text:
Chemical Structure of PLA soaked in Ethyl acetate
I am trying to find a way to separate my PLA samples and ABS samples. Both of these are 3D printing plastics (Polylactic acid and Acrylonitrile butadiene styrene)
Last night I've observed:

When soaked in Ethyl Acetate:

ABS disintegrates in about an hour,
PLA loses "fancy" properties, ( such as green/yellow color-changing filament, went blue. )
All PLA, fancy or standard, turns very flexible.

I would say this could be a good start for my process, as the ABS is separated from the PLA, but the PLA has changed now to a fairly unusable form, (the ABS especially so)
How can I understand this "flexible PLA" and the solution? I can see that the PLA has been chemically altered, but how? and can it be corrected? What happened to the Ethyl Acetate and the ABS?
I'm still quite noobish in chemistry, so I'm happy for anyone to throw a book at me or where I should start to learn about this.
In the meantime, I'm planning to make more flexible samples of PLA to put into the dehydrator later this week in a vain effort to see if that might "evaporate" and correct it, but I'm hoping someone would be able to give me a bit of a leg-up on understanding this chemical change and how to account for it. The end result is for the PLA to be grounded and melted into usable forms.
The plastics are visibly similar, which makes it a bit difficult to separate by eye.

What is PLA  ? Is it Polylactic acid ? And ABS ?

PLA Polylactic acid
ABS Acrylonitrile butadiene styrene


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How to get the strongest reaction out of 1kg of guncotton in a space of 15*15*15 meters without using an additional gas source
It would be burnt outside at about 2 degrees Celsius with about 75 percent air humidity, 1400 meters above sea level. Even though that's not exactly true, I'll treat it like pure guncotton. I'm asking myself questions like if I should use an initial explosion to heat it up and/or spread it out.

I don't know which grade of Nitrocellulose you obtained, but the 12.2% nitrogen is used in gelatine dynamites and the 13.5% nitrogen takes the name of guncotton. I would avoid make exploding 1 Kg of one or another.

@jojostack it's guncotton. Of course I wouldn't use it as a weapon. (Guns are way easier to use and harder to trace) I'm living next to a forest and I'm used to playing around with fireworks and smaller amounts of guncotton. Can you give me a link or something?

We allow questions about chemical weapons on our site, so this wouldn't be much of an issue. However, your question is somewhat thinly rooted in chemistry (the interesting part you have already accomplished). It is quite impossible to judge the quality of your product and from that point it is almost impossible to answer your question. That may attract the downvotes.

@Martin-マーチン Thanks a lot. I'll edit the question.

Fire requires three things: heat, fuel and oxygen. When burning a cotton ball, a match will provide heat, the cellulose fibers will provide fuel and the air will provide oxygen. It is the amount of oxygen in the air that causes the cotton to burn relatively slowly. By soaking the cotton in Sulfuric Acid and Nitric Acid, the hydrogens that were attached to the "spokes" of the carbon wheel are broken off and replaced by nitrogen and oxygen. Simply put, the spokes of the carbon wheel get more oxygen atoms stuck to them after a bath in strong acids. With all this extra oxygen, the cellulose can burn much more efficiently and will produce a quick, clean burning flame!
The name nitrocellulose gives a pretty good idea of the chemistry of the gun cotton. The "nitro" refers to the nitrogen groups that attach to the spokes of the glucose carbon wheel. Each of the nitrogens brings extra oxygens with it when it attaches. The "cellulose" of nitrocellulose refers to the cellulose fibers of the cotton. Hence the name "nitrocellulose".
Nitrocellulose is often found as a product called flash paper when 100% cotton paper is subjected to the same reaction. This paper is often used in magic shows as the magician is attempting to redirect the attention of the audience. It is available here and it is wiser to buy flash paper than to try to make your own!

This is a nice summary of how gun cotton works, it doesn't answer the question though.

source: Stack Exchange

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Basic question about an expression for Gibbs free energy
I'm getting into thermodynamics, and I have a basic question about the following expression for Gibbs free energy:
$G = \sum \mu_i N_i$
Is it correct that this equality only holds for constant temperature and pressure, as follows from integration over N of the expression for $dG$:
$dG = -SdT + VdP + \sum \mu_i dN_i$

No, it is not constrained to just constant temperature and pressure.  It follows from the definition of partial molar properties.

Both Chet Miller and Buck Thorn have provided good answers, and explained that
$$G = \sum \mu_i N_i \tag{1}\label{1}$$
is indeed a function of $T$ and $P$. One way to show that is as follows.
As you say, we can derive (1) from the Fundamental Theorem of Thermodynamics for a closed system in which composition can change:
$$ dG = V dP -S dT + \sum_i \mu_i dN_i \tag{2}\label{2}$$
Now take the differential of (1):
$$dG = \sum \mu_i d N_i + \sum N_i d\mu_i $$
Equate that differential with the Fundamental Theorem (2), to give
$$\begin{align}
dG = \sum \mu_i d N_i + \sum N_i d\mu_i &= V dP -S dT + \sum_i \mu_i dN_i \\
\sum N_i d\mu_i &= V dP -S dT
\end{align}$$
which shows that a change in either $T$ or $P$ at constant composition necessarily results in a change in some chemical potential $\mu_i$.  Hence, $\mu_i$ is a function of both $T$ and $P$, as is $G$ in equation (1).
For a system composed of a pure substance,
$$\begin{align}
N d\mu &= V dP -S dT \\
d\mu &= \overline{V} dP -\overline{S} dT
\end{align}$$

In the equation
$$G = \sum \mu_i N_i$$
$G$ is the absolute value of the Gibbs free energy at some point in state space. While not explicitly stated, the values of the chemical potentials $\mu_i$ are in general functions of composition and other state variables (e.g. T and p), rendering $G$ a function of these same state variables.
On the other hand, the equation
$$dG = -SdT + VdP + \sum \mu_i dN_i$$
describes an infinitesimal difference in the free energy between two (infinitesimally close) points in that space.
Neither equation puts constraints on T and p, that is, both are general.

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text:
Pressure vs volume plot for real gas and ideal gas
Background
Recently I have been studying about the states of matter and came to the topic of ideal gases and real gases, and the laws related to them. While studying it from my textbook, I saw the plot of pressure vs volume for real gas and ideal gas. The problem is that the interpretation I am getting from the graph is quite contrary to what I have learnt till now (this might be because of some conceptual misunderstanding of mine. Sorry if the question looks foolish).
Question Explained

In the above image of the plot, there is a point where real gas behaves the same as ideal gas (where the curves intersect). From this point, as we decrease the volume in the graph we find that pressure exerted by the real gas is greater than that by the ideal gas. This seems to be contrary to what I have learnt, which is as the volume of a real gas decreases the pressure exerted by a real gas becomes lesser than ideal gas at the same volume (due to increased intermolecular attraction). In reality even this isn't the case as decreasing the volume much further causes the intermolecular repulsion to overcome the attraction at some point and hence increase the pressure. But as far as I know the graph has been made with the use of the van der Waals equation (I checked it myself via plotting the graph in Desmos and it looked the same as shown above). But the assumption with which the equation (and hence the graph) were derived say something different. The equation I'm talking about is $P_\mathrm{ideal} = P_\mathrm{real} + \dfrac {an^2}{V^2}$, and as far as I can comprehend it says pressure by an ideal gas is greater than pressure by a real gas at a given volume. Hence the question:
Why does the equation $\left(P_\mathrm{ideal}=P_\mathrm{real}+\dfrac {an^2}{V^2}\right)$ and the graph (shown above) say totally different thing about the pressure at a given volume?
Please do point out if I am going wrong anywhere.
Thanks in advance!
Here in this crash Course video the same logic as mine is given for decrease in pressure for real gas.

Could you please provide a reference or link to the source of that figure. It is misleading and has appeared in numerous other posts. It would be great if we could track down the culprit.

@BuckThorn Please see pg. 14 https://ncert.nic.in/ncerts/l/kech105.pdf

Summary
At sufficiently low volumes, the excluded volume effect causes the van der Waals (VDW) pressure of all gases to exceed the ideal gas (IG) pressure.   At higher volumes, the VDW pressure may or may not cross below the IG pressure.  This depends on the temperature, and the relative size of the VDW a and b terms. If the temperature is sufficiently low, and if the a term is sufficiently large relative to the b term, you will see a crossover.  Otherwise, you won't.
Notes:
(1) The OP asks for a discussion of VDW gases, and thus most of this analysis concerns the predictions of the VDW model.  However, the curves presented in the OP's screenshot are for real gases and thus, at the end, I discuss their behavior.
(2) Below the VDW critical temperature (which equals $\frac{8a}{27bR}$, and is thus  gas-specific), the VDW isotherms (the curves on a  $p-V$ graph) begin to show oscillations.  Those oscillations are beyond the scope of this question, but they should be noted for completeness.  This answer assumes we are above the VDW critical temperature for each gas.
The VDW model
The VDW model assumes the gas molecules are hard spheres that interact only attractively.  Unlike the case with the attractive forces, there are no long-range repulsive forces in the VDW model.  Indeed, there are no repulsive forces at all between the gas molecules (except insofar as they are hard spheres that can't interpenetrate).
Instead, the molecules are assumed to have a physical volume, and the effect of this volume is to reduce the free volume left for the gas.  The volume of one mole of molecules (or atoms) is given by the VDW "b" term.
This should help illustrate how the physical volume of the gas particles affects the pressure:
Suppose you are at a volume where the hard spheres themselves take up half the container.  If there were no attractive forces, the pressure of the VDW gas would be double that of the ideal gas, because only half as much free volume is left for the VDW gas, such that p will be a monotonically decreasing function of V.
Physical explanation of difference between pressure of VDW gas and ideal gas
At very low volume (low relative to the volume taken up by the hard spheres), the molecular volume term dominates, and thus p_VDW > p_IG (where IG = ideal gas).
What happens at higher volumes, however, is gas- and temperature-specific.  And thus whether or not there is a crossover is gas- and temperature-specific.
Case I:  Lower temperature
At a sufficiently high volume, the pressure-increasing effect of molecular volume becomes less than the pressure-reducing effect of the attractive force, and thus p_VDW < p_IG.  Hence there is a point at which the VDW and IG curves cross over.
Again, let me emphasize that what constitutes "lower" vs. "higher" temperature is gas-specific. For instance, while $T = 273 \, \ce{K}$ is in the low-temperature regime for $\ce {N2(g)}$, it's in the high-temperature regime for Ne.   Thus, at this temp, you would see a crossover for $\ce {N2(g)}$, but not for Ne.
More specifically, $\ce {Ne(g)}$, which has a small a value relative to b, loses its crossover behavior above $149.8 \, \ce{K}$, while $\ce {N2(g)}$ doesn't lose its crossover behavior until the temperature exceeds $425.8 \, \ce{K}$.
Case II: Higher temperature
As explained above, as the volume increases, the effect of molecular volume decreases.  However, the intermolecular spacing also increases, lessening the average attractive force between the particles.  Further at, high temperatures the kinetic energy of the gas is higher, thus making the effect on pressure of intermolecular attraction relatively less.  If the attractive force is sufficiently low, and the temperature is sufficiently high, then the attractive force is never able to overcome the molecular volume effect.  Hence, for these gases at these temperatures, p_VDW > p_IG at all volumes.
Note: the above discussion is referring to the VDW model, not necessarily to how real gases actually behave.
Mathematical explanation of difference between pressure of VDW gas and ideal gas
We can also see this from the mathematical structure of the VDW equation:
$$p=\frac{RT}{V_m-b}-\frac{a}{V_m^2}$$
In the limit as $V_m \rightarrow b$, $(V_m-b)\rightarrow 0$, causing the first term to dominate over the second.
As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b.
In the limit as $V_m \rightarrow \infty$, for all gases, the VDW equation reduce to the ideal gas equation:
$$p=\frac{RT}{V_m}$$
Plots, Case I, lower T ($\ce{N2(g)}$ at $\pu{273K}$)
Comparative pressure vs. volume plots of VDW and ideal gas
Now let's take a look at what the actual $p-V$ graphs look like, using $\ce {N2(g)}$ as an example.  The VDW a and b terms for $\ce {N2(g)}$ are:
$$a=1.37 \ce{ \frac{bar L^2}{mol^2}}$$
$$b=0.0387 \ce{ \frac{L}{mol}}$$
Assuming $T=273 \,\ce{K}$, we can determine crossover point by setting pIdeal = pN2VDW (the VDW pressure for $\ce{N2(g)}$), and solving for $p$.  The crossover occurs at $V = 0.108 \,\ce{ L}$.  But if we plot  pIdeal and pN2VDW from about 0.1 L to 10.0 L, we find they are nearly on top of each other:

