is a significant change in energy. Also, When KI dissolves, the water cools down at least 10 degrees.


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Why does Pb normally have an oxidation number of +2?
Why does $\ce{Pb}$ have an oxidation number of $2^{+}$ while carbon and other elements in the same group have an oxidation number of $4^{+}$?
Furthermore, why doesn't carbon also have a $2^{+}$ oxidation number, as can't it also give up its two electrons in the p-suborbital to have a full s-suborbital? Why does it prefer to gain four electrons instead to fill the p-suborbital?
Similarly, why can't $\ce{Pb}$ gain four electrons just like carbon instead of giving up two?
Finally, doesn't this mean that $\ce{Pb}$ forms ionic bonds? Why doesn't carbon do the same?

This is usually called the inert pair effect, but there is more to it than just that. It is a consequence of relativistic corrections to the behavior of atoms, which occur more strongly for atoms far down in the periodic table.

For other answers discussing some aspects of relativistic effects, check here, here, and here, and for a very thorough analysis of relativistic effects, check here and here.

Why can't lead gain four electrons? The lower you go on the periodic table, the lesser you will find elements that have a more non-metallic character (so with this I mean elements that will have a lower tendency to gain electrons). Just visualize a lead atom with a very small nucleus and a large area surrounding the nucleus, filled with electrons. Now, the more you go away from the nucleus, the weaker the positively charged nucleus can 'pull' on the outer , negatively charged, valence electrons. Elements that haven't got a very tight grip on these valence electrons, will lose their valence electrons. However, elements with a small atomic radius (such as Fluorine) will have such a tight grip on its valence electrons, that it will even be able to snatch electrons from other atoms.
Of course the electron configuration of Radon is stable, because it has fully filled orbitals and it has a noble gas configuration, but you need to keep the nuclear forces in mind. This you can also see by looking at the electronegative value of lead : it is 1.87 (Pauling scale). Now compare it to carbon, which has 2.5. The lower you go in the same group, the lower the electronegativity goes and the more metallic character you will get. Elements with more metallic properties (such as lead) will mostly form ionic bonds and elements with more non-metallic properties (such as carbon) will mostly form covalent bonds. There exists a rule to predict (however it isn't always correct) if a compound will be covalently/ionically bonded : if the difference in electronegative value is higher than 1.66, then you have an ionic bond and if it is lower than 1.66 then you have a covalent bond.
Why does lead have an oxidation number of 2+ while carbon and other elements in the same group have an oxidation number of 4+? Just to correct you on this : lead does have an oxidation state of 4+ and elements above such as carbon and tin also have 2+ as an oxidation state. Just look at the electron configuration : the outer $\ce{s^2 p^2}$ orbitals will be able to lose 2 or even 4 electrons, or even gain 4 electrons. In fact these are all possible oxidation states of the elements in the carbon-group:
   Carbon 4, 3, 2, 1, 0, −1, −2, −3, −4
   Silicon 4, 3, 2, 1 -1, -2, -3, -4
   Germanium 4, 3, 2, 1, 0, -1, -2, -3, -4
   Tin  4, 3, 2, 1, -4
   Lead   4, 3, 2, 1
   (source : wikipedia)

Please do note the tendency to have more positive oxidation states, the lower you go in the group. (the reason why? see the beginning of the answer) Note : not all of these oxidation states are (very) stable.
Also have a look at this picture:

Notice the metallic (shown by red) character increases the lower you go on the periodic table.

The preference of $\ce{Pb}$ for the lower oxidation state is an example for the so-called inert pair effect. Due to relativistic contraction of the outermost $s$ orbital, its 2 electrons are closer to the nucleus and thus more tightly bound than the electrons in the $p$ orbitals. More energy is required to use the $s$ electrons for covalent or ionic bonds. Stable compounds with $\ce{Pb}$ in the oxidation state $\ce{+IV}$ are therefore mostly formed with very electronegative elements like $\ce{O}$ or $\ce{F}$.

[Carbon cannot shift the electrons from the S orbital to the P orbital because there is a significant difference in energy, unlike lead.] - oops, I was wrong
Lead also has a +4 oxidation state. Look at the chemistry of a lead acid battery. Lead doesn't form ionic bonds since the difference in electronegativity between lead and the anions is too small.
NaCl - large difference in electronegativity
CO2 - small difference in electronegativity
PbO2 - small difference in electronegativity


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Why is it OH- and not HO-?
I am told that in a chemical equation the metal comes first and then the non metal. for example MgO, ZnSO4,etc. But when both the elements are non metals or metals, the one with the lower atomic number would come first. for example:- H2O,CO2,HF,etc. But why is it that hydroxide is OH- and not HO-? Also why is methane CH4 and not H4C?

OH- and HO- are both acceptable.  The charge is on the oxygen, of course.  One can argue cation then anion (salts) versus alphabetical (e.g., AuBe, which is brittle and weak rather than ductile, suggesting ionicity).
Organic formulas are C, H, then alphabetical.  Order may be altered to conform to structure (e.g., propane as $\ce{H3CCH2CH3}$) or convenience $\ce{H3C^{-}Li^{+}}$ (which is nowhere near the real structure of methyllithium, but is how it reacts).


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A reversible exothermic reaction between a molecule in an excited triplet state and a reducing agent
Say I promote a molecule to a triplet state via repeated electronic excitation events.  Here, this is going to mean that an electron from the HOMO level, originally with a spin-UP and spin-DOWN electron, will be promoted to the LUMO level such that its spin is the same as the electron left-over in the HOMO level (we can then use the Pauli exclusion principle to partially understand that the triplet state will generally have a longer lifetime than the typically higher energy singlet state).  The classic example of a molecule in a triplet state is oxygen (though here, this is also the ground state, which is somewhat unusual):

Prior to returning to the ground state via phosphorescence or a reverse intersystem crossing event, let's say that the triplet $T_1$ state molecule reacts with a reducing agent via an exothermic bond formation process, and that the reaction is reversible.  Consider that there are numerous instances where triplet state reactivity is higher than ground state reactivity.
The following two things are unclear to me:
(1) When the reducing agent dissociates, is the molecule in the $T_1$ triplet state or the $G_0$ ground state?  If the molecule is still in the triplet state, can it enter the ground state while remaining bound to the reducing agent?
(2) Call the energy gap between the relevant $T_1$ triplet state and the $G_0$ ground state $E_1$.  Call the energy released into a surrounding thermal bath by the reducing agent reacting with the molecule in the $T_1$ triplet state $E_2$.  Must $||E_2|| > ||E_1||$?
My guess is that the pair of electrons donated by the reducing agent will necessarily set  $||E_2|| > ||E_1||$ and that the molecule will be left in the ground state upon dissociation of the reducing agent since the electrons will be reallocated to properly satisfy spin-pairing requirements.

