USNCO question - Finding pH of salt solution from pKa of its parent acid The question from 2017 USNCO is as below:

What is the $\mathrm{pH}$ of a $\pu{0.25 M}$ solution of $\ce{NaCN}$? (The $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is $9.21$.)
(A) $4.91$
(B) $8.61$
(C) $11.30$
(D) $13.40$

The answer sheet states (C).
My understandings/reasoning:

salt hydrolysis: solution of salt derived from strong base and weak acid is basic;
$\ce{NaCN}$ is derived from the base $\ce{NaOH}$ (strong) and acid $\ce{HCN}$ (weak according to given $\mathrm{p}K_\mathrm{a}$);
Solution is basic, so (A) is incorrect.

However, I am unable to proceed from here.

Sodium cyanide is a salt formed by a strong base and a weak acid, so the reduced ionic equation of hydrolysis is

$$\ce{CN- + H2O <=>[$K_\mathrm{h}$] HCN + OH-}$$

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]} \qquad \right|\cdot \frac{[\ce{H+}]}{[\ce{H+}]}\\
&= \frac{[\ce{HCN}][\ce{OH-}][\ce{H+}]}{[\ce{CN-}][\ce{H+}]}\\
&=\frac{K_\mathrm{w}}{K_\mathrm{a}} \label{eq:1}\tag{1}
\end{align}
$$

Sodium cyanide is highly soluble salt, so one can neglect auto-protolysis of water, and also express hydrolysis constant (assuming $[\ce{OH-}] = [\ce{HCN}]$ and $C$ – concentration of sodium cyanide) as follows:

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{C-[\ce{OH-}]} \label{eq:2}\tag{2}$$

Equating \eqref{eq:1} and \eqref{eq:2}:

$$\frac{10^{-14}}{10^{-9.21}} = \frac{[\ce{OH-}]^2}{0.25-[\ce{OH-}]}\, ,\label{eq:3}\tag{3}$$

and solving the quadratic equation \eqref{eq:3} for $[\ce{OH-}]$, one can find that $[\ce{OH-}] \approx \pu{2e-3 M}$, $\mathrm{pOH} = 2.70$, hence

$$\mathrm{pH} = 14.00 - \mathrm{pOH} = 14.00 - 2.70 = 11.30$$


How can magnesium hydroxide be a strong base and be sparingly soluble at the same time? A strong base means it completely dissociates forming $ OH^- $ ions when added to aqueous solution.
Sparingly soluble means it dissolves very little when added to aqueous solution.
Yet magnesium hydroxide is said to be both. These two properties seem to contradict each other. How can this be?
A very similar question was asked here, the most relevant answer states that

Some strong bases like calcium hydroxide aren't very soluble in water.
That doesn't matter - what does dissolve is still 100% ionised into
calcium ions and hydroxide ions. Calcium hydroxide still counts as a
strong base because of that 100% ionisation.

But I don't understand the difference - how can something dissolve and not be ionised? (wouldn't that imply a covalent structure like glucose?)

I think we should not conflate "solubility" with "dissociation". They are different phenomena. The former has to do with a molecule breaking away from the other solute molecules and forming interactions with the solvent. The latter is a process which occurs after dissolution, whereby the solute molecules break apart to form ions, separately solvated by the solvent molecules. For a simple covalent substance, the two phemomena are clearly different.
For an ionic solid, the two may seem to occur together since when a soluble ionic salt is placed in water, its ions are in fact pulled apart by water molecules. In this sense, it is dissolving and dissociating at the same time. Hence, by the Arrhenius definition of acids and bases, insoluble hydroxides such as $\ce {Mg(OH)_2}$ may not even be considered to be strong bases. Arrhenius theory would tell us that the strong base is the hydroxide ion, that is produced from the dissolultion and dissociation of the salt.
Actually, in particular for some metal oxides such as $\ce {MgO}$, the first reaction it undergoes upon contact with water is $\ce {MgO + H2O -> Mg(OH)2}$. Hence, we can rightfully say $\ce {MgO}$ is not an Arrhenius base, but it is the precursor to the weak Arrhenius base $\ce {Mg(OH)2}$.
I would like to disagree with the part of the answer you have referenced in your post because as aforementioned, dissolution and dissociation are the same phenomena for ionic salts. It is thus not correct to say that "what does dissolve is still $\ce {100}$% ionised into [ions]" and hence, say that insoluble (or sparingly soluble) ionic salts are strong bases.
To generalise, Arrhenius theory would consider soluble ionic hydroxides such as $\ce {NaOH (s)}$ to be strong bases while it would consider sparingly soluble ionic hydroxides such as $\ce {Ca(OH)2}$ to be weak bases. In other words, the strength of these metal hydroxides as bases is directly linked to their solubility and this makes perfect sense since dissolution and dissociation for an ionic salt are the same process.


How are solubilities of hydrate salts determined when competing equilibria are unavoidable? When you calculate the solubility product constant for a salt and one of its hydrates, you sometimes get drastically different numbers! I don't know how this can be because they both dissociate into the same ions. For example, sodium sulfate heptahydrate $\ce{Na2SO4.(H2O)7}$ is more soluble than anhydrous sodium sulfate $\ce{Na2SO4}$. If you work through the math, the discrepancy in solubility is not completely explained by the larger mass of the heptahydrate (see calculations below).

I can explain the differences in solubility by assuming that the water molecules within the crystal structure of sodium sulfate heptahydrate lower the enthalpy of crystallization, which makes solvation energetically more favorable than the anhydrous salt.

What confuses me is the logical series of events that would happen when you add the solid hydrate into solution. The hydrate will dissolve into $\ce{Na+}$ and $\ce{SO4^2-}$ions. The concentration of those ions will increase until the solution is saturated (with respect to the anhydrous salt). Anhydrous $\ce{Na2SO4}$ will begin to precipitate out, keeping the ion concentrations below $K_\mathrm{sp}(\ce{Na2SO4})$ The hydrate will dissolve completely, because the solution will never reach saturation (with respect to the heptahydrate). How, then, was the solubility of the hydrate ever experimentally determined?

Using $\ce{Na2SO4}$ as an example:

Solubility of anhydrous $\ce{Na2SO4}$ = $\frac{4.76\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)
Solubility of $\ce{Na2SO4.(H2O)7}$ = $\frac{19.5\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)

Calculating saturated concentration:

Solubility of anhydrous $\ce{Na2SO4}$ = $$\frac{4.76\ \mathrm g\ \ce{Na2SO4}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4}}{142.04\ \mathrm g\ \ce{Na2SO4}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.335\ \mathrm M\ \ce{Na2SO4}$$

Solubility of $\ce{Na2SO4.(H2O)7}$ = $$\frac{19.5\ \mathrm g\ \ce{Na2SO4.(H2O)7}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4.(H2O)7}}{268.1491\ \mathrm g\ \ce{Na2SO4.(H2O)7}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.727\ \mathrm M\ \ce{Na2SO4.(H2O)7}$$

Calculating $K_\mathrm{sp}$:

Anhydrous $\ce{Na2SO4}$: $$\ce{Na2SO4 <=>2Na+ + SO4^2-}$$  $$K_\mathrm{sp}(\ce{Na2SO4})=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.335\times 2\right)^2\times\left(0.335\right)=0.150$$

$\ce{Na2SO4.(H2O)7}$: $$\ce{Na2SO4.(H2O)7 <=>2Na+ + SO4^2- +7H2O}$$  $$K_\mathrm{sp}\left(\ce{Na2SO4.(H2O)7}\right)=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.666\times 2\right)^2\times\left(0.666\right)=1.18$$

Let there be three species: a hydrated solid $H$, an anhydrous solid $A$, and a complete solvated (liquid) aqueous solution, $S$.

Your question is about differences in the two equilibria $\ce{H <=> S}$ and $\ce{A <=> S}$.  Can $H$ really have a different solubility than $A$?

I think the answer is yes, but only if the equilibrium $\ce{H<=>A}$ cannot and is not established.  In the presence of an "infinite" kinetic barrier that blocks conversion of the hydrate to the anhydrous form and vice versa, it makes sense to treat $\ce{H<=>S}$ and $\ce{H<=>A}$ and independent equilibria which might have different equilibrium constants.

In a comment to your question, I made it seem like "any" precipitation from a (super)saturated solution of $S$ would be the hydrate.  On further reflection, I think that is true, unless (i) the equilibrium $\ce{A<=>H}$ cannot be established due to kinetic barriers, and (ii) there is solid phase $A$ present in the solution.  In that case, the supersaturation could just result in "precipitation" of solid onto the existing crystal surface of already-present $A$, a process that would increase the mass of $A$ (solid) in the system but which would not require formation of any crystallites or nucleation intermediates.

  How, then, was the solubility of the hydrate ever experimentally determined?

I think the answer is that the hydrate solubility can differ from the anhydrous solid solubility only in cases when there is no mechanism by which equilibrium can be established between the hydrate and the anhydrous solid.


Why is disilyne bent? Why is disilyne $\ce{Si2H2}$ bent when the steric number of silicon is 2?

The following disilyne has been prepared and found to be stable to ~100 C.  An X-ray crystal structure found that the two silicon atoms in the triple bond, Si(1) and Si(1'), along with the two attached silicon atoms, Si(2) and Si(2') are coplanar and the Si(1)-Si(1')-Si(2') angle is 137.44 degrees ("trans-bent").

The full text of the article describing the preparation, characterization and reactivity of the compound can be found here.  See figure 4 in the paper for a MO explanation as to why the trans-bent geometry occurs. Basically, the authors suggest that, "the bending is thought to be the result of the mixing of an in-plane $\ce{\pi}$-orbital with a $\ce{\sigma^{\ast}}$ orbital whose energies are close enough to cause the interaction."


How does one calculate chemical shift in a gauge independent manner? The literature on ab initio calculations of chemical shift in NMR experiments usually provide Lamb's and Ramsey's formulae as the solution. Yet the expressions explicitly depend on vector potential gauge choise. Some authors claim that the theory is intrinsically gauge invariant, giving no proper proof yet. Can the solution be expressed in an explicitly gauge invariant form? Some literature on this.

NOTE: I know about London orbitals theory, yet it's still not fully understood by me, how it can deliver a type of expression I want. Some suggestion on where this is clarified are very welcome.

The references below describe the answer.

Prog Nucl Magn Reson Spectrosc. Author manuscript; available in PMC 2012 May 1.
Published in final edited form as:
Prog Nucl Magn Reson Spectrosc. 2011 May; 58(3-4): 176–201.
Published online 2010 Dec 15. doi: 10.1016/j.pnmrs.2010.10.003
PMCID: PMC3058154
NIHMSID: NIHMS248338
Chemical shift tensors: Theory and application to molecular structural problems
Julio C. Facelli

Nucleic Acids Res. 2014 Dec 16; 42(22): e173.
Published online 2014 Nov 17. doi: 10.1093/nar/gku1006
PMCID: PMC4267612
Accurate ab initio prediction of NMR chemical shifts of nucleic acids and nucleic acids/protein complexes
Andrea Victora,1 Heiko M. Möller,2 and Thomas E. Exner1,3,*

A comparison of models for calculating nuclear magnetic resonance
shielding tensors
James R. Cheeseman, Gary W. Trucks, Todd A. Keith, and Michael J. Frisch

J. Chem. Phys. 104 (14), 8 April 1996


How to tell if a molecule will participate in hydrogen bonding with water? I was recently doing some organic chemistry work and came across a set of problems.

"Which of the following compounds will participate in hydrogen bonding with water? For each compound, indicate whether it will be a hydrogen bond donor, hydrogen bond acceptor, or both."

I do not have an idea of how to approach the first part of the question. Can someone help me with the first molecule: (a) $\ce{CH3CH2OCH3}$ in order of connectivity.

