Adding hydrogens to multiple pdb files using Openbabel GUI software I am  completely new to using OpenBabel software.

I have about 9000 pdb files to which, I want to add hydrogens. It will be a very hectic task to open all of them 1 by 1 and then convert each of them. Is there any way I could convert all of them simultaneously.

Also, can you please tell me how to correctly select the output file? It asks for a destination file to which, it appends/overwrites the output. Is there a way I could choose the destination where it would create another .pdb file and save the output in it? I guess I am missing something here while choosing the output file.

Thanks in advance. Also, I asked a similar question before which was closed. Considering I am new with OpenBabel, if any further information is required or if there is some issue, please tell me so that I may improve my question.

If you got an installation of openbabel's GUI, you equally have an installation of openbabel for the terminal (e.g., in Linuxes) / command line cmd.exe (in Windows).  There is nothing wrong using the GUI for an edit on one file or a few data -- your instructions set are right -- however using the CLI is closer to the engine, thus often more powerful.

Enter the directory containing the .pdb to work with from the command line.
The instruction babel *.pdb -opdb *_h.pdb -h will request babel to look for all *.pdb there and write new .pdb with hydrogens added.  Each file by name in pattern of example.pdb will have its edit recorded in a file named in pattern of example_h.pdb.  While the CLI will report the progress of this conversion, I recommend to slice the task into smaller batches than just one of 9k.  How many .pdb per batch depends on the computing power available on your machine.
The new files written are both discernable by name and time stamp from the retained input.  So you have two criteria to sort and discern the files.  Assuming you subsequently moved the new ones into their own folder, you may rename them en bloc back to the pattern of example.pdb.  If you work with Windows, a program like Ant Renamer may considerably accelerate this task.

Commands and grammar openbabel uses are documented in depth.  For commands accessible from the CLI, just a few are compiled here.  As perspective: note its use may be automated (scripted), for example with Python like the portable WinPython.


GGA functionals requiring second derivatives? In "Lecture Notes in Quantum Chemistry II" from the European Summer School in Quantum Chemistry, the authors note that when discussing the LYP functional and other GGA functionals, that

Thus one can immediately see that an evaluation of $v_{xc}$ demands an evaluation fo the second derivatives of basis functions.
(pp102).

Indeed, when (for a functional like B88) I start with the functional itself and compute the functional derivative (either by hand or with a CAS), I arrive at a (messy) expression that involves second derivatives with respect to space. However, libraries like libxc only need scalar functions of the gradient to be passed, in particular $\nabla \rho \cdot \nabla \rho$ , in order to compute $v_{xc}$. Now libxc contains the expressions for its functionals in Maple and then performs symbolic differentiation on them and then uses the maple-to-C translator; thus the actual C code in libxc is uninterpretable by a human.
Is there a known expression for $v_{xc}$ for B88 or another GGA-like functional that might give me a hint as to how to simplify the functional derivative to only depend on these parameters?

It turns out that my original reference is now somewhat out of date, and it is quite possible to compute the relevant properties without resorting to second spatial derivatives. [Johnson1993] has a nice overview of the relevant math, including clean derivations of both YLP and B88 functional terms, and states:

Note that Eq. (7) does not require evaluation of the spin density second de- rivatives as is necessary in many other formulations, a major computational advantage [48]. The fact that these can be avoided has been previously noted by Kobayashi et al.

Johnson, B. G., Gill, P. M. W., & Pople, J. A. (1993). The performance of a family of density functional methods. The Journal of Chemical Physics, 98(7), 5612–5626. https://doi.org/10.1063/1.464906
and
Kobayashi, K., Kurita, N., Kumahora, H., & Tago, K. (1991). Bond-Energy calculations of Cu2, Ag2, and CuAg with the generalized gradient approximation. Physical Review A, 43(11), 5810–5813. https://doi.org/10.1103/PhysRevA.43.5810


What solutes will give the largest reduction in volume when dissolved in water This question is inspired by another question.

If you take some common salt (NaCl) and some water and mix them, then the volume of the solution is less than the volume of the constituents.

I did some calculations with;

5.75 L salt
44.25 L water
total initial volume 50 L

Density of salt is 2.170 kg/L so resulting solution is 22% concentration, which has a density of 1.164 kg/L. Therefore resulting volume is 48.73 L. This means the solution has a volume 1.27 L lower than the constituents.

My question is in two parts;

Are my calculations correct?
Is there another solute that would give a solution with volume 4 L lower than the constituents, given the volume of the constituents must be 50 L.

(1) Your calculations are correct.

(2) $\ce{CaCl2}$ has a density of 2.16 kg/L

           kg        vol(L)
  CaCl2    18.00     8.38
  H2O      41.62    41.62
           -----    -----
           59.62    50.00 L

But when dissolved the solution has a density of 1.300 kg/L, so the total volume is 45.86 L. (Data about $\ce{CaCl2}$ solutions.)


Does the carbon dioxide dianion exist? I have heard that the dianion of carbon dioxide, $\ce{CO2^{2-}}$ was named carbonite.

Does such a species really exist? Basically, is it stable enough to be a plausible molecule? If so, under what conditions is this true?

This question is very confusing to me, I understand the concept somewhat but I don't know what information I need in order to figure out the answer.

The obvious Star Wars puns aside, I have never heard about the dianion of a carbon dioxide dianion and/or that it was named carbonite.
However, the carbon dioxide radical anion, $\ce{CO2^{.-}}$, is experimentally available.
Typically, it is generated by hydrogen atom abstraction from formate, $\ce{HCOO-}$, using $\ce{HO^.}$ radicals. The $\ce{HO^.}$ radicals, in turn, are obtained from vacuum ultraviolet (VUV) irradiation of water at a wavelength of $\lambda = 175~\mathrm{nm}$.

UPDATE
Apparently, I was wrong about the dianion! As Mithoron pointed out in his comment, species like $\ce{(M^+)_2CO2^{2-}}$, formed in the reaction carbon dioxide with alkali metals in argon or nitrogen matrixes, have been detected by IR spectroscopy. See Inorganic Chemistry, 1984, 23, 177-184 for details.


Sulfur Doped Titanium Dioxide I'd like to know why my $\ce{TiO2}$ sol gel did not get coated nicely on my 200 gsm fiberglass? Here my solution, 5 g of tween 80, 40.15 ml isopropanol, 3.4 ml titanium isopropoxide, 0.62 ml sulphuric acid (0.02 mol).  Beside that, my $\ce{TiO2}$ sol gel can't calcinated at 500°C.  It will turn into powders and get burned into a black colour.  What should I do to overcome this problem?

Regarding why your ceramic material did not get coated nicely on the fiberglass, I think there's a problem of adherence between the fibers and the the ceramic. You need to prepare the surface of the fibers by etching for example before coating. They are often shielded.
Regarding why your  sol-gel can't be calcinated at $500°\mathrm{C}$. You need to figure out the proper temperature of calcination by monitoring the phase appearance of the ceramic materials by XRD diffractometry. You start with the XRD diffractogram of your as-prepared sample(normally, it should be amorphous). Then, you treat your sample at different  temperatures within different periods of times (you have two parameters to find: Temperature and Time). Unless you are working on a published article that has addressed all these issues.


Reduction of methylene blue Why does methylene blue (which is initially blue) reduce to form a colorless solution?

I was told by my teacher that it has to do with the presence of an additional double bond on methylene blue, but I can't figure out or find any source on the internet that claims so.

The color change results from the reversible oxidation-reduction reaction of the methylene blue indicator.  In alkaline solutions, glucose is oxidized to D-gluconic acid or alpha-D-gluconolactone.

$\ce{HOCH2(CHOH)4CHO + 3OH- -> HOCH2(CHOH)4CO2 + 2H2O + 2e-}$

In the course of this reaction, methylene blue is reduced from the blue (oxidized) form to the colorless (reduced) form.

Shaking the flask dissolves $\ce{O2}$ in the solution, which oxidizes the indicator back to the blue (oxidized) form.


Can thionyl bromide produce alkyl bromides? Thionyl chloride ($\ce{SOCl2}$) is used with alcohols to produce alkyl chlorides. Similarly, can thionyl bromide ($\ce{SOBr2}$) be used with alcohols to produce alkyl bromides?

Yes, thionyl bromide can be used with alcohols for synthesis of alkyl bromides.
No, the process is not as efficient.  The reasons for the inefficiency include are primarily (a) thionyl bromide is much less stable than thionyl chloride, and (b) that a bromide-tribromide redox equilibrium is far more relevant than a chloride/trichloride equilibrium, which leads to undesired side pathways instead of the desired reaction.

For more information, see this 1955 paper from Frazer & Gerrard.

  On the other hand, pyridinium tribromide is readily formed during the addition of thionyl bromide to base hydrobromide, and its intervention weakens the analogy between the thionyl bromide and the thionyl chloride system.

In the same paper, a yield of 75% is reported for the synthesis of 2-bromobutane from butan-2-ol using pyridinium tribromide (which itself would be synthesized from thionyl bromide and pyridinium bromide).


Is it possible for an acid and base to react to form an even stronger base or acid? Is there a combination of an acid and a base that when mixed results in a solution that is more basic or more acidic than the initial base or acid, respectively?

If so, do such combinations have a name?

$\ce{6HF + Al(OH)3 (s) -> H3[AlF6] + 3H2O}$
$\ce{6HCN + Fe(OH)2 (s) -> H4[Fe(CN)6] + 2 H2O}$
$\ce{TlOH + HOCl + H2O -> Tl(OH)3(s) + HCl}$

May be one of these will work. Even if they don't, you should get the idea where to search.


Water becomes cold on mixing energy drink I noticed that the temperature of water decreases noticeably when I mix an energy drink (Glucon-D) in it. The ingredients mentioned on the pack are majorly glucose (52%) and sucrose (45%). I believe that this is due to the inversion of sucrose being an endothermic reaction
$$\ce{C12H22O11 + H2O + \Delta -> \underset{glucose}{C6H12O6} + \underset{fructose}{C6H12O6}}$$
However, I could not find any sources on the internet to validate this. Please confirm/disprove this hypothesis and also explain the correct reason for this.

EDIT: After everyone's recommendations of sticking a thermometer in my drink, I did so and came up with the following observations:
\begin{array}{|c|c|c c|c|}
\hline
m_\text{drink} & V_\ce{H2O} & T_i & T_f & \Delta T\\
\hline
\pu{14.85 g} & \pu{200 ml} & \pu{20.3°C} & \pu{19.5 °C} & \pu{0.7 °C}\\
\hline
\end{array}

Some elementary thermochemistry gives me the result that $\Delta H_\text{solution}$ for the energy drink is $\approx \pu{39.6 J g-1}$.

Well, the solution enthalpy of sugars is positive.
I found these numbers on the internet

$\ce{C12H22O11}$ (sugar(sucrose)) : 5.4 kJ/mol
$\ce{C6H12O6}$ (glucose) : 11 kJ/mol
$\ce{C6H12O6·H2O}$ (glucose monohydrate) : 19 kJ/mol

So if your "energy drink" is a dry powder (and not a readymade drink in an aluminum can), this could explain your observation.

You should however put a thermometer into your experiment, and get us some numbers. As is, my above is just another piece of guesswork.


How can an aromatic ketone be reduced in the presence of a nitro group? Is it possible to reduce an acyl group to an alkane, but still keep a nitro group intact without its reduction to an amino group? A Clemmensen reduction would reduce the nitro group, and it seems like other reducing agents will also reduce the nitro group to an amino group. Would some other mechanism be better suited for this reaction?