Since you stated it is an exothermic reaction, then  ||E2||>||E1||.

@LDC3 Sorry, I should clarify that I meant "exothermic" with respect only to the reaction between the molecule in the $T_1$ state and the reducing agent.  I.e. if we had a population of molecules in the triplet state and introduced the reducing agent, a thermal bath containing these molecules would heat up.  We're not accounting here for the energy required to originally "push up" the molecule to the triplet state (i.e. $E_1$).

As to your terminology and notation: What do you mean with "fluorescent excitation"? Emission from $S_1$ or higher does not play a role here: The excitation is mostly just $S_0 \rightarrow S_1$, followed by ISC to a triplet state with lower energy. And what do you mean with "repeated events". I've never seen $G_0$ to denote a ground state.

@KlausWarzecha I should have just said "optical excitation", I didn't mean to imply that there was a relevant emission event.  We're just pushing something like a dye from the ground state to the excited state repeatedly until we have an intersystem crossing event to the triplet state.

@KlausWarzecha I don't know if $G_0$ is exactly standard notation, but I've seen it in the literature.  Also texts, for example, it's used repeatedly here: http://www.springer.com/physics/theoretical,+mathematical+%26+computational+physics/book/978-3-540-22298-9

@KlausWarzecha By "repeated events" I just mean that ISC has a probability of less than $1$ per excitation event event, so you need to keep absorbing and emitting photons (if that's how you're exciting the molecule and if the decay process is radiative) to enter the triplet state.

Say "electronic excitation" and the photochemical community is happy ;)

@KlausWarzecha Done and done. :)

If you want to check if the electron transfer will occur use $\Delta G^O = E^O_{donor} - E^O_{acceptor} - E^$ where $E^$ is the energy of the excited state.

We're looking at a case of PET (photoinduced electron transfer) with electron back transfer between an (electron) acceptor A and a donor D here.
For organic dyes, the ground state typically is a singlet state ($S_0)$, excitation of A yields an excited singlet state (typically $S_1$).
$\ce{A ->[h\nu]\  ^1A^{\ast}}$
Quinone based dyes (or aza-BODIPY derivatives) often have a high triplet quantum yield, the following ISC (intersystem crossing) to a triplet state is efficient then.
$\ce{^1A^{\ast} ->[ISC]\  ^3A^{\ast}}$
In the excited (triplet) state, electron transfer from a (ground state) donor may occur by filling the lower SOMO of the excited species:
$\ce{^3A^{\ast} + D -> A^{\bullet-} + D^{\bullet+}}$
Is the acceptor radical anion still in a triplet state? No, since there no longer an electron pair with opposite spins.
Q: So, what is this good for?
A: By electronic excitation, we managed to facilitate a redox reaction that would not have been possible in the ground state and generated a radical ion pair.
Radical ions are reactive intermediates.
For a spectroscopist, it can be favourable if the only reaction of the radical ion pair is  electron back transfer:
$\ce{A^{\bullet-} + D^{\bullet+} -> A + D}$
This regenerates acceptor and donor in their former spin states, which - as mentioned before - are typically singlet states in the case of organic molecules.
Practically, this means that no permanent bleaching occurs. You can excite the sample with laser pulses again and again to measure the transients in time-resolved uv-vis spectrosopy.
+
From a preparative perspective, this is the worst case: there is no product being formed.
We have examined more productive cases of PET in the context of PET-initiated radical cyclizations of isoprenoid polyalkenes a while ago and published about these, e.g. in J. Phys.Chem A, in J.Chem. Soc., Faraday Trans. and in Pure Appl. Chem.. (The last two should be freely accessible).

In the course of these reactions, an alkene is oxidized to the corresponding radical cation by the excited state of an electron acceptor. The radical cation
traps a nucleophile (water, alkanol), immediately deprotonates and thus form a radical, which undergoes 5-exo-trig cyclization to a stabilized dicyanomethyl radical. The latter is reduced to its corresponding anion by $A^{\bullet-}$ and eventually protonated to yield the product (as a pair of separatable diastereomers).

So in any case, we lose the triplet state, meaning that $||E_2|| > ||E_1||$ right?  Could you explain a little bit more about what you mean by: "From a preparative perspective"?  As you might guess, my background is not in chemistry.

@1Q86 I will explain the spectroscopic and preparative aspects a bit later - at the moment I'm busy enjoying the wonderful weekend weather :)

Please enjoy the nice weather!  It is raining here unfortunately.

Also, I think I know what you mean by preperative in this case, you mean the situation where we're aiming to use PET to generate a population of radical dye/etc. anions by irradiating this dye groups in the presence of a reducing agent.  As such, electron back transfer will reduce the yield.  Is this right?  Just to say, my interest is mostly to understand what's going on, I don't have a particular reaction in mind that I'd like to optimize.  As such, given what you've said, a confirmation for the answer $G_0$ / "no" for my 1st question, and "Yes" for my 2nd question would be very helpful.

Can we assume that the radical anion acceptor has lower free energy (i.e. is more thermodynamically stable) than the original singlet ground state of the acceptor prior to interacting with the donor?


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What is the product of CO + 2H2 under normal pressure?
Under the pressure of $\pu{340 atm}$
$$\ce{CO + 2H2 <=> CH3OH}$$
But what is its product during normal pressure of $\pu{1 atm}$? I think it would be a single displacement reaction
$$\ce{CO + 2H2 <=> CH4 + O2}$$
but I'm not sure. Thanks in advance.

This is all very much dependent on the catalyst and temperature that you use. At 340 atm it is also easily possible to get CH4 out of the reaction. The CO, H2 mixture is called syngas and the type of reaction you are looking at is the Fischer-Tropsch process. You might want to look into that!

@Michiel why don't you post an answer? Answer questions in  comments is not a good practice! We will stay in beta forever!:-)

@GM I don't consider what I wrote an answer t all. It is merely a bit of 'pointing in the right direction'. I didn't have the time to do the search myself, hence the comment. (with success because the OP found the reaction himself)

Thanks a lot  Michiel I got the answer in the Fischer-Tropsch process. During normal atmospheric pressure the following reaction occurs :
$$\ce{CO + 3H2 <=> CH4 + H2O}$$
Also could someone tell me the type of reaction this is? I think it is partially a displacement reaction for $\ce{O}$ is displaced by $\ce{H}$ and gives $\ce{CH4}$, but then $\ce{O}$ is left out and reacts with the remaining two hydrogen and gives $\ce{H2O}$.