I figured out the answer. If there is a lone pair of electrons belonging to nitrogen, oxygen, or fluorine, then it is possible to bond to water. Therefore, it is can accept a hydrogen. If there is a hydrogen bonded to one of the three elements listed above, then it able to donate.


Entropy increase when heating a gas What is the entropy change when the temperature of 1.00 mol of ideal gas is raised reversibly from 10 °C to 50 °C at constant pressure, constant volume, and assuming the heat capacity is independent of temperature?

This is all the information given. I am assuming that the formula $\Delta S = C \ln (T_2/T_1)$ is used. However, I do not know how the heat capacities are supposed to be obtained from the given information. I do not have the enthalpy or internal energy of the process. I cannot calculate the heat capacity from pressure or volume.
Please say if my reasoning is incorrect.
(These aren’t the values of the original question but this is based on Atkins Chemical Principles 5th ed. Ch 8 # 5 and 6)

With constant volume heat capacity, it is defined as, for an ideal gas, that has the internal energy equation
$$dU=TdS-\rho dV= \frac{\nu}{2} Nk_{\beta} T\ (1)$$ which is the Ideal Gas Law, obtained from the first law of thermodynamics. Using moles, (1) is equivalent to
$$ \frac{\nu}{2}nRT $$
Now, onto heat capacities.
$C_V$ is defined as,
$$C_V=(\frac{\partial U}{\partial T})_V= \frac{\partial}{\partial T}\left(\frac{\nu nRT}{2}\right)=\frac{\nu}{2} nR $$

Constant pressure heat capacity is defined as
$$C_{\rho}=C_V + nR=\frac{\nu+2}{2} nR$$, where $n$ is the amount of substance and $R$ is the universal gas constant. $\nu$ is the degrees of freedom of the molecule, which is easily obtained by thinking about a monoatomic ideal gas and the ways it can move.

Below is some fun theory surrounding the heat capacity for constant volume.
Note that constant-pressure specific heat is also defined as
$$C_{\rho}=(\frac{\partial U}{\partial T})_{\rho}=dU+\rho dV+Vd\rho=
(\frac{\partial H}{\partial T})  $$
Where $H$ is the enthalpy of heating the gas, which is
$$\mathrm{d}H=\mathrm{d}U+p\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S-\rho\mathrm{d}V+\rho\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S+V\mathrm{d}\rho$$
Enthalpy is commonly defined with a constant pressure process, thus $\mathrm{d}\rho=0$.
So, for a constant pressure, we have $$\mathrm{d}H=T\mathrm{d}S$$,
exploting the definition of entropy for a reversible change,
$$\mathrm{d}S=\frac{\mathrm{d}Q}{T} \rightarrow \mathrm{d}Q=T\mathrm{d}S $$
so, we finally get $$\mathrm{d}H=\mathrm{d}Q$$
which is a standard thermodynamic identity, and can also be used to get to the answer at the beginning of the post (though it is admittedly longer and more complicated to do so.)

Hope these hints help you!


Copper Chloride: neutralizing and disposal I plan on etching some PCBs with hydrochloric acid and hydrogen peroxide, which will therefore produce some kind of copper chloride (green color) which is highly toxic if released into the environnement (it may be useful as an herbicide, but I don't have any garden right now).

While cupric acid (obtained from hydrochloric acid, hydrogen peroxide and some PCBs etching) may be bubbled with air/oxygen to re-activate it to etch again other PCBs I'm not sure this may what I want and I seek suggestions in case I need to dispose of this solution.

Basically I heard (though none were actually on chemistry sites, except the 1st one):

Acid/Base neutralization: pour some baking soda to take care of the chloride
Pour some alluminium balls to get back some of the copper (didn't really understand how this works)
Copper electrolysis: basically get solid copper at the cathode, creating gaz at the anode and leaving (?) some residues that can be neutralized with baking soda (?)

Any insights on this? I think once the copper and chloride are taken care of, it shouldn't pose a huge problem to pour it down the drain or leave it outside to evaporate.

Specifically:

Which metal should I choose for the anode / cathode ? In the first source it speaks about reactivity (and since copper is less reactive than hydrogen, copper is produced at the cathode). No idea about the metal I should use though.
Would adding alluminium balls into the solution help make it more "eco-friendly" ?
Any other suggestion is welcome

Sources:

Electrolysis
How to etch PCBS - see comments
Recovery of copper using electrolysis

Regarding #2: Adding $\ce{Al}$ or $\ce{Mg}$ metal to $\ce{CuCl_2}$ solution reduces the $\ce{Cu_2^+}$ to $\ce{Cu}$, and you then have $\ce{Cu}$ metal (filter out and recover) and $\ce{AlCl_3}$ or $\ce{MgCl_2}$, respectively, in solution. Both solutions are safer in small quantities to flush down the drain.

Regarding #1: $\ce{2NaHCO3 + CuCl2 -> CuCO3 + 2NaCl + CO2 + H2O}$. $\ce{CuCO3}$ is insoluble, but should still be disposed of at an appropriate facility.

In both cases, you get an insoluble copper compound (or metal) rather than a soluble one.


Calculation of an equilibrium reaction quotient For the equilibrium

  $$\ce{2 SO2(g) + O2(g) <=>  2 SO3(g)}$$

  $$K_c = 245\ (\text{at}\ 1000\ \mathrm K)$$

  The equilibrium concentrations are $[\ce{SO2}] = 0.102$, $[\ce{O2}] = 0.0132$, and $[\ce{SO3}] = 0.148$. Suppose that the concentration of $\ce{SO2}$ suddenly doubled. Calculate the $Q$ and use it to show that the forward reaction would take place to reach a new equilibrium.

I created an ice table so then I was able to find $x$:

$245=(0.148+2x)^2$ divided by $(0.0132-x)(0.204-2x)^2$

$x=0.01$

So the equilibrium concentrations are: $0.184$ for $\ce{SO2}$, $3.2\times10^{-3}$ for $\ce{O2}$ and $0.168$ for $\ce{SO3}$.
To check if the forward reaction is favored, $q<k$. The answer key states that $q=61.6$ but I don't know how they got it.

The reaction quotient, Q, is calculated the same way you would calculate the equilibrium constant, K. That is, Q is the product of the concentrations of the products of the reaction divided by the product of the concentration of the reactants,

$$
Q = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}
$$

Plugging the concentrations in $([SO_2] = 2 \times 0.102, [O_2] = 0.0132, [SO_3] = 0.148)$ should give you the answer for $Q$. I think you may have the concentrations written down incorrectly $([SO_3] = 0.184$ gives $Q = 61.6$, for instance$)$.


Why is distilled water such a poor conductor of electricity? Water is composed of hydronium ions and hydroxide ions. Since ions do conduct electricity, why is distilled water such a poor conductor of electricity?

You are mistaken.

Water is not composed of these ions. There is ongoing autodissociation equilibrium reaction$$\ce{2 H2O <<=> H3O+ + OH- }$$ shifted strongly toward left, with concentration of ions  $\pu{1e-7 mol/L}$ at $\pu{25 ^{\circ}C}$ .

Therefore conductivity of water itself is very low.


Electronegativity of Potassium (Pauling) Just a little question, I was taught that the electronegativity of potassium[K] is 1, but I recently discovered most internet resources say 0.82. Some even say both. What is the actual electronegativity of potassium, 0.82 or 1?

It depends which electronegativity scale you choose to use.  There are a lot of different scales for comparing electronegativities of the elements.  I've included a list of 11 of them below if you're interested.  In any case here are the electronegativities reported for potassium using 5 different electronegativity scales.  All values are in "Pauling units" so that the numbers can be compared.  The Pauling scale is the most commonly used, so 0.82 would be a reasonable estimate for the electronegativity of potassium.

Pauling 0.82
Sanderson 0.45
Allred Rochow 0.91
Mulliken-Jaffe 0.73
Allen 0.734

On the Pauling scale, electronegativities range from around 0.7 to 3.98 (fluorine), so as expected, potassium is at the electropositive end.  None of these scales report a value of 1.

List of 11 Different Electronegativity Scales

Pauling Scale: (1932)  Obtains values by thermochemical methods
Mulliken Relation: (1934)  Defines a relation that depends upon the
orbital characteristics of an atom in a molecule. Mulliken
electronegativity is the numerical average of the ionisation
potential and electron affinity
Gordy Scale: (1946)  Defines electronegativity in terms of the
effective nuclear charge and the covalent radius. (Zeff)e/r
Walsh Scale: (1951)  Relates electronegativity to stretching force
constants of the bonds of an atom to a hydrogen atom.
Huggins Scale: (1953) Alternative to Pauling's thermochemical
procedure
Sanderson Scale: (1955)  The ratio of the average electron density of
an atom to that of a hypothetical "inert" atom having the same number of electrons. This ratio is a measure of the relative compactness of the atom
Allred-Rochow Scale: (1958)  Defines electronegativity in terms of the effective nuclear charge and covalent radius. Like the Gordy scale but uses (Zeff)e/r2
Jaffe Scale: (1962)  Uses the electronegativity of orbitals rather
than atoms to develop group electronegativities for molecular fragments (e.g. $\ce{CH3}$ vs $\ce{CF3}$) that take into account the charge of a group, the effects of substituents, and the hybridization of the bonding orbital
Phillips Scale: (1968)  Defines electronegativity in terms of the
dielectric properties of atoms in a given valence state
Martynov & Batsanov Scale: (1980) Obtained by averaging the
successive ionisation energies of an element's valence electrons
Allen CE Scale: (1992) Configuration energy (CE), the average
one-electron valence shell energy of the ground-state free atom, is
used to quantify metal-covalent-ionic bonding


Does benzene react with steam to create an alcohol? Benzene has a molecular formula of $\ce{C6H6},$ right? Surely it has some alkene-like properties because of the double bonds, so can it undergo a hydration reaction to create alcohol?

Strange as it may seem, benzene can be reacted with steam over a suitable catalyst [1].  The result is a complete breakdown to simpler molecules like methane and carbon dioxide.  Typical conditions are 0.055 atm partial pressure benzene, 0.60 atm partial pressure steam, and 370 °C temperature.

The cited reference also considers the breakdown of alkylbenzenes with steam under similar conditions.

Reference

Masahiro Saito, Yoshio Sohda, Michiaki Tokuno, Yoshiro Morita, Reactions of Benzene or Alkylbenzenes with Steam over a Silica-Supported Nickel Catalyst, Bulletin of The Japan Petroleum Institute, 1972, Volume 14, Issue 1, Pages 54-63.


What is the difference between λ and Λ with regard to molar conductivity? I came across these equations (boxed in red) in a book:

  $$
\begin{align}
&\therefore & t_\mathrm{c} &= \frac{u_\mathrm{c}}{u_\mathrm{c} + u_\mathrm{a}}\\
&\text{and} & t_\mathrm{a} &= \frac{u_\mathrm{a}}{u_\mathrm{c} + u_\mathrm{a}}
\end{align}
$$
  At infinite dilution,
  $$\text{Also,} \quad t_\mathrm{c}^\infty = \frac{u_\mathrm{c}^\infty}{u_\mathrm{c}^\infty + u_\mathrm{a}^\infty}$$
  Since, $\lambda_\mathrm{c}^\infty = u_\mathrm{c}^\infty\times F$ and $\lambda_\mathrm{a}^\infty = u_\mathrm{a}^\infty\times F$
  $$
\bbox[3px,border:1px solid red]{
\begin{align}
&\therefore & t_\mathrm{c}^\infty \times \Lambda_\mathrm{elecrolyte}^\infty &= \lambda_\mathrm{c}^\infty\\
&\text{and} & t_\mathrm{a}^\infty \times \Lambda_\mathrm{elecrolyte}^\infty &= \lambda_\mathrm{a}^\infty
\end{align}
}
$$

Where $t$ refers to the transport number of an ion, and the subscripts $\mathrm{a}$ and $\mathrm{c}$ refer to an anion and cation, respectively.

