Electrolysis with baking soda solution, a penny as the cathode and graphite for the anode I set up an electrolysis experiment to show my kids and wanted to explain to them what is happening.
My setup has a penny as the anode and graphite for the cathode. The solution is water with baking soda mixed in.
I see bubbles at each of the electrodes which I assume are hydrogen and oxygen bubbling out. I also see some cloudy substance coming off the penny, which I assume is some sort of copper compound, but I'm not sure.
Can someone help me figure out what's going on in the setup at the two electrodes?
Edit : corrected the electrode descriptions. I earlier stated penny as the cathode and graphite as the anode.

Yes, I agree you are getting hydrogen an oxygen as you expected.
However, the copper anode is undergoing anodic corrosion and forming CuO (which is characteristically black) and possibly also some Cu2O (yellow or red, depending on the size of the particles, where the color combination could be described as orange).
Now, in the presence of HCO3- and O2, a naturally occurring mixed salt, copper patina (which forms on the copper roofs of houses) could also be forming. Wikipedia on patina, to quote:

patina is a coating of various chemical compounds such as oxides, carbonates,... formed on the surface during exposure to atmospheric elements (oxygen, rain, acid rain, carbon dioxide,...

Patina has been described as a blueish-greenish naturally forming on a copper surface (here, your copper penny electrode).
A closely related version of your electrolysis experiment has been previously discussed elsewhere.


Is the refractive index of a solution of four substances the median of the four substances' refractive index? I need to determine the purity of a solution and we need to discuss about the  'theoretical' refractive index of the substance that we are supposed to have.

There is no single equation that accurately predicts the refractive index of a mixture.  This reference discusses 9 approaches to estimating the refractive index of mixtures and provides the formula for each method. The simplest method is the Arago-Biot approach.  It is reasonably reliable and is described by the following formula

$$n = n_1 \phi_1 + n_2\phi_2$$

where $n$ is the predicted refractive index of the mixture, $n_x$ is the refractive index of the $x^{th}$ component, and $\phi_x$ is the volume fraction of the $x^{th}$ component.  The volume fraction of the first component is given by

$$\phi_1 = x_1 V_1 / \Sigma~ x_i V_i$$
Here $x_i$ is the mole fraction of the $i^{th}$ component and $V_i$ is the molar volume (the molecular weight divided by the density) of the $i^{th}$ component. The more ideal the liquids, the more accurate the predictions.


Why does ethane release more energy than ethyne when burned? My book explains that since ethane contains a single bond, ethene contains a double bond, and ethyne contains a triple bond, and that when burned ethane releases the most energy because "the fact that the lowest energy is released by ethyne is taken to indicate that this compounds bonds are already in the highest energy configuration. The high energy of  combustion for ethane on the other hand indicates that its bonds have lower energy to start with and can release more energy on combustion." This seems counter intuitive to me. Wouldn't the double and triple bonds release more energy because they're stronger and it takes more energy to break them? Any insight would be appreciated!

@a-cyclohexane-molecule has it right. Ethane generates more water. Here is a diagram that helps explain the issue in more detail. The black numbers and arrows are the heats of formation of the three hydrocarbons. Ethane is the only one more stable than the elements from which it is formed. Blue numbers and arrows are the heats of formation of CO2 and H2O from graphite and hydrogen. The red arrows and numbers are the heats of combustion of the three hydrocarbons. Although acetylene and ethylene have a headstart in combustion because the are less stable than ethane, all three compounds produce two CO2's but ethane generates more water than acetylene and ethylene. Ergo, ethane generates the greatest amount of heat. All values are in kcal/mol.


How to get $\Gamma_{vib}$ for hexafluoroethane? An exercise asks me to find $\Gamma_{vib}$ for that molecule.
I know that: $$\Gamma_{cart}=\Gamma_{vib}+\Gamma_{translations}+\Gamma_{rot}$$

Since the molecule I'm considering belongs to the $D_{3d}$ group (in the staggered configuration), from the character table you get:
$$\Gamma_{translations}=A_{2u}+2E_{u}$$
$$\Gamma_{rot}=A_{2g}+2E_{g}$$
and then:
$$\Gamma_{vib}=\Gamma_{cart}-2E_{u}-2E_{g}-A_{2g}-A_{2u}$$
My question is: does exist a fast way to find out $\Gamma_{cart}$?

The only method that I know is to express each operation of the group in terms of matrices using the Cartesian basis, so I should write down 12 matrices 24x24, and then find out the corresponding reducible representation, but I'd take ages. I'm pretty sure that there must exist a faster method.

To transcribe what is given in levineds's linked answer and remove the parts only relevant to linear molecules, here are the steps you can apply to determine the reducible representation of the vibrations.

Determine how the molecule is changed by all symmetry operations. If an atom is moved by a symmetry operation, it contributes it doesn't contribute to $\Gamma_\text{atoms}$. If the atom remains in place, add +1 to the column of that symmetry operation.

For hexafluoroethane, this gives:
$$\small \begin{array}{c|cccccc} \hline
D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6  & 3\sigma_d  & \\ \hline
\Gamma_\text{atoms} & 8 & 2 & 0 & 0 & 0 & 4 &\\
\end{array}$$

Multiply each column of $\Gamma_\text{atoms}$ by the trace of that operation's matrix representation.
$$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

$$\small \begin{array}{c|cccccc} \hline
D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6  & 3\sigma_d  & \\ \hline
\Gamma_\text{cart} & 24 & 0 & 0 & 0 & 0 & 4 &\\
\end{array}$$

Use reduction formula to determine irreps. $n(i)=\frac{1}{h}\sum_R N\cdot\chi_r(R)\cdot\chi_i(R)$ where $n(i)$ is the number of the $i^{\text{th}}$ irrep, $R$ is a symmetry operation, $h$ is the order of the group, $N$ is the coefficient in front of the operation, and $\chi(R)$ is the character of the operation in the reducible/irreducible representation.

I won't go through the calculation (there are online calculators that you can use to do this), but this gives. $$\Gamma_{xyz}=3A_{1g}+1A_{2g}+4E_g+1A_{1u}+3A_{2u}+4E_u$$
which if we subtract off the translational and rotational irreps you have already obtained gives:
$$\Gamma_{vib}=3A_{1g}+3E_g+1A_{1u}+2A_{2u}+3E_u$$
(You had small error with your rot and trans, the $E$ irreps are degenerate and so count for 2 degrees of freedom. This means in each case, you would only have a single $E$ irrep.)


Orbital overlap and size "mismatch" Why is atomic size mismatch a sign of poor bond strength? Wouldn't orbital overlap be proportional whether the two atoms are small or the two atoms are big?

For example I heard someone comment that an argument that the C-I bond is weaker than the C-F bond based on orbital overlap made no sense to them, because why would the size of the individual orbitals matter in overlap? Small atom, big atom and small atom, small atom - shouldn't their overlaps be proportional? Or something along the lines of that. I'll have to pull up the post again ...

Small atom, big atom and small atom, small atom - shouldn't their
  overlaps be proportional

No.  Look at the figure below and let's consider s-orbital overlap, just to keep it simple.  Remember that these are spheres with the electron density distributed throughout the entire volume.  In the "small-small" case there is effective overlap of the electron density in both 1s orbitals; we would expect the resultant bond to be strong.  In the "small-big" case, the small 1s orbital involves a significant fraction of its electron density in overlap, but not much of the electron density in the large s orbital is involved in overlap.  In the "small-big" case the overlap of electron clouds is ineffective and any resultant bond would be weak.


Reaction of glucose with 2,4-dinitrophenylhydrazine (DNPH) In its open-chain form, glucose possesses an aldehyde group; however, glucose does not test positive with 2,4-dinitrophenylhydrazine (which typically forms a yellow/orange/red precipitate with carbonyl compounds).

Why is this so? Is this a reflection of the cyclic hemiacetal form (where the aldehyde group is not present)?

Both forms are in equilibrium so I'd expect it to give a positive test.

Consider also that glucose gives a positive test in both the Tollens' and Benedict's test. While it is true that less than one in hundred exists in the aqueous solution as the cyclic form, this is not sufficient to explain why it does not give a positive test with 2,4-dinitrophenylhydrazine but yet gives a positive test in Tollens' test and also in Benedict's test.

The test is based on imine formation. The formation of imine from carbonyl compound occurs in equilibrium ergo the principle of Le Chatelier is applicable to the test. According to this principle the increase of the concentration of at least one of the products results in the decrease of degree of conversion of the reactant.
I guess that you are trying to perform the test in aqueous solution, because glucose has poor solubility in ethanol. The second product of the conversion of carbonyl to imine is water. So, performing the reaction in the tremendous excess of water, you are in fact pushing the degree of conversion toward zero in accordance with the principle of Le Chatelier.
That's why the expected precipitate is not formed.


How to recycle Lithium-Polymer battery? I can't manage to find any information about the recycling process and the chemistry behind it. The only thing I see everywhere on the internet - is a guide like this:

Step 1 - Discharge the battery.
Step 2 - Store at room temperature.
End of the article.

I have a DELL XPS15 battery, slightly swollen. And when I asked local repair service about recycling - they said: "nobody recycling Li-Pol here" (in my city).
So I decided to investigate this topic. Any idea where I can start?

Lithium ion batteries are pretty poorly recycled as is and so when I was looking through literature reviews I found only one mention of lithium ion polymer batteries being recycled. The paper that was using this method was also recycling lithium ion batteries the same way though so it wasn't clear if anything different had to be done.
The main thing that seems to be different with lithium ion polymer batteries is just the electrolyte. The electrolyte is often not recycled anyway since the main focus is typically on recycling the cathode because it contains cobalt and/or nickel and is profitable to recycle.
If you want a more detailed overview of the different processes lithium ion batteries are recycled this review is open access and explains the ways different companies currently recycle lithium ion batteries:
https://www.mdpi.com/2313-0105/5/4/68
There's no mention of lithium polymer batteries in the review but this patent mentions recycling lithium ion and lithium polymer batteries together so I assume some of the same methods can be used:
https://patents.google.com/patent/US8979006B2/en
There's no standard method for recycling lithium ion batteries but usually methods involve mechanical, pyrometallurgical and/or hydrometallurgical processing. Usually the batteries are first discharged and then shredded or dismantled.
If mechanical processing is carried out it usually relies on different physical properties in the materials that allow it to separate out components. For example, the electrodes are a fine powder so if you put the shredded battery through a sieve you can separate the electrodes from the other materials. The cathode and anode can also be separated using froth filtration since they have different hydrophobicities. This can allow the cathode to be isolated and further processed to obtain metals like cobalt if the battery has a lithium cobalt oxide cathode. The review goes into more detail about the other mechanical steps that can be carried out but there's quite a lot of them.
Pyrometallurgical processes involve putting the battery in a furnace and you end up with an alloy of metals (containing cobalt, nickel and iron usually) and a slag (containing aluminium, lithium, manganese and some other metals). The metals in the alloy can then be extracted.
Hydrometallurgical processes involve using solvents to leach metals. This is generally combined with other methods like mechanical processing so the battery components you have are slightly more homogenous and easier to process. Sulfuric acid is the most common acid used to leach cobalt from lithium cobalt oxide. The leached metals can be precipitated out and recovered.
It's a bit difficult to clearly explain how lithium ion batteries are recycled since it does seem like every company does it differently. This just mentions some of the processes that are being used at the moment but there's lots more as well that are being researched. I know this doesn't clearly mention how lithium polymer batteries are recycled but I hope it's of some help anyway!


