text:
Can carbonic acid form a salt with pH < 7?
Please read edit before answering
Background
I am currently studying in 10th grade. This question appeared in my science exam -
Which acid does not form an acidic salt?

Phosphoric acid
Carbonic acid
Hydrochloric acid
Sulphuric acid

I understood that all acids other than carbonic acid are strong, and the question is expecting me to answer carbonic acid, as at our level we are only taught
SA + SB -> Neutral salt
WA + SB -> Basic salt
SA + WB -> Acidic salt
My Efforts
Because there was no mention of WA + WB reaction in my book, while this was being taught in class out of curiosity I searched online and found out that to find the nature of salt of WA and WB, it is required to compare their dissociation constants (correct me if I am wrong here)
I felt that there must be a base weaker than carbonic acid so that their reaction would result in an acidic salt. I have tried to find such bases online but because of my lack of understanding of this topic, I am unable to do so.
Question
Can carbonic acid form an acidic salt? If yes, then upon reacting with which base?
Edit: By acidic and basic salts, I mean salts with pH < 7 and pH > 7 respectively. I did not know acidic and basic salts mean something else in chemistry. At our class level we are taught the meaning of acidic and basic salts in this context only
Edit 2: Phosphoric acid is not a strong acid, my bad for writing it as such

Phosphoric acid is not strong, all 3 acidic hydrogens are weakly acidic (even is the first one not much weak). Similarly, only the first H in sulfuric acid is strongly acidic, The second one is very strong weak acid.

Both ammonium and sodium bicarbonate solutions are basic, as the anion $\ce{HCO3-}$ is a stronger base than is the strength of the acid $\ce{NH4+}$, not even talking about $\ce{Na+}$.
Carbonic acid could form a mildly acidic salt in sense of forming acidic solution with a base weaker than $\ce{HCO3-}$, e.g. with pyridine or aniline. But their water solutions would be highly hydrolyzed to the respective base and $\ce{H2CO3(aq)/CO2(aq)}$, as such bases need very acidic pH. Carbon dioxide would be gradually escaping as gas.
$$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \tag{1}$$
$$\ce{H2CO3(aq) + H2O(l) <=> HCO3-(aq) + H3O+(aq)}\tag{2}$$
$$\ce{C5H5N(aq) + H3O(aq) <=> C5H5NH+(aq) + H2O(l)}\tag{3}$$
Very formally, the salt formula can be written as $\ce{(C5H5NH)HCO3}$, as it would in large extent hydrolyze to the left side of reactions (2) , (3) to pyridyne and carbon dioxide.

An acidic salt – aside of pH of its solution – means a salt where not all acidic hydrogens of a multiprotic acid are neutralized. Like

$\ce{KHSO4}$ (very acidic solution)
$\ce{KH2PO4}$ (acidic solution)
$\ce{NaHCO3}$ (mildly basic solution)
$\ce{Na2HPO4}$ (basic solution)

As the anion of such partially neutralized acid is both an acid ($\ce{HA- <=> H+ + A^2-}$) and a base ($\ce{HA- + H+ <=> H2A}$), pH of its solution depends on which of them is stronger.
Similarly, basic salts are combination of a salt and a hydroxide, like the bleaching powder is a complex combination of hydrated calcium chloride, hypochlorite and hydroxide.
Wikipedia:

Commercial calcium hypochlorite consists of anhydrous $\ce{Ca(ClO)}$2, dibasic calcium hypochlorite $\ce{Ca3(ClO)2(OH)4}$ (also written as $\ce{Ca(ClO)2·2Ca(OH)2}$), and dibasic calcium chloride $\ce{Ca3Cl2(OH)4}$ (also written as $\ce{CaCl2·2Ca(OH)2}$).

Or there is $\ce{CuCO3 . Cu(OH)2}$ in the mineral malachite.

How can someone figure out which of the bases is stronger between $\ce{HCO3-}$ and Pyridine. And if we have the assertion that pyridinium bicarbonate will be acidic in nature, then is the reason that pyridine is a weaker base than $\ce{HCO3-}$ hence pyridinium ion will be stronger acid than $\ce{HCO3-}$?

The values of acidity/basicity constanst can be found tabulated. The most available ways is getting them from Wikipedia pages for respective compounds, or various chemistry dedicated sited have listed them for many compounds.
If there is an equilibrium
$$\ce{HA(aq) <=> H+(aq) +  A-(aq)}$$
or
$$\ce{B(aq) + H+(aq) <=> BH+(aq)},$$
then acidity of $\ce{HA}$ or $\ce{BH+}$ and basicity of $\ce{A-}$ or $\ce{B}$ are linked by the equation
$$K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w},$$
resp. (at $\ce{25 ^{\circ}C}$)
$$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$$.
If there are two bases, then the stronger one forms by neutralization a weaker acid. Pyridinium with $\mathrm{p}K_\mathrm{a} = 5.23$  is  a stronger acid than CO2(aq) with $\mathrm{p}K_\mathrm{a1}=6.35$, therefore pyridine is a weaker base than $\ce{HCO3-}$.

Can you please tell the reaction of Carbonic acid and pyridine? I cannot find it online

Thanks for responding to my edits. I have one more query, will the reaction of Carbonic acid and Pyridine be an example of acid-base reaction? I am asking because at my level we only know that acid-base reactions give salt + water, but in the carbnoic acid and pyridine example I cannot understand how water will be produced.

Forming water is optional. // Many things that are taught in early classes are highly simplified in expense of distorted knowledge of more complex reality.  Like in physics, mechanical work is taught as multiple of force and length $W=F.L$.  Later, there is taught $W=\vec F \cdot \vec L=|\vec F|\cdot |\vec L| \cdot \sin{\alpha}$. and yet later is taught $W=\int_L{\vec F \cdot \mathrm{d}\vec L}$

Thanks for answering, I finnally get it

The question is poor or even foolish. It relies on a nebulous definition of acid or acidic salt. Most seem to think it refers to the acidity of a solution of a salt. This definition means that the student must know the pH of every solution of every possible salt of the acids. The more astute realize it might mean can the acid form an ion that is still an acid in the Bronsted sense and supply an additional proton [or 2].  The test writer and possibly the class deserve an F. but the student is graded. Bicarbonate, bisulfate, dihydrogenphosphate, etc.  are both Bronsted acids and bases.

@jimchmst Students are not guilty for quality of questions at hand to be answered. Advising the answer that is expected requires sometimes suppression of all you know and feel.

@Poutnik I have two more questions. How can someone figure out which of the bases is stronger between HCO3- and Pyridine. And if we have the assertion that pyridinium bicarbonate will be acidic in nature, then is the reason that pyridine is a weaker base than HCO3- hence pyridinium ion will be stronger acid than HCO3-?

Yes. Carbonic acid $\ce{H2CO3}$ can make an acidic salt, and it does in the compound $\ce{NaHCO3}$ which is sodium hydrogen carbonate, sometimes called sodium bicarbonate. It can react with a base like $\ce{NaOH}$ according to the equation :$$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$$
Every acid containing more than one acidic $\ce{H}$ atom in its formula can make acidic salts. Examples : $\ce{H2SO4, H2CO3, H3PO4}$. And of course $\ce{HCl}$ and $\ce{HNO3}$ cannot make acidic salts.

It needs to be said, though, that the nominally acidic salt NaHCO3 gives a basic solution.


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Why hyperconjugation is called as σ to π* donation for alkenes and alkynes?
Why hyperconjugation is called as σ to π* donation for alkenes and alkynes if for C2 lowest unoccupied molecular orbital is σ2p not π*2p?

source: Stack Exchange

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sort: votes


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The existence of a principal quantum level or energy shell after the loss of an electron from the outer shell of an atom
With the loss of an electron from $\ce{Li}$ ($\mathrm{1s^2\ 2s^1}$), we get $\ce{Li+}$ ($\mathrm{1s^2}$), then the  2nd shell where the electron had been revolving in $\ce{Li}$ will get disappeared or remain there around the nucleus but empty after the loss of the electron?
I'm actually asking for all the other atoms as well, like generally, are shells constructed already upto the very highest ones containing f orbitals in them and just remain empty when there are no electrons in them?

Chem+Math formatting reference:
MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles.

Well, the shell keeps existing in peoples minds, so nothing changes - it's only an abstraction.

It might be easier to think of  a ladder of energy levels which can still exist even if they  contain zero, one or two electrons

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Isothermal expansion of an ideal gas calculated two ways
Consider the following problem:

$\pu{1.0 mol}$ of a monoatomic ideal gas is expanded from state 1 to state 2 as shown in the figure. The magnitude of the work done for the expansion of gas from state 1 to state 2 at $\pu{300 K}$ is ..... $\pu{J}$.  (Nearest integer)

(Given: $R=8.3 \,\mathrm {J\,K^{-1}\,mol^{-1}}$, $\ln 10 = 2.3$, $\log 2 = 0.30$)

I tried to solve this question with 2 methods but my answers were different.

I used the equation for isothermal processes
$$  w = 2.303nRT\log(v_2/v_1) \tag{1} $$
Considering $v_2$, $p_2$ (final volume and pressure) and $v_1$, $p_1$ as initial. Solving this way I got $\pu{1718.1 J}$.

It's simply that I substituted $nRT$ in the above equation as $p_1V_1$ as $p_1V_1 = p_2V_2= nRT$ at constant temperature (isothermal). My answer came to be around $\pu{9397 J}$.

I'm not sure what I did wrong. Is my concept wrong?

Show us the steps. If the term $\ln \frac{V_2}{V_1}$ was used in both, and the numerical value of the factor you are multiplying with is the same, you should get the same answer.

Also, I'm surprised you have a correction factor of 2.303 in there, given that you are using the natural logarithm.

I mistyped I'm sorry. It's actually w=2.303nRTlog(V2/V1). The steps: since process is isothermal so T constt and work= 2.3×1×8.31×300 ×0.3 = 1720J (approx).   Method 2 : P1=6 bar v1=22.7 so W=p1v1×2.3×log2=93.97 bar-L. To convert to J ×100 so 9397 J

Four questions. First, the expansion looks isothermal. Is it ? Nobody mentions it.  Second, the curve should be a hyperbola. It is not. Why ? Third, P1 = 2·P2. Can we derive that V2 = 2·V1 ? Fourth, how can the volume of 1 mole gas be 22.7 L at 300 K  and 6 bar ?

The given data are wrong. I explain. The state 1 is : n = 1 mol, P = 6 bar, T = 300K, V = 22.7 L. With these values, the product pV (= 6 10^5 ·0.0227 = 13620 J ) is not equal to nRT (= 8.314 300 = 2494 J).  pV is 6 times too high !.

@Maurice That is the correct answer. Do you mind posting it? Or maybe the question will be closed before anyone has a chance to post an answer, which would make sense as well.

What is the correct answer ? My calculated $pV$ or my calculated $nRT$, as both are VERY different ? How should I know ?


text:
Potassium permanganate and hydrogen peroxide decomposition
I saw a reaction by chemicalforce on YouTube where he decomposed $\ce{H2O2}$ into $\ce{H2O}$ and $\ce{O2}$ using potassium permanganate. The permanganate is a catalyst in the reaction.
I am wondering if this decomposition occurs because the initial $\ce{H2O2}$ decomposition reacts with the $\ce{KMnO4}$ and reproduces $\ce{H2O2}$ which then again decomposes?
How is the permanganate a catalyst?