For instance, how can I selectively reduce 1-(3-nitrophenyl)propan-1-one to 1-nitro-3-propylbenzene?

You will most likely be successful if you convert the ketone to the corresponding tosylhydrazone and reduce that with $\ce{NaCNBH3}$ (DOI).

A previous study (Bull. Chem. Soc. Jpn., 1974, 47, 2323-2324) has shown that the tosylhydrazone of cyclohexanone can be reduced to cyclohexane with $\ce{NaBH4}$ in polar aprotic solvents.

$\ce{NaCNBH3}$ is an even milder hydride transfer agent, that has been widely used in the reduction of imines and iminium salts. A review by C.F.Lane, Sodium Cyanoborohydride - A Highly Selective Reducing Agent for Organic Functional Groups in Synthesis, 1975, 3, 135-146 gives some background information on the versatility of this reagent.

The nitro group is definitely safe with $\ce{NaCNBH3}$. However, considering the high toxicity and the price for this reagent, you might want to run a small scale reduction of your tosylhydrazone with the much cheaper $\ce{NaBH4}$ first.


Thermodynamic Derivation of Ideal Gas Law The ideal gas law is perhaps the best-known equation of state, and admits both a derivation via the kinetic theory of gases and via statistical mechanics. But these are both microscopic theories, and the ideal gas law is a macroscopic equation.

Can the ideal gas law be derived from macroscopic thermodynamics alone, without recourse to microscopic theories? If so, how? If not, why not?

Assume now that $$U = \frac{3}{2}Nk_\text{B}T,$$ as is the case for a monatomic ideal gas.

Can we now derive the ideal gas law?

Perhaps something to consider regarding the second question. From thermodynamics, $$P = -\left(\frac{\partial U}{\partial V}\right)_{S,N} = -\frac{3}{2}Nk_\text{B}\left(\frac{\partial T}{\partial V}\right)_{S,N} \stackrel{?}{=} \frac{Nk_\text{B}T}{V} \quad \Longleftrightarrow \quad \left(\frac{\partial T}{\partial V}\right)_{S,N} \stackrel{?}{=} -\frac{2}{3}\frac{T}{V},$$ where $\stackrel{?}{=}$ is read "is to be shown equal to." This is an interesting identity, I suppose, but I'm not sure if or where I can proceed from here. As usual, any help is appreciated.

Classical thermodynamics does not admit that molecules exist thus if the gas law is derived it must be without recourse to any molecular ideas regarding the nature of matter.

The chemical potential of a perfect pure gas at temperature T and pressures p and $p^{\mathrm{o}}$ is defined as
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p/p^{\mathrm{o}})$$
We can choose $p^{\mathrm{o}}$ to be unity then
$$ \mu = \mu^{\mathrm{o}} + RT\ln(p)$$
This is still dimensionally correct as $p^{\mathrm{o}} = 1  \pu{~bar}$ and as we have chosen the temperature, $\mu^{\mathrm{o}}$ is the standard chemical potential which is a function of temperature only.

Differentiating the chemical potential at constant T gives
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T = \frac{RT}{p}$$

To find another expression for the derivative we can use
$$dG=-SdT+ Vdp +\Sigma\mu_idn_i$$
 for $dn_i$ moles of species i gives. For a pure species we can drop the subscript i and replace the summation with $\mu dn$ then
$$ \left(\frac{\partial \mu}{\partial p}  \right)_T=\frac{\partial}{\partial n}\left(\frac{\partial G}{\partial p}  \right)_T= V_m$$
where $V_m$ is the volume per mole.

Comparing these two derivatives gives $V_m=RT/p$ or
$$pV=nRT$$

Thus the definition of the chemical potential implicitly contains the perfect gas law.


Chiral carbon in carvone molecules This is the structure of carvone, with the chiral carbon noted by *.

I don't understand why this is a chiral carbon, to my understanding a chiral carbon is a carbon with 4 different chemical groups attached to it.

In the molecule, the chiral carbon is surrounded by $\ce{H}$, $\ce{C(CH3)=CH2}$ and two $\ce{CH2}$ groups.

Why are both $\ce{CH2}$ groups different and thus making that carbon chiral?

If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!

From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.

Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!


Estimation of pressure and kinetic energy density of stellar interior using kinetic theory of gases In Physical Chemistry we are working a problem that is confusing. We started by calculating the pressure half way to the centre of the sun assuming that the interior consisted of ionized hydrogen atoms. We determined this to be $\pu{3.59e6 kPa}$. However, the third part of the question asks:

  What is the kinetic-energy density half way to the centre of the Sun? Compare with the (translational) kinetic energy density of Earth's atmosphere.

Can someone expand on what kinetic energy density is, if it is different from kinetic energy, and what the formula is. I might just be overthinking the problem.

I would like to first point out that the pressure you've determined is too low, most likely because you used only molar mass of protons $\ce{H+}$ instead of average mass for star matter in order to fulfill the condition of plasma quasineutrality:

$$\bar{M} = \frac{M(\ce{H+}) + M(\ce{e-})}{2} = \frac{\pu{(1 + 0) g mol-1}}{2} = \pu{0.5 g mol-1} = \pu{5e-4 kg mol-1} \tag{1.1}$$

Referring to the question from Atkins' Elements of Physical Chemistry [1, p. 40], where this question has been taken from, we obtain necessary data to complete the solution:

  In the Sun [...] the density is 150 times that of liquid water at its centre and comparable to that of water about half-way to its surface

and

  (a) Calculate the pressure half-way to the centre of the Sun, assuming that the interior consists of ionized hydrogen atoms, the temperature is $\pu{3.6 MK}$, and the mass density is $\pu{1.20 g cm−3}$ (slightly higher than the density of water)

the pressure of star plasma (that can be treated as an ideal gas) can be found as following:

$$p = \frac{\rho RT}{\bar{M}} = \frac{\pu{1.2e3 kg m-3} \cdot \pu{8.314 J K-1 mol-1} \cdot \pu{3.6e6 K}}{\pu{5e-4 kg mol-1}} = \pu{7.18e13 Pa} \label{eqn:1.2}\tag{1.2}$$

Now for the second part, kinetic energy density is defined as kinetic energy per volume:

$$\rho_\text{k} = \frac{E_\text{k}}{V} \label{eqn:2.1}\tag{2.1}$$

Kinetic energy for a set of entities (ion, atom, molecule) is

$$E_\text{k} = \frac{Nm\bar{v^2}}{2} \label{eqn:2.2}\tag{2.2}$$

where $N$ is the number of particles, $m$ is the particle's mass, $\bar{v}$ is an average (root mean square) particle velocity. From the kinetic model of an ideal gas, the pressure obtained from \eqref{eqn:1.2} is also

$$p = \frac{Nm\bar{v^2}}{3V} \label{eqn:2.3}\tag{2.3}$$

Equation \eqref{eqn:2.3} implies $Nm\bar{v^2} = 3pV$, so kinetic energy from \eqref{eqn:2.2} can be rewritten as $E_\text{k} = \displaystyle\frac{3pV}{2}$. Finally, kinetic energy density, taking \eqref{eqn:2.1} into account, is

$$\rho_\text{k} = \frac{3p}{2} = \frac{3 \cdot \pu{7.18e13 Pa}}{2} = \pu{1.08e14 J m-3} \quad (\pu{1 J} = \pu{1 Pa}\cdot \pu{1 m3})$$

Comparison between kinetic energy densities $\rho_\text{k}(1/2r_\odot)$ (halfway to the Sun's center) and $\rho_\text{k}(\text{atm})$ (Earth's atmosphere) is really a matter of comparing the pressures:

$$\frac{\rho_\text{k}(1/2r_\odot)}{\rho_\text{k}(\text{atm})} = \frac{p(1/2r_\odot)}{p(\text{atm})} = \frac{\pu{7.18e13 Pa}}{\pu{101325 Pa}} = 7.09\cdot10^8$$

References

Atkins, P. W.; De Paula, J. Elements of physical chemistry, 5th ed.; Oxford University Press: Oxford; New York, 2009. ISBN 978-1-4292-1813-9.


Is the vapor pressure of a liquid dependent on the pressure of the atmosphere in which it is measured? The figure below shows how vapor pressure of a liquid is measured. Initially the flask is at atmospheric pressure. Then the liquid is dropped in the flask. Once equilibrium is reached, the pressure from the vapor created is recorded. My question is - if the flask were initially at a different pressure - let's say 5 atm - would the vapor pressure of the liquid be different? My thinking is that the additional pressure in the flask would keep more of the liquid molecules in liquid form.

Yes, in principle it could be different, but the effect would be very small unless the increase in pressure were very large.  To put it in simplified terms:  The increase in ambient pressure compresses the liquid, bringing it to a higher-energy state, causing a higher proportion of the liquid molecules to want to escape from the liquid, and into the gas state, thus causing the vapor pressure to increase.  This assumes ideal gas behavior, i.e., that the increased ambient pressure affects only the liquid, not the gas.

DETAILS:

Let's call your liquid "$X$".  At equilibrium, the chemical potential of $X_{(l)}$ equals the chemical potential of  $X_{(g)}$.  The pressure of $X_{(g)}$ at which that occurs is called the vapor pressure.

Now suppose you increase the ambient pressure at constant temperature. Consider what happens to the chemical potential of the liquid: $d\mu = -S_mdT +V_m dP$, where $\mu$ is the chemical potential, and $S_m$ and $V_m$ are the molar entropy and volume, respectively.  Thus, at constant $T$, $d\mu = V_m dP$.  Hence the chemical potential of $X_{(l)}$ increases.  This effect, by itself, would cause the vapor pressure to increase (to get the chemical potential of $X_{(g)}$ to match the increased chemical potential of $X_{(l)}$).

But what about the chemical potential of $X_{(g)}$?  If the gas behaves ideally, then it is unaffected by the presence of other gases, and thus its chemical potential is unaffected as well.  I.e., as with the liquid, its chemical potential is determined by its pressure.  But unlike the case with the liquid, the pressure of $X_{(g)}$ is independent of the ambient pressure -- it depends only on the partial pressure of $X_{(g)}$ itself.  Thus the only effect to consider would be the increased chemical potential of the liquid, and the vapor pressure would increase.

However #1: At high concentrations, gases tend not to behave ideally.  Thus there are two possibilities: the increased partial pressure of the other gas increases the chemical potential of $X_{(g)}$, or decreases it.  If it's the former, and the effect is sufficiently strong, the vapor pressure will decrease.  If it's the latter, the vapor pressure will increase.

However #2: If the ambient gas is soluble in the liquid, increasing the ambient pressure will increase the concentration of the ambient gas in the liquid, which could reduce the chemical potential of the liquid, thus reducing the vapor pressure!

Again, let me emphasize that these are very small effects.


Is the pyrolysis of tetrachloroaluminate(III) anion possible? Wikipedia says that lithium tetrafluoroborate decomposes to lithium fluoride and boron trifluoride when heated. Does the same hold for sodium tetrachloroaluminate (all anhydrous) as well?

Unlike lithium tetrafluoroborate $\ce{LiBF4},$ sodium tetrachloroaluminate $\ce{NaAlCl4}$ doesn't decompose upon melting, and doesn't show any signs of decomposition or phase transition up to approx. $\pu{800 °C}$ [1, p. 688]:

FIGURE 3. Calculated $(\ce{NaCl + AlCl3})$ phase diagram at $\pu{0.1 MPa}$ (dotted lines are liquid–liquid miscibility gap boundary and monotectic at $P > \pu{0.1 MPa}).$

Molten sodium tetrachloroaluminate is used as electrolyte in ZEBRA batteries with normal operating temperature range of $\pu{270–350 °C}.$
References

Robelin, C.; Chartrand, P.; Pelton, A. D. Thermodynamic Evaluation and Optimization of the $(\ce{NaCl + KCl + AlCl3})$ System. The Journal of Chemical Thermodynamics 2004, 36 (8), 683–699. https://doi.org/10.1016/j.jct.2004.04.011.