Solute mole fraction for hydrated magnesium sulfate dissolved in water Problem: 70 grams of pure, hydrated $\ce{MgSO_4}$ is dissolved in 90
  grams of $\ce{H_2O}$. If the mole fraction of the solute is 0.039,
  find the molecular formula of the solute.

My flawed attempt at a solution:

Let the solute be $\ce{MgSO_4.xH_2O}$. In $70~\mathrm{g}$ of solute, the amount of solute is $$\frac{70}{120+18x}~\mathrm{mol}.$$

Likewise, in $\pu{90 g}$ of solvent, we have $$\frac{90}{18} = 5~\mathrm{mol}.$$

Plugging these values into the formula for mole fraction and equating it to 0.039, I thought I would get the value of $x$ as 7. Instead, I seem to be getting approximately 12.

Could somebody please help me solve this question? In general, when we have a hydrated solute in an aqueous solution, do we include the water of crystallisation in the total mass of available water while calculating the number of moles of water?

Alright, let’s take this one step at a time.

First of all, you want to determine a molecular formula, and among your known data you have a mole fraction. This suggests it’s a good idea to work with moles to solve the problem. As such, the first step is to convert the amount of water added and magnesium sulfate hydrate used. Calculating the number of moles of water added is easy:

$$90\ \mathrm{g}\ \text{water added} \times \frac{1\ \mathrm{mol}\ \ce{H2O}}{18\ \mathrm{g} \ \ce{H2O}} = 5\ \mathrm{mol}\ \ce{H2O}$$

The number of moles of $\ce{MgSO4.$x\,$H2O}$ is unknown because even though you know how much mass was dissolved, you don’t know its formula completely. However, $\ce{MgSO4}$ contributes $120\ \mathrm{g\ mol^{-1}}$ to the molar mass, and each additional molecule of water in the hydrate stoichiometry adds a further $18\ \mathrm{g\ mol^{-1}}$, so that the molar mass of $\ce{MgSO4.$x\,$H2O}$ is $120\ \mathrm{g\ mol^{-1}} + x \times 18\ \mathrm{g\ mol^{-1}} = (120+18x)\ \mathrm{g\ mol^{-1}}$. With this molar mass, the number of moles of $\ce{MgSO4.$x\,$H2O}$ is:

$$70\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O} \times \frac{1\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}}{(120+18x)\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O}} = \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}$$

From the definition of mole fraction:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{total}}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}$$

Now, one is tempted to write the following and solve for $x$:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}\ }{\frac{70}{120+18x}+5 \ }=0.039$$

However, this equation is incorrect. The problem is that the following process was not taken into account:

$$\ce{MgSO4.$x\,$H2O(s) \xrightarrow{water} MgSO4(aq) + $x$\ H2O(l)}$$

In other words, when the magnesium sulfate hydrate dissolves in water, the water molecules contained in the solid are released and become an indistinguishable part of the solvent; the final mixture is a simply a solution of the solute $\ce{MgSO4}$ in water. Therefore, the number of moles of solvent is not simply $5\ \mathrm{mol}$ as calculated at the start, but rather a larger value, which should include the number of moles of water released when the solid dissolves.

From the dissolution equation, we see that $1\ \mathrm{mol}$ of $\ce{MgSO4.$x\,$H2O}$ generates $1\ \mathrm{mol}$ of $\ce{MgSO4}$ and $x\ \mathrm{mol}$ of $\ce{H2O}$ from crystallized water. However, we know the amount of moles of $\ce{MgSO4.$x\,$H2O}$. This means the following:

$$n_\text{solute}=n_{\ce{MgSO4}}=1 \times n_{\ce{MgSO4.$x\,$H2O}}= \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4}$$

$$n_\text{solvent} = n_\text{added water} + n_\text{crystallized water} = n_\text{added water} + x \times n_{\ce{MgSO4.$x\,$H2O}} = \bigg(5 + \frac{70x}{120+18x}\bigg)\ \mathrm{mol}\ \ce{H2O}$$

Now we can finally apply these values to the mole fraction equation:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}}{\frac{70}{120+18x} + \bigg(5 + \frac{70x}{120+18x}\bigg)}=0.039$$

Solving for $x$ provides the answer $x=7.03$, which means there are seven moles of water per mole of $\ce{MgSO4}$, and thus the formula for the solid is $\ce{MgSO4.7H2O}$, representing magnesium sulfate heptahydrate.


Non-halogen intermediate between alkane and alcohol under basic conditions? Using organic chemistry I have failed to find a method that doesn't involve halogens (gasseous especially).

Please ignore the enzyme mono-oxygenase, which adds $\ce{O2}$ directly to alkanes to form alcohols.

[edit] Regarding selectivity the reaction should preferably stop at the alcohol stage.

The most promising method I have found involves Boranes, but only works for terminal alkanes. The termial organoborane can then be converted to an alcohol.

I found it using a google search inspired by the answers below: "photochemistry of alkanes."

I think that you're asking if there is a way to convert an alkane to an alcohol without using halogens in the process.

Alkanes are relatively unreactive and so high energy species (like free radicals) are most often used to carry out reactions on alkanes. The  reaction of alkanes with halogens that you've mentioned is an example of this free radical process.

Other high energy molecules that can be used to convert alkanes to alcohols include ozone and (photo-) chemically generated singlet oxygen.  Just like with the halogens, these reactive molecules show little selectivity, so if primary, secondary and tertiary hydrogens co-exist in your alkane, then a mixture of primary, secondary and tertiary alcohols result.  However in molecules such as cyclohexane, where there is only one type of hydrogen, somewhat useful results can be obtained.  I say "somewhat useful" because ozone and singlet oxygen can oxidize the resultant alcohol further to a carbonyl.  Here is a reference to an abstract on the use of ozone to functionalize alkanes.

Reagents such as potassium permanganate and chromic acid also oxidize alkanes by a free radical mechanism to produce alcohols.  Again the issues of selectivity and further oxidation remain.

Here is a link to a nice review entitled "Alkane Hydroxylation." It covers a wide variety of reagents.  Most of the molecules depicted contain functional groups which can help improve selectivity, but there are a number of hydrocarbon examples as well.  Perhaps you'll find an example that is similar to your molecule of interest.


What chemicals could absorb/adsorb ambient atmospheric methane? From an article in the New Scientist in 2010,

  Since pre-industrial times, methane levels in the atmosphere have risen from around 715 parts per billion to nearly 1800 ppb, claims Folberth. "If methane could be taken back to pre-industrial levels it would bring about half a degree of cooling and would be equivalent to removing one-third of anthropogenic carbon dioxide," says Folberth. Furthermore, the cooling would take place within 10 years.

Klaus Lackner and others have developed technology such as the moisture-swing sorbent to capture ambient $\ce{CO2}$

Where might be the chemical options for a material that would absorb or adsorb ambient methane?

There is none. Alkanes are very hard to activate. They require either free radicals, or coordinated unsaturated sites either in the molecule or on surface of solid. But in both cases, the atmosphere contains molecules that will react more readily.

However, the average life-time of methane in atmosphere is quite small, less than 10 years.


When do I use the hydronium ion in an acid-base equation? I have a question about when to use H3O+ and when to use the actual compound. I'm not really sure how to word this question so I'll show you an example of what I mean.

Example of using H3O:

"A titration for a sample of 25mL of 0.1 M NH3 by 0.1M HCl is conducted. Calculate the pH after the addition of 12 mL of 0.1M HCl"

$\ce{NH3 + H3O -> NH3 + H2O} $
(etc)

Example using the compound:

"Calculate the pH of 75 mL of 0.1M HC2H3O2 and .2M of NaC2H3O2 to which 9.5mL of 0.1M HCL is added"

$\ce{C2H3O2- + HCl -> HC2H3O2 + Cl-}$
(etc)

My question is, when are we supposed to use HCl in the equation and when are we supposed to use H3O+? Is it only when you are using a buffer that you are to use HCl? If so, why is this?

In both your cases, only H+ actually participates in the reaction. Remember that HCl is a strong acid, and dissociates completely into H+ and Cl-. For example, consider the net ionic form of your second equation:

$\ce{C2H3O2- + HCL -> HC2H3O2 + Cl-}$

$\ce{C2H3O2- + (H+) + Cl- -> HC2H3O2 + Cl-}$

$\ce{C2H3O2- + H+ -> HC2H3O2}$

For your first equation, also note that, similarly, although the entire H30+ molecule participates in the reaction, only the H+ is transferred. (Also, I think you meant NH4+ instead of NH3 as a product.) So:

$\ce{NH3 + (H3O)+ -> NH4+ H2O}$

$\ce{NH3 + H+ -> NH4+}$

So basically, they both use the same compound (H+). Usually water is the main producer of H+ (through self-ionization). You just need to consider factors such as the common ion effect that would make other compounds such as HCl the main producer of H+ instead of water.


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

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

I'm stuck with this question. What I know:
$$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$$
$$\ce{NaH2PO4 + NaOH -> Na2HPO4 + H2O}$$

Finally 50 millimoles $\ce{Na2HPO4}$ and 50 millimoles $\ce{NaH2PO4}$ remains.
Should I apply acid buffer equation?

The answer to your question does not need a single calculation.

You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.

Now what is between those two volumes, at your desired volume of 150 mL?
A simple buffer system that can be described using the Henderson–Hasselbalch equation:
$$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$

As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).

$\hskip3.5cm$


How to make artificial diamond? Carbon can be turned to diamond artificially, but while watching a video I learned that a diamond can be made from the ashes of a man, how is this possible?

First off, take a look at the Wikipedia article: Synthetic diamond. Basically, there are two ways to artificially generate the heat and pressure needed. First is the "conventional" high-temperature, high-pressure method, which uses a press to put a sample of pure carbon graphite under pressure of about 5 gigapascals (1GPa = about 10,000 atmospheres or about 145,000 psi) and about $\pu{1500 ^\circ C}$. This combination encourages the flat hexagonal lattice structure of the graphite to morph into the trihexagonal 3-D lattice of diamond. Second is the chemical vapor deposition method, in which a substrate of diamond crystal is exposed to sublimated carbon plasma (temperatures exceeding $\pu{3000 ^\circ C}$) in a hard vacuum (less than .0001 torr). The sublimated carbon attaches to the crystal substrate, "growing" it.

Both of these can be used to create so-called "memorial diamonds" from the cremated remains of a human. To do that, first the ash is chemically filtered to separate the carbon from "pollutants" naturally occurring in the body that don't combust into gaseous substances during cremation, such as calcium, phosphorous, iron etc. Then, this "amorphous carbon", much like any other carbon, can be used in HTHP or CVD processes, along with additional carbon as may be required to grow the diamond to the proper size. Usually, the HTHP process is used, as it makes the best use of the available carbon (CVD requires sublimating a lot of carbon, more than would be left behind in the ash, and most of which ends up on the chamber walls and not the substrate), and requires less energy (it's easier to maintain mechanical pressure once applied than to maintain temperature).


Benzene as a pi-acceptor? In coordination chemistry, we classify ligands as sigma donors, pi acceptors and donors. Benzene is a pi donor, as is the cyclopentadienyl anion, but do these two possess any pi acceptor character? (Can a ligand be both pi acceptor and pi donor?)

Aromatic rings that complex "face on" to a transition element are acting as pi acceptors all the time, including cyclopentadienide complexes.  And they act simultaneously as pi donors too.