Permanganate is not exactly a catalyst here, it gets destroyed as well.

https://chemistry.stackexchange.com/questions/43256/why-is-the-reaction-between-potassium-permanganate-and-hydrogen-peroxide-spontan

Potassium permanganate is involved in two ways.
It oxidizes hydrogen peroxide to oxygen:
$$\ce{2 MnO4- + 3 H2O2 -> 2 MnO2 + 2 OH- + 3 O2 + 2 H2O}$$
The formed $\ce{MnO2}$ acts as the catalyst of reaction:
$$\ce{2 H2O2 ->[MnO2] 2 H2O + O2}$$

Hello Poutnik! Is it a mistake to call the first reaction a disproportionation? The oxidation state of oxygen goes down (to $-2$ in $\ce{H2O}$) and goes up (to $0$ in $\ce{O2}$). Or calling it a redox reaction is more accurate in this case?

You would not need permanganate for it.


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Mass of gaseous ammonia required to prepare ammonia solution of a given pH

How many grams of gaseous ammonia do you need to make $\pu{200 mL}$ aqueous ammonia solution with $\mathrm{pH} = 11.34?$

As the mole ratio is the same for $\ce{H+}$ and $\ce{NH3},$
$$\ce{NH3 + H+ -> NH4+}$$
$$n(\ce{NH3(aq)}) = (\pu{10^{-11.34} mol L^-1})(\pu{0.200 L}) = \pu{9.14E-13 mol}$$
$$m(\ce{NH3(g)}) = (\pu{9.14E-13 mol})(\pu{17.031 g mol^-1}) = \pu{1.6E-11 g}$$
The answer seems way too small. What have I not taken into the account?

https://chemistry.stackexchange.com/questions/139089

As the mole ratio is the same for H+ and NH3, - It is not. Not even close. Off by multiple orders.

@Poutnik Isn't the ratio for H+ and NH3, 1:1, seen the equation? NH3 + H+⟶ NH4+

That is ratio for the reaction, not for the occurence, There is more than $\pu{e10}$ times more NH3 than H+.

@Poutnik Thank you, I know at least one mistake now

Search for calculation of pH of solutions of weak bases. Internet is full of related content.

@Poutnik pOH + pH = 14,
14 – 11,34 = 2,66,
10^(-2,66) x 0,200 = 0,00218 x 0,200 = 0,000436 mol OH-,
Mole ratio reaction -> OH- : NH3 = 1 : 1,
0,000436 x 17,031 = 0,0074 gram = 7,4 x 10^(-3) g gaseous ammonia. Would you say this is correct?

Amount ratio OH- : NH3 is not 1 : 1 either. You are closer, but still few orders off. You are guessing. // Kb = 10^(-4.75)=[NH4+][OH-]/[NH3]

This problem is an equilibrium problem, and namely the following equilibrium : $$\ce{NH3 + H2O = NH4^+ + OH-}$$ The tables give the equilibrium constant : $$\ce{$K$_b = \frac{[NH4^+][OH-]}{[NH3]} = 1.8 10^{-5}}$$ Let's define $c$ as the total concentration of $\ce{NH3}$ brought into solution (included the part transformed into $\ce{NH4^+}$), of which the amount $\alpha$ (< $c$) has been reacted and transformed into $\ce{NH4+ and OH-}$. The problem is to find the value of $c$. Now $c$ - $\alpha$ is the residual concentration of $\ce{NH3}$ in solution.
As the final pH is $11.34$, $\pu{[OH^-]}$ = $\alpha$ = $\pu{\frac{10^{-14}}{[H^+]} = 10^{-14 + 11.34} = 10^{-2.66}}$. This $\ce{[OH-]}$ is also equal to $\ce{[NH4^+]}$ Introducing this value in $K_b$  yields : $$\ce{$K$_b = 1.8 10^{-5} = \frac{\alpha^2}{c - \alpha} = \frac{{10^{-5.32}}}{c - 10^{-2.66}}}$$ The value of the concentration $c$ is :  $$c - \pu{10^{-2.66} = \frac {10^{-5.32}}{1.8· 10^{-5}} = 0.05555} \mathrm{mol/L}   $$ As $\ce{10^{-2.66} = 0.002188}$ it gives : $$c = 0.0577 \mathrm{mol/L}$$ $$n\ce{(NH3) = 0.2 \mathrm{L} · 0.0577} \mathrm {mol/L} = 0.01154 \mathrm{mol}$$ This corresponds to a volume of gaseous $\ce{NH3}$ equal to $$\ce{V(NH3) = \frac{$nRT$}{$P$} = 0.01154\frac{$RT$}{$P$} = 0.282 L}$$

I just calculated it myself and saw your comment, I'm very glad it seems to be the same as you have deduced. It's convenient to be able to remove some of my doubts, much appreciated!


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Why enthalpy change equals to heat transfer
I know that at constant pressure, and when all work is external work, we can write
$$ \mathrm{d}H= q - p \,\mathrm{d}V + p \,\mathrm{d}V \tag{1} $$
and the two last expressions cancel out. But here, I matter that $p$ in the first of these expressions is the external pressure, and $p$ in the second one is the internal pressure. Considering that, we can't cancel out those terms, because we didn't make any consideration that internal and external pressures must be equal.
I know that I'm wrong, but where?

Who says the external- and internal pressures are not equaL?  Also, your first equation should read dH=dq+Vdp.

H =E+PV =  Q-W+PV; at constant pressure work is usually PdeltaV  so H=Q if only work is PV work.

As already explained briefly by the comments under your question:
Let us recall the functional form of $H$:
$$
H=U+pV
$$
Where p is the external pressure on the system. In equilibrium, we get:
$$
dU=\delta q -pdV
$$
$$
dH=dU+pdV+VdP =\delta q-pdV+pdV+VdP = \delta q +Vdp
$$
And, at constant external pressure:
$$
dH=\left(\delta_q\right)_p
$$
Please, remember that in equilibrium external pressure should be equal to the internal pressure of the system, anyway, but not necessarly constant.

Thanks, so the enthalpy describes only equilibrium processes, reversible?

In general, all of classical thermodynamics in the formulation used by chemists mostly deals with equilibrium processes. In this particular case, it is an equilibrium process and enthalpy correctly encodes all of the information relative to the heat transfer. Anyway keep in mind that in every situation in which is possible to define the state variables, it is still possible to use enthalpy as a state function, it is just not possible to state that it is equal to the heat exchanged in the process.


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Can I Detect Hydroxyl Radicals Using Methylene Blue Spectrophotometry?
I am going to be pulsing radiation into a chamber containing a solution of water and both ferric and ferrous iron. The irradiation event should cause a Fenton reaction to ensue and I'd like to trap the radicals that are produced as a result. I do have a few questions, however, as I'm not super familiar with radical chemistry:
What amount of methylene blue should I add to make sure all radicals are accounted for?
How long is the methylene blue - OH radical complex stable for?
Will analysis in a spectrophotometer provide a crude quantitative measurement of the radicals?
Thank you for any advice and suggestions, I really appreciate it!


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In distillation at what stage does purification happen?
A novice question: is it purified during evaporation or elsewhere? I’m getting a water distiller for drinking water after re-adding minerals for health reasons. I described how it works to my son. He postulated that the tube where the condensation happens will get full of contaminants and will produce dirty water that’s  bad to drink.

There is no information to decide if the son's statement is true or not.
Evaporation strips water from all nonvolatile components. But since its condensation, it can get new ones.
For more, details of specific scenario is needed, including guaranteed purity of minerals.
Generally, distillation to make drinkable water is overkill, economically and energetically.

With distilling, "purification" will depend on the boiling point of the water compared with whatever else is in the pot.  You can measure the temperature of the vapor before it goes into the condenser.  Water vapor will be around 100 C or 212 F.
If you are distilling from a salt solution or hard water, everything else will stay in the "pot". That's what will need to be cleaned occasionally ("scale" can be cleaned off with vinegar).
The condenser should remain clean.  To be sure, throw away what comes out of it at first, and keep the rest.  Try not to let your pot get completely evaporated.  It may get too hot if everything boils off.
If there is any chance other volatiles are in the water, such as alcohol or benzene, they will be boiled off and condensed too, contaminating your "distilled" water.  In that case your son is right and distillation will not reliably purify the water.


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Formation of carbocation on reaction with strong acid in presence of halogen

Shouldn't the carbocation be formed away from the -I effect of the Cl atom for more stability. Backbonding is achieved in a later step after hydride shift.

That would result in a primary carbocation. Due to substituent groups stabilizing the carbocation's empty p-orbital in what's known as hyperconjugation, primary carbocations are less stable than secondary.
Here are some heats of formation from the  Active Thermochemical Tables (version 1.130):
t-Butylium (tertiary carbocation): 710.72 $\pm$ 0.98 kJ/mol
sec-Butylium (secondary carbocation): 769.0 $\pm$ 1.6 kJ/mol
n-Butylium (primary carbocation): 803.1 $\pm$ 1.2 kJ/mol
While tertiary carbocations are more stable than secondary ones, an electron-withdrawing group such as a chloro group destabilizes the carbocation.

Right. So basically we can say that here we get more number of alpha H so more hyper conjugation hence, inspire of greater -I, carbocation will be formed there due to compensation by hyper conjugation. Thanks a lot.

Hyperconjugation stabilizes the carbocation by electron donating groups (with p orbitals) adjacent to the empty p orbital of the positive formal charge carbon, not the destabilizing effect of alpha hydrogens.


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Does manganese dioxide disperse differently in bases than in water?
As part of a project in uni I've been given a task in my chemistry laboratory to synthesise manganese dioxide powders.
I prepared a precipitate of it by reacting manganese chloride with sodium hydroxide and hydrogen peroxide. I got quite a lot of dark brown precipitate and I've been trying to wash it all to be neutral. For this, I'm using a centrifuge. I washed it in water, then ethanol, then water, then ethanol, and then in water until neutral.
The first few washes came out fine - I ran the centrifuge for two minutes each time and the supernatant was very clear. However after the third wash I've had to run the centrifuge for longer and longer to get a clear supernatant. Is this because manganese dioxide disperses more finely in neutral water than in a basic solution?

Different pH may lead to different particle surface structure and charge, affecting tendency to disperse and to coagulate.


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Finding the pH level of a liquid process
A photo of 15ml of distilled water accompanied with 3 drops of both universal indicator and bromothymol blue were mixed together and taken (white background was put into the back of the beake). The color of the resultant liquid in RGB is 140, 177, 185. The color space is sRGB IEC61966-2.1. Taken with an iPhone 14 camera. How would you find the exact number of pH level that the very distilled water measures? I understand that you need documentation to the color space and camera and figuring out many more stuff, but I couldn't put them together into math or any other tangible form and solve it

Welcome to Chemistry StackExchange! You could certainly, in principle, make a simple colorimetric pH meter with your equipment, namely a universal indicator and a camera. However, the crucial ingredient you are missing is the calibration. You would have to prepare multiple solutions with known pH, then "measure" they color by taking a photo. With this, you would create a sort of calibration curve to be used for determination of pH of some other solution. Moreover, there is a big issue with using phone cameras, because they usually do automatic ...

... brightness/exposure/white point control, which means that you might not get reproducible measurements. You would have to use the "manual mode", and keep the camera parameters the same for all measurements. Long story short, given just one RGB color point, you unfortunately don't have any good chance of determining the pH.