Thus to see the crossover, we need to focus on the region around $V = 0.108 \,\ce{ L}$:

Plot of p_VDW–p_ideal vs volume,  using $\ce{N2 (g)}$ as the VDW gas
We can further improve our understanding by plotting the difference between pIdeal and pN2VDW vs $V$.  This allows us to see, directly, how the deviation between the VDW equation and the ideal gas equation changes with volume.
Let's start at high volume, and move to the left.  At high volume, the attractive term dominates.  As the volume decreases, the molecules get closer together, so the attractive term has even more effect, increasing the difference between pIdeal and pVDWN2. This continues until $V=0.154 \,\ce{L}$ [I determined this minimum by setting the derivative of (pN2VDW - pIdeal) equal to zero, and solving for $V$.]  Beyond this, the increase in pressure due to the diminishing free volume becomes greater than the decrease in pressure due to the increasingly strong attractive forces.  Eventually, at $V=0.108 \,\ce{L}$ the effect of the hard sphere volume just balances that of the attractive forces, and the curve crosses above the zero line.

Plots, Case II, higher T ($\ce{N2(g)}$ at $\pu{500 K}$)
Assuming $T = 500 \,\ce{K}$, if we solve for the crossover point as we did above, we get a non-physical answer of  $V = -0.222 \,\ce{ L}$.  This indicates the curves don't cross over.  We can see this from the plots:

Real Gases
The easiest way to see how the isotherms for real gases compare with those for an ideal gas is to look at the measured compressibility factor, $Z$:
$$Z=\frac{p V_b}{R T}$$
$Z = 1$ for an ideal gas.  Thus, for a given $V_b$ and $T$, $Z>1 \implies p_{real} >p_{ideal}$, and visa-versa.
From https://en.wikipedia.org/wiki/Compressibility_factor#/media/File:Z_Overview.png, we have the following diagram for $\ce{N2(g)}$.  It appears to show behavior similar to what the VDW equation predicts— at lower temperatures, we see a crossover from
$p_{real} > p_{ideal}$ at high $p$ (low $V$), to $p_{real} < p_{ideal}$ at low $p$ (high $V$).
However, as the temperature increases, $p_{real}$ reduces its tendency to dip below $p_{ideal}$ (see red curve).  However, a direct examination of the numerical data would be needed to determine if, at high $T$, there is no crossover at all.

Can even at very high pressure $V_m$ $\ce{->}$$b$ ?

+1 for your effort but I am confused as there are contradictory answers. Few confusions: (1) From the VDW equation you mentioned, when $V_m$$\ce{->}\infty$, it reduces to ideal gas expression but from the graph, $P_{real}<P_{ideal}$. $P_{real}=P_{ideal}$ point is at a much lower $V_m$. (2) Assuming the graph do cross, can we say that at very high $V_m$, a real gas is more $perfect$ than an $ideal$ gas?

It looks like there is something wrong here: at volume 0.25 dm$^3$/mol the vdw pressure for nitrogen is 85.56 bar and ideal gas  90.8 at 273.15 K using your constants.  You seem to have used a V high temperature in your plots, are you sure the constants used apply when this is the case ? Also look at the scales in you crossover graph they appear to show no difference $\delta x=\delta y=0$ ? What value for gas constant did you use?

You sometimes write 'van der Walls' instead of the correct 'van der Waals'. If you again edit it, you might want to change that.

Please have a look at the MathJax Guides here and here. You are using it wrong. The macro \ce{…} needs the curly braces to work properly, your way of writing it \ce N_2(g) is equivalent to \ce{N}_2(g), however \ce{N2 (g)} is what you want. Don't use it for units either. Incorrect: a=1.37 \ce{ \frac{bar L^2}{mol^2}} Correct: a = \pu{1.37 bar L2 // mol2}.

@Apurvium (1) According to the VDW model, the pressure goes to infinity in the limit as  Vm -> b.  Physically, it's much more complicated, because gases aren't hard spheres, and they will typically become solids before extreme pressure pushes them into a repulsive regime. I.e., energetically, gas interactions are always attractive, and it's entropy that keeps them apart.

@Apurvium  (2)For all the plots that go to higher volumes you can see, from the RHS of these the plots that, as Vm increases, the VDW and IG curves get closer.  The curves never exactly overlap on these plots, because they only become the same as Vm->infinity, and you can't plot to infinity.   The only point at which the WDW and IG curves show identical pressure is at the crossover (if there is a crossover).

@porphyrin Thanks, I should have caught that. Turns out it's tricky to use units when plotting in Mathematica, so I just made my quantities dimensionless, which eliminated the scaling error (other functions, like Solve, handle dimensioned units without issue). Also caught another error. Please take another look.

Can we conclude a final mathematical expression between $p_{real}$ and $p_{ideal}$? The one which is written in the question or assuming $V>>nb$, $p_{real}=p_{ideal} +(n^2/V^2)(bRT-a)$ as posted by Lllt?

@Apurvium  Yes, I agree with the expression dervied by Lllt for the high-volume regime ($V_b>>b$).  Assuming you are above the VDW critical temperature ($\frac{8a}{27bR}$), if there is a $T$ below which $a<bRT$ then, at high volumes,  the attractive forces will win, and you will see a crossover.  Otherwise, you will not, because the VDW pressure will always be greater than the IG pressure.

@Apurvium The accurate expression is the one which I wrote in my answer initially, however using certain approximations I tried to explain the graph mathematically.

@Apurvium  See the note I added at the end about the behavior of real gases. They also show a crossover at lower temperatures.

The lower right part of graph indicates that real gases exert less pressure than ideal gas for a fixed volume. This happens because the ideal gas molecules are not attracted by surrounding molecules (therefore if an ideal gas molecule move at velocity $V_1$, It will strike the wall with same velocity, provided it doesn't collide with another molecule) while real gas molecules face attraction from surrounding molecules as they move towards the wall ( thus if a real gas molecule moves with a velocity $V_1$, it will strike the wall with somewhat less speed as nearby molecules' attractive force will slow it down)
In the upper left part of graph, you can see that to exert same pressure , we need more of a real gas then an ideal gas or similarly at a certain volume of both , the pressure exerted by real gas is more than that of ideal gas. This can be explained by fact that molecular size of ideal gas particles is negligible while that of real gas particles is considerable. If you take equal volume $V$ of both, the ideal gas has occupied complete volume $V$ while real gas has occupied effectively somewhat less volume (which is $V'=V-nb$). Therefore even if we have same number molecules of real gas as that of ideal gas, real gas molecules have lower volume to move into thus they exert more pressure.
The Van der Waals equation of state is
\begin{align}
  p &= \frac{nRT}{V-nb} -\frac{an^2}{V^2}\\
    &= \frac{\frac{nRT}{V}}{1-\frac{nb}{V}} - \frac{an^2}{V^2}\\
    &= \frac{nRT}{V}\left({1-\frac{nb}{V}}\right)^{-1}
       -\frac{an^2}{V^2}\\
    &= \frac{nRT}{V}\left({1+\frac{nb}{V}}\right) - \frac{an^2}{V^2}
       & \text{assuming } nb &\ll V\\
    &= \frac{nRT}{V} + \frac{n^2bRT}{V^2} - \frac{an^2}{V^2}\\
    &= p_\mathrm{ideal} + \frac{n^2}{V^2}(bRT-a)
\end{align}
Depending on whether $bRT$ is less than or greater  than $a$, you have $p$ less than or greater than $p_\mathrm{ideal}$.
Also note that  when $V$ is very large the second term can be neglected and in that case real gas tends to ideal conditions.

Please compare your answer with the equation also for both situations, i.e. large and small volumes.

@Apurvium If repulsive forces become dominant then why should pressure be less than $P_{ideal}$, it should be more

Yes, you are right, I will repost.

The $T$ is constant & I think the $V$ is volume of container. The only way we can increase the $P$ is by reducing the volume of container by a piston or so. When we do so, the repulsive forces also become significant for real gas. Now, the pressure exerted at the walls of the container should be more than $P_{ideal}$. This is not according to the equation between $P_{ideal}$ and $P_{real}$ but yes, according to the graph. At high volume the situation is vice versa.

$\frac{1}{1-a} \ne 1+a$  //  $\frac{1}{1-a} \approx 1+a$ for $|a| \ll 1$. So  there is the condition $nb \ll V$

@Poutnik yes, Thanks for pointing out, I edited my answer

(1) Here, the assumption is $V>>nb$ for all cases (which I don't know, but may be true). (2) When $bRT>a$, $P_{real}>P_{ideal}$ and when $bRT<a$, $P_{real}<P_{ideal}$. Is there any intuitive explanation for this? The units of $a$ and $bRT$ are $\pu{J m3 mol-1}$, which does not give any insight. (3) When $V$ is $very~very~large$, $P_{real} <P_{ideal}$ from the graph but $P_{real}$$\ce{->}$$P_{ideal}$ from the equation.