Let's look at the familiar cyclopentadiende pi orbitals from a different perspective.  We know that there are three bonding orbitals that should be occupied and two antibonding orbitals that shouldn't be -- until we introduce the d orbitals of a transition element.  As illustrated by https://www.chemtube3d.com/orbitalsm-cyclopentadienyl/, now overlap occurs with the possibility of all the ring orbitals bonding to the metal:

So, in a complex such as ferrocene, the filled cyclopentadienyl pi orbitals share their electrons with three of the transition element d orbitals and the latter element, in turn, donates electrons back to the two unoccupied pi orbitals of the ring.

Of course these interactions sacrifice aromaticity in the cyclopentadienyl rings, but it's a bargain with the bonds forming between carbon and the transition element.  Aromaticity is not so much broken as it is superseded by a further bond deolcalization.  In the especially well-known case of ferrocene, the energy bargain is especially good because when all is said and done with the iron/cyclopentadienyl electron exchange, you've exactly filled the bonding orbitals and left the antibonding orbitals empty.  The diagram from this source shows the molecular orbital structure in the case of a $3d$ metal.  For ferrocene where the iron(II) center provides six additional electron, they fill the orbitals in the box right up to the metal $3d$ orbital level (the HOMO is the $a_{1g}$ orbital inside the box) but not beyond.  Note that the filled $e_{2g}$ orbital has a component from the antibonding orbitals of the cyclopentadienide, showing the pi acceptor action of the latter.

Now compare this outcome with magnesocene.  Magnesium has no valence d orbitals or d electrons.  So the pi electron exchange between pi and d orbitals is not happening.  In this case the cyclopentadienyl ion can still act as a donor to the metal s abd p orbitals, but this does not make for as good bonding because (1) the overlap is not as good and (2) magnesium, like most Group 1 and 2 metals, has relatively diffuse valence orbitals that make for relatively poor overlap anyway.  Unable to develop the full covalent bonding network seen in transition element cyclopentadienide complexes such as ferrocene, magnesocene has greater ionic character -- so much as to form ions in some solutions -- but much less stability.


Brackets in chemical formulas When you have brackets such as $\ce{Zn(NO3)2}$, if there is a coefficient around the formula such as $\ce{2Zn(NO3)2}$, how would you read this? Would there be 2 zinc atoms, 4 nitrogen and 12 oxygen atoms if I am multiplying that out correctly?

$\ce{2Zn(NO3)2}$ can be interpreted differently, depending on the context.
The '2' outside of the parenthesis could either refer to two molecules or two moles of substance. In either case, the chemical formula tells you the same thing: that there are two units of $\ce{NO3-}$ for every unit $\ce{Zn^2+}$.

I would write it as zinc (II) nitrate, (read as 'zinc-two-nitrate) The roman numeral (II) refers to the oxidation state of the zinc ion, which is $+2$.


Coordination isomerism variation The school book and Wikipedia state that coordination isomers are:

In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change.
For example, a solution containing $\ce{[Co(NH3)6]^3+}$ and $\ce{[Cr(CN)6]^3−}$ is a coordination isomer with a solution containing $\ce{[Cr(NH3)6]^3+}$ and $\ce{[Co(CN)6]^3−}$.

So in this definition, consider $\ce{[Co(NH3)5(CN)][Cr(CN)5(NH3)]}$. Is it coordination isomer of $\ce{[Co(NH3)6][Cr(CN)6]}$ or not?
My teacher says yes, but I'm not quite sure according to Wikipedia.

Yes, it is.
In both the cases, the number of ligands remains same, which is according to the definition in Wikipedia.

For any type of structural isomerism, the premise is that the primary valence and secondary valence (as defined by Werner) of central metal/ ion remains same for all the isomers.

In this case, the primary valence is 3 and secondary valence is 6, which remains same for all isomers.

Other possible coordination isomers are:

$\ce{[Co(NH3)4(CN)2]+[Cr(NH3)2(CN)4]-}$

$\ce{[Cr(NH3)4(CN)2]+[Co(NH3)2(CN)4]-}$

$\ce{[Cr(NH3)5(CN)]^2+[Co(NH3)(CN)5]^2-}$

(cation is always written first followed by anion)

I don't think $\ce{[Co(NH3)3(CN)3][Cr(NH3)3(CN)3]}$ can exist because, now both parts become neutral.


Are diamonds really forever? Common saying. Diamond possesses:

ultra hardness, (10 on the Mohs scale; 10000 HV on Vicker's Hard Test (iron merely 30-80))
hyper thermal conductivity, ($2320~\mathrm{W\, m^{-1}\, K^{-1}}$, or over ten times better than the heatsink in your computer!)
extreme pressure resistance, (withstands a crushing 600 gigapascals; or around 2 times the pressure at the center of the earth, enough to snap carbon nano-tubes and graphene or create metallic oxygen or overcome copper's electron degeneracy pressure, making the maximum chamber pressure of a firing pistol seem literally like popping popcorn... I digress)
and excellent luster (what do you expect, it's a diamond) combine to make the gemstone coveted by all.

Diamonds are the stuff of awesome.

But do they really exist forever?

Wikipedia notes that,

  Diamond is less stable than graphite, but the conversion rate from diamond
   to graphite is negligible at standard conditions.

Huh. But Wikipedia doesn't mention how long. So how long would it take for this super-material to convert to the stuff I scribble with?

(If you doubt the claims about diamond's seemingly unbelievable properties, check out the link on Wikipedia about diamond and this and this.)

Great point Joe made, that $10^{80}$ is just forever to us puny humans. Being a geek I can't resist the urge to compare the time length $10^{80}$:

Makes the entire lifespan of a red dwarf star seem like the Planck second.
Enough time for you to sift through all the atoms in the entire UNIVERSE at a rate of one atom per second.
Getting $67 worth of US quarters and flipping them, one per second, to get all heads-up.
Chance of macroscopic quantum tunneling!! (I don't know precisely how much, but quite large)

...and this is $10^{80}$ seconds I'm talking about...

how long would it take for this super-material to convert to the stuff
  I scribble with?

No, despite the fact that James Bond said "Diamonds are Forever", that is not exactly the case. Although Bond's statement is a fair approximation of reality it is not a scientifically accurate description of reality.

As we will soon see, even though diamond is slightly less stable than graphite (by ~ 2.5 kJ/mol), it is kinetically protected by a large activation energy.

Here is a comparative representation of the structures of diamond and graphite.

(image source: Satyanarayana T, Rai R. Nanotechnology: The future. J Interdiscip Dentistry 2011;1:93-100)

(image source)

Note that diamond is composed of cyclohexane rings and each carbon is bonded to 2 more carbons external to the cyclohexane ring. On the other hand, graphite is comprised of benzene rings and each carbon is bonded to only 1 carbon external to the benzene ring. That means we need to break 6 sigma bonds in diamond and make about 2 pi bonds (remember it's an extended array of rings, don't double count) in graphite per 6-membered ring in order to convert diamond to graphite.

A typical aliphatic C–C bond strength is ~340 kJ/mol and a typical pi bond strength is ~260 kJ/mol.  So to break 6 sigma bonds and make 2 pi bonds would require ~((6*340)-(2*260)) ~ 1500 kJ/mol. If the transition state were exactly midway between diamond and carbon (with roughly equal bond breaking and bond making), then we might approximate the activation energy as being half that value or ~750 kJ/mol.  Since graphite is a bit more stable than diamond, we can refine our model and realize that the transition state will occur a bit before the mid-point. So our refined model would suggest an activation energy something less than 750 kJ/mol. Had we attempted to incorporate the effect of aromaticity in graphite our estimate would be even lower.  In any case, this is an extremely large activation energy, so, as we anticipated the reaction would be very slow.

An estimate (see p. 171) of the activation energy puts the reverse reaction (graphite to diamond; but since, as noted above, the energy difference between the two is very small the activation energy for the forward reaction is almost the same) at 367 kJ/mol. So at least our rough approximation was in the right ballpark, off by about a factor of 2. However, it appears that the transition state is even further from the midpoint (closer to starting material) than we might have guessed.

This activation energy tells us that at 25 °C, it would take well over a billion years to convert one cubic centimeter of diamond to graphite.

Note 04/17/20: As mentioned in a comment, the original "estimate" link became  defunct and was replaced today with a new estimate link.  However the original article and estimate can can still be seen on the Wayback Machine and it estimates the activation energy to be 538.45 kJ/mol, reasonably close to our estimate.


Acetic acid or ethanoic acid - trivial names vs systematic names and correct formula representation My teacher likes to use household materials to do experiments. Today, we did an experiment that showed the reaction of vinegar and eggshell.

When we were doing the analysis of the experiment, I realized that my teacher used the term acetic acid ($\ce{CH3COOH}$) instead of ethanoic acid. So, my questions are:

What is the accepted usage for the terms acetic acid and ethanoic acid?
Which formula is the accepted representation of acetic acid, $\ce{CH3COOH}$ or $\ce{C2H4O2}$?

While I agree that the systematic name ethanoic acid makes a lot more sense, the name acetic acid has been used for so long that it's the accepted IUPAC name and is used almost universally. Both refer to the same compound.

Molecular formulas like $\ce{C2H4O2}$ are often seen when doing elemental analysis and the like, but give no structural information. $\ce{CH3COOH}$ makes it clear that it's an acid and shows the carbonyl group.


Why was the definition of chiral carbon changed? Before 1984, the definition of chiral carbon was,

  carbon atom that is attached to four different types of atoms or
  groups of atoms

In 1984, the definition was changed to

  Any carbon molecule that is not superimposable on its mirror image.

Why was this change made? Don't both definitions mean the same or is/are there cases where the former is false. Preferably, please provide an example

The IUPAC definition of asymmetric carbon is still:

  The traditional name (van't Hoff) for a carbon atom that is attached to four different entities (atoms or groups), e.g. Cabcd.

The definition of chirality centre is:

  An atom holding a set of ligands in a spatial arrangement which is not superposable
  on its mirror image. A chirality centre is thus a generalized extension of the concept
  of the asymmetric carbon atom to central atoms of any element, for example N+abcd,
  Pabc as well as Cabcd.


How does the steric effect affects the electronic density of the central atom? For example, if you have Nitrogen attached to some methyl groups, and other attached to some ethyl groups. Why the electron releasing effect on the ethyl will be lower than for the methyl?

Both methyl and ethyl groups have +I effect and its obvious that the ethyl groups have more electrons than the methyl groups, so in this case the present of more electron density on nitrogen atom with ethyl groups lessen the -I effect of the nitrogen atom toward the ethyl groups compared to that on the methyl groups. I think this should be the reason.


Why does hydroxide attack the carbon atom of methyl iodide (in an Sn2 reaction)? As the electronegativity difference between carbon and iodine is negligible, why does $\ce{OH-}$ attack the carbon atom (in an $\mathrm{S_N2}$ mechanism) even though carbon doesn't seem to be electrophilic? (Since the $\ce{C-I}$ bond is not polar enough to give a sufficiently positive partial charge on carbon)

Of course, just looking at methyl iodide and the partial charges, one would assume that there is no real reason for any nucleophile to attack the carbon atom for the reasons you stated. I’ll just point out and leave as an exercise to the reader that there is equally hardly any reason for the nucleophile to attack iodide, though.