What are the ways to make radioactive water drinkable again? Let's say survivors in the area contaminated by radioactive fallout need to somehow process the water from rivers, rain or pipes. How could they go about it? Is it even possible to 'remove radioactivity' from the water?

The water itself isn't radioactive, rather particles dissolved in it, because water is a quite good solvent. Therefore, filtration (e.g. with activated carbon) or precipitation can remove such particles. Also a ion exchanger can remove radioactive species. But this can take quite long, especially if the needed amount is large.


How does Grignard reagent form? So I've read that during the formation of the Grignard reagent, The R-X bond cleaves homolytically forming free radicals.
My question is how is this happeneing? Due to the high electronegativity difference between Carbon and Halide group shouldn't it result in R+ and X- ?
To be clear, I have no problem following through how the electrons move around. My only doubt is how exactly the Alkyl-Halide bond resulted in a free radical instead of an ion.

The electronegativity difference between carbon and the halogen used for Grignard reagents is actually not all that high.  Alkyl fluorides are generally un-reactive towards magnesium, and other halogen whose alkyl compounds are more reactive have electronegativity ranging from only a few tenths above carbon to a few tenths below carbon.

In any event the neutral alkyl halide molecule does not directly cleave into either radicals or ions.  According to Wikipedia, the first stage of the reaction involves a one-electron reduction of the alkyl halide to a radical anion; the magnesium is oxidized to a radical cation.  The alkyl halide anion, not the original molecule, cleaves off its halide ion to form an alkyl radical.  Thus, in effect, after the initial electron transfer you get "a little bit of both" -- the alkyl group is a radical but the halogen (even with its modest electronegativity) is an anion.  The alkyl radical, the magnesium radical cation, and the previously displaced halide ion then combine to get the Grignard reagent molecule (the combination process finishes off the oxidation of magnesium to the +2 state).  At no point in the given mechanism do you have a free radical halogen atom.


How to explain disagreement between Le Châtelier's principle and the simplified Gibbs free energy equation? For example, for the dissolution of a salt in water that is exothermic, heating the solution would drive the reaction towards the solid form of the salt according to Le Châtelier's principle.

However, according to the simplified Gibbs equation $(\mathrm dG = \mathrm dH - T\mathrm dS),$ the reaction would be spontaneous towards the products due to the increase in entropy and the negative enthalpy.

To me, these seem to contradict each other. How is this reconciled? Am I making a mistake somewhere?

The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ >0$ and $\Delta G^\circ<0$ must have a larger $K$ at a higher temperature because $\Delta G^\circ$ is more negative. If that were true, we would have a relationship of the form $\Delta G^\circ = -cK$, where $c$ is a constant. Instead, the key relationship is

$$\Delta G^\circ = -RT\ln K.$$

So $\Delta G^\circ$ is proportional to $-T\ln K$. In the case above where $\Delta G^\circ <0$ and $\Delta S^\circ>0$, as T increases, $\Delta G^\circ$ increases in magnitude (becomes more negative), but so does $-RT$, so we don't necessarily need to have a larger $K$ to satisfy the equation. To figure out the temperature dependence of $K$, we need to substitute $\Delta G^\circ$ with $\Delta H^\circ - T\Delta S^\circ$ and then rearrange things:

$$\Delta H^\circ - T\Delta S^\circ=-RT\ln K$$

$$\frac{\Delta H^\circ}{T}-\Delta S^\circ=-R\ln K$$

From that equation, hopefully it is clear that if $T$ increases (which reduces the magnitude of the $\frac{\Delta H^\circ}{T}$ term), $K$ will only increase if $\Delta H^\circ > 0$. If $\Delta H^\circ < 0$, K will have to decrease with increasing $T$ to maintain the equality. $\Delta S^\circ$ is a constant term that does not affect the change in $K$. Thus, our result is completely consistent with both Le Chatelier's principle and with the van't Hoff equation analysis.


Which are the differences between resonance Raman and fluorescence? If the fluorescence is the re-emitting of a photon with a larger wavelenght due to the transition from a higher energy state to a lower energy state the case of resonance Raman (where there aren't any virtual states involved) seems be equal to the fluorescence. Which differences are there?

The main difference between Raman scattering and fluorescence is the excited state lifetime. Fluorescence excited states are longer-lived than the 'virtual' states associated with Raman scattering.

In fluorescence, absorption of light excites an electron to a higher energy state. The lifetimes of these excited states are long enough where the geometry of the molecule can relax to accommodate this new electron configuration. However, this new geometry is not equivalent to the lowest energy structure, so when the molecule returns to the ground state, the energy of the system will be slightly higher than where it started. This means that the fluorescence energy is always lower than the excitation energy (this is true for molecules, not atoms).

For Raman scattering, light absorption promotes the molecule to a very short-lived 'virtual' excited state. When this state relaxes, it can return to a different vibrational or rotational sublevel of the ground state than where the molecule started. This leads to slightly different energies than the elastically scattered light, which is at the same energy as the incoming radiation source.

Resonance Raman is a special case where the excitation energy is tuned to a particular electronic transition of a molecule. This has the advantage that vibrational transitions associated with the electronic change are substantially enhanced compared to other Raman transitions, which simplifies the Raman spectrum. Because you are exciting a specific electronic transition, though, it is possible to get a substantial fluorescence background as well.

This J Chem Ed article has a nice description of Raman versus fluorescence. ChemWiki has a page devoted to the differences between Raman and Resonace Raman.


The effect of the groups at para position on SN2 reactivity Compare the SN2 reactivity order for:

The benzyl group stabilises the transition state of SN2 reaction using the π system of the benzene ring for conjugation with the p- orbital of the electrophilic carbon which is electron deficient. (reference: Clayden, page 341)

So in my opinion, groups increasing the electron density of the ring should make the SN2 reaction go faster as compared to groups decreasing it.

Thus the order should be $2 > 3 > 1 > 4$ but the answer given in my worksheet is the exact opposite of this. I would like to know the reason for that and the flaw in my reasoning.

This is what I found in March:

  Holtz and Stock showed that electron-withdrawing groups increase the
  rate of SN2 reactions. This can be ascribed to stabilization of the
  transition state by withdrawal of some of the electron density.

I find this contradictory to the content given in Clayden that the TS is "electron deficient". In March its also given that for alkyl groups at para position the Baker Nathan order is followed (methyl > ethyl > t-butyl). (Hyperconjugation is most in methyl so it increases electron density of the ring) (Another  apparent contradiction)

So how exactly do I determine the effect of the p-substituents in SN2.

This is what I found from your reference to Clayden. You just had to read the next paragraph ( I have highlighted the point ). You can see that March and Clayden have agreed on the same point, that electron withdrawl from the phenyl/allyl group increases the effective conjugation in the transition state.
So I hope You can understand what will be the correct order and why.

P.S. I know this question is old, but it was unanswered, and someone may benefit from an answer.

Reference: Jonathan Clayden, Nick Greeves, S.Warren, P.Wothers -Organic Chemistry (2000) - Pg 422.


What is the Natural abundance of Potassium The three naturally occurring isotopes of potassium are $\ce{^39K}$ ($38.963707\:\mathrm{amu}$); $\ce{^40K}$ ($39.963999\:\mathrm{amu}$); and $\ce{^41K}$.

  The percent natural abundances of $\ce{^39K}$ and $\ce{^41K}$ are $93.2581\%$ and $6.7302\%$ respectively.

  a) What is the natural abundance of $\ce{^40K}$?

  b) Determine the isotopic mass of $\ce{^41K}$.

So I started to utilize this equation:

$\mathrm{atomic\:mass = (fractional\:abundance\:isotope\:1
\cdot mass) + (fractional\:abundance\:isotope\:2 \cdot mass)}$

But then I realized that I did not have some values. $\ce{^39K}$ is the only isotope which has two values. $\ce{^40K}$ only has the mass weight, however, if I were to multiply $\ce{^40K}$'s mass with ($1-0.932581$) would that help me arrive at an answer?

Also, would that answer be accurate since it only takes into account that there are only to values that add up to 100%.

A). If you know the percentages of the abundance of $\frac{2}{3}$s of the isotopes then finding the percentage of the $3^{rd}$ is as simple as

$ 100\% - (\%_{\ce{^39K}}+\%_{\ce{^41K}}) = \%_{\ce{^40K}} $

B). Finding the mass of the isotope should be pretty simple. It would be

$(\mathrm{Mass\ of\ proton} \cdot \mathrm{{\#\ of protons}} )+(\mathrm{Mass\ of\ a\ neutron} \cdot \mathrm{{\#\ of neutrons}} ) \approx \mathrm{Mass\ of\ isotope}$

You neglect the weight of electrons because the mass is small relative to the others.


How to calculate the amount of substances product produces from the masses and amounts of starting materials? Calculate the amount of carbon dioxide formed when we burn 1 mole of carbon in the presence of 16 g of dioxygen.
  $$\ce{C + O2 -> CO2}$$

1 mole of carbon means 12 g. Therefore amount of carbon dioxide is 12 + 16 = 28. Am I correct?

From the equation,
$$\ce{C + O2 -> CO2},$$
we can see that one mole of $\ce{C}$ reacts with one mole of $\ce{O2}$.

Therefore, let us calculate the amount of each reactant present.

The amount of $\ce{C}$ is already given: $n(\ce{C}) = \pu{1 mol}$.

The molar mass of $\ce{O2}$ is $M(\ce{O2}) = \pu{32 g//mol}$, so $\pu{16 g}$ of $\ce{O2}$ are $\pu{0.5 mol}$.

Since one mole of $\ce{C}$ only reacts with one mole of $\ce{O2}$, there remain $\pu{0.5 mol}$ of unreacted $\ce{C}$, while $\pu{0.5 mol}$ of $\ce{C}$ react with $\pu{0.5 mol}$ of $\ce{O2}$.

In formal terms, oxygen is called the limiting reactant as it limits the amount of product formed.

From the equation, one mole of $\ce{CO2}$ is produced per each mole of $\ce{C}$. Therefore, $\pu{0.5 mol}$ of $\ce{CO2}$ is produced.


Degenerate orbitals in the Hydrogen atom I came across the following link:
Which orbitals of the hydrogen atom are degenerate for $n=3$

And all the answers said that for the hydrogen atom, the energy of an orbital depends only on n. Is this really true? And also, isn't it true that for other atoms (at least according to those who answered the other question) the energy depends upon both $l$ and $n$?

It is true that if you solve the Schroedinger equation the solutions depend only on the principal quantum number $n$ and the energies (often called eigenenergies or eigenstates) have the values predicted by the Bohr formula, $E=-hcR_\infty /n^2$ and energy differences as predicted by the Balmer formula. The calculated energy does not depend on $l$; this degeneracy of wavefunctions with different $l$ is a peculiar and special feature of the symmetry of the Coulomb potential.