From what I've learnt, the uppercase lambda $\Lambda$ refers to molar conductivity. But then what does the lowercase lambda $\lambda$ stand for?

Different sections within the same book give different definitions for $\lambda$, one being 'ionic conductance' and another being 'ionic conductivity', and I'm pretty sure 'conductance' and 'conductivity' are not the same thing.

I did try looking it up online, but Wikipedia wasn't much help.

Look in the Wikipedia article Ion transport number.

Lower case lambda, $\lambda$, is the molar ionic conductivity

Upper case lambda, $\Lambda$, is the molar conductivity.

$\lambda = t \dfrac{\Lambda}{\nu}$

Where $\nu$ is the number of ions for the cation or anion in the molecule. So $\ce{Na3PO4}$ would have 3 cations per molecule, but only 1 anion per molecule.


Ultrasonic bath operation time I use an ultrasonic bath in my lab for some chemical treatment of samples. It usually takes several hours with short breaks or even without it. I have a timer at the bath with a limit 90 min. So I have to run it every time again after it ends. My colleague told me that if I'm not giving a rest to the bath after every 90 min run, then the ultrasonic generator will be broken very soon. So, in general, how much time should be between the runs? Can external chilling enhance the safe treatment duration? It is a 3 liter, 100 W, 40 kHz Finnsonic bath with a heater.

Beside Jon Custer's comment, to be taken with a grain of salt (as the op mentions a heating unit implemented in the ultrasonic bath), if the manual allows, it depends.

You will recognise that by simple use (even if nothing but water is in the trough), that the temperature of the water will increase simply bu use.  Hence, to keep the results reproducible from one batch, to the next, you power on the bath until this temperature stabilises.  (It is better to briefly switch off the sonification to record the temperature with a thermometer.)

In terms of duration of sonification: while one aims to keep the reaction time under sonification short(er) than by conventional heating (alone), like

(source, doi 10.1016/S0008-6215(00)90155-1)

there are examples, where the sonification reportely has to last much longer than $\pu{90 min}$, for example

(source)

I've seen colleagues run reactions from morning till noon, for the sake of selectivity of the reaction intentionally cooling the bath by intermittent addition of water ice into the trough.

To attenuate the cavatational noise the bath may generate, esp. if your flat bottom vessel hovers with some distance just above the resonators (the hot spots), either consider a different room (or at least, keep the sash closed), or have a "noise-proof shoe box" like

(source)

that you may design by yourself, too, covering your bath durin the operation.  Instead of the foam-lining, a wooden box lined with polysterene is even cheaper, but offers less "relief". It is even better to place the bath (if weight permits) at the same time on a sheet of expanded polysterene that you get "for a smile" at at your departemental chemical shop, or appliance shop.

But do not forget the ultrasound bath underneath the box / in the other room, especially if your heating unit is used.

(source)

Beside this, you may interested in periodicals like Ultrasonics and Ultrasonics Sonochemistry, or Luche and Capelo-Martínez as secondary references, too.


Identifying chiral carbons Looking at this structure, I would say this molecule has five chiral carbons. However, its IUPAC name (3R,4R,5S,6R)-3-amino-6-(hydroxymethyl)oxane-2,4,5-triol suggests that there are only four. Which is it?

The structure as drawn indeed contains 5 chiral carbon atoms.  However, the one carbon that is bound to two oxygen atoms is called the anomeric carbon.  It is part of a hemiacetal moiety in the structure you have drawn.  That means that at room temperature and in aqueous solution, it will equilibrate reasonably quickly between:

One enantiomer of hemiacetal
An open-chain non-cyclic molecule where the anomeric carbon is an aldehyde moiety and is thus not chiral.
The other enantiomer of hemiacetal.

Thus, for many applications it is probably better to think of this molecule as having only four "fixed" stereocenters.


Adiabatic mixing of liquids One mole of benzene at $\pu{20^\circ C}$ is adiabatically mixed with 2 moles of toluene at $\pu{40^\circ C}$. The heat capacity of benzene is $\pu{84 J K-1 mol-1}$ and the heat capacity of toluene is $\pu{96 J K-1 mol-1}$. If benzene and toluene form an ideal solution, compute the entropy change for the process.

I tried to use

$$\Delta S = n C \ln{(T_\mathrm f/T_\mathrm i)}$$

where $C$ is the heat capacity and $T_\mathrm f$ and $T_\mathrm i$ are the final and initial temperatures of the liquids, respectively. However, I am unsure of what the final equilibrium temperature of the mixture comes to.

Let's say the final temperature is $T$. Also we assume the system is isolated and the process is adiabatic, as you said. Also all energy that is released by toluene or gained by benzene is heat.

The absolute amount of heat then released by toluene equals to the absolute amount of heat gained by benzene:

$|Q_\mathrm t| = |Q_\mathrm b|$

Considering signs, it will look like this:

$Q_\mathrm t = -Q_\mathrm b$

By definition, $Q = nc\Delta T$. Thus,

$Q_\mathrm t = 2\cdot96\cdot(40-T)$, and

$Q_\mathrm b = 1\cdot84\cdot(20-T)$

Plug in the latter two equations into the second one, and here's what you get:

$2\cdot96\cdot(40-T) = -1\cdot84\cdot(20-T)$

Now just solve it for $T$. And plug that in your equation as $T_\mathrm f$.

Note that it's not really important to do the temperature in kelvin, since we are dealing with changes in temperature. And yeap, your $T$ should be between $20\ \mathrm{^\circ C}$ and $40\ \mathrm{^\circ C}$, because there are such things as thermodynamics laws. So if it's not, you know for sure you did something not really correct.


Can I use one alkaline battery to recharge another? Unfortunately, I don't have batteries to waste to test this myself, but I am curious...

Let's say you have two identical batteries. AA Duracell.

Now let's say you run one of the batteries through an LED until the LED starts emitting at half the intensity it was emitting at when you first started (effectively draining the battery to half voltage)

Now, lets' say you touched the positive ends of both batteries for a reasonable amount of time, and then did the same for the negative ends.

Given that the used battery would have a much lower voltage, raw electrical current would travel from the unused battery into the used battery.

So this brings me to my question -- Would the batteries now effectively have the same amount of energy in them? I am not just talking voltage, but chemically too, or would the similarity in voltage on both batteries immediately disappear if either of them were used (because while, yes, either end of the battery does contain equal electrical charge to its counterpart, either end of each battery does NOT contain equal chemical potential)

To simplify this, would touching two batteries as described above effectively drain the unused one, and charge the used one?

First, some batteries are rechargeable, but some are not. However, even if we talk about rechargeable one, then
1) some degradation during recharge usually occurs
2) deep equality will require very long time to achieve


Compounds forming between silicon gallium and nitrogen What are the possible compounds that can form between silicon, gallium and nitrogen?

The ASM Alloy Phase Diagram Database (ASM Database if your institution has access) shows GaN and Si$_{3}$N$_{4}$ as the only compounds in that system. Standard binary alloy references agree. No ternary compounds are listed. Given the difference in crystal structures between all end members and the two nitride phases, mutual solubility of any two phases seems unlikely (little to none exists in the binaries, even Ga in Si has a pretty low solubility).


How can enthalpy change of a system be negative while entropy change is positive? $$\Delta G = \Delta H_\text{system}-T\Delta S_\text{system}$$

$$
\begin{array}{ccc}
\hline
\text{Sign of}~ΔH & \text{Sign of}~ΔS & \text{Spontaneity} \\
\hline
+ & + & \text{The reaction is spontaneous at high temperature} \\
+ & - & \text{The reaction is never spontaneous} \\
- & - & \text{The reaction is spontaneous at low temperature} \\
- & + & \text{The reaction is always spontaneous} \\
\hline
\end{array}
$$

When enthalpy change is negative, the reaction is exothermic, which means it releases energy into the surroundings. If the system is losing energy, shouldn’t the entropy of the system always decrease? I understand mathematically that $\mathrm{d}S = \mathrm{d}Q/T$, and if there is heat exchange, then entropy change can be positive.

But intuitively, if energy is taken away from the system, shouldn’t the entropy of that same system decrease? For some reason, I always thought that more energy means more entropy.

For example, suppose you have a solid block of TNT.

It explodes and releases much energy. $\Delta H$ is negative.

Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.


Reduction of nitro group to amine In the above reaction, I do know that $\ce{SnCl2/HCl}$ reduces $\ce{-NO2}$ group to $\ce{-NH2}$ but which one of the two $\ce{-NO2}$ will be reduced first by $\ce{SnCl2/HCl}$? Why so?

And how does $\ce{NH4HS}$ react and again which one of the two $\ce{-NO2}$ will be reduced first by $\ce{NH4HS}$ and why so?

NOTE: 1eq represents 1 equivalent.

Hyperconjugation accounts for the somewhat greater negativity of the 4-nitro-group compared with the 2-nitro-group in 2:4-dinitrotoluene; in consequence, selective reduction of the former by alkaline sulphides produces mainly 2-nitro-4-aminotoluene, although some 4-nitro-2-aminotoluene is also formed. In acid medium the 2-nitro-group becomes relatively negative in comparison with the 4-nitro-group, owing to the inductive (+I) effect having the chief role, with resultant preferential salt formation to give the very positive $\ce{-NO2H}$ group at 2-position and preferential reduction of it by acid stannous chloride.

Quoted from Coloration Technology 62 (1946) 114

As for how ammonium sulphide reacts see this answer:  Selective nitro reduction of poly nitro compounds


Are IR signals affected by nearby atoms? E.g. If there is a methyl from an aldehyde vs a methyl at the end of a hydrocarbon chain, will the difference in neighboring atoms have an effect on the location of the IR signal?

As was already said,

  IR works on bond lengths. So if the difference is negligible, instrument noise is going to dwarf anything useful you'll pull out.

There will be a difference, and you can definitely detect it computationally, but by normal experimental means (benchtop FT-IR or ATR, not a gas-phase experiment) background noise might the same order of magnitude as any frequency shift you'd see.

For example, the carbonyl stretching frequency and carbonyl bond lengths calculated using B97-D/6-31G(d):

$$
\newcommand{\AA}{\unicode{x212B}}
\begin{array}{llll}
           & \text{bond length} \, (\AA) & \text{frequency, calc.} \, (\text{cm}^{-1}) & \text{frequency, expt.} \, (\text{cm}^{-1}) \\ \hline
\text{pentanal}   & 1.218 & 1772 & 1727 \\
\text{2-hexanone} & 1.223 & 1753 & 1718
\end{array}
$$

Technically, you can distinguish between these in an experiment, but one would be more interested in the presence/absence of a large aldehyde band. This works due to the sharpness of the carbonyl stretch.

Closer to what I think you're asking about is the influence on vibrational frequencies of groups far from the area of substitution. These particular examples are tough because the normal modes of hydrocarbon-like species tend to couple the motion of many $\ce{C-H}$ stretches together. With that in mind, here are the highest-frequency $\ce{C-H}$ stretching modes, which happen to belong to the terminal $\ce{CH3}$ groups. The bond length is for the hydrogen eclipsed with the carbonyl.

$$
\newcommand{\AA}{\unicode{x212B}}
\begin{array}{lll}
           & \text{bond length} \, (\AA) & \text{frequency, calc.} \, (\text{cm}^{-1}) \\ \hline
\text{pentanal}   & 1.102 & 3044 \\
\text{2-hexanone} & 1.102 & 3038
\end{array}
$$

"Wow, that's amazing, you can resolve it!" Not this time. Compare the full spectra for pentanal/valeraldehyde (top) and 2-hexanone (bottom) and try to perform the peak assignment. Is it it the 2960/2962 pair? 2935/2936?