However, the reaction also has a product side. Remember that what is happening is the following:

$$\ce{OH- + CH3-I -> HO-CH3 + I-}$$

While the attack onto the only very slightly polarised carbon in itself may not be particularly beneficial, iodide is a very good leaving group as it is able to distribute the resulting negative charge well. (Large size, many electrons, small charge per volume ratio.) Likewise, if we consider thermodynamics, the $\ce{C-I}$ bond is not a particularly strong bond while a $\ce{C-O}$ bond is in the region of $\ce{C-C}$ or $\ce{C-H}$ bonds — so overall we are gaining a much stronger bond for a rather weak one.

Consider what had happened had oxygen attacked the iodide side of iodomethane. While oxidising iodide is something rather common in chemistry, the leaving group would have to be $\ce{CH3-}$ — a terribly non-stabilised leaving group, the conjugate base of an extremely weak acid; something that simply does not leave. Thus, there is no way hydroxide would attack the iodide side in an $\mathrm{S_N2}$ reaction — even aside from the reasons above left for the reader to figure out.


Direct conversion of amides to esters Rarely does the literature describe the synthesis of esters from amides. I understand that amides are a relatively unreactive functional group but they can easily be hydrolysed to give carboxylic acids under basic or acid conditions. Why can't they be converted to methyl esters under similar conditions using methanol in place of water?

Surely the mechanism would be identical to hydrolysis? Additionally, the pKa of water is very similar to methanol so I cannot justify it on the basis of nucleophilicity/basicity. The only thing I can think of is steric hindrance but methanol isn't that bulky compared to water.

You are right about the rarity of amide to ester conversion. Amide to ester conversion has remained a challenging and underdeveloped synthetic transformation. It is often due to the fact that amides are sufficiently stable because of its resonance-stabilized $\ce{C–N}$ group that imparts a planar geometry to amides and as such, esterification is difficult and requires the use of harsh acidic or basic conditions, while employing a large excess of nucleophile (e.g. using alcohol nucleophile as solvent). In contrast, esters are generally reactive under a variety of mild condition and therefore the opposite conversion of ester to amides is quite normal.
The most promised method of achieving amide to ester conversions is Keck’s methylation/hydrolysis although this method is limited to the synthesis of methyl esters and thus researchers has developed some method to expand this conversion's feasibility and efficiency.

Hie et. al. has achieved amide to ester synthesis using nickel complex catalyst. They studied the conversion of various aryl and heteraryl amides. One studied reaction is conversion of benzamide to methyl benzoate using $\pu{10 mol{\%}}$ $\ce{Ni(cod)2}$, $\pu{10 mol{\%}}$ SIPr, 2 equivalents of methanol, and toluene as solvent at $\pu{110 °C}$ for 12 hours. (Ref. 1)

An efficient two step transformation of 1° and 2° amides to methyl and ethyl esters has been developed using trimethyl- and triethyloxonium tetrafluoroborates, and dilute acid. This method was applied to 1-benzyl-4-phenylamino-4-piperidinecarboxamide, a precursor in the synthesis of carfentanil, to produce the methyl ester in $\pu{60{\%}}$ yield and the ethyl ester in $\pu{80{\%}}$ yield. (Ref. 2 and 3)

Conversion of amides to ester has been achieved using $\ce{CeO2}$ catalyst and HY zeolites. A variety of metal oxides as catalyst and ester yields has been discussed. (Ref. 4)

References

Hie, L., Fine Nathel, N., Shah, T. et al. Conversion of amides to
esters by the nickel-catalysed activation of amide C–N bonds. Nature
524, 79–83 (2015). DOI: 10.1038/nature14615 (PDF link)
Amide to ester conversion: a practical route to the carfentanil class of analgetics
Douglass F. Taber and Mohammad Rahimizadeh, The Journal of Organic Chemistry, 1992, 57 (14), 4037-4038, DOI: 10.1021/jo00040a061
Anthony J. Kiessling & Cynthia K. McClure (1997) The Conversion of Amides to Esters with Meerwein'S Reagent. Application to the Synthesis of a Carfentanil Precursor., Synthetic Communications, 27:5, 923-937, DOI: 10.1080/00397919708004212
Conversion of Amides to Esters via Selective Cleavage of Amide C–N Bonds over a CeO2 Catalyst by Shimizu et. al. (PDF)


Finding simplest formula of a compound with following specifications I ran into the following problem, wherein I'm asked to find the simplest formula of a compound containing $50 \%$ of element $X$ (atomic wt. 10) and $50\%$ of element $Y$ (atomic wt. 20).

I know about few laws of chemical combinations. I think Law of Definite Proportions may come into play.

$$\text{atomic wt. }X:\text{atomic wt. } Y=1:2\implies XY_2$$

Am I correct in my reasoning? If not kindly point out the flaws, if any? Thanks


What are isodiaphers? We know that isodiaphers have same difference in neutrons and protons, but I encountered another definition:

  Atoms which have same isotopic axis are called isodiaphers.

I don't know what it means by "isotopic axis". Please try to explain in simple words.

These are basic terms of radiochemistry:

Isotopes are nuclides with a common number of protons $Z$.
Isobars are nuclides with a common number of nucleons $A$.
Isotones are nuclides with a common number of neutrons $N$.
Isodiaphers are nuclides with a common neutron excess $N-Z=A-2Z$.

Thus, in the chart of the nuclides, isodiaphers are found in diagonals from bottom left to top right. For example, the mother and daughter nuclides of alpha decay are isodiaphers.


Water and Ethanol Vapor Phase diagram for specific range I am looking for a Vapor Phase diagram for Water and Ethanol where Ethanol would be from 0 to 5%.

The diagrams I have found are for Ethanol concentrations of 0% to 100% and this doesn't allow me the accuracy for the range I have in my process.

If anyone could point me to a table of values that the 0 to 100% diagrams are generated from I can generate a diagram for the range I am looking for.

The Dortmund Data Bank might be a resource to start searching.

They do seem to have a page for vapour-liquid equilibrium data for the ethanol/water mixture.

On a first view, however, the data seem to be for less than atmospheric pressures only and are compiled from Ind. Eng. Chem., 1942, 34, 1501–1504.


Stereoisomerism in 1,2-cyclobutanediol For 1,2-cyclobutanediol, it's given in my book that it has 3 stereo isomers.
Here's the passage:

My questions:

How can it have stereo-isomers?
What will the meso compounds look like?

These two isomers are enantiomers, i.e. mirror images, of each other:

$\hspace{50 mm}$

This one has a plane of symmetry through the molecule and is therefore meso:

$\hspace{60 mm}$


Equipment Suggestion Small Burette? Is anyone familiar with a piece of equipment that can be purchased like a small burette? I need to find a way to dispense a liquid at a constant rate, however, the amount of liquid that I need to dispense will probably be less than 5 mL. A syringe is good in that regard, but it requires that I am able to push down with a constant pressure (which I doubt I can do). I've heard that they make burettes that are smaller (i.e around 25mL). Does anyone know if they make them smaller or at least not 2 feet long. Thank you for any useful information.

The usual commercially available burettes are standardizes in the international standard ISO 385:2005 Laboratory glassware – Burettes. This standard includes small burettes with capacities of 1, 2, 5, 10 and 25 ml. The very small versions are typically advertised as “micro burettes”.

$$
\begin{array}{lllll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} & & \text{Max. length}\\
& & \text{Class A and AS} & \text{Class B} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} & \pm\ \mathrm{ml} & \mathrm{mm}\\
\hline
1 & 0.01 & 0.006 & 0.01 & 575 \\
2 & 0.01 & 0.01 & 0.02 & 650 \\
5 & 0.01 & 0.01 & 0.02 & 820 \\
5 & 0.02 & 0.01 & 0.02 & 800 \\
10 & 0.02 & 0.02 & 0.05 & 820 \\
10 & 0.05 & 0.03 & 0.05 & 620 \\
25 & 0.05 & 0.03 & 0.05 & 820 \\
25 & 0.10 & 0.05 & 0.10 & 620 \\
50 & 0.10 & 0.05 & 0.10 & 820 \\
100 & 0.20 & 0.10 & 0.20 & 870 \\
\hline
\end{array}$$


Converting mm Hg to mm H2O So here's my problem:

What is the osmotic pressure, noted in mm of $\ce{H2O}$, of a solution with 125 micrograms of vitamin B12 ($\ce{C63H88CoN14O14P}$) in 2.50 mL of $\ce{H2O}$ at 25 degrees Celsius.

The "hint" that comes with the problem says: Weak osmotic pressures are often measured and given in mm of $\ce{H2O}$. Use here the density of water (1,00 g/mL) and of $\ce{Hg}$ (13,6 g/mL). In this case, the constant $R$ has a value of $\pu{62,36 mm Hg * L / mol * K}$.

Here:

$$
2.50 \,\text{microgram} = 0.000125 \,\text{grams}
$$

$$
pi = cRT
$$

$$
\frac{(\pu{0.000125 g * 1 mol / 1355.3652 g})}{(\pu{2.50 mL / 1000})} * \pu{62.36 mm \ce{Hg} * L / mol * K * 298.15 K} = \pu{0.685890194 mm \ce{Hg}}
$$

So, I blocked here and I saw that the solution was:

$$
\pu{0.685890194 mm \ce{Hg} * 13.6 g/ml / 1.00 g/ml = 9.32 mm \ce{H2O}}
$$

So, here's two questions:

Why did I have to use R = 62.36 and not the usual R = 8.31?
Also, what's happening in the conversion exactly? How was I supposed to figure out what they did to convert from mm $\ce{Hg}$ to mm $\ce{H2O}$?

You CAN use R=8.31 but it has the (standard) unit of (m3 Pa K−1 mol−1), simply convert your volume from liter (aka dm3) to meter cube (m3) and the pressure will be in Pascal (Pa)

If you use R=62.36, which has a unit of (L mmHg K−1   mol−1), as long as your   volume is in liter, the pressure will be in mmHg
But the question ask for mmH2O.

Pressure is simply force on an area (F/A), weight on an area (mg/A), density * volume on a area (rho*V /A) = density * height (rho*h)

Thus P = density (Hg) * h (Hg) = density (water) * h (water)

h (water) =  (13.6 / 1.00) * h(Hg)


Can we make Gold in a Lab? Can we make Gold in a Lab? Is it possible to make changes in metals that will create Gold? and How?
I have tried to search something related to this but didn't get any good results.
Thanks in Advance

Yes, gold has been made from mercury.

See "Transmutation of Mercury by Fast Neutrons" Phys. Rev. 60, 473


C + O2 is equal to C + O, how is that possible C + O = CO2 . This is because carbon has valency of 4 while oxygen has valency of 2. When they react the valencies are criss crossed which means we will get C2O4 but this is simplified to CO2.

But my textbook say that C + O2 = CO2.

So how is it possible that C + O2 and C + O both get the same product of CO2?

Also you can say that the reaction of C + O = CO2 is not balanced but the balanced chemical reaction would be C + 2O = CO2 not C + O2 = CO2 which is the reaction that my textbook says.

When thinking about chemical reactions it is very important to know which chemicals may react with each other. Jun-Goo Kwak already pointed out the nature of oxygen.

A quick reminder: The ground state of elementary oxygen is the triplet biradical $\ce{o2}$, which is a gas. This is what we have on the surface of the earth. Carbon however comes in many  different forms in nature. The most popular and often is graphite. Other forms include diamond, fullerenes and graphene.
At one point in your life you haw almost certainly come in contact with graphite: coal. As the principle repeating unit is carbon itself, its formula will be written as $\ce{C}$.