Thank you so much, okay, so I have some calibrated distilled water images an their pHs, but don't know how to do the graphing thing, there are of course lots of math going on but I don't know anything about them. Anyways, what do I do next?

AFAIK, the universal indicator already contains bromthymol blue. Why to add it explicitly?
RGB value is not enough to determine pH due multiple factors.
Aside of that, such a measurement is useless, as pH is set by the used indicators that are all weak acids or bases.

Can you further elaborate on why a measure ment is useless and what are those "used indicators". I'm really just a beginner, don't know much, so correct me if I'm wrong

Try to localize, what exactly you do not understand, as the answers to both questions are already present above. // By measurement of pH as in your question, you do not measure pH of distilled water, but of diluted indicator solution. It is like if you measured temperature of water by thermometers much more massive than the water sample. // Used indicators = indicators in the universal indicator mixture (link) + bromothymol blue.

Tap water would be different as there is usually present CO2/HCO3-  pH buffer.


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Selective chemical absorbent for N2 gas removal
Are there any chemical absorbants for selective  N2 removal from a gaseous mixture ?
I am thinking of something similar to say Pyrogalkol for O2 removal or KOH for CO2 etc.
Of course, N2 being relatively inert I can imagine this would be rather hard but still.
The particular application requires removal of small amounts of N2 ( say  less than 1% ) from a gas stream that is mostly Helium.
Alternatively an adsorbent bed is also an option. But do we have adsorbants that would selectively trap N2 but not He?
Otherwise a cryogenic condensor seems the only option? Any other ideas?

There is probably possible to have a molecular sieve, based on molecule geometry, similarly as is used for oxygen (as alternative to cryoseparation). (Big molecules do not fit, small ones do not stay).

@Poutnik Anyone know if such a Molecular sieve or zeolite is commercially available? I could not find any.

Why do you want to remove 1% $\ce{N2}$ from helium?

My Google is not better than yours, sorry.

Why not to use helium of guaranteed sufficient purity? // I assume using metallic lithium to form Li3N is not applicable.

The great thing about cryogenics is (a) there is no component that gets “used up” and requires reactivation or replacement and (b) it removes practically any non-He contaminant, thus solving possible future problems before they happen.

There are various techniques to trap $\ce{N2}$ by adsorption in a molecular sieve, or behind a semi-permeable membrane, rather than chemically combining it.
Though one could do as Cavendish, Rayleigh and Ramsay did, reacting the nitrogen with added oxygen in an electrical arc and absorbing the reaction products, or use a metal getter, as used in radio tubes (valves, if you prefer), those would be inefficient and costly.
look at this ACS article that discusses the economics of helium purification.


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Inductive Effect (R-NH3+ and R-NH2)
Textbooks I've read state that R-NH3+ has exhibits a greater -I inductive effect than R-NH2 does.
However, I've also read in places that the + charge in R-NH3+ only represents a formal charge and has no physical significance.
If this is the case what explains the greater inductive effect shown by R-NH3+?

Does a proton have a formal charge without physical significance? As the extra proton is the difference between R-NH2 and R-NH3+.

I'm actually unsure. I've read in textbooks that compounds like NH4+ and R-NH3+ are formed as a result of coordinate covalent bonds, in which we only represent the + and - "formal" charge to denote exclusively, the donor and acceptor atoms

All N-H bonds are equivalent there, partial delocalized positive charge distributed on all of them.

Thanks for your interesting question,
To my knowledge, depending on the solvent you use, you can have a protontaed or a non protonated species. if you are using an aliphatic amine in water, for example, the compound is mostly protonated. For sure the protonated form has a stronger electron-withdrawing effect than its non-protonated form. To answer your question, sure! the charge on the RNH3 group is important and it has a physical significance. It makes the groups more able to attract negative charge through single bond thus increasing the inductive effect on the carbon skeleton of your compound.


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Temperature and zeta potential
What is the temperature effect on zeta potential?
My problem was assigned by my teacher for the final exam preparation: "When the temperature is increased, and given in this temperature range, there is no desorption of potential-determining ions on the surface of colloidal particles occurs, we will have:
A. Zeta potential increases, surface potential decreases
B. Zeta potential and surface potential remain constant
C. Zeta potential decreases, surface potential remains constant
D. Zeta potential and surface potential both increase
*Thought: It is stated in the below publishment that the zeta potential decreases while temperature is increased at low salinity (0.01M NaCl). The opposite trend could be seen for high salinity (0.5M NaCl).
Temperature dependence of the zeta potential in intact natural carbonates


text:
How can the double-layer capacitance be constant in the Helmholtz model?
The differential capacitance of ideal electrochemical double layers is constant, as the surface charge $\sigma$ linearily increases with the applied surface potential $\Psi$ relative to the point of zero charge
$$ C = \frac{\mathrm{d}\sigma}{\mathrm{d}\Psi} \rightarrow
       \sigma = C (\Psi - \Psi_\mathrm{ZPC}) \tag{1} $$
So, in a cyclic voltammetry experiment in which there is only capacitive current and no faradaic current due to charge transfer, the current will be constant
$$ i = C\frac{\mathrm{d}\Psi}{\mathrm{d}t } \tag{2} $$
This is because in a CV experiment the scan rate $\mathrm{d}\Psi/\mathrm{d} t$ stays constant (e.g. $\pu{10 mV/s}$). Experimentally it looks like this (not ideal, but close enough):

But, how is this consistent with the Helmholtz idea of the electrochemical double layer? Physically, there needs to be a linear increase in surface charge for higher potentials on the electrode. But, the surface-active Helmholtz plane is limited to the inner Helmholtz layer, with the rough diameter of one solvent molecule and one charged ion:

In my imagination, at low potentials, the whole electrode is already covered in positive excess charge, and the anions of the electrolyte already adhere to the surface.
As the inner Helmholtz plane is already filled up by anions, and the potential further away has already dropped to zero, I cannot understand how the surface charge can still increase to higher potentials as is needed for Eq. (1) to be valid.

I have to flip this around. At low voltages, why can't the surface charge density in the conductor change with voltage? Why would incomplete surface coverage be incompatible with a Helmholtz model?

Maybe you can find further information in in A. J. Bard, L. R. Faulkner, H. S. White, Electrochemical Methods: Fundamentals and Applications, 3rd Edition, Wiley (2022)

To start forget for a second about double layer models and particularly details of the electrolyte or other medium adjacent the electrode. Assume you have a perfect conductor which you hooked up to a voltage source and has a flat surface in contact with a vacuum. Classical electromagnetic theory (specifically Gauss' law) tells you that the electric field inside the perfect conductor is zero (there is no current) but that there is a field outside due to charge at the surface. If the surface is large and flat the surface charge generates a uniform field perpendicular to the surface. The field is related to the surface charge as
$$\sigma = \epsilon_0 E$$
It is this field of course that attracts countercharges from whatever medium you place next to the surface. If you substitute the vacuum with a dielectric* this just becomes
$$\sigma = \epsilon E$$
In this case a layer of counter-charges forms adjacent the surface due to polarization (reorientation) of molecules or distortion of the charge distribution including concentration of counterions. This polarization extends all the way to the opposing pole due to the voltage drop in the medium. A similar analysis leads to similar equations for a parallel plate capacitor.
The takeaway is that an electrolyte can be considered a dielectric* that is polarized with a net countercharge next to the surface of a charged conductor. That countercharge can in effect be zero, as it is in the case of vacuum. The extent of polarization depends on the surface charge of the conductor and the dielectric constant. The surface charge of the conductor depends on the voltage $\Delta \phi$ applied. It also depends on the geometry and the nature of the dielectric; for a parallel plate capacitor $\Delta \phi=Ed$ so charge depends on the plate separation $d$:
$$\sigma = \epsilon \Delta \phi/d \tag{1}$$
For an ideal conductor you can alter the surface charge by tuning the applied voltage, meaning it can be arbitrarily small. For small surface charges the counter-charge density will also be small and a linear variation with the applied voltage is not inconceivable. This linear relationship is expressed in terms of an electric susceptibility $\chi_{el}$ (1), in the present case of one component of the full tensor:
$$P = \chi_{el} \epsilon_0 E$$
where $P$ is the polarization of the dielectric, the density of induced dipoles per unit volume, which can be shown equal to the surface density of counter-charges adjacent the electrode:  $P=\sigma_{pol}$. The counter-charges also generate a field that partially cancels the field due to the surface charges. Applying Gauss's law then leads to
$$E = \frac{\Delta \phi}{d} = \frac{\sigma_{\textrm{free}}}{\epsilon_0(1+\chi)}$$
Turning back again to equation (1), Since the capacitance is $C=Q/\Delta \phi$ and $Q=A \sigma$, it follows that
$$C=\frac{\epsilon A}{d}$$
where $\epsilon=\epsilon_0(1+\chi)$.
In the Helmholtz model the dielectric contains ions* which respond to the applied electric field by moving to the oppositely charged electrode. The field then drops to zero some distance from the electrode and, as described in the problem, a counterlayer with the same total charge Q forms next to the conductor, representing a virtual parallel plate a very short distance away. The basic assumptions still hold, however, in essence that a linear electric susceptibility apply. This assumption requires above all that the applied voltage be low.

*A dielectric does not have free charges and cannot carry a current beyond that sustained while it is being polarized. The analysis here applies to an electrostatic situation, for instance a capacitor that has achieved a fully charged state and for which the current is zero.

Reference

Alberty and Silbey, Physical Chemistry, 3rd Edition, Wiley 2001.


text:
Help interpreting IR Spectrum of Vanillin ("aldehyde bands")
I'm not sure about the attributions of 3 bands of my ATR FT-IR  spectrum of an extract of vanilla sugar, that probably contains mainly vanillin and some impurities. The extraction solvent was ethyl acetate and was evaporated with a rotary evaporator. The final extract was a dark yellow liquid and vanillin is normally a clear solid. These impurities were probably due to the fact that the vanilla sugar uses "natural" vanilla extract.
Vanillin has an aromatic ring, an aldehyde, an ether and aromatic alcohol.
So back to the IR spectrum :

Vanillin is an aldehyde and elongation of C-H bond of aldehyde usually appear as a two bands (asymmetric and symmetric elongation) between $\pu{2650}$ and $\pu{2800 cm-1}$ (from my IR table). Before I determined the positions of the band with my ruler and my square (so it's less precise then what a computer would do), I thought the "aldehyde bands" were the $\pu{2911 cm-1}$ and $\pu{2727 cm-1}$ bands.
But $\pu{2911 cm-1}$ I think is too high to be an aldehyde band, whereas the band of weak intensity at $\pu{2727 cm-1}$ is in the right position.
But that $\pu{2727 cm-1}$ band seems of too weak intensity. Then the $\pu{2911 cm-1}$ band would be the $\ce{C-H}$ elongation from the methoxy group ($\ce{O-CH3}$) of vanillin.
So which is the right one? If neither, can aldehyde shows just one band at $\pu{2839 cm-1}$ instead of two (for asymmetric and symmetric elongation)?

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations and with pu{} one for a non-breakable  space between figure and unit.