Please post your second question as a new question. Otherwise it will not get any attention and it is not useful for the community.


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Stability of cyclobutyl methyl carbocation
I know that cyclopropyl methyl carbocation is exceptionally stable compared to say, benzyl carbocation.
But I want to know how stable is cyclobutyl methyl carbocation compared to say, tertiary/secondary carbocations? Or, will it go rearrangement breaking the ring?

DOI: 10.1021/ja01473a028
DOI: 10.1002/anie.196706661 http://thesis.library.caltech.edu/656/1/Silver_ms_1959.pdf
http://thesis.library.caltech.edu/4757/1/Cox_ef_1955.pdf The original literature arises from these two theses.

Did you mean a 3-methyl cyclopropene carbocation? I believe that's the aromatic compound that is exceptionally stable (read: still not that stable) - a regular methyl cyclopropane would be very strained, and with a charge I think it would be even less so.

@qwersjc No I am not talking about that.

George Olah and coworkers have examined the cyclobutyldicyclopropylmethyl and the cyclobutylmethyl cation.
(Yes, this is an awfully short answer and I can't really provide more information on the topic.)

Doesn't that require to buy the content?

@Awesome Yes, unless you have access via a university or research facility.

I don't have access. Could you just tell whether it will go rearrangement breaking ring or not?


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Aldol of 7-oxooctanal
I want to do intramolecular aldol condensation of 7-oxooctanal:

I am talking about the major product. Clearly, a 6-membered ring would be a major product, but there are 2 possibilities: taking hydrogen from $\ce{C^2}$ or $\ce{C^6}$. Which one would be preferred?

The more stable enol would involve $\ce{C^6}$, with the aldol condensation you would get a six member ring with $\ce{-COCH3}$ attached to the ring.
With the enol involving $\ce{C^2}$, the product would have a methyl and $\ce{-CHO}$ attached to the ring.
I don't know the distribution between the two products; it could be 51:49 or as much as 99:1. You should do the reaction to determine this. Factors that may effect the product ratio: temperature, solvent, reaction time, concentration, etc.

How do you say that first enol is more stable?

The C8 enol would be less stable than the C6 enol. With the hydrogen on C1, I would think the C2 enol would be less stable than the C6 enol. But like I said, I don't know what the distribution actually is.

If in doubt, draw:

Deprotonation at C-6 under thermodynamic control results in the more stable enolate (1c) and leaves the aldehyde (= the centre with the higher carbonyl activity) intact for the aldol reaction.
As already outlined by Uncle Al, the product is likely to undergo elimination to furnish acetylcyclohexene.
However, if that is your target molecule, addition of acetylene to cyclohexanone, followed by an acid-catalyzed rearrangement of the propargylic alcohol would be an alternative route.

Curious: Why did you not consider the other two dehydration steps for the first and second path?

We expect the aldehyde to be more reactive toward nucleophilic attack for its smaller steric hindrance (kinetics).  The product ratio may vary with time, kinetics vs. thermodynamics at equilibrium.  Attack followed by elimination helps lock in the product.

Am I correct in thinking that major product would be that of 6 membered ring not attack of $C_8$?

I expect cyclohexene with a conjugated methyl ketone hanging off.  NMR product.  Methyl ketone versus aldehyde is obvious for the methyl group around 2 versus an aldehyde proton around 9.5 or lower (conjugation).  I vote for the kinetic product. The tetrasubstituted endocyclic olefin is probably the thermodynamic product.


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Discrepancy in current inside an electrolytic cell
Suppose A(+)B(-) type fused salt is present in an electrolytic cell (not electrochemical cell). Let n electrons be entering the cell per second. Since each A(+) requires one electron to discharge, so n electrons will discharge n A(+) per second. Similarly, n B(-) are also discharged per second because the number of electrons entering and leaving the cell must be same. Now, each A(+) has unit positive charge and hence n A(+) means ne(+) charge flowing towards cathode. Also, n B(-) means ne(-) charge flowing towards anode or we can say ne(+) charge towards cathode. Does that mean total 2ne(+) charge is flowing towards cathode per second ? However, from above only ne(+) charge is entering the cell.

Are you confusing the flow of current with the flow of charge?  They are different entities.

I think current is charge flowing per unit time.

Electrons are negative.  Current and charge flow in opposite directions.  Ben Franklin could have done a better job.


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Which branches of chemistry and in what order to learn (self study)?
Hi I have decided to give chemistry another shot (it didn't take so well way back in high school). My main interest is for reading up on brain chemistry and pharmacology.
I currently have a textbook on inorganic chemistry [Introduction to Organic Chemistry 3rd edition: William Brown.Thomas Poon] that I picked up and am working my way through - well I have made it through chapter 1 at least [Covalent Bonding and Shapes of Molecules]. And am also making use of The UC Davis (Organic) ChemWiki.
So after finishing the chapter today I decided to take a break and play around with an experiment I sore on youtube to make copper nanoparticles using copper sulphate and ascorbic acid. In the video he was pretty lenient with the measurements [1g sulphate, 5g ascorbic, 200ml water] and I wanted to scale the reaction up a bit so I had a go at trying to figure what the reaction was so I could get the true ratio... Obviously that duck wouldn't fly (I am only just starting on chapter 2 after all) so I just went ahead and multiplied his recipe by a factor of 5.
So where I was going with that last paragraph is that even after I finish reading the organic chem textbook I'm not seeing how that's going to help me with something like $CuSO_4.5H_2O$ since i'ts got nothing to do with carbon. Should I put the organic book to the side and start of with something like the (Inorganic) ChemWiki? Is it necessary to go through both of them or would it be enough to learn one subject and fill the gaps as there encountered [this exercise is a personal mission - I don't envision any possible future where I become a full fledged working chemist]? Better yet are there any broader texts that teaches a unified chemistry that doesn't make artificial distinctions? Or should I go ahead and invent my own syllabus?
Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.

Organic chemistry is a second your course, so it assumes you have some knowledge of chemistry. I would suggest you find a senior high school, or first year collage course to start with.

