text:
How did Planck reach the conclusion that energy comes in packets based on black body curves?
I understand that if wavelength continues to decrease energy emitted by an atom would get to infinite, but how does the max wavelength at a particular temperature translate to discrete energy packets. Please explain.

Planck empirically found an expression that fit the black body radiation curve, particularly for high frequencies. One way to rationalize the form of the expression was that energy emissions had to be discrete.

I'm voting to close this question as off-topic because this question seems to be more suitable for SEPhysics.


text:
Synthesis of phenethyl alcohol from benzene
How can I make phenethyl alcohol $(\ce{C6H5CH2CH2OH})$ from benzene?
I have got information that phenethyl alcohol is made from benzene by Friedel-Crafts reaction between benzene and ethylene oxide in the presence of aluminium trichloride as follows:
$$\ce{C6H6 + CH2CH2O + AlCl3 → C6H5CH2CH2OAlCl2 + HCl}$$
But I have no idea how it reacts and proceeds to work.
Can anyone help me understanding the mechanism of synthesis and reaction of making phenethyl alcohol from benzene?

$\ce{AlCl3}$ is a Lewis acid and will bind to the oxygen of the ethylene oxide giving a species that functions as $\ce{[R2O-AlCl2]+}.$
This cationic species will be attacked at carbon by the electron-rich aromatic ring in much the same way as a standard Friedel-Crafts reaction intermediate generated from an alkyl halide to give $\ce{[Ph-CH2CH2-OAlCl2]+}.$ This then loses $\ce{HCl}$ to give $\ce{PhCH2CHOAlCl2}$ which is stable until aqueous workup which gives phenethyl alcohol.

I can understand well about the mechanism of making phenethyl alcohol and how it works as Friedel-crafts reaction with Lewis acid and and make alkyl benzene!

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Weathering steel
I'm building something out of weathering steel (A606). It is chemically formulated to rust until it becomes chemically stable and stops rusting. Its composition is: Carbon: 0.22%; Manganese: 1.25%; Sulphur: 0.04% and Copper: 0.2% (minimum).
I want to accelerate the "weathering" process rather than wait for Mother Nature to do it. What chemicals should I use?

Now that's tricky. Chemicals might form a subtly different kind of rust which never becomes stable, but just continues until it eats your steel all through.

I'd ask the manufacturer. You're surely not their first customer who wants to sell a product in it's final optics.

To accelerate "weathering" , give it more weather - spray it with water each day . You could add a little salt ( NaCl) initially  but then let the water spray remove it over time.
The steel also has some P in it and a bit of Cr ( like 0.1 % ) . The Mn is also to improve strength as A 606 is "higher" strength steel ( higher than ordinary hot rolled ). It doesn't exactly "stop" rusting ; it builds a somewhat protective layer of rust. In most environments this works well   . It does not work on a seacoast or in an industrial environment.  Another situation that may be a problem is above concrete or light color stone , as some red rust will slowly run off the steel over the years ( leaving drip marks on the concrete)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Can a Titanium Grade 5 alloy produced by SLM have greater 'surface area' exposed?
I was doing an electrochemical test (LPR) and Titanium Grade 5 produced by AM respassivated oxide layer faster than the specimen produced with traditional manufacturing under identical experimental conditions. I am wondering if this can be explained with 'porosity' and being exposed to more 'surface area' as a result?
Regardless of the the porosity level, will the SLM sample be air tight? Only then would my guess here be wrong. I am assuming that the layers below get exposed to oxygen as well which is why I see the plateau (a flatter curve) for SLMed specimen.
I am trying to find a reason for the slightly better performance of the same material from a corrosion point of view when submerged in a specific liquid.

Now which, SLM or AM? There can be all kinds of differences in surface structure or trace impurities. You've got their data sheets.

Alloyed titanium ( grade 5 = 6 Al, 4 V) has lower corrosion resistance in corrosive environments than the unalloyed grades. I never saw porosity on any titanium surface with a microscope.

AM = SLM for the purpose of this question. So please treat the AM as SLM in the graph.


text:
Additional example determining Marcus parameters searched
Background / theory: The electron transfer in organic materials was formalized by the Marcus theory, introducing $|H_{ab}|$ to describe the how molecules (or more generally: sites) are coupled with each other, and reorganization energy $\lambda$, about how costly it is to to accommodate a charge after the carrier moved.  Especially if the atomic coordinates of the material already are known, it is possible to access these parameters by quantum chemical calculations.
Background / software: NWChem seems to be one of the freely available programs allowing such an additional characterization.  New to this program, I found both the standard entry in the manual (here), as well as a post in the user forum (here) by 2012, comparing the result to a literature reference in Phys Rev B 2007 as "work[ing] pretty well".  It is possible for me to to replicate the computation filed there under "Sep 19th 2012", too.
The feeling: It seems dangerously simple to use this script for an other structure, just replacing the first / second / third block of coordinates by the $(x,y,z)$ extracted from X-ray diffraction analyses about molecule A, hoovering neighbor B, and the close couple / dimer (AB).  Indeed, my first tests (again, as a beginner) failed; the program launched, yet eventually shut down -- likely because I do not know well enough how to proceed.
The question: Does one of you know about an additional tutorial, starting from accessible atom coordinates of molecular materials, concluding with the Marcus parameters of the electron hopping?

I'm not entirely sure what you're asking. It sounds like you want to run a particular NWChem calculation, but change the coordinates. If so, it would help to know the specific errors you're getting, rather than just 'eventually shut down.'

