text:
Why Is Hydrogen in HSO3- Connected to Oxygen Instead of Sulfur?
One of the questions in my homework was to draw $\ce{HSO3-}$, which I thought to be pretty easy. Sulfur is the least electronegative (except hydrogen but it can't be the central atom), hence it is the central atom. Then I drew out the structure with the hydrogen single bonded, 2 oxygen double bonded and a single bonded oxygen.
When I went to look it up to see if it matched my structure, I saw all if the results with oxygen binding to hydrogen. Is there a reason why hydrogen cannot bind to sulfur?
All the search results showed me why hydrogen is not considered the central atom, which I already knew.

It's actually both, at least in solution. Bisulfite ion in solution tautomerizes, existing as an equilibrium of the isomer with hydrogen bonded to sulfur and the isomer with hydrogen bonded to oxygen.

Solid bisulfite salts are surprisingly hard to achieve, because the bisulfite ion tends to dimerize producing the metabisulfite ion, $\ce{S2O5^{2-}}$. Caesium bisulfite, $\ce{CsHSO3}$, has been obtained as a solid and the structure of this solid characterized; in this salt the hydrogen is bonded exclusively to sulfur [1].
Reference

Johansson, Lars-Gunnar & Lindqvist, O. & Vannerberg, N.. (1980). The structure of cesium hydrogensulfite. Acta Crystallographica Section B Structural Crystallography and Crystal Chemistry. 36. 2523-2526. 10.1107/S0567740880009351.

While in the specific case of bisulfite you were technically correct (at least some of the time, as Oscar Lanzi explains), the reason most sources show the hydrogen only bonded with the oxygen is because that's the way most oxy-acids work.
Because the oxygen nuclei are identical (up to isotopes), it doesn't make sense to localize the extra bonds, so a resonance structure arises, analogous to a benzene ring. This wreaks havoc with the "least electronegative besides hydrogen is central" principle, and the hydrogen is more attracted to the resonant structure of the oxygens than the sulfur. It does not, however, choose a particular oxygen to bond to, anymore than the electrons did, causing the hydrogen to move around the molecule in a similarly resonant way. It's this looseness that makes the oxy-acid a proton donator.
In the specific case of bisulfite, I think the only reason a hydrogen-sulfur bond is possible is because oxygen and sulfur are both chalcogens.

 As per recent studies, there is very little experimental evidence for low concentrations of $\ce{HSO3−}$ in aqueous medium, where the $\ce{S\bond{-}H}$ bond has been considered more favorable. In gas phase, however, formation of $\ce{S\bond{-}H}$ bond is considered energetically favorable, and is agreed upon by several researchers.

To add to the excellent answer by Oscar Lanzi, according to Golding R. M. (1960),$^\text{1}$ based on UV spectroscopy results in aqueous solutions, the equilibrium is dependent on concentration, and dimerization was also observed in aqueous solutions:

Controversy Surrounding Hydrogensulfite Structure
However, tautomerization remained, and remains, a topic of debate. For example, Baly E. C. C. and Bailey R. A. denied (1922) any isomerization of bisulfite.

No isomerism of sulphite molecules has been detected. The adsorption bands of sulphurous acid, bisulphite, and sulphite molecules lie in the extreme ultra-violet region.

Since, for long, tautomerization of bisulfite has been debated, I sought out for more recent works. The following are excerpts by Townsend, T. M., Allanic, A., Noonan, C., and Sodeau, J. R. (2012).$^\text{3}$

Initial calculations indicated that the hydrogen sulfonate form, $\ce{HSO3−}$, is the most stable form of the tautomers. However, Hoffmann has argued that the chemical reactivity of bisulfite ions in aqueous solutions is consistent with $\ce{HO\bond{−}SO2−}$ being the major reactive tautomer, which is in rapid equilibrium with $\ce{HSO3−}$. Baird and Taylor suggested that the $\ce{H\bond{−}S}$ bonded tautomer is the most energetically favorable. In contrast, ab-initio calculations by Stromberg indicated that the tautomeric form, bisulfite, is lower in energy. In a later study, Connick et al. proved the existence of the two tautomers in solution by examining the $\ce{S\bond{−}O}$ stretching region of Raman spectra of aqueous hydrogensulfite solutions.

Further confusion has arisen regarding the assignment of bands in the vibrational spectra of hydrogenosulfite solutions
containing both tautomers. There is a common misconception in the literature that, in the $\ce{S\bond{−}O}$ stretching region, the $\pu{1052 cm^{−1}}$ band is assignable to $\ce{S2O5^{2−}}$ and the $\pu{1021 cm^{−1}}$ band to $\ce{HOSO2−}$. However, utilizing both UV/vis and Raman spectroscopies, Connick et al. proved that the $\pu{1052 cm^{−1}}$ band is a composite of both $\ce{S2O5^{2−}}$ and one of the hydrogensulfite tautomers; the $\pu{1021 cm^{−1}}$ band is attributable to just one of the tautomers. However, the experiments were unable to distinguish between each tautomer. Risberg et al. concluded from an X-ray absorption and vibrational spectroscopic study that the $\pu{1052 cm^{−1}}$ peak was attributable to the $\ce{SO2}$ symmetric stretching mode of $\ce{HOSO2−}$, while the $\pu{1021 cm^{−1}}$ band was assigned to the $\ce{SO3}$ symmetric stretch of $\ce{HSO3−}$ only.

Zhang and Ewing have made a very tentative proposition for the existence of another dimer in solution with an $\ce{S\bond{−}S}$ bond, whose chemical formula is the same as Golding’s dimer ($\ce{H2S2O6^{2−}}$) but for which the structure is different. Like Golding’s dimer, the newly proposed complex is said to undergo condensation to form $\ce{S2O5^{2−}}$, whereas in contrast to Golding’s dimer, it is formed from two $\ce{HOSO2−}$ ions. It was stated that this structure is more plausible than the structure of Golding’s dimer because it can eliminate water from its structure more easily. However, this proposition is based on the assumption that the $\ce{HSO3−}$ concentration is very low, a statement for which there is little experimental evidence.

From this discussion, I infer that the major equilibrium existing in aqueous medium is the dimerization equilibrium, whatever the structure of dimer maybe:
$$
\ce{HO\bond{-}SO2- <=>[H2O] H2SO6^{2-}}
$$

And that in gas phase, the most stable isomer is $\ce{H\bond{-}SO3-}$.
References

Golding, R.M., (1960). Ultraviolet Absorption Studies of the Bisulphite-Pyrosulphite Equilibrium. J. Chem. Soc., 3711-3716. 10.1039/JR9600003711
Baly E. C. C., Bailey R. A. (1922). The equilibria in aqueous solutions of the alkali metal bisulphites. J. Chem. Soc., Trans., 121, 1813-1821. 10.1039/CT9222101813
Townsend, T. M., Allanic, A., Noonan, C., and Sodeau, J. R. (2012). Characterization of Sulfurous Acid, Sulfite, and Bisulfite Aerosol Systems. J. Phys. Chem. A, 116, 4035-4046. 10.1021/jp212120h.
Doctoral thesis by Mubarak Elsayed Osman at the University of London. Spectroscopic and thermodynamic properties of some oxy-sulphur species. Available as PDF.


text:
At what temperature is bismuth the densest?
Both water and bismuth have denser liquids compared to solid forms, and we know water is the densest at the temperature of about 4°C.
Does the same thing apply to bismuth? Of course, I know the melting point of bismuth is about 271° C. So the temperature of interest should be around it I think.

The reasons that water has that density anomaly do not apply to bismuth. NIST has tabulated density values to peruse at your leisure.

So what's the reason for bismuth?

I guess I'm a bit confused - do you mean the liquid being denser than the solid, or that water is densest at about 4C?

Both. And in the case of bismuth it's densest at about 310C based on the paper in Nilay Ghosh's answer. While in the case of antimony there's no such an anomaly but its liquid form is also denser.

For water at 4C, see https://chemistry.stackexchange.com/questions/81756/why-is-density-of-water-maximum-at-4-degree-celsius/105390#105390. For the rest, you need to understand that thermal expansion coefficients of many (well, all really) materials are not a constant and can even change sign. Why? the shape of the potential well the atoms find themselves in.

If all materials have a non-constant thermal expansion coefficients then why do most materials have a denser solid?

Density change on a phase change does not correlate with thermal expansion coefficients of a specific phase. But, at a high level, atoms in liquids want to move around so need more elbow room. Note that Bi (and Si, Ge, Sb, ...) are not close packed crystals. Si goes from a 4-fold coordinated semiconducting diamond cubic solid to an 8-fold coordinated metallic liquid. That one is easy to see how it could happen.

The densities of Sb and Bi melts were investigated by an improved Archimedean method. The results show that the density of the Sb melt decreases linearly with increasing temperature, but the density of the Bi melt firstly increases and then decreases as the temperature increases. There is a maximum density value of $\pu{10.002 g/cm^3}$ at 310 ℃ ,about 39 ℃ above the melting point. The temperature dependence of the Sb melt is well fitted with the expression $\pu{ρ = 6.8590 - 5.8105 × 10^{-4} T}$ , and that of the Bi melt is fitted with $\ce{ρ
= 10.3312 - 1.18 ×10^{-3} T}$.

Source: Geng, Haoran & Sun, Chunjing & Wang, Rui & Qi, Xiaogang & Zhang, Ning. (2007). Temperature dependence of densities of Sb and Bi melts. Chinese Science Bulletin. 52. 2031-2034. DOI: 10.1007/s11434-007-0309-7
Also see:

Greenberg, Y. & Yahel, Eyal & Caspi, E. & Benmore, Chris & Beuneu, B. & Dariel, M.P. & Makov, Guy. (2009). Evidence for a temperature-driven structural transformation in liquid bismuth. EPL (Europhysics Letters). 86. 36004. DOI: 10.1209/0295-5075/86/36004.
Tournier, R.F. Multiple Glass Transitions in Bismuth and Tin beyond Melting Temperatures. Metals 2022, 12, 2085. DOI: 10.3390/met12122085

Why does this say the density of liquid bismuth is 10.067 g/cm$^3$ which is bigger than 10.002? Also, there is another article saying the number as 10.022.

@SnackExchange You should check latest papers which has improved measurement techniques. Density of bismuth around melting point is just above 9.9 g/cm3 which goes to maximum value of 10.002 g/cm3 at around 310 C (the ref.2 has a better graph). The Wikipedia article needs to have the values corrected.

When I plugged temperatures in the density formula of bismuth, I found the temperature should be about 280 K in order to have a density of about 10 g/cm$^3$. But when we put 583 K (310 C) the answer is about 9.6! How can it be explained?

@SnackExchange If you follow the paper, it says that in the range of m.p. and 310 C, the equation is $\mathrm{ρ = 9.44903+1.73×10^{−3} T}$ giving density value within 1.3% error. Above 310 C, the equation is $\mathrm{ρ =10.3312−1.18×10^{−3} T}$

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text:
Possibility of carbocation rearrangements in SNi reactions
According to several sources, the $S_{N}i$ reactions follow a mechanism allowing carbocation formation while others disagree. Precisely, for example, according to A Guidebook to Mechanism in Organic Chemistry - Peter Sykes, it states that

Whether the breaking of the C-O and the S-Cl bond occurs simultaneously, or whether the former occurs first, is still a matter of debate.

Thus, it's unknown if a carbocation formation occurs or not. Now my question is if carbocation formation occurs, wouldn't that imply hydride and methyl shifts are possible within the cation? However, in the case when the reaction is concerted, such rearrangements are not possible. Take the case of the following reaction for example,
$$\ce{3-Methylbutan-2-ol + SOCl2 \to 2-Chloro-3-methylbutane (Major)}$$
Is the above more likely or the following after carbocation rearrangements:
$$\ce{3-Methylbutan-2-ol + SOCl2 \to 2-Chloro-2-methylbutane (Major)}$$
Now shouldn't the latter imply the carbocation formation and the former reject it? Which reaction occurs in reality and why is it the reaction mechanism undetermined exactly?