"brain chemistry and pharmacology"  Organic chemiisry, biochemistry; inorganic for foundation - pH, equilibrium, reduction and oxidation.  You'll need some physical chemistry, classical thermodynamics, to appreciate what should happen (thermodynamics) and what does happen (kinetics).  Diamond (the denser phase) is unstable versus graphite at room pressure, and calculably so,

Diamond is stable to 1000 C in air, 1500 C under hydrogen. CVD diamond is made in vacuum.  Know where Official Truth has footnotes.  You must be conversant with species, ions, molecules, reactions before you touch organic (which is all that plus wickedly structural).
https://www.khanacademy.org/science/chemistry
   Look at the Khan Academy.  Begin with inorganic chemistry.
https://www.khanacademy.org/science/organic-chemistry
  And then organic.  Use LCAO to understand geometry, structure, hybridization, reactions.  LCAO is wholly ad hock and, point of fact, is wrong.  It sill works - splendidly.  Molecular orbitals are irrelevant until Woodward-Hoffmann.
Hint:  The folks who drool advocacy the loudest are clueless about the science, thereby not willfully lying.  Government is corrupt, ignorant, and mean.  Industry is all that plus desperate.
If you want to play with organic, buy HyperChem Lite.  Diddle the molecule together, add hydrogens (drop down menu), assign atom types (same menu), then minimize to at least ten times the recommended iterations plus much smaller acceptable error.  Learn how to see into stereograms (HyperChem genertes them, Display/Rendering).  2D representations get sloppy real fast,

Organic chemistry is a second your course, so it assumes you have some knowledge of chemistry. I would suggest you find a senior high school, or first year collage course to start with.

Your answer is a little convoluted.

I forgot all about khanacadamy! good tip - ill just have to fast forward through some of the sections since the sites style can sort of drag on sometimes (have used it in the past for mathmatical stuff).  Having the molecular CAD in front of me will help also thanks for the tips [no you didn't scare me off ;) ]

To do a job you need good tools. To supervise a job you need good people who need good tools. Somebody must be competent. Competence is knowing and successfully doing. Add efficiency (engineering) and you have beauty. Beauty needs no excuses.

I agree with LDC3 that inorganic chemistry should definitely be your starting point; it builds foundations that apply to all of chemistry, which is why they teach it first; organic chemistry is just a branch, albeit a very large one, of chemistry that can also be very interesting once you know the fundamentals. Unfortunately using conventional textbooks etc, you will most likely be constricted to this order of learning, because of the level of these courses in the education system, where inorganic chemistry is very traditionally taught first, with organic chemistry only afterwards.

Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.

I don't know if there are any shortcuts that you are seeking, Wikipedia can be pretty advanced because it is (usually) written by experts who know exactly what they're talking about; and especially on those high-level concepts, there are a lot of "things" (for lack of a better word) that you have to already know to understand fully what the big picture is. Trying to pick and choose would be maybe even more time-consuming than learning things in a progressive order.
Even though this sounds pessimistic, learning chemistry can be rewarding if you work at it, not to mention you have this site and tons of other online resources to help you out along the way.
-Best of luck should you choose to continue, and I hope you do!

So the linear route it is then - thanks for your feedback and yes I will be continuing - expect more questions on the site once i have crawled through the basics :)

To ask further: I similarly want to read Wikipedia's pharma pages, having taken chem. I read through a little bit of OChem. Between that and pharma, I'm guessing there is biochem? What other books should i be looking at?

I tell a true story about myself. In my senior year of college, I took as an elective, a course in Pre-engineering Physics. At that point in my studies, I had already taken a lot of math (Advanced Calculus, Linear Algebra, Statistics, Theory of Differential Equations,..., and even graduate-level credit in Numerical Analysis, Numerical Methods,...).
To my surprise, much of the latter was favorable to my mastering of the course material in physics.
Math has long been a building block for many sciences.
In the case of chemistry, such a math background is not immediately obvious. But, if you ever get to the more advanced research, you may feel remiss for not having taken a course in many of the above, or even topology.

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text:
Boiling point of a liquid increasing with temperature?
In my textbook, a statement is given, as follows:

Boiling point of a liquid increases with increase in temperature

However, I was wondering, isn't this wrong? The boiling point of a liquid always remains the same, it doesn't change.
Hence, my question is:
Is the above statement wrong? If it is, then under what conditions does the boiling point of a liquid change?

Could you add a source for this statement, just to have a reference if somebody else stumbles across this.

@Martin-マーチン Hey Martin! Sorry but the post is too old, and I have completely forgotten which book it was :(

It sounds that they actually meant:

Boiling point of a liquid increases with increase in temperature pressure.

Thanks for clarifying :) One thing more, can you please point me to an explanation (or give an explanation) of why does this happen ?

@GaurangTandon Think about it like this: if you have liquid under atmospheric pressure, it takes quite a bit of energy to get that liquid to excite and turn to gas. However, if it was 2x atmospheric pressure, it would take a lot more energy to excite the molecules in the liquid because there are more molecules pushing down on the liquid.

I agree with @Klaus Warzecha's answer but I would like to add a little more to it.
They probably mean to say

Boiling point of a liquid increases with increase in pressure.

Since a liquid vaporises when molecules moves faster and farther apart but while increasing pressure, you are applying an external force o hold molecules closer to each other. Hence, molecules will require more energy to overcome this resistance thus this increases boiling point of liquid..!!
or

Boiling point of a liquid may also increase with decrease in volume of container.

I think there is no need to explain this further.

Given a pot of boiling water at one atmosphere pressure: Turn up the temperature of the burner.   More heat (energy) is transferred to the pot down the temperature gradient as the water continues to boil at 100 C.  Latent enthalpy of phase transition of one gram of water at 100 C to one gram of steam at 100 C is 540 cal/g.  Faster heat input means faster rate of boiling.
Note that a 100 C boiling point only obtains at the surface.  The bottom of the pot has hydrostatic head and therefore a higher boiling point.  Sufficiently high temp at the bottom will get that boiling, too, resulting in faster heat transfer than mere convection.  Too high a temp will obtain the Leidenfrost effect, suddenly dropping heat transfer to near zero.  This is a tremendous problem in heat exchangers.
https://engineering.purdue.edu/BTPFL/BTPFL%20Publications/81.pdf
  Leidenfrost temperature versus liquid, surface, and surface composition.
The world in non-linear.  Do not believe engineers who think a line behaves as it travels to the right.  Oil storage tank fires fought with water accumulate heating water at the tank bottom, under the hydrostatic head of the tank.  The water superheats util it pushes the oil above it away, then it flashes to steam, expanding a thousand-fold, removing more hydrostatic head...resulting in a BLEVE.  The expanding cloud of oil mist and steam, if ignited, is a fuel-air explosion.  People get infrared thermal burns hundreds of meters distant.