There's also an NWChem forum, which may be helpful: http://www.nwchem-sw.org/index.php/Special:AWCforum

@GeoffHutchison Yes I would like to use NWChem to determine Marcus parameters, eventually with coordinates about other materials.  Like as the post by "St" (Oct 25th 2015) I was wondering about the large coupling (0.92 eV) even if the antracenes in the preceedingly post form a sandwich, only 3 \AA apart from each other.  Other typical organic electronics e.g., the better polymorph of rubrenes are reported just around 0.1..0.2 eV (with far less superposition, of course).  So it still is a search for «more realistic» coordinates, with partial overlap of the $\pi$-systems only and determined MPs.

@GeoffHutchison  As I recall, many, yet not all «shut downs» were caused by input errors of mine -- which anyway I'll check again tomorrow, and I already deposit a question in their forum prior to ChemSE, too.


text:
Producing soap from sodium carbonate and oil
Can we get soap by mixing $\ce{Na2CO3}$ from seaweeds with oil?
I am not good at chemistry and saw this in the anime Dr. Stone. I just want to know if it is true or not.

It does not work. Oil cannot be converted into soap with sodium carbonate $\ce{Na_2CO_3}$ or with ashes. A stronger base is needed, like $\ce{NaOH}$

What do you mean by "oil"? Crude oil and petroleum products are very different chemically from plant or animal oils (which are what soap is derived from).

@Matt_black. Rodrigo and me, we both mean that oil means vegetable or animal oil. Petroleum oils cannot be converted into soap

In nature fat is stored as triglycerides (TGs): Three fatty acid (FA) chains bound together by a glycerol backbone. Olive oil is mainly composed of TGs, though it also contains small amounts of free fatty acids (FFAs).
An example of Triglyceride (from Wikipedia)

Soap is commonly made via the reaction of triglycerides  with a strong base: usually Sodium Hydroxide ($\ce{NaOH}$) or Potassium Hydroxide ($\ce{KOH}$). Though $\ce{Na2CO3}$ is a base, it is too weak of a base to be efficiently used in the traditional production of soap: It may saponify triglycerides but only at a very slow rate, if at all. So you are unlikely to produce much soap by mixing $\ce{Na2CO3}$ (of any source) with oil.
Soap can however also be made by reacting a weaker base with free fatty acid (FFA) chains; that is, FAs that have already been broken away from the TGs. $\ce{Na2CO3}$will form salts with FFAs, and it will do so at a much faster rate than it would saponify TGs. It can thus be used to make soap that way instead (and will, for instance, react with the small amount of FFAs in your oil).
You could try to find a way to separate the FFAs from the rest of the oil (though it would be a highly inefficient process as you'd be left with the dominant TG portion), you could in theory also separately break down TGs into FFAs prior to saponifying them (using enzymes for instance), or (highly recommended) you could go the traditional route of using a strong base which will efficiently and (relatively) rapidly saponify both TGs and FFAs in your oil.
Splitting triglycerides is the "hard work" from a chemical point of view.

Sources
https://en.wikipedia.org/wiki/Saponification
https://en.wikipedia.org/wiki/Triglyceride
https://en.wikipedia.org/wiki/Fatty_acid
https://en.wikipedia.org/wiki/Sodium_carbonate
https://www.soapmakingforum.com/threads/how-much-sodium-carbonate-to-use-for-soapmaking.60798/

This related question says otherwise, at least for potassium carbonate.


text:
Why doesn't Grignard reagent attack alkyl halides
When we add magnesium to
 alkyl halide, Grignard reagent is formed. $$\ce{R-X + Mg -> RMgX } $$ It is highly nucleophilic in nature.  Now at this time the electrophilic alkyl halide is also present in the solution why doesn't grignard reagent simply attack the alkyl halide again ?? $$\ce{ RMgX + R-X  -> R-R + MgX2}$$ Why doesn't this reaction take place ??

The above does contain the answer, although the question itself has a few more points.

Because what happens is mainly exchange between the initial Grignard and the alkyl halide. If you want coupling you need to add Cu(I), even then its not that good.

Alternatively, how is it possible to make a Grignard reagent if coupling is an issue?


text:
Can there be geometrical isomerism in resonating structures
If a double bond which has two different substituents on each of it's two ends is showing resonance with one of the substituents would the double bond exhibit Geometrical isomerism?..if yes how so since the electron cloud is dispersed over them?..Moreover can the distances and relative configurations of the substituents even change because what exists would actually be a hybrid which would not even have a double bond?

Resonating structures always have the same fixed sequence and relative position of nuclei, it is just the electron distribution around them that changes. That analysis must be done on the two isomers separately. But from beginning, as the idea and fact is that they are two different molecules!

I am taking N-methyl-N-phenylacetamide as an example. The amide bond exhibits resonance. The lone pair on nitrogen is in conjugation with carbonyl group. In its resonance structure, it acquires $\ce{C=N}$ around which geometrical isomerism can be exhibited.
In N-methyl-N-phenylacetamide, $\ce{N-C}$ bond undergoes conformational rotation (as shown in pink, figure 1). This gives rise to another conformer. Both the conformers undergo resonance exhibiting different geometrical isomers.
That said, cyclic double bonds (here lactums) with 3-7 atoms in the ring are fixed in the cis or Z structure (figure 3). Even after resonance lactums with 3-7 atoms in the ring do not exhibit geometrical isomerism since the strain in the compound is very high.1

Reference

http://www.siue.edu/~tpatric/Ch%2003%20Stereochem%20H%20T%20I.pdf

I've noticed this in other posts by you, so I thought I should point it out: The space comes after the punctuation mark.