The freely accessible SDBS holds three records of transmission IR spectra for vaniline (CAS 121-33-5) as record SDBS 726: a $\ce{KBr}$ disc, in solution of $\ce{CCl4}$, and in nujol.  Only the recording in tetra displays absorption bands one can attribute to $\ce{C-H}$ vibration and these are (in comparison to the other ones) of weak to moderate intensity only:

Compare this with the one in nujol:

Especially if there are complementary data by NMR and MS available, I would not invest too much time to assign each band in an IR spectrum.  With keyword "aldehyde", I associate the motif of $\ce{C=O}$ and hence seek in the spectrum for the presence of its absorption band around $\pu{1725 cm-1}$ (value from Wikipedia's compilation) of at least moderate if not  strong intensity.
The position of IR absorption bands depends a bit on the chemical environment, so an alkyl $\ce{R-CHO}$ may show at higher numbers, than an aldehyde bond to an aryl, or a $\alpha$,$\beta$ insaturated chain.  There equally is some influence in position by recording sample in nujol or tetra solution (an apolar matrix) or in a pellet of $\ce{KBr}$ (polar matrix) in transmission IR spectroscopy, and the neat sample (typical for ATR).  Second (and sometimes operators forget this), the absorption intensities recorded on an ATR spectrometer have to be corrected for the difference in refractive indices of the sample and the window of the spectrometer (e.g., Ge, Si, Diamond) because the penetration depth of the evanescent wave varies over wavelength.  This limits the comparison absorption bands' intensity between data recorded by transmission, and one by ATR.
Discounting information in the fingerprint region, the one on $\pu{1697 cm-1}$ (your reading) possibly is the signal indicative enough for the presence of the $\ce{C=O}$ motif.  There may be an individual assignment of the bands in the literature but if the extract still is a yellowish liquid instead of a colorless solid (mp $\pu{81 ^\circ{}C}$, property box of Wikipedia) your sample might not yet be pure enough to merit this investment.

If you can influence the recording and subsequent plot of the IR spectrum, seek to split the scale of the abscissa e.g., around $\pu{2000 cm-1}$ for a subsequent expansion (cf. the SDBS spectra).


text:
Unattainable equilibrium
I read a question on a textbook that asked:

For the reaction
$$ \ce{CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)} \tag{1}$$
determine whether or not a mixture of $\ce{CaCO3(s)}$ and $\ce{CO2(g)}$
at a pressure greater than the value of $K_p$ can attain the above
equilibrium.

The answer it gives is no, because the partial pressure of $\ce{CO2}$ won't be able to decrease to $K_p$.
However, I'm confused about what happens to the mixture if equilibrium isn't attained. Will it go to completion? What if it is a mixture of $\ce{CaO}$ and $\ce{CO2}$ at a pressure less than $K_p$?

From my memories, I think the reaction stays out of equilibrium and just can't progress further so it stops. I think it's the also the case for dissolution before saturation. But maybe someone else can verify this because I'm not 100% sure.

This is an equilibrium reaction. If CO2 (or CaO) were all removed more carbonate would form products until equilibrium is re-established and vice versa.

I assume that we could formally take disappearing of CaO(s) as its activity going from 1 toward 0, being out of scope of the standard definition of equilibrium constants involving solids.

In chemical reaction problems attaining equilibrium or completing a reaction means reaching a state at which the reaction ceases. Here it never started, how can it cease? The system begins in a state of dynamic equilibrium (the posted problem is poorly worded) and the net amounts of the species will not change. The reaction can be regarded as having already gone to completion - in the reverse direction.
The high pressure of $\ce{CO2(g)} (> K_p)$ prevents net decomposition of $\ce{CaCO3(s)}$. The term "net" is important because there is a dynamic equilibrium: $\ce{CO2(g)}$ might not interfere with the forward reaction and some  $\ce{CaCO3(s)}$ may decompose, but $\ce{CaO(s)}$ thus produced will react back to carbonate, the high pressure of the gas exhausting all $\ce{CaO(s)}$.
On nomenclature: thermodynamic changes ($\Delta G^\circ$, $\Delta H^\circ$ etc) refer to processes between equilibrium states, an initial equilibrium state and a final equilibrium state. This is why the field is called equilibrium thermodynamics. For instance, a starting state of unmixed reagents (in separate containers, say)  might be regarded as an equilibrium system (under the imposed constraints it is stationary, will not change). That is analogous to the meaning of equilibrium here. No change in the amounts of the species (or in T, p, etc) will occur under the given conditions.
Another post with a similar question explains how to analyze such a problem mathematically in terms of sums of chemical potentials of the components. It can be used to demonstrate that dissociation of carbonate will increase the total free energy:
$$\begin{align} \qquad &\ce{CaCO3(s) &<=>& CaO(s) + &CO2(g)}\\ \qquad &\ce{A &<=>& B + &C} \\ \mathrm{(initial)} \qquad & n_A &\qquad& 0 \quad &n_C\\ \mathrm{(final)} \qquad & n_A -\alpha &\qquad& \alpha \quad &n_C+\alpha \\ \mathrm{(change)} \qquad & -\alpha &\qquad& \alpha \quad &+\alpha\end{align}$$
The free energy change is
$$\Delta G = - αμ^∘ _A + αμ^∘ _B + αμ^∘ _C + RT (n_C + α) \ln p_{fin} - RT n_C \ln p_{ini} \\ = α\Delta G^∘ + α RT \ln p_{fin}  + RT n_C \ln \left(\frac{p_{fin}}{p_{ini}}\right) \\ = - αRT \ln K_p + α RT \ln p_{fin}  + RT n_C \ln \left(\frac{p_{fin}}{p_{ini}}\right) \\ = α RT \ln \left(\frac{p_{fin}}{K_p}\right)  + RT n_C \ln \left(\frac{p_{fin}}{p_{ini}}\right) $$
If carbonate decomposes $α>0$ and $p_{fin}> p_{ini}$ so that $\Delta G>0$.

Thank you. I also read that if there is some CaO and CO2 at the pressure of Kp they would react to form some CaCO3, thereby lowering the partial pressure to below Kp when equilibrium is reached. Is that incorrect then?

That doesn't seem right. Where did you read this?

"Chemistry: The Central Science" 12th ed. by Brown and others. It wasn't stated in the main text but was from an exercise.

Again, that doesn't seem right. I can include a mathematical explanation later.

If the partial pressure of $\ce{CO2}$ is greater than $K_p$, some $\ce{CaO}$ will react with the excess of $\ce{CO2}$ and produce more $\ce{CaCO3}$. At the end, the system contains less $\ce{CaO}$ and more $\ce{CaCO3}$, so that the final pressure of $\ce{CO2}$ is equal to $K_p$
If the partial pressure of $\ce{CO2}$ is smaller than $K_p$, some $\ce{CaCO3}$ will get decomposed to produce more $\ce{CaO}$ and more $\ce{CO2}$, so that the final pressure of $\ce{CO2}$ is equal to $K_p$
Edit : I have admitted that the amounts of solids is at least an order of magnitude greater than the amount of gas. With this assumption, I will never run out of one of the solid species.

Unless you run out of one of the solid species, in which case you don’t reach equilibrium.

Yes Karsten. But I have admitted that the amounts of solids is an order of magnitude greater than the amount of gas. It will be added in my text.

The question says they are starting with a mixture of carbon dioxide and calcium carbonate. This implies that there is no calcium oxide at all, and making some is in the direction away from equilibrium. But I agree with your answer and it is important to know what happens when all three species are initially present.

This is an interesting question! Lets maintain a CO2 pressure of one atmosphere at a temperature below 800C sufficient that CaCO3 will decompose to a pressure much less than 1 atmosphere say 0.1 atmosphere. If a sufficient amount of pure CaCO3 is introduced and maintained at that temperature would not a dynamic equilibrium be established between CaCO3 and CaO +CO2. If our definition of Kp is reasonable it seems that our definition of the activities of the solids is questionable. I suggest the possible formation of a nonstoichiometric compound so that the ratios of activities are such that the ratio will correspond to that needed to satisfy Keq.
A quick web search failed to find any studies of CaCO3 decomposition at lower temperatures and controlled CO2. This requires a library search.  Possibly Xray diffraction of CaCO3 decomposed under high CO2 pressures has been studied.

Why the downvote I known I went out on a limb. please explain the error in my thinking or if you have access to actual data.


text:
Why are buffer solutions not neutral
I am very confused about buffer solutions and I have lots of ideas about them which don’t integrate together so I really can’t tell which are correct and which are wrong. That being the case it’s likely the title question is the wrong one to ask. Here is my understanding of buffer solutions and the issues that arise from it.
To clarify: I am only considering buffers made from weak acids and their conjugate bases, I know that basic buffers exist but to simplify matters I am only considering acidic ones in the rest of this question
The acid has the following dissociation equation:
$$\ce{HA <=> H+ + A-}$$
I believe that the position of equilibrium would lie very far to the left as it is a weak acid and so will scarcely dissociate.
This would mean that the associated equilibrium constant,  $\ce{K_a}$,  would be very small as the numerator would be very small compared to the denominator.
I hope that is correct so far, if not please do tell me what I have misunderstood!
Now if the conjugate base, $\ce{A-}$, is added, I would say that that is an increase in the concentration of $\ce{A-}$ so the reaction quotient would also increase and would no longer be equal to  $\ce{K_a}$. I would then apply Le Chatelier’s principle for increasing the concentration of a species in a system at equilibrium to find how the position of equilibrium would change to make the reaction quotient once again equal   $\ce{K_a}$.
The increased concentration of the species on the right side of the equation, $\ce{H+ + A-}$, would cause the position of equilibrium to shift to counteract the change. The only way that the concentration of the species of left side of the equation could decrease would be for $\ce{H+}$ and $\ce{A-}$ to react to form the original acid, $\ce{HA}$ thus decreasing the left concentration and increasing the right concentration until the ratio of the two concentrations (the reaction quotient) returns to being equal to the equilibrium constant, $\ce{K_a}$.
Is that correct so far?
The result of this new equilibrium position would be a decrease in $\ce{[H+]}$ but an increase in $\ce{[A-]}$.
Now to make buffer solutions you add rather large concentrations of the conjugate base. I would assume that the number of conjugate base ions added would by far outnumber the $\ce{H+}$ already in the system.
Therefore, I would predict that the conjugate base ions would react with all the $\ce{H+}$ that are present (as I have explained above). I would expect that there would be no hydrogen ions left. This brings two problems that (at least!).
Firstly:  zero hydrogen ions would mean that all the non-water species in the buffer solution have a pH of 14, and so the pH of the buffer would not be acidic, (would it not be basic?) In my textbook it says that these buffer solutions are slightly acidic, due to the slight dissociation of the acid. That is a contradiction!
Secondly: If you add equal molar quantities of the acid and the conjugate base and the acid only slightly dissociates, surely it will never be possible for the quotient to equal $\ce{K_a}$, because you have increased the left hand concentrations so much and there aren’t enough hydrogen ions to react to shift it back. Would that mean that equilibrium is never established because the reaction quotient can never equal $\ce{K_a}$?
Or perhaps for some reason the position of equilibrium wouldn’t change as I have described above. Perhaps it would move to counteract the change a little but not so much that all the hydrogen ions would be ‘removed’ (as part of the acid). Why would this be the case and how could one possibly predict that Le Chatelier’s principle would apply only partially in this situation?
Perhaps there is another explaination. I hope so!
I hope I have made my confusion clear. As you can see I have come across many stumbling blocks and any help would be very much appreciated. I am studying A level chemistry so I can only understand chemistry at that level. I am aware that there are lots of simplifications made at A Level level and perhaps that is why I am confused. Please do tell me if it is just something that I have to accept rather than understand, like Le Chatelier’s Principle, that would be most helpful!
Thank you so much for reading my question!