text:
Which temperature is meant in the Gibbs free energy equation?
In the equation for Gibbs free energy change, $\Delta G = \Delta H - T\Delta S$, does $T$ refer to the temperature of the system or the surroundings?
I know we have to calculate Gibbs free energy of the system but the criterion for spontaneity says $\Delta S_\mathrm{total}$ should be greater than zero. When we relate it to Gibbs free energy to show that Gibbs free energy change should be always negative, we keep both system and surrounding temperature same with pressure also constant. Then how can the process occur? Wikipedia says that it is chemical potential that undergoes changes there but what about Gibbs free energy change?
From Wikipedia:

Thus, Gibbs free energy is most useful for thermochemical processes at constant temperature and pressure: both isothermal and isobaric. Such processes don't move on a P—V diagram, such as phase change of a pure substance, which takes place at the saturation pressure and temperature. Chemical reactions, however, do undergo changes in chemical potential, which is a state function. Thus, thermodynamic processes are not confined to the two dimensional P—V diagram. There is a third dimension for n, the quantity of gas

Actually, it is $\Delta G$ that needs to be negative for a spontaneous reaction.

If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment.
You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this must occur equally in both directions, otherwise a temperature gradient would settle between system and environment. We could describe thermal equilibrium as the result of "a zero balance of rate of heat transfer between system and environment". That's what goes on during a phase transition (e.g. solid to liquid, where entropy of the system increases due to heat given by the environment, whereas temperature stays constant).
Now, applying the second law of thermodynamics to an ensemble system + environment, undergoing an infinitesimal transformation, we write:
$$\mathrm dS_\text{syst} + \mathrm dS_\text{env} \geq 0 \tag1$$
As said above, the entropy of the environment changes if there's a  reversible heat exchange with the system:
$$\mathrm dS_\text{env} = \frac{\delta Q_{env}}{T}=-\frac{\delta Q_{syst}}{T} \tag2$$
Merging eq. $\text{(1)}$ and $\text{(2)}$, you write:
$$\mathrm dS_\text{syst} - \frac{\delta Q_{syst}}{T} \geq 0 \tag3$$
Now the first law allows you to write:
$$\delta Q - P\mathrm dV_\text{syst}=\mathrm dE_\text{syst} \tag4$$
and then
$$T\mathrm dS_\text{syst} - \left(\mathrm dE_\text{syst}+P\mathrm dV_\text{syst}\right) \geq 0  \tag5$$
Now (and this is crucial), assuming both pressure and temperature constant, you end up with
$$\mathrm d{\left(E_\text{syst}+PV_\text{syst}-TS_\text{syst}\right)_{T,P}} \leq 0 \tag6$$
The expression within the brackets is a state function of the only system and corresponds to the free energy. Quoting Wikipedia:

A state function describes the equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state.

Considering that $E+PV=H$, we write:
$$\left(\mathrm dG_\text{syst}\right)_{T,P} = \left(\mathrm dH_\text{syst}-T\mathrm dS_\text{syst}\right)_{T,P} \leq 0 \tag7$$
Now, while equation $\text{(1)}$ is much more general, equation $\text{(7)}$ applies only to the situation where the work is expansion or contraction and for constant temperature and pression. Nonetheless, since for most situations of chemical interest, those conditions are satisfied, eq $\text{(7)}$ maintains a well deserved relevance and is very convenient to establish whether a process is spontaneous or not, within the limitation of constant temperature and pressure.

A chemical reaction occurs down to the point where $\left(\Delta G\right)_{T, P}=0$.
For a general reaction:
$$\ce{$a$A + $b$B <=> $l$L + $m$M}$$
For constant temperature and pressure:
$$\left(\Delta G\right)_{T, P}=l\mu_{\ce L} + m\mu_{\ce M} - a\mu_{\ce A} - b\mu_{\ce B}  \tag8$$
being $\mu$ the chemical potential for each of the species involved.
The chemical potential relates to activity, according to:
$$\mu=\mu^\circ+RT\ln a \tag9$$
Equation $\text{(8)}$ and $\text{(9)}$ shows that free energy varies with the composition (expressed as activity) of our system: that's exactly what happens during a chemical reaction.
Proceeding further, you will end up with Van't Hoff isotherm:
$$\left(\Delta G\right)_{T, P}=\left(\Delta G^\circ\right)_{T}+RT\ln\frac{a_{\ce L}^{l}a_{\ce M}^{m}}{a_{\ce A}^{a}a_{\ce B}^{b}}$$
Activities vary until a chemical equilibrium is established, so that $\left(\Delta G\right)_{T, P}=0$.

I already know everything that you wrote .My weakness is that I am strong in explaining my question and answer while speaking and not when writing because question here is much theoretical and I am not from English Speaking Country.Well My Question is actually regarding meaning of Paragraph from wikipedia in second edit.Please Look EDIT2

Above wikipedia passage tells that system and surrounding temperature are same and not isothermal for each of system and surrounding.

I can't just respond to you if you respond to me.Please consider me if made response after 12 hours .I am in GMT+5:30 zone.

@VishvajeetPatil edited, hope that helps

@mannaia Why system and surroundings must be in thermal equilibrium when a reaction takes place? Is it because in such a way we can maintain a constant temperature for the system? Also when we say chemical, mechanical or thermal equilibrium should we say with respect to surroundings or not? Is equilibrium a property of the system or a property of both system and surroundings?

In the equation for Gibbs free energy change, ΔG=ΔH−TΔS, does T refer to the temperature of the system or the surroundings?

The temperature $T$ of the Gibbs energy of a system is the temperature of the system.  There is a related thermodynamic state function called the "exergy" or "availability" and sometimes given the symbol $B$. This function, unlike the Gibbs energy, is defined relative to the temperature of the surroundings.
$dB = dH - T_0 dS$
The Gibbs energy change for a process represents how much useful work could have been extracted from that process, if it were run at constant temperature.
The "exergy" change for a process represents how much useful work could have been extracted from that process, if after running it at some constant system temperature, a series of reversible heat engines was used to cool the system down to the temperature of the surroundings.

It depends on what you want to calculate. For the Standard Free Energy, $\Delta G°$, the temperature is 25°C. If you want to calculate the Free Energy at another temperature, $\Delta G$, it is usually provided.
https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm

You aren't clear

In essence, you choose the temperature you want to do the calculation at.