Please note that the symbol for pressure is a lower case $p$.

It's an assumption , certainly not true for all cases

intuitive explanation is already given in above paragraphs, I try to frame mathematical explanation also

Yes $P_{real}<P_{ideal}$ and tends to or approaches $P_{ideal}$

When the pressure tends to infinity, the volume of the ideal gas tends to zero, and the volume of the real gas does not. It tends to the proper volume of the molecules, which is greater than zero. This is exactly what you see on the upper part of the picture.
And when the pressure is extremely low, the volume increases but not as much as calculated with the ideal law, because there is some small attraction between the molecules. This is exactly what you see on your picture if you draw a horizontal line at its lower part

At very large volumes and low pressures, the behavior of a real gas should approach that of an ideal gas.

Sorry but I have a confusion. If that's the case then aren't the both graphs (real as well as ideal) showing the fact of non zero volume (reason being that both look like a rectangular hyperbola to me)?

The ideal gas is a hyperbola. The real gas has a non analytical curve.

@maurice At a constant $V$, if we compare the pressures vis-a-vis the equation between $p_{ideal}$ and $p_{real}$?

Related

The ideal gas pressure is always above the 'real' gas pressure as given by the van-der Waals eqation
$$p=\frac{RT}{V_\mathrm{m}-b}-\frac{a}{V_\mathrm{m}^2},$$
where the volume is the molar volume $V_m$. The molar volume varies from say $0.1$ to $\pu{5 dm^3/mol}$ and by experiment constant $a$ in units $\pu{dm6 bar mol-2}$  is numerically greater than $b$ in units $\pu{dm3/mol}$ and the gas constant $R=\pu{0.08314 dm3 bar mol-1 K-1}$.
At large volume the second term tends to zero and the first to $RT/V$ the ideal gas value. At small volumes the second term becomes larger and negative and reduces the pressure from the ideal behaviour. Thus the ideal and real curves do not cross and the ideal is always greater than the real.
The interaction between molecules ensures that the real pressure is lower than in the ideal case. (You can imagine this in a hand waving sort of way by imagining that a small fraction of molecules form dimers so that the expected pressure is not measured because some molecules are paired up.)
Thus the figure in the question is wrong on a number of points.

I'm adding some notes here about the vdw equation as there is confusion in the comments.
First it is clear that the van-der-Waals equation can produce negative pressures. It is an empirical equation that produces largely qualitative results. All that has to happen to get negative pressures is that the second term exceeds the first in the equation above. Negative pressures are, of course, unphysical and this shows that this equation has to be used with great care otherwise completely misleading results will be obtained.
Second, at large volumes the $V_m-b$ term is smaller than $V$ because $b$ is positive, so the pressure will exceed that given by the ideal gas but the effect is small. A typical molar volume of $0.25\;\mathrm{ dm^3/mol}$ vs a  typical value of $b=0.02\;\mathrm{dm^3/mol}$, so small as to be physically meaningless but mathematically possible, what I mean is that we would normally take $V_m-b \to V_m$ at large $V_m$.
The important point about the two curves in the figure is that the interaction between molecules, (due to polarisability/ dipoles ) makes the experimental pressure less than that of the ideal, e.g. methane at 273.15 K pressure is 78.6 bar by experiment but 90.8 by ideal gas law and 72.9 by van-der-Waals. The van-der-Waals eqn. tries to model the experimental results and does so quite well but should not be pushed to the mathematical limits as here it fails.

Your answer is definitely enlightening ,are you sure that those curves never cross each other (I am asking because the figure taken above is from the national book of India)

They never cross, you can see that in the limit when V is very large the ideal gas law is produced to that they tend to the same value. Try this with some numbers, say at 300 K for methane  a=2.3 , b=0.043 in the units I gave and V=1 vs V=10 etc.

I am afraid the national book of India is wrong !

@Maurice please see figure at pg.14 https://ncert.nic.in/ncerts/l/kech105.pdf

@porphyrin It is true, the van der waals gas curve is always below the ideal gas curve

@Theorist see my comment above, you seem to have used a v high temperature which would be the effect of using 8.314 instead of 0.08314 dm$^3$bar/mol/K for the gas constant

Correcting my earlier comment: Sometimes the curves do cross over, sometimes they don't. As $V_m$ gets large, whether the attraction can win out over the excluded volume is determined by the magnitude of T, and the relative values of a and b.

I don't have the rep to downvote this, but it is plain wrong. While the second term indeed pulls the pressure below that of ideal gas, the first term produces pressure higher than that of ideal gas (RT/V), because the denominator is smaller. See the answer by "theorist" for a practical example.

@TooTea It is not 'plain wrong' as you write, the curves are not sketched but  calculated using the vdW and ideal gas equations using experimental constants for methane. I have explained with some extra text in my answer why taking the vdw equation to its limits leads to an incorrect understanding, i.e. the difference between strict mathematical usage and using a scientists physical understanding of what the equations are describing.

@porphyrin Sure, for methane at ambient temperatures, the curves never cross. What's wrong with your answer is that you then generalize this to all gasses. See the answer by "theorist" for counterexamples where the curves do cross for real gases under non-extreme temperatures.

@Tootea, When you compare the ideal gas law with the empirical vdw you can get negative pressures and crossings etc. What one is doing  is only comparing two equations nothing else - no physics/chemistry is involved. The original question showed real gas vs ideal gas I used vdw to model a real gas because I don't have  any real gas data at hand.  Within the limits I used what I said is correct. It is true that I should have added that the vdW eqn can give negative pressures so is limited in what it can predict. The main point is that real molecule interact with one to reduce pressure vs ideal.

@porphyrin My point is that at high temperatures and/or densities, the repulsion always wins (at least for real gas, even though the vdW parameters for some gasses might be unable to capture this) and thus the real pressure becomes higher than ideal (compressibility factor exceeds 1). This is inevitable because ideal gas means no repulsion, no molecular volume at all. If you compress any (supercritical) gas hard enough, at some point the molecular volume will overcome any other interaction (Pauli repulsion always trumps everything with shrinking interatomic distance).

@TooTea,  quite so, I agree, but this is at an extreme.  The figure in the original question must relate to the gas at sort of normal conditions one might meet as its a from a general textbook, it is very misleading for students otherwise, i.e. to show extreme case (as it does) and ignore more general case.  My comments related to that general position, interactions between molecules lead to a reduced pressure vs ideal. From this position you can then go on to extremes.

Mathematically, the curves of real and ideal gas equations can intersect.
$(P + \dfrac{an^2}{V^2})(V - nb) = nRT$
Plugging $P= \dfrac{nRT}V$ gives  $V = \dfrac{anb}{a-bRT}$.
Thus the curves will intersect (in the first quadrant) if $a>bRT$, otherwise they won't.
Graphs intersect:

Graphs don't intersect:

Desmos


text:
Is there a way to dehydrate hydrated aluminum chloride for use as a Lewis acid?
I heard that one can use thionyl chloride to dehydrate it, but I'm not really sure if it would be available for amateur chemists since it's listed on the UN Convention for (the Elimination of) Chemical Weapons.
And no, I'm NOT thinking of getting cryolite either.

I think obtaining anhydrous aluminium(III) chloride from hexahydrate $\ce{AlCl6 · 6 H2O}$ isn't worth it, both economically and time-wise.
Thionyl chloride might work, but, as you said, it's an unpleasant reactant to work with, which is also likely more expensive than dry $\ce{AlCl3}$ (or is on par, depending on the purity of aluminium trichloride).
A common way to prepare water-free $\ce{AlCl3}$ in the lab is to heat pure aluminium metal with the excess of dry $\ce{HCl}$ gas at $\pu{150 °C}.$
The product is then sublimated (above $\pu{180 °C}$) and collected in the condenser (air-cooled).
The quirk here is you have to make sure pure aluminium is used, not an alloy; but this is usually evident from the color of the product, which should be white/colorless.
If you already have $\ce{AlCl3}$ labelled as anhydrous, but you suspect it was slightly moisturized, it can be regenerated via sublimation [1, p. 636]:

Aluminium chloride (anhydrous) […] Sublime it several times in an all-glass system under nitrogen at 30-50 mm pressure.  It has also been sublimed in a stream of dry $\ce{HCl}$ and has been subjected to a preliminary sublimation through a section of granular aluminium metal.

An extensive description is provided in Experimental part in [2, p. 1230]):

Aluminum Chloride.—The aluminum chloride (Baker Chemical Company) was purified by subliming three times down a straight Pyrex tube (with constrictions) which contained a small amount of nitrogen gas (50 mm.). In the first sublimation, the aluminum chloride mas sublimed through a section of granular aluminum metal, which was supported with plugs of glass wool. After the final sublimation and before sealing the tube off, aluminum chloride mas sublimed back down the tube until the remaining material formed a loose plug.
In order to remove the aluminum chloride from the tube, the tube is scratched and then the scratch touched with a hot rod in such a manner that the tube cracks but does not open. The tube may then be weighed, the tube tapped slightly to open. and after removing the aluminum chloride, the empty tube weighed. Using these techniques, it is possible to prepare very pure samples of aluminum chloride and weigh and transfer them without bringing them into contact with air.

Remember to store anhydrous $\ce{AlCl3}$ in an air-tight vessel and perform all operations with it in a fume hood.
Also, keep in mind that you can purchase $\ce{AlCl3}$ dissolved in organic solvents such as nitrobenzene, THF, or even $\ce{AlCl3}$-functionalized silica gel.
References

Armarego, W. L. Purification of Laboratory Chemicals; Elsevier: Boston, MA, 2017. ISBN 978-0-12-805457-4.
Jensen, F. R. Substitution of Polynuclear Aromatic Compounds. I. The Friedel-Crafts Benzoylation of Naphthalene. J. Am. Chem. Soc. 1957, 79 (5), 1226–1231. DOI: 10.1021/ja01562a051.