A buffer solution is not necessarily made of a salt of a weak acids and their conjugate base. For example, mixtures of NaOH and NaHCO3, both about 0.05 M, are excellent buffers for pH values between 10 and 11. Mixtures of NaOH and KCl about 0.1 M make excellent buffers between pH 12 and pH 13.

You should try to evaluate it quantitatively, applying acidity and basicity constants. If an acid is weak, it's salt is a weak base. pH of equimolar mixture depends on which one is weaker. If Ka > 10^-7 (pKa < 7) at 25 Dec C, than solution of equimolar mixture of acid and it's conjugate base is acidic. If Ka < 10^-7, it is basic.

A mix of bicarbonate and hydroxide results in a buffer of acid HCO3- and base CO3=.  More concentrated solutions of strong acids or strong bases are their own buffers because slight additions of acid or base result in a slight change in pH. Water is the conjugate base or acid respectively.

I suppose the biggest issue is this: doesn’t adding the conjugate base change the reaction quotient of the acid dissociation so that it is no longer equal to the equilibrium constant? If so, doesn’t the system change to change the quotient back into the K_a If not then is it not at equilibrium? If so then how come it doesn’t go to equilibrium?
If the system does change, how can it change enough to restore the equilibrium constant value if there has been such a huge increase in the numerator?

Well, the system definitely does change to keep the equilibrium. If you add a little, it changes a little, and if you add a lot, it changes a lot. It is as simple as that.

@QuantumHamster  All you do is speculative reasoning, but I have not seen any calculation yet.

@poutnik ah yes that is because I can see how it works in calculations but I am interested in understanding it fully and the calculations don’t enable me to do that. Perhaps there is no way to visualise and deduce the situation without calculations but I am curious to see whether there is :)

Therefore, I would predict that the conjugate base ions would react with all the H+ that are present (as I have explained above). I would expect that there would be no hydrogen ions left. This brings two problems that (at least!).

Be aware that a small relative mutual conversion within a conjugate pair of a buffer leads to a big relative change  of H+ concentration, leading to easy managing of dissociation equilibrium.
Imagine
$$K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} =  \frac{(c_\ce{0,H+} - x)(c_\ce{0,A-} - x)}{c_\ce{0,HA} + x}$$
with $$c_\ce{0,H+} \ll  c_\ce{0,A-} \approx c_\ce{0,HA}$$
Mutual conversion HA <=> A- restablishes the equailibrium by a figurative finger snap.

The alpha and omega is the equation of dissociation equilibrium for weak acids
$$\ce{HA(aq) <=> H+(aq) + A-(aq)} \tag{1}$$
that is for diluted solutions approximately (thermodynamic activities approaximated by concentrations)
$$K_\text{a} \approx \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}, \tag{2}$$
respectively in the rearranged logarithmic form, known as the Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {\frac{[\ce{A-}]}{[\ce{HA}]}} \tag{3}$$
where $K_\mathrm{a}$ is the respective acidity constant  and $[\ce{X}]$ denotes amount concentration of $\mathrm{X}$.
There are two other important equilibrii
$$\ce{A-(aq) + H2O <=> HA(aq)  + OH-(aq)}$$
and
$$\ce{H2O <=> H+(aq)  + OH-(aq)}$$
All three equilibrii are related by the equation
$$K_\mathrm{w} = K_\mathrm{a} \cdot K_\mathrm{b}$$
The same is valid for weak bases, if we consider their conjugate acids and their acidity constants.
For pure solutions of just the acids, there are possible conditional simplifications:
$$K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \overset{\text{All H+ from acid}}{\approx}\frac{[\ce{H+}]^2}{c_0 - [\ce{H+}]} \overset{\text{low dissociation}}{\approx}\frac{[\ce{H+}]^2}{c_0} \implies \\
\implies [\ce{H+}] \approx \sqrt{K_\text{a} \cdot c_0} \implies \text{pH} \approx \frac 12 (\text{p}K_\text{a} - \log{c_0}) \tag{4}$$

If there are mixed both a weak acid and its conjugate base to form solution of significant concentrations of both then changes of their concentration due equilibrium reactions are negligible.
the equations (2), (3) above become:
$$K_\text{a} \approx [\ce{H+}] \cdot C,\tag{5}$$
respectively
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {(C)}\tag{6}$$
where $C$ is the concentration ratio $\dfrac{[\ce{A-}]}{[\ce{HA}]}$ .

Thank you very much! I am not sure I understand all of it but I will go through it and try to!

I believe that the position of equilibrium would lie very far to the left as it is a weak acid and so will scarcely dissociate.

Only if you start with weak acid in pure water. If you start with the weak base in pure water, you get a basic pH.

Therefore, I would predict that the conjugate base ions would react with all the H+ that are present (as I have explained above).

Water can always make more H+ by dissociating. Alternatively, you can think of the weak base ($\ce{A-}$) react with water instead of with $\ce{H+}$:
$$\ce{A- + H2O <=> AH + OH-}$$
This would be the combination of
$$\ce{A- + H+ <=> AH}$$
and
$$\ce{H2O <=> H+ + OH-}$$

[…] would have a pH of 14.

That’s impossible because it would be more basic than a weak base. The weak acid alone is a bit acidic, the weak base alone is a bit basic, and the mixture (the buffer) somewhere in between.

[OP in comments] Oh so are you saying that the addition of the conjugate base would increase the dissociation of water? Are the hydrogen ions common to both equilibria, that of water dissociation and acid dissociation? I didn’t consider that perhaps it is the source of all my confusion! Would there not be an increase in pH if water dissociates as although the ‘lost’ hydrogen ions are replaced hydroxide ions are also formed? Wouldn’t there be a net increase in alkalinity?

Yes, the hydrogen ions are common to both equilibria, which makes it a bit difficult to think through these situations in our head. A math engine simply solves the system of coupled equations and spits out the answer. This does not help you understand what is going on conceptually, though. Again, yes, if you take away most of the hydrogen ions, they would be replaced by water dissociating, also producing hydroxide. This is how a weak base like ammonia can get a pH higher than 7. The hydroxide ions come from water, and most of the hydrogen ions that ammonia accepts to become ammonium also come from water (as a neutral solution contains just $\pu{1E-7 mol/L}$ of $\ce{H+}$ to begin with).

Oh so are you saying that the addition of the conjugate base would increase the dissociation of water? Are the hydrogen ions common to both equilibria, that of water dissociation and acid dissociation? I didn’t consider that perhaps it is the source of all my confusion! Would there not be an increase in pH if water dissociates as although the ‘lost’ hydrogen ions are replaced hydroxide ions are also formed? Wouldn’t there be a net increase in alkalinity? Thank you so much for your help so far!

I added your followup questions to the main answer, and commented there.

that is very helpful thank you very much!

In aqueous solutions the acidity is mostly governed by the concentration of hydronium ions[H3O+ aka H+] and usually denoted by its negative logarithm, pH. Although H3O+ is most important, weak acids and weak bases present have their own properties. Some reactions happen at an optimum pH and pH control is advantageous. This can be accomplished in several ways. The easiest is to simply use a stoichiometric excess of a strong acid or base or even a weak acid or base if that gives the appropriate acidity. This method is used in permanganate or dichromate oxidations and many reactions. When an intermediate pH is necessary the direct method fails because any change in acid or base gives a large change in H3O+. The answer is to use an excess of the weak acid or base and add the right amount of the conjugate base or conjugate acid to give the proper pH. When the proper amount of buffer is used the reaction pH will stay near the optimum pH. If acid is consumed the weak acid ionizes. If acid is formed the weak base accepts it. The equilibrium conditions satisfy both the ionization quotient of the acid and Kw.
LeChatlier's principle starts with a system at active equilibrium. The buffer is perturbed by a slight change in acidity; either water or the weak acid will change to restore equilibrium. Agreed the rates are not equal. Remember the properties of H3O+ and weak acids differ. Instead of an extreme shock treatment use small equilibrium steps.
example: Reaction uses some acid, solution needs acid to reestablish equilibrium so water and weak acid both ionize to give more H3O+ each to its ability[acid probably ionizes faster that is why it is an acid]. Since H3O+ was consumed the overall solution is now at a higher pH. Since the solution is buffered the change in pH is less than if the acidity were a trace amount of strong acid. In a nonbuffered solution, when a reaction requires acid, as acid is consumed the reaction slows, all the acid is not consumed. Concentrations do not reach zero [pH infinity not 14].


text:
Calculate IC50 from fluorescence kinetic
I have an assay in which a fluorescence signal is generated when the enzyme reaction progresses. The fluorescence signal increases until a maximum is reached. Than it keeps constant. The first 1 h appears to be linear.
My data therefore looks in principle like this

(taken from another publication, my exact data points look different of course)
Now I want to titrate an inhibitor into the solution which inhibits the enzyme. I expect, that the initial speed of the enzymatic reaction decreases and therefore the slope of the initial linear part. If I get the idea right, I compare the slope of the linear part of all inhibitor concentrations and I compare it further it to the slope of the control without inhibitor.
Therefore I determine how much of activity is left in percent in comparison to the control in which no inhibitor is added. If the slope is 59 units is comparison to 100 units of the control the activity would would be 59 percent and the inhibition 41 percent if I am right.
When you check the literature, there are figures that look like this, where Inhibition is plotted against concentration of the inhibitor. A paper calls for the fitting of the curve by this equation:
$I =\frac{I_{max}C}{IC_{50}+C}$
where $I$ is the percentage inhibition $I_{max}$ is 100% inhibition, $C$ is the concentration of inhibitor, and
$IC_{50}$ is the concentration of inhibitor for 50%.
I am wondering about multiple things here:

What is meant by $I_{max}$ ?
Is it the absolute value of fluorescence at 100 % Inhibition? Or is this the max inhibition for a specific concentration? But then it would be the same as $I$?

Is there any guidance/tutorial to realize this plotting for a software like graphpad or a good resource you can recommend?

I still need to draw a conventional dose response curve (activity against lg c) or is the the plot inhibition against concentration sufficient?

Hello!!! (1) It is mathematically impossible for your equation to have a value of $I_\mathrm{max}$, but it tends to that value when $C \rightarrow \infty$. For practical purposes, when $C$ is several times higher than $IC_{50}$, then essentially $I \approx I_\mathrm{max}$. (2) If your university has money use Origin or MATLAB, if not use Octave or code in Python. (3) Rearrange your equation, invert it, and check that a plot of $C/I$ vs $I$ gives a linear plot, in which the slope is $1/I_\mathrm{max}$ and the y-ordinate is $b = IC_{50} / I_\mathrm{max}$.

The equation looks similar to.the Michaelis Menton equation. I am wondering, is $I_{max}$ always the same value in a titration row? For example always 98 percent inhibition, regardless of the inhibitor concentration?

The solution I found appears to be straight forward. Feel free to comment if you have any insights about this.