Please check Edit

I would if you would point out what you think is wrong. I'm not a mind reader.


text:
How to rationalise that when we dilute a weak acid, the pH increases but also the dissociation increases?
When water is added to a weak acid like ethanoic acid, the number of ethanoic acid molecules that dissociate increases, however, pH increases (less acidic), too. Why is this so?

did you have a look here ?http://chemistry.stackexchange.com/questions/7648/does-the-number-of-h-ions-in-solution-go-up-on-dilution-of-a-weak-acid

Dilution increases faster than dissociation outputs excess protons in kind.  "8^>)

It's simply due to dilution.
If it's true that adding water increases dissociation (Gerard has done a nice effort in Does the number of H+ ions in solution go up on dilution of a weak acid?), concentration of $\ce{H3O+}$ inevitably decreases. As a consequence, $\mathrm{pH}$, defined as $-\log[\ce{H3O+}]$, increases.
Start with $\pu{1 L}$ acetic acid concentration $\pu{1 M}$,
$$\ce{[H3O+]} = \sqrt{K_c} = \pu{4.22E-3M},$$
that corresponds to $\pu{4.22E-3mol}$ of $\ce{H3O+}$.
If you double the volume ($\pu{2 L}$),
$$\ce{[H3O+]}=\sqrt{K_c/2}=\pu{2.98E-3M},$$
that corresponds to $\pu{5.96E-3mol}$ of $\ce{H3O+}$.
The amount of substance is higher, but concentration is less: dilution does play a role here.

Let's first examine the chemical reaction between our acids and bases. Ethanoic acid, commonly referred to as acetic acid reacts with our base, water. Water is amphiprotic, in that it can act as both a base and acid.
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+} $$
However, the acetate ion undergoes hydrolysis with water which now instead acts as an acid, reforming acetic acid as well has hydroxide ions:
$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$
Acetic acid is weak acid, however, the conjugate base, acetate, is a strong base/salt. Water acts as a strong base, however the hydronium ion is a weak acid, so we can except the pH to increase.
What many people often forget to realize is that once an acid/base reaction occurs, a salt is created along with water. However, a hydrolysis reaction between the salt and water can occur.
Taken from Chemical Principles: The Quest for Insight by Atkins et. al on page 494 in the tenth edition:

Calcium acetate, $\ce{Ca(CH3CO2)2 (aq)}$ is used in medicine to treat patients with a kidney disease that results in high levels of phosphate ions in the blood. The calcium binds with the phosphates so that they can be excreted. If you are using calcium acetate for this purpose, it is important to know the pH of the solution to avoid complications in treatment.

Say we have a $\pu{0.15 M}$ solution of $\ce{Ca(CH3CO2)2 (aq)}$.
Since the conjugate base of acetic acid is strong, we can expect the pH to increase.
Our expression can be written as this:
$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$
Calcium is a spectator ion and thus can be ignored.
$$K_b = \ce{\frac{[CH3COOH][OH-]}{[CH3COO-]}}$$
Constructing our RICE table:
$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$
\begin{array} {|c|c|c|c|c|}
\hline
&\ce{CH3COO^-} & \ce{H2O} & \ce{CH3COOH} & \ce{OH-}\\\hline
\text{Initial conc.} & 0.30 & - & 0 & 0 \\
\hline
\text{Change conc.} & -x & - & +x & +x\\
\hline
\text{End conc.} & 0.30 - x & - & x & x \\
\hline
\end{array}
\begin{align}
K_b &= 5.6\times10^{-10} = \frac{[x][x]}{[0.30-x]}\\
x &= 1.3 \times 10^{-5}\\
\text{pOH} &= -\log([\ce{OH-}])\\
\text{pH} &= 14 - \text{pOH} = 14 - 4.89 = 9.11
\end{align}

On dilution, ionisation increases, although pH increases due to the fact that, volume increases dominantly over dissociation. Hence concentration of $\ce{H+}$ ions decreases instead of the extent of ionisation increases.

Welcome to Chemistry.SE! Your answer describes something already mentioned in the other answers to this post. If you could edit to elaborate a bit further, that would make it stand out among the rest. Please spare some time to go through the tour and the help center. Here'a a (+1) in advance!

I think you are confusing the dissociation rate with acidity and pH. Lets take a weak acid for example, CH3COOH. So the dissociation rate of that acid is very little, about 4% max if my memory serves. However, it can only dissociate in water, so the more diluted the acid is, the more molecules of the acid is dissociated. However, this does not mean that the dissociation rate changes. Now lets take a strong acid, HCl. Clearly hydrochloric acid would be a much stronger acid that ethanoic acid. It therefore has a high dissociation rate, with it fully dissociating in water. So the more water is added, the more ions it produces, but it also gets more diluted, so the pH increases. Im not sure if you see my point here, but basically, (percentage of dissociation is dilution. The rate of dissociation is the measure of how strong an acid is.) Percentage of dissociation simply means that it is more diluted, and therefore it has a lower pH. Man im bad at explaining things sorry my bad. In short, my main point is the bracketed part.

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text:
Deriving goldman equation
So I've been trying to follow the derivation of the Goldman-Hodgkin-Katz equation on Wikipedia (https://en.wikipedia.org/w/index.php?title=Goldman_equation ) - I can follow it through after it sets up the initial model.
However, I'm having trouble using the Einstein-Stokes relationship to derive the initial model in the first place.
There is the flux from diffusion down the chemical gradient, $-D_{A}\frac{d[A]}{dx}$. The negative sign is there because I'm imagining the ions being "lost" from the side I'm considering.
Then we have the electrical field pushing ions the other way. This won't depend upon the rate of change of concentration because the electric field is uniform and so there's effectively a constant force on each of the particles. The flow will instead depend upon the constant velocity that these particles will be pushed at. So the flow will be $[A]v_d$ - the moles per unit volume being multiplied by the speed in the direction towards the membrane will give me the number of moles per unit area passing through the membrane per second.
So all I need to do is work out $v_d$ I'm guessing. So I use the electrical mobility equation, $D_A=\frac{\mu_{q}k_BT}{q}$.
$q$ will be the charge, so $nzF$ (where $z$ is the valance, $nz$ gives the moles of electrons and multiplying by $F$ gives the charge).
$\mu_q$ will be $\frac{v_d}{E} = \frac{v_d L}{V_m}$
$k_B = \frac{R}{N_A}$ where $N_A$ is Avogadro's number.
So $D_A=v_d \frac{L}{V_m} \frac{RT}{zF} \frac{1}{N_A n}$, thus $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n=v_d$
Putting this back in to get the ion flux, $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n [A]$.
Adding my two fluxes together to get the total flux, I get: $j_a = -D_A(\frac{d[A]}{dx} - \frac{V_m}{L} \frac{zF}{RT} N_A n [A] )$ - this is not quite what the Wikipedia article has - a stray $n N_A$ has crept in, somehow!
Please help me by pointing out where I'm making a mistake!