Experimental values/images taken from the Spectral Database for Organic Compounds.


Finding the Freundlich adsorption isotherm from the graph of log(x/m) v/s log(p) The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

The mass of gas adsorbed, $x$, per unit mass of adsorbate, $m$, was measured at various pressures, $p$. A graph between $\log\frac xm$ and $\log p$ gives a straight line with slope equal to $2$ and the intercept equal to $0.4771$. The value of $\frac xm$ at a pressure of $\pu{4 atm}$ is: (Given $\log 3 = 0.4771$)

I know the Freundlich adsorption isotherm formula, but I decided to write the equation of given line $( y = c + mx)$ as,
$$ \log \frac xm = \log 3 + 2\log p $$
So, the relation comes out to be,
$$ \frac xm = 3p^2 $$
On plugging $p = 4$, we get,
$$ \frac xm = 3(16) = 48 $$
But, the answer was wrong as per the official key, which mostly has genuine answers. Where am I wrong?

The answer given is,

 6

One version of Freundlich adsorption isotherm equation is:
$$ \frac xm = Kp^{\frac12}, $$
which can also be written as:
$$ \log \frac xm = \log K + \frac12\log p $$
This is a straight line equation of type $( y = c + mx)$ as given in the question. However, I think the question has made a mistake saying the slope is $2$, but instead it should be $\frac12$. Accordingly, the equation with given numeric values should be:
$$ \log \frac xm = \log 3 + \frac12\log p $$
Or without logarithm:
$$ \frac xm = 3p^\frac12 $$
On plugging $p = 4$, You'll get:
$$ \frac xm = 3 \times 4^\frac12 = 3 \times 2 = 6 $$
Hence, you get the given answer.


How do you Synthesise Glycosides of flavonoids and stilbenoids? I would like to know how you go about synthesising glycosides of flavonoids and stilbenoids from their respective sugars and flavonoids/stilbenoids. I read http://www.netsci-journal.com/97v1/97005/005p20.htm but I found it too specific (as it only applied to aliphatic compounds) and required an enzymatic catalyst that is impossible to predict. I would like a general formula that if it does contain an enzymatic catalyst the enzyme is either constant or at least usually predictable.

I suggest to forget about the enzymes.

Pepare for a probably lengthy synthesis - you will need patience and a clever protection/deprotection scheme.

The following is definitely not applicable to all starting materials but might give some directions.

If the flavonoid contains a catechol moiety, just like quercetin (1), the vicinal phenolic $\ce{OH}$ groups can be protected as an acetal by heating the starting material with $\alpha,\alpha$-dichlordiphenylmethane in diphenylether (DOI). The protecting groups is easily removed by hydrogenation over $\ce{Pd/C}$ at 1 atm.

Note that the PMB (4-methoxybenzyl) group is another protecting group for (phenolic) $\ce{OH}$ that gets off by hydrogenation.

As far as selectivities for the protection of the other $\ce{OH}$ groups in 1 are concerned, have a look at the 1962 paper by Leonard Jurd, who described that the methylation of quercetine pentaacetate in dry acetone predominantly furnishes 7-O-methylquercetin. There are quite some more recent articles citing this!


Does the bond dissociation enthalpy vary with temperature? The $\ce{H-H}$ bond enthalpy is $436 \mathrm{kJ/mol}$:

$$\ce{H2 (g) -> 2H (g)} \qquad \Delta H = +436 \mathrm{kJ/mol}$$

Does this only apply at the standard temperature of $298 \text{K}$, i.e. would the bond enthalpy vary if I decrease or increase the temperature?

And is it practically just so that, for a temperature $T$, as long as you pump the bond enthalpy’s worth of kilojoules into the substance, it will be atomized?

The bond dissociation enthalpy, like any enthalpy of reaction, will have some variation with temperature. This change in enthalpy can be determined using Kirchoff's Law: $$H_{T_f}=H_{T_i}+\int_{T_{f}}^{T_{i}}\Delta c_p(T')dT'$$
Where $$\Delta c_p(T')=\sum_i\nu_ic_{p,i}(T')$$ so that $c_{p,i}$ is the constant pressure heat capacity of the $\mathrm{i^{th}}$ compound in a reaction and $\nu_i$ is the the reaction coefficient of the $\mathrm{i^{th}}$ compound ($\nu_i$ is positive for products and negative for reactants).

If none of the $c_{p,i}$ have a temperature dependence, we can rewrite the $1^{st}$ equation above as $$H_{T_f}=H_{T_i}+\Delta c_p\cdot (T_f-T_i)$$ If they do have a temperature dependence, you will have to know the functional form of $c_{p,i}$ with respect to temperature to integrate them.

So, in regards to your specific question, the bond enthalpy for an $\ce{H-H}$ bond will be different at different temperatures. However, it is worth noting that for highly energetic reactions like this, the contribution due to $c_{p,i}$ is usually fairly small in comparison.


Can benzaldehyde undergo chlorination (aka halogenation)? I've read articles online and they rarely mention about chlorination of benzaldehyde but more on bromination.
Also, I would like to know the catalyst used for this is FeCl3 or AlCl3?
Thanks for your help!:)

The experimental process of chlorination of benzaldehyde to form benzoyl chloride can be found here:

  A round bottom flask is fitted with a reflux condenser and a gas inlet
  tube. 100 grams of cold and dry benzaldehyde are added to the flask
  and saturated with the current of dry chlorine. The chlorine gas is
  readily absorbed and the evolution hydrogen chloride takes place. When
  the reaction has moderated, the external heat is applied by
  keeping gentle boiling. The stream of chlorine is continued until no
  more hydrogen chloride is evolved. A stream of dry air or carbon
  dioxide is then passed through the apparatus in order to remove
  an excess of chlorine. Benzoyl chloride is obtained by distillation as
  a colorless fuming liquid with a very irritating smell. Yield almost
  quantitative.

  Preparation of organic compounds, E. de. Barry Barnett, 52, 1912

There is also a German paper written by Freidrich Wohler and Justus von Leibig which mentions an early preparation of benzoyl chloride by the action of chlorine over benzaldehyde.


Polarity of alcohols and their miscibility in water I am an AP Chemistry student and there is a small detail concerning polarity of molecules that I am confused about.

My review book (Princeton review) says that ethanol has stronger intermolecular forces than methanol because it has a large molecular mass and is therefore more polarizable and more soluble in water than methanol.

Yet, my chemistry textbook says that both ethanol and methanol are miscible in water, but the more the carbon chain increases, the less miscible it becomes. Further supporting this statement is the fact that methanol is less soluble in hexane, a nonpolar substance, than ethanol. Therefore ethanol is less polar. Plus, heptanol is hardly soluble in water, but according to the Princeton Review's logic, it should be even more soluble in water than methanol because it has a larger molecular mass.

So, which is it? Does the addition of CH3s make a molecule more or less polar, and more or less miscible in water?

If your book says exactly that, then it says nonsense. Sure enough, the addition of CH2's makes a molecule less polar. As for the solubility in water, methanol is miscible - i.e., infinitely well soluble in it. So is ethanol. How would you compare the two infinities?

That being said, the bigger molecule of ethanol is indeed more polarizable (not more polar, mind you!), and has more intermolecular interactions per molecule than methanol, which accounts for the higher boiling point.


hydrated halide structure What is the structure of $\ce{LiCl.2H2O}$?
I think when $\ce{LiCl}$ is put in aqueous medium, it will break into $\ce{Li+}$ and $\ce{Cl-}$. Then $\ce{H2O}$ will give LP to Li and hydrate it. But why 2 $\ce{H2O}$ molecules? 2 means 2 coordinate bonds? Octet is not filled? Then why does this happen?

The radius ratio of $\ce{Li+}$ and $\ce{Cl-}$ is very small, this means the size of $\ce{Li+}$ is so small than $\ce{Cl-}$ that the crystal lattice of $\ce{LiCl}$ becomes unstable. So to make the size of $\ce{Li+}$ bigger $\ce{H2O}$ molecules surround the caption bigger and make the structure stable.


What is an exocyclic double bond? I have been tasked to find the $\lambda_{\text{max}}$ of 3,4-dimethyl-1,2,3,4,4a,5-hexahydronaphthalene.

This was supposed to be fairly easy. I used the base value for homoannular diene ($\pu{253 nm}$) and added the increments for the three ring residues attached to the diene ($\pu{+(3\times 5) = +15 nm}$). This should get me a value of $\lambda_{\text{max}}=\pu{268 nm}$. However, checking the solution this question, they have also added the increment for an exocyclic double bond, and have got the answer to be $\pu{273 nm}$ instead.

As far as I know, an exocyclic double bond means exo+cyclic, where exo means "outside" and cyclic means, well "cyclic", or more simply, a double bond that is outside the ring. I'm looking at this compound, and I dont see any exocyclic double bond at all.

Am I dealing with a key mistake here or is there something that I'm not aware of?

Surely you are aware that the calculation of this sort is but a very rough estimate, yet still I feel the urge to stress that once more. Having said that, let's look at the double bond in the first cycle: isn't it exocyclic with respect to the other cycle?


Sunglasses that darken in sunlight I have seen several of my (rich) friends and teachers having glasses that are colorless indoors, but take up a darker shade when exposed to sunlight.

Upon searching for the chemistry of this, I did find out that silver was being reduced in the sunlight, and reverted back to silver ions indoors.

If silver was getting reduced, what's getting oxidized? And how is it that indoor lights (fluoroscent tubes etc.) do not affect the lens?

As mentioned in the comments, this topic can become exceedingly complex if all the aspects of current-gen photochromic lenses are brought to the table (and I'm sure a lot of it is locked away in trade secrets). My goal here is to provide some insight into the chemical/physical mechanisms at work.

Photochromic sunglasses are a smart new application of very old knowledge. Silver salts have been isolated and manipulated for centuries, so it's surely been known they have a tendency to darken in light for a long time. This simple fact (plus a lot of auxiliary processes) was the basis of the main forms of photography all throughout the 19th and 20th century.

Because of this, the classic examples of photosensitive compounds are the silver halides ($\ce{AgX}$), particularly $\ce{AgCl}$, $\ce{AgBr}$ and $\ce{AgI}$. These compounds are colourless to slightly yellow when pure. Upon exposure to photons of appropriate frequency, the idealised sequence of events is:

$$1) \quad \ce{2 X- \xrightarrow{h\nu} X_2 + 2 e-}$$
$$2)\quad \ce{2 Ag+ + 2e^- -> 2 Ag^0}$$

So essentially, illumination of the silver halides triggers the production of elemental (i.e. metallic) silver. This is a photoredox process; we say the halide anions undergo photooxidation, and the silver cations undergo photoreduction. An interesting way to think of it is that the halides normally are not strong enough reductants to trigger the reduction of $\ce{Ag+}$ to $\ce{Ag^0}$. However, this is only true in the electronic ground state of these species - the halide ion $\ce{X-}$ in an electronic excited state has different chemical properties during its short life, and in particular it becomes a much stronger reductant, and now can reduce silver directly. This leads to another description, which can be more accurate in other contexts

$$1) \quad \ce{2 X- \xrightarrow{h\nu} 2X^{-*}}$$
$$2)\quad \ce{2 Ag+ + 2X^{-*} -> 2 Ag^0 + X_2}$$

Back to the silver. Instead of forming a shiny reflective film, the silver is formed as a multitude of tiny disconnected particles. As the photons bounce around chaotically in the maze of silver clusters, a large fraction of the incident light ends up getting absorbed (and turned into heat). On the macroscopic scale, this makes the material appear dark.