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text:
Why does sodium hydrogen carbonate decompose into sodium carbonate, water vapour, and carbon dioxide?
Why is the reaction for sodium hydrogen carbonate:
$$\ce{2 NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$$
but not:
$$\ce{4 NaHCO3(s) -> 4 Na + 2 H2O(g) + 4 CO(g) + 3 O2(g)}$$
or:
$$\ce{2 NaHCO3(s) -> 2 NaOH(s) + 2 CO2(g)}$$
And how could you determine this without actually performing the experiment? I was thinking it may have to do with the fact that carbonate is a polyatomic ion and therefore stays that way, but I'm not sure.

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

Could you elaborate your reasoning?

Related:  https://chemistry.stackexchange.com/questions/141689/why-is-chemistry-unpredictable

This practical discusses exactly this with an additional reaction $\ce{2NaHCO3 (s) -> Na2O (s) + 2CO2 (g) + H2O (g)}$

The stability of carbonate anion (it being a polyatomic anion) is not the factor at play here, since, if I may be a little vague, in the thermal decomposition of sodium bicarbonate, carbonate anion does change to carbon dioxide moiety.

Releasing $\ce{CO2}$ from $\ce{Na2CO3}$ and especially $\ce{O2}$ from $\ce{Na2O}$ or $\ce{CO2}$ requires much higher temperature than temperature of $\ce{NaHCO3}$ decomposition.
There are three major ways to know that:

You learn the chemistry and behaviour patterns of elements and substances, so you know it without the need to memorize it.
You calculate the energetic outcome (reaction enthalpy) and thermodynamic preference (reaction Gibbs energy) to predict if such reaction can be expected.
With no such knowledge,  you have to memorize it or search for it.

Generally, there is no shortcut to learn chemistry.

Summary: all these reactions are possible under the right conditions (for example, temperature, pressure, phase, and catalyst); however, for thermal decomposition at atmospheric pressure, the first reaction to become spontaneous (at about $\pu {400 k}$, called the thermal decomposition temperature of $\ce{NaHCO3(s)}$), is the decomposition of $\ce{NaHCO3(s)}$ into $\ce{Na2CO3(s)}$, $\ce{H2O(g)}$, and $\ce{CO2(g)}$.

There is no (accurate and entirely) theoretical way to predict a reaction (as of now). However, we can check if a reaction is possible based on thermodynamic-, to what extent it occurs based on equilibirum-, and weather it happens in real-time based on kinetic-analysis.
Note: This are all semi-empirical analysis methods since the values of thermodynamic potentials, equilibrium constants and kinetic rate coefficients are often obtained experimentally.
I will do the thermodynamic analysis here.

Thermodynamic Analysis
First, let us convert all your reaction-components into standard states in the simplest stoichiometries.
Reactions
At $\pu{298.15 k}$, the reactions are.
$$
\ce{2 NaHCO3(s) -> Na2CO3(s) + H2O(l) + CO2(g)} \tag{1}
$$
$$
\ce{4 NaHCO3(s) -> 4 Na + 2 H2O(l) + 4 CO(g) + 3 O2(g)} \tag{2}
$$
$$
\ce{NaHCO3(s) -> NaOH(s) + CO2(g)} \tag{3}
$$
At $\pu{400 k}$, the reactions are:
$$
\ce{2 NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)} \tag{1}
$$
$$
\ce{4 NaHCO3(s) -> 4 Na + 2 H2O(g) + 4 CO(g) + 3 O2(g)} \tag{2}
$$
$$
\ce{NaHCO3(s) -> NaOH(s) + CO2(g)} \tag{3}$$
This is because of the phase transition $\ce{H2O(l) \xrightarrow{T>\pu{373.15 K}} H2O(g)}$
Data
The following table shows the standard enthalpies and free energies of formation for each component at $\pu {298.15 K}$ and $\pu {400 K}$.$^1$ All values are in $\pu{kJ mol^{-1}}$

Chemical
$\Delta_\text{f} H^\circ$($\pu {298.15 K}$)
$\Delta_\text{f} H^\circ$($\pu {400 K}$)
$\Delta_\text{f} G^\circ$($\pu {298.15 K}$)
$\Delta_\text{f} G^\circ$ ($\pu {400 K}$)

$\ce{O2(g)}$
$\pu{0}$
$\pu{0}$
$\pu{0}$
$\pu{0}$

$\ce{H2O(g)}$
—
$\pu{-242.847}$
—
$\pu{-223.951}$

$\ce{H2O(l)}$
$\pu{-285.830}$
—
$\pu{-237.141}$
—

$\ce{Na(s)}$
$\pu{0}$
$\pu{0}$
$\pu{0}$
$\pu{0}$

$\ce{NaHCO3(s)}$
$\pu{-950.810}$
$\pu{-953.885}$
$\pu{-852.851}$
$\pu{-819.095}$

$\ce{NaOH(s)}$
$\pu{-425.931}$
$\pu{-428.299}$
$\pu{-379.737}$
$-363.788$

$\ce{Na2CO3(s)}$
$\pu{-1030.768}$
$\pu{-1135.717}$
$\pu{-1048.005}$
$\pu{-1019.363}$

$\ce{CO(g)}$
$\pu{-110.541}$
$\pu{-110.129}$
$\pu{-137.180}$
$\pu{-146.354}$

$\ce{CO2(g)}$
$\pu{-393.505}$
$\pu{-393.580}$
$\pu{-394.364}$
$\pu{-394.646}$

Calculations
Now, let us calculate heats and free energies of reactions (1), (2), and (3) using $\Delta_r H^\circ = \sum_\text{products} \Delta_f H^\circ - \sum_\text{reactants} \Delta_f H^\circ$ and $\Delta_r G^\circ = \sum_\text{products} \Delta_f G^\circ - \sum_\text{reactants} \Delta_f G^\circ$ (considering the stoichiometry). All values are in $\pu{kJ mol^{-1}}$

Reaction
$\Delta_r H^\circ$ ($\pu{298.15 k}$)
$\Delta_r H^\circ$ ($\pu{400 k}$)
$\Delta_r G^\circ$ ($\pu{298.15 k}$)
$\Delta_r G^\circ$ ($\pu{400 k}$)

1
$\pu{119.517}$
$\pu{135.626}$
$\pu{26.192}$
$\approx \pu{0}$

2
$\pu{1657.56}$
$\pu{1755.526}$
$\pu{1359.666}$
$\pu{1249.894}$

3
$\pu{131.374}$
$\pu{132.006}$
$\pu{78.750}$
$\pu{60.661}$

Results and Discussion
This tells us that all the reactions are endothermic and non-spontaneous except for reaction (1) at $\pu{400K}$ for which $\Delta_r G = 0.230 \approx 0$. This is (nearly) called the thermal decomposition temperature of $\ce{NaHCO3}$ to convert to $\ce{Na2CO3}$.
This also tells us is that the reactions are endothermic and non-spontaneous at ambient temperature. This why $\ce{NaHCO3}$ is required to be heated for the decomposition. According to reference$^2$ the thermal decomposition temperature of $\ce{NaHCO3}$ is $\pu{373.70 k}$, but our value is closer to $\pu{400 k}$. You can obtain even better (negative values of $\Delta_r G$ for reaction (1) at higher temperatures.
References

Thermodynamic Data of Pure Substances. Ihsan Barin, Gregor Platzki. Third edition. VCH Verlagsgesellschaft mbH, Weinheim (Federal Republic of Germany), VCH Publishers, Inc., New York, NY (USA)
Thermal Decomposition of Sodium Hydrogen Carbonate and Textural Features of Its Calcines. Miloslav Hartman, Karel Svoboda, Michael Pohořelý, and Michal Šyc. 10.1021/ie400896c

P.S. Could someone check the data and calculations once more (because the decomposition temperature didn't match with that of reference [2]). I just want to be sure. What is causing the $\Delta_r G$ to be slightly positive even above the thermal decomposition temperature$^2$ of $\ce{NaHCO3}$?


text:
Is aqua-meth-myoglobin (Fe3+) paramagnetic?
Is aqua-meth-myoglobin (Fe3+) paramagnetic?  In this case, the iron has all 6 coordination positions occupied by ligands: 4 nitrogen atoms from the pyrrolic rings that compose the porphyrin, 1 nitrogen atom in the axial position from the imidazolic system of the distal histidine and in the other axial position is water (or OH  - in the case of hyroxy-meth-myoglobin).  Being different ligands, I don't know whether to consider that the ligands are strong or weak field and implicitly I don't know how the configuration of the iron will be after coordination.  My assumption is that it remains with 5 unpaired electrons, if the ligands are weak field.


text:
Doubt in chirality and pseudo chirality
So a carbon atom is chiral if it has 4 different group attached to it, and it is pseudo chiral if it has 3 different groups of atoms attached to it, where the 2 similar groups have different configuration (such as R/S or cis/trans). For the pseudo chiral carbon atom, it must have 2 and only 2 similar groups on it. (according to the IUPAC Goldbook)
But what if the carbon atom has only 2 different group, but both with different configuration (like 2 similar groups of cis/trans and 2 similar groups of R/S) like this one :

This carbon atom can't be pseudo chiral as it does not have "only 2" similar groups on it. But it is not chiral because it does not have 4 different substituents.
This atom does not have a plane of symmetry or centre of symmetry or alternate axis of symmetry (correct me if I am wrong). So this molecule is chiral.
But is the carbon atom chiral or pseudo chiral or some other terminology is used in this case(the middle one, with 4 different groups).

The question, I think, is pretty well presented, but the title would benefit from an improvement, e.g. see I have a doubt about "doubt" in doubt titles.

I don't actually have anything to put in the title, so if you could suggest something, it would be helpful

I have simplified your structure to the dienes 1 and 2. The "pseudo-asymmetric" carbon in diene 1 is stereogenic and achirotopic and labeled with a lower case descriptor. The CIP rules dictate that Z>E and R/S and that Z/E>R/S. The priorities in 1 are Z>E>R>S. The stereochemistry at the "pseudo-asymmetric" carbon is "s". There is a plane of  symmetry passing through Z-s-E and R mirrors S. The compound is optically inactive.
The central carbon in diene 2 is non-stereogenic (no descriptor) and chirotopic. The S,S-enantiomer is optically active as is its enantiomer.
For related posts on this topic on ChemSE go here and here.

So if I understand correctly, you are saying, the carbon labelled s in 1 can be optically active when the S and R enantiomers become S and S (it will be non stereo genic then). So what would the terminology of that carbon be? Like it can't be "pseuodochiral" because it does not have "only two" same groups with different configuration (and not chiral because it does not have 4 different groups), right?

Atoms are not chiral, their environments are. Hypothetically, if all of the bonds were removed and the atoms were held in place, the carbon atom (a sphere?)  in question in structure  2 would find itself in a chiral environment. I avoid using the term "pseudo-whatever" and prefer the nomenclature of Mislow and Siegel. If you switch the attachment of the E and Z groups to the central carbon in 2 and rotate this structure by 180 degrees about a vertical axis, you will still have the original structure. continued

...There is no stereochemical change but there is a chiral environment owing to the atoms  with the  S-configuration. If you do the same group switch in structure 1, you obtain a diastereomer that is also achiral. Read this.

Thank you for the explanation and the reference, I understood what you tried to say, but I still actually have a doubt pending, according to iupac, what would the central carbon atom be called? I understand that you avoid the iupac nomenclature here, but my original doubt remains, so sorry if I can't verify your answer


text:
Why is there a limit to the amount of water vapor in the air even though the atmosphere is so vast?
I have read that temperature and pressure dictate how much water vapor can exist in the air.
Air is a mixture, so shouldn’t the addition of water vapor just add to the mixture regardless the temperature and pressure? And since mixture don’t need a specific amount of each of it’s components, why is there a limit even though the atmosphere is so vast?
I have also read that when air is ‘’saturated’’ it means that it reaches the maximum amount of water vapor it can hold at that temperature and pressure
How can water dissolve in air if air is a neutral mixture and water is polar, considering that polar only dissolves polar?