You start with $c=0$ and increasing inhibitor concentration to $c_{max}$
Generate linear equations $y=mx+b$ for the first linear segment of the fluorescence signal, in my case I used the first 60 minutes from the fluorescence over time graphs. In this case the change of fluorescence is equivalent to the change of product concentration, and therefore the reaction speed over time:

$\frac{dF}{dt}=\frac{dP}{dt}$

The uninhibited protein equals 100 percent activity. If the slope is 40 % of the uninhibited turnover speed, this equals 60 % inhibition.
Inhibition over concentration values are imported into Graphpad and non-linear regression is chosen.
I applied the Michaeils kinetic, because the equation for the inhibition, is of the same structure than the Michaeils equation.
The calculated Km value equals the $IC_{50}$


text:
What is the correct reaction mechanism for the following :
The question is :

Now, this is what I have done :

So I dont know how to proceed to the final product. In some book they tell we need to break the cyclopropane to get the desirable answer but how ?
Please help.

This paper (JCS Perkin 1 (1975) 1638-40) reports that dichlorocarbene reacts with cyclopentadiene at 180C to give chlorobenzene. The strained cyclopropyl ring opens to give insertion with loss of chloride anion.  I see no reason why dibromocarbene should give a different outcome, so the answer is C.
See this SE answer How does chloroform react with cyclopentadiene in base?

Here OTs being a good leaving group will be eliminated by alc KOH making cyclopent-1,3-diene, t-BuO- being a solvent will make :CBr2 which will act as a Base and abstract proton from C5. Hence option D will be the answer.

Do you have a literature reference for this?


text:
If water is placed in an isothermal system, with the rest of it being vacuum, what would be the final pressure of the system?
Given that a certain volume of water is placed in an isothermal system with the rest of it being vacuum, I believe the pressure would come from the vapor pressure of water at that temperature. My approach to this was by using the equation for work done in an isothermal process:
$$ W = nRT\ln\left(\frac{V_f}{V_i}\right) \tag{1} $$
Equating this to the latent heat of vaporization of the total mass of water present, but I couldn’t reach an answer that was viable.
To show my working:
\begin{equation}
    \pu{(536cal/g)}(10^5 \; \pu{g}) =
    (1 \cdot 10^5/18 \; \pu{mol}) (1.987) (303)
    \ln\left(\frac{V_f}{V_i}\right) \tag{2}
\end{equation}
where $V_i = 10^5 \; \pu{cc}$ as given in the problem.
On solving this my answer comes out to an absurdly large volume. And I am not sure that will help me find the right final pressure (for which I thought of using $P_\mathrm{i}V_\mathrm{i} = P_\mathrm{f}V_\mathrm{f}$ taking $P_\mathrm{i}$ as the vapor pressure of water at $\pu{303 K}$).
What’s wrong with my approach?

With any liquid water left, the final pressure depends only on temperature, raising with it roughly exponentially.

The final pressure is the vapor pressure of water at this temperature, which is $3167$ Pascal at $25$°C, and $101325$ Pascal at $100$°C. These values can be found in Tables. This is all to be done. The work has nothing to do here.

@Maurice in that case, given that initially the system is a vacuum, what would be the initial pressure? Since it needs to satisfy PiVi=PfVf

There is no initial pressure, as the initial system is a vacuum.  2) The expression PV = constant is only valid when the amount of gas is constant. In the vacuum, the vacuum does not contain any gas. So pV is not constant, when introducing water in the system.

Is the system open or closed? As it is not explicitly stated. For the former is the result trivial, as the answer suggests. But I think vacuum was assumed as just initial state (plus water) of the closed system.

You can't do work expanding against a vacuum. You may want to quote the entire problem or cite the source.

Since it's isothermal, and the evaporating water removes heat, then heat energy is being supplied.

Since it's in vacuo, all evaporated water is lost; i.e., the volume is considered infinite.

At any temperature, eventually all the water will evaporate. The only question is how long it takes. At the cosmic background temperature, a ball of a few liters of ice would likely last a billion years, but at 303 K, in a few days, there would only be left the $\ce{H2O}$ adsorbed on the surface of the container.
The volume of a gas is undefined in vacuo. Solids and liquids have defined volumes, not unconstrained gas.

Yep, you can't do work expanding against a vacuum. So the water would just go whoosh! Although evaporative cooling will lead it to freeze, so it might be  little less dramatic than that.

What you are describing is physically incomprehensible. At 303K water has a vapor pressure P. If you place 5Kg of water in a container with a moveable piston applying enough force and adding enough heat to maintain 303K and pressure P eventually the water will all evaporate with a final volume V= nRT/P. Collect all the proper units and do the arithmetic. If the applied pressure is different from P a new equilibrium point[T] will be established or one of the phases will vanish.

This might be more appropriate as a comment (I did not downvote btw).

I am used to mysterious downvotes with no explanation. I tried to give a reasonable physical process to the questioner. The question unanswered is if work is done  in expanding vapor pressure against itself my moveable piston, opposed to placing the water in a vacuumjar of final volume V.

We are scratching our heads about an incomplete problem, methinks. Incomplete information. You'd have to provide either an external pressure or final volume. In the first case there would be work, not in the second.

2 scenarios: I. Place water in evacuated container of volume V gas occupies at T and VP. add sufficient heat to evaporate liquid isothermally ideal gas Vi = 0 dE=0 dH= PV. 2. Place water in Container with piston of force equal to vapor pressure. add heat to vaporize liquid doing work against piston isothermal dE=0 = Q-W  Q=W dH = PV. Ideal gas law loses 5L of volume. E and H state functions. Each gives same value[Total dE = latent heat of vaporization]


text:
Is this evidence that solubility of sodium bicarbonate<sodium carbonate?
Someone introduced $\ce{CO2}$ into saturated $\ce{Na2CO3}$ solution, and observed that white precipitate is produced. He therefore claims that $\ce{NaHCO3}$ has less solubility than $\ce{Na2CO3}$.
His claim is of course true, but does his experiment suffice to support this claim? The solubility of a solid substance refers to the mass of a substance that dissolves when it reaches saturation in $100~\mathrm g$ of water.
The related reaction is
$$\ce{Na2CO3 + CO2 + H2O}=\ce{2NaHCO3}\downarrow.$$

My doubts include: 1. The reaction uses up water; 2. The mass of $\ce{NaHCO3}$ produced is greater than the mass of $\ce{Na2CO3}$ reacted.

The essential info is the stage when the precipitation occurs. If it occurs at very beginning, then yes, it would be proof it is less soluble. If it occurs when most of Na2CO3 reacted, it would not be the proof.

The solubility of $\ce{NaHCO3}$ is $0.82$ M ($69$ g/L) at $0$°C, $1.95$ M ($164$ g/L) at $60$°C, and the salt start being decomposed at higher temperatures.
As a comparison, the solubility of $\ce{Na2CO3}$ is rather strange. The stable hydrate contains $10$ $\ce{H2O}$ at $\pu{T}$ < $30$°C, and only $1$ $\ce{H2O}$ at $\pu{T}$ > $30$°C. Expressed in moles, the solubility does not depend on the number of $\ce{H2O}$ crystallized with $\ce{Na2CO3}$. This is why we will reason with molar units.
The decahydrate $\ce{Na2CO3·10 H2O}$ has a solubility of $0.245$ $\pu{M}$ ($70$ g/L) at $0$°C, and $1.357$ $\pu{M}$ ( $388$ g/L) at $30$°C. At T > $30$°C, the solubility of the monohydrate $\ce{Na2CO3·H2O}$  is higher but it decreases with T.  It is $4.075$ $\pu{M}$ ($505$ g/L) at $30$°C, and $455$ g/L ($3.66$ M) at $100$°C).
As a consequence, if we compare their molar concentrations,   $\ce{NaHCO3}$ is more soluble than $\ce{Na2CO3}$ at $0$°$\pu{C}$. This tendency is reversed at $30$°C, where $\ce{NaHCO3}$ has a solubility $1.32$ $\pu{M}$, which is slightly lower than the solubility of $\ce{Na2CO3·10H2O}$ which is $1.357$ $\pu{M.}$  Above $30$°C, $\ce{Na2CO3·H2O}$ is much more soluble than $\ce{NaHCO3}$.

But the question is not if it is less or more soluble but if the precipitation is the sign of the lower solubility. Even if solubility of NaHCO3 was by 50% higher than of Na2CO3, it would still start precipitation if reacted stoichiometrically, because of higher mass and slighly less of water.

@Poutnik. I don't understand what you say. In my opinion, precipitation is the sign of a lower solubility, on the molar scale. Higher mass and small amount of water do not play the role of solubility. Precipitation happens if the solubility is low. That's all. Except in cases of supersaturation.

Solubility is evaluated on mass scale, not amount scale. Na2CO3 has from the solubility table on Wikipedia value about 21 g/100 mL. If reacted stoichiometrically with CO2, NaHCO3 would precipitate if having solubility around 35 g/100 mL or less (has less than 10 g/100 mL).   So precipitation is NOT a selective sign of lower solubility, as it would precipitate as well if it was somewhat more soluble than Na2CO3.

Yes. Precipitation IS a sign of low solubility. It is even the best sign of a low solubility. Or maybe you don't like the word "precipitation". Would you prefer another word ?

In my opinion, precipitation is the sign of a lower solubility, on the molar scale. -  That assumes 1:1 molar ratio. But 1 mol of Na2CO3 forms 2 mol of NaHCO3. So hypothetically and illustratively, if molar solubility would have been  0.2 mol/100 mL for Na2CO3 and 0.3 mol/100 mL for NaHCO3, it would have precipitated even with the higher molar solubility of NaHCO3, as 0.4 mol NaHCO3/100 mL would not stay dissolved ( not speaking about water bonded in HCO3-)

Your reasoning is correct, but it is a new approach. I perfectly know the $1$ mol $\ce{Na2CO3}$ forms $2$ mol of $\ce{NaHCO3}$. This does not change much in saying that $\ce{NaHCO3}$ makes a precipitate when $\ce{CO2}$ is bubbled into a concentrated solution of $\ce{Na2CO3}$ at $30, 40$ or $50$°C

Let us continue this discussion in chat.


text:
Why is the lysine side chain more basic?
As per PubChem: $pK_1 = 2.18$; $pK_2 = 8.95$; $pK_3 = 10.53$ at 38 °C, indicating that the side chain amino group is deprotonated at a higher pH, indicating it is more basic. I tried to justify this by analysing electronic effects in the conjugate acid:
Attempt I) The carboxylic acid group, at a higher pH would exist as $\ce{-COO}^{-}$ and exert a +I effect on nearby atoms. Then the alpha-amino group should be more stable in the protonated form and be more basic.
Attempt II) The carboxylate anion and $\ce{-NH_3}^{+}$ group may form hydrogen bonds. This again leads me to conclude that the alpha-amino group should be more basic.
But the experimental data contradict this. Perhaps the carboxylate anion H-bonds with (and stabilises) the side-chain amino group instead, but this seems unlikely (9-membered ring?).
Could anyone provide a decent theoretical explanation of the $pKa$ values?

https://chemistry.stackexchange.com/questions/63181/does-the-carboxylate-group-show-a-positive-or-a-negative-inductive-effect#:~:text=the%20carboxylic%20acid%20group%20always,due%20to%20excess%20negative%20charge.

Do you have a source for COO- being a +I group?

@GauravSaiMaddipati Solomons/Fryhle/Snyder adapted by MS Chouhan, page 93

We must have different editions, my page 93(2016) is about Lewis acids and bases, btw all the best for your jee

@GauravSaiMaddipati It is a few sections after that in the 2019 edition. "Relationships between structure and acidity". And thank you!

Perhaps the alkyl groups provide a better stabilizing effect, i don't know about the strength of the induction of the carboxylate.