However, nature cares little for our equations and experiments show that in fact the energy of the $l$ levels are different in energy to one another. This is called 'fine structure' and the energy shifts are very small as the name suggests. The first cause is explained as being due to the fact that the electron has an intrinsic quantum property called 'spin', which does not mean that it is literally spinning, but crudely speaking that it can be described by the same equations as angular momentum. As the electron is charged its spin produces a magnetic field  which interacts with the magnetic field of its orbital motion, which in turn depends on $l$. This interaction is called spin-orbit coupling, and technically it is a relativistic effect.The second effect is called the Darwin term, for electrons with $l=0$, and the third effect is the Lamb Shift, which needs explanation beyond relativistic effects and requires Quantum Electrodynamics.

The fine structure due to spin-orbit coupling can be observed easily in many atoms so the $l>0$ levels are split for each $n$ and so are no longer degenerate. The common example is the two lines of Sodium 'D' emission. The difference is small, the two transitions are 589.59 nm to 588.99 nm. This is the yellow colour associated with sodium easily seen if you sprinkle salt into a gas cooker's flame and in some types of street lights. The splitting can be measured with a simple spectrometer and is a common teaching lab experiment. The other effects mentions are far,far more difficult to measure and in a Chemistry course you are unlikely ever to meet them.


Is there an abbreviation for "something bulky"? I want to draw a scheme that indicates the difference between the products formed in a reaction with a molecule that will form one thing when it has methyl or hydrogen substitution, but another thing when it has any substituent more bulky than isopropyl.

Is there any abbreviation (like "R" or "X") for "bulky"?

I've never come across one, but mostly those I come across will have "R" or "X", but somewhere in the reaction scheme specify what R is, like:

R = Me, Et, Pr, iPr...

So in your case, you could specify something like:

R = iPr or bulkier


What is the mass fraction of trichloroethene in the liquid phase using vapour pressure and mole fraction in gas phase? Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene ($\ce{C2HCl3}$) and trichloromethane ($\ce{CHCl3}$) is established at $\pu{25 °C}$. The mole fraction of $\ce{CHCl3}$ in the vapour phase is $0.73$. What is the mass fraction of $\ce{C2HCl3}$ in the liquid phase?  Round your answer to two significant figures.

The vapour pressures of trichloroethene and trichloromethane at $\pu{25 °C}$ are:

$$\begin{align}P_\text{vap}(\ce{C2HCl3}) &= \pu{73.0 mmHg}\\[0.5em]
P_\text{vap}(\ce{CHCl3}) &= \pu{199.1 mmHg}\end{align}$$

So, what I did was I found mole fraction of $\ce{C2HCl3}$ and then used the two mole fractions along with the vapour pressures to find the total pressure of the solution.

$$P_\text{vap}= \frac{0.73}{199.1}+\frac{0.27}{73} = \pu{165.053mmHg}$$

Then, from Raoult's Law I know that the mole fraction in liquid phase is equal to mole fraction in vapour phase, multiplied by vapour pressure, divided by total pressure. From that, I found the mole fraction of both things in liquid phase. I use the mole fraction to find mass of both, and then did mass of $\ce{C2HCl3}$ divided by the total mass that I calculated. I got an answer of

$\frac{15.68}{120}=0.13$ but it says that it's wrong. I'm not sure where I messed up?

As far as the question is concerned, you do not need the vapor pressure. In order to calculate the mass fraction of $\ce{C2HCl3}$, we first have to calculate the molar mass of $\ce{C2HCl3}$ and $\ce{CHCl3}$.
$$M(\ce{C2HCl3}) = \pu{131.38 gmol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 gmol-1}$$
With these two calculated we can calculate the absolute mass relative to $\pu{1 mol}$, due to the fact of $\frac{\pu{0.73 mol}}{\pu{1 mol}}$ being relative to a total of $\pu{1 mol}$ of molecules. Now if we calculate the absolute masses, we get:
$$M(\ce{C2HCl3}) = \pu{131.38 gmol-1} * \pu{0.27\frac{mol}{mol}} = \pu{35.47 gmol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 gmol-1} * \pu{0.73\frac{mol}{mol}} = \pu{87.14 gmol-1}$$
And the total mass in the system results in:
$$m_\text{total}=\pu{35.47 gmol-1}+\pu{87.14 gmol-1}=\pu{122.61 gmol-1}$$
And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows:
$$w(\ce{C2HCl3})=\frac{\pu{35.47 gmol-1}}{\pu{122.61 gmol-1}}=0.29$$
$$w(\ce{CHCl3})=\frac{\pu{87.14 gmol-1}}{\pu{122.61 gmol-1}}=0.71$$


Good acid for cleaning marine toilets Problem:
On my boat I have a marine toilet. It is operated with salt water.
The problem is, on the interior of the outflow tubing it accumulates some sort of hard grime. I am looking for a good way to dissolve this grime as it is impossible to replace the tubing.
From experience it is known that citric acid work - but not very well. Also it leaves a "sandy" residue that is difficult to flush.
Does anybody have an idea of what this "grime" consists of - and thus how best to dissolve it?
Materials:
On hand I have:

citric acid
56% phosphoric acid
23% HCl

Materials:

tubing: PVC
toilet itself: porcelain
pump: ABS or polypropylene with selected parts in acetal resin, with 316 passivated stainless steel fastenings, brass weights and neoprene seals and gaskets
valve: grey, harder plastic

I know this quite vague but am hoping somebody can help anyway.

A cheaper and easier to find solution than EDTA (permeakra's answer) is citric acid. Citric acid reacts with metal carbonates (bicarbonates, phosphates, chlorides, etc.) to produce the soluble citrates. The citrate ion keeps the metals in solution by chelation, the same process that EDTA (ethylene diamine tetra acetate) does.

Citric acid can be purchased in the grocery store or convenience store as one of the principle components for any sour powdered drink mix (like Tang).

You could also use any substance or cleaner designed to remove hard water deposits or lime scale, such as CLR, which contains lactic and gluconic acids as chelating agents.


Why do gummy bears explode when added to hot potassium chlorate? This link shows that a gummy bear explodes when in contact with heated potassium chlorate, $\ce{KClO3}$. But what in a gummy bear creates this reaction?

Also, do other foods (fruit, icing sugar...) react as violently with potassium chlorate?

Potassium chlorate is a source of oxygen. After heating, it decomposes to $\ce{O2}$ and $\ce{KCl}$:
$$\ce{4 KClO3 → KCl + 3 KClO4}$$
$$\ce{KClO4 → KCl + 2O2}$$

The gummy bear is mainly composed of sugar and other carbohydrates. Those carbohydrates will react with oxygen, combustion occurs. For example, glucose will react in this manner:
$$\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O}$$

If there is any material present which does not burn, such as $\ce{H2O}$, the temperature will not rise as high. For gummy bears the reaction works spectacularly because they are mainly carbohydrates (>70%).

An apple, for example, has only ~13% carbohydrates – unless you dry it, of course. On the other hand, this video on YouTube is an example of how sugar itself reacts violently with potassium chlorate.


How did they assign absolute configuration to these cis and trans 2-methylcyclohexanols? From [1, p. 56]:this is the 4/B rule I was talking about
This is how I practiced. In the textbook, the absolute configuration of one of the enantiomers (I am referring to the the first compound) is written as 1R,2S. But when I tried to redo it, I arrived at 1R,2R. I didn't understand where I went wrong. I think the way I am visualising the compound is not correct.

4.3.5 Nomenclature of polysubstituted cyclanes
Polysubstituted cyclanes, e.g., cyclopentanes, cyclohexanes etc. exist in a number of stereoisomers (diastereomers and enantiomers) depending on the number and nature of the substituents. The disubstituted cyclanes give two diastereomers which can be conveniently and unambiguously described as cis and trans isomers, e.g., cis- and trans-2-methylcyclohexanols (XXVII) and (XXVIII) (Figure 4.13). Each of the diastereomers in turn gives two enantiomers which can be named, following the R, S nomenclature, as R, S and S, R for the cis and S, S and R, R for the trans isomer (use 4/B rule) as shown in the Figure.

Figure 4.13 Disubstituted cyclohexanes

When I practice on this picture I always arrive at the same configuration both on C1 and C2. Am I assigning the priorities in a wrong way?
Reference

Nasipuri, D. Stereochemistry of Organic Compounds: Principles and Applications; New Age International, 1994. ISBN 978-81-224-0570-5. (Google Books)

You have somehow decided that the front side of the ring is actually the rear side … or something. So your drawings show exactly the wrong stereochemistry. For the hydroxy-carbon, you thus would, in theory, get the incorrect assignment (S) but you use the rule you reference to reverse that into (R). Naturally, because your initial assertion of which atoms are in which plane was incorrect, your application of the rule was also incorrect. For the methyl carbon, your inverted ring reproducibly gives you the inverted assignment (R) and you do not apply any incorrect correction.

Figure 1: the molecule in question.

In figure 1, I have reproduced the molecule in the more ‘standard’ chair configuration which gives a better approximation of the bond angles. The wedged/bold portion of the ring is in front of the paper plane, the thin solid portion behind. I have also rotated the ring by $60°$ so that the hydroxy function is on the rightmost carbon atom for easier discussion.

It should be a rather simple exercise to determine that the hydroxy-carbon (carbon 1) indeed does have an (R) configuration: the hydroxy group has highest priority, the front side of the ring second-highest, the back side third and the hydrogen pointing downwards fourth. Therefore, we must look from above the ring plane to put the lowest substituent to the back; the rotation from hydroxy to methyl side to non-methyl side is thus clockwise, therefore (R).

The same analysis performed on carbon 2, the methylated one, should give you:

carbon 1 ($\ce{CHOH}$) is highest priority; going to the back right and slightly downwards
carbon 3 ($\ce{CH2}$) is second; going to the left, slightly downwards while staying in the plane
the methyl group is third; going up while staying in the plane
the unmarked hydrogen is fourth; this goes to the front right and slightly downwards.

Thus, we should look from inside the ring as the smallest substituent is pointing towards us. However, we can also look from the smallest substituent and invert the assignment; the way we see the three groups is again clockwise, so if we were to look from inside the ring (or behind it) we would see them anticlockwise, thus (S).

Note that the wedged and bold bonds — both in the paper’s drawing and in my figure — are not there to indicate a positioning from the stereocentre but to indicate which half of the ring is in front of the paper plane and which is behind. Thus, you cannot — like you did — assume the bolded bond to be in front; it actually is in your paper plane when you are looking at the two stereocentres on either side of it.


Why does saliva clean so well? I notice that saliva seems to clean well, much better than plain water. For example, tar spots on the car can be difficult to get off. I typically use really powerful solvents to get them off, like 95% acetone and things like that. Yet, saliva works too.

Why is this?

Saliva has alpha-amylase (ptyalin), which hydrolyzes carbohydrates (so starch starts to taste sweet in just a few seconds), kallikrein, an enzyme that breaks down protein, and lingual lipase, which helps disintegrates fats (lipids).

So next time you clean off a splattered insect from your wind-screen (windshield, should you prefer), think of it as external digestion, as practiced by flies and by starfish. Or perhaps you might not want to consider that thought...


Clausius–Clapeyron relation for states of water and vapor I've seen many questions about this topic on this site, but I still have some issues trying to understand this aspect. Here's what I believe I understood
looking at the phase diagram for a pure substance (water):

If you put water in a container with a piston pushing $\pu{1 atm}$ directly on the water with no extra space for inert gases, it will stay 100% liquid as at room temperature and $\pu{1 atm}$ liquid is the stable phase.
If the water is put at room temperature in a closed box where the volume is more than the volume of the water (creating a vacuum), the molecules will evaporate in a number so that the water vapor will be at the equilibrium pressure corresponding to room temperature, which can be calculated by the phase diagram.