I recall a paper where heating the hydrate in a stream of $\ce{HCl}$ gas avoided the formation of a basic salt.
This may work for hexahydrate $\ce{AlCl6⋅6H2O}$ also.
References

https://books.google.com/books?id=Pef47TK5NfkC&pg=PA1147&lpg=PA1147&dq=dehydration+with+a+stream+of+HCl&source=bl&ots=W8NJaeEg0E&sig=ACfU3U3dCuWuFJzRiOnxsqGoX7UT28jvxw&hl=en&ppis=_c&sa=X&ved=2ahUKEwj7jeeR0LHmAhUDw1kKHUegAx4Q6AEwC3oECAgQAg#v=onepage&q=dehydration%20with%20a%20stream%20of%20HCl&f=false (ebook)
https://patents.google.com/patent/US3346333 (patent)

This may work in theory (the fact that it could be done for $\ce{MgCl2·6H2O}$ doesn't imply it also works for $\ce{AlCl3·6H2O}$), but it requires a substantial excess of dried HCl, takes a long time and requires a complex apparatus (this is reflected in the patent). It would be much more easier to react this dried HCl directly with aluminium metal instead.


text:
When MgCl₂ · nH₂O is heated to the boiling point of H₂O
When magnesium chloride hydrate is heated, does $\ce{Mg(OH)2}$ form or what? I'm curious, because Wikipedia says that it decomposes to release $\ce{HCl}$ gas.

Wikipedia pretty much has it all already.
There are numerous magnesium chloride hydrates $\ce{MgCl2 · x H2O}$ $(x = 1, 2, 4, 6, 8, 12),$ however in the temperature range from 0 °C to 115 °C the most stable one is hexahydrate $\ce{MgCl2 · 6 H2O}.$
Whether $\ce{MgCl2 · 6 H2O}$ decomposes to hydroxo-salt or only dehydrates depends on the conditions.
When heating takes place in atmosphere of $\ce{HCl},$ anhydrous $\ce{MgCl2}$ is formed:
$$\ce{MgCl2 · 6 H2O (l) ->[>\pu{115 °C}][HCl(g)] MgCl2(s) + H2O(g)}$$
Otherwise, magnesium hydroxochloride is formed [1, p. 522] and $\ce{HCl}$ gas is released:

The product obtained is always the hexahydrate, $\ce{MgCl2 · 6 H2O}.$ It is dehydrated to anhydrous magnesium chloride by spray drying and heating with dry hydrogen chloride gas. In the absence of $\ce{HCl},$ heating hexahydrate yields the basic salt, $\ce{Mg(OH)Cl}:$
$$\ce{MgCl2 · 6 H2O -> Mg(OH)Cl + HCl + 5 H2O}$$

References

Patnaik, P. Handbook of Inorganic Chemicals; McGraw-Hill handbooks; McGraw-Hill: New York, 2003. ISBN 978-0-07-049439-8.

Andeselisk has pointed out that hydrared magnesium chloride decomposes to the basic salt $\ce{Mg(OH)Cl}$ unless the atmosphere is supplemented with hydrogen chloride gas. This basic salt decomposes upon further heating to give the oxide, a reaction that parallels the decomposition of magnesium hydroxide upon heating. From Wikipedia:

The anhydrous forms decompose when heated above 450-500 °C by decomposition of the hydroxide and chloride anions, releasing water and hydrogen chloride and leaving a magnesium oxide residue, by the reactions:[1]
$\ce{2HO^- -> O^{2−} + H2O}$
$\ce{H2O + 2Cl− -> O^{2−} + 2HCl}$

The driving force for the hydrolysis is the stronger bonding of alkaline earth cations with oxide or hydroxide ions than with chloride ions, plus the favorable entropy of evolving hydrogen chloride gas at high temperature. The effect is most notable with magnesium, whose small ion size compared with heavier alkaline earth cations magnifies the difference in electrostatic attraction to the various counterions. However, at higher temperatures calcium chloride may be reacted with water vapor or oxygen to give calcium oxide, too[2].
In the case of calcium chloride, Reference [2] reports that the thermohydrolysis to form the oxide is seen above the melting point, so any basic salt that might form would probably be part of the melt solution rather than a separate solid. Calcium oxide, the observed product which melts at a much higher temperature, is what precipitates out.
Cited References

W. F. Cole and T. Demediuk (1955): "X-Ray, thermal, and Dehydration studies on Magnesium oxychlorides". Australian Journal of Chemistry, 8, 2, pages 234-251. doi:10.1071/CH9550234

G. Fraissler, M. Jöller, T. Brunner, I. Obernberger (2009). "Influence of dry and humid gaseous atmosphere on the thermal decomposition of calcium chloride and its impact on the remove of heavy metals by chlorination", Chemical Engineering and Processing: Process Intensification,
48, 1, Pages 380-388. doi:10.1016/j.cep.2008.05.003


text:
Calculating pH of rain water

The concentration of $\ce{SO2}$ in the atmosphere over a city on a certain day is $\pu{10 ppm}$. Given that solubility of $\ce{SO2}$ in water is $\pu{1.3653 mol L-1}$ and $\mathrm{p}K_\mathrm{a}$ of $\ce{H2SO3}$ is $1.92,$ estimate $\mathrm{pH}$ of rain on that day.

$\pu{10 ppm}$ means there is $\pu{10^-2 g}$ of $\ce{SO2}$ in $\pu{1 L}$ of air. This accounts to $\pu{1.25E-4 mol}$ of $\ce{SO2}$ in $\pu{1 L}$ air, which is less than its solubility, which means that all this amount will be dissolved in rain water to make $\ce{H2SO3}.$
Now, using the expression $\frac{4c^2α^3}{(1-α)}$ for $K_\mathrm{a}$ expression of $\ce{H2SO3},$ and substituting $c$ as $\pu{1.25E-4},$ I get $α$ to be approx. $0.99.$
Then $[\ce{H+}]$ is $2cα,$ which comes out to be $\pu{2.475E-4}.$ So $\mathrm{pH}$ is $3.606.$ But the answer given is $4.865.$ Why I am wrong?

Use Unicode symbol α (works everywhere) or macro \alpha (in math mode, e.g. inside $…$ or $$…$$). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Also, one thing that is not immediately obvious: it's often a good idea also to enclose the following punctuation marks in math mode: this way they won't become orphans if the inline math expression is placed on the end of the line. For example, $\alpha^2.$ is more bullet-proof than  $\alpha^2$..

Note that H2SO3 does not exist even in solution. The equilibrium equation is SO2(aq) + (2) H2O <=> HSO3- + H+ (H3O+)

You are wrong when you say that "10 ppm means there is 0.01 g SO2 in 1 L air". One liter air weighs about 1.3 g. So 0.01 g SO2 would be about 0.8%. 10 ppm means 10 mg in 1 kg air, which is about 0.8 cubic meter.

Maurice, I have learned that ppm of something means 1 mg of the substance in 1L of that something.

@dhanesh Vijay ppm stands for "parts per million,i.e 1 in 1 000 000. Unless further specified, it is ambiguous as it can be mass/mass ( w/w ), volume/volume ( v/v ), Mass/volume ( w/v ), molar amount/molar amount ( n/n) and few less common combination. For this ambiguity, it is no longer recommended to be used.

So finally, what concentration should I take for H2S2O3 in equilibrium equation.

Do not expect there H2S2O3 either, it would produce colloidal sulphur, SO2(aq) and H2O. In air context, V/V ppm (sometimes written as ppmv ) as cm^3/m^3 is usually implied. w/v ppm as mg/L is used in context of water solutions(implying density 1 kg/L), not for gas composition.

Note also a gas water solubility is the water concentration of the dissolved gas for water in equilibrium with gas.  It is proportional to the gas partial pressure.

only slightly related: Is there any reason to fear personal exposure to rain with a pH of 3.1?

I agree with Buck Thorn's explanation on $\ce{SO2}$ concentration in aqueous phase. Thus, when dissolve in water (or when is added to water), the initial reaction of $\ce{SO2}$ with water is shown in the following reaction (Ref.1):
$$\ce{SO2 (g) + H2O (l) -> H2SO3 (aq)}$$
Then, formed $\ce{H2SO3}$ would stabilize following equilibrium:
$$\ce{H2SO3 + H2O <=> H3O+ + HSO3-}$$
Suppose initial concentration of $\ce{H2SO3}$ is $c$ and at equilibrium, $[\ce{H3O+}]$
is $\alpha$. Thus, concentration of $\ce{HSO3-}$ is $\alpha$ as well.
$$\therefore \; K_\mathrm{a1} = \frac{[\ce{H3O+}][\ce{HSO3-}]}{[\ce{H2SO3}]}=\frac{\alpha \cdot \alpha}{c-\alpha}= 10^{-1.92}$$
When simplify this equation, you'd get:
$$10^{1.92}\alpha^2 + \alpha - c = 0 \tag{1}$$
Since $c$ is $1.3653 \times 10^{-5}$ (see Buck Thorn's  answer elsewhere), you can solve $(1)$ for $\alpha$, which is equal to $2.7308 \times 10^{-5}$. Thus $\mathrm{pH}$ of solution is $4.564$.
References:

A. Sathasivan, B.S. Herath, S. T. M. L. D. Seneviratne, G. Kastl, “Chapter 14: Dechlorination in Wastewater Treatment Processes,” In Current Developments in Biotechnology and Bioengineering: Biological Treatment of Industrial Effluents; Duu-Jong Lee, Veeriah Jegatheesan, Huu Hao Ngo, Patrick C. Hallenbeck, Ashok Pandey, Editors; Elesvier: Boston, MA, 2017, pp. 359-380.

Wikipedia "Raman spectra of solutions of sulfur dioxide in water show only signals due to the SO2 molecule and the bisulfite ion."

How did u find c?

I started my answer agreeing with Buck Thorn. So I use his solution for $c$, to avoid work it again.

The value $\pu{1.3653 mol L^{−1} atm^{−1}}$ is the solubility constant (or Henry's law solubility constant), not the solubility. The solubility is defined as the maximum possible concentration (the saturation concentration) of a solute under given solution conditions (e.g. temperature and pressure), whereas the solubility constant $H^{cp}$ defines how solute partitions between the gas (concentration in gas here given as the partial pressure in atmospheres — although this is not stated in the OP it can be inferred from the expected answer) and solution phases (concentration in $\pu{mol/L}):$
$$H^{cp} = \frac{c}{p}= \pu{1.3653 mol L^{−1} atm^{−1}}$$
Assuming atmospheric pressure of 1 atm then $\pu{10 ppm}$ translates into a partial pressure of $\pu{1e-5 atm}$ (if the atmospheric pressure is $p_T$, then the partial pressure of $\ce{SO2}$ is $p = p_T \cdot 10^{-5}$) and a solution concentration $\pu{1.365e-5 M}$ $\;(c=H^{cp} p)$. Proceeding from the definition of $\mathrm{p}K_\mathrm{a}$ the degree of dissociation $(\alpha=0.99886)$ can be computed and from this the $\mathrm{pH} = 4.865,$ which agrees with the expected result.
You should check solubility limits when evaluating this sort of problem, but since the solubility is not given you can ignore this and assume that the limit has not been reached.