Rather, temperature dictates how much of water vapor can exist in the space. This amount in g/m3 decreases with temperature roughly exponentially, typically to half by T decrease by 10 Deg C. Vapor need not air at all. Air is just a bystander, contributing to the pressure. By other words, wet air is not like a solution of salt in water. Vapor need not to be dissolved. Their coexistence in gaseous phase is merely like ignoring each other. Air is a gas, vapor is a gas too. Vapor need not other gas to be a gas. All what vapor needs is space without condensed phases.

As stated in the first comment water vapour does not dissolve in air. Water molecules will evaporate from the surface of liquid water and so form the vapour (whether or not air is present), however, some of these molecules will re-enter the liquid water by collisions with its surface. This evaporation/condensation always occurs even when equilibrium is reached. The equilibrium between liquid water and its vapour is determined by the temperature among other properties of water.

Air consists mostly of gases that can never be liquids at room temperature or pressure. Gases like that are completely miscible. A vapour sometimes behaves like a gas but is in equilibrium with its corresponding liquid. So it will condense out of air if too much is present (and the point is does that is the saturation point). Vapours can't be thought of as equivalent to gases.

[OP] I have read that temperature and pressure dictate how much water vapor can exist in the air.

Considering water+air as a system of two components in vapor-liquid equilibrium, the number of intensive variables to specify its thermodynamic state is given by the Gibbs phase-rule for non reacting systems
\begin{align}
  F &= 2 + \text{[Num. of phases]} - \text{[Num. of components]} \\
  &= 2 + 2 - 2 = 2 \rightarrow F=2 \tag{1}
\end{align}
So, you are correct, and we need two variables (which you may choose $p$ and $T$, but we can choose other ones). Note that this derivation is "lumping" air in one component, but this is not true! Air is made of many components, but since this approximation yields good results, we get away with this (tiny) lie.

[OP] Air is a mixture, so shouldn’t the addition of water vapor just add to the mixture regardless the temperature and pressure? And since mixture don’t need a specific amount of each of it’s components

You can add all the water you like to the mixture, but only a part of it can exist in the gas phase. We will determine rapidly this amount accepting two hypotheses:

The liquid and the gas phase behave as an ideal solution and ideal gas, respectively.
The air is almost insoluble in the liquid phase, so that the molar fraction of water in the liquid phase is essentially $1$. You could relaxate this hypothesis and model air according to Henry's law, but is not important right now.

Denoting $1$ as water and $2$ as air, VLE equilibrium for 1 resumes to Raoult's law
\begin{align}
  y_1 p &= x_1 P_1^\pu{sat}(T) \approx P_1^\pu{sat}(T) \\
  y_1   &= \frac{P_1^\pu{sat}(T)}{p} \tag{2} = \text{function}(p,T)
\end{align}
Eq. (2) is in agreement with Eq. (1). If you want to know the molar fraction in the gas phase, you need to give me two things: $p$ and $T$. However, in everyday life, you check the news and it seems at first glance that temperature determines this. This is so, because the news may take the ambient pressure as a fixed value throughout the centuries, so we just "need" to know $T$.
Here, $P_1^\pu{sat}(T)$ is the saturation pressure of water as a pure substance. We can plot Eq. (2) if we have data of this value, which you may find in the references. I present you the results for temperatures representative of planet earth:

Pressure decreases as you go up in the atmosphere, so what we say is approximately true at sea-level. At $\pu{25} \;^\circ$C, if you could watch the molecules through atmosphere, $\ce{97} \%$ is made of air and only $\ce{3} \%$ of water. Say your atmosphere has $n$ moles, then, at most $0.03n$ moles is water. What happens to the rest? They end up accumulating in the liquid phase.

[OP] Why is there a limit even though the atmosphere is so vast?

If you think in terms of amount of substance, well, for the atmosphere $0.03n$ is a lot to count, and you are correct. However, in terms of relative proportions of water and air, the proposition "the atmosphere, in its vast extension, is scarce of water relative to air" is also true regarding what we have just said.

[OP] I have also read that when air is ‘’saturated’’ it means that it reaches the maximum amount of water vapor it can hold at that temperature and pressure

This is what I showed in the graph.

[OP] How can water dissolve in air if air is a neutral mixture and water is polar, considering that polar only dissolves polar?

This is a famous rule that has it shortcomings. But here in this case I think it applies rather well, and your statement in reverse is true: air is non-polar, water is polar, so air doesn't let water come in easily. As you can see in the Figure, this is true.
References
The saturation pressure of water was obtained as a function of temperature in the following book:

The Properties of Gases and Liquids, J.M. Prausnitz, J. P. O'Connell, McGraw-Hill, New York, 2001.

that has an Antoine's equation functionality:
$$
  \ln\left(\frac{p^\pu{sat}}{\pu{kPa}}\right) = \pu{16.3872} -
  \frac{3885.79}{T/{^\circ}C + 230.170} \hspace{1 cm}
   (0 \; ^\circ C \leq T \leq 200 \; ^\circ C)
$$


text:
Why does ammonium chloride form an ionic bond?
Why does $\ce{NH4Cl}$ form an ionic bond if the difference in electronegativity between nitrogen and chlorine is $0$?
Shouldn't it be at least 1.7 in order for an ionic bond to form? If not what is the reason behind this?

Ionic bond bond is between ammonium cation and chloride anion, there's no bond between N and Cl.

I don't understand the downvotes, do people downvote everything they see?

I think you should rather be happy, that you got any answers, because the question is hardly outstanding.

@Mithoron I am happy with the answers, its just every post at least two people downvote. And anyway, what makes a question ''outstanding''?

Well, compare your post with Ananta's answer, it doesn't look remotely similar, doesn't it? And yes, questions can look like that, not only answers. If they did, upvotes would follow.

I recommend making the following edits: (1) Make sure it is error free; 'o' should be replaced with $\pu{0}$, '??' should be replaced with '?'. (2) Research and cite sources; why should we believe that the electronegativity difference between two atoms should be at least $\pu{1.7}$ to form an ionic bond? Of course, we cross-checked it, but knowing your sources is important.

@ananta part 2 is rather complicated, I can't really cite sources because that's what my professor told me, I taught it was a well known concept. But I fixed part 1, thank you for the tip.

@NeaguCristian you could have mentioned that your professor told you this. Nonetheless, regarding a citation, you could start here and here. The second citation provides a (somewhat unclear) formula based on Pauling's ideas, that is "amount of ionic character" = $1 - \mathrm{e}^{\frac{\Delta \chi}{4}}$. Pauling's "The Nature of the Chemical Bond" is also an interesting read if you want to know more about it.

It looks to me you may consider NH4Cl similar to CH3Cl. But there is huge difference. There is direct covalent bond C-Cl, but no bond N-Cl. N atoms cannot form 5 bonds.

Quick Answer

Explicit valence of $\ce{N}$ is four, $\ce{N}$ of $\ce{NH4+}$ cannot form another (covalent) bond with $\ce{Cl-}$.

The chemistry of $\ce{NH4+}$, owing to similar size, charge, and polarizing power, is similar to that of alkali metal ions, especially $\ce{K+}$; $\ce{NH4+Cl-}$ is ionic just like $\ce{K+Cl-}$ is, with about $\pu{80\%}$ ionic character.

The gap between HOMO of $\ce{Cl-}$ and LUMO of $\ce{NH4+}$ is about $\pu{5 eV}$ making any covalent interaction rather weak; $\ce{NH4+}$ doesn't have available empty orbitals to make a covalent bond with $\ce{Cl-}$.

The Details
Maximum Valency and Complete Octet of $\ce{NH4+}$
It is commonly known that $\ce{N}$ does not form more than four bonds. A similar question (How many bonds can nitrogen form?) on chemistry stack exchange answered this. Since, $\ce{N}$ in $\ce{NH4+}$ forms four bonds with four $H$, another covalent bond is not possible; moreover, it is forming a complete octet, which is especially stable for elements with atomic number $Z \leq 10$.

Similarity between $\ce{NH4+}$ and Alkali Metal Ions
$\ce{NH4+}$, due to its size and charge, as you mentioned, often shows similarity to alkali metal ions, especially $\ce{K+}$. By this analogy, since alkali metals for ionic bonds with halides, so does the $\ce{NH4^+}$ cation.
According to reference$^1$

(Thus,) the room-temperature $\ce{NH4+Cl-}$ bond length estimated from
equation (5) is $\pu{3.28} \overset{\circ}{\text{A}}$ , corresponding
to the $r_\ce{NH4+}$ value of $\pu{1.47} \overset{\circ}{\text{A}}$.

Comparing with values given in reference$^2$, this value falls within the (Pauling) ionic radii of $\ce{K+}$ ($\pu{1.33} \overset{\circ}{\text{A}}$) and $\ce{Rb+}$ ($\pu{1.48} \overset{\circ}{\text{A}}$). Also note the Gourary and Adrian Electron-density (ionic) radius of $\ce{K+}$ is $\pu{1.49} \overset{\circ}{\text{A}}$.$^2$. Thus, $r_\text{ionic}(\ce{NH4+}) \approx r_\text{ionic}(\ce{K+}), r_\text{ionic}(\ce{Rb+})$.
Especially in Biochemistry, chemistry of $\ce{NH4+}$ has often been compared with that of $\ce{K+}$. For example, in reference$^3$, it is stated:

(We conclude) that a unique transporter, a potassium-ammonium
($\ce{K+}$/$\ce{NH4+}$) antiport, is responsible for $\ce{NH4+}$
transport in renal inner medullary collecting duct cells. This
antiporter is sensitive to verapamil and Schering 28080, is
electroneutral, and is selective for $\ce{NH4+}$ and $\ce{K+}$ as
substrates. The $\ce{K+}$/$\ce{NH4+}$ antiporter may play a
significant role in acid-base regulation by excretion of ammonium and
elimination of acid.

Citing reference$^3$, it is stated in reference$^4$

Since host–guest chemistry may also include non-metallic cations, it
is interesting to note that the K+ ion has been stated to be similar
to the ammonium ion ($\ce{NH4+}$).

Large gap between HOMO of $\ce{Cl-}$ and LUMO of $\ce{NH4+}$
Furthermore, as calculated by MolCalc, the difference between the HOMO of $\ce{Cl-}$ and LUMO of $\ce{NH4+}$, both lying at energies greater than zero ( $\pu{2.73 eV}$ for $\ce{Cl-}$ and $\pu{7.21 eV}$ for $\ce{NH4+}$), is about $\pu{5 eV}$, making any covalent interaction rather weak (not occurring in this case).
References

On the effective ionic radii for ammonium. Sidey, V. (2016). Acta Cryst. B72, 626-633. 10.1107/S2052520616008064

Conway, B.E., Ayranci, E. Effective Ionic Radii and Hydration Volumes for Evaluation of Solution Properties and Ionic Adsorption. Journal of Solution Chemistry 28, 163–192 (1999). 10.1023/A:1021702230117

Hassane Amlal, Manoocher Soleimani,
K+/NH4+ antiporter: a unique ammonium carrying transporter in the kidney inner medulla, Biochimica et Biophysica Acta (BBA) - Biomembranes,
Volume 1323, Issue 2, 1997, Pages 319-333, ISSN 0005-2736, 10.1016/S0005-2736(96)00200-3

Kim, Y., Nguyen, TT.T., Churchill, D.G. (2016). Bioinorganic Chemistry of the Alkali Metal Ions. In: Sigel, A., Sigel, H., Sigel, R. (eds) The Alkali Metal Ions: Their Role for Life. Metal Ions in Life Sciences, vol 16. Springer, Cham. 10.1007/978-3-319-21756-7_1

+1, very nice answer, but perhaps "a cannon for sparrows".

@Poutnik I do try to write as succinctly as possible, but the more I read, the more curious I become, and the more I read on, and I feel sharing knowledge is essential. Perhaps this is my writing style.