I can't understand why the air (and its pressure) doesn't work like a piston, making sure no water is in gas state at $\pu{1 atm}$.

In particular, there was this exercise where a piston with $p = p_\mathrm{tot}$ was pushing down on a mixture of vapor $\ce{A}$ and inert gas above pure liquid $\ce{Ar}$, so that:

\begin{align}
p_\mathrm{tot} &= p_\mathrm{inert} + p_\mathrm{a}\\
\text{Equilibrium:}\qquad \ce{A(l, $p_\mathrm{tot}$) &= Inert(g, $p_\mathrm{inert}$) + Vapor(g, $p_\mathrm{a}$)}
\end{align}

Why doesn't the vapor $\ce{A}$ 'feel' the $p_\mathrm{tot}$ and get condensed into liquid $\ce{A}$?

Was it because $p_\mathrm{tot}$ is less than the vapour pressure of pure $\ce{A}$ at that temperature? If it was $p_\mathrm{tot} = p_\mathrm{vap}$ ($T = T_\mathrm{exercise}$), then the equilibium would be

$$\ce{A(pure, l, $p_\mathrm{tot}, T$) = Inert(g, $T, p_\mathrm{tot}$)?}$$

This doesn't look like an everyday case to me though where water is in gas form even if the pressure is $\pu{1 atm}$.

The answer may be somewhat counterintuitive. Yes, an inert gas does very much exert pressure on a solid or liquid. The problem is that the chemical potential of the substance in the gas phase is not affected by the inert gas (if the gas is assumed ideal).
Assume you have a substance (liquid or solid) in a rigid isothermal but otherwise empty box. The result is as expected evaporation or sublimation of the substance until the pressure of the vapour has increased to an equilibrium value (the saturation vapor pressure). We assume there is plenty of substance so that some remains in the condensed phase while the rest fills the box as vapour. What drives evaporation/sublimation? The fact that the chemical potential of the substance is (initially) lower in the vapour. Once equilibrium is achieved the chemical potential is the same in both phases.
Now assume an alternate scenario: in addition to the substance (solid or liquid) there is now some inert gas in the box. As suspected this gas exerts a pressure on the liquid and thereby affects its chemical potential (free energy). But the free energy of the substance in the gas phase is unaffected by the inert gas (assuming the gases behave ideally). And here the somewhat counterintuitive result: the substance again evaporates/sublimes until reaching the saturation vapour pressure, but precisely because the inert gas exerts a pressure on the condensed phase and increases its chemical potential in this phase, but does not alter the chemical potential of the vapor, it increases the saturation vapor pressure. The increased vapour pressure raises the chemical potential of the vapour until it matches the heightened potential of the condensed phase.
And the short answer to the question is that the inability of vapor A to 'feel' the total pressure including the inert gas is part of the definition of an ideal or perfect gas, which fails if the pressure is raised sufficiently.


Aqua regia and nascent chlorine Aqua regia has the famous property of being able to dissolve noble metals, like gold and silver. I read that this has to do with the generation of nascent chlorine. How does this happen and how does it help to dissolve such metals?

Aqua regia is a concentrated mixture of nitric and hydrochloric acids. The two acids react to generate soluble nitrosyl chloride, water and nascent chlorine: $$\ce{HNO3 + 3HCl -> NOCl + 2H2O + 2Cl}$$

Secondly, the gold is oxidised by nitric acid to gold(III) ions: $$\ce{Au <=> Au^{3+} + 3e-}$$

This reaction is normally highly unfavourable and so very few gold(III) ions are formed if you put gold into only concentrated nitric acid. However, the nascent chlorine is highly reactive as it has an unpaired electron in a high energy orbital and it is able to react with the gold(III) ions to form the soluble tetrachloroaurate(III) complex. This removes the gold(III) ions from the equilibrium, driving the reaction to completion.

Overall the reaction is: $$\ce{Au + 3HNO3 + 4HCl -> [AuCl4]- + 3NO2 + H3O+ + 2H2O}$$


Why are magnesium chloride and calcium chloride more soluble than sodium chloride? I read that $\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ in water. Solubility of $\ce{MgCl2}$ is $\pu{543 g/L}$ and that of $\ce{NaCl}$ is $\pu{360 g/L}$ (both at $20^{\circ} \pu{C}$).

I think that $\ce{NaCl}$ should be more soluble due to its higher ionic nature. $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ are more polarizing and have more covalent nature and thus should be less soluble.

I want to know the reason for why the opposite is happening.

I don't think there is any simple answer.There are some rules of thumb, all of which have exceptions. Most chlorides are soluble. That includes both MgCl2 and CaCl2.

Solubility is the result of the balance between competing interactions: between the ions in the salt and between those ions and the water.

Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.

The longer answer is to do with something called the "enthalpy of solvation".'

you can think it this way that the lattice enthalpy in MgCl2 and CaCl2 will be less and can be easily be overpowered by hydration enthalpy


Why are Kc and Kp interchangeable in the free energy-equilibrium equation? In this equation that relates Standard free energy and the equilibrium constant:

$\Delta Gº=-RT \ln K_{eq}$

My textbook (and my teacher) say both $K_p$ (constant related to pressure) and $K_c$ (constant related to concentration) can be used in the place of $K_{eq}$.

When using the equation that relates $K_p$ with $K_c$

$K_p=K_c(RT)^{\Delta n}$

We would get two different $\Delta Gº$ values, which shouldn't be possible.

My question is, why can we use both constants if their values are different?

The equilibrium constant is defined by activities:

The activity of a species in a liquid solution can be approximated to: $a=c/c^0$ where c is the concentration of the species and $c^o$ is the standard concentration ($1M$). This approximation leads to $K_{eq}=K_c$. The activity of gasses is called fugacity and it can be approximated to: $f=p/p^0$ where p is the pressure of the gas and $p^0$ is the standard pressure ($1$ $bar$). This leads to $K_{eq}=K_p$. Therefore, if a reaction takes place in e.g. water, one should use $K_c$, but if the reaction takes place in the gaseous state, $K_p$ should be used when calculating $dG$.


Whose solubility increases with increase in temperature? Solubility of which among the following substances in water increases slightly with rise in temperature?
(a)$\ce{KBr}$
(b)$\ce{KCl}$
(c)$\ce{KNO3}$
(d)$\ce{NaNO3}$

Answer is (b).
I don't know why.
According to me if temperature is increased then Bonds between any of these four must be broken due to supplied energy & solubility must decrease. And what about the other three. Do they too also show the same trend.

All four substances have higher solubility at higher temperatures in the range from 0 to 100 degrees Celsius.

$\ce{KBr}$ and $\ce{KCl}$ both roughly double their solubility from 0 to 100 degrees Celsius.

To me the question does not make sense.


Equivalent Cyclohexane Structures? Are these structures equivalent?

Image 1:

http://i.minus.com/jnojtCx6wYvqD.png

Image 2:

Yes, the two structures you've drawn are the same; both substituents are in axial positions.  The two structures shown in the link you provided are not the same; they are conformational isomers.  The one on the left is the same as the one you've drawn with both substituents in the axial position.  The one on the right has both substituents in the equatorial position.  These two conformational isomers exist in equilibrium with the (e,e) isomer favored.  The two conformers interconvert by a cyclohexane ring flip.


Why is benzoic acid a stronger acid than acetic acid? Why is benzoic acid ($\mathrm{p}K_\mathrm{a} = 4.20$) a stronger acid than acetic acid ($\mathrm{p}K_\mathrm{a} = 4.76$), even though the conjugate base in case of benzoic acid is destabilized due to electron donation through resonance. Whereas there is nothing of that sort operating in acetic acid; only inductive and hyperconjugative electron donation.

It is all about the electron withdrawing/ donating nature of the acid's functional groups.  The more electron withdrawing the group, the stronger the acid will be.  The more electron donating the weaker the acid will be.  A phenyl ring is electron withdrawing, while a methyl group is electron donating.  I don't know why resonance would destabilize an organic molecule or ion.  if you are referring to the carboxylate donating electrons to the phenyl ring, I don't see how this could be possible and not violate the octet rule.


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

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


Achiral universe - asymmetric synthesis Would it be somehow possible to conduct asymmetric synthesis if all chiral molecules would exist only as racemates in nature?

Yes! There is a simple possibility. Take one racemate and do what Pasteur did:

be lucky, in that both enantiomers crystallise separately;
be lucky again in that they form macroscopically distinguishable crystals (in this case probably: mirror images);
collect the different enantiomers. Voilà, you have enantiopure substances and can use them in synthesis.

(Thankfully, using tartaric acid immediately gives access to the Sharpless asymmetric epoxidation.)

Once you have any single enantiopure compound, you are able to generate further ones. For example, reacting a racemate with an enantiopure compound in a way that preserves the respective stereocentres will result in diastereomeric products. Diastereomers can theoretically always be separated (it may be harder for some and some may interconvert under ambient conditions but the theory stands). You could also generate diastereomeric salts, by taking an enantiopure anion (say (R)), and a racemic cation. You will receive the diastereomers (R,S) and (R,R) which can then be separated and so on and so forth.


Where does Oxygen come from in KO2 reaction with CO2? $\ce{KO2 + CO2}$ gives $\ce{K2CO3 + O2}$

I think it should be from $\ce{KO2}$ as $\ce{2KO2}$ can split to $\ce{K2O}$ and $\ce{3/2O2}$. This $\ce{K2O}$ being unstable reacts with $\ce{CO2}$ to form $\ce{K2CO3}$.

Only problem is, is the first step possible?
Is $\ce{O2}$ released from $\ce{KO2}$ or $\ce{CO2}$?

Yes, the oxygen indeed comes from the decomposition of potassium superoxide ($\ce{K2O}$) but the other decomposition product is potassium peroxide ($\ce{K2O2}$) and not potassium monoxide($\ce{K2O}$).(Here)

$$\ce{2KO2 ->[290°C, vacuum] K2O2 + O2~~~~~~~~~~(1)}$$
Now, this potassium peroxide reacts with carbon dioxide to form potassium carbonate and oxygen.(Here)

$$\ce{2K2O2 + 2CO2 → 2K2CO3 + O2~~~~~~~(2)}$$

Adding $\ce{(1)\times2}$ and $\ce{(2)}$, we get full reaction. (Here)

  $$\ce{4KO2 + 2CO2 → 2K2CO3 + 3O2}$$

If reaction proceeds in aqueous medium, it gives potassium hydrocarbonate and oxygen.(Here)

$$\ce{4KO2 + 4CO2 + 2H2O → 4KHCO3 + 3O2}$$


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

Answer choices:

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

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

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

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

EDIT

Deprotonation and subsequent benzylation of the $\beta$-ketoester occur at the most CH-acidic position, i.e beween the keto and the ester group. As a side note, benzylation at the methyl group adjacent to the carbonyl group is possible when using 2 equivalents of a strong base.

Using ethoxide rather than methoxide as a base makes sure that transesterification does not lead to a mixture of products.

The resulting benzylated $\beta$-ketoester is cleaved under alkaline conditions to the corresponding carboxylate. Under acidic conditions, this from a $\beta$-ketoacid.

Upon heating, $\beta$-ketoacids undergo decarboxylation (= they loose $\ce{CO2}$) to form an enolate which tautomerizes to the final methyl ketone.


Is there a free online resource where I can get the solubility of a solute in a solvent? I'm looking for a online resource that lets me find the solubility at of any solute (e.g Choline Chloride, Sodium Hydroxide, Dextrose) in any solvent (e.g. Ethylene Glycol, Oleic Acid, Isopropanol, Water). Preferably, I'd like a webpage that lets me enter solute and solvent, then gives the answer, but even just lookup tables would be helpful.