Can you elaborate how u got the concentration from ppm.

@dhaneshvijay Ok, done...

Sorry I am not aware of Henry law constant yet. Is there anyother way to get the solution concentration without this concept?

@dhaneshvijay you don't have to call it Henry's law constant, you can call it solubility constant or partition constant if you like. But no, you can't solve the problem without using the concept afaik. Basically it tells you in what concentration ratio the solute (SO2) will segregate between the gas and liquid phases. Without this concept you would not know to what extent the gas transfers into solution.

I agree. But $c$ in the equation represents the concentration of solute in aqueous phase, isn't it?

@MathewMahindaratne yes, see e.g. https://en.wikipedia.org/wiki/Henry%27s_law#Henry's_law_solubility_constants_%7F'%22%60UNIQ--postMath-0000000E-QINU%60%22'%7F


text:
Why lithium looses electron more easily than sodium according to electrode potential values?
As lithium is smaller in size than that of sodium, then sodium should loose electron more easily than lithium

Sodium does lose electron more easily than lithium; check the ionization energies and you'll see just that. Electrode potentials are more complicated.

Of course, the Lithium atom is smaller than Sodium. But the Sodium nucleus attracts its outer electron much stronger than Lithium, at it has 11 protons, and Lithium only 3. So both effects (size and charge) have to be taken into account. Apparently the size effect is not dominant.

Consider also the electron shielding effect. https://en.wikipedia.org/wiki/Shielding_effect?wprov=sfla1   https://en.wikipedia.org/wiki/Slater%27s_rules?wprov=sfla1


text:
Does pure iron not rust?
In my school science textbook, written more than twenty years ago, it said that pure iron does not rust. Accompanying was a photograph of an ancient iron statue situated outside, which had not rusted.
Is this correct?

Related: Is there pure iron in nature? As for the statue, it's by no means made of pure iron and is protected by passivation. Meteorite iron is also not chemically pure and typically contains nickel.

The purity of the iron is less important here than the conditions of the atmosphere and other surroundings.  Rusting of iron involves the formation of hydrated oxides and so requires the presence of moisture, either as water vapor or as liquid.  In a dry environment iron is indeed more resistant to corrosion because it forms anhydrous oxides which, as long as moisture is kept away, do provide some passivation.

So it does rust then?

Again, it depends on keeping moisture away. What we recognize as rust needs moisture, as can be seen by clicking the link.

Pure iron is more corrosion resistant than regular steel. Also there is an alloy of iron named wrought iron which has very low carbon content, and that makes it more resistance to corrosion, but steel has better strength. I think also Eiffel tower was made out of wrought iron.
From the Wikipedia article about wrought iron:

Wrought iron is an iron alloy with a very low carbon content (less than 0.08%) in contrast to that of cast iron (2.1% to 4%). It is a semi-fused mass of iron with fibrous slag inclusions (up to 2% by weight), which gives it a "grain" resembling wood that is visible when it is etched or bent to the point of failure. Wrought iron is tough, malleable, ductile, corrosion resistant, and easily welded.


text:
Is redox potential a valid term?
I mean doesn't every compound has its own different reduction and oxidation potential ? Or its just one potential that can be compared to understand which one will get reduced and which one will get oxidised ?

I mean doesn't every compound has its own different reduction and oxidation potential ? Or its just one potential that can be compared to understand which one will get reduced and which one will get oxidised ?

A compound has neither. You need to specify the half reaction. For example, take an aqueous solution of $\ce{Fe^2+}$ and hydrogen peroxide, among others. What are possible half reactions for $\ce{Fe^2+}$?
$$\ce{Fe^2+ -> Fe^3+ + e-}\tag{1}$$
$$\ce{Fe -> Fe^2+ + 2e-}\tag{2}$$
and maybe some more exotic ones. For hydrogen peroxide, you would have e.g.
$$\ce{H2O2 -> O2 + 2e- + 2H+}\tag{3}$$
$$\ce{2H2O -> H2O2 + 2e- + 2H+}\tag{4}$$
Notice that these are all written in the direction of reduction, and are associated with a reduction potential. To see which reactions are feasible, you would compare reduction potentials and come up with a redox reaction combining a half reaction as written with a half reaction (the oxidation half reaction) in the reverse direction. For example, hydrogen peroxide could undergo disproportionation:
$$\ce{2H2O2 -> O2 + 2H2O}$$
This is a slow reaction, but the enzyme catalase (which happens to have an iron(III) bound to its active site) increases the rate substantially, as you can see from the oxygen evolving when you put some hydrogen peroxide on a freshly cut slice of potato.

Is redox potential a valid term?

Yes, under certain circumstances it makes sense to talk of redox potential as a property of an aqueous solution, similar to discussing its pH. For example, if a biological cell contains glutathione at fairly high concentration in the reduced and oxidized form (abbreviated as GSH and GSSG), this will determine the redox state of other substances in the cell. If a strong oxidation agent is introduced, it will react with GSH. If a strong reduction agent is introduced, it will react with GSSG. Like mainining a pH with a buffer, the redox potential of the cell is maintained through the ratio of GSH and GSSG (and we could call it a redox buffer).
The analogy between pH buffer and redox buffer only goes so far. The biggest difference is in the role of water. While water acts as an acid and base at any pH (i.e. whenever acids and bases are added), it does not undergo reduction and oxidation for many redox reactions in aqueous solution. Any pair of acid and base will react (because water provides the "reservoir" of hydrogen and hydroxide ions). Many pairs of possible oxidizing and reducing agents will not react, even when thermodynamically favored because there is no good mechanism for the electrons to be transferred (a question of kinetics).

There is the table of standard electrode potentials. These are values which is relative to the standard hydrogen electrode. They were measured at certain, constant for all, conditions.
To measure actual potentials one can use Nernst equation
$$E_\mathrm{Red} = E_\mathrm{Red}^\circ + \frac{RT}{nF}\ln\frac{a_\mathrm{Red}}{a_\mathrm{Ox}}.$$
Sometimes you just have to compare standard electrode potentials to understand which one is stronger oxidant or reductant.

Please avoid embedding illustrations from the third party services (use SE uploader instead to prevent link rot), and don't use images for the textual information. There is Markdown and MathJax support to assist you with scientific typesetting, leaving your post searchable and looking uniformly decent across all devices. For more information, visit this page, this page and this one.

Is redox potential a valid term?

Yes it is a valid thermodynamic term. Also oxidation potential is a valid term. These terms differ only by a sign. If a reduction potential of a given half cell is -0.73 V with respect to hydrogen cell. its oxidation potential is +0.73 V. So you have to stick to one term for comparative purposes.
After a century of arguments, electrochemical societies decided to write all the half cells as reduction e.g., $\ce{Fe^{3+} + e- -> Fe^{2+}}$ and call the potential of this reaction as the reduction potential with respect to hydrogen half cell which is arbitrarily taken as zero.

This does not address the question which seems to be shall we say, "reduction (or oxidation) potential" or can we use redox as passpartout? I am oriented to say that is either reduction or oxidation, at least in a pedantic way.

The post title was a decent question. The text of the question does not make any sense. Also, what do you mean by "passpartout"? Only the signs differ between reduction potential and oxidation potential.

Ok but you have treated the term redox as if it is red... I am not saying you wrote something wrong. Just the question wasn't addressed, while it seems so for the answer by @Karsten Theis. Passpartout in the sense that often we indeed say or hear of redox potential instead of a more correct reduction (or oxidation) potential, especially when a value is then given.


text:
Why is latent heat of vaporization not exactly proportional to boiling point?
Latent heat of vaporization and boiling point are both chemical properties related to the bond strength of the molecule. In general, the higher the boiling point is, the higher the heat of vaporization will be. However, there are some molecules that break this trend. For example, ammonia has a higher heat of vaporization than you’d expect based on its boiling point relative to other molecules with similar boiling points. See table here: https://en.m.wikipedia.org/wiki/Enthalpy_of_vaporization. Why does ammonia (and other substances) break this trend?

Im guessing this has something to do with the change in heat capacity between liquid and gas. Different rotational/vibrational degrees of freedom. ?

Hm, the proportionality is not so very good. (data from the wikipedia link above)

The line is a linear regression through all ten data points.
The answer to your question is given on the same wp article. The quotient of enthalpy of vaporisation and temperature is the difference in entropy between the liquid and gaseous phase.
According to Trouton´s rule, it is about 85 to 88 J/molK (ca. $10.5 R$) for most liquids. https://en.wikipedia.org/wiki/Trouton%27s_rule explains the reasons for deviations, like hydrogen bonding, small rotational excitation, dimerisation in the gas phase etc.

In the plot graph above, why does ammonia have a higher latent heat of vaporization than butane but butane has a higher boiling point?

Th logscale plot looks a lot more convincing ... ;)

Some of those values from the wp article seem to be wrong, btw. Definitely butane and propane.

Also isoprop and ethanol are wrong.

Latent heat of vaporization is a physical property of a substance, which is defined as the heat required to change one mole of liquid at its boiling point under standard atmospheric pressure. Hence, it is usually expressed as $\pu{kJ mol-1}$.
The physical properties of a substance (e.g., its boiling point, melting point, latent heat of vaporization, etc.) will be effected by the types of intermolecular forces (e.g., H-bonding, dipole-dipole interactions, van der Waals forces, etc.) that occur in a substance. For instance, substances with weak intermolecular forces will have low melting and boiling points since only less thermal energy is needed to overcome these forces. On the other hand, those with strong intermolecular forces (e.g., water and alcohols)  will have high melting and boiling points as more thermal energy is required to overcome these forces. Similarly, latent heat of vaporization is very much depends on the strength of these intermolecular forces.
For example, let's look at first ten linear alkanes, which have the least intermolecular forces (strong intermolecular forces such as H-bonding or dipole-dipole interactions are absent).
$$
\begin{array}{lrr}
\hline
n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline
\text{Methane} & 112 & 8176\\
\text{Ethane} & 184 & 14640\\
\text{Propane} & 231 & 18832\\
n\text{-Butane} & 273 & 22390\\
n\text{-Pentane} & 309 & 26352\\
n\text{-Hexane} & 342 & 28850\\
n\text{-Heptane} & 372 & 31800\\
n\text{-Octane} & 399 & 33972\\
n\text{-Nonane} & 424 & 36910\\
n\text{-Decane} & 447 & 38750\\
\hline
\end{array}$$
Data sources: NIST Chemistry WebBook; The Engineering Tool Box.
When you plot these values, you'd get nice correlation:

A nice linear fit, but not a proportionality. ;)

@Karl: I'm not physical chemist, so I didn't understand why there is a negative intercept here. I thought to beat van der Waals forces (and/or London forces), we  need positive input. Do you understand why? I'd love to hear your opinion.