As there is no bond between nitrogen and chlorine in ammonium chloride, it cannot be either ionic either covalent. Nitrogen bonding in an ammonium cation is fully saturated by four covalent bonds to four hydrogen atoms.
It has the same symmetry but with opposite charges as potassium perchlorate. There is no bond between potassium and chlorine either.

Ionic bonds don't have to be bonds. Theoretically, a 100% ionic bond is merely an electrostatic attraction without any orbital overlap (or bond).

@ananta I do agree.  Even practically. I mentioned it in context of evaluation of ionic percentage of eventual mentioned N-Cl bond.

@ananta I'd rather say chemical bond needs something more than just electrostatics. That's true for hydrogen bonds, agostic interactions, etc. Related https://chemistry.stackexchange.com/questions/86918/the-impossibility-of-100-ionic-bond

@Mithoron I agree. I only meant a theoretical (perhaps hypothetical would have been a better word) $\pu{100 %}$ ionic bond would mean only electrostatic attraction or electrostatic potential minimization. Since this is not possible, at least in chemistry, and perhaps even in physics (because, otherwise, opposite charges would come infinitesimally close to each other, which doesn't happen), there is always some interaction that cannot be attributed to electrostatics alone.

@Mithoron It depends on the point of view. All chemical bonds are merely just electrostatics, somewhat complicated by quantum mechanics phenomena. So if 2 or many ions are bound by their net charges, it is as the specific kind of electrostatic attraction a specific kind of a chemical bond.

Think of it this way. Electronegativity is defined as the tendency of an atom to pull electrons towards itself. So, say you have an atom having a positive charge, it will obviously having more power to pull electrons towards itself.
Having a difference of 1.7 is not something applying everywhere. There are ionic compounds with difference less than that.

This value defined to $1.7$ is an order of magnitude, valid for about $50$% of all the bonds. It does not prevent ionic bonds existing between $\ce{N}$ (in the center of $\ce{NH4^+}$) and $\ce{Cl^-}$.

Does N and Cl always form an ionic bond? Or is this a particular case?

@Neagu Cristian. No ! N and Cl can make a molecule called $\ce{NCl3}$ which is made of 3 covalences $\ce{N-Cl}$.


text:
What is alkalimetry?
I've found two answers. One says its the determination of concentration of unknown alkali solution with the help of standard acid solution.
The other says it as the determination of concentration of unknown acid solution with the help of standard alkali solution.

IUPAC's current terminology for the kinds of titrations is based on the following classification [1, p. 580]:

nature of end-point detection (coulometric, potentiometric, amperometric, conductometric, thermoanalytical, optical, radiochemical titrations);
nature of the reaction between analyte and reagent (acid-base, precipitation, complexometric, redox titrations);
nature of the titrant (acidimetric, alkalimetric, iodometric,  coulometric titrations).

So, alkalimetry is the term used to reflect the nature of the titrant, and the historical IUPAC's definition [2] is still correct, and shall be viewed as a type of acid-base titrimetric analysis:

2. Alkalimetry. Determination of a substance by titration with a base.
[…]
27.03. ALKALIMETRIC. An acid-base titration in which an acid is titrated with a standard solution of an alkali.

As for your confusion, it may be explained by the following footnote [2, p. 429]:

The term acidimetry has opposite meanings in different countries.
For instance, in Britain and the U.S.A. it is used in both senses, i.e. determination of acid and determination with acid.
In France acidimetry generally means measurement of acid, and this appears to be the original meaning of the term.
However, all other usages of similar terms imply titration with, e.g. argentimetry; accordingly, to maintain consistency it is recommended that acidimetry means titration with.
The same remarks apply to alkalimetry.

References

Camões, M. F.; Christian, G. D.; Hibbert, D. B. Mass and Volume in Analytical Chemistry (IUPAC Technical Report). Pure Appl. Chem. 2018, 90 (3), 563–603. DOI: 10.1515/pac-2017-0410. (Free Access)
Sandell, E. B.; West, T. S. Recommended nomenclature for titrimetric analysis. Pure Appl. Chem. 1969, 18 (3), 427–436. DOI: 10.1351/pac196918030427. (Free Access)

To quote the theatre play written by the Czech fictitious author Jára Cimrman, "Thank you, my chieftain, for supporting my case!". :-)

Practically, the right meaning is the meaning used and/or assumed in given context, circumstances, community. Use, what is more usual in your region, or perhaps better, replace it by more universal and less controversial/confusing acid-base volumetry.

I assume that the meaning of alkalimetry/acidimetry has the similar fate as the work sign convention schisma in the 1st law of thermodynamics for thermodynamic systems (work done on  a system
$\Delta U = Q + W$  versus work done by a system $\Delta U = Q - W$ ).
Some communities in some periods used terms acidi/alkalimetry for volumetric acid-base titration of acids/bases, as AChem suggests in his etymology analysis as measuring of acids/bases.
Some other communities in some periods used terms acidi/alkalimetry for titrations by acids/bases, like manganometry by permangate, complexo/chelatometry by EDTA etc. Manganometry or chelatonetry are not obviously volumetric analysis of manganese or EDTA contents.
Both notions have some logic in them, similarly as the mentioned opposite work sign conventions. So the right answer may be context dependent and it is always good -- as for TD work -- to explicitly write what you mean by the terms.

If you look at the dates in the OED, the meaning has not changed since 1817. I agree that some people took the opposite meaning but that is not what a classically trained analytical chemist would understand. There are books with the title "Alkalimetry" and all they mean is measurement of base contents by standard acids.

@AChem It may depend on regions where classical analytical chemists were trained. I cannot speak for EN speaking countries. I did near all my titrations before I have learnt English.

But German authors also used the same def as given the EN language dictionary. I thought Czech and German sciences went hand in hand (or heavily influenced each other).

@AChem But CS/CZ science had strong hand in used terminology. Even you must admit inconsistency that acidbase titrations are OF and other ones BY. Some chemists may have noticed and started to use BY sense. In practical life, the term was not used much and chemists might not be sure which was correct.

This is why all these terms are no longer used in modern analytical chemistry...I saw this term after a long time.

@Achem Partly because there is not many titrations these days, aside of basic parameters in water analysis.

It is an obsolete term but nevertheless an analytical chemist should be familiar with it. Break the word into two: alkali + -metry. Measurement of an alkali (content). In other words, you use a standardized acid during titration, and the base concentration is unknown. The opposite is acidimetry.
Please note Wikipedia definitions of acidimetry and alkalimetry are incorrect. If you look at original books from 1890s or early 1900s even from respectable German authors, they define alkalimetry as defined in the answer. I pasted the screenshot of the unabridged version of the Oxford Eng. Dict. because it is not available to public. Maybe someone started the inconsistent definitions in some handbook and some people  followed the same without giving a second thought.
https://en.wikipedia.org/wiki/Acid%E2%80%93base_titration

Why obsolete? What is used instead of it? I think you got it confused, it takes the name after the standard solution, similar as bromatometry, manganometry or chelatonetry, not after what is determined

Acidimetry is the specialized analytic use of acid-base titration to determine the concentration of a basic (synonymous to alkaline) substances using standard acid. Alkalimetry, is the same concept of specialized analytic acid-base titration, but for an acidic substance using standard base.

@Poutnik, Since when Wikipedia became your reference for consulting good quality science? This is just the opposite what classical analytical chemists would say. I will add more details.

It is not reference for knowledge, it is source of handy quotes, not to type it myself. // The funny thing is I am finding both ways by searching. // In case the way you have provided is right, volumetry methods have naming inconsistency.

My CZ book Small Encyclopedia of chemistry from 1976 says(translated) alkalimetry is volumetric method to determine content of acidic substances by titration with solution of hydroxide, usually NaOH  and then enumerates acidic primary standards.  // Unless EN and CZ chemists used the opposite meanings for alkalimetry and acidimetry.


text:
What is the math behind the formula for the concentration of the strong titrant when calculating theoretical buffer capacity of a diprotic system?
I have a question related to finding the theoretical buffer capacity of a diprotic system, specifically about the formula for the concentration of the strong titrant. In this post:
What is the formula for theoretical buffer capacity for a diprotic buffer system?
The top answer, given by grsousajunior, gives the equation for a buffer capacity, calculated by taking the derivative of the concentration of strong base as a function of pH. All of the math makes sense to me, except for the given equation for the concentration of strong base, $C_\mathrm{B}$. Here’s how my math for that equation worked out (most of this is reiterated in grsousajunior’s answer):
The first ionization is represented by the equation:
$\mathrm{H_2A+H_2O\rightleftharpoons HA^-+H_3O^+}$, and therefore has a $K_\mathrm{a1}$ of $K_\mathrm{a1}=\mathrm{\frac{[H_3O^+][HA^-]}{[H_2A]}}$
The second ionization is represented by the equation:
$\mathrm{HA^-+H_2O\rightleftharpoons A^{2-}+H_3O^+}$, and therefore has a $K_\mathrm{a2}$ of $K_\mathrm{a2}=\mathrm{\frac{[H_3O^+][A^{2-}]}{[HA^-]}}$
Shifting around the equation for $K_\mathrm{a1}$ to find $\mathrm{[HA^-]}=\frac{K_\mathrm{a1}\mathrm{[H_2A]}}{\mathrm{[H_3O^+]}}$ and putting that into the equation $\mathrm{[A^{2-}]}=\frac{K_\mathrm{a2}\mathrm{[HA^-]}}{\mathrm{[H_3O^+]}}$, $\mathrm{[A^{2-}]}=\frac{K_\mathrm{a1} K_\mathrm{a2}\mathrm{[H_2A]}}{\mathrm{[H_3O^+]^2}}$, at least by my math.
As far as I can tell, grsousajunior after that used the equation for the charge balance:
$\mathrm{[H_3O^+]+[B^+]=[OH^-]+[HA^-]+2[A^{2-}]}$
Which I think is correct, solved it for $\mathrm{[B^+]}$, and got $\mathrm{[B^+]}=C_\mathrm B=\mathrm{[OH^-]-[H_3O^+]+[HA^-]+2\,[A^{2-}]}$. Again, this is just speculation on my part. This brings me to my source of confusion. In the answer, grsousajunior gives the equation $C_\mathrm B=\frac{K_\mathrm w}{\mathrm{[H_3O^+]}}-\mathrm{[H_3O^+]}+\frac{K_\mathrm{a1}\,\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)}{\mathrm{[H_3O^+]^2}+K_\mathrm{a1}\mathrm{[H_3O^+]}+K_\mathrm{a1}K_\mathrm{a2}} $. To me, it looks like $\mathrm{[OH^-]}$ from the charge balance equation became $\frac{K_\mathrm w}{\mathrm{[H_3O^+]}}$, which makes perfect sense.
What I don’t get is how $\mathrm{[HA^-]+2[A^{2-}]}$ became $\frac{K_\mathrm{a1}\,\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)}{\mathrm{[H_3O^+]^2}+K_\mathrm{a1}\mathrm{[H_3O^+]}+K_\mathrm{a1}K_\mathrm{a2}} $.
When I add the equations that I have for those two concentrations, doubling the latter, I instead get $\frac{K_\mathrm{a1}\,\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)}{\mathrm{[H_3O^+]^2}}$. Everything is the same, except for the denominator. To me, it looks like the denominator somehow ended up as a quadratic, and I have know idea why. My best guess is that the concentrations of hydronium are separate between the two ionizations, but I don’t know how to wrangle out the math for that.
Does anyone know if or how the math I did is wrong? Thank you!
P.S. - Everything after this step makes sense to me, but the equation for buffer capacity that I would end up with using my equation for the concentration of strong base would look a bit different.

The equation that you have derived (the last one you show)
\begin{equation}
  \frac{K_\mathrm{a1}\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)} {\mathrm{[H_3O^+]^2}} = \color{blue}{\left(\frac{\ce{[H2A]}}{\ce{[H3O^+]}^2}\right)}
  K_\mathrm{a1}([\ce{H3O+}] + 2K_\mathrm{a2}) \tag{1}
\end{equation}
is correct. However

$C_\ce{H2A}$ refers to the initial concentration of the species
$\ce{[H2A]}$ refers to the equilibrium concentration of the species. This is the term that appears in the equilibrium constants.