It's interesting to point out here that the element silver alone is not responsible for the darkening effect. The process starts with the absorption of light to generate a strongly reducing species, from which the silver ions then steal an electron to become silver metal. In principle, therefore, it is possible to tune this effect depending on what else is mixed with the silver ions. Indeed, as the halides get heavier, ionisation becomes easier (or alternatively, it becomes easier to reach the strongly reducing photoexcited state), and the silver halides become more photosensitive. However, it's possible to mix in something else entirely, such as electron-rich organic compounds, which could then surrender their electron to the silver ions when exposed to light. It's also not fundamental to use silver at all - all metals are dark when dispersed into very tiny clusters.

So we now put this knowledge together into one of the simplest examples of photochromic lenses. This technology may have been used in the past, but is now likely supplanted.

Like before, we start with a silver halide. For example, during the fabrication of glass, a small amount of $\ce{AgCl}$ can be added. Though it would look like otherwise ordinary glass, when struck with light, it can potentially trigger the formation of $\ce{Ag^0}$ and darken the glass.

The problem now is to make this process reversible. In the ideal reactions above, the formation of $\ce{Ag^0}$ would also require the formation of $\ce{Cl_2}$. These two compounds can in principle react to regenerate $\ce{AgCl}$ (after all, $\ce{AgCl}$ was the thermodynamic minimum, it had to be hit with a photon to turn into something else!). However, $\ce{Cl_2}$ could potentially trigger a number of other reaction, or move around in the glass (it's a gas in standard conditions), which makes the process unlikely to be reproducibly reversible.

To solve this problem, it's possible to add another pair of electrochemically active species. In particular, it turns out $\ce{CuCl}$ is a nice choice. Why? $\ce{AgCl}$ and $\ce{CuCl}$ can be mixed into glass without reacting, once again generating a colourless material. However, upon exposure to light, the following (idealised) reactions happen:

$$1) \quad \ce{ Cl- \xrightarrow{h\nu} Cl^{0} + e^-}$$
$$2) \quad \ce{Ag+ + e^- -> Ag^0 }$$
$$3) \quad \ce{Cu+ + Cl^0 -> Cu^2+ + Cl-}$$

Now the chloride is essentially a spectator species. The process could be rewritten as:

$$\text{Net ionic equation}: \quad \ce{Cu+ + Ag+ \xrightarrow{h\nu} Cu^2+ + Ag^0}$$

Over time, the tendency is for the $\ce{Cu^2+}$ to reoxidise the silver metal back completely to $\ce{Ag+}$, thus making the glass transparent again. This process is comparatively slow, however, and therefore while there is light shining on the glass, there is a buildup of $\ce{Ag^0}$, making it dark. All in all, very convenient, indeed.

Lastly, I'll briefly address why this process happens in sunlight, but not indoors. Unsurprisingly, this is an entirely intentional property. Good eye-wear glasses block invisible and damaging UV photons present in sunlight, but shouldn't block visible photons. Meanwhile, a good visible light source should create visible photons, and avoid wasting energy on harmful UV photons. The design of the materials used in both photochromic glasses and fluorescent lamps has therefore naturally lead to the situation where photochromic lenses should generally not darken with indoor light sources.

For a more hands-on approach to this whole topic, I highly recommend this excellent video.


Stuck with derivation of the integrated rate equation for a pseudo first order equilibrium reaction I was reading about the integrated rate law. However I have problems to follow the solution.
I have an equilibrium reaction:

$$\ce{A + B<=>[{k_{on}}][{k_{off}}]AB}$$

with forward and back reaction. I approximate the measurement to a pseudo first order reaction. The complex can be measured.
B can be calculated:

$$[\ce{B}]\overset{\text{def}}{=}[\ce{B_0}]-[\ce{AB}]$$

Therefore the equilibrium is

$$0 = k_\mathrm{on} \cdot [\ce{A_0}] \cdot ([\ce{B_0}]-[\ce{AB}])-k_\mathrm{off} \cdot [\ce{AB}]$$

which can be expressed also like this:

$$0 = k_\mathrm{on} \cdot \ce{[A_0]} \cdot \ce{[B_0]} - (k_\mathrm{on} \cdot A_0 - k_\mathrm{off})F$$

To solve the differential equation I simplified to this:
\begin{align}
\frac{\mathrm{d}F(t)}{\mathrm{d}t} &= c_1 - c_2 \cdot F(t)\\
c_1 &= k_\mathrm{on} \cdot A_0 \cdot B_0\\
c_2 &= k_\mathrm{on} \cdot A_0 - k_\mathrm{off}\\
\end{align}

So far the theory was easy and I am sure everything is correct, but now I am stuck. The solution of this differential equation would be straightforward:

$$F(t) = \frac{c_1}{c_2}+k\cdot \mathrm{e}^{-(c_2\cdot t)}$$

However, the solution to fit and simulate the kinetic should be (in the simplified writing). This equation is published in http://afm1.pharm.utah.edu/pnscourse/Anal_Biochem_1995.pdf.

$$F(t) = \frac{c_1\cdot (1-\mathrm{e}^{-(c_2\cdot t)})}{c_2}$$

I really would like to know, where is my mistake and how the derivations of this formula has to be.

You have made no mistake. Your solution and the solution from the paper are practically identical. The solution from the paper is

\begin{align}
  F(t)=c_1 \frac{1−e^{−c_2 t}}{c_2}=\frac{c_1}{c_2} - \frac{c_1}{c_2} e^{−c_2 t}
\end{align}

which is exactly the form your equation has, i.e.

\begin{align}
  F(t)=\frac{c_1}{c_2} + K e^{−c_2 t} \qquad \text{with} \qquad K = - \frac{c_1}{c_2} \ .
\end{align}

The integration constant $K$ is not defined the way it is but fixed via the boundary conditions of the reaction. In this case you get $K = - \frac{c_1}{c_2}$ by requiring that at the start of the reaction, i.e. $t=0$, the reactants have not yet reacted with each other and there is no product $\ce{AB}$ present initially, i.e. $F(t\!=\!0) = 0$. This leads to the desired result:

\begin{align}
  F(t\!=\!0) \overset{!}{=} 0 &=\frac{c_1}{c_2} + K e^{0} \qquad \Rightarrow \quad K=- \frac{c_1}{c_2} \ .
\end{align}


Carbon-Carbon double bond rotation I was solving this question and I thought the answer was B>C>A as more the resonance, more the pi electrons are delocalized so more free movement.

However the answer mentions it as B>A>C.

What is wrong with my reasoning?

Here you need to consider two factors.

The one which follows $4n+2$ rule.
The stability of the ring.

When we see the IIIrd one, the structure don't follow the $4n+2$ rule when the double bond breaks. That means, that the double bond will be reluctant towards breaking. Hence its stability will be least.

COMING TO NEXT ONE

The order of stability of the ring is in the order: 6−member > 5−member > 7−member > 4−member > 3−member.

So when we combine these facts we find the correct order is B>A>C.


What are methods of identifying which amino acids are in a sample of protein? Are there any methods out there to identify which amino acids are in a sample of protein? Perhaps using NMR?

For a destructive technique that works with any mixture of proteins (i.e. to assess dietary questions), you could first hydrolyze the proteins into the amino acid building blocks. Then, you could separate them using chromatography (TLC, HPLC, GC), either as they are or after derivatizing functional groups to make them less polar.

If you are asking about a sample containing a single protein coming from an organism that has been sequenced, you could also enzymatically hydrolyze the protein (for instance with trypsin) and then subject fragments to mass spectroscopic analysis. A database search would identify the protein

For non-destructive techniques, you could start out with a natural abundance 2D NMR experiment (1H 15N HSQC), which gives you limited information (e.g. number of Asn/Gln). If you are interested in aromatic amino acids, the UV spectrum will be of interest. If you want to know whether there are free cysteins, there are coupling reagents you could use.


Placing oxygen molecule into electric field I am trying to understand what happens inside various types of air ionisers.

Some of them are sold for their purported health effects (which I do not want to discuss here), another type are used in industry to control electrostatic charges on surface of materials. Also there are some natural processes where oxygen form is modified by electric current namely thunderstorms.

  Can an oxygen molecule be broken by placing it into a strong external electric field?

If yes how is it possible because as we know the oxygen atoms in molecules are bonded by a non-polar bond?

What are possible by products of this process (even temporary one)?

I assume that air is composed only from oxygen (no nitrogen, argon, carbon dioxide etc.).

Yes, an oxygen molecule (much like any other molecule) can be broken by electric field alone.
No, it won't be broken into separate oxygen atoms, precisely because the atoms are neutral and the bond between them non-polar, like you said. Instead, it would be broken into $\ce{O2+}$ and an electron.
This is not important anyway, since it would require electric field much stronger than those present in air ionizers.

What happens in air under voltage, then? Well, air always contains some ions from whatever sources (cosmic rays, background radiation, etc). As you apply voltage, your positive and negative ions start speeding up in the opposite directions. Once the voltage is high enough, they accelerate to the point that they might break neutral molecules by bumping into them. In doing so, they produce more ions, and those do the same, so pretty soon there is a breakdown going on. You see a blinding spark, the fuses blow up, the voltage shuts off, and the ions start to recombine into molecules. As luck would have it, sometimes they combine into $\ce{O3}$ - that's an important source of ozone both in nature and in the lab.

In fact, the situation is much more complicated. Gas discharge does not have to be that dramatic. It comes in various forms, some of them pretty controllable.


What is the structural difference between ketogenic amino acids and glucogenic amino acids? We know some amino acids could go in ketogenosis path way ,some in glucogenic and some in both. I 'm wondering to know what the differences are between the amino acids that make this situation.

“Standard” amino acids are degraded to one of seven metabolic
  intermediates: pyruvate, α-ketoglutarate, succinyl-CoA, fumarate,
  oxaloacetate, acetyl-CoA, or acetoacetate.

The amino acids may therefore be divided into two groups based on their catabolic pathways:

Glucogenic amino acids;

whose carbon skeletons are degraded to pyruvate, α-ketoglutarate, succinyl-CoA, fumarate, or oxaloacetate and are therefore glucose precursors.

Ketogenic amino acids;

whose carbon skeletons are broken down to acetyl-CoA or acetoacetate and can thus be converted to ketone bodies or fatty acids.

The amino acids degraded to  acetyl-CoA and acetoacetyl-CoA are used in the citric acid cycle, but mammals cannot synthesize glucose from acetyl-CoA. This fact is the source of the distinction between glucogenic and ketogenic amino acids. Glucogenic amino acids can be converted to glucose, with oxaloacetate as an intermediate, but ketogenic amino acids cannot be converted to glucose.

Strictly speaking, whether an amino acid is regarded as being glucogenic, ketogenic, or both depends partly on the observer. This classification is not universally accepted, because different quantitative criteria are applied will identify the degradation pathways by the entry point into metabolism.

I would also attribute to things like availability of feasible pathways, enzymes etc (for instance alanine is glucogenic because its transamination product, pyruvate can be converted to glucose via gluconeogenesis)

Some amino acids have more than one pathway for catabolism, which explains why four of the amino acids are listed as both glucogenic and ketogenic examples are Isoleucine, phenylalanine, threonine, tryptophan, and tyrosine):

Below I will include snippets of degradation products of named amino acids:

Five amino acids, alanine, cysteine, glycine, serine, and threonine, are broken down to yield pyruvate:

Asparagine and Aspartate Are Degraded to Oxaloacetate:

Arginine, Glutamate, Glutamine, Histidine, and Proline Are Degraded to α-Ketoglutarate:

The reason I included these pathways, is to allow you to visualise structure, feasible metabolic pathways and enzymes so you have a clear picture on the differences between these amino acids.

And so on..

Hope this helps..