If you add a line through zero and decane, you will find it has a slope of 86 or so, like Trouton predicts. The smaller alkanes deviate, because kT at their boiling point is too small to allow (reaching equillibrium between translational and) rotational excitation.

Btw. the values I copied off wikipedia deviate a lot (for butane + propane, while methane is identical) from yours, and I think you got the correct ones. I´ll have to check the rest of my table too ...

When comparing entropy changes, which are dominated by the large value for the gas, account should also be made for the different Tb. This can explain away some of the differences (deviation from Troutons rule), but Karl's explanation is good for the lighter alkanes.

Thank you both for correct explanation in this regard. :-)

@BuckThorn It´s getting late here ... ;) Tb?

@Karl Boiling temperature :-) Not tuberculosis


text:
Which is more precise? 328.4 or 0.0008?
What is more precise? 328.4 or 0.0008?
I want to know if precision is based on the number of significant figures or the number of places past the decimal?

It depends, if you are interested in absolute or relative precision. Also, precision is related to uncertainty of the value, otherwise, it is rather related to  resolution.

Questions about significant figures and dimensional analysis in general are off-topic on Chemistry.SE, unless they are chemistry-related. From Meta post: "Questions about dimensional analysis that do not refer to a specific chemistry question or problem should be closed. They can be migrated to mathematics.se as they allow such kinds of questions in their homework policy."

I'm voting to close this question as off-topic because general questions about numerical analysis are off-topic on Chemistry.SE.

The number 328.4 has 4 significant figures. It is very precise.
The number 0.008 has only one significant figure. It is not precise at all.
In order to define the precision, the number of zeros necessary for representing the first significant figure, does not count. This number depends on the unit. For example, 0.0008 kg is as precise as 0.8 g. Both values have the same level of precision : one significant figure.


text:
What could be causing a small uptick in FCS curve at approximately 10 microseconds
I work as an RA in an academic lab. I've been working with FCS fitting for an experiment, and took FCS curves for fluorescein and calmodulin, displayed with residuals below:
]2
I fitted both to an autocorrelation function
$$G(\tau)=1+\frac{T}{(1-T)} \frac{\exp(-\tau/\tau_{trip})}{V_{eff}(c)(1+\tau/\tau_D)\sqrt{1+p^2(\tau/\tau_D)}} $$
As you can see, the fit isn't too bad for preliminary results, but there's a weird uptick at about 1E-5 seconds that isn't accounted for by the model. That it looks about the same for both molecules suggests to me that it may be a small instrumental artifact. I collected both curves at about 6 uW power, 1 nM sample, and averaged five curves for each sample after subtracting an afterpulsing correction found by taking an FCS curve of nanopure water on a slide. I am using a confocal microscope setup. Does anyone know what could be causing an uptick in that timeframe?

I see undulations in both directions, particularly evident in the residuals. I find it hard to see what you are worried about. Maybe you can draw a circle around the detail that you think stands out. I don't know how the variation in uncertainty in the data points is described in the fits but if the trend of dropping rms noise as you go to longer times is not properly described then the fit routine might try hard to fit the noisy early part (overfitting it) at the expense of the less noisy latter part (which ends up being "misfit").

It probably won't make much of a difference to the answer, but you might want to report what parameters in the equation are fixed and which are fit (floating).


text:
Mechanism of the reaction of cyclohexanol and concentrated nitric acid to form adipic acid
If someone could please give me the hint to how the initial attack is carried out, it'd be a huge help.

From the article by van Asselt and van Krevelen [1]:

Cyclohexanol is oxidized to cyclohexanone.
The cyclohexanone reacts with the nitrous acid to yield isonitrosocyclohexanone producing 2-nitro-2-nitrosocyclohexanone by reaction with nitric acid.
With attendant ring opening and uptake of water the latter compound is transformed into 6-hydroxyimino-6-nitro hexanoic acid.
More than 90% of it is converted to adipic acid.

References

van Asselt, W. J.; van Krevelen, D. W. Preparation of Adipic Acid by Oxidation of Cyclohexanol and Cyclohexanone with Nitric Acid: Part I. Reaction Mechanism. Recl. Trav. Chim. Pays-Bas 1963, 82 (1), 51–67. DOI: 10.1002/recl.19630820107.


text:
Is there any non-toxic and non-solvable-in-air liquid?
I want to heat the air with non-aqueous, non-toxic and non-solvable-in-air liquid (with very low relative pressure in air). Is there any such liquid exists?
It would be great if the boiling point of the liquid would be 110 °C.

Water seems to be fine. What do you mean by non-solube in air?

Why not just get an oil filled heater? An electric coil heats the oil in a sealed metal radiator which then radiates the heat into the room. (The oil can't get hot enough to catch something else on fire....)

As I want to heat the air in direct contact with the fluid, accordingly I concern about the liquid.

@Alchimista As we know the moisture content of air can increase by water evaporation until reaching the saturation level. I want a fluid which air does not absorb it like water vapor.

If the boiling point of the liquid is $\pu{110^\circ C}$, then by definition the liquid’s partial pressure at $\pu{110^\circ C}$ is $\pu{1013mbar}$ which means its partial pressure at lower temperatures will typically be far, far away from not noticable.

Additionally, the dependence of water partial pressure on temperature  is steeper than for most liquids with comparable boiling point, so such a liquid with the b.o. 110 Deg C may be the same or even more volatile as/than water.

I think you are under the impression that vapor pressure is a function of it's "solubility" in the head-space gas. This is not the case. For example, water has the same boiling point under air as it does under helium. Confusion is often caused here by the terminology that at 100% humidity, the air is "saturated" with water or that the air can't "hold" any more water. This makes it sound like air absorbs water vapor, as you say, which it does not.

Thanks all. I don't need a non-aqueous, non-toxic and non-solvable-in-air liquid with exactly 110 °C. I state this temperature because it's the minimum temperature which satisfies my conditions. so the boiling point of the liquid could be 500 °C or above.

I do also concern about the cost of the liquid.

Do you think the liquid desiccants would be a proper choice?

High viscosity of the liquid is a negative point for my work.

See @airhuff comment. Gases are always miscible, there is even not the concept of a solubility limit. There is not absorption of vapours by the overhead atmosphere.

You may look at silicon Si and its derivatives. For example, the ethoxytriethylsilicane, or triethylsilicol ethyl ether is boiling at 153 °C. Or the tetraethylsilicane is also boiling at 153 °C. This last substance is used as a reference in NMR spectroscopy.


text:
I have a solution of citric acid, tartaric acid, water and glucose. How do I find the concentration of the citric acid present in the solution?
If I try to find the concentration of citric acid through back titration or reverse titration, the base I would use, like NaOH, would react with both acids to form sodium citrate and sodium tartrate, which are both white compounds that are colourless in aqueous form. How do I find the concentration of citric acid?

You cannot  do this by ordinary classical titrations until and unless you know a way to separate tartaric acid. You can find total acidity but not individual components. Modern methods rely on chromatography or infrared spectra.

Use 1H NMR spectroscopy and quantify the different components present.

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off.
If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving you an estimate of the concentration. The biggest systematic error is that the apparent $\mathrm{p}K_\mathrm{a}$ is dependent on ionic strength, which you don't know unless you know the concentrations in the first place.

"Adjusting" pH to 6.4 would be difficult. Titrating with NaOH solution and simultaneously measuring pH so as to plot a curve would allow calculation, using 6.4 pH as the reference even if you didn't actually hit 6.4 on the nose.

@JamesGaidis That's a good idea: Just add NaOH continuously and monitor the pH. Then, get the midpoint of the titration empirically, and proceed to estimate the citric acid concentration.

@andselisk Thanks for a year of improving the typography (and spelling) of my posts. You would think I have \mathrm{p}K_\mathrm{a} down by now (or figured out the shortcut button). Anyway, hats off to you!

Buffer capacity is already something that is wonkey to determine even with pure solutions. I doubt this method will give you an accurate enough reading. I would say precipitation and dissolution is a safer analytical choice.

I am afraid the measurement of the pH at half titration does not give a useful information on the pKa.
There has been a detailed series of measurements of the pH at half titration for different concentrations of phosphoric acid. See Cecile Canlet, BUP 938, 1, p. 1129, November 2011. For the first H atom, the theoretical pKa should be 2.12. But the pH measured at half-titration is 1.99 at 0.2 M, 2.06 at 0.1 M, 2.41 at 0.01 M.
For the second H atom, the pKa is 7.21 in the tables. But the measured pH at half titration is is 6.69 at 0.2 M, 6.71 at 0.1 M, and 6.99 at 0.01 M.
So the pH at half-titration may be different from the pKa by a factor as high as 0.4 pH unit. The reason is to be found in the ionic strength.

This is probably a homework-type theoretical problem. In theory there is no difference between theory and practice. But, in practice, there is.

You could try reacting the mixture with excess calcium hydroxide to precipitate insoluble calcium citrate/tartarate.
You could then filter off this mixture to obtain calcium citrate/tartarate.
Reacting these with just enough sulfuric acid till the original pH is reached would give you citric acid/tartaric acid (and calcium sulfate which can be filtered off).
Then dry the solids and weigh them. Redissolve the citric/tartaric acid mixture and titrate it against NaOH.
You could then calculate the amount of citric acid/tartaric acid in the solution.

Would this work or would it be very inefficient?


text:
Does aniline react with FeBr₃?
I have read that aniline does not undergo Friedel–Crafts reaction because it complexes with Lewis acid $\ce{AlCl3}.$ Will it also react with $\ce{FeBr3}$ during bromination (to receive a 3-bromo substituted benzene)?

I was just about to ask a similar question as to why phenol doesn't react with Alcl3 and shows F.C reactions whereas o-cresol reacts with Alcl3(just like aniline) and doesn't show F.C reactions

Have you tried reacting the aniline with acetyl chloride or acetic anhydride first?  Conversion to an acetamide is one way to get avoid acid-base side reactions when doing EAS on an aromatic amine.

Oscar Lanzi - its a theoretical question, i do not do these reactions

related https://chemistry.stackexchange.com/questions/87145/why-doesn-t-aniline-undergo-friedel-crafts-alkylation


text:
What is considered room temperature in Celsius?
Recently, I have seen room temperature listed at 20 °C and 25 °C. Which one is more accurate for use in chemistry problems?

Before the 1970s energy crisis, it was 25, afterwards 20. There is no way of knowing unless it is specified explicitly.

In high school and college, we used standard temperature & pressure, 59F, 29.92 inches of Mercury (sea level) for calculations.