You need to narrow down your question.  Currently, the potential solutes include NaOH (inorganic salt), dextrose (organic neutral compound), and choline chloride (organic quaternary ammonium salt) on one hand, and on the other hand a wide range of potential solvents.  It is unlikely a single database will (ever) cover such a  broad range of possibilities.

Nevertheless, there are some resources that may be useful to consult.  For example solubility tables like this or the properties tables for some of the compounds in wikipedia, like here.  If you know both solvent, and solute, NIST-IUPAC's solubility database here may be worth to visit.

If your institution / library provides access to Reaxys/Beilstein (a commercial database by Elsevier), you may find solubility data in the category of physical properties. This database covers both organic and inorganic (including metal organic) compounds and will provide you both the numerical value, as well the primary reference.


Transition metal and complexation When can we say that such metal forms a complex?

Are $\ce{Pd(OAc)2}$ and $\ce{Pd(PPh3)2Cl2}$ complexes?

Both of these compounds are coordination complexes. The label complex is preferred by many chemists because despite the appearance of some of these compounds as simple ionic compounds, they often display much more complex bonding motifs.

For example in $\ce{Pd(PPh3)2Cl2}$, there are two neutral ligands, $\ce{PPh3}$ that interact with the metal center through dative bonds. There is also evidence that the $\ce{Pd-Cl}$ bond is more covalent than ionic, since palladium forms a complex ion $\ce{PdCl4^-}$ with chlorine.

Palladium acetate man seem like a more clear cut case until you look at the crystal structure. Each acetate anion bridges two palladium, centers in a polymeric chain:

  It is worth noting that IUPAC does not recommend using the term complex to describe these compounds given that complex can refer to a number of other, more loosely bound, species. IUPAC prefers coordination entity, though I have never heard a practicing chemist or seen a textbook use that phrase.


What drives a molecule/atom towards stability? I'm very familiar with the concept that chemical reactions occur because the reactants a state of stability via the reaction. It's a very satisfying and fulfilling statement. But my question has always been - "How do the molecules know what to do? Do they have a certain amount of intelligence? Are they living?" I'm really sorry that the question isn't properly framed because this topic is just a mess inside my mind.

It's like the reactants already know what to do in order to attain stability.

The question arises because we are taught chemistry in a way relatable with moral values and actions humans perform under a given circumstance. So who is to be frowned upon - the way we are taught, or the lack of understanding within the students?

Thanks for going through the mess :)

It boils down to thermodynamics. No, molecules do not possess intelligence. But they respect the laws of thermodynamics, which in a general look state:

Energy of a closed system is constant.
The entropy of the universe is always increasing.
You can't get to absolute zero temperature! (Ok, this one is less important for this analysis)

The second law is the key: in order to augment the entropy of the universe a certain transformation can do one of two things: increase its own entropy or the entropy of its surroundings. The latter can be achieved by releasing heat, which is something exothermic reactions do. The sum of these effects is described by the system's Gibbs free energy:

$$G = H - TS$$

For a chemical reaction occurring at a constant temperature and pressure, $G$ has to decrease and becomes minimized at equilibrium. We then say the system is (thermodynamically) "stable". So when you hear someone say "chemical reactions minimize their energy", it would be better spoken as "... minimize their free energy".

Now you might wonder why the second law is true. Why does entropy has to increase? Why does the free energy has to be minimized? It certainly doesn't want to.

The truth lies in the molecular interpretation of entropy: it measures the amount of micro-states at which system can be found. Put together a bunch of blue balls with another bunch of red balls in a box, shake it all and look at how they end up. They surely won't be found in separate sides of the box, they will mix! The scenario where they are separated corresponds to a very low-entropy state: you can think of the red ones to one side or vice-versa: not that many possibilities of arrangements. The mixed state though has a much higher entropy: think of how many arrangements are possible by exchanging the positions of red and blue balls!

Since molecules in a closed system are very very much like tiny balls being shaken inside a box - not by a mechanical source but by their own thermal energy -, their behavior with respect to entropy is the same. Again, it's not that they want to increase the entropy of universe, there just is a higher probability that they do rather than don't. Given that we are usually observing systems of sextillion molecules or more, statistics kicks in and tell us probabilities mean more or less certainty: we are pretty much certain that chemical reactions will keep on trying to maximize the entropy of the universe.


Hydroxyl group to free amine in polyethylene glycol using CDI? For converting $\ce{OH}$ of methylated polyethylene glycol. The first step is converting hydroxyl group to imidazole carbamate as an active intermediate. But what about the next step?

If I will add primary amine containing group to form carbamate linkage, how can I convert it to a free amine?

Reference: Zhou, S.; Wang, J.; Wang, G.; Jiang, Z.; Ren, H. An approach to developing high dielectric constant nanocomposites based on polyrotaxane derivative. European Polymer Journal 2017, 90, 312-322. DOI: 10.1016/j.eurpolymj.2017.03.037.

Conversion of Hydroxyl group to free amine in polyethylene glycol: This is a case of converting a primary alcohol to a primary amine

$$\ce{R-CH2OH -> -> -> R-CH2NH2}$$

To my understanding, converting hydroxyl group to imidazole carbamate as an active intermediate is not the way to do this. First thing come to my mind was a named reaction called the Mitsunobu reaction. The Mitsunobu Reaction allows the conversion of primary and secondary alcohols to esters, phenyl ethers, thioethers and various other compounds including amines. The reaction uses triphenylphosphine ($\ce{Ph3P}$) and diethylazodicarboxylate ($\ce{DEAD}$) to activate hydroxyl group. Then, activated alcohol can undergo substitution reaction with a suitable nitrogen nucleophile (either phthalimide or hydrogen azide can be used) followed by reduction (in the case of azide formation) or acid or base hydrolysis (if phthalimide were used as nucleophile) to give primary amine groups.


When calculating enthalpy changes of reaction using, why is only the amount of substance of one reactant used? When calculating the thermal energy released during an acid-base neutralisation using the equation $Q = mc\,\Delta T$, and then using $\Delta H = - Q/n$ to calculate the enthalpy change of neutralisation, why is the amount of substance of one reactant used ($n$ in the second equation) rather than the total amount of substance of acid and base reacting together?

Consider the following reaction ($\Delta H_{\text{rxn}}$ is the enthalpy change for one mole of the reaction, or the amount of moles of each substance found in a balanced equation: in this case, 1 mole $\ce{A}$, 1 mole $\ce{B}$):

$$\ce{A->B}\qquad\Delta H_{\text{rxn}}=~?$$

You are given 5 moles of $\ce{A}$, and you measure the change in heat to be $\pu{-100 kJ}$. However, this $\pu{-100 kJ}$ is actually just the $\Delta H$ of this reaction: $$\ce{5A->5B}\qquad\Delta H=\pu{-100~kJ}$$

Looks the same, right? That's because it is the same. However, if you want to calculate the enthalpy change of the first reaction, you are going to have to divide the measured enthalpy change, $\pu{-100 kJ}$, by 5. Thus, the $\Delta H_{\text{rxn}}=\pu{-20kJ}$.

Now consider this:
$$\ce{A +B->C}\qquad\Delta H_{\text{rxn}}=~?$$
If you were again given 5 moles of $\ce{A}$, along with 5 moles of $\ce{B}$, and you measured the $\Delta H$ to be $\pu{-100 kJ}$, you would get something similar to the first scenario:
$$\ce{5A + 5B -> 5C}\qquad\Delta H=\pu{100kJ}$$

Looks familiar right? Now, if you wanted to calculate the $\Delta H_{\text{rxn}}$, you wouldn't divide 100 by 10; you would divide it by 5, since the second reaction is just $5x$ the first one.

Side-note: This same reaction would be written if we have more than 5 moles of $\ce{B}$. Since $\ce{A}$ and $\ce{B}$ react in a 1 : 1 ratio, adding more $\ce{B}$ than $\ce{A}$ would mean that there is excess $\ce{B}$.


What purpose do chain reactions serve? I just learned what chain reactions are, but I can't seem to wrap my head around them because it seems utterly useless and doesn't accomplish anything. At least that's how I understood it because the textbook doesn't have any real world applications for it.
So what I am asking is do chain reactions serve any purpose. If so, it would be great if you could support your answer with a real world example or application.

[...] it would be great if you could support your answer with a real world example or application.

Free radical polymerizations of monomers with $\ce{C=C}$ double bonds are not completely useless ;)

Let me give some examples:

Polymethyl methacrylate (PMMA) is obtained by polymerization of methyl methacrylate (MMA). Think acrylic glass.
Styrene can be polymerized to polystyrene (PS). Think Styrodur and food containers.
Vinyl chloride is polymerized to polyvinyl chloride (PVC). Think pipes, insulation for electric cables, flooring.


Is there a particular reason for the way these structures are represented? I know the question might seem a bit elementary but i was was curious whether there is a particular reason behind the way hydrogens are represented in a hydrocarbon chain. Is there a difference between representing them with straight lines or with slant lines as in the two images? Is one preferred over the other? If so under which conditions?

There is no real difference. Sometimes this comes about from the drawing software people use or personal preference. In some texts the authors or typesetters are using LaTeX packages to draw organic molecules, this can be rather tedious and it is easier to use 90° angles.

The angled hydrogens are a bit closer to what a 2d drawing of the 3d object would look like, but not to the point where any information is conveyed. It can make visualizing the compound in 3 dimensions a bit easier for me when they are drawn this way.

For the double bonds the cis/trans or E/Z information is conveyed, although this is irrelevant in your example.


What is the chemical composition of the human body? What is the average percent composition of each element in the average human body?

oxygen 61.3533
carbon  22.8291
hydrogen 9.9877
nitrogen 2.5683
calcium 1.4268
phosphorus 1.1129
potassium 0.20
sulfur 0.20
sodium 0.14
chlorine 0.14
magnesium 0.03
iron 0.01
fluorine 0.004
zinc 0.003
silicon 0.001
rubidium 0.001
strontium 0.0005
bromine 0.0004
lead 0.0002
copper 0.0001
aluminum 0.0001
cadmium 0.0001
cerium 0.0001
barium 0.00003
iodine 0.00003
tin 0.00003
titanium 0.00003
boron 0.00003
nickel 0.00002
selenium 0.00002
chromium 0.00002
manganese 0.00002
arsenic 0.000010
lithium 0.000010
cesium 0.000009
mercury 0.000009
germanium 0.000007
molybdenum 0.000007
cobalt 0.000004
antimony 0.000003
silver 0.000003
niobium 0.000002
zirconium 0.000001
lanthanium 0.000001
gallium 0.000001
tellurium 0.000001
yttrium 0.0000009
bismuth 0.0000007
thallium 0.0000007
indium 0.0000006
gold 0.0000003
scandium 0.0000003
tantalum 0.0000003
vanadium 0.0000002
thorium 0.0000001
uranium 0.0000001
samarium 0.00000007
beryllium 0.00000005
tungsten 0.00000003

Reference: Emsley, John, The Elements , 3rd ed., Clarendon Press, Oxford, 1998


Alternative to Hanging Drop Crystallization Forgive my ignorance if I am way off, but I was having a look at the crystallization of a number of different substances (e.g. proteins) through the hanging drop vapour diffusion method.

The examples given use a precipitating agent (e.g. $10~\%\ \ce{NaCl}$) at the bottom of the well and then you put the liquid protein you are looking at upside down on the top cover and the water moves from the protein into the precipitating agent, slowly crystallizing the protein.