The relation between both can be found by a mass balance, and replacing $\ce{[HA-]}$ and $\ce{[A^2-]}$ with the expressions that you have obtained
\begin{align}
  C_\ce{H2A} &= \ce{[H2A]} + \ce{[HA-]} + \ce{[A^2-]} \\
  &= \ce{[H2A]} + \frac{K_\mathrm{a1}\ce{[H2A]}}{\ce{[H3O^+]}} +
  \frac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2A]}}{\ce{[H3O^+]}^2} \\
  &= \frac{\ce{[H3O^+]}^2\ce{[H2A]}}{\ce{[H3O^+]}^2} +
     \frac{\ce{[H3O^+]} K_\mathrm{a1}\ce{[H2A]}}{\ce{[H3O^+]}^2} +
     \frac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2A]}}{\ce{[H3O^+]}^2} \\
  &= \frac{[\ce{H2A}]}{\ce{[H3O^+]}^2} (\ce{[H3O^+]}^2 + K_\mathrm{a1}
  \ce{[H3O^+]} + K_\mathrm{a1}K_\mathrm{a2}) \\
  \color{blue}{\frac{[\ce{H2A}]}{\ce{[H3O^+]}^2}} &=
  \frac{C_\ce{H2A}}{{\ce{[H3O^+]}^2 + K_\mathrm{a1} \ce{[H3O^+]} +
  K_\mathrm{a1}K_\mathrm{a2}}} \tag{2}
\end{align}
and combining Eqs. (1) and (2) in blue
\begin{equation}
  \frac{C_\mathrm{H_2A}K_\mathrm{a1}([\ce{H3O+}] + 2K_\mathrm{a2})}
  {{\ce{[H3O^+]}^2 + K_\mathrm{a1} \ce{[H3O^+]} + K_\mathrm{a1}K_\mathrm{a2}}}
\end{equation}
which is the term that you are searching for.


text:
Comparison of atomic radii of boron and beryllium
Let me write down a couple of facts and this will be easier for you to know the essence of the question.

Radius of boron is smaller than beryllium.
Ionization energy of boron is smaller than beryllium.

I see that boron has 1 electron at 2p. We know that 2p orbital have lower ionization energies because they are on average a little further from the nucleus. Due to this, it's easy to see why ionization energy of boron is smaller than beryllium.
Now, the fun part is that it's a fact that boron is smaller in size, which I don't get why. If 2p on average is further away than 2s and 1s, it means boron's radius must be bigger. Note that with Slater's rule, I can see it holds to true as effective nuclear charge for boron is bigger, hence more attraction, hence smaller in size
What I'm trying to understand is why being 2p further away than 2s or 1s don't suggest that boron has to be bigger in size ? I think the hint here is with the word: "average", but I still don't get it to be happy with it.

Not related to downvote but there is general shrinking tendency along periods. Also orbital energy has indirect relation to size. The 2nd radial probability maximum of 2s orbitals is farther from nucleus than 2p maximum.

@Poutnik I took your suggestion and made my answer considerably shorter :)

The Answer

 The quantum mechanical model (shown at the end) explains both the facts: radius of boron is smaller than beryllium; ionization energy of boron is smaller than beryllium.

The electrostatic model and your reasoning fall short because these don't account for inter-electron interactions, which is why we had to develop quantum mechanical model in the first place.
So, we don't say an orbital is further away, or closer to the nucleus. It doesn't make sense because an orbital is not a point, it is a wavefunction. Nonetheless, there are a few ways to go about investigating your question.
Your reasoning is incorrect because, according to you, $\ce{B}$ should have a higher ionization energy than $\ce{Be}$ because $\ce{B}$ has a lower size than $\ce{Be}$; the latter is correct, but doesn't imply the former. According to the electrostatic model too, the ionization energy of $\ce{B}$ should be higher than that of $\ce{Be}$, which it is not. Therefore, you need to think in quantum mechanical terms to understand this.
Actually, as I reasoned in The Fun Part section, ionizing an electron from the singly filled $\mathrm{2p}$ orbital of $\ce{B}$ is easier than ionizing an electron from the fully filled $\mathrm{2s}$ orbital of $\ce{Be}$, just like ionizing an electron from quadropoly filled $\mathrm{2p}$ orbital of $\ce{O}$ is easier than ionizing an electron from the triply filled $\mathrm{2p}$ orbital of $\ce{N}$; fully and halvely filled orbitals are more stable reducing the ionizing energy required to attain such configurations.
Electrostatic Model
Radii of $\ce{Be}$ and $\ce{B}$
We assume the valence electron is orbiting (with potential and kinetic energy) at a distance $r$ from a positive charge ($Z_\text{eff}$ of the nucleus) of charge $Z_\text{eff}$. We can obtain $r_\ce{Be}$ and $r_\ce{B}$ using the Bohr theory:
$$
r_n = \pu{5.29 \times 10^{-11}} \dfrac{n^2}{Z}\pu{m}\\
\text{may }\implies (r_n)_\text{eff} = \pu{5.29 \times 10^{-11}} \dfrac{n^2}{Z_\text{eff}}\pu{m}\\
\implies \dfrac{(r_2)_{\text{eff},\ce{B}}}{(r_2)_{\text{eff},\ce{Be}}} = \dfrac{Z_{\text{eff},\ce{Be}}}{Z_{\text{eff},\ce{B}}} = 0.787
$$
Therefore, the valence electron of $\ce{B}$ is situated closer to that of $\ce{Be}$ because $\ce{Be}$ has a lower effective nuclear charge than $\ce{B}$, as you correctly reasoned.
Ionization Energies of $\ce{Be}$ and $\ce{B}$
Assume that the electron, situated at distance $r$ from the nucleus is orbiting, the energy of this electron may be approximated using Bohr theory as:
$$
E_n = -13.6\dfrac{Z^2}{n^2}\pu{eV}\\
\text{may }\implies (E_n)_\text{eff} = -13.6\dfrac{Z_\text{eff}^2}{n^2}\pu{eV}\\
\implies \dfrac{(E_2)_{\text{eff},\ce{B}}}{(E_2)_{\text{eff},\ce{Be}}} = \dfrac{Z_{\text{eff},\ce{B}}^2}{Z_{\text{eff},\ce{Be}}^2} = 1.613
$$
Therefore, the ionization energy of $\ce{B}$ should be more than that of $\ce{Be}$ because $\ce{B}$ has a higher effective nuclear charge than $\ce{Be}$, but this isn't the case. The electrostatic model doesn't account for inter-electron interactions, which is why we had to develop quantum mechanics in the first place.
The Fun Part
First Things First
The Data
The data (obtained from sources 1, 2, and 3) for the atomic radii, ionization energies, and effective nuclear charges of Group 2 elements is:

The Graphs
The following is graph for elements of Group 2, that is $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$, $\ce{N}$, $\ce{O}$, $\ce{F}$, and $\ce{Ne}$.
Let us also see how the radius varies with ionization energies and effective nuclear charges.

Let us also compare the

So, even though, effective nuclear charges are increasing on going from $\ce{B}$ to $\ce{Be}$, we see that the ionization energies are decreasing. A similar anomaly is observed in going from $\ce{N}$ to $\ce{O}$. This is because extracting the valence electron makes the configuration halvely filled ($\mathrm{2p^3}$) for $\ce{O}$ or fully filled ($\mathrm{2s^2}$) for $\ce{B}$, which are considered stable.
Quantum Mechanical Model
I wanted to compare the orbital energies of $\ce{B}$ and $\ce{Be}$ to see which atom has the highest energy orbital. I created a simple model in ORCA 5.0.4 where $\ce{B}$-$\ce{Be}$ distance was large, $7 \overset{\circ}{\mathrm{A}}$, to avoid any significant interaction between the two elements. The highest-occupied molecular-orbital (HOMO) of the system will show us which atom is easier to ionize.
Results
The HOMO of the system, as shown below, corresponds to $\ce{B}$ $\mathrm{2p}$ orbital from where it will be easiest to ionize. Thus, Ionization energy of $\ce{B}$ is less than that of $\ce{Be}$.

Notice that the size of $\ce{Be}$ 2s orbital is larger than that of $\ce{B}$ 2s orbital, verifying that $\ce{Be}$ is larger than $\ce{B}$.
Conclusion
The quantum mechanical model explains both the facts:

Radius of boron is smaller than beryllium.
Ionization energy of boron is smaller than beryllium.

Hello ananta. If you are using Excel (I think you are), and you have two magnitudes that share the same x-axis value (elements in this case), you can use a secondary axis. Selecting the series, you can go to format, and on the left you click on series options→secondary axis (I am translating from Spanish to English, but it should be something like that). That way, the orange curve in the first two graphs will be more clear, because the magnitude of $Z_\pu{eff}$ is much lower than $r$. A humble tip.

Good answer, but unfortunately it doesn't state the reason why it's actually like this. in your ionizing energy section of B and Be, you use Bohr model which affirms my viewpoint, but true, bohr model is wrong in its own, I get that, but explanation of what i'm. asking is not stated in your answer. Seems like Until I fully learn schrodinger equation, i won't be able to understand this at all. I knew about half filled and fully filled rule, but still, no good explanation why that is. Again, i mightn't be understanding it as i don't understand schrodinger

@ananta Hello ananta. There are things in this answers that seem to me completely wrong or maybe I am crazy, and you can help me to clarify. (1) The value $E_n$ there is the ground state energy of the $\ce{H}$ atom, there is no such thing as the energy of the second, third, and so on electron when you get approximate solutions of the the Schrödinger's equation.

(2) The quantum number $n$ makes sense in the $\ce{H}$ atom, because it appears when we solve the radial part of the differential equation (for those who like solving it with series, we need to kill that series or it will diverge). For another atoms, it doesn't make any sense of speaking of any $n$ number (what is the $n$ number of $\ce{Be}$?), we rather speak of the lowest energy eigenvalue which are equal to the ground state. So, when you take the ratio of $E_n$ of $\ce{Be}$ or $\ce{B}$, what are you actually dividing there?

(3) Even if you apply the Bohr's model, you cannot use that formula to estimate the radius of any other element, there are two many electrons and you get nonsensical results. Even if you use $Z_\pu{eff}$, you can try that formula: for $\ce{B}$ the exp. value is $85$ $\pu{pm}$, with the Bohr's formula (using $n=1$, which still doesn't make any sense to talk about $n$ in another atom rather than $\ce{H}$) you get $11.30$ $\pu{pm}$.

(4) One thing is the ionization energy and another thing is the energy eigenvalues. We can get approximate values of $E_1$, $E_2$, .... But this has nothing to do with the energy needed to extract the "easiest one" electron in an element. Clearly $\ce{He}$ has a higher ground state energy than $\ce{H}$, $\ce{Li}$ has a higher ground state energy than $\ce{He}$, and so on, but how do you link this $E$'s with the ionization energy? You are connecting again the $E_1$ with the valence electron!

The trend of the atomic radius across the period is decreasing of course but if we consider the Hund's rule which states that the "completely filled or completely half filled orbitals" are more stable than the incompletely filled, incompletely half filled orbitals and the more stable orbitals usually tend to have lower energies.
If we look into the configurations of the II and XV th groups which are ns2 and ns2,np3 where IInd group is completely filled s-orbital and the XV th group completely half filled p-orbital as they are more stable compared to the groups before and after they have less energies comparatively more stability and hence comparatively a bit larger size than the surrounding groups.
Using the same thing Boron falls under the III rd group which has a configuration of 1s2,2s2,2p1 which is a completely filled 1s and 2s -orbitals but an incompletely filled 2p orbital making it have comparatively more energy, less stability and more radius.

Atomic radius decreases as the group increases. Consider that a Group I element, such as Li or K, has a lone electron ready to "drop off". Halogens, though, are ready to "pull in" that last electron to approximate a noble gas shell.

Well, the actual problem is with slater rule, boron is smaller, but my question is that if Boron's valence electron is in 2p(which is farther away from nucleus than 2s, then this contradicts the slater rule and suggests boron should be bigger in radius).