References

Voet and Voet biochemistry
J.Berg Biochemistry
Campbell and Farrel Biochemistry


How to relate the conductance of an aqueous electrolyte to its conductivity In this answer on Physics Stack Exchange, the author states that conductance $G$ is related to conductivity $\sigma$ by $$G = \sigma\frac{A}{l}$$
where $A$ is the cross-sectional area perpendicular to the direction of electric current and $l$ is the length of the conductor.
How does this relate to the conductivity of solutions such as aqueous electrolytes?
Papers in the literature often give the conductivity of aqueous electrolytes in units of $\pu{S/m}$ or $\pu{mS/cm}$.  Here is an example from Wu et al., J. Electrochem. Soc., 2015 where the authors plot conductivity in $\pu{mS/cm}$ versus molar concentration in $\pu{M}$:

But, in looking for conductivity values for sulfuric acid as a function of molar concentration, I found this report.  It states the conductance, rather than conductivity, of aqueous electrolytes (Or so it seems).
On page 7 of that pdf, for example, the authors provide a plot of $\pu{micromhos}$ versus percentage by weight:

I think that $\pu{1 micromho} = \pu{1 S}$.  What information would I need to determine the conductivity (in units of, for example, $\pu{S/m}$ or $\pu{S/cm}$) of an aqueous solution of sulfuric acid from the above plot?

The answer is quite simple, if you look at the equation carefully. Conductance and conductivity are related by cell constant, $\displaystyle\frac{\text{Area}}{\text{Distance}},$ where the area is the area of the electrodes and the distance is the distance between the electrodes.
$$G = \sigma\frac{A}{l}$$
In conductivity measurements, the cell constant was either given by the manufacturer or it was determined experimentally by making a certain concentration of $\ce{KCl}$, the conductivity of which was well established and standardized.
In short, you cannot convert micromho to conductivity because you do not know the cell constant. And weight by weight conversion to molarity is not that trivial. You need to know exact density of each solution.
Did you see the end pages of the manual from you posted the old graph, there is a useful chart (for future readers).


Solution to the rate law of first order reaction followed by second order reaction (consecutive) I'm trying to derive the rate law of a first order followed by second order consecutive reaction:

$$ \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=-k_1[\ce{A}]$$
$$ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_1[\ce{A}]-k_2{[\ce{B}]^2}$$
$$ [\ce{A}](0)=[\ce{A}]_0,\ce{[B]}(0)=0$$

The solution for A is trivial:
$$ \ce{[A]}=\ce{[A]}_0\exp(-k_1t)$$

The problem is B, whose rate law ODE is not linear:
$$ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_1[\ce{A}]_0\exp(-k_1t)-k_2{[\ce{B}]^2}$$

This is a nonlinear first order ODE, but can be rewritten to a linear second order ODE by substituting [\ce{B}] with:
$$[\ce{B}]=\frac{\mathrm{d}y}{\mathrm{d}t}\frac{1}{k_2y}$$

Which gives:
$$ -\frac{\mathrm{d}^2y}{\mathrm{d}t^2}+k_1k_2[\ce{A}]_0\exp(-k_1t)y=0$$

The solution is quite complicate and contains bessel functions of both the first and the second kind:

$$ [B] = \frac{C_1\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,J_1\left(\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-\ce{A}_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)\,\sqrt{-A_{0}\,k_{1}\,k_{2}}-\ce{C}_2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,Y_1\left(-\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{2}\,\left(C_1\,J_0\left(\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)+C_2\,Y_0\left(-\frac{2\,{\mathrm{e}}^{-\frac{k_{1}\,t}{2}}\,\sqrt{-A_{0}\,k_{1}\,k_{2}}}{k_{1}}\right)\right)}$$

There are two arbitrary constants but I only have one condition:
$$\ce{[B]}(0)=0$$

So my questions are:

Is my approach solving the rate law correct ?
If so, how do I get rid of both constants ?
Is there any approximated solution to this rate law ?

[update]:
The ode is solved numerically with Runge-Kutta method. The symbolic solution seems correct.

The  following code is in Python and uses numpy, scipy and matplotlib that are usually packaged with python.  The code will solve the equations $\mathrm{A} \underset{k_3}{\stackrel{k_1}\leftrightharpoons}  2\mathrm{B} \stackrel{k_2} \longrightarrow \mathrm{ P}$ but you can set $k_3$ to zero if you want. (its called km1 in the code). I've used log time spaced points so that a wide time range can be seen. The x axis is log in the plots.

I have used a built in solver, odeint(), buy you could write your own based on the Euler or Runge -Kutta methods.

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

def dX_dt(X, t, const):     # returns rates dEf/dt, only used in odeint()

    k1, km1, k2 = const     # unpack constants needed for calc'n
    A = X[0]
    B = X[1]
    dAdt=    -k1*A  +  km1*B*B                 #  rate equations here
    dBdt=     k1*A  - (km1+k2)*B*B

    f = np.array( [(dAdt), (dBdt)] )           # evaluate equations;
    return f

C0 = np.zeros(2,dtype=float)
C0[0] = 1.0e-3    # C0= [ A(0), B(0)   ]      # initial concentrations
numt = 200
t = np.logspace( -9,-4,  numt )    # time as initial power, final power, num points
k1  = 1e7  # 1st order rate const s^(-1)
km1 = 1e6  # 2nd dm^3mol^(-1)s^(-1)
k2  = 1e9  # 2nd  ....
const = [ k1, km1, k2 ]                            # rate constants for calc'n

soln = odeint(dX_dt, C0, t, args = (const,)  )     # solve equations
Aval, Bval = soln.T     # extract data

plt.plot(t,Aval,color='red',  label='A')
plt.plot(t,Bval,color='green',label='B')
plt.plot(t,C0[0]-Aval-Bval,color='blue',label='P')
plt.xscale('log')
plt.legend()

plt.tight_layout()
plt.show()

The figure shows the populations vs (log) time


What household compound will neutralize naphthalene-stained washer (for clothes)? Please help me tag this post.

Googling provided no answers except for vinegar, which did not work.

I washed this blanket for the first time that was purchased off of ebay.  There was a smell emitting from the washer as it was washing clothes.

This is no ordinary "mothball smell".  It is extremely stinky, like 30 magic markers were opened in the wash.  It's almost as if the blanket was DIPPED in naphthalene.

I tried running it empty with all the vinegar we had which was about 6-10 cups, together with powdered laundry soap, and both hot and cold wash cycles.  Will it help if I added some throw-away cloth in it to clean it out?  I even put it on a 2-hour wash.  The smell is as potent as when we started!

Will airing it out reduce the smell?  Ideally, I would like to completely remove the naphthalene, not just the smell.

Is cold better than hot water for this?  And lastly, what household compound can I buy that will neutralize the naphthalene?

Thanks.

Napthalene is not water soluble, so the temperature doesn't matter as much as the detergent. What makes you think it's napthalene, though? It could be some organic compound in which case you need a good de-greaser (Dawn is the easiest to get, but there are better industrial ones like what cleaning crews may use). If what you're dealing with is an aromatic hydrocarbon (which napthalene is), airing it out will go a long way. I would see how far you can get with a fan blowing it out before you wash the rest out with a good de-greaser.


Understanding secular equation derivation In many basic derivations of the secular equations for the energy of two molecular orbitals from a linear combination of atomic orbitals, we see the following reasoning which I find confusing.

We have a trial wavefunction $\phi = c_a\psi_a + c_b\psi_b$

The energy of the trial wavefunction is

$$\begin{align} \langle E\rangle &= \frac{\langle \phi | H | \phi \rangle}{\langle \phi | \phi \rangle}\\
&= \frac{c_a^2H_{aa}+2c_a c_b H_{ab} + c_b^2 H_{bb}}{c_a^2 S_{aa} + 2c_a c_b S_{ab} + c_b^2 S_{bb}} \end{align}$$

Where $S_{aa}$ and $S_{bb}$ equal 1. Now per the variational principle, we take the partial derivatives with respect to the coefficients $c_a$ and $c_b$ and set them equal to zero to find the optimal coefficients to minimize $\langle E \rangle$.

But instead, things follow a different route. We take the equation and move the denominator over:

$$
\langle E \rangle (c_a^2 + 2c_a c_b S_{ab} + c_b^2) = {c_a^2H_{aa}+2c_a c_b H_{ab} + c_b^2 H_{bb}}
$$

And then it's stated that this equation is differentiated with respect to $c_a$ and $c_b$ such that

$$
\langle E \rangle(2c_a + 2c_b S_{ab}) + \frac{\partial\langle E\rangle}{\partial c_a} (c_a^2 + 2c_a c_b S_{ab} + c_b^2) = 2c_a H_{aa} + 2c_b H_{ab}
$$

with a similar expression for differentiation wrt $c_b$.

Now the partial derivatives are set to 0, such that you end up with

$$
c_a(H_{aa}-E) + c_b(H_{ab} - ES_{ab}) = 0 \\
c_a(H_{ab}-ES_{ab}) + c_b(H_{bb} - E) = 0
$$

which then you can solve for E.

Is my understanding correct that we've obtained the minimum energies in a single step by first rearranging the equations and then differentiating? Of course we do not obtain values for the coefficients. How is the approach valid?

THe final equations are called the secular equations, which for linear systems isa way of finding the eigenvalues of the system by $A - \lambda I = 0$. But here we have off-diagonal terms $-ES_{ab}$, so I do not see the connection to linear algebra here.

It doesn't matter how you obtain an expression for the partial derivative, as long as you obtain one, and set the partial derivative to zero. The method you described above is probably easier than differentiating the fraction. All the steps that you did are mathematically valid, so your final result is valid. (The "differentiate both sides" step is valid because the equality holds for all values of $c_a$ and $c_b$.)

If you want to, you can differentiate the fraction and you will end up with the same result (i.e. the secular equations). I actually typed the maths out before, so here you go: https://chemistry.stackexchange.com/a/42636/16683 The notation is a tiny bit different, but you should be able to understand it.

As for the linear algebra parallel, the secular equations are only analogous to $Ax = \lambda x$ if you make the additional assumption that $S_{ab} = 0$, in which case $H-ES$ is directly analogous to $A - \lambda I$ (since the overlap matrix is now equivalent to the identity matrix). In this case, the secular equations reduce to $Hc = Ec$.

If you do not make that assumption, then the secular equations take the form of $Hc = ESc$ and the parallel you should draw is to the generalised eigenvalue equation $Ax = \lambda Bx$. The eigenvectors $x$ are said to be the "eigenvectors of $A$ with respect to $B$". There is a lot more information online if you are interested - simply google "generalised eigenvalue".


Mass of oxygen in He+O2 mixture A sample of $\pu{1.42 g}$ of helium and an unweighted quantity of oxygen gas are mixed in a flask at room temperature. The partial pressure of helium in the flask is $\pu{42.5 torr}$, and partial pressure of oxygen gas is $\pu{158 torr}$. What is the mass of the oxygen in the container?

As volume is not stated, I assume it's constant, and temperature is a constant too. Therefore, from the ideal gas law

$$\frac{p}{n} = \frac{RT}{V}$$

For helium and oxygen gas, the above formula is applicable, and the mass of oxygen can be determined:

\begin{align}
\frac{\pu{0.5592 atm}}{1.42/4.0026} &= \frac{\pu{0.20789 atm}}{m(\ce{O2})/31.99} \\
\to m(\ce{O2}) &= \pu{4.22 g}
\end{align}

This is my attempt on finding the mass. However, I believe it's not right. Why is that the case?