Room temperature in my labs in Germany was somewhere just above $\pu{20^\circ C}$. Room temperature at my previous lab in Japan was $\pu{25^\circ C}$ in summer and $\pu{20 ^\circ C}$ in winter. Room temperature at my current lab in Japan was around $\pu{30^\circ C}$ in summer and is now around $\pu{15 ^\circ C}$. Yes, I died in summer.

Considering you've read Wikipedia article on room temperature (r.t.), you probably already have an idea that there is no uniformly defined value due to the subjective matter the term is defined upon.
There are numerous standards across industries defining temperature ranges, most of which lead to the value for r.t. of $\pu{(20 ± 5) °C}.$
In natural sciences there are several well-defined standard conditions describing the combination of exact values for temperature, pressure and humidity.
Room temperature is not among these, so ideally, an exact value must be specified alongside, if relevant, e.g. "the parameter has been determined at r.t. $(\pu{20 °C})$".
German Wikipedia article presents an observation that in physical chemistry and statistical physics in European region the typical value taken for r.t. is $\pu{20 °C},$ whereas in U.S.A. the majority of textbooks refer to r.t. as to $\pu{25 °C}.$
In nuclear chemistry, thermal neutron energy $E_0 = \pu{25.26 meV}$ is fixed as characteristic energy at r.t., which is a reference point for tabulated parameters such as neutron cross section.
Using $E = k_\mathrm{B}T$ relation, an exact value for r.t. can be determined, which is usually taken as $\pu{20.43 °C}$ (see e.g. [1, p.  11-173] and references therein).
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.

I agree with your argument. But why only USA use the value $\pu{25 ^\circ C}$? For example, we always teach the room temperature as $\pu{298 K}$, which has coveted to $\pu{25 ^\circ C}$.

@MathewMahindaratne I didn't attribute any geographical exclusivity to one value or another, only a note on a trend which I fairly loosely translated from German article. And since your profile says you are from Texas, USA and you teach 25 °C, whereas I'm from Russia (European part, so I think it counts:) ) and I usually would attribute r.t. to 20 °C, this trend seem to be accurate.

I think it should be a fixed value internationally, regardless of what part you are form, for better understanding. Thasnks for the explanation.

The (broadly international) community for thermodynamics of phases and alloys, heavily favors 298K as the reference temperature.


text:
CO and H2 gas converting to methane question
Why is it when you heat and pressurize CO and H2 it converts to methane? Is it because when you heat the molecules the bonds get weaker and the pressure breaks the bonds and allows them to flow and create new molecule?
Thank you

You are correct that there are aspects to consider:

The increase of pressure equates to more particles per unit of volume.  It is like filling a metro with more and more people during rush hour, the propability of  collisions increases.
The increase of temperature equates to a higher kinetic energy of the molecules, so if they hit each other, than they do with more momentum.  Gradually, this is like knocking gently at a door (as synonym of low temperature), to pound against a door (warmer), or even to smash across the door (hot, passing the energy barrier).

For the collision theory, it is the chemical bond which is to break.  The increase of either one of the two parameters increases the chance that bonds break. Their influence is synergistic, too.  Of course, there are different bond strengths, requiring different amounts of impact to break.
Try to answer this question -- first for your own:  And what happenz after the collison, where is the energy of the two molecules now (key word conservation of energy)?  Why the product molecules don't fall apart?

 Of course there are product molecules clashing into each other, regenerating
 the initial molecules (microscopic reversibility of back- and foreward
 reaction).   This is expressed by the equilibrium constant $K$.
 Yet you equally recall that for a given kinetic energy $E_\mathrm{kin}$ a
 heavy molecule moves slowier, than a lighter one, $$E = \frac{1}{2} mv^2$$
 while some of the impact of collision will be dissipated, e.g., by vibration.

Thank you for explaining that really clarified it for me!


text:
Why are only two bond structures considered for ozone and not three?
We were taught in chemistry that:

Formal charges can be determined by drawing the Lewis dot structure, then subtracting the nonbonding electrons and half of the bonding electrons from the number of electrons ordinarily in that atom's valence.
Molecules tend to be in the state with the lowest formal charge on the central atom; in the event that there are multiple such forms, it will take on an in-between state called a resonance hybrid.
Three-dimensionally there's VSEPR theory which determines molecular geometry. We haven't gotten to how to calculate more complex angles just yet.

Consider ozone. There are two resonance structures for it: $\ce{O-O=O}$ and $\ce{O=O-O}$, both of which have equivalent formal charges: $(1)$ on the central atom, and $(-1)$ and $(0)$ on the axial atoms. This makes a resonance hybrid of $\ce{O\bond{~-}O\bond{~-}O}$, with bond order $\frac32$ for each bond.
Why isn't a third structure accounted for? Consider ozone structured like a triangle, with a single bond attaching each oxygen to the other two, and four unbound electrons on each oxygen:

The formal charge for every oxygen would be $(0)$.
The bond order would be $1$ for every bond; if merged as a resonance structure with the above two, it would create a structure with a bond order of $\frac43$ between the top oxygen and each side oxygen, and a bond order of $\frac13$ between the bottom two oxygens.

Yet the empirical evidence supports a bond order of $\frac32$ as a bent molecule, not a bond order of $\frac43$ as a triangular molecule. Why not?

Also, if anyone can help out with mhchem formatting to write out my triangular ozone rather than my comparatively ugly computer sketch, that would be great.

Does Can ozone have a triangular structure? answer your question?

As for the illustration, I'm afraid this isn't a task for mhchem. An external tool is required to draw the structure, which is then added as a raster image.

@andselisk That’s a shame that mhchem doesn’t support that. Anyway, that link does seem to ask the same question, but the chemistry in the answer is a bit above my pay grade. You can close as a duplicate.

Well, you can re-formulate the question focusing on the aspects you don't understand and put a link to the current duplicate.

In the meantime I've redrawn the image with the font size used for the text. If it looks too tiny for you, here is a doubled version.

@andselisk Thanks for the image - where can I find software like that to make such an image myself in the future? For the time being I think it’s probably better to close this as a dupe and once I’ve learned a bit more chemistry I can come back and revisit the topic.

@DonielF No prob. This one is made with chemfig, I also occasionally use MarvinSketch. For a  more complete and less opinionated list check out software. Closing the question as a dupe then.


text:
Is there a rule of thumb to predict when coupled cluster might dip below the variational limit?
Coupled cluster is a non-variational method, meaning that it can give energies that are below the true FCI energy (the "variational limit" for variational methods).
Often coupled cluster still gives energies far above the true FCI energy, and sometimes it gives energies that dip below the variational limit.
Are there some guidelines on when this will happen and when we are safe to assume it will not? I'm aware that coupled cluster generally works best when there is one dominant determinant (single-reference cases), so I am asking about guidelines within the single-references case, when we might expect to get energies that violate the variational principle.

This doesn't exactly address your question, which is a very good one.  But, Sherrill and colleagues wrote a 2005 paper on diatomic carbon as a benchmark to compare the validity of CCSD(T) and MR-CI.  Reviewing the abstract reveals that coupled clusters provided good agreement with MR-CI between 1.0 and 2.0 angstroms:
https://doi.org/10.1063/1.1867379
This older presentation of Meissner has an excellent graph of deviation of CCSD and MR-CI from FCI, which could be considered as an exact solution to the equation:
http://www.academia.edu/download/42658253/Meissner1.pdf

@Bertram, the second link leads (for me) to the message "Oops! It looks like you're in the wrong aisle". As for the Sherrill paper, I really don't see any connection between that paper and my question, at all.

Let us begin with the following principle: the CC ansatz always returns an appropriate electronic wavefunction (meaning a combination of Slater determinants). Still, at least in a non-relativistic approach, the Hamiltonian does commute with the Spin operator
\begin{equation}
S^{2}=S_{+}S_{-}-S_{z}+S_{z}^{2}
\end{equation}
(I'm using atomic units here)
and we want the wavefunction to be an eigenfunction of total spin. The difficulty in doing so is due to the fact that the excitation operator to ensure spin purity is not easy to parametrize unless you are lucky enough to have a $S^{2}=0$ state and your system wavefunction is represented by a $S^{2}=0$.
So, when you're starting with a wavefunction which is not a spin eigenfunction (for example the ground state of the nitrogen atom with the unpaired electrons), CC might not be able to recover the spin purity and, by allowing superposition of functions with different spin, you might get an energy lower than the correct one because you're looking for the minimum in a space bigger than the one you are physically allowed to .
Hope I was clear

Great answer. I'm sorry it took so long to see it. This question was asked way back in 2019. Now there's a better place for coupled cluster questions: https://materials.stackexchange.com


text:
Fitting multiple pKa's
I have three histidine residues that are next to each other in a protein. When I compute the fraction of unprotonated states for each individual histidine as a function of $\mathrm{pH}$ my titration curve has a non-Henderson–Hasselbalch (HH) shape. Hence I can't fit it using the HH equation for 1 site:
$$\frac{1}{1 + 10^{n(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}$$
in order to solve for the Hill coefficient $n$ and $\mathrm{p}K_\mathrm{a}.$
Obviously the three histidine $\mathrm{p}K_\mathrm{a}$'s are dependent on each other since they are very close to one another and are probably sharing the proton. So how do I go about fitting the curves? Should I sum the total unprotonated states in all three histidines and fit the curve with the three site equation
$$\frac{1}{1 + 10^{n_1(\mathrm{p}K_\mathrm{a1} - \mathrm{pH})}} + \frac{1}{1 + 10^{n_2(\mathrm{p}K_\mathrm{a2} - \mathrm{pH})}} + \frac{1}{1 + 10^{n_3(\mathrm{p}K_\mathrm{a3} - \mathrm{pH})}}?$$
When I use this equation I still don't get a good fit and I get unreasonable $\mathrm{p}K_\mathrm{a}$'s $(-1.4, 7.7, -1.3).$ Any suggestions?

As requested below, the matlab code I used for the fit is the following:
ft=fittype('(1/(1+10.^(a1*(p1-x))))+(1/(1+10.^(a2*(p2-x))))+(1/(1+10.^(a3*(p3-x))))', 'independent','x','dependent','y');
opts = fitoptions('Method','NonlinearLeastSquares');
[fitresult, gof] = fit( xData, yData, ft, opts );
h = plot(fitresult,xData,yData)

I posted values for p1, p2, p3 which should report $\mathrm{p}K_\mathrm{a1},$ $\mathrm{p}K_\mathrm{a2}$ and $\mathrm{p}K_\mathrm{a3}.$

I suggest you post the code for the matlab function, at least. The fitting parameters you reportedly obtained do not match the shape of the curve in the figure.