@Martin - マーチン , I will keep it in mind while posting my answers in future .


text:
Why won't this reaction work?
I'm trying to do a reaction with phenyl piperazine, triphosgene, and 4-bromophenol. However, every time I do this reaction I seem to get the carbonyl di-4-phenylpiperazine as a major product (based on the LCMS masses). I tried to reduce the triphosgene step to 5min, which seemed to yield trace amounts of the intended product, but I still can't get my intended product as the major product. This should be such a simple reaction - what am I doing wrong?

Try reversing the order. First, add 1: 1 triphosgene to 4-bromophenol (with stronger base) and then add phenyl piperazine.

You are forming the urea because there is always an excess of phenylpiperazine present to react with the carbamoyl chloride you are forming. You need to add the phenylpiperazine to the triphosgene. The paper by Varjosaari et al. [1] has a procedure (described on Organic Chemistry Portal) that should work.
@Mathew Mahindaratne's suggestion will also work:

Try reversing the order. First, add 1: 1 triphosgene to 4-bromophenol (with stronger base) and then add phenyl piperazine.

References

Varjosaari, S. E.; Suating, P.; Adler, M. J. One-Pot Synthesis of O-Aryl Carbamates. Synthesis 2016, 48 (01), 43–47. https://doi.org/10/gf4wh2.

Ahh yes that makes sense. I’ll retry and report back!

So I redid the reaction but did not use pyridine. I thought it was too messy for a reaction where high yield was not an issue. Instead, I started with triphosgene in DCM (same 1/3 eqv.), then added the piperidine diluted in DCM dropwise to it, all at -10C. I let that stir for a while and added catalytic DMAP, and then bromophenol dissolved in DCM with 2eqv of Et3N. I do see the product in the LCMS, but the yield is probably only around 20%.

Did you have a base (Et3N or similar) mixed with the piperidine or in with the triphosgene?

I used 2 eqv of triethylamine which I added to the bromophenol, I did so because I assumed the puperadine was nucleophic enough for the phosgenation.

You need Et3N in with the piperazine. The reaction with triphosgene produces HCl which will protonate further piperazine as you add it so it will not react.


text:
Will mixing hydrogen and oxygen without a spark result in combustion?
I plan to do the electrolysis of dilute sulphuric acid. This will produce hydrogen gas and oxygen gas at the anodes. I do not plan to collect these gases, but to simply let it form and release into the air without separating them. Would this carry any risk whatsoever, considering that an electric setup is nearby?

How many anodes does your system have ;-) Seriously though, your cell will have an anode and a cathode. I'll let you look up which gas is produced at which electrode. As far as there being "any risk whatsoever" of an explosion we'd have to say yes, though it is small and depends on the construction of your apparatus.

Can you provide some more detail about the power source you'll use? Unless you're using a lot of power, you're unlikely to reach explosive levels of hydrogen in air.

For hydrogen to burn or explode in air, its concentration needs to be at least 4%.
Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing into the air as it goes.
If you're doing this using a battery, or a small line-operated power supply, there's very little danger of explosion.

I'll be using a maximum of 6V from a power supply and I plan to use a 2M solution of sulfuric acid for no more than 10 mins at a time. I suppose this would not create enough Hydrogen to potentially become dangerous.

It's not the voltage that's important, but the current. How much current can the power supply provide? Unless it's in the tens of amps, you'll be fine -- and if it is in the tens of amps, you'll have more trouble from boiling your electrolyte due to excess power. At one amp, you'd be producing a few cc's of hydrogen per minute, far too little to pose any serious risk.

I asked a somewhat similar question a few weeks ago and Poutnik made a smart suggestion that I simply mix the H2 with nitrogen to keep it safely below explosive concentrations, which I set out to do prior to extracting it outside with a fan. So the nitrogen ensure that concentrations are low enough for the fan engine's sparks not to light it on, and the fan extraction avoids both hydrogen and nitrogen accumulation in the building.
You can buy the fans and tubing in stores made for hydroponic gardening. The tubing looks like this:

Also keep in mind that every time you turn an electric switch on or off, a voltage peak occurs which ionises a small amount of air between the contacts and creates a tiny spark, so an electric setup nearby that is turned on / off may be hard to separate from having a "spark", which may be as dangerous as any other sparks if the gaseous mix penetrates the switch area.

Sources
How can I neutralise H2 gas without burning it?
https://www.quora.com/What-causes-sparking-in-the-switches-when-light-is-put-off


text:
How to make 1,1-diphenyl-1-butene from benzophenone and 1-bromopropane?
I would like to know the method of syntheses of making
1,1-diphenyl-1-butene $(\ce{C16H16})$ from benzophenone $(\ce{(C6H5)2CO},$ generally abbreviated $\ce{Ph2CO})$ and 1-bromopropane $(\ce{CH3CH2CH2Br}).$
I think that the carbonyl group which is a functional group composed of a carbon atom double-bonded to an oxygen atom $(\ce{C=O})$ of benzophenone will react to the alkyl halide.
Because oxygen is more electronegative than carbon, carbonyl compounds often have resonance structures which affect their reactivity.  This relative electronegativity draws electron density away from carbon, increasing the bond's polarity, therefore making carbon an electrophile (i.e. slightly positive).
But I have no idea how it reacts and proceeds to work. Can anyone help me understand the detail mechanisms and  reaction procedures to make 1,1-diphenyl-1-butene from benzophenone and 1-bromopropane?

I present another (almost) two-step synthesis, involving the famous Wittig reaction. The first step details the preparation of a phosphonium ylide, which subsequently is allowed to react with the carbonyl compound (benzophenone in our case) to yield the final product, 1,1-diphenyl-1-butene.

For mechanistic details, visit NotEvans.'s answer to Which is the currently accepted mechanism of a Wittig reaction?.

Thank you for useful information about Wittig reaction.

But I can’t understand well about this action procedures.

@Jan Do you mean mechanism?