So my question is, for simplicity would it be possible to use a divided container (e.g. petri dish) which has two compartments that share the same atmosphere to achieve the same thing, that is have a container that has a salt solution on one side, and the solution I wish to crystallize on the other?

Would this likely work?

The process you are describing is essentially the sitting drop crystallisation.

Yes, it is used. Frequently, sitting drop is used in robotic setups, because robots have a much harder time creating hanging drops. With a sitting drop, it becomes simple; just pipet a drop into the sitting drop reservoir.

Gerhard mentioned that the hanging drop method is the preferred one for manual handling since it becomes easier to handle the crystals. Possibly because they sit at the bottom of the hanging drop rather than stick to the ground of a sitting drop’s reservoir.


Why is p-methoxyphenol less acidic than phenol? They have slightly different $\mathrm{p}K_\mathrm{a}$ values of 9.9 (phenol) vs 10.2 (p-methoxyphenol).
They both have the same number of resonance structures too.

My guess is that the resonance structure for p-methoxyphenol with the lone pair on the para carbon contributes less to the stabilisation of the negative charge since it's in an electron rich area (near the lone pairs of the methoxy group oxygen).

Is this a valid explanation? If not, what is the true reason?

Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.

You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)


How does the pH of solutions of salts in which the cation is acidic and anion basic, depend only on the relation of their constants? My book is not very explanative about it, I'm trying to get the equation:
$$[\ce{H^+}]=\sqrt{K_\mathrm{w}\times\frac{K_\mathrm{a}}{K_\mathrm{b}}}$$
but I didn't succeed, can someone explain the theory behind and how to get the formula comparing the two ionization constants?
Translation of the actual page of the textbook (Cacace Fondamenti di stechiometria (in Italian)):

Salts which cation has acid features and anions has basic features. A salt like ammonium acetate, $\ce{NH4Ac}$, release after the following (complete) dissociation:
$$\ce{NH4Ac -> NH4+ +Ac-}$$
A cation will behave as an acid and an anion will behave as a base.
In cases like this one, acid-base equilibrium in solution depends only by the strength of the cation compared to the anion. It can be proved

(but it does not, that's why I'm here)

that in first approximation , the concentration of $\ce{H+}$ in those solution is indipendent from the concentration of the salt and it depends only from the comparison between ionization constants of the weak acid and the weak base from which the salt has been formed:
$$[\ce{H^+}] = \sqrt{K_\mathrm{w}\times\frac{K_\mathrm{a}}{K_\mathrm{b}}}$$
From which:
$$\mathrm{pH} = \frac{1}{2}(14 + \mathrm{p}K_\mathrm{a} - \mathrm{p}K_\mathrm{b})$$

It seems to be unclear. I would like to understand the mathematics of the equation that brings the comparison of the two constant became like the one I already wrote. Why is independent from the salt concentration?

Disclaimer: as we will see later, the formula

$$[\ce{H+}] = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{1}\tag{1}$$

is only applicable to highly soluble salts with equimolar rests of monovalent weak acid and weak base such as $\ce{NH4OAc}$, $\ce{NH4HCO3}$, $\ce{NH4HSO3}$, $\ce{NH4NO2}$ and cannot be used for polyvalent electrolytes, poorly soluble salts and when hydrolysis is irreversible (e.g. $\ce{Al2S3}$).

Let's derive formula \eqref{1} in for the salt of the general $\ce{BHA}$ formula. As we agreed, the salt is highly soluble (also true for your textbook example, ammonium acetate), so the main approximation here is that we assume this salt dissociates completely in solution:

$$\ce{BHA → BH+ + A-} \label{R1}\tag{R1}$$

Each ion participates in hydrolytic equilibrium (\eqref{R2}, \eqref{R3}), plus there is autoprotolysis of water, \eqref{R4}. Subtracting \eqref{R4} from \eqref{R2} and \eqref{R3} we get a brutto equation for hydrolysis \eqref{R5}:

$$
\begin{align}
&+\quad&\ce{BH+ + H2O &<=> B + H3O+} \label{R2}\tag{R2}\\
&+\quad&\ce{A- + H2O &<=> HA + HO-} \label{R3}\tag{R3}\\
&-\quad&\ce{2 H2O &<=>[$K_\mathrm{w}$] H3O+ + HO-} \label{R4}\tag{R4}\\
\hline
&&\ce{BH+ + A- &<=>[$K_\mathrm{h}$] B + HA}, \label{R5}\tag{R5}\\
\end{align}
$$

where $K_\mathrm{h}$ is hydrolysis constant. According to the law of mass action, it can be written for \eqref{R5} as such:

$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} \label{2}\tag{2}$$

Now let's multiply both divider and denominator in \eqref{2} by $[\ce{H3O+}][\ce{HO-}]$ so that

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]}\qquad \right|\cdot \frac{[\ce{H3O+}][\ce{HO-}]}{[\ce{H3O+}][\ce{HO-}]}\\
 &= \frac{\color{blue}{[\ce{B}]}\color{red}{[\ce{HA}]}[\ce{H3O+}][\ce{HO-}]}{\color{blue}{[\ce{BH+}]}\color{red}{[\ce{A-}][\ce{H3O+}]}\color{blue}{[\ce{HO-}]}}\\
 &= \frac{[\ce{H3O+}][\ce{HO-}]}{\color{red}{K_\mathrm{a}}\color{blue}{K_\mathrm{b}}}\\
 &= \frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}\label{3}\tag{3}
\end{align}
$$

where $K_\mathrm{a}$ and $K_\mathrm{b}$ are dissociation constants of the conjugated acid and conjugated base, respectively:

$$
\begin{align}
\ce{HA + H2O &<=>[$K_\mathrm{a}$] A- + H3O+} \label{R6}\tag{R6}\\
\ce{B + H2O &<=>[$K_\mathrm{b}$] BH+ + HO-} \label{R7}\tag{R7}\\
\end{align}
$$

Since the dissociation of the salt is complete (see \eqref{R1}), its initial concentration $c_0$ equals to the concentrations of $\ce{BH+}$ and $\ce{A-}$ in solution:

$$c(\ce{BH+}) = c(\ce{A-}) = c_0 \label{4}\tag{4}$$

and the corresponding equilibrium concentrations of the ions participating in hydrolysis are

$$[\ce{BH+}] = [\ce{A-}] = (1-α)c_0, \label{5}\tag{5}$$

where $α$ is the degree of hydrolysis showing the ratio between the amount of hydrolyzed salt $n_\mathrm{h}$ to the total amount of salt $n_0$ ($α = n_\mathrm{h}/n_0$; $α\in [0;1]$). The equilibrium concentrations of the weak acid and base formed during hydrolysis are

$$[\ce{B}] = [\ce{HA}] = αc_0 \label{6}\tag{6}$$

and eq. \eqref{2} for $K_\mathrm{h}$ can be rewritten using degree of hydrolysis:

$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} = \frac{α^2c_0^2}{(1-α)^2c_0^2} = \frac{α^2}{(1-α)^2} \label{7}\tag{7}$$

Note that hydrolysis constant doesn't depend on salt's concentration. Since we are interested in $\ce{pH}$, it is convenient to express $[\ce{H3O+}]$ from $K_\mathrm{a}$:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]};$$

$$[\ce{H3O+}] = K_\mathrm{a}\frac{[\ce{HA}]}{[\ce{A-}]} = K_\mathrm{a}\frac{αc_0}{(1-α)c_0} = K_\mathrm{a}\frac{α}{1-α} \label{8}\tag{8}$$

From \eqref{7} $α/(1-α) = \sqrt{K_\mathrm{h}}$ so that \eqref{8} finally becomes the first equation you asked about:

$$[\ce{H3O+}] = K_\mathrm{a}\sqrt{K_\mathrm{h}} = K_\mathrm{a}\sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}} = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{9}\tag{9}$$

An equation for $\mathrm{pH}$ is obtained trivially by taking negative decimal logarithm on both parts of \eqref{9}.


Rate law for breakdown of ozone to oxygen For the following problem:

  Example 3: Write the rate law for $\ce{2 O3 -> 3 O2}$ given that the first step is fast and reversible, and the second step is slow.

  $$
\begin{align}
&\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[$k_1$][$k_{-1}$] O2 + O} \\
&\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[$k_2$] O2 + O2}
\end{align}
$$

  The rate is determined by the slowest step.

  The rate of formation of $\ce{O2}$ is equal to $2$ times the rate of the slow step $(k_2[\ce{O}][\ce{O3}]),$ since two molecules of $\ce{O2}$ are formed.

  Thus, rate of formation of $\ce{O2} = 2k_2[\ce{O}][\ce{O3}],$ but $“\ce{O}”$ is an intermediate, solve for $“\ce{O}”$ in
  terms of products and reactants and rate constants.

  Since the first step is fast and reversible and the second step is slow, the first step is in equilibrium and we can write

  $$\frac{[\ce{O2}][\ce{O}]}{[\ce{O3}]} = \frac{k_1}{k_{-1}} = K_1 \quad\text{or}\quad [\ce{O}] = \frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}]}$$

  Substituting:

  $$
\begin{align}
\text{rate} &= \frac{2k_2k_1[\ce{O3}]^2}{k_{-1}[\ce{O2}]} \\
\text{rate} &= k_\mathrm{obs}\frac{[\ce{O3}]^2}{[\ce{O2}]}
\end{align}
$$

The solution indicates that the rate of formation $\ce{O2}$ is $2k_2[\ce{O}][\ce{O3}]$. Why isn't it $2k_2[\ce{O}][\ce{O3}] + k_1[\ce{O3}]$, because the first reaction also produces $\ce{O2}$?

You should write down the rate equations and set [O] as an intermediate so that $\ce{d[O]/dt} = 0$. This is the steady state approach and leads to $\displaystyle \frac{d\mathrm{[O_2]}} {dt}=2\frac{k_1k_2\mathrm{[O_3]}^2}{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$ which differs from your answer because when you assume a fast equilibrium you also assumed that $k_{-1} \gg k_1$. You should make this assumption after you have the full equation, if necessary, but not before you do the calculation.

To answer your last point you still have the intermediate [O] which is an unknown and so has to be eliminated as shown by using the steady state approach used above.


How to calculate pH of basic salt solution mixed with strong base? Calculate $\mathrm{ pH }$ of $\pu{ 0.05 mol }$ of $\ce{Zn(CH3COO)2}$ and $\pu{ 0.025  mol }$ of $\ce{ NaOH }$ dissolved in $\pu{ 1 L }$ of distilled water?

I tried to use the systematic method by forming the following equations to extract cubic equation which can  be solved by computer :
\begin{array}{   }
[\ce{ CH3COO- }]_I &= 2\times {0.05} = \pu{ 0.1 M }\\
[\ce{Zn^{2+}}] &=\pu{ 0.05 M }\\
[\ce{Na+}] &=\pu{ 0.025 M }\\
0.1 = [\ce{CH3COO-}]_I &= [\ce{CH3COO-}]_e + [\ce{CH3COOH}]_e\\
2\ce{[Zn^{2+}] + [Na+] +[ H+]_e &= [CH3COO-]_e +[ OH-]_e} \\
K_\mathrm{a} &=\frac{ [\ce{ H+}]_e\cdot{ [\ce{CH3COO-}]_e}}{[\ce{CH3COOH}]_e}\\
[\ce{OH-}]_e &=\frac{ K_\mathrm{w}}{[\ce{H+}]_e}
\end{array}

First question: Are the equations correct?

Second question: Is there a simple approximate method to solve the above problem? How?

I appreciate any help?