I second the comment above - please provide more detail of what and how you are doing the fitting? Are you fitting to the n values and allowing them to vary freely? It would make more sense to fit directly to the concentrations of 3H, 2H, H and no H with the constraint that they must all sum to 1. I would also worry about the effect of very high and very low pH on your protein. Is it stable across that range? The behavior should be similar to what one gets with a triprotic acid such as citric acid. For example, look at the chemistry libretext entry on "titration of a polyprotic weak acid".

As pointed out in one of the answers, your equation as written should reach a maximum of three at high pH not one. You need to put a correction factor to get from three to a scale of 0 to 1. More generally, fitting to that exponential form is extremely difficult with non-linear regression, even for a single HH equation. Replacing those denominators with the relevant expressions in terms of Ka’s and [H] will make the fit more robust. Even with that, fitting six independent parameters accurately with that small dataset will be hard. I would try setting all Hill coeffs to 1 on first try.

Do you have additional measurements than the 13 blue points represent?  I'm not a biochemist, but to determine out three $pK_a$ (and not only one) with one measurement per unit of pH appears to me as coarse data, especially with the jumps for pH ~5 (in comparison to pH ~4 and ~6) and for pH ~7 in reference to the interpolation in red.  As a suggestion: What about three runs each with measurements in increments of 0.1 units of pH, ploting both a) fraction of unprotonated in function of pH and b) the  first derivative (change of fraction of unprotonated over change of pH value?

Given your hypothesis that the three $\mathrm{p}K_\mathrm{a}$'s are dependent on one another and that experimentally you do not seem to be able to identify which histidines are protonated and which are not at any given time, it would seem that the simplest model to use is that of a triprotic acid such as citric acid. The triple HH equation model you used is only appropriate for three completely independent $\mathrm{p}K_\mathrm{a}$'s (such that the $\mathrm{p}K_\mathrm{a}$ at site 1 is unchanged by the protonation state of site 2).
In the dependent $\mathrm{p}K_\mathrm{a}$ model, we have four species (vs the eight in your model): $\ce{H3P},$ $\ce{H2P},$ $\ce{HP}$ and $\ce{P},$ where "$\ce{P}$" is the protein. There are three "apparent $\mathrm{p}K_\mathrm{a}$'s" none of which represents a true $\mathrm{p}K_\mathrm{a}$ of a specific site. They instead correlate to the combination of possible reactions that give a change in the overall protonation state. For example, the first $\mathrm{p}K_\mathrm{a}$ represents the change in state from $\ce{H3P}$ to $\ce{H2P},$ regardless of which of the three sites loses a proton.
We can then write four equations starting with mass balance:
$$
\begin{align}
[\text{total protein}] &= [\ce{H3P}] + [\ce{H2P}] + [\ce{HP}] + [\ce{P}]\tag{1}\\
K_\mathrm{a1} &= \frac{[\ce{H}][\ce{H2P}]}{[\ce{H3P}]}\tag{2}\\
K_\mathrm{a2} &= \frac{[\ce{H}][\ce{HP}]}{[\ce{H2P}]}\tag{3}\\
K_\mathrm{a3} &= \frac{[\ce{H}][\ce{P}]}{[\ce{HP}]}\tag{4}
\end{align}
$$
We can also define the fraction of sites unprotonated as
$$\phi = \frac{3[\ce{P}]+2[\ce{HP}]+[\ce{H2P}]}{3[\text{total protein}]}\tag{5}$$
It's relatively straightforward to use the four above equations to get expressions for each of the species concentrations in terms of $[\ce{H}]$ and $[\text{total protein}]$ only. Substituting those expressions into the bottom equation will give you an equation to which to fit your data.
If you post your data in tabular form, others can play around with different fitting approaches and give feedback.

I understand that it might be an improvement but I am not sure this is entirely appropriate because the protons are not quite equivalent (the microscopic pKas are not identical at any given pH).

They don’t need to be equivalent. In fact, if they are completely independent and have widely spaced pka’s, the apparent pka’s will converge on the microscopic pka’s

I took a liberty to brush up formatting a little and unify notations such as "[total protein]" and "total protein concentration", HH and H-H etc. for clarity. Also, I changed $[\ce{H3}]$ to $[\ce{H3P}]$ in eqn. (1).

It makes no sense to solve for different pKa´s. They are constants specific to the acid groups of Histidine. Furthermore, the HH-equation describes the fraction of (in your case) dissociated acid in one acid/base equilibria. So if you sum the three steps, your Y-axis for one thing, has to go from 0 to 3. Just off the top of my head, it seems a better solution to multiply the HH-equations: At a given pH, a fraction of His has dissociated the first proton, and out of this fraction another fraction has given of one more proton etc.

The data look as if you have simply done a titration of histidine, with a first pKa at around 6, and another one at around 8.5 - 9. This is exactly what is expected from the tabulated values for histidine : pKa = 6.04 and 9.09. Why are you not satisfied about these measurements ? Your results are in perfect agreement with predictions form the tabulated values.
It is no use trying to fit your points with the strange analytical curve that you propose.

I believe the 9.09 pKa you are referring to is for the amino terminus. Within a polypeptide chain, this proton is not acidic, as the amine is now part of an amide bond.


text:
What is the Lewis structure of the azide ion?
On a chemistry homework, we're asked to draw the Lewis structure of  $\ce{N3-}$.
My answer:

Nitrogen ordinarily has five electrons in its valence; $\ce{N-}$ has six. This makes a total of 16 electrons overall.
If we place the negative charge on the central atom, we get a structure of $\ce{N=N^-=N}$

This has a formal charge of $(-1)$ on each of the side nitrogen atoms and $(2)$ on the central nitrogen atom once you've accounted for the extra electron, so the formal charges cancel out only when you account for the added electron.

This answer is supported by a bunch of online sources (1, 2).

Homework software response:

The correct answer is $\ce{N#N+-N^2-}$

That's all it says. My understanding of why this is valid is that this gives a formal charge of $(0)$ for every atom, as opposed to averaging out to that.
I suppose the second form makes sense, as it's a lower formal charge on the central atom. If so, why do online resources prefer the first? Is there some resonance going on here, and the actual structure has bond orders of $\frac52$ and $\frac32$?

If you place the negative charge on the central atom of your structure, you'll exceed the octet. Instead, you might want to place a positive charge there.

@IvanNeretin Why will I exceed an octet? Nitrogen normally has five electrons in its valence; adding an electron gives it six. It makes a double bond with each side nitrogen and donates electrons to fill their valences in that version.

https://chemistry.stackexchange.com/questions/19904/what-is-the-hybridization-of-the-nitrogen-in-the-azide-ion https://chemistry.stackexchange.com/questions/33101/can-the-structure-of-an-azide-ion-contain-a-coordinate-bond

Your proposed structure is wrong. Nitrogen does not exceed the octet in any of its known compounds (and even if $\ce{NF5}$ will be discovered it will not exceed the octet according to everything we know now). However, if you have a formal negative charge that means an additional electron added to the 5 nitrogen usually has; if four of those six electrons are used to build the double bonds, there is still a lone pair on nitrogen for a total of 10 electrons.
If you are having trouble determining Lewis structures, there are four quick calculations that you can perform to help you:

Add up all valence electrons the atoms are bringing into the compound.
Each nitrogen has five electrons plus there is one negative charge (additional electron) so:
$$3\times5+1=16\tag{1}$$
Add up how many valence electrons would be needed so that each atom has an octet (for hydrogen: dublet) of its own.
Each nitrogen would want eight electrons so:
$$3\times8=24\tag{2}$$
Take $(2)-(1)$. This represents the number of electrons the atoms must share, i.e. the number of bonds.
$$24-16=8\tag{3}$$
Take $(1)-(3)$. This represents the number of electrons that do not have to take part in bonds; these must then be distributed as lone pairs.
$$16-8=8\tag{4}$$

Then, start drawing but make sure you have as many lone pairs and bonding electrons as the equations state. Ignoring the lone pairs, we can get the following possible structures for $\ce{N3-}$:
$$\ce{N#N-N}\qquad\qquad\ce{N=N=N}\qquad\qquad\ce{N-N#N}$$
(The exercise of distributing four lone pairs across the three nitrogens so that each ultimately has eight valence electrons is left to the reader because I am too lazy to open ChemDraw to draw the structures.)
After you have done that, you need to take a look at potential formal charges. For that, split each bond homogenously (i.e. give each atom one of the bonding electrons) and count. Compare that count to what an atom should have; the difference corresponds to the atom’s formal charge. (Because electrons are negative, an additional electron corresponds to a charge of $-1$.) When done for those three structures, we arrive at:
$$\ce{N#\overset{+}{N}-\overset{2-}{N}}\qquad\qquad\ce{\overset{-}{N}=\overset{+}{N}=\overset{-}{N}}\qquad\qquad\ce{\overset{2-}{N}-\overset{+}{N}#N}$$
In each of those cases, the formal charges sum up to the overall charge of the molecular ion ($-1$) which is an indication that we have done it correctly. (Again, I have sheepishly left out the lone pairs; you can use my formal charges to determine where they should have been and how many.)
There is no principle of zero formal charges. However, when debating between different structures, a structure with less formal charges is often (not always!) more ‘favourable’. (The actual term should be ‘contributes more to the overall picture’ but that may confuse too much at this stage.)
But which of the three is correct? They all are! In fact, this is what is known as mesomery: we have a number of (resonance) structures that all explain the actual compound a little bit but neither holds the absolute truth. To show this, resonance arrows are usually drawn between the depictions:
$$\ce{N#\overset{+}{N}-\overset{2-}{N} <-> \overset{-}{N}=\overset{+}{N}=\overset{-}{N} <-> \overset{2-}{N}-\overset{+}{N}#N}$$
The key difference between correct structures and your proposition is that the central nitrogen atom can never carry a negative formal charge as it needs to accomodate four bonds to its neighbours which is only possible for $\ce{N+}$.
As for the answer given in the homework response: It is not strictly correct because it is incomplete. All three structures should be marked as correct – until the concept of resonance has been formally introduced at which point only a combination of the three should be.

But… but… you don't need ChemDraw when it can be done with el3g4nt MathJax syntax: $$\ce{:!!N#\overset{+}{{N}}-\overset{2-}{\overset{\Large.!!.}{\underset{\Large.!!.}{N}}}!!:}$$ *Overconfident alcoholic*

@andselisk I was, in fact, contemplating drawing them with MathJax but then decided today was not the day for that rabbit hole. But thanks for a hearty laugh! =D

Okay, I see where I went wrong at least. Resonance has been introduced in class, making a number of these homework problems even more head-scratching because there’s more than one equally correct answer.