Try this in that case: https://www.masterorganicchemistry.com/2018/02/06/wittig-reaction/

Thank you so much for your information!  I can understand the mechanism of Wittig reaction that ylide attacks aldehyde or ketone and makes alkene and triphenylphosphine oxyde.

Two steps:

Form the Grignard reagent with 1-bromopropane and magnesium metal. This can be done in a variety of ethereal solvents, THF or $\ce{Et2O}$ are most commonly used. This species is nucleophilic through carbon and will add to the carbonyl group.

Cool the Grignard solution in an ice bath under nitrogen with stirring, slowly add a solution of 0.9 eq of benzophenone. Allow it to stir until TLC shows consumption of the benzophenone starting material.
Add a solution of aqueous ammonium chloride to quench residual Grignard. Add organic solvent $(\ce{Et2O}$ or $\ce{EtOAc})$ and collect the organic phase, dry and concentrate. This will form 1,1-diphenyl-butan-1-ol.

Take the crude product from step 1 and dissolve in toluene. Add a catalytic amount of p-toluene sulfonic acid (PTSA) and heat under Dean-Stark reflux to remove the water. The PTSA will protonate the tertiary alcohol you have formed in step 1. This gives a cation stabilised by the two phenyl groups so it forms readily. This eliminates water to give the required product cleanly.

When the system has ceased to generate water allow it to cool, wash with diluted aqueous sodium carbonate solution, dry and concentrate. This should result in fairly clean 1,1-diphenyl-1-butene.
To produce cleaner material use chromatography on silica with hexane/ethyl acetate (0 to 10%).
Further reading on Grignard reagents: Master Organic Chemistry.

Step 2, why is conc H2SO4 not preferred?

PTSA is more soluble in Toluene and not oxidising

I can understand fairly well about Grignard agent and its detail reaction procedures

Thank so much for your great teaching! I can study from this site organic chemistry fairly well!


text:
Why is the y intercept of a G vs X curve treated as chemical potential?
Whenever I am to prove the fact that the $y$ intercept to the $G$ vs $X$ curve is basically my chemical potential, there are two equations that need to be assumed. One is
$$G=\mu _{A}X_{A}+\mu _{B}X_{B},$$
and the other is the total differential expression of $\mathrm{d}G'$. However they are treated as two independent equations. However, are they not related? What is the  intuition to the geometrical interpretation of the chemical potential?

What have you done so far?  Are you familiar with the Gibbs Duhem equation?

I have done that but I really do not know the intuition behind the equation.

See Smith and van Ness, Introduction to Chemical Engineering Thermodynamics, Chapter 11

Thanks @ChetMiller it was just what I was looking for.


text:
Reaction of Lead(II) Chloride and Ammonia
While searching how to identify cations in a solution I found that ammonia could be used to distinguish between $\ce{AgCl}$ and $\ce{PbCl2}$ precipitates after $\ce{HCl}$ had been added.
Why does the addition of ammonia cause $\ce{PbCl2}$ to dissolve while $\ce{AgCl}$ doesn't? (With chemical equations if possible.)

Are you sure you have this right?  I thought the silver compound was the one that dissolves.

I used to do this lecture demo: add NaCl soln to silver nitrate soln to precipitate AgCl. Then add concentrated ammonia soln to dissolve the AgCl. Then add KBr soln to precipitate AgBr. Then add sodium thiosulfate soln to dissolve the AgBr. Then add KI soln to precipitate AgI. Then add KCN soln to dissolve the AgI. (I skipped the last step in the lecture hall, for safety reasons.) I suggest looking up the various silver complexes than can form. Also, if I remember correctly, there are both silver and lead chloro complexes and HCl was used initially. Hope this helps and +1 to Oscar Lanzi.

@Oscar Lanzi is correct ,Silver ion in ammonia solution forms a soluable complex.


text:
How to derive the conductivity titration curve which accounts for salt formation
I derived an equation that gave me the volume of base required to get a certain $\mathrm{pH}$:
In a titration between a weak acid $\ce{HA}$ and weak base $\ce{B}$ (adding base into acid solution) the following equations apply:
$$
\begin{align}
\ce{HA &<=> H+ + A- & K_{a}&=\frac{[H+][A-]}{[HA]}}\\
\ce{B + H2O & <=> HB+ + OH- & K_{b}&=\frac{[HB+][OH-]}{[B]}}\\
\ce{H2O & <=> H+ + OH- &K_{w}&=[H+][OH-]=10^{-14}} \\
\ce{Deg[A-] &= \frac{K_{a}}{K_{a} +[H+]} &Deg[HB+]&=\frac{K_{b}}{K_{b} + [OH-]}}
\end{align}
$$
These can be substituted into a charge balance and solved for $V_{b}$ (volume of base added):
$$
\begin{align}
\ce{[A-] + [OH-] &= [H+] + [HB+]}\\
\ce{\frac{K_{a}}{K_{a} + [H+]}\times\frac{C_{a}V_{a}}{V_{a} +V_{b}} + \frac{K_{w}}{[H+]} &=[H+] + \frac{K_{b}}{K_{b} + \frac{K_{w}}{[H+]}}\times\frac{C_{a}V_{a}}{V_{a} +V_{b}}}\\
\cdots&\cdots\\
\ce{V_{b} &=\frac{\left([H+] - \frac{K_{w}}{[H+]}-\frac{C_{a}K_{a}}{K_{a} + [H+]}\right)\times V_{a}}{\frac{K_{w}}{[H+]} -[H+] - \frac{K_{b}C_{b}}{K_{b} + \frac{K_{w}}{[H+]}}}}
\end{align}
$$
I used this to generate a pH curve and conductivity curve for the titration using the concentration of the various chemical species in solution:

For calculating the conductivity of the solution:
$$
\kappa_{\rm{soln}}=\lambda_{\ce{A-}}\times\left[\ce{A-}\right] + \lambda_{\ce{HB+}}\times\left[\ce{HB+}\right]+\lambda_{\ce{H+}}\times\left[\ce{H+}\right] + \lambda_{\ce{OH-}}\times\left[\ce{OH-}\right]
$$
Where $\lambda\chi$ is the conductivity constant for chemical species $\chi$
I was told (by a person we shall call $X$) that this model is incorrect as the weak acid/weak base titration does not have a sharp gradient change at the equivalence point but is rather sloped/gently curved due to the following equilibrium:
$$
\ce{HBA <=> HB+ + A-}\tag{$K_{\rm{sp}}=[\ce{HB+}][\ce{A-}]$}
$$
Is there a more mathematical explanation and is it possible to incorporate this equilibrium expression into the titration curve.
Please note that I know that a strong acid/base-weak base/acid conductivity curve does not have a well defined equivalence point (being rather curved), and upon asking $X$, she said that the weak acid/weak base conductivity curve was also curved at equivalence and not as distinctly sharp as my model predicts

It is possible to calculate reasonable curves, but I'm too lazy to sort all this out. I'll ask one question however. Did you account for the volume change? 100 ml of 0.1 molar HCl netralizes 100 ml of 0.1 molar NaOH, but the result is 200 ml of 0.05 NaCl.

Yes I believe I have hence the $V_{a}+V_{b}$ term. It can also be seen in some of the other graph I produced using this model, one such example is in a strong acid-strong base titration, the conductivity curve plateaus when large volumes of base are added, which is a result of volume increase

Ok, I'll nibble on this some more what are the acid and base pK values? Looks like the acid has a pKa of about 3.8. The top of the titration curve is pretty flat so I'm guessing a pKb of 2 or less. So to me the titration curve looks like a weak acid being titrated with a relatively strong base.

All the values are at the left of the image of the graph, but you can change them ass appropriate and it's not relevant to what I'm asking. I would like for the values of ka and kb to be variable rather than set values and i don't see how setting them to a certain value helps.

The values are on the very left in the first column. So you can see that ka and kb for both is 0.0002 which means it is a weak acid-weak base titration. But just to clarify, please note that my question is asking for an alternate derivation which includes the equilibrium described in OP.

If $Q_\ce{sp}=\ce{[HB+][A-]}$ through titration don't exceed $K_\ce{sp}$ value, don't will be need incorporate the salt formation equilibrium and the curve don't be affected.

@GRSousaJr but if I did how would I incorporate it into the question, that is the basis of my question

@EdV the pdf was not all that helpful from what I saw, is there a specific page or section you were referring to. Regarding the conductivity curve and the flat line, it is not necessarily horizontal (this depends on the $K_{\rm{b}}$ value) after equivalence and curves very very slightly.
In the hypothetical case where $\ce{HBA}$ is completely insoluble I believe the conductivity would decrease to 0 at equivalence and then graduall increase (and plateau)

@EdV nonetheless even if it is indeed correct, I was looking for a titration curve which incorporated the said equilibrium into consideration. While I know that the overall qualitative shape is correct, the purpose of this question is for a quantitative answer. On another note, I wanted a more mathematically grounded reason as to why it is sharp at equivalence rather than just a figure from a PDF from an online source. It may be the case that the source uses the same model as I do (which is a very standard model used by many) and the actual experimental results may differ. Thank you tho!

I believe have found a answer.
  It's valid mathematically, make sense physically but I don't know if chemically is true.
  I posted to community appreciation.
  There we go!

The reactions

ionization of weak acid:
$$\ce{HA + H2O <=> H3O+ + A-}\qquad K_\ce{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}
\label{eq:KWeakAcid}$$
ionization of weak base:
$$\ce{B + H2O <=> BH+ + OH-}\qquad K_\ce{b}=\frac{\ce{[HB+][OH-]}}{\ce{[B]}}
\tag{2}
\label{eq:KWeakBase}$$
self-ionization of water:
$$\ce{2 H2O <=> H3O + OH-}\qquad K_\ce{w}=\ce{[H3O+][OH-]}
\tag{3}
\label{eq:KWater}$$
equilibrium of poorly soluble salt:
$$\ce{HBA <=> HB+ + A-}\qquad K_\ce{sp}=\ce{[HB+][A-]}
\tag{4}
\label{eq:KSalt}$$

Charge balance
$$\ce{[H3O+] + [HB+] = [OH-] + [A-]}\tag{5}\label{eq:ChangeBalance}$$

Mass balance
To mass balance we can use an expression based on amount of substance:

Weak acid:

\begin{align}
{n_\ce{HA}}_\text{total}&={n_\ce{HA}}_\text{solution}+{n_\ce{A-}}_\text{solution}+{n_\ce{A-}}_\text{ppt}\\
C_\ce{HA}V_\ce{HA}&=\left(\ce{[HA] + [A-]}\right)V_\text{solution}+{n_\ce{A-}}_\text{ppt}\\
\\
\ce{[HA] + [A-]}&=\frac{C_\ce{HA}V_\ce{HA}-{n_\ce{A-}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{6}
\label{eq:AcidMassBalance}\\
\end{align}

Weak base:

\begin{align}
{n_\ce{B}}_\text{total}&={n_\ce{B}}_\text{solution}+{n_\ce{HB+}}_\text{solution}+{n_\ce{HB+}}_\text{ppt}\\
C_\ce{B}V_\ce{B}&=\left(\ce{[B] + [HB+]}\right)V_\text{solution}+{n_\ce{HB+}}_\text{ppt}\\
\\
\ce{[B] + [HB+]}&=\frac{C_\ce{B}V_\ce{B}-{n_\ce{HB+}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{7}
\label{eq:BaseMassBalance}\\
\end{align}