In aqueous solution, the solvation of the COO- group ensures that it maintains an I- effect. This is clearly observed in a comparison of the ammonium group pKa on glycine (~9.8) vs methylamine (~ 10.7). This effect is much stronger for the a-amino group of lysine vs the e-amino group, whose pKa is close to what is typical for an alkyl amine

@Andrew that is what I was suspecting, but why does the text mention COO- as a +I group? Is that with respect to vapour phase?

@Andrew seems plausible, if you can cite sources saying the same I can accept that explanation. Since some texts directly contradict that.


text:
Storing 'Tollens Reagent'?
I have read that tollens reagent is unstable and hence must be freshly prepared before testing. However my school has a bottle of 'tollens reagent' on the shelf for use in organic qualitative analysis and when i asked my teacher she said they don't need to replace it and it's a special bottle. (it has a brown tint)
I have tested for aldehydes and gotten a fairly clear silver mirror with this solution so i am confused as to what is going on here.

Is this actually tollens reagent or just an AgNO3 solution? Or something else totally?

Can you get a clear silver mirror with just an AgNO3 solution?

Why exactly is tollens reagent unstable and is this magical bottle story
feasible atleast temporarily?

Is this actually tollens reagent or just an AgNO3 solution? Or something else totally?

It is an aged bottle of Tollens reagent.

Can you get a clear silver mirror with just an AgNO3 solution?

If just silver nitrate is used (without ammonia) then the silver forms more quickly, as a suspension rather than coating on the surface of the glass. leading to a black and cloudy appearance. This is why ammonia is used to form the diamminesilver(I) ion before reaction with an aldehyde. This ion is harder to reduce than silver ions, resulting in a slower, more controlled production of silver during the reaction. (Source)

Why exactly is tollens reagent unstable and is this magical bottle story feasible atleast temporarily?

Tollens reagent is sensitive to light. Also, Tollens reagent is never stored but it is prepared in-situ due to the danger of forming silver nitride. See: Is Tollen's test safe?

Tollens' reagent is sensitive and decomposed by violet and ultra-violet rays. If the bottle glass stops these rays, the reagent in such a bottle will resist decomposition. But it is not possible to obtain a nice silver mirror with just an $\ce{AgNO3}$ solution

What is "violet" rays?

@NilayGhosh   Rays of the wavelength 400-430 nm. UV as ultraviolet is just ultra violet.


text:
Could I expect any adverse reactions with this ink mix?
Our newest R&D project for a permanent, outdoor ink includes:

$\pu{25 g}$ of solvent blue 70.
$\pu{25 g}$ of nigrosene.
$\pu{??? g}$ of silver nitrate in a denatured alcohol base.
Several "natural" binders of the shellac variety as well.

My question really has to do with the silver nitrate. I've never had the chance to experiment with it, so I'm coming to the experts to make sure it's safe enough to at least try.

Adding silver nitrate to any mixture or solution will produce a solution that will slowly turn grey-black and irregular permanent grey-black, due to the effect of sunlight. This effect will occur on all places were the smallest drop of silver nitrate was in contact, everywhere, and even on the skin or trousers. So silver nitrate is safe, but the dark spots due to silver are more or less dark grey-black and irregularly distributed.


text:
Does tropylium chloride exist in ionic form or does it exist as a covalent molecule?
My teacher taught me that in tropylium chloride, the chlorine atom posses negative charge to make the ring aromatic.
I wonder that then will this compound will exist in ionic form due to very high polarity or will it exist as covalent molecule only?

Tropylium bromide was deduced to be a salt by Doering and Knox in 1954 by analysis of its infrared and ultraviolet spectra. The ionic structures of tropylium perchlorate and tropylium iodide  have been confirmed by X-ray crystallography. Wikipedia

https://scripts.iucr.org/cgi-bin/paper?S0365110X55002053

The linked source in Mithoron's comment seems not to give a definitive answer, due to the instability/reactivity of the tropylium chloride/7-chloroheptatriene.
We do know that hydroxy-tropylium ($\text{cyclo-}\ce{C7H6OH^+}$) forms an ionic compound with chloride ion [1]. The hydroxyl group is a pi donor, so this case is not exactly equivalent to a simple tropylium ion ring forming an ionic chloride.
Reference

Jandl C, Pöthig A (2017 Oct 1)."Hydroxytropylium chloride: the first crystal structure of an unfunctionalized hydroxytropylium ion". Acta Crystallogr C Struct Chem. 73(Pt 10):810-813. doi: 10.1107/S2053229617013183. Epub 2017 Sep 27. PMID: 28978788.


text:
Electrolytic reduction of nitrobenzene in differing media
In my lecture notes it is given that in a weakly acidic medium, nitrobenzene on electrolytic reduction gives aniline, which makes sense to me because there is ample H+ in the medium to continue adding electrons and H+ to fully reduce nitrobenzene past nitrosobenzene and phenyl hydroxylamine to aniline.
However, in strongly acidic medium electrolysis apparently gives p-aminophenol via rearrangement of the hydroxyl-amine.
Google only showed me this result, and I couldn't find any convincing reason for this.

Why does the reduction not go all the way to aniline?
Why does this rearrangement take place and why is it favored by strongly acidic media?


text:
Freezing of a beverage
I am trying to generally describe the process of freezing beer and how the different ingredients influence the freezing process. My general approach right now is to observe a beer as 90% water and 10% beer substance, which would make a binary system, since the composition is way too complex to describe otherwise. In Germany there exists a beer specialty, that is called "Eisbock" and in basically a freeze concentration of a beer.As stated above, I am trying to describe this process.
I have only little expertise in working on thermodynamic topics. So any help regarding this I appreciated. My general goal would be to bridge the gap from describing beer to water.
I have since found out that colligative properties play a role, the freeze kinetics and the heat/ mass transfer.
Also, I have been looking through the internet for a paper that describe to me how liquid/ solid phase transition works or is influenced. But the process has not been easy. That's why I am now trying to get some input via this forum.
Thanks a lot.

That 10% "beer substance" would be before fermentation, where most of approx 5% of formed CO2 would escape during the process.

That would be after fermentation. And my goal is first to take a broad look. Then the influence of certain substances

@Poutnik Besides ~4% of ethanol there's similar amount of sugars, proteins etc.

@Mithoron I have just meant that the usual marking of Czech beers - 10% or 12% is the most typical, means 10 or 12% of fermentable extract from the malt, giving at hypothetical (never that much) 100% fermentation around 5-6% of alcohol and about that much CO2. There is 0.3 - 0.5% of proteins.

Freezing soda, tea, cider, OJ, beer, usually results in pure ice coated with a concentrated liquid containing everything else until the eutectic is reached, if the container does not rupture. This was a common way to make applejack in very cold climates. Ferment some cider and place the barrel in the outside cold until it froze, then tap the center.

thanks for the info with the eutectic point @jimchmst


text:
Aqueous state on CO2 and O2 in the Photosynthesis Chemical Equation
While reading through a book I came across the equation for photosynthesis:
$$ \ce{6CO2(aq) + 12H2O(l) -> C6H12O6(aq) + 6O2(aq) + 6H2O(l)}\tag{1} $$ in the presence of sunlight + chlorophyll.
For some reason it seems counter-intuitive to me. I always thought that $\ce{CO2}$ would be present in the gaseous state. Similarly, $\ce{O2}$ is released, so I always thought that it would be in the gaseous state. I'm not able to understand why there are aqueous labels next to $\ce{O2}$ and $\ce{CO2}$. Does carbon dioxide get converted upon reaching the cell or something?

It's about reactions in the tissues - that's what aq reflects - all this happens in solution!

I want to know exactly how all this happens. CO2 is taken in as a gas and as you've said the reaction occurs in the tissues. Can somebody elaborate on how the aqueous O2 turns into gaseous O2 as it is expelled? And whether writing the states as gas would be correct.

In this case even respiration and all other reactions that occur in our body should have (aq) and (l) states, right? Pls answer.

https://en.wikipedia.org/wiki/Respiratory_system#Plants

This synthesis is a long series of chemical and physical reactions. The given reaction is a summary of some of them, but not all. There is at least one reaction occurring before it, and one after it. Before the written reaction, the following physical  phenomena (which is not a chemistry reaction) should happen : $$\ce{CO2 (gas) -> CO2 (aq)}$$ And, at the end of the written reaction, the following phenomena has to occur : $$\ce{O2 (aq) -> O2(g)}$$ although it is not a chemical reaction.

Photosynthesis happens inside of cells. So does glucose oxidation as the reverted process during aerobic energy release. You surely do not expect there are bubbles of CO2 or O2 in respective cells, being transported between cells and air.

The formula you were given is greatly simplified from the full Calvin-Benson photosynthesis cycle. If you want to know the details, it is a multi-step process, involving energy transfer among many molecules. "[I]t takes six turns of the Calvin cycle to fix six carbon atoms from CO2. These six turns require energy input from 12 ATP molecules and 12 NADPH molecules in the reduction step and 6 ATP molecules in the regeneration step."
Not only that, but the process differs amongst different organisms. There are C3 and C4 plants, cacti have yet another path, and some bacteria yet another!

I always thought that CO2
would be present in the gaseous state. Similarly, O2
is released, so I always thought that it would be in the gaseous state.

As the other answer states, this is a net reaction (or a black box with inputs and outputs). Depending where you draw your box, different physical states are appropriate.
If you look at an entire leaf, you would provide carbon dioxide in the gas phase, and release oxygen in the gas phase. If you look at a single chloroplast inside a plant cell, the aqueous physical state would be appropriate. In aqueous solution at pH~8 (in the stroma), carbon dioxide would mostly be in the hydrogen carbonate form (see Bjerrum_plot), but it seems to bind to the relevant enzyme, Rubisco, in the form of carbon dioxide (doi: 10.3390/catal11070813).

The CO2 and water never come into contact. All this rather poor statement about photosynthesis means is that so many CO2 and H2O molecules form carbohydrate and oxygen.
There are two main complex sets of reactions, light and dark (Calvin Cycle). In the light reaction one of many chlorophylls (chl) in a protein chl complex  absorbs a photon and transfers the energy to a 'special pair' of chl molecules from where electron transfer occurs (in the reaction center). The electron released ultimately is used to provide the energy reduce CO2 to carbohydrate. The 'special pair' (now positively charged) drags an electron off a water molecule via a metal complex, thus making water the fuel and the process is repeated.
Most of the structures of the complexes used are now known via x-ray diffraction of the proteins. Ultra-fast (picosecond/femtosecond) visible spectroscopy has determined the electron pathways and rate constants for many of these events. There is a summary here https://www.rsb.org.uk/images/15_Photosynthesis.pdf


text:
How to chemically remove niobium from boron carbide?
Previously I asked about chemical removal of CaO from boron carbide. I keep considering some refractory materials for use as sacrificial ones, so now I wonder if and how a layer of niobium (about 1.5 mm thick) could be chemically removed from a layer of boron carbide (about 0.15 mm thick). Thank you for your help.

Maybe https://patents.google.com/patent/US4266008A/en ?

@JonCuster : Maybe, but it is not clear if their etchants remove boron carbide as well, which would not be good.

Unless someone has specifically tried etching of niobium on BC, you are likely going to need to experiment some.

@JonCuster : Thank you.