Having a partially(or fully) filled p orbital increases the size of a atom if the nuclear charge is the same which it isnt in your case.The size of a atom is determined by the wavefunction of the valence electron(s).As the nuclear charge is increase the rate of change of the complex wavefunction $\frac{d\Psi(r,\theta,\phi)}{dr}$ is increased as we slowly start moving away from the nucleus and reaches its absolute maximum faster then is descreased faster than in the case of a nucleus with less nuclear charge resulting in a wavefunction which probability density is located much closer to the nucleus which means by definition atom radius decrease.

My worry is that with slater rule, boron is smaller, but my question is that if Boron's valence electron is in 2p(which is said to be farther away from nucleus than 2s, then this contradicts the slater rule and suggests boron should be bigger in radius). Isn't there any way to understand this without schrodinger ?

hmm atomic structure is described by quantum mechanics and quantum mechanics is described by Schrodinger equation , Slater's rules are rules of thumb and cannot explain the Why?

@Volpina Hello Volpina. Why choose $(\partial \Psi/\partial r) (r,\theta,\phi)$ instead of $\partial (\Phi^* \Psi)/\partial r (r,\theta,\phi)$? Would you extend your argument to a molecule also?

@MetalStorm because we find $\Psi(x)$ from the Schrodinger equation not the probability density.

@Volpina Yes but analyzing the wavefunction like that is extremely misleading. I will give a simpler case than the N-electron wavefunction. For the 1D particle-in-the-box, $\psi_2(x) = \sqrt{2/L}\sin(2\pi x/L)$. The highest rates of change of $\psi_2$ are in $x \in  (0,L/2,L)$. Guess the probability of finding the particle here? They are all zero. Do you agree with me?


text:
Does anybody know anything about this patented, late-19th-century process for producing white lead?
I debated on whether to put this question in the History or Engineering Stack Exchange. Ultimately, I decided on the Engineering Stack because I believe the people on this site will understand the process and possibly know modern uses more readily than History-oriented people on the History Stack. When I did post it on there, it was suggested that I try on the Chemistry Stack Exchange. Since, it sat over on the Engineering Stack for several days without any help, I removed it from there and am asking on here.
I came across this 1 January 1892 edition of the Rocky Mountain News (RMN) that talks about a new process for producing white lead by the Denver White Lead company (DWL). It says:

(...) The managers claim that with a cheap process at their command they can control the Western market and ship as far East as the Missouri line, invading the borders of that great lead producing city and underselling the lead corroders of St. Louis in Kansas City.

The Morris-Bailey process, which corrodes 96 per cent in three days as against 65 per cent by the Dutch process in four months, will be used. It has been tested by experts, and notwithstanding the quick process of corroding the lead is pronounced by English and American experts superior to any lead in the world.

Despite owning the patent to this supposedly revolutionary process, the business was defunct by October of 1900. I find a 19 October 1900 edition of the RMN with an article claiming that DWL was put out of business by the “a lead trust”. Says the article:

(...) No sooner did the Denver company attempt to place its white lead on the market than the company discovered there was a lead trust. The trust lowered its price until the rate was less then [sic] the cost of crude material to the Denver firm. The Denver crowd held a meeting, and, by the way, George Ross-Lewin, Sam Landon and Dr. Morris, were members of the company, and decided to compromise with the lead trust. Their advances were met with a sneer. As a last resort the company meekly offered to sell out to the trust. They were met with a laugh and up to the present day the Denver organization lies as dead as a mackerel. Every dollar invested in the local plant was lost.

Now, this is entirely possible as DWL was started in 1891 with an initial capital of 50,000k (a single source claims the capitalization was at 150K, but I have chosen to go with the multitude of other sources, some local, which I believe are accurate). Certainly nothing to sneeze at; however, the National Lead Trust (right-hand side of page 8 of the .pdf) was capitalized in 1887 at $90,000,000, absolutely dwarfing DWL’s funding. That’s almost 2,000 times more funding. And, I remember a few years back when Saudi Arabia drastically reduced prices below production costs for several months, driving several smaller competitors into selling out. So, it is certainly possible. That being said, what happened to this supposedly superior process?
First, what exactly is this process? The Morris-Bailey process was patented by Dr. N. K. Morris and John W. Bailey. I find a description of the process in the 1892 Volume 13 of Paint, Oil and Chemical Review:

By the new process the metallic lead does not have to be made up into “buckles.” Plain pig lead is placed in cylinders and melted till a mass four feet deep is produced. At the bottom of the cylinders are dies with minute perforations. Now, the force of four feet of molten lead is equal to that of fifty feet of water, and the hydraulic miner knows what tremendous pressure a column of water fifty feet high can exert. A stream of water from a pipe under such pressure can support in the air a rock weighing a ton, and is equal in exertion to 100 horse power [sic].

Under this pressure the molten lead is forced through the minute perforation in the dies at the bottom of the cylinder and comes out in fibers hardly the one-hundreth [sic] part of an inch in diameter. It falls into a rack fitted with trays very much like a chest of drawers, in fact, but so arranged with flanges that each tray is perfectly air-tight. These trays, or drawers, are also perforated at the bottom, so as to allow gases to penetrate the fibers. Then the tray is dipped in acetic acid, replaced in the rack, closed up air-tight and the carbonic acid gas turned on from the bottom by means of suction from a boiler.

But tanbark is scarce in Colorado, and is not needed by this process anyhow. The carbonic acid gas is manufactured from common petroleum, or residuum oil, and in this way is less expensive than the mere handling of tanbark. When the trays or stacks are all ready, and the carbonic acid gas is turned on, the same chemical change is produced in the metallic lead, and it also comes out as white lead, but in this case it is a mass of beautiful fibers. These are washed, ground up with linseed oil and the product is ready for the market.

The main saving of time by the new process is brought about by the lead being reduced in fibers. In the “buckles” a mass half an inch thick and two or three inches in diameter is presented for the gases to work upon. The fibers present 10,000 times more surface in the same amount of metal, and the gases attack them with the greatest ease. The work which, under the old process, required nearly four months [sic] time to complete, is thus accomplished in two days.

The claim is made for the Denver white lead that it is a purer article than any made with the aid of tanbark, because other gases are produced from tanbark, which contaminate in some degree the metal.

I tried to find this patent by looking up the patent authors. First, I find the first name of Morris mentioned in an 1887 directory of Denver. Then,  Norman Kelsey Morris and John Winfield Bailey were both granted a patent on 13 December 1892 for “Method of and apparatus for producing lead carbonate”, lead carbonate being white lead. Further, two days later, a patent was granted to Morris alone for “Method and apparatus for the manufacture of lead fibre”. Sadly, these patents contain no information (or perhaps I am not using Google Patents correctly?).
So, my question is, what happened to this process? Reducing a processing time down from four months to two days would be one of those rare examples when an industry makes a quantum leap forward overnight. It would be surprising if this process, if it is as revolutionary as they claim, disappeared. So, can anyone tell me if this process is still used? Was it ever used outside of the Denver White Lead company? Was it superseded by a new process?
I can’t find where I read it, but somewhere I read that Morris and Bailey had their process for a while but they couldn’t get anyone to invest in their work. Then, they went to London and made a small-scale showing, which interested investors back in Denver, who again had them (Morris and Bailey) repeat the process, only after then was the company capitalized at $50,000. Further, I find the business was definitely open at least until 1898. So, they definitely conducted business. Maybe, being such a small company competing with an absolute titan doomed them. Still, I would imagine someone would have purchased the patent. And, even if not, the patent has been free for a VERY long time, so I imagine someone, somewhere must have put it into production. Or was the new process all hype and its improvements not as dramatic as claimed? I appreciate any insight anyone could provide. Thanks.
Some notes that MAY assist anyone who looks into this, is that DWL was initially Western White Lead company, but they changed their name when they realized a company in Pennsylvania already had that name. Also, a place I always like to start to find biographical information that may help my research, especially with places a person may have lived, (and it can be interesting to see if/how you are related to other people if you have attached your tree to theirs) is FamilySearch.org. Morris’s page can be found HERE and I am unsure if Bailey has one. If he does, the information on it is too little to be able to search for him.
Lastly, lead paint was banned in the late 1970's, so information on white lead in general is hard to come by. Despite this, I have found a recent paper on the small-scale production of white lead. I believe, such small-scale production precludes the use of large amounts of molten lead, like that needed for the Morris-Bailey process to make the fibrous lead. That paper does mention a mid-19th-century Spanish process for producing white lead, but I can't find anything on that. Also, just from reading the Wikipedia page on white lead, I believe the Morris-Bailey process is a special variation on the Dutch process, but I am not sure, I might be misunderstanding.

Please visit a good university library and check the Kirk Other's Encyclopedia of Chemical Technology, similarly Ullmann's Encyclopedia of Industrial Chemistry is worth consulting. It will have some history.

The is also History of Science and Mathematics https://hsm.stackexchange.com

@AChem Thanks. The UCCS library has the first. I'll check that out tomorrow. The other one is up in Denver, not worth the trip. I'll see about something through ILL. Thanks for that. Crazy how expensive those sets are. Cost of a cheap car.

Here are the pdfs of the full US versions: https://patentimages.storage.googleapis.com/7f/c4/09/fab8e7ee4eb07c/US467042.pdf https://patentimages.storage.googleapis.com/cd/38/29/5b072efc8fbd9e/US493106.pdf

@Andrew thank you. That's awesome. Can I ask how you found those? Were they on any of the pages I linked?

Your links were to CA (Canadian) patents. I think I just searched on the same inventor names for US patents. Or maybe I found the US patent numbers first and then used the numbers to find full text. I don't really remember.


text:
Why does silver have more affinity for chlorine and bromine rather than fluorine in Swarts reaction?
Why do silver and other metals used in Swarts reaction $(\ce{AgF},$ $\ce{HgF2},$ $\ce{CoF2},$ $\ce{SbF3})$ have more affinity for chlorine and bromine even though fluorine is more electronegative?
$$\ce{R-X + AgF -> R-F + AgX(s)} \\ (\ce{X} = \ce{Cl/Br};\quad \ce{R} = \text{alkyl})$$

Same reason the fluorides are water soluble, while chlorides and bromides are not.

One more reason is that this heavy metals have the capacity to form mixed halide products which can be the active component in the reaction:https://chemistry.stackexchange.com/a/7799/17368

Silver actually does prefer fluorine over heavier halogens. But this preference involves less (free) energy difference than the preference for the alkyl groups for fluorine. As a result, transfer of fluorine away from the silver has a net favorable energy output.
Let's look at some enthalpies of firmation (I cannot readily find free energy data for fluoromethane, so I had to use enthalpies).
$\ce{AgF}: -206\text{ kJ/mol}$ Source
$\ce{AgCl}: -127\text{ kJ/mol}$ Source

Silver fluoride preferred by $79\text{ kJ/mol}$.

$\ce{CH3F}: -234\text{ kJ/mol}$ Source
$\ce{CH3Cl}: -102\text{ kJ/mol}$ Source

Fluoromethane preferred by $132\text{ kJ/mol}$.

By this comparison the transfer of fluorine to the methyl group and chlorine to the silver is exothermic because of the stronger preference for the methyl group for the fluorine, the difference here being $53\text{ kJ/mol}$. The methyl group basically wins the alley fight for the fluorine because its muscles (energy for preferentially forming the fluoride) are stronger, leaving the silver to settle for second place (the chlorine).
A similar situation occurs naturally on Earth, with Group 16 elements. Practically all metals, even such metals as copper and lead which are usually found as sulfide ores, energetically prefer oxygen over sulfur. But their preference is less than what is seen with some other metals such as calcium, magnesium or even iron. So the latter metals are favored to form oxides, leaving sulfur predominantly for less thermodynamically selective metals like copper and lead.


text:
Does nitration of phenol with nitric acid yields 2,6-dinitrophenol?
I've learnt that the $\ce{OH}$ group is 2- and 4-directing, but this has caused some confusion. If $\ce{HNO3}$ reacts with phenol, would there be any 2,6-dinitrophenol produced?
I know that the first nitro group will add to position 4 or 2, but if one is added to 2, can a second add to 6? Or must position 4 be added prior to 6?