The general equation
Combining ($\ref{eq:KWeakAcid}$) and ($\ref{eq:AcidMassBalance}$) equations we have:
$$\ce{[A-]}=\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}\frac{C_\ce{HA}V_\ce{HA}-{n_\ce{A-}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{8}
\label{eq:Anion}$$
Combining ($\ref{eq:KWeakBase}$), ($\ref{eq:KWater}$) and ($\ref{eq:BaseMassBalance}$) equations we have:
$$\ce{[HB+]}=\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}\frac{C_\ce{B}V_\ce{B}-
{n_\ce{HB+}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{9}
\label{eq:Cation}$$
From ($\ref{eq:KSalt}$) equation we have ${n_\ce{HB+}}_\text{ppt}={n_\ce{A-}}_\text{ppt}=
n_\ce{HBA}=n_\text{ppt}=n$ and replacing ($\ref{eq:KWater}$), ($\ref{eq:Anion}$) and
($\ref{eq:Cation}$) equations in ($\ref{eq:ChangeBalance}$) equation we have:
$$\ce{[H3O+]} + \frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}
\frac{C_\ce{B}V_\ce{B}-n}{V_\ce{HA}+V_\ce{B}}
= \frac{K_\ce{w}}{\ce{[H3O+]}} + \frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}
\frac{C_\ce{HA}V_\ce{HA}-n}{V_\ce{HA}+V_\ce{B}}
\tag{10}
\label{eq:TitrationCurve}$$
Besides that combining the ($\ref{eq:KSalt}$), ($\ref{eq:Anion}$) and ($\ref{eq:Cation}$) equations we have:
$$K_\ce{sp}=\overbrace{
\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}\frac{C_\ce{B}V_\ce{B}-n}{V_\ce{HA}+V_\ce{B}}
}^{\ce{[HB+]}}
\overbrace{
\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}\frac{C_\ce{HA}V_\ce{HA}-n}{V_\ce{HA}+V_\ce{B}}
}^{\ce{[A-]}}
\tag{11}
\label{eq:Kppt}$$

Reproducing your data
If we ignore the poorly soluble salt formation, the ($\ref{eq:TitrationCurve}$)
equation is exactly the same which you deduced.
From your image we have
$V_\ce{HA} = \pu{1.00 L}$,
$C_{A} = \pu{0.01 mol L^{-1}}$,
$K_\ce{a} = \pu{2.0E-4}$,
$C_\ce{B} = \pu{0.20 mol L^{-1}}$,
$K_\ce{b} = \pu{2.0E-4}$, and
$K_\ce{w} = \pu{1.0E-14}$
for which the titration curve would be:

The plot of the concentrations of $\ce{[HB+]}$ and $\ce{[A-]}$ through titration would be:

Using the conductivity equation of the solution
$$\kappa_\ce{solution} = \lambda_\ce{A-}\ce{[A-]} + \lambda_\ce{HB+}\ce{[HB+]}
+ \lambda_\ce{H3O+}\ce{[H3O+]} + \lambda_\ce{OH-}\ce{[OH-]}
\tag{12}
\label{eq:ConductivitySolution}$$
and
$\lambda_\ce{A-} = \pu{ 90.0E-4 m^2 S mol^{-1}}$,
$\lambda_\ce{HB+} = \pu{172.0E-4 m^2 S mol^{-1}}$,
$\lambda_\ce{H3O+} = \pu{849.6E-4 m^2 S mol^{-1}}$, and
$\lambda_\ce{OH-} = \pu{398.0E-4 m^2 S mol^{-1}}$
from your image is possible to plot the conductivity curve:

If we look to contribution of each species to conductivity curve
we can understand its final shape.

Some considerations
To take account the formation of $\ce{HBA}$, suppose that $K_\ce{sp} = \pu{7.5E-5}$.
The $Q_\ce{sp}$ plot through titration would be:

While $Q_\ce{sp}<K_\ce{sp}$ we have $n=0$, because $\ce{[HB+][A-]}$ is less than
critical concentrations to start precipitation and the titration curve follows as usual.
When $Q_\ce{sp}=K_\ce{sp}$ we have the precipitation eminence and $n=0$.
When $Q_\ce{sp}>K_\ce{sp}$ I'm not sure what will happen.
The $V_\text{critical}$ in which $Q_\ce{sp}=K_\ce{sp}$ can be found graphical or analytical way.
The last way in not so easy, but I believe then can be done.
Personally I will use the graphical way.
Solving ($\ref{eq:TitrationCurve}$) equation to $V_\ce{B}$ (with $n=0$) we can write:
$$V_\ce{B}=\frac{V_\ce{HA}\left(\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}-\ce{[H3O+] +
C_\ce{HA}\alpha_\ce{A-}}\right)}{C_\ce{B}\alpha_\ce{HB+} -
\left(\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}-\ce{[H3O+]}\right)}
\tag{13}
\label{eq:Vtit}$$
Again, the ($\ref{eq:Vtit}$) equation is exactly the same which you deduced.
Solving ($\ref{eq:Kppt}$) to $V_\ce{B}$ (with $n=0$) we have:
$$V_\ce{B}=\displaystyle\frac{-\left[V_\ce{A}\left(2 -
\displaystyle\frac{C_\ce{HA}\alpha_\ce{A-}C_\ce{B}\alpha_\ce{HB+}}{K_\ce{sp}}\right)\right]-
\sqrt{\left[V_\ce{A}\left(2 - \displaystyle\frac{C_\ce{HA}\alpha_\ce{A-}C_\ce{B}\alpha_\ce{HB+}}
{K_\ce{sp}}\right)\right]^2 - 4V_\ce{HA}^2}}{2}
\tag{14}
\label{eq:Vcritical}$$
where
$${\alpha}_\ce{HB+}=\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}$$
and
$${\alpha}_\ce{A-}=\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}$$
The $V_\ce{B}\ vs.\ \ce{pH}$ plot to ($\ref{eq:Vtit}$) and ($\ref{eq:Vcritical}$) equations
allows to find $V$ and $\ce{pH}$ for which the curves cut across each other.