You might consider chemical or electrolytic etching, e.g., https://www.steeldata.info/etch/demo/list/nb.html , depending on conductivity of the boron carbide, https://www.sciencedirect.com/science/article/abs/pii/S1293255822001820 and its chemical resistance. As @JonCuster states, you'll need to experiment for your needs.


text:
Looking for a practical book on IR spectroscopy techniques (LIF, TLDAS, FTIR, etc)
I am looking for a book that would ideally delve into the more practical and applied side of the IR spectroscopic techniques. Anyone can recommend me a a good book about this topic?
I have somewhat of a background in quantum mechanics and optics side of things, I am more interested in the actual implementation of these techniques.
As for the abbreviations used above:

LIF: Laser-induced fluorescence
TLDAS: Tunable diode laser absorption spectroscopy
FT-IR: Fourier-transform infrared spectroscopy

Though it isn't a substitute for a literature search/visit of a research library, the entries in Wikipedia added to translate the acronyms can be one entry.  Well, perhaps with the exception of FT (hard to find a modern IR spectrometer which doesn't use it, even old editions of the Silverstein book (resources) cover it).


text:
liquid (solution or polymer) with a density of more than 1.35g/cm3 and with element atomic numbers less than 9
Search for a liquid (solution or polymer) with a density of more than $\pu{1.35 g/cm3}$ and containing only elements with an atomic number less than 9.
For example, a saturated solution of sugar in water is $\pu{1.35 g/cm3}$. There are liquids with high density (for example, sulfuric acid), but it contains sulfur with atomic number 16. Liquid polymers known to me have a density of less than $\pu{1.1 g/cm3}$ to $\pu{1.2 g/cm3}$.

Can you be more specific? By sugar you mean sucrose? Would honey be ok, or corn syrup?

corn syrup has the same density. I found one substance - nitroglycerin. but it's explosive

Can you elaborate the intended purpose of the solution/liquid with a density greater than $\pu{1.3 g/cm3}$?  E.g., trifluoroacetic acid, $\ce{C2HF3O2}$ would fit the requirement ($\rho = \pu{1.489 g/cm3}$ at $\pu{20 ^\circ{}C}$) though definitively requires some precaution to handle.

this substance is needed as a sample for testing in CT. Ideally, this substance should be safe to use, inexpensive, transparent, odorless and have a density of more than 1.5. Perfluoropolyether CAS 69991-67-9 is almost suitable, but has too much fluorine and is slightly high in density. I have not found how to dilute it to obtain the desired density.

The density of honey typically ranges between 1.38 and 1.45 kg/L at 20 °C.[43] (Wikipedia) . I assume traces of heavier elements would not matter.

65-68% HNO3 has density around 1.4 g/mL. 100% around 1.5.

If it doesn't need to be liquid at room temperature, PET should meet the requirement without even exceeding atomic number 8.

If fully perfluorinated polyether polymers (such as Fomblin® PFPE Lubes from Solvay, density $\pu{1.88-1.90 g/cm3}$) are too dense or contain too much fluorine (how much fluorine is too much?) you might try a polyfluorinated polyether, for instance Galden-H (Density $\pu{1.65 - 1.67 g/cm3}$ from Solvay).

I found the density of Galden in an MSDS from Canon (also other websites), not directly from Solvay or PubChem.

I need to get a liquid with an equivalent atomic number less than 7.5. I don’t want to burden you with formulas for calculating the equivalent atomic number, but approximately this is the arithmetic mean of the atomic numbers of the elements included in this substance. Florine-containing liquids suit my density, but have a somewhat high equivalent atomic weight. I need to dilute (dissolve) them with a low atomic number component. or... find another solution to the problem

Computing the equivalent atomic number seems pretty trivial. Just guessing what it is gave me a number very close but perhaps slightly above 7.5 . Unfortunately perfluorinated compounds tend to be immiscible with anything else so finding a solvent to get the average AN of polyfluorinated polyether down does sound like a challenge. Maybe there is a solubility chart somewhere.

To my regret, as I understand it, this class of compounds dissolves (can serve as a solvent) only in fluorinated solvents and the previously proposed corn syrup does not dissolve in it; it matches my atomic number, but has a low density. Maybe there is another solution to this problem. (I'm not a chemistry expert). I will be grateful.

You've set some tight constraints. Excluding fluorine doesn't make it any easier!

You may want to check this post: https://chemistry.stackexchange.com/questions/179885/what-liquid-mixture-has-largest-negative-excess-volume/179973#179973


text:
Acidic strength of 4-(tribromomethyl) benzoic acid vs 4-(trichloromethyl) benzoic acid
From what I leart from my teacher is that 4-(tribromomethyl) benzoic acid is more acidic than  4-(trichloromethyl) benzoic acid. But I'm not sure what the reason is.According to me 4-(trichloromethyl) benzoic acid should be more acidic since Cl will exert much stronger -I effect and also hyper conjugation effect than Br.


text:
What is meant by "the position of equilibrium lies left (or right)"?
I am a high school student currently learning about equilibrium.
I was told that when there are more products of a reaction formed compared to reactants when the reaction reached the equilibrium, the position of equilibrium lies to the right (the product side) and vice-versa if more reactants are formed compared to products.
But I want to know why we say the position of equilibrium is located where there are more quantities of products/reactants? How is the quantity of reactants and products formed related to the position of equilibrium?
I apologize if my question is unclear. But thank you in advance.

Related, and maybe a duplicate: https://chemistry.stackexchange.com/q/8609

the equilibrium reaction notation has the left and the right side, why not to use it to describe which of the reactions (forward one from L to R, backward one from R to L) takes upper hand, taking the reaction state to the left or right? // As analogy, imagine collective ball sports with the left and right team like football/soccer, with the "game equilibrium" on the left or right, depending which team has the upper hand in the moment.

This answer is not so much about chemistry as about history of science. When you run google ngram on "equilibrium position", you get the following:

For the German term "Gleichgewichtslage" in the German corpus, you get:

If you search for the early uses of the term, you get articles concerned with mechanics. Objects can have one or multiple positions of equilibrium, for instance a cube (or a die) can stably sit on each of its 6 faces. This is different from chemical equilibrium, where there usually is a single state of lowest Gibbs energy called the equilibrium state.
If you do an internet search in 2024, you get a mix of physics (i.e. mechanics) and chemistry hits.
There is a mechanical analogy to a chemical equilibrium involved two wells, table-tennis balls and shaking. You can also implement this as a snowball fight. In the two-well analogy, you add some balls to either side, and shake the surface. After a while, you will find that more balls are on the right, and that we reach a dynamic equilibrium (i.e. some balls will go from left to right, others from right to left, without changing the numbers much). If you change the intensity of shaking (the temperature), you will get a different result. There might be a video of a physical model of this somewhere, but I only found the corresponding PhET simulation.

I think this is the picture folks have in mind when they say "the equilibrium is on the right side". There are several inaccuracies:

In the real world, reactants and products occupy a common space, so "position" does not quite capture the situation, "state" might be better.
In a chemical equilibrium, there is always some reactant and some product. So it is not all the way on the right or all the way on the left.

Tangential: interesting to read i) Waage spent time with Bunsen in Heidelberg (ref) hence possibly knew German (perhaps as his brother in law Guldberg, too). ii) They presented the results once 1864 (possibly contributing to the first peak in the second trace), followed by van't Hoff 1877, and a twice by Guldberg and Waage in 1872 (ref, the French Wikipedia equally has the English translation of the 1864 paper by JChemEduc here).

@Buttonwood I suspect it might not be lost on you that the english translation of "Waage" is scale. That would make the term equilibrium almost eponymous.

The peak in usage of "equilibrium position" around 1960 is somewhat strange. I wonder why it peaked and what alternative terms might have replaced it. As a side note "Silent Spring" was published in 1962 and a nascent environmental movement might have led to a decrease in enthusiasm for chemistry around that time.


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How do positive ions help in conducting electricity?
I understand that positive ions usually move towards the negative electrode and gain back their electron(s). But how is this considered as the positive ions helping to conduct electricity?

Same as negative ions do.

Two things that may help you: (1) the 'electricity' you mention is exactly the motion of the ions across the electrolytic solution, they are the same thing, (2) what 'helps/not helps' conducting the electricity is the potential differences established across the interfaces in an electrochemical cell; no potential differences and we get no movement of ions, huge potential differences and we get huge density currents. You may get some ohmic loss in the bulk of the electrolytic solution, that will affect the movement too.

You might look at: https://chemistry.stackexchange.com/questions/144010/do-ions-still-conduct-electricity-if-a-physical-connection-made-by-aqueous-ele?rq=1

As an external macroscopic observer, how to you distinguish a positive charge moving to the right from a negative charge moving to the left? They both yield the same net charge flow (i.e. current).

A definition of conduction is "moving charges from one place to another." Positive ions are charged particles. They move. That is conduction.
As an aside, in metals, conduction is like the action in Newton's cradle. Electrons bump other electrons, rather than each migrating the whole distance; the drift velocity is far slower than the speed of signal propagation. The same can be true of ions in liquids, but ions in a rarefied gas may travel the whole distance.


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Definition of alpha-carbon
What is an $\alpha$-carbon, and how can I identify one?
The other questions because of which my questions has been termed as duplicated is asking about alpha carbons in alkyl halide , while my question addresses simply ,"what an alpha carbon is and not in any particular kind of carbon compound", I didn't even knew if alpha carbon is supposed to be connected to functional groups or halogens or double/triple bond, While the other post asks for alpha carbon in "alkyl halide".

It is a descriptor of the position of one group relative to another. The most common situation would be for a carbonyl group, where the α-carbon is the one immediately adjacent to the carbonyl itself:

It is for this reason that naturally occurring amino acids are often called α-amino acids - the amine is attached to the α-carbon.

so what you are trying to say is that  $\alpha carbon$ is a carbon that's attached to the carbon that's attached to  a functional group?

Not quite - I'd say the α-carbon is the one that's attached to the functional group. For example, the "ketone" functional group in the molecule above includes both the oxygen and the carbon. So the α-carbon is the first carbon attached to the ketone


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I can't understand a fragment in the Mass-Spectrum of Benzophenone
Summary:
Took multiple Mass spectra of Benzophenone; There always is a peak at $\pu{152.2}$ that I don't understand.
I made some Measurements of Benzophenone with a GC-MS. Always using EI (electron impact ionization). The spectra at an energy of $\pu{70eV}$ is pretty easy to interpret at first.

However taking a closer look the peak at 152.2 doesn't make any sense to me. I see it as well in the spectra with other ionization energies.

The same peak also appears in spectra that can be found on the Internet. E.g. Here or on NIST and even in this advert.
I can't find to think of a reasonable explanation for this peak. A book of mine (1) tells me that the characteristic mass difference of Δm/z = 30 is typical for $\ce{H2C=O}$ fragments. However I like I said I can't explain how this would make any sense.
Can anyone clear this for me?
(1) Strukturanalytik organischer und anorganischer Verbindungen; M. Reichenbächer, J. Popp; Teubner Verlag, first edition; Wiesbaden 2007

Why reject the choice you mentioned that a CH2O fragment is removed resulting in (C6H4)2 biphenylene?

An argument in favor of the loss of a fragment $\ce{CH2O}$ is the rather low intensity of the resulting ion $152.2$, about $1-2$%. So this loss with production of biphenylene should be rare.

Another argument is the large 77.1 peak which I imagine as phenyl groups of some sort able to recombine with each other at low yield to form some mix of biphenyl at 154 and biphenylene at 152.