I don't see any flaws in your argumentation. I agree that $R, T, V = \mathrm{const}$, hence

$$\frac{p(\ce{He})}{n(\ce{He})} = \frac{p(\ce{O2})}{n(\ce{O2})}$$

$$m(\ce{O2}) = \frac{p(\ce{O2}) \cdot M(\ce{O2})}{p(\ce{He}) \cdot M(\ce{He}) } \cdot m(\ce{He})$$

Just one minor thing: you incorrectly converted partial pressure of helium: $\pu{42.5 torr}$ is $\pu{0.056 atm}$, not $\pu{0.559 atm}$, therefore your resulting mass should be 10 times higher.

In fact, there was no need to convert $\pu{torr}$ to $\pu{atm}$ in the first place. Don't do unnecessary steps if you can avoid them. At the same time don't drop any units and always remember to include them in your final calculation.

$$m(\ce{O2}) = \frac{\pu{158 torr} \cdot \pu{31.99 g mol-1}}{\pu{42.5 torr} \cdot \pu{4.00 g mol-1}} \cdot \pu{1.42 g} = \pu{42.22 g}$$


Which chemicals can counter the corrosion of tamperproof hardware encryption chips? I'm moving from software implementations to hardware. I need to reverse engineer silicon chips, though the tech that I want to study may have a corrosive chemicals that's supposed to destroy the chip upon touching.

What chemical gets sprayed onto the hardware and which chemical reactions occur?

If there's one substance that causes destruction then there must also be one that prevents it. Does anyone know what this chemical is?

This question in currently going on in the back of my head. Chemistry isn't really my thing; but I know enough to understand the chemical reactions to make various drugs.

I'm not funded by government. This is independent research. The field of cryptography is somewhat of a front line of engineering. If there are no answers to the main question, chemistry resources in the context of silicon and the relating chemicals used in production as well as the securing in the post production will be accepted as the answer.

Hydrogen fluoride protects the chip by converting into hydrofluoric acid upon contact with moisture or human tissue.

I was able to conclude this through this wiki article:
Hydrogen Fluoride

  Upon contact with moisture, including tissue, hydrogen fluoride immediately converts to hydrofluoric acid, which is highly corrosive and toxic, and requires immediate medical attention upon exposure.[31] Breathing in hydrogen fluoride at high levels or in combination with skin contact can cause death from an irregular heartbeat or from fluid buildup in the lungs

The substance that can neutralize the Ph levels of the hydrofluoric acid is calcium gluconate.

Calgonate 2.5% calcium gluconate gel neutralizes HF

This can be used to help clean up accidental contact with hydrofluoric acid and could possibly be used to prevent any chemical reactions from occurring upon tampering with the chip. The best bet in this situation is to conduct experiments in a dry lab environment and wear gloves. You want to have a fume fan as well as the caluium gluconate in the event of failure.


Regarding acid-base reactions, how does one know when a hydrophobic base substance is fully protonated? When there's a weak base substance (hydrophobic powder), and it needs to be completely protonated by adding an acid (eg: HCl), how does one know when it's completely protonated? Since the powdered base substance isn't soluble until protonated, doesn't that mean that it won't be reflected when measuring the pH (via the commonly used pH test strips)?

I'm under the impression that when the base is suspended in water, as HCl is added and given time to react, the pH would float round 7 or 8. Then when it's fully protonated, the pH will start to dip below 7, which would mean there's now an excess of HCl since there's nothing for it to bind with, which would be too much.

I don't exactly have a specific substance base substance in mind. This is more a general question about acid-base reactions involving a solid base.

Thank you

Yes, that is correct.
As you add HCl, the pH will remain more or less constant at around 5, since in solution there is a salt of a strong acid and a weak base.
Then when all the base has been protonated, the pH will fall rapidly.

If you need to know how much insoluble base you have, it must be done through back titration. You add an excess of HCl to the sample and then titrate the remaining free HCl.


Steam distillation- how it actually works? To obtain phenylamine after it is formed from the reduction of nitrobenzene with tin, we carried out steam distillation.

This is what I learned in class:

  "During the process,steam is injected into the flask (containing the reaction mixture) to create artificial vapour pressure on the surface of the liquid. When vapour pressure is equal to atmospheric pressure, the liquid starts to boil at 95 degrees"

This didn't make much sense to me. So I searched it up on the net and I  got another explanation (source):

  As the hot steam passes through the mixture it condenses, releasing heat. This will be enough to boil the mixture of water and phenylamine at 98°C provided the volume of the mixture isn't too great

Here, it is saying that the heat produced from the condensation of the steam will allow the mixture to boil. I am really confused, I mean I carried out steam distillation but how is it different from normal distillation and how is phenylamine collected at a lower temperature through this process?

You boil the water and the product co-distills.  There is an anomalous selective interaction between steam and many hydrophobes.  If you heat the pot you deplete the water contained and may pyrolyze contents on the hot wall.  Blowing in steam limits temp to near 100 C, moves a lot of energy into the flask and out (540 cal/g enthalpy of vaporization), and maintains the water content.

You could filter off the solids, acidify the aqueous raffinate,  extract non-basics, then alkalinize and extract the aniline.  Since aniline is air sensitive to oxidation and extremely toxic by skin contact, you can appreciate advantages of steam distillation.


Does the temperature during ice melting change? I did an online test for physics and one of the questions was: Salt was added to ice. Because of it the ice melted. What was happening with temperature DURING the process of melting?   The right answer was that temperature was lowering.
Can please someone explain why? I always thought that during melting the temperature is constant and after all ice is melted then the temperature changes.
Thank you very very much

Here is a naive interpretation, that may help you understand this phenomena.
In the presence of salt, water is attracted and "wants" to be adsorbed around $\ce{Na+}$ and $\ce{Cl-}$ ions. But this operation requires some energy for water molecules to separate the positive and negative ions. This is why the dissolution of $\ce{NaCl}$ in water is endothermic.
Now if you put salt in contact with ice, water in ice "wants" also to be attached to surrounding ions. But it cannot move. It first have to be a liquid. And there is no heat in the surroundings to make ice melt. Nature has found a way. Every piece of ice is divided in two parts. One part gives heat to the second part, that melts, makes water to dissolve the salt. But the remaining part has lost heat. So it gets cooler and cooler. The temperature of the remaining ice decreases from $0$°C to about -$20$°C.


Will magnesium undergo a redox reaction with zinc(II) or copper(II)? Magnesium can do a redox reaction with $\ce{Cu^2+}$ as well as with $\ce{Zn^2+}$, since Mg has the lowest reduction potential $E_0$. (I am aware that a reaction may or may not be feasible as redox, and this is according the reduction potential of the two half-equations)

\begin{array}{l|l|r}
\ce{Cu^2+ & Cu} &\pu{0.34 V} \\
\ce{Zn^2+ & Zn} &\pu{-0.76 V} \\
\ce{Mg^2+ & Mg} &\pu{-2.37V} \\
\end{array}

I'm supposing we have in one beaker, a solution of $\ce{Cu^2+}$ ions and $\ce{Zn^2+}$ ions and a piece of magnesium. So the possible reactants are: $\ce{Cu^2+}$, $\ce{Zn^2+}$, and $\ce{Mg}$.

The potential difference between $\ce{Mg}$ and $\ce{Cu^2+}$ is greater than with $\ce{Zn^2+}$.
So what's the answer?

My guess is "with $\ce{Zn^2+}$", since by comparing with another example, ethanol ($\ce{CH3CH2OH}$) reduces $\ce{MnO4-}$ to give $\ce{CH3CHO}$ instead of $\ce{CH3COOH}$, if the quantity of $\ce{MnO4-}$ was limited.
\begin{array}{l|l|r}
\ce{MnO4- & Mn^2+}     & \pu{1.51 V} \\
\ce{CH3CHO & CH3CH2OH} & \pu{0.19 V} \\
\ce{CH3COOH & CH3CH2OH}& \pu{0.05 V}
\end{array}

So $\ce{CH3CH2OH}$ reacted with $\ce{MnO4-}$ to give $\ce{CH3CHO}$, that leads me to deduce that in the case of multiple possibilities (i.e. feasible redox reactions), the reaction with the least potential difference takes place (since $1.51 − 0.19 < 1.51 − 0.05$).

From another hand, I guess it's familiar that the other option of having $\ce{Mn^2+}$ reacting with $\ce{Mg}$ is true as being held between the most oxidant which is $\ce{Mn^2+}$ and the most reductant which is solely $\ce{Mg}$ in our case.

Let’s assume for a second that magnesium reacts only with the zinc. The reaction would occur according the following equation:

$$\ce{Mg + Zn^2+ -> Mg^2+ + Zn}$$

You have already identified, if I read your question correctly that the redox potentials allow this and that the reaction is spontaneous etc. Assume that you have equimolar amounts of magnesium, zinc and copper for the rest of this thought experiment.

Now our new solution would contain magnesium(II), copper(II) and zinc(0) (okay, the zinc would deposit itself somewhere, but you get the picture). Notice anything? Yes, zinc and copper can react according to the well-known and often-displayed electrochemical reaction (depositing copper on a blank piece of zinc metal). The equation for that one is:

$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$$

Again, check out the standard potentials to see that this reaction is possible and spontaneous.

If we add the two reactions together, we get an effective reaction:

$$\ce{Mg + Zn^2+ + Zn + Cu^2+ -> Mg^2+ + Zn + Zn^2+ + Cu}$$

$$\ce{Mg + Cu^2+ -> Mg^2+ + Cu}$$

You notice that the zinc is little more than a spectator ion.

Now how does this differ from the ethanol case? Well, permanganate can (and will) react with ethanol to form ethanal, and it can (and will) react with ethanal to form ethanoic acid — more rapidly. However, if permanganate (and therefore: electrons) is the limiting reagent, we end up with a mixture of ethanol, maybe ethanal and ethanoic acid. By a reaction analogous to the reverse Cannizarro reaction, these can comproportionate:

$$\ce{CH3CH2OH + CH3COOH <=>> 2 CH3CHO}$$

Therefore, the overall reaction seems to be one where ethanol only reacted with permanganate to only form ethanal, while in fact ethanoic acid was formed, but in the end the equilibrium favoured ethanal.


Learning chemistry in a new language I'm from Slovakia, 16 and like chemistry a lot. I want to pursue it in the future but there aren't any good universities here where you can get a Ph.D. in chemistry so I want to go abroad, preferably to America or the UK. But there's a problem: I don't even know how to name molecules in English. I want to prepare for the future so I want to learn that before I go to UNI. And there's another problem. I can't find a source that I can learn from. I just want to know the basics: oxides, salts, acids, and hydroxides. If anybody could send me a link to a site that I can learn it from I'd be very thankful.

P.S.
I don't know if it's done in English the same way, but we use oxidation numbers to get the chemical formula from the name and in reverse. That's what I want to know but in English.

EDIT: I just want to know the basics RN and I don't want to start off with organic chemistry and stuff like that. I just want to know how to take a simple molecule, for example, AlCl3 and know how to say its name. Is it aluminum (III) chloride or aluminum trichloride? Or something like SO4 and SO3. Is it sulfate and sulfane or what? I just want to know how to name molecules and write down their chemical formula.

Beside the resources for learning chemistry Poutnik indicated, you may learn some of the chemical names and concepts from wikipedia, too. Frequently, articles are available in more than one language and sometimes a version in an other language offers additional details, too.  You find the switch at the left side of the pages:

Typically, you find some references at the bottom of these pages which may be accessible by you without going to a university library.

As side note: You mention that you did not yet start to study chemistry at a university at all.  Hence, it is too early for you to judge if a Slovak (or Czech, given the language) university is good or bad for your aspiration of a PhD in chemistry; because you likely do not know yet which specialty will attract your interest, nor potential co-workers and supervisors.  Remain active to obtain a solid chemical education first -- a chemistry department with classes about 30--60 students per year may be more helpful for laying such a foundation, than a larger institution where you become "just a number".  Do not exclude smaller departments (and countries) just because they are smaller, or less frequently in the scientific media, than others when thinking about a move to an other institution is an opportunity.