The mistake is in the charge $q$, that you have put in:
$
D_A=\frac{\mu_{q}k_BT}{q}
$
$q$ is simply the particle charge (and not the whole charge, as you've unfortunately considered) and should then be expressed as:
$q=\frac{zF}{N_{A}}$
with $F$ the Faraday constant (charge / mole ), ${N_{A}}$ the Avogadro's number (number of particle / mole) and $z$ the charge valence.
${N_{A}}$ will finally cancel out when you divide Boltzmann's constant $k_B$ by the the charge $q$, in the definition of diffusion coefficient. That will allow you to recover the Goldman equation found on Wikipedia.
By checking all that through an always useful dimension analysis for $D_A$ and using the defintions you have given in your question for ($\mu_q$ = $\frac{v_d}{E} = \frac{v_d L}{V_m}$ and $k_B = \frac{R}{N_A}$):
$
\frac{\mu_{q}k_BT}{q}=\frac{m/s\,\times m}{V}\times \frac{J/(mol\times K)\,\times K}{{N_A}\times C/mol}\times{N_A}=\frac{m^2}{s}
$
you find the correct units for a diffusion coefficient.


text:
What is this wave function in Chemical bonding?
Was studying chemical bonding(Valence bond theory) but there came this topic called wave equation but  have not studied quantum mechanics,so finding hard to understand these terms.Please help!!!

$\Psi$ is the wavefunction describing an electron's motion.  $\Psi^2$ is the probability density function for the electron. $\Psi^2$ is usually interpreted as the probability of finding the electron at a given set of coordinates.
In the case that $\Psi$ is a complex function (in that it contains imaginary numbers and not that it is complicated, which it usually is), $\Psi^2$ is defined as the product of $\Psi$ with its complex conjugate: $\Psi \Psi^*=|\Psi|^2$ so that the probability density function contains only real numbers.
Consider a simple case of $\Psi=a+bi$. The complex conjugate would be $\Psi^*=a-bi$.
$$\Psi \Psi^* = (a+bi)(a-bi)=a^2+b^2$$
Often the probability density function is normalized so that its integral over all space is 1 (i.e. there is 100% chance of finding the electron in all of space). In the equation below, $A$ is the normalization constant.
$$1=A \int_o^{2\pi} \int_0^{2 \pi} \int_o^\infty \Psi \Psi^* dr d\theta \phi$$

The probability density is given as $\vert\Psi\vert^2$, i.e. the modulus squared.

@KlausWarzecha Surely you meant absolute squared, not modulus?

@tschoppi Taking into account that $\Psi$ may be complex, I thought that modulus is the correct term. But I might be wrong ;)

@KlausWarzecha I've confused modulus with modulo. Learned something again, what a great day today is :)

A theoretical chemist wants to mathematically describe electron density, energy, and directionality in space from first principles, ab intio, pure math.  One can completely ignore atoms and do it with orthogonal polynomials and spherical harmonics.  This proves to be unworkable for needing unending curve fitting (like epicycles for solar system orbits before Copernicus; or SUSY, or dark matter) for things like hybridization, molecular orbitals, and solid state band structure.
One instead derives the math specifically from atomic, molecular, and sold state systems beginning with the Shroedinger equation via Hamiltonian or Lagrangian approaches, as quantum mechanics and its wavefunctions, thus wave equations.  At atomic and molecular scales, electrons being low mass and low energy, the de Broglie equation says there are no little pellets.
If you are doing organic, you survive and prosper by returning to the 1930s with LCAO (linear combination of atomic orbitals - purely descriptive), plus the Woodward-Hoffmann rules as appropriate. LCAO is an empirical black box.  In goes the problem, crank it, out comes the correct answer.  If you are doing pharma, crude LCAO lock-and-key rapidly gives way to detailed calculations for the very many subtle ways electron densities interact.  Pharma requires fit plus selectivity.


text:
Predict the product (organic chemistry)
I think that the first reaction would add into the ketone carbon, but I don't know where to go after that.
A step-by-step explanation of the answer would be greatly appreciated!

Answer choices:

If the ketone is attacked, what is the intermediate result?

The second carbon would have an -OH group and an -OCH2CH3 group with everything else the same

Not quite, the oxygen would be negatively charged, balanced by the sodium (since we haven't added water yet). If you look at the possible results, this doesn't look good as the possible pathway. So what else is possible?

Would it involve something with the ester carbon?

The sodium ethanoate and ethanol would be suitable for a transesterfication (if there wasn't already an ethanol on the ester). So, when the benzyl chloride is added, the ethanol is displaced. Hold on a moment, I need to check something out. I thought I knew what was going to happen.

Actually, you are correct in that the ketone is attacked and the ethanol will be displaced by benzyl chloride to give a tertiary alcohol.

How does the benzyl chloride react? I thought it would be some sort of SN2 reaction, but this route doesn't seem consistent with the answer choices.

The rest of the workup removes a methyl group and reduces the alcohol, so the product is ???

I'm not certain what the process is, but when I looked for the reagents, it suggested that this (http://www.sigmaaldrich.com/catalog/product/aldrich/w242306?lang=en&region=US) is a product.

I'm not sure I understand...

Look at http://www.chtf.stuba.sk/~szolcsanyi/education/files/Organicka%20chemia%20II/Prednaska%205_Aldehydy%20a%20ketony_Reakcie%20II/Doplnkove%20studijne%20materialy/Alkylation%20of%20enolates_Clayden.pdf. When the sodium ethanoate is added, the oxygen get a negative charge and a double bond is formed (with the inner carbon). CC(O-)=CC(=O)OCC

On the second page it states "A strong base can be chosen to deprotonate the starting material completely. There is complete
conversion of the starting material to the anion before addition of the electrophile, which is added
in a subsequent step" which is what we have.

Do you have access to a chemical drawing program? I can point you to one on the web. If you enter this SMILES, you will see the intermediate: CC(=O)C(C(=O)OCC)CC1=CC=CC=C1

Yes, I have chemdraw

Can I get the web reference from you? I can't figure out how to convert it in chemdraw without additional Excel downloads

Unfortunately, I had a problem with my computer and I don't have Chemdraw loaded so give me a moment and I'll load it.