Another (graphical) way could be the $Q_\ce{sp}$ plot.
Any way the values found are $\pu{45.20 mL}$ and $4.685$ respectively.
What happens with $\ce{pH}$ when $V_\ce{B}$ is greater than $V_\text{critical}$?
Some hypothesis can been proposed.

Hypothesis #1

$\ce{HBA}$ will precipitate and the titration curve don't be affected.

In this case the ($\ref{eq:Kppt}$) equation can be solved to $n$:
$$n=\displaystyle\frac{\left(n_\ce{HA}+n_\ce{B}\right)-
\sqrt{\left(-\left(n_\ce{HA}+n_\ce{B}\right)\right)^2 -
4\left(n_\ce{HA}n_\ce{B}-
\displaystyle\frac{K_\ce{sp}\left(V_\ce{HA}+V_\ce{B}\right)^2}
{{\alpha}_\ce{HB+}{\alpha}_\ce{A-}}\right)}}{2}
\tag{15}
\label{eq:nppt}$$
where $n_\ce{HA} = C_\ce{HA}V_\ce{HA}$ and $n_\ce{B} = C_\ce{B}V_\ce{B}$.
The ($\ref{eq:nppt}$) equation allows calculate the amount of precipitate is formed
given a $V_\ce{B}$ and $\ce{pH}$ value. But given a $V_\ce{B}$, what is $\ce{pH}$ value?
I believe that, if this hypothesis is true, the $\ce{pH}$ value can be calculated
from ($\ref{eq:TitrationCurve}$) equation and the titration curve don't be affected.
The the $n\ vs.\ V_\ce{B}$ plot of ($\ref{eq:nppt}$) equation is

and the $n\ vs.\ \ce{pH}$ plot of ($\ref{eq:nppt}$) equation is

The effect of precipitation can be seen in plots below.
The $\ce{[HB+]}$ and $\ce{[A-]}$ plot

The $Q_\ce{sp}$ plot

The plot of contribution of each species to conductivity

The total conductivity plot

Hypothesis #2

$\ce{HBA}$ don't precipitates and the $\ce{pH}$ rises.

In this case the $V_\ce{B}\ vs.\ \ce{pH}$ plot gives some clue.
The ($\ref{eq:Kppt}$) equation can be rearranged to give:
$$\ce{[H3O+]}^2 - \ce{[H3O+]}\left(
\frac{K_\ce{a}C_\ce{HA}V_\ce{HA}C_\ce{B}V_\ce{B}}
{K_\ce{sp}\left(V_\ce{HA}+V_\ce{B}\right)^2} - K_\ce{a} -
\frac{K_\ce{w}}{K_\ce{b}}\right) +
\frac{K_\ce{a}K_\ce{w}}{K_\ce{b}}=0
\tag{16}
\label{eq:TitrationCurvePpt}$$
Note that, using one of the roots from ($\ref{eq:TitrationCurvePpt}$) equation,
the $\ce{pH}$ jumps from $4.685$ to $9.315$ (see titration curve below) when
$V_\ce{B}=V_\text{critical}$ and, while $V_\ce{B}$ increases, the titration curve
is look like as usual.
Mathematically in this hypothesis no precipitate is formed, but the
solution cross the region where there should be precipitate.
The consequence this is the concentrations of $\ce{[A-]}$ and $\ce{[HB+]}$,
which depends on $\ce{pH}$ values, jumps too.

Hypothesis #3

$\ce{HBA}$ don't precipitates and the $\ce{pH}$ falls.

Again the $V_\ce{B}\ vs.\ \ce{pH}$ plot gives some clue.
Using the other root from ($\ref{eq:TitrationCurvePpt}$) equation, the titration curve
have a discontinuity at $V_\text{critical}$ and then the $\ce{pH}$ values falls
while $V_\ce{B}$ increases. No jumps, neither on concentrations or pH, are observed.

If hypothesis #3 is true, the plot of the concentrations of $\ce{[HB+]}$ and $\ce{[A-]}$
through titration is

the $Q_\ce{sp}$ plot is

the conductivity curve to each species

and finally the conductivity curve is

Addendum
To those interested follows the gnuplot code to produce these graphs. All equations needed to reproduce this answer are available.
The present code creates several graphs used in this answer but no all. Some of them are combination of two or more data sets or a graph from an equation available.