As for the binary combination of oxygen and carbon, there are also many different modifications. The most important of them are carbonmonooxide ($\ce{CO}$) and carbondioxide ($\ce{CO2}$). Like Uncle Al stated there are also suboxides known, which are usually byproducts of incomplete combustion (if not targeted explicitly).

Having said all that, if you burn coal, the following main reaction will happen (1):

$$\ce{C + O2 -> CO2}$$

However, given the right conditions (excess carbon) also carbonmonoxide  may be formed (sum of reaction, 2):

$$\ce{2C + O2 -> 2CO}$$

The reaction itself will detour via the Boudouard reaction, which is very important in blast furnace processes. First forming carbondioxide via 2 and then convertin excess carbon to carbonmonoxide via 3:
$$\ce{C + CO2 <=> 2CO}$$


In helium-2, would each electron experience a single positive charge from the nucleus? If we have an atom of hydrogen-1, we know that there exists one proton with one electron and in helium-2 two protons with two electrons. My question is that do protons get shared equally with electrons? For example, is it true that each electron in helium experiences the same force from the nucleus as each electron in hydrogen, due to the proton to electron ratio being 1:1 in both atoms?

To answer this question, I am going to assume that a $\ce{^2He}$ nucleus actually exists long enough for two electrons to surround it — this is probably not a valid assumption since Webelements.com does not list $\ce{^2He}$ while it does list the isotopes $\ce{^6He}$ and $\ce{^8He}$ (although they admittedly have long half-lifes measured in tenths of seconds). But for our case, let us assume so.

The charge radius of a proton — our closest experimental measurement of its diameter, if ‘diameter’ is even a valid term at subatomic scales — is given on Wikipedia as $\pu{8.75e-16 m}$ or $\pu{0.875fm}$. At this point, take a step back to remember that the Bohr radius, approximately the most probable distance between the electron and the proton in a hydrogen atom, is $\pu{5.29e-11 m}$ or $\pu{52.9pm}$. This is a difference in scale by a factor of $10^5$.

Therefore, a reasonable assumption for the $\ce{^2He}$ nucleus — and indeed for practically any nucleus — is that it can be approximated by a dimensionless point of $+2$ charge and not any sphere, ovaloid, double-sphere or other three-dimensional object. Therefore, if we want to use quantum mechanics to calculate the wave function of $\ce{^2He+}$, we assume a spheric potential caused by a positive charge in the origin of a coordinate system; the value of the positive charge is $2e$ with $e$ being the elementary charge.

A two-electron system cannot, to the best of my knowledge, accurately be calculated. However, we may approximate it as just taking the wave function solutions of $\ce{^2He+}$ and adding another spin-opposite electron. (Then, we slightly adjust the energy values to account for electron-electron repulsion.) This way of approaching the problem shows that the helium atom should really better be considered a central positively charged dot with the outer space occupied by the negatively charged electrons, each of which see a $+2$ charge in the centre and an additional negative charge ‘buzzing around them’ (Note: physically incorrect macroscopic picture).

Replacing $\ce{^2He}$ (as is present in your question) with either of the stable isotopes $\ce{^3He}$ or $\ce{^4He}$ would not change the picture; the size of a neutron (see my note in the second paragraph on subatomic scales) is roughtly that of a proton. If anything, adding one or two neutrons to the two protons will make the nucleus smaller since neutrons increase the intranuclear attractive forces.

In many introductory chemistry books, the core and shell model of an atom and the relevant size differences are exemplified by macroscopic comparisons. My chemistry book offered: if you enlargen an atom so that it would fit the Munich Olympic Tower inside it, the nucleus would inflate to the size of a cherry stone, match head or something of that size. This again shows how good the assumption of poiut-sized charges in the nucleus is.


MO theory methods other than LCAO What methods other than linear combination of atomic orbitals does MO theory use? Which one is the most accurate and comprehensive?

Well, this gets into the concept of basis sets.

LCAO implies that the number of molecular orbitals will be the same as the number of atomic orbitals involved.

In fact, we usually use more basis functions than just the simple number of atomic orbitals. We might add diffuse or polarization functions to better describe electron density away from the nucleus.

Moreover, rather than the physically correct Slater-type orbitals:

$$R_l(r) = A(l,\alpha) r^l e^{-\alpha r}$$

We usually use Gaussian-type orbitals to express the radial component.

$$R_l(r) = B(l,\alpha) r^l e^{-\alpha r^2}$$

Now GTOs don't express the radial part of the wave function well -- they decay too quickly as you get far from the nucleus. So we use a linear combination of multiple Gaussian orbitals to approximate the physically-correct Slater terms. Still, it's much faster to use a set of multiple GTOs.

Beyond STOs and GTOs, people do use other types of orbitals. For example:

Wavelets
Plane waves
Muffin-tin


Mechanism of chlorine disproportionation It is a common fact that $\ce{Cl2}$ reacts to form $\ce{HCl}$ and $\ce{HOCl}$ when dissolved in water. However, how exactly does the reaction proceed?
The first step could either be that the oxygen atom in $\ce{H2O}$ attacks $\ce{Cl2}$, forming $\ce{Cl-}$, and $\ce{H2OCl+}$, which then loses a proton to become $\ce{HOCl}$.

But it could also be that chlorine attacks the $\ce{H}$ in $\ce{H2O}$, forming $\ce{Cl+}$, $\ce{HCl}$, and $\ce{HO-}$. The first reaction seems more probable (based on the products), but I am not sure.

According to this paper,
the attacking species is $\ce{OH-}$ rather than $\ce{H2O}$ on $\ce{Cl2}$

The paper also says that:

  The proposed mechanism also provides a good explanation for the high initial rate of reaction. At the instant the strong chlorine solution is mixed with the distilled water the hydroxyl ion concentration is considerably higher than that calculated from $\ce{kw}$ and the hydrogen ion concentration, for it takes a finite time for the hydrogen and hydroxyl ions present to react to reduce the ion product to the equilibrium value. During this period chlorine molecules also react with hydroxyl ions at approximately the same rate as hydrogen ions are reacting. So for this initial period a higher rate of hydrolysis of chlorine is expected than that found after the hydroxyl ion has reached its equilibrium concentration.

It also says that:

  During the course of the reaction the equilibrium concentration of hydroxyl ion is only about 10^(-12) M. Since about 10^-(3) mole per liter of chlorine must react before final equilibrium is reached, new hydroxyl ions must be supplied by the dissociation of water molecules. This could easily complicate the reaction kinetics, if the dissociation rate were not rapid enough. However, calculations assuming that the dissociation of water is a normal bimolecular reaction with an activation energy equal to the endothermicity of the reaction show that the total hydroxyl ion required may be formed in 10^(-5) sec. This is so rapid that the steady state concentration of hydroxyl ion should not be measurably different from the equilibrium value.

What this stuff basically means is that using $\ce{OH-}$ as the nucleophile, as it can explain the initially high rate of reaction(at the back of our minds, we know that $\ce{OH-}$ is a better nucleophile than $\ce{H2O}$). When $\ce{Cl2}$ is added to the distilled water, we find that $\ce{[OH-]}$ is more than that what would be expected at equilibrium value, so $\ce{OH-}$ and $\ce{H+}$ keep combining to reduce the ion product, while at the same time, $\ce{OH-}$ also reacts with $\ce{Cl2}$. So at this time, rate of reaction will be quite high.
Also, the ionization of water to provide the $\ce{OH-}$ and $\ce{H+}$ will be keeping up with the consumption. Also, the reaction rate tells us that collisions in this reaction are very effective, and lead to reaction almost all the time

So, essentially, to explain the initial high reactivity, we take the reaction mechanism to be a coupled equilibria of $\ce{Cl2 + OH- <--> HOCl + Cl-}$ and $\ce{ H2O <--> H+ + OH-}$


Is photosynthesis endothermic and exothermic at the same time? ... somehow? Photosynthesis is obviously an endothermic reaction, I mean, what would be the point otherwise, right? This is probably just sheer and utter stupidity on my behalf, but why does Wikipedia say the heat of formation of glucose is negative? As you can probably see I'm on a quest of understanding enthalpy as a concept and I'm still quite new, but some sort of ultranoob clarification would be very much welcome.

Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :

$
\ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)}
$

Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$

$
\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}
$

You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:

$
\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}
$

Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.

So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.


If change in free energy (G) is positive, how do those reactions still occur? I was doing a couple of problems for homework:

Calculate $K_\mathrm{sp}$ of $\ce{AgI}$ at $55.0\ \mathrm{^\circ C}$

Calculate $K_\mathrm{b}$ of $\ce{NH3}$ at $36.0\ \mathrm{^\circ C}$

I have to use $\Delta G^\circ= -RT\ln K$ and $\Delta G= \Delta H-T\,\Delta S$

When I did this $\Delta G^\circ$ is positive ($89.59\ \mathrm{kJ/mol}$ and $28.037831\ \mathrm{kJ/mol}$ respectively), yet $K_\mathrm{sp}$ for $\ce{AgI}$ is $5.5\times10^{-15}$ and $K_\mathrm{b}$ for $\ce{NH3}$ is $1.8\times10^{-5}$, indicating that there are some products and the reactions do occur. Plus, $1.0\ \mathrm M$ $\ce{NH3}$ in solution has a $\mathrm{pH}$ of $11.6$ so it must react a little.

According to the second law of thermodynamics, if $\Delta G$ is positive, the reaction is not spontaneous, right? But clearly, they, in fact, are to a certain extent. What is going on?

$\Delta G^\circ_m$ is the difference in molar Gibbs free energy between the reagents and products in their standard states (in the case of $\ce{AgI(s)}$, the standard state for the reagent is the pure solid and that for the product is the solute in a $\pu{1 molal}$ ideal solution, all at $p^\circ =\pu{ 1 bar}$).

If $\Delta G^\circ_m>0$ you cannot convert the reactants completely into products without doing some accompanying work (for instance, applying an electrical voltage, in the case of an electrochemical cell).

If you want to see where the reaction is headed, for instance, if you start with only reactants (say $\ce{AgI(s)}$), then you should compute the following:

$$\Delta G_m=\Delta G^\circ_m+RT\log(Q)$$

For a reaction $\ce{A<->B}$ this is illustrated in the following plot (generated assuming an equilibrium constant $K_{eq}=10$, and $\Delta G$ in arbitrary units):

Here $\zeta$ represents a progress coordinate ($\zeta=Q/(1+Q)= \chi _B$).

Note that some authors place a "tilde" following $\Delta G_m$ to emphasize that it is a partial molar quantity, that is, it measures the change in free energy with reaction progress.
At the start of the reaction, when there is no solute dissolved, Q=0 and $\Delta G_m=- \infty$. In other words, the driving force for the reaction to occur is theoretically unlimited. At equilibrium that force goes to zero. Past the equilibrium point, the reverse reaction becomes more favorable.

See "What is the difference between ∆G and ∆G°?" for an elaborate answer addressing a similar question.


How to identify the aromatic compound with the sum formula C7H6Cl2? An aromatic compound (A), $\ce{C7H6Cl2}$, gives $\ce{AgCl}$ on boiling with alcoholic $\ce{AgNO3}$ solution and yields $\ce{C7H7OCl}$ on treatment with $\ce{NaOH}$. (A)  on oxidation gives a mono-chlorobenzoic acid which affords one mono-nitro derivative. Find the Compound (A).