Yes, there would be.

In my textbook it only shows 2- and 4- nitrophenol, along with 2,4,6-trinitrophenol, is that because it's hard to isolate 2,6-nitrophenol, and it's much more likely that all the possible hydrogens have been substituted?

It's too minor of a product.

The products of nitration of phenol depend on various factors. For example, different nitrating reagents gave different ratio of products.

See here for example.

As mentioned in the comments, the products depends on various factors like nitrating agent, concentration etc. If you use dilute nitric acid, you will get a mixture of mononitrophenol but when you use concentrated nitric acid, you will get picric acid. Note that although picric acid is the targeted product, you will get all sorts of dinitro- and mononitro- isomers as well. To get pure picric acid, steps are taken to distill of the DNPs and MNPs or make the reaction more selective. So, 2,6-dinitrophenol is also a product of the reaction. It's just that it's too minor of a product to mention.
Side information:
Pure 2,6-DNP can be prepared by separate sulfonation and nitration of o-nitrophenol. It is also a byproduct in the synthesis of 2,4-DNP (2,4-Dinitrochlorophenol is heated with 6% aq. $\ce{NaOH}$ at 95-100 °C for 4 hours which hydrolyzes it. 2,4-DNP is the product and is precipitated by adding acid and is filter off. The solution contains the more soluble 2,6-DNP which is separated by adding excess of acid).
Source: Toxicological Profile for Dinitrophenols, U.S. Department of Health and Human Services, 1993

Yes, nitration of phenol with nitric acid can yield 2,6-dinitrophenol.
Here is a diagram showing the possible products of the nitration of phenol with nitric acid:
        OH
        |
       C6H5

Nitration at position 2:
        NO2
        |
       C6H4OH

Nitration at position 4:
        NO2
        |
       C6H5OH

Nitration at position 6:
        NO2
        |
       C6H4NO2OH

2,6-dinitrophenol:
        NO2
        |
       C6H3(NO2)2OH

The site allows to upload illustrations in e.g. the .png file format. For clarity of your answer, structural formulae would be of advantage in comparison to your diagrams. Get familiar with a couple of them (a list, and a review). In absence of a (personal/group/uni campus) license, visit their demo pages (for example ChemDraw, Marvin, ChemDoodle).


text:
Why would there be a non-zero Gibbs energy of mixing for ideal gases?
When two ideal gases $\ce{A}$ and $\ce{B}$ mix, they don't interact. The chemical potential of each gas is independent of the other, e.g. for $\ce{A}$:
$$\mu_\ce{A} = \mu_\ce{A}^\circ + R T \ln \frac{p_\ce{A}}{p^\circ}\label{eqn:q1}\tag{1}$$
So if the two gases are independent, why is there a Gibbs energy of mixing? Shouldn't there be a requirement for an interaction energy in order to see a difference in Gibbs energy before and after mixing?
However, we know that the Gibbs energy of mixing of two ideal gases is non-zero, as it is given as:
$$\Delta_\mathrm{mix}G = n R T (x_\ce{A} \ln x_\ce{A} + x_\ce{B} \ln x_\ce{B}) \label{eqn:q2}\tag{2}$$
where $x$ is the molar ratio. In a related question about a problem set involving Gibbs energy of mixing, an answer uses \eqref{eqn:q2} in the first part and \eqref{eqn:q1} in the second.
In other words, how are \eqref{eqn:q1} and \eqref{eqn:q2} related to each other? Another way of asking is: if the Gibbs energy of mixing is non-zero even for ideal gases, how could you turn the process of mixing into a reversible process and show the maximal work that is done?

When you joint the same containers with 2 gases, at the same T and p, than partial p of both gases decreases to 1/2 of their initial p, therefore their chem. potential decreases as well.

For a more physical interpretation, enthalpy indeed cannot account for the change in free energy in the mixing of ideal gasses, so it must be due to entropy. Mixing is always entropically favourable - there are necessarily more mixed microstates than separate microstates - so we should expect some non-zero Gibbs free energy change.

If you devise a process to separate the two components into separate containers each at the same pressure as the total pressure of the mixture, such a process will require you to do work on the gases.

When we talk about mixing, we usually mean that two components are in different parts of a container before mixing and then they share the entire container after mixing. In other words, the total volume does not change.
To show how you can make mixing reversible while capturing the Gibbs energy of mixing as work, you need two pistons with semi-permeable membranes. In the animation below, the green piston does not let the "green" gas particles pass, and the red piston does not let the "red" gas particles pass.
So even though the two different particles don't interact with each other, mixing them can be used to do work, as both expand from half the volume to the full volume of the container (and the partial pressures decrease, as mentioned by Poutnik in the comment to the question).

Thus, there is an outward force on the pistons. If you provide an equal opposing force (or infinitesimally smaller force) on the pistons from outside, you can use the mixing of the gases to do work. Conversely, if you use just the right amount of inward force, you can "unmix" the two gasses by pushing the pistons inward (similar to water purification by reverse osmosis, which uses a single semi-permeable membrane).
To show that the way we understand mixing matters, here is a related contraption that keeps the partial pressure of the mixed gases constant (while changing the total pressure - not clear how this is defined in this case?). If this is surrounded by vacuum, the Gibbs energy change during this process is zero, somewhat surprisingly.

From (1) to (2)
So how do we get from equation (1) describing the chemical potential of each component to equation (2) describing the entropy of mixing. In principle, we could start with pure $\ce{A}$ and add $\ce{B}$, integrating over the two chemical potentials. This would be pretty complicated. Instead, we can place our mixture into a syringe in a surrounding that has the same pressure, and slowly push out the mixture until nothing is left. Or we can start with an empty syringe, and push in the two components at the correct mixing ratio from reservoirs at the same pressure. In either case, the chemical potential would stay constant over the entire process, and the integrals are just the change in amount times the chemical potential in the mixture.
Because this is done at constant temperature and constant pressure, the terms $V\,\mathrm{d}p$ and $S\,\mathrm{d}T$ are zero, and we only have to consider the change in amount and the chemical potentials.
For the mixture, assuming the total pressure is $p$, we get:
$$\begin{align}
G &= n_\ce{A} \left(\mu_\ce{A}^\circ + R T \ln \frac{p_\ce{A}}{p^\circ}\right) + n_\ce{B} \left(\mu_\ce{B}^\circ + R T \ln \frac{p_\ce{B}}{p^\circ}\right) + C \\
&= n_\ce{A} \left(\mu_\ce{A}^\circ + R T \ln \frac{x_\ce{A}p}{p^\circ}\right) + n_\ce{B} \left(\mu_\ce{B}^\circ + R T \ln \frac{x_\ce{B}p}{p^\circ}\right) + C \tag{3}
\end{align}$$
where we've expresssed the partial pressure of each component $p_\ce{A}$ as the mole fraction $x_\ce{A}$ times the total pressure $p$. This is true for an ideal gas.
For the state before mixing (with the same reference point for zero), both sections of the mixture have the same pressure $p$, and we get:
$$G = n_\ce{A} \left(\mu_\ce{A}^\circ + R T \ln \frac{p}{p^\circ}\right) + n_\ce{B} \left(\mu_\ce{B}^\circ + R T \ln \frac{p}{p^\circ}\right) + C \tag{4}$$
To get the Gibbs energy of mixing, we have to subtract $(4)$ from $(3)$:
$$\Delta G_\mathrm{mix} = n_\ce{A} R T \ln x_\ce{A} + n_\ce{B} R T \ln x_\ce{B}$$
Factoring $n R T$ with $n = n_\ce{A} + n_\ce{B}$, the total amount of particles, we get:
$$\Delta G_\mathrm{mix} = n R T \left(\frac{n_\ce{A}}{n}\ln x_\ce{A} + \frac{n_\ce{A}}{n} \ln x_\ce{B} \right)$$
The ratios of amounts can also be written more succinctly as mole fractions $x$:
$$\Delta G_\mathrm{mix} = n R T (x_\ce{A} \ln x_\ce{A} + x_\ce{B} \ln x_\ce{B})$$
to arrive at expression (2).

Really nice animation. Found or created?

@Poutnik Powerpoint has a new morph slide transition where all objects that change positions are moved along a linear path during the transition. This makes some types of animations really easy to create (and then you turn them into an animated GIF with free or expensive software).

@Karsten Very nice. A couple of comments: your second- and third-last equations are the same. I guess the point is that $p_\ce{A}$ and $p_\ce{B}$ before and after are not the same. Also, I think it'd be great to show that 'bit of reformatting', i.e. explicitly expressing the change in partial pressures in terms of the mole fractions. I'll offer to do the edits if you like, or leave it to you.

@orthocresol I changed it to standard pressure to make it easier, fixed the second expression, and added the missing bit. Good to hear from you, by the way, doc!

@Poutnik Here is the file the with two figures.

@Karsten No, it was good to see this Q&A. :) I still felt the maths was a bit inaccurate, or to be precise, I think it was making an implicit assumption that the container pressure $p$ was equal to standard pressure $p^\circ$. That assumption isn't needed to arrive at the right result, so I made some changes to generalise it a bit. I hope I didn't take too many liberties with your answer, but if you feel I did, you know what to do!

With your reply you seem to have answered part of your question. I would like to add a bit about the other part. I will have to do it mathematically first, there is no other shortcut, but then we will come back and try to connect the idea. I will be precise on where the thermodynamic properties are evaluated, using parenthesis everywhere.

[OP] In other words, how are (1) and (2) related to each other?

The connection between Eqs. (1) and (2) was established by Gibbs himself, it is named Gibbs theorem. Simply put, he was trying to connect the properties of the pure species (as ideal gases) to the properties of the mixture (still an ideal gas). One possible formulation is the following:
A partial molar property of a species in an ideal-gas mixture is equal to its molar property as a pure substance, evaluated at the temperature of the mixture but at its partial presure in the mixture. The exception of this is the partial molar volume.
In mathematical form, for a generic thermodynamic property of species $j$ named $M_j$, we have
$$ \bar{M}_j^\pu{ig}(p,T) = M_j^\pu{ig}(p_j,T) \tag{1} $$
Eq. (1) is almost all that we need to obtain the Eq. (2) of your post. The other one is the summability relation that connects a thermodynamic property $M$ (of the whole mixture) to the partial molar property of all the species in the mixture
$$ M(p,T,y_1, y_2,..., y_N) = \sum_k y_k \bar{M}_k(p,T,y_1, y_2,..., y_N) \tag{2} $$
Eq. (2) is general and is not restricted by the behaviour imposed over the mixture. The derivation of this equation is not that important here.
We will calculate the partial molar enthalpy, partial molar entropy, and then obtain the partial molar Gibbs energy by its definition.