We will need to use an online program since I didn't get the download yet. Goto https://www.sigmaaldrich.com/catalog/search/substructure/SubstructureSearchPage  You will probably see a message from Java. If you get an error, click on Start, All Programs, Java, Configure Java: select the Security tab, Click Edit Site List, Add and enter https:/ /www.sigmaaldrich.com (without the space), click OK twice and reload the web page.

Okay, that works! So the first step would form an enolate ion that is resonance stabilized with the double bond of the ester. The double bond then attacks the benzyl chloride in an SN2 fashion. Next, does the base react with the ester?

Well, the final product is either C or E. I thought it might be E, but now I'm confused. Either the ethanol is removed to give a carboxylic acid, then CO2 is removed to give C, or CH3CO- (I don't know how) is removed to give E. So I am betting on C.

Folks, if you are going to discuss the possible answers, please make a chatroom instead of drawing out the comments section.  Thanks!

Ah okay, I think I understand that now. Thank you very much, LDC3! (Sorry, jonsca, will keep that in mind for next time.)

Looks like a classic to me. But since this is homework, I will not provide a full reaction scheme and only give some hints.

Your starting material is a $\beta$-ketoester (1,3-dicarbonyl compound).

Where is the most CH-acidic position in this molecule?
Benzylchloride is an alkylating agent. What will happen when it reacts with the anion of your starting material?
For extra credit: Why is sodium ethoxide in ethanol a reasonable choice when you want to isolate, purify and characterize the product?

Your starting material did have an ester group. Does it still have it after the first step? If so, what will happen to it in aqueous alkaline solution?
(This step would not work with the saponification product of a simple ester.)
But you started with $\beta$-ketoester! Heating the product of step 2 in acidic media will let it loose some atoms. The mechanism may be drawn using a cyclic intermediate, in which the ketone has an assisting role.

EDIT

Deprotonation and subsequent benzylation of the $\beta$-ketoester occur at the most CH-acidic position, i.e beween the keto and the ester group. As a side note, benzylation at the methyl group adjacent to the carbonyl group is possible when using 2 equivalents of a strong base.
Using ethoxide rather than methoxide as a base makes sure that transesterification does not lead to a mixture of products.
The resulting benzylated $\beta$-ketoester is cleaved under alkaline conditions to the corresponding carboxylate. Under acidic conditions, this from a $\beta$-ketoacid.
Upon heating, $\beta$-ketoacids undergo decarboxylation (= they loose $\ce{CO2}$) to form an enolate which tautomerizes to the final methyl ketone.


text:
Ratio of conjugate bases of a tricarboxylic acid

A tricarboxylic acid $\ce{H3A}$ has three different types of $\text{p}K_\text{a}$ $(\text{p}K_\text{a, 1} =2.0,$ $\text{p}K_\text{a, 2}=5.0,$ $\text{p}K_\text{a, 3}=8.0)$.

Identify the ratio of the three charged components($\ce{H2A-}$,$\ce{HA^{2-}}$ and $\ce{A^{3-}}$) in $0.2 ~\text{M}$ solution of tricarboxylic acid $\ce{H3A}$.

$\pu{0.55 M}$ of the strong base $\ce{NaOH}$ is added into a solution of $\pu{0.20 M}$ tricarboxylic acid $(\ce{H3A})$. Calculate the final pH. (Assume that the addition of $\ce{NaOH}$ will not influence the volume of the solution.)

I managed to get $\pu{0.04 M/L}$ for the $\ce{H2A-}$ species but I do not know how to find the concentration of the acids.

For question 1, what is the pH, or are we to assume that it was added to water and we are looking for the equilibrium when it dissolves?

Let's use citric acid as an example. Although the $pK_a$ values of citric acid might not necessarily be the same as those above, it will be useful for demonstration.
$$\ce{C6H8O7 + H2O <=> C6H7O7- + H3O+ <=> C6H6O7^2- + H3O+ <=> C6H5O7^3- + H3O+}$$
Throughout the polyprotic acid titration, two protons are donated. The way this problem is solved is you have to use successive $\mathrm pK_\mathrm a$ values and concentrations.
Primary proton transfer equilibrium:
$$\ce{C6H8O7 + H2O <=> C6H7O7- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} & 0.20 & - & 0 & 0 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final}& 0.20 - x & - & x & x \\
\hline
\end{array}
$$K_\mathrm {a_1} = 0.01 = \frac{[\ce{C6H7O7-}][\ce{H3O+}]}{[\ce{C6H8O7}]} $$
$$ K_\mathrm{a1} = 0.01 = \frac{x^2}{0.20-x} $$
Unfortunately, we can't disregard the x in the denominator so we have to solve the quadratic:
$$ x = 0.04 \ \mathrm{or} \ x = -0.05$$
We disregard our negative root. From $[\ce{C6H7O7}]^- = [\ce{H3O+}]$:
$$[\ce{C6H7O7}-] = \pu{0.04 mol L-1}$$
Now we use our second $\mathrm pK_\mathrm a$ to find the concentration of $\ce{C6H6O7^{2-}}$. Since $K_\mathrm {a_2} << K_\mathrm {a_1}$, we can assume that the hydronium ion concentration is unchanged by the second deportonation.
$$\ce{C6H7O7- + H2O <=> C6H6O7^2- + H3O+} $$
\begin{array} {|c|c|c|c|c|}
\hline
C_\mathrm{initial} & 0.04 & - & 0 & 0.04 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final} & 0.04 - x & - & x & 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_2} = .00001 = \frac{x(0.04 + x)}{0.04-x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 9.995 \times 10^{-6} $$
Ignoring the negative root:
$$ [\ce{C6H6O7^2-}] = \pu{9.995 \times 10^{-6} mol L-1}$$
Finally, assuming that the concentration of hydronium ions and the conjugate base of citric acid after the 2nd deprotonation is unaffected by additional deprotonation, we can write up our equilibrium table again:
\begin{array} {|c|c|c|c|c|}
\hline\
C_\mathrm{initial}& 9.995*10^{-6} & - & 0 & 0.04 \\
\hline
\Delta C & -x & - & +x & +x\\
\hline
C_\mathrm{final} & 9.995*10^{-6} - x & - & x & 0.04 + x \\
\hline
\end{array}
$$ K_\mathrm {a_3} = .00000001 = \frac{x(0.04 + x)}{9.995 \cdot 10^{-6} -x} $$
$$ x = -0.04002 \ \mathrm{or} \ x = 2.49875 \times 10^{-12}$$
$$ [\ce{C6H5O7^3-}] = \pu{2.49875 \times 10^{-12} mol L-1}$$