$\large \text{ --------- Maxw Notes ---------}$

Editing the answers with all the markup is painfully slow, so I added different analysis in a number of different answers to this problem. No doubt that will be confusing to someone looking at this problem. So let me summarize the answers.

  Note: I am going to carry "extra" significant figures in the  calculations to gauge the mathematical exactness, but this is of course is ridiculous for the chemistry since the equilibrium constants are not known to that precision. If the true solution values are within 5% of the mathematically calculated values that's really good.

Exact Solution

Setup an ICE table and solved the nasty linear equations with Wolfram Alpha to get an "exact solution."

Iterative Solution

Setup an ICE table and then simplified the problem to create a "Simplified ICE Table". Went through some head pounding to create an iterative solution. Realized the original ICE table could be used for iterated solution so calculated that solution too.

Reducing the Problem to a Single Variable

I had tried to solve the problem first with what I am calling the Iterative Solution. I could tell all the species other than $\ce{[Na+]}$ were independently pH controlled, but I couldn't figure out how to iterate the solution. Using the charge balance finally occurred to me so I tabulated the data to show that the approach would work.

Solution with Cubic Equation

Used charge balance of solution, and one assumption to solve the simplified problem with a cubic equation with $\ce{[OH-]}$ as the variable. Solved the cubic equation with Wolfram Alpha.

Assumption:

$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$

Solution with Quadratic Equation

Used charge balance of solution, and two assumptions to solve the simplified problem with a quadratic equation with $\ce{[OH-]}$ as the variable. Solved the quadratic equation with Wolfram Alpha, but is reasonable to assume that solution could have been done by hand.

Assumptions:

$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$
$\ce{[CH3COO-]_{equil} \gg K_b }$
That the normalization tweak works. (I know that it did from the exact solution, but the calculations showed that it worked too.)

Exact Solution

I've fouled up this problem badly. Here is a more carefully thought out solution.

So let's start at the beginning with the ICE table. Making the normal assumptions:

Concentrations will be used instead of activities.
The $\ce{Zn(CH3COO)_2}$ dissociates completely in aqueous solution, thus the $\ce{Zn^{2+}}$ and $\ce{CH3COO-}$ ions can be considered separately.
The $\ce{NaOH}$ dissociates completely in aqueous solution, thus the $\ce{Na+}$ and $\ce{OH-}$ ions can be considered separately.
The $\ce{Na+}$ is a spectator ion and is left out of the table.

The problem only calls for the pH of the final solution, but a solution to the whole table will be calculated as a check.
Initial ICE Table
Let's let:

$x = $ the moles of $\ce{Zn(OH)2}$ formed. (The numerical value is the same as the change in molarity of $\ce{Zn^{2+}}$ since there is 1.0 liter of solution.)

$y = $ the molarity of the $\ce{CH3COOH}$.

\begin{array}{c | c c c c c}
               &\ce{[Zn^{2+}]}     & \ce{[CH3COO-]}        & \ce{[CH3COOH]}   & \ce{[OH-]}                         & \ce{Zn(OH)2(s)}  \\\hline
\text{I}  &\pu{0.050000 M}     & \pu{0.100000 M}           & \pu{0.000000 M}               & \pu{  0.025000 M}               & \pu{0 moles}  \\
\text{C} &  -x                           &  -y                               &   y                              & \pu{- 2x + y M}                  & \pu{x moles}  \\
\text{E} &\pu{0.050000 - x M} & \pu{0.100000 - y M}     &   y                              & \pu{0.025000 - 2x + y M}   & \pu{x moles}   \\
\end{array}
Applicable equilibriums
Wikipedia gives the $K_{sp}$ of zinc hydroxide is $K_{sp}=3.0\cdot10^{−17}$. So:
$K_{sp}=3.000\cdot10^{−17} = \ce{[Zn^{2+}][OH-]^2}$
Wikipedia gives the $\pu{pK_b}$ for sodium acetate as 9.25, which yields $K_b = 5.623\cdot10^{−10}$.  So:
$K_b = 5.623\cdot10^{−10} = \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
Exact Solution
Starting with the $K_{sp}$ equation for zinc hydroxide and making the appropriate substitutions for the final concentrations:
$K_{sp} = \ce{[Zn^{2+}][OH-]^2}$
$3.000\cdot10^{−17} = (0.050000 - x)\cdot(0.025000 - 2x + y)^2$
Starting with the $K_b$ equation for the acetate anion:
$K_b =  \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
$5.623\cdot10^{−10} = \dfrac{(y)\cdot(0.025000 - 2x + y)}{0.100000 - y} = \dfrac{0.025000y - 2xy + y^2}{0.100000 - y}$
Wolfram Mathworld's solution is:

x = 0.0134623
y = 0.00192466

$\ce{[Zn^{+2}]_{equil}} = 0.0500000 - 0.0134623 = 0.0365377$
$\ce{[OH-]_{equil}} = \sqrt{\dfrac{ K_{sp}}{\ce{[Zn^{2+}]_{equil}}}} = \sqrt{\dfrac{3.000\cdot10^{−17}}{0.0365377}} = 2.86543\cdot10^{-8}$
$\pu{pOH} = 7.542810$
$\pu{pH} = 14.000000 - 7.542810 =  6.457190 \ce{->[Round to 2 SF]} 6.46$
$\ce{[CH3COO-]_{equil}} = 0.100000 - 0.00192466 = 0.09807534$
There are two equations and two unknowns. Solving the equations yielded an exact solution. This is no problem with a computer but a mess even for a good calculator.


Why there is charge on the elements, in the dative bonding? According to the definition of dative bonding electrons are shared which are given by one of the atoms, so if the electrons are shared so how come there is charge on the elements in the dative bonding?

To begin with, the term "dative bond" is not used anymore. It has been replaced by "coordinate bond" :

Let's have an example of this kind of bond.

As you can see, this kind of bond occurs when an atom is giving two electrons from itself to form the bond. The typical coordinate bond is the $\ce {N}$ already bonded with $\ce {3H}$ forming a bond with a proton. Since the proton has no electron, the $\ce {N}$ atom has to fully give one of his electron from his remaining doublet to form the bond. But to respect the charge equality, you had a positive charge at the begining, no electrons were added or removed from the system, so the positive charge should remains on the final molecular edifice.

The question is, where is located this positive charge ? One could start thinking this charge is located on the $\ce {N}$ atom since it is giving an electron. But actually, the $\ce {N}$ atom has a greater electronegativity than the $\ce {H}$ atom, that's say its hability to attract electrons is greater. Therefore the positive charge remaining should be equaly distributed over all the $\ce {4H}$ since they are equivalent.

But, altogether, the net charge of the edifice will be $+1$.


notation on MO diagrams I wanted to ask a question about notations on MO diagrams.

Our first lesson about MO diagrams was today and I was taken through an example with $\ce{F2}$ as shown below:

but one thing that the lecturer did not explain, and I couldn't find over the internet with explanation, was the notation $\ce{3\sigma _g}$, $\ce{1\pi _u}$, $\ce{1\pi _{g}^{*}}$ and $\ce{3\sigma _{u}^{*}}$.

Is this notation relevant to the symmetry of the orbitals, or is there something else?

I haven't found any information that can explain this thus far.

There is likely a question that addresses this on the site already, but I couldn't find it either, so I'll add an answer here.

You are correct that these symbols relate to the symmetry of the molecular orbitals that they label. In chemistry, we often label molecules by their point group symmetry, which describes how the molecule behaves with respect to particular symmetry operations (e.g. rotations, reflections, inversions).

If a molecule has a particular point group symmetry, its orbitals (and some other properties like vibrations) can be divided up into irreducible representations, which are essentially just different ways in which the overall symmetry can occur. One can find these irreducible representations (often shortened to irreps) in a character table for that point group.

$\sigma_\mathrm{g},\sigma_\mathrm{u},\pi_\mathrm{g},\pi_\mathrm{u}$ are all irreps of the $D_{\infty\mathrm{h}}$ point group, which you will most commonly encounter as the symmetry of a homonuclear diatomic. Since you are just being introduced to these, you won't likely use a character table and will just be given these MO diagrams.

At the level you will need for the moment, $\sigma$ orbitals come from the overlap of a single lobe from each atomic orbital and $\pi$ orbitals come the overlap of two lobes from each atomic orbital. The $\mathrm{g}$ and $\mathrm{u}$ labels tell you whether the orbital is symmetric or antisymmetric under inversion, that is, if you move every point $(x,y,z)$ to $(-x,-y,-z)$, does the sign of the orbital remain the same or does it become the opposite sign. The * label is used to describe whether an orbital is bonding (overlap is between atomic orbitals/lobes of the same sign) or antibonding (overlap is between atomic orbitals/lobes of opposite sign).


Can linear SOCl2 exist? Specifically, instead of its normal structure, this Lewis structure seems to be valid as a constitutional isomer:

$\ce{Cl-O-S-Cl}$

So what prevents this from existing, or if it already exists how is it made?

I highly doubt that it could exist. The co-linear arrangement, $C_{\infty\mathrm{v}}$, of $\ce{Cl-O-S-Cl}$ is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the $S=0$ potential energy surface. It has two imaginary bending modes, $\nu_1=371.39\mathrm{i}$, $\nu_2=108.23\mathrm{i}$. (Please note that they are actually perpendicular to each other.)

That basically means, that distortion along these modes will lead to an arrangement of the nuclei which is lower in energy. Since this is a second order saddle point, this arrangement can itself only be a transition state (first order saddle point), which also cannot be considered stable. (Compare also here.)
Therefore distortion among $\nu_1$ would probably lead to $\ce{Cl-O}$ and $\ce{S-Cl}$ fragmentation. While distortion along $\nu_2$ might preserve the $\ce{S-O}$ bond, fragmenting the chlorine atoms/ chloride ions - but that is pure speculation of a hypothetical case.

Let's try a very tempting ad-hoc explanation of why this is the case. In that case let us assume all atoms are $\mathrm{sp}$ hybridised and the bonding axis is $z$. The Lewis structure that seems to be valid has therefore three lone pairs ($\mathrm{sp}$, $\mathrm{p}_x$, $\mathrm{p}_y$) at each chlorine and two lone pairs ($\mathrm{p}_x$, $\mathrm{p}_y$) at the oxygen and sulfur. The three σ bonds are formed by the $\mathrm{sp}$ orbitals.
That would result in four lone pairs that are co-aligned in the same plane. Overlap is inevitable, and in the most simplest terms of VSEPR theory, we would call that lone pair repulsion. (Please note that for strictly VSEPR theory we would have to assume tetrahedral hybridisation.)
In other words, same spin electrons have to avoid occupying the same space and since the orbitals are so close together this will rise the energy a lot. You can probably call it Pauli repulsion.

The MO picture is too messy, so that I won't show it here. Basically most of the occupied orbitals are anti-bonding with respect to one of the bond. The HOMO is of π symmetry and anti-bonding with respect to every bond.


Classification of the Inside of a Grapefruit In the text Chemistry: The Central Science, the inside of a grapefruit is classified as a homogeneous mixture. Unless I’m missing something, to me it would be a heterogenous mixture owing to pulp, seeds, and ultimately the fact that cells are not homogenous. Is this a mistake in the text, and, if not, what am I missing?

According to Wikipedia:

  In chemistry, a heterogeneous mixture consists of either or both of a)
  multiple states of matter or b) hydrophilic and hydrophobic substances
  in one mixture

The inside of a grapefruit is clearly made up of multiple states of matter (solid and liquid), and is thus heterogeneous. Text books sometimes contain errors, and that appears to be the case here.