Here from $\ce{C7H6Cl2}$ we can say it is a benzene derivative. Now as only one chlorine atom reacts with $\ce{AgNO3}$, one chlorine atom is directly connected to benzene and one is connected to methyl group which is connected to benzene. But how to find where the $\ce{-CH2Cl}$ is situated, i.e. ortho, meta or para?

Hints to find the structure of (A):
(1) The given compound (A), is aromatic, hence, it is highly likely to have only one benzene ring (deduced from it's index of hydrogen deficiency/degree of unsaturation)
(2) Only those halides which can give a stable carbocation at the carbon to which the halogen atom is attached, can give precipitate of AgCl with AgNO3 solution. Therefore, one of the chlorine atoms is benzylic, as you mentioned.
(3) Now, oxidation of (A) yields a monochloro-benzoic acid, which on nitration, yields only one mono-nitro derivative. Out of the possible three options: ortho, meta, and para. now, chlorine atom on a benzene ring is ortho-para directing, while the COOH group is meta directing. Therefore, if the initial compound (A) would have -CH2Cl situated para to the Cl, it's obvious that, on oxidation, the COOH group would be formed from the CH2Cl group. Now, there is only two symmetric positions, at which both COOH and Cl will direct NO2+, hence, as both are symmetrical, they can be considered to be the same molecule. For that very reason, it only makes sense for (A) to be para substituted.


Organic nomenclature of substituted diamines Nomenclature of substituted amines is done with the N-locants, with the N being used to indicate that a group is connected to the methyl group. I was wondering, therefore, on how to properly name a substituted diamine.

I found this source from the University of Calgary, here, which recommended to use N, N', N'', etc locants for compounds with multiple substituted amines. However, their examples are all with symmetrical diamines, and it is apparent that using the N, N' locants only works because of the molecule's symmetry. Thus, my question is: how can we name substituted diamines properly, such as the one below?

Would it be N,N,N'-trimethyl,3,5-diaminopent-1-ene or N,N',N'-trimethyl,3,5-diaminopent-1-ene, or something else?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), superscript arabic numbers are used to differentiate the nitrogen atoms.

  P-16.9.2 Superscript arabic numbers, which are the locants of the parent structure, are used to differentiate the nitrogen atoms of di- and polyamines, di- and polyimines, di- and polyamides, except for geminal amines, imines, and amides.

In particular, note that the superscript arabic numbers correspond to the locants of the parent structure. Since the parent structure of the compound that is given in the question is pent-4-ene-1,3-diamine, the locants are 1 and 3. Therefore, the complete name for the substituted compound is N1,N3,N3-trimethylpent-4-ene-1,3-diamine.


Are there bonds which share single electron? Are there bonds which share single electron?

During chemical bonding, I asked our teacher why there are only 2 electrons shared in a bond. He said that because it's that one orbital can accommodate only two electrons. Two electrons make it stable. He said that there are bonds which share a single electron which has some term I don't remember and is out of our understanding for now. So, I am curious whether there are such bonds, what are they called, some examples, and are they unstable due to the reason aforementioned?

There exists an interesting molecule dicarbon with bond order two and two pi bonds instead of the usual one sigma and one pi. This is the case when 2s-2p mixing is allowed in molecular orbital theory. However when the aforementioned condition is not allowed, dicarbon exists as a molecule with one sigma and two pi bonds of, interestingly, one electron each. So in a way, one electron bonds do exist.


Mechanism for the Beilstein test? What is a possible mechanism for the Beilstein test for halogens ? I can't seem to find a arrow pushing mechanism for this reaction.

When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide:

$$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$

Then when you mix it with the halide and heat it, the higher reduction potential makes it displace the oxygen.

Here's an example with sodium chloride:

$$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$

Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green.


Master equation for catalyst and substrate binding? I recently asked this on Biology.SE, and need to start with the master equation for the reaction:
$$\ce{E + S -> ES}$$
Where $\ce{E}$ is our enzyme and $\ce{S}$ the substrate to get a probability density function for the binding time of a substrate with one specific enzyme to be:
$$f{\left(t\right)}=k\left[\ce{S}\right]\mathrm e^{-k\left[\ce{S}\right]t}$$
I have read up on master equations, but really have no idea where to start in this derivation, any help would be appreciated?

The way to approach this is not how you have mentioned in your answer.

The representation of the master equation is correct and the generalized form of the chemical master equation (CME) [1] would be:

$$\frac{\partial P(x,t)}{\partial t}=\displaystyle\sum _{\mu=1}^M a_\mu(x-v_\mu)\ P(x-v_\mu,t) - a_\mu(x)\ P(x,t)$$

Where:

$x$ denotes the state of the chemical system (number of molecules of each species)
$\mu$ denotes a type of reaction
$a_\mu$ denotes the probability of happening of reaction-$\mu$ in a small interval (also called reaction propensity)
$v_\mu$ denotes the change in the state due to the reaction-$\mu$
$M$ denotes the total number of possible reactions

As far as I can understand (also from your question on Biology.SE), you are interested in knowing the distribution of the time interval when no reaction happens. A chemical reaction can be modelled as a Poisson process which is based on 3 main postulates:

$\lim\limits_{h\to0+}\dfrac{P(N_h=1)}h=\lambda$  i.e. the probability of occurrence of one event in a very small interval of time is equal to the macroscopic rate or intensity ($\lambda\,$).
$P(N_h\geqslant2)=o(h)$ i.e. the probability of occurrence of more than one events in an infinitesimal interval is essentially zero.
Consecutive events are independent of each other.

A chemical reaction seems to follow all these postulates. The detailed derivation of the CME can be found in [1]; I am trying to present a short derivation based on Poisson postulates.

  The probability that system is in state $x$ at time $t+\Delta t$ would
  be the probability that system is in state $x-v_\mu$ at time $t$ times
  the probability that reaction $\mu$ happens in the interval $\Delta t$, or the probability that system is in state $x$ at time $t$ times probability that no reaction happens in the interval $\Delta t$.

$$P(x,t+\Delta t)=\sum_\mu P(x-v_\mu,t)\ P(\mu) + P(x,t)\left(1-\sum_\mu P(\mu) \right)$$

when you rearrange this equation, divide by $\Delta t$ and take the limit of $\Delta t \to 0$, you will end up with the CME. Note that, as per the first postulate: $$\lim\limits_{\Delta t\to 0}\frac{P(\mu)}{\Delta t}=a_\mu$$

Now, the probability that $k$ reactions would have happened in a time interval $\tau$ follows the Poisson distribution (see [2] for the derivation).

$$P(N=k)=\frac{(\lambda\tau)^k e^{-\lambda \tau}}{k!}$$

This is a well known property of the Poisson distribution that the time interval between reactions will follow exponential distribution. The probability that no reaction has happened in time interval $\tau$ would be:

$$ P(N=0)=\frac{(\lambda\tau)^0 e^{-\lambda \tau}}{0!}=e^{-\lambda\tau}$$

in this case, $\lambda=\displaystyle\sum _{\mu=1}^M a_\mu$ i.e. probability of any reaction.

Conversely, the probability that some reaction would happen in time $\tau$ would be $P(r\leq \tau) = 1-e^{-\lambda\tau}$. Where $r$ denotes the first reaction time. This is the cumulative distribution function for the exponential distribution.

In other words, the reaction time between intervals would follow an exponential distribution such that the probability that no reaction would happen will exponentially reduce with the size of the interval $\tau$.

In your specific case there are two reactions — association and dissociation, the propensities of whose will be:

$$a_f=\frac{k_f}{V}n_E.n_S\\[1em] a_r=k_r n_{ES}=k_r(n_{E_0}-n_E)$$

Where $k_f$ is the rate constant of association, $k_r$ is the rate constant for dissociation (as per mass action kinetics), $n_i$ denotes the number of molecules of species-$i$ and $V$ is the volume (of the reaction vessel). See [3] for details. Note that the amounts of the species change with time, and so will the reaction propensities. You can calculate the exact reaction time distribution when the system is at the deterministic steady state (using the steady state concentrations).

I think this answers your question on [biology.se] as well.

References:

Gillespie, Daniel T. "A rigorous derivation of the chemical master equation". Physica A: Statistical Mechanics and its Applications, 188 (1992): 404-425.
Hogg, Robert V., and Allen T. Craig. Introduction to mathematical statistics. New York: Macmillan, 1978.
Gillespie, Daniel T. "A general method for numerically simulating the stochastic time evolution of coupled chemical reactions." Journal of computational physics 22.4 (1976): 403-434.


Cyanogenic glycosides and HCN I'm researching cyanide for a biology project and noted that one of the largest sources of carbohydrates in the tropics is the cassava root -- which I previously researched as a food that contains a cyanogenic glycoside that is metabolized into hydrogen cyanide.

Most of the sources I reviewed claimed that, while boiling the cassava, the cyanogenic glycosides were broken down into components such hydrogen cyanide, which made the plant safer to consume. I have a limited understanding of chemistry and can't resolve why this is considered a safe way to cook the plant.

My research indicates that inhalation of hydrogen cyanide is significantly more toxic than the oral route, so why would this method be considered beneficial? Not sure why this is considered safer, unless the information I've found is not correct. Any help in putting this together would be appreciated. Thanks!

OK, I was too lazy to look up LD50 and LC50 of $\ce{HCN}$ for you ;)
But if you have a chance, have a look at

G. Padmaja & Keith H. Steinkraus,
Cyanide detoxification in cassava for food and feed uses,
Critical Reviews in Food Science and Nutrition, 1995, 35(4), 299-339 [DOI]

Seemingly, the processing of cassava (manioc) involves several steps, such as drying in the sun, grating, watering and cooking, each of which either facilitates the enzymatic or thermal cleavage of the linamarin (i.e. the toxic compound) to glucose or supports the release of $\ce{HCN}$.
In total, the processing minimizes the risk of exposition through inhalation. None of the steps is perfomed in confined space, watering helps to wash out the $\ce{HCN}$ which in water dissociates a bit ($\ce{HCN}$ is a weak acid).
Thus, digestion of the untreated roots is likely to be fatal, while the slow processing is apparently not.

Ok, I digged a bit and maybe you have you have access to this one:

Bala Nambisan
Strategies for elimination of cyanogens from cassava for reducing toxicity and improving food safety Food and Chemical Toxicology, 2011, 49, 690–693 [DOI]


Is there a formula to determine the total number of constitutional isomers? Is there a trick or a formula that given a molecular formula, allows you to know exactly how many constitutional isomers can be formed with that many atoms? Or is it more of a trial-and-error technique?

I'm looking for something along the lines of $2^n$ of maximum number of possible stereo-isomers, where n is the number of stereogenic centers present in a compound.

There seems to be some confusion here : in order to determine the number of stereo-isomers of a molecular formula (say C4H11N) you would need to first know the number of constitutional isomers as you need to know how many stereogenic centers across all the structures.

There are a bunch of existing questions on constitutional isomers (a, b, c, ...) but the summary is : "it is tricky without software for anything more than very small formulae".

There are mathematical formulae for determining the count of constitutional isomers, although the most powerful techniques are non-trivial (such as those based on the Pólya enumeration theorem). To efficiently list a non-redundant set of compounds from an elemental formula is possibly even harder, as it requires a mechanism to avoid duplicates in the output.