Is this poor laboratory safety? I'm a formulation chemist. My co-worker, is also a formulation scientist but has no previous formal training in chemical safety or lab chemistry settings.
I always work in the hood unless I’m just mixing fruit extracts or something non-volatile. My coworker doesn't ever work in the hood as they believe that since this goes in food it must be safe.
The lab is small and our work stations are pretty close. My coworker opens up reagent bottles and performs all other tasks on the bench top.
I don't think their reasoning is correct since in food items, the chemicals we work with are present in the ppm to ppb ranges whereas when a pure bottle of them is opened on a lab bench, they are in high concentrations. In addition we face prolonged exposure  ($8$ hr work days,$5$ days a week). The two scenarios are very different.
Also, I feel like I’ve had a sore throat ever since I’ve started working there.
I just don’t think these are prudent lab practices. Am I overreacting or does my coworker need to start working in the hood?

This is more a comment than an answer, but too long for the former....
There are two things for you (and your company) to consider:

The SDS of the chemicals being handled - that will give you guidance for how best to handle a chemical. Flavouring chemicals like benzaldehyde, cinnamaldehyde, limonene all have irritation warnings, and need to be handled in well ventilated areas; others like diacetyl and allyl hexanoate are straight up nasty. And I don't mean to be patronising with this list of flavouring chemicals 101. I'm sure your list of chemicals is extensive, but they will all have SDS that will dictate how you should be handling them. Use that to base your arguments for change, not the opinions of an online discussion group.
Relevant building codes for laboratories that will specify the required number of air turnovers per hour. If your workplace is old, there is a good chance that ventilation is not quite what it should be. This is not an easy (cheap) problem to solve, but if the company is aware of any potential problems it will be easier to improve workplace practices of working in fumehoods.


Memorising exceptions for soluble salts My daughter has a chemistry exam on Tuesday. She is doing quite well with her revision except for the following table which she is having difficulty remembering.

Does anyone know of an easy way to remember this? Or perhaps if someone could explain why these exceptions are insoluble that will help her instead?

Understanding why it happens unfortunately will not help her to remember. This is the classic case where you have to find a dirty way to remember them. Maybe you can make some practical experiment I remember most of them with a practical application of the salts.

There are a lot of online resource regarding Mnemonic Tricks to remember solubility rules, these are two video on youtubes.

Memonic tricks

Song!

Quoting the song:

  This is solubility it's so crazy, don't be a fool and follow these rules!


Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide Dehydration of carboxylic acids with phosphorus pentoxide ($\ce{P2O5}$) yields acid anhydrides, and in a similar reaction, amides dehydrated with phosphorus pentoxide yield cyanides. Can an arrow pushing mechanism be given for these reactions? If there isn't a mechanism, what, at least, gives phosphorus pentoxide the ability to dehydrate carboxylic acids and amides?

Phosphorus pentoxide itself reacts with water molecules to form phosphoric acid. That enables it to remove free water molecules by a simple hydrolysis reaction, but here of course there must be some other electron movement since the water being removed is not free water.

Here is a mechanism for the dehydration of an acid to an anhydride using phosphorus pentoxide.

The $\ce{OH}$ group is not a very good leaving group.  Even in simpler reactions like the dehydration of an alcohol to an olefin we often first convert the hydroxyl group into a better leaving group by protonation with acid or conversion to an inorganic ester (for example using thionyl chloride or phosphorus pentoxide).  The same thing takes place here - we are converting the acid's hydroxyl group into a better leaving group.

Why is $\ce{HPO3}$ such a good leaving group? Look at all the resonance structures you can draw for the anion $\ce{PO3-}$.  It's stability makes the reaction exothermic and strongly drives the dehydration process to the product side.

For completeness, here is a link to a drawing and discussion of the mechanism for the dehydration of an amide to a nitrile using phosphorus pentoxide. Same mechanism and principles (good leaving group, exothermic) as discussed above.


Oxidation of hydroxyl groups of glucose In all oxidation reactions of glucose, it seems that the aldehyde group alone gets oxidised and none of the hydroxyl groups. In one reaction with nitric acid, the aldehyde group and the terminal hydroxyl group get oxidised but none of the remaining hydroxyl groups. I was wondering why this happens. Are the hydroxyl groups less susceptible to oxidation for any reason? If so why would that be? And also is there a difference is ease of oxidation of the terminal hydroxyl and remaining hydroxyl groups?

Aldehydes are easier to oxidize via radical mechanisms because of the ease with which one can break the C-H bond. The resulting radical is stabilized via delocalization of an oxygen lone pair.
Such stabilization is absent in an alcohol. The alcohol shows a similar moiety, but I think the stabilization in the aldehyde is greater because of the shorter C-O bond, which improves orbital overlap.
In a laboratory setting, alcohols tend to be more stable to air, whereas aldehydes can be prone to air oxidation.


Is the cyclopropenyl anion antiaromatic? According to my book, a compound is antiaromatic if it is cyclic, planar, and possesses a fully conjugated system of p-orbitals with $4n$ π-electrons.

However, I have also been told that the cyclopropenyl anion is neither aromatic nor antiaromatic.

How is this so? It has four π-electrons and therefore should satisfy the criteria listed above.

When you have a ring that has $4n$ π electrons it's anti-aromatic if you force it into a fully conjugated cycle.

Often such rings find a way to break the unfavorable (or at least, less favored) anti-aromatic conjugation.  Large ones like cyclooctatetraene are likely to bend so all ring atoms are no longer in the same plane. Small ones like the cyclopropenyl anion shown here can't bend the ring out of planarity easily (or do it at all with only three atoms), but they can still break the conjugation by using unequal bond lengths or moving a ligand out of the plane giving that ring atom a pyramidal bonding geometry.

The pyramidal bonding geometry seems to be what cyclopropenyl anion actually does.  A computational result is shown at http://comporgchem.com/blog/?p=2987; the top carbon atom has its bonds directed towards the observer indicating a pyramidal geometry.


Any differences in metal ion emission spectra? If I measure the emission spectra of potassium using the following solutions

1) $0.1~\mathrm{M}$ of $\ce{KNO3}$

2) $0.1~\mathrm{M}$ of $\ce{K2SO4}$

will I see any significant differences?

I think there will be no significant differences because the spectrum does not depend on whether in 1) I am using $\ce{K}$ while in 2) I am using $\ce{K2}$ (the emission spectrum should be independent). Am I right?

$0.1~\mathrm{M}$ $\ce{KNO3}$ dissociates to give $0.1~\mathrm{M}$ $\ce{K+}$ ions.

$0.1~\mathrm{M}$ $\ce{K2SO4}$ dissociates to give $2 \times 0.1~\mathrm{M}$ $\ce{K+}$ ions.

Hence, in terms of emission intensity, $\ce{K2SO4}$ should give twice the signal if other parameters are same and if the detection response is linear.

Very sensitive instrument might detect the effect of counter-ion but that will be very difficult since the cation itself is always surrounded by water.


interaction of grape juice or wine with marble I have a marble cup which I drink wine from. If I leave a little bit of wine in the cup and wait maybe 30 minutes, the wine becomes discolored and appears black (photos attached). My conjecture is that this is the acid from the wine interacting with the calcium carbonate of the marble to produce calcium choloride? If so, is it toxic to continue drinking wine from the cup? Thanks.

CaCl2 and magnesium are not toxic at these levels. In fact NIH suggests around 1 g calcium per day: https://medlineplus.gov/magazine/issues/winter11/articles/winter11pg12.html

Marble itself is predominantly calcite or dolomite, that is carbonates of calcium and magnesium. Marble also contains other elements. We can't say if these other chemicals are safe just by a photo because composition of marble varies.

When you bought this cup - did they say it is safe to drink from it? Supplier should know if the material is safe or not. If you didn't get assurance that it is safe then you should assume it is not.


Can an organic compound undergo ring contraction to give more stable carbocation? Throughout my journey in organic chemistry I’ve only seen ring expansions by a compound in order to attain more stability, but yesterday I thought about the following reaction.

I feel that ring contraction resulted in a more stable carbocation.

Can someone tell whether the idea behind this reaction is correct or not?

Your idea of ring contraction to give more stable carbocation is correct and similar contraction has observed before (see example 4 below):

Courtesy of Department of Chemistry, Michigan State University, East Lancing, MI: link to webpage


How can an α,β-unsaturated ketone tautomerise? I was taught that for aldehydes and ketone to exhibit keto-enol tautomerism, it is essential that there must be at least one α-hydrogen.

For example, acetophenone and butan-2-one show keto-enol tautomerism, but benzaldehyde and benzophenone don't show keto-enol tautomerism, since they don't have any α-hydrogens.

However, I was subsequently told that the cyclic enone 6,6-dimethylcyclohex-2-en-1-one also exhibits tautomerism:

How is this possible? It does not have any α-hydrogens (the vinylic proton cannot be enolised as it is orthogonal to the C=O π* orbital).

If you have an α,β-double bond next to your carbonyl, a γ hydrogen can also take part in tautomerism. Take crotonaldehyde $\ce{H3C-CH=CH-CHO}$:

$$\ce{O=CH-CH=CH-CH3 <<=> HO-CH=CH-CH=CH2}$$

In general, a double bond is able to ‘extend’ the mesomeric properties of a carbonyl compound across two additional atoms. This is known as vinylogous; crotonaldehyde can form a vinylogous enol.

In the specific case of 6,6-dimethylcyclohex-2-en-1-one, these are the two tautomers:


Good Alternative for Plasticized PVC In my current job we use gloveboxes, but unlike most we use disposable PVC bags on the entry ports instead of fixed gloves. The bags have an O ring which is stretched onto a spigot proud of the porthole and are then clamped into place. They get disposed of and replaced after every use.

^ Like this but our glovebox is permanent ^

The problem with the PVC bags is that they are releasing hydrogen chloride. I'm in a team investigating a replacement material; however none of us are chemists or material scientists!

The material needs to be able to fit over 6” and/or 10" spigots, allow hydrogen to pass, and contain no halides.

So far we have considered using HDPE + plasticizer, and are trying to find a way to make it, although we are not sure it will work until we try it. We have also looked into using polyester, but we couldn’t find a clear and stretchy version.

I think what you should try is thermoplastics elastomers. For general descriptive overview, read: https://en.wikipedia.org/wiki/Thermoplastic_elastomer

DUPONT has a long line of Chemical Resistant Elastomers, but may be expensive for your use, yet I mention it here because you said your team may consider custem made units to fix it in your glove boxes, so it could be quite expensive. Following is a cheaper option, but you can find more by just googling under "thermoplastics elastomers":
Kent Elastomer Products, Inc., PO Box 668, Kent, OH 44240 (1-800-331.-4762); email: info@kentelastomer.com (https://www.kentelastomer.com/cost-of-cheap/)


Are there any actual uses of isodiaphers? While studying atomic structure, I came across the terms isotopes, isobars, isoelectronic species, isotones and isodiaphers. While I can accept that the classification of isotopes, isobars, isoelectronic species and isotones may be useful, I do not understand what could be the use of species with the same difference between number of neutrons and protons (isodiaphers).
Is there any place in chemistry where they have a practical use (like in experiments, theories or laws), or is it just a useless classification term?

In practical radiochemistry, the term is rarely needed but it is not useless. In particular, isodiaphers are used in radiochemistry to describe chains of alpha decays.
I would expect, the relative frequencies of the practical use of the related terms is about
"isotopes" > "isobars" > "isodiaphers" > "isotones".
Isobars are used in radiochemistry to describe chains of beta decay, therefore their importance might be similar to isodiaphers for alpha decay. However, isobars are also very important to describe fission yields so that in total they might be used more often than isodiaphers.