#
#  Implements a lot of equations and create a couple of graphs showing what
#  happens with pH, concentrations, conductivity, and Qsp when a poorly soluble
#  salt is formed during a titration.
#
#  To more details see the original question/answer
#
#  Question: https://chemistry.stackexchange.com/q/117984/80491
#  Answer:   https://chemistry.stackexchange.com/a/118760/80491
#
#  Author: GRSousaJr (https://chemistry.stackexchange.com/users/80491/grsousajr)
#  gnuplot 5.2 patchlevel 7
#

# ========== INITIAL CONFIGURATION ===============================================
reset               # return all graph-related options their default values
set encoding utf8   # selects a character encoding

# Save each graph as pdf file
set terminal pdfcairo size 8 in, 6 in font "Times New Roman,18" enhanced

# Colors of lines
set style line 1 lc "#e41a1c"   # red
set style line 2 lc "#377eb8"   # blue
set style line 3 lc "#4daf4a"   # green
set style line 4 lc "#984ea3"   # purple

# Margins
set lmargin screen 0.110    # left margin
set tmargin screen 0.940    # top margin
set bmargin screen 0.130    # bottom margin

set grid ls -1 lc "gray"    # style for grid
set tics out nomirror       # tics marks on axis
set samples 1000            # number of points to each curve
set dummy pH                # "pH" as "x" for all plot-commands
unset key                   # turn-off the legend (key) exhibition

# ========== INPUT PARAMETERS ====================================================
# ---------- Weak acid (HA) ------------------------------------------------------
VA    = 1.0     # L, HA volume (mandatory)
CA0   = 0.01    # mol/L, HA concentration (mandatory)
pKa   = 3.69897 # Ka = 0.00020 (mandatory)

# ----------- Weak Base (B) ------------------------------------------------------
CB0   = 0.2     # mol/L, B concentration (mandatory)
pKb   = 3.69897 # Kb = 0.00020 (mandatory)
VBmax = 120.0   # mL, B volume (optional)

# ----------- Poor soluble salt (HBA) --------------------------------------------
pptn  = "True"  # "True/False" (mandatory) indicates if occurs precipitation
Ksp   = 7.5E-5  # HBA poor soluble salt (mandatory if pptn = "True")

# ----------- Solvent ------------------------------------------------------------
pKw   = 14.0    # H2O self-ionization

# ----------- Conductivity constants ---------------------------------------------
array lambda[4]
lambda[1] =  90.00 # (E-4) A-
lambda[2] = 172.00 # (E-4) HB+
lambda[3] = 849.60 # (E-4) H+
lambda[4] = 398.00 # (E-4) OH-

# ========== DERIVATE PARAMETERS =================================================
# ---------- Equilibrium constants -----------------------------------------------
Ka     = 10**(-pKa)     # Ka from pKa
Kb     = 10**(-pKb)     # Kb from pKa
Kw     = 10**(-pKw)     # Kw from pKw
# ---------- Logical test to VBmax------------------------------------------------
if (!exists("VBmax")){
    VBmax = 2.0*VA*CA0/CB0  # Double that required for neutralization
} else{
    VBmax = VBmax/1000.0    # Converting mL to L
}

# ========== FUNCTIONS ===========================================================
# ---------- H+ and OH- concentrations (as a function of pH) ---------------------
H(pH)  = 10**(-pH)      # H+ from pH
OH(pH) = 10**(pH-14.0)  # OH- from pH

# ---------- Ionization fractions (as a function of pH) --------------------------
aA(pH)  = Ka/(H(pH) + Ka)       # A-
aB(pH)  = Kb/(Kw/H(pH) + Kb)    # HB+

# ---------- B volume (as a function of pH) --------------------------------------
VB(pH) = (VA*(Kw/H(pH) - H(pH) + aA(pH)*CA0))/(aB(pH)*CB0 - (Kw/H(pH) - H(pH)))

# ---------- Amount of substance (as a function of pH) ---------------------------
nA       = CA0*VA         # HA
nB(pH)   = CB0*VB(pH)  # B

# Amount of precipitate formed
nppt(pH) = ((nA + nB(pH)) - sqrt((-(nA + nB(pH)))**2 - \
    4*(nA*nB(pH) - (Ksp*(VA + VB(pH))**2/(aB(pH)*aA(pH))))))/2.0

# Logical test to know if precipitation must be taken account
if (pptn eq "True" ){
    n(pH) = nppt(pH) <= 0 ? 0 : nppt(pH)  # Disregards negative values of 'n'.
    sufix = "_with_ppt"
} else {
    n(pH) = 0   # No occurs precipitation
    sufix = "_without_ppt"
}

# ---------- A- and HB+ concentrations (as a function of pH) ---------------------
CA(pH) = aA(pH)*(nA - n(pH))/(VA + VB(pH))       # [A-]
CB(pH) = aB(pH)*(nB(pH) - n(pH))/(VA + VB(pH))   # [HB+]

# ---------- Qsp (as a function of pH) -------------------------------------------
Qsp(pH) = CB(pH)*CA(pH)

# ---------- Individual and Total conductivities (as a function of pH) -----------
kappaA(pH)   = lambda[1]*CA(pH)
kappaB(pH)   = lambda[2]*CB(pH)
kappaH(pH)   = lambda[3]*H(pH)
kappaOH(pH)  = lambda[4]*OH(pH)
kappaSol(pH) = kappaA(pH) + kappaB(pH) + kappaH(pH) + kappaOH(pH)

# ---------- Initial and final pH ------------------------------------------------
pH0   = -log10((-Ka + sqrt(Ka**2 + 4*CA0*Ka))/2.0)    # Initial pH
pHmax = pKw + log10((CA0*2*VA - CA0*VA)/(VA + VBmax)) # Estimate for pHmax at VBmax

#---------- Finding pHmax by iteration -------------------------------------------
while (VB(pHmax) > VBmax || VB(pHmax) < 0){
    pHmax = pHmax - 0.0001
}

# ========== CREATING THE GRAPHS =================================================
# ---------- Titration curve -----------------------------------------------------
filename = "Titration_Curve" . sufix # Define a filename to pdf file

# Plotting using a data block (like a temp file)
set table $Titration_Data
    plot [pH0:pHmax] 1000*VB(pH)
unset table

set xlabel "Volume of titrant / mL" # name for x-axis (the same to all plot-commands)
set ylabel "pH"                     # name for y-axis
set xrange [0:VBmax*1000