Cp-Cv For a gas following Van Der Waal equation of state How do you derive the relation

$$C_p-C_v= R\left(1 + \frac{2a}{RTV}\right)$$

for a gas obeying van der Waals equation of state? Any leads?

You could start with $C_V=(\partial U/\partial T)_V=T(\partial S/\partial T)_V$  and H and p instead of U and V for $C_p$ as appropriate. Then generate an expansion for S as $dS=(\partial S/\partial V)_T dV+(\partial S/\partial T)_VdT$ and differentiate wrt T. You should then get an expression in $C_V$ and $C_p$ plus other terms that you can find using the vdw equation.


Why is the thioester bond weaker than a regular ester bond? What are the contributing factors to the fact that thioesters are less stable than their oxygen counterparts?

Why is the thioester bond weaker than a regular ester bond?

Sulfur is a much bigger atom than oxygen or carbon. As a result, resonance donation from the sulfur into the carbonyl carbon is limited, and the C-S bond doesn't have as much multiple-bond character as the carbonyl carbon-oxygen bond in an ester.

  What are the contributing factors to the fact that thioesters are less stable than their oxygen counterparts?

Kinetically or thermodynamically?

Stable with respect to what?

If we're talking about thermodynamics, and stability with regard to hydrolysis, then yes, thioesters are generally less stable than esters.

It is more favorable for water to attack a thioester's carbonyl carbon because it has a greater partial positive charge than a corresponding ester's. This is due to the resonance issue discussed above. A thioester also has a more stable leaving group, as the larger size of sulfur allows negative charge to be distributed better.


What does my textbook mean by saying this about petroleum? It reads:

  Petroleum consists of crude oil and natural gas and contains a mixture of up to 300 hydrocarbons, including sulfur and nitrogen compounds. It is formed over thousands of years from the remains of fossilised marine organisms.

Is petroleum a single entity that is simultaneously a gas and a liquid oil compounds? How does that work? It is also a mixture of up to 300 hydrocarbons, so it is a liquid and gas in one? I don't understand? Do they mean that petroleum can be either gas OR it can be either liquid? What about the 'mixture of 300 hydrocarbons bit'. Does that mean you will never extract petroleum of a single hydrocarbon eg pure $\ce{C6H14}$ etc? I haven't revised this in a long time so I forgot..

Petroleum is a mixture of many different compounds.
Some of those compounds are gas.  Some are liquid.  Some are solid.
Many of them are hydrocarbons, and there are many different hydrocarbons included in the mix (e.g. different numbers on the C and H).
Petroleum is generally refined into more specific products before use, at a refinery.
Another example to consider might be "mud" - that's a mixture of some liquids, maybe some dissolved gases, and a bunch of different solids, usually including clay, maybe a bit of gravel, and some organic materials formed from decaying plants and animals.
If you take mud and compress it underground for thousands of years, then you'll have petroleum.

The second sentence seems clear enough.


Phosphoric Acid Problem What is the $\ce{[HPO4^2-]}$ of a solution labeled "$\pu{0.10 M}$ Phosphoric Acid"? Given the $K_\mathrm{a}$ values of $7.1 \times 10^{-3}$, $6.3 \times 10^{-8}$ and $4.2 \times 10^{-13}$ for the first, second and third ionizations, respectively.

Some sources have shown that the answer is $6.3 \times 10^{-8}$, but following through the long calculation through each ionization to get the equilibrium concentration of $\ce{HPO4^{2-}}$ I get an answer of $3.8 \times 10^{-5}$. Hoping someone can enlighten me on this matter.

Attempt at solving:

So first what I did was to individually find the concentrations using the given $K_\mathrm{a}$ values. Working from the top, I calculated for the concentration of $\ce{H2PO4-}$ from the given concentration of $\ce{H3PO4}$. Had to go quadratic with this one since the first ionization constant is a relatively big number. After getting the concentration of $\ce{H2PO4-}$, went on to get the concentration of $\ce{HPO4^2-}$.

After this one, used the final ionization constant to get the concentration of $\ce{PO4^3-}$ produced, which I found to be so small that the $\ce{HPO4^2-}$ from earlier isn't all that much affected. And after all that I get the answer of $\pu{3.8e-5 M}$ for $\ce{HPO4^2-}$.

Let's list what's going on here. Orthophosphoric acid dissociates stepwise:

\begin{align}
\ce{H3PO4 + H2O & <=>[$K_\mathrm{a1}$] H3O+ + H2PO4–} \tag{R1}\\
\ce{H2PO4- + H2O & <=>[$K_\mathrm{a2}$] H3O+ + HPO4^2–} \tag{R2}\\
\ce{HPO4^2- + H2O & <=>[$K_\mathrm{a3}$] H3O+ + PO4^3–} \tag{R3}
\end{align}

Mass balances for protons and phosporous, respectively:

\begin{align}
C_\ce{H} & = [\ce{H3O+}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + [\ce{H3PO4}] = [\ce{H3O+}] + \sum_{n=1}^3 [\ce{H_$n$PO4^$(3-n)-$}] \tag{1}\\
C_\ce{P} & = [\ce{PO4^3-}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + \ce{H3PO4} \tag{2}
\end{align}

Now we compare the equilibrium constants:

\begin{align}
K_\mathrm{a1} & = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.1 \times 10^{−3} \tag{3}\\
K_\mathrm{a2} & = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 6.3 \times 10^{−8} \tag{4}\\
K_\mathrm{a3} & = \frac{[\ce{H3O+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} = 4.2 \times 10^{−13} \tag{5}
\end{align}

Notice that $K_\mathrm{a1} >> K_\mathrm{a2}$ (more that $10^5$), therefore by the first dissociation step phosphoric acid behaves as an acid of medium strength, producing equal amounts of $\ce{H3O+}$ and $\ce{H2PO4-}$ (considering charge balance):

$$[\ce{H3O+}] \approx [\ce{H2PO4-}] \tag{6}$$

which directly leads us to the answer. Using expression for $K_\mathrm{a2}$:

$$[\ce{HPO4^{2-}}] = K_\mathrm{a2} \times \frac{[\ce{H2PO4-}]}{[\ce{H3O+}]} = K_\mathrm{a2} \tag{7}$$

That's it, $[\ce{HPO4^{2-}}] = \pu{6.3 \times 10^{−8} (mol/L)}$

Now for the fun part, if you really want to solve quadratic equation, we can also find solution pH and $[\ce{PO4^{3-}}]$.

Based on the values of equilibrium constants, it is safe to assume that pretty much only first dissociation step defines solution's pH. We can rewrite equation for $K_\mathrm{a1}$ as follows:

$$K_\mathrm{a1} = \frac{[\ce{H3O+}]^2}{C_0 - [\ce{H3O+}]},$$

where $C_0$ is the concentration of the solution ($\pu{0.10 M}$). Using the given numbers we have

$$7.1 \times 10^{-3} = \frac{[\ce{H3O+}]^2}{0.10 - [\ce{H3O+}]}$$
$$[\ce{H3O+}]^2 + 7.1 \times 10^{-3} [\ce{H3O+}] - 7.1 \times 10^{-4} = 0 $$
$$[\ce{H3O+}] = \frac{-7.1 \times 10^{-3} + \sqrt{(7.1 \times 10^{-3})^2 + 4 \times 7.1 \times 10^{-4}}}{2} = \pu{2.33 \times 10^{-2} (mol/L)}$$

$$\mathrm{pH} = - \log{[\ce{H3O+}]} = 1.63$$

WolframAlpha thinks the same.

Now we know $[\ce{H2PO4-}] = \pu{2.33 \times 10^{-2} (mol/L)}$ and we can finally find $[\ce{PO4^3-}]$ from the expression for $K_\mathrm{a3}$:

$$[\ce{PO4^{3-}}] = K_\mathrm{a3} \times \frac{[\ce{HPO4^2-}]}{[\ce{H3O+}]} = 4.2 \times 10^{−13} \times \frac{6.3 \times 10^{−8}}{2.33 \times 10^{-2}} = \pu{1.1 \times 10^{-18} (mol/L)}$$

This demonstrates that in pretty concentrated solution of phosphoric acid anions $\ce{PO4^3-}$ are practically non-existing.


Using Distribution Coefficient to Determine Mass and Percent of Acid Extracted The $K_\mathrm{D}$ (distribution coefficient) of coffee water at $25\,°\mathrm{C}$ is $7.8$ in a methylene chloride/water system. If there is $2.0\,\mathrm{g}$ of caffeine in $200\,\mathrm{mL}$ of water, calculate the grams and the % of caffeine that can be extracted with a single $100\,\mathrm{mL}$ portion of methylene chloride.

My attempt:

$$\frac{x}{\frac{2\,\mathrm{g}}{200\,\mathrm{mL}}}=7.8$$

$$\frac{x}{0.01\,\mathrm{g\,mL^{-1}}}=7.8$$

$$x=0.078\,\mathrm{g\,mL^{-1}}$$

$$x=0.078\,\mathrm{g\,mL^{-1}}$$

$$0.078\,\mathrm{g\,mL^{-1}}\cdot 100\,\mathrm{mL}=7.8\,\mathrm{g}$$

But I think I'm wrong. Am I supposed to maybe use the molar mass of caffeine ($194\,\mathrm{g\,mol^{-1}}$)?

You got it almost right ;-)

The distribution coefficient gives the ratio of equilibrium concentrations in a biphasic mixture.

It is crucial to realize that a caffeine concentration of $2.0\,\mathrm{g}$ in $200\,\mathrm{mL}$ of water is the initial concentration, but not the concentration after a single extraction step! The latter will be lower, since a part of the caffeine went into the organic layer.

It doens't matter whether you express the concentrations in $\mathrm{g\cdot L^{-1}}$ or $\mathrm{mol\cdot L^{-1}}$ since $K_D$ is a dimensionless constant.

$$K_D = \frac{{[\mathrm{caffeine}]_{\ce{CH2Cl2}}}}{{[\mathrm{caffeine}]_{\ce{H2O}}}} = \frac{x}{10-x}\ \mathrm{g\cdot L^{-1}} = 7.8$$

$$x = 7.8\cdot (10-x)\,\mathrm{g\cdot L^{-1}} = \mathbf{8.86} \,\mathrm{g\cdot L^{-1}}$$

Note that $x$ is the concentration in $\mathrm{g\cdot L^{-1}}$. In order to calculate the absolute amount of caffeine in the organic layer or the remaining amount of caffeine in the water phase, you have to consider the volumes.

By the way, there's an article by G.S. Klebanov, L.N. Mednikova, and A.D. Ovcharova on this topic: Extraction of caffeine from aqueous solutions, published in
Pharmaceutical Chemistry Journal, 1967, 1, 221-223 (DOI).


Normalized standard calibration curve I have a dynamic assay whose absorbance changes with time. I recorded it’s spectrum initially after preparing it. Then I added my analyte and mixed the contents. Then I waited for 2 min and recorded the second absorbance soectrum. I generated a normalized standard calibration curve by dividing the absorbance maximas (captured 2 min post analyte addition) at different analyte concentrations (1M, 2M, 3M...)  by an average value of control (initial absorbance maxima before analyte addition). Is this approach correct? My ordinate is normalized absorbance and abscissa is concentration of analyte.

I'm not sure whether calibration for a dynamic assay is any different from one that is stable. But let's think of the situation like this. What you have done is no different from what an operator would do to obtain a calibration curve with an internal standard (IS) correction (by plotting concentration against the ratio of the intensities of analyte/IS) except you have used the absolute intensity of your background signal (no analyte) instead of an internal standard for correction. This is not correct by any standard since the background is a bad choice for correction (the background is NOT stable, and has a terrible RSD due to low counts). What you should do is to plot your calibration curve without normalising it to the background.