The enthalpy is easy, because for an ideal gas the enthalpy is independent of the pressure, and by Gibbs theorem we have
$$ \bar{H}_j^\pu{ig}(p,T) = H_j^\pu{ig}(p_j,T) = H_j^\pu{ig}(p,T) \tag{3} $$
Applying Gibbs theorem to the entropy yields
$$ \bar{S}_j^\pu{ig}(p,T) = S_j^\pu{ig}(p_j,T) \tag{4} $$
We would like to have evaluated the right-hand side at $p$ rather than $p_j$. This can be done by remembering the entropy of an ideal gas as a function of pressure and temperature, and evaluating the change for an isothermal process at temperature $T$
\begin{align}
  \require{cancel}
  dS_j^\pu{ig} &= \frac{c_\pu{P}^\pu{ig}}{T} \; dT - \frac{R}{p} \; dp \\
  dS_j^\pu{ig} &= -\frac{R}{p} \; dp \\
  \int_{S_j^\pu{ig}(p_j,T)}^{S_j^\pu{ig}(p,T)} dS_j^\pu{ig} &=
  -R\int_{p_j}^{p} \frac{dp}{p} \\
  S_j^\pu{ig}(p,T) - S_j^\pu{ig}(p_j,T) &= -R\ln\left(\frac{p}{p_j}\right) \\
  S_j^\pu{ig}(p_j,T) &= S_j^\pu{ig}(p,T) + R\ln\left(\frac{p}{p_j}\right) \\
  S_j^\pu{ig}(p_j,T) &= S_j^\pu{ig}(p,T) + R\ln\left(\frac{\cancel{p}}{y_j \cancel{p}}\right) \\
  S_j^\pu{ig}(p_j,T) &= S_j^\pu{ig}(p,T) - R\ln(y_j) \tag{5}
\end{align}
In the last line I just related the partial pressure to the pressure of the mixture. Combining Eqs. (4) and (5)
$$ \bar{S}_j^\pu{ig}(p,T) = S_j^\pu{ig}(p,T) - R\ln(y_j) \tag{6} $$

We can obtain the partial molar Gibbs energy just by definition, multiplying both sides by the total number of moles $n$, and then differentiating both sides at constant $p$, $T$, and all moles except the species $j$
\begin{align}
  G &= H - TS \\
  nG &= nH - TnS \\
  \left[\frac{\partial (nG)}{\partial n_j}\right]_{p,T,n_k\neq j} &=
  \left[\frac{\partial (nH - TnS)}{\partial n_j}\right]_{p,T,n_k\neq j} \\
  \bar{G}_j &= \left[\frac{\partial (nH)}{\partial n_j}\right]_{p,T,n_k\neq j}
  -T\left[\frac{\partial (nS)}{\partial n_j}\right]_{p,T,n_k\neq j} \\
  \bar{G}_j &= \bar{H}_j - T\bar{S}_j \tag{7}
\end{align}
This equation is general, not resctricted to an ideal-gas behaviour. Now we can combine Eqs. (3), (6), and (7) to obtain the partial molar Gibbs energy (note that between lines one and two we use Gibbs theorem over the enthalpy and entropy)
\begin{align}
  \bar{G}_j^\pu{ig}(p,T) &= \bar{H}_j^\pu{ig}(p,T) - T\bar{S}_j^\pu{ig}(p,T) \\
  \bar{G}_j^\pu{ig}(p,T) &= H_j^\pu{ig}(p_j,T) - TS_j^\pu{ig}(p_j,T) \\
  \bar{G}_j^\pu{ig}(p,T) &= H_j^\pu{ig}(p,T) -
  T[S_j^\pu{ig}(p,T) - R\ln(y_j)] \\
  \bar{G}_j^\pu{ig}(p,T) &= [H_j^\pu{ig}(p,T) -
  TS_j^\pu{ig}(p,T)] + RT\ln(y_j) \\
  \bar{G}_j^\pu{ig}(p,T) &= G_j^\pu{ig}(p,T) + RT\ln(y_j) \tag{8}\\
\end{align}
Finally we combine the summability relation Eq.(2) with Eq. (8), since we are summing, now we convert the subscript $j$ to $k$
\begin{align}
  G^\pu{ig}(p,T, y_1, y_2, ..., y_N) &= \sum_k y_k [G_k^\pu{ig}(p,T) + RT\ln(y_k)] \\
  G^\pu{ig}(p,T,y_1, y_2, ..., y_N) &= \sum_k y_k G_k^\pu{ig}(p,T) + RT\sum_k y_k\ln(y_k) \\
  G^\pu{ig}(p,T, y_1, y_2, ..., y_N) - \sum_k y_k G_k^\pu{ig}(p,T) &= RT\sum_k y_k \ln(y_k)
\end{align}
the left side, by definition is the Gibbs free energy of mixing. We are subtracting the value of the Gibbs energy of the mixture and the mole-fraction-weighted Gibbs energy of all the species involved. Finally
$$ \boxed{\Delta_\pu{mix}G = RT\sum_k y_k \ln(y_k) <0} \tag{9} $$

[OP] So if the two gases are independent, why is there a Gibbs energy of mixing? Shouldn't there be a requirement for an interaction energy in order to see a difference in Gibbs energy before and after mixing?

Even in the absence of intermolecular forces between the species, there is a change in the Gibbs energy when we mix them. IMHO (I cannot prove this), this was what Gibbs foresaw before doing all this:

He knew that for an ideal gas the enthalpy will not be modified. Thus, for a mixing at constant $p$ and $T$, there will not be heat exchanged with the surroundings.
He knew that the mixing could not violate the second law.

Therefore, he visioned $N$ non-interacting particles separated at $p$ and $T$, and then they got mixed at $p_k$ and $T$. This decrease in pressure is the effect wanted by Gibbs, because it inevitably increases the entropy. As a consequence, it inevitably decreases the Gibbs free energy (his energy), and mixing of ideal gases is always a spontaneous process. Likewise, unmixing is also always a non-spontaneous process. This is really near to what Poutnik said in the comments when he made a difference between both pressures:

[Poutnik] When you joint the same containers with 2 gases, at the same T and p, than partial p of both gases decreases to 1/2 of their initial p, therefore their chem. potential decreases as well.

[OP] How could you turn the process of mixing into a reversible process and show the maximal work that is done?

You visualized this in the images in your post. The calculation of the maximum work is automatic. The maximum work that can be done by the system to the surroundings at the same temperature $T$ as the system is
$$ \boxed{W^\pu{ideal} = \Delta_\pu{mix} G^\pu{ig} < 0} \tag{11} $$
so mixing of ideal gases can be used to obtain work from the outside.
Final Comment When I said that there was a connection between Eqs. (1) and (2), it is implicit in here, but we can show it more clearly. For a constant temperature process and a pure substance we have
\begin{align}
  dG_j^\pu{ig} &= V_j^\pu{ig}dp - S_j^\pu{ig}dT \\
  dG_j^\pu{ig} &= V_j^\pu{ig}dp \\
  dG_j^\pu{ig} &= \frac{RT}{p} dp \\
  \int dG_j^\pu{ig} &= RT
  \int \frac{dp}{p} \\
  G_j^\pu{ig}(p,T) &= G_{j,0}^\pu{ig} + RT\ln(p) \\
\end{align}
The jump to your Eq. (1), is by Gibbs theorem in order to change $p$ for $y_jp$
\begin{align}
  \bar{G}_j^\pu{ig}(p,T) &= \mu_{j,0}^\pu{ig} + RT\ln(y_jp) \\
  \mu_j^\pu{ig}(p,T) &= \mu_{j,0}^\pu{ig} + RT\ln(y_jp) \\
\end{align}
If not, there is no way we can write your equation (with $p_\mathrm{A}$ in your case), and arrive at an expression for the partial molar Gibbs energy, also known as the chemical potential.


text:
When HI reacts with H2SO4, why is the sulphate ion reduced to hydrogen sulphide instead of sulfur dioxide?
When $\ce{HI}$ reacts with $\ce{H2SO4}$, it can be represented by the following chemical equation:
$$\ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}$$
However, when $\ce{HBr}$ reacts with $\ce{H2SO4}$, the sulphate ion is reduced to sulphur dioxide instead:
$$\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}$$
Could anyone account for the difference here? Is there any more cases where the sulphate ion is reduced to $\ce{H2S}$ instead of $\ce{SO2}$?

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

Apparently iodide ion is a better reducing agent. Any table of EMF potentials that includes nonmetal reactions will tell you that.

The book intitled Qualitative Analysis emitted by F. P. Treadwell in $1924$ mentions in the chapter Iodine (p. 323) that the reaction depends on the relative amount of $\ce{H2SO4}$. In the presence of a great excess of $\ce{H2SO4}$, its reaction on $\ce{HI}$ or iodides produces $\ce{SO2}$ apart from $\ce{I2}$. In the presence of an excess of iodides, the reaction produces $\ce{H2S}$ as mentioned by Jelly Qwerty.

Is there any explanation to this? Or is it purely empirical

If an explanation is to be found about this phenomena, it may be said that the first reaction between $\ce{H2SO4}$ and $\ce{HI}$ (or $\ce{KI}$) reduces sulfur from +$6$ to +$4$. The equation is  $$\ce{H2SO4 + 2 HI -> SO2 + I2 + 2 H2O}$$ Then, in case of an excess of $\ce{HI}$, the Sulfur is further reduced from +$4$ to -$2$ according to the following equatoion : $$\ce{SO2  + 6 HI  ->  H2S  + 3 I2  +  2 H2O}$$

As @Oscar said, hydrogen iodide (or iodide ion) is much stronger reducing agent than hydrogen bromide (or bromide ion). So the sulfuric acid reduce through a rather complicated pathway than just ending up with $\ce{SO2}$. The reduction pathway is somewhat like this:
$$\ce{H2SO4 -> SO2 -> S -> H2S}$$
Sulfurous acid is an intermediate product in the reaction which takes part in the transient reactions forming sulfur, hence driving the reduction reaction forward.
$$\ce{SO2 (or H2SO3) + 2H2S -> 3S + 2H2O}$$
...and yes, if one of the reagent is a very strong reducing agent, you can end up with hydrogen sulfide instead of sulfur dioxide.
You can read the following articles:

https://www.chemguide.co.uk/inorganic/group7/halideions.html
Sulphuric Acid and Hydriodic Acid, Florence Bush, The Journal of Physical Chemistry 1929 33 (4), 613-620, DOI: 10.1021/j150298a010


text:
Which group takes CIP priority CH2OH or CH2OCH3?
I am confused why this structure is (2R, 3R) when CH2COMe takes priority over CH2OH which takes priority over CH3, surely it should be (2S, 2R)?

Isn't it 3-ol? So what group has the highest priority?

Presumably you mistyped (2S,3R) as (2S,2R).
In fact (2R,3R) is correct because the "tie" between the top two groups is actually broken earlier than you suppose. When you get to the first carbon atom from C-2 the one in the group on the lower left is attached to $\ce{O,\color{blue}{H},H}$ and the one in the group on the lower right is attached to $\ce{O,\color{blue}{C},H}$. (What you think is $\ce{CH2OH}$ is actually $\ce{CH(C2H5)OH}$.) The difference between the blue atoms favors the right-hand group according to the CIP rules, and then the orientation at carbon atom 2 is rendered as R.


text:
Does BCl4- exist?
Is [BCl4-] a good Lewis acid as I according to Jerry March's Advanced Organic Chemistry ,the reactivity order of Lewis acids in Friedel Craft reactions , BCl3 is mentioned which means that BCl4- is a Lewis acid . Then why on hydrolysis of BF3, HBF4 is formed and on hydrolysis of BCl3 [BCl4-] does not show up?

Could you provide the page numbers (and edition) from Jerry March's Advanced Organic Chemistry or any other authentic sources for (1) Friedel-Craft Reaction with $\ce{BCl3}$ as Lewis acid and (2) Hydrolysis of $\ce{BF3}$?

There are several dozens of refined crystal structures with $\ce{[BCl4]-}$, so the answer to the title question is apparently yes. I'm not sure though I follow the logic in the question body: how exactly Friedel–Crafts reaction is involved in this (maybe cite the exact page and edition of the book)? Hydrolysis of $\ce{BF3}$ is heavily temperature-dependent, and $\ce{H[BF4]}$ exists only in diluted $\ce{HF}$ as the product above 20 °C. Also, Lewis is the scientist's name.

You may want to try a different solvent. Boron-chlorine bonds tend to be broken up by water but not by liquid sulfur dioxide[1]. In sulfur dioxide solvent the formation of $\ce{BCl4^-}$ accounts for the increased solubility of potassium chloride when boron trichloride is present. In the reference this synthesis is carried out by adding successively larger amounts of boron trichloride to an equilibrium mixture of solid potassium chloride with the saturated solution of this salt in the liquid sulfur dioxide at $0°$C.
Reference

D. E. Burge, H. Freund, and T. H. Norris (1959). "The Formation of Tetrachloroborate Ion in Liquid Sulfur Dioxide Solution". J. Phys. Chem. 63, 11, 1969–1971. https://doi.org/10.1021/j150581a040