Is [Co(NH₃)₄Cl₂]Cl paramagnetic or diamagnetic? $\ce{NH3}$ is known to be a strong field ligand, while $\ce{Cl}$ is known to be a weak field ligand.

Is $\ce{[Co(NH3)4Cl2]Cl}$ a high spin complex or a low spin complex?

I assumed this to be a high spin complex. Oxidation state of $\ce{Co}$ is $+3$.
And electronic configuration of $\ce{Co(III)}$ is $\mathrm{[Ar]~3d^6}$.

Since I assumed this to be a high spin complex — pairing of electrons of $\mathrm{3d}$ orbitals will not happen. The $6$ pairs of electrons from the ligands must be included in the $\mathrm{4s}$, $\mathrm{4p}$, $\mathrm{4d}$ orbitals — which leaves electrons unpaired in the $\mathrm{3d}$ orbital, making this a paramagnetic complex, with hybridisation $\mathrm{sp^3d^2}$.

Answer given to this question says otherwise. Hints are welcome!

This question is an excerpt of a question that appeared in the JEE(A) - 2016

The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $\ce{Co^3+}$ is $\mathrm{d^6}$, and virtually all "octahedral" $\mathrm{d^6}$ complexes are low spin - essentially some complexes of $\ce{Fe^2+}$  and a very small number of fluoro complexes of $\ce{Co^3+}$ are the only exceptions to the rule that all $\mathrm{d^6}$ octahedral complexes are low spin.

Why is this? Well let's use a bit of Crystal Field Theory (much of below stolen from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy). One can consider the crystal field stabilization energy which is how stabilized an ion is compared to the free ion when in a given electronic state electronic and geometry. Here we are interested in high and low spin, and octahedral geometry. Taking a table of these values from the above link:

In the above $\Delta_\circ$ is the splitting energy of the orbitals. We can see that low spin $\mathrm{d^6}$ has the largest possible stabilization energy of any electronic configuration compared to the high spin case ($2\Delta_\circ$-P). Thus unless the splitting is very small octahedral $\mathrm{d^6}$ prefers low spin. So how do we get small splitting to see high spin $\mathrm{d^6}$? Well small splitting is favoured by

Low charge first row transition metals - for instance $\ce{Fe^{2+}}$
Very weak field ligands

And what were our exceptions above?


Why does hydrogen burn spontaneously in fuel cell at room temperature? It's well known that if you mix $\ce{H2}$ and $\ce{O2}$ in ambient conditions, the $\ce{H2}$ will be stable indefinitely, and will only react with $\ce{O2}$ to form water when ignited by a flame. The reason being that for the reaction

$$\ce{2H2 + O2 -> 2H2O} \label{rxn:1}\tag{1}$$

to occur, first the bonds must be broken in $\ce{H2}$ and in $\ce{O2},$ which is an insurmountable activation energy at room temperature.

So I am confused why a hydrogen fuel cell works at room temperature without ignition, as it is powered by the same chemical reaction \eqref{rxn:1}.

In particular, why can the strong double bond in $\ce{O2}$ suddenly be broken in this context without a flame?

My best guess is that somehow the electron sea in the copper wire acts as a catalyst. If this is true, what are the details of this mechanism that allows the $\ce{O2}$ bond to break at room temperature?

You guessed it right that hydrogen and oxygen mixtures will remain stable. Someone in Harvard waited for >30 years and found very little water if any. As you said, there is a energy barrier. What type of materials can lower this barrier? Catalysts.

The electrode materials require some catalysts such as platinum or nickel. Of course, making cheaper and sustainable catalysts is an extensive area of research. If a person discovers a better catalyst which is far more common than Pt, then that person certainly deserves a big award. Copper does not act as a catalyst here.


Does 1,3-butadiene in bromine/water do addition twice? Considering 1,3-butadiene (below) added to bromine water (considering it as $\ce{Br2(aq)}$):

I've seen online that the products are 1,2-dibromobutene and 1,4

but what I don't understand is why wouldn't the other double bond undergo substitution again? They're on a carbon chain like normal so shouldn't they react with the bromine water too? Is something stopping it?

If reaction is carried out in 1:1 ratio of Br2 and 1,3 Butadiene, then the above two products will be formed. If more Br2 is taken, then surely 4 positions will be brominated.


which are the possible ways to detect impurity in drugs? I want some information regarding the protocol for determining any type of impurity in pharmaceutical substances or drug product.
Which type of impurities can be identified by HPLC and MS?
what is the difference between related substances and impurity?

There are a lot of types of impurity. Basically, much as any plant that's growing where you don't want it is a weed, any material in your sample that isn't what you want to have in it is an impurity.
Since you asked a question in simple terms I will attempt to answer it in equally simple terms:

Byproduct contamination - The theoretical reaction is usually not the only possible one, especially in organic chemistry such as pharmaceutical work. Areas of the reaction which didn't have just the right mixture of reactants, or wasn't quite at the right temperature, could have produced different products. Additionally, devations from the ideal when adding reactants can result in unreacted quantities of those reactants.
Various forms of chromatography, such as HPLC, CC, GC, MS, etc can detect and even separate these various byproducts from the ideal product. How effective any one of these techniques is depends on the type and combination of the products and byproducts. One of the most basic aspects of any drug research chemist's job is to isolate their product from all of the byproducts of their reaction.

Structural isomer contamination - Two compounds are structural isomers if they have the same molecular formula (number and type of atoms), but have different molecular structures. Tetramethylbutane, for instance, is based on a 4-carbon alkane, to which two additional methyl groups each are added to the carbons in the "2" and "3" positions of the alkane. The resulting molecule has all the same carbons and hydrogens as the totally linear octane molecule, but its structure causes very different behaviors (for instance, it has a much narrower range of temperatures at which it is a liquid).
Your average petrochemist doesn't care too much about the difference between any of the structural isomers of these hydrocarbon alkanes; their primary reaction of interest, combustion, only cares about the required stoichiometric ratio of fuel to air, and the resulting rate and heat of combustion. However, such isomers pose problems in pharmacology for the above reasons; a reaction that was theorized to produce a molecule with one arrangement may in fact produce an entirely different isomer, with similar physical properties as the desired molecule, but much different reaction chemistry including biochemical effects. Chromatography methods based on molecular weight or physical properties like boiling point or solubility may or may not be able to tell you whether you have the right isomer, but other methods might show a difference in the physical properties of the chemical allowing separation.

Chirality and Racemization - Even if two isomers have all the same atoms, and each atom is bonded to the "correct" adjoining atoms, the molecules can have different shapes because the atoms are bonded to different relative positions of their adjoining atoms. These "spatial isomers" or "stereoisomers" have many forms, from structures that exhibit the same "mirror-image" symmetry that your left and right hand do (enantiomers), or that have certain functional groups transposed by rotation around a "chiral center" (diasteromers, including cis-trans and most detro-levo isomers). Again, like structural isomers, these stereoisomers can cause problems for pharmacologists because the different isomers can have different reactive properties in the complex biochemical soup that is the human body.
Sometimes these chiral isomers can be separated from each other; enantiomers and other spatial isomers are often "optically active", meaning they rotate polarized light in opposing directions depending on their chiral arrangement. Certain "chiral resolving agents" can also separate these isomers based on a differing affinity for one isomer over another, and these agents are often used as "stationary phases" of column chromatography including HPLC that can separate and thus "purify" the desired isomer from its chiral mirror. However, a few troublesome spatial isomers will "racemize"; that is, they will readily rearrange between the two stereoisomers when given the right environment, and so any attempt to produce a pure sample of one enantiomer is ultimately pointless.


How to calculate the amount of citric acid to neutralise the pH within wastewater? We have an effluent tank, with a consent to discharge at between 5.5 - 11 pH, the tank holds 100000 liters of dirty water.

When the pH reaches 10.5 we add citric acid to bring the pH down. How many liters of citric acid with a pH of 2.2 should we add to the system to bring the pH down to around 7. We always end up putting too much in then add caustic to bring it back up.

Gotta start with my favorite formula:
$$Q = CV$$

$$C_\textrm{eff}V_\textrm{eff}+C_\textrm{acid}V_\textrm{acid} = C_\textrm{out}V_\textrm{out}$$

$$(10^{-10.5}~\mathrm{mol/L})(10^5~\mathrm{L})+(10^{-2.2}~\mathrm{mol/L})V_\textrm{acid} = (10^{-7}~\mathrm{mol/L})(10^5+V_\textrm{acid})$$

$$V_\textrm{acid} = 1.51~\textrm{L}$$


Why do liquids and solids have constant concentrations? I don't understand why liquids and solids have constant concentrations. Can anyone explain this to be in simple terms?

If you think about what concentration, it usually refers to the molar concentration in $\mathrm{mol/l}$ (at least in the field of chemistry). To calculate this concentration we need to find how many moles of a certain chemical are in 1 liter of the liquid or solid.

I will start with an example and describe the general principle afterwards. Let's say we have pure water at room temperature and atmospheric pressure. This means that the density is $998\,\mathrm{g/l}$. We also know that the molar mass of water, consisting of 2 hydrogen and 1 oxygen, is $18.016\,\mathrm{g/mol}$. If you look at the units of the molar mass and the density you can already see that we can combine these into something that has units $\ce{mol/l}$. If we do that we get that the concentration of water is $\frac{998\, \mathrm{g/l}}{18.016\,\mathrm{g/mol}}=55.4\,\mathrm{mol/l}$. So the concentration of water is $55.4\,\mathrm{mol/l}$. Notice that I have only used the density and the molar mass of water, which are (at least for equal temperature and pressure) constant.

In general you can do this for any solid or liquid (or gas) and find the concentration as $c=\frac{\rho}{M}$, where $\rho$ is the density, $M$ the molar mass and $c$ the molar concentration. For a given chemical, the molar mass is constant.

Now about the ‘fact’ that the molar concentration is constant for liquids and solids: strictly speaking it is not. That is because the density of liquids and solids is temperature and pressure dependent. The reason that the molar concentration is often called constant is twofold:

the density of liquids and solids has a much weaker dependence on the temperature than gases do therefore it can be regarded as approximately constant
if you look at solutions, the variation of concentration can be
orders of magnitude, whereas for e.g. water the concentration
variation when going from $0$ to $100\,\mathrm{^\circ C}$ is only a few percent.
Therefore, again, the concentration of the liquid is approximately
constant.


How to make a CCl3F Fluorine NMR standard? I am running some fluorine NMR and would like to make a bulk solution of $\ce{CCl3F}$ in $\ce{CDCl3}$ to aid in assessing the amount of fluorinated product present by having my NMR solvent act as a reference. Can anyone suggest a good ratio of the $\ce{CCl3F}$  standard to $\ce{CDCl3}$ NMR solvent?

You need the concentration of your reference peak to be the same order of magnitude as your solute. Accuracy of your integration depends on something called dynamic range, which refers to the relative intensities of the peaks being integrated. The greater the dynamic range; the greater the difference in peak intensities, the bigger the error.

You might want consider that for most medium-high field NMR applications, CDCl3 is used with TMS at concentrations of about 0.03%v/v. Low field and bench top applications use TMS at 0.5-1%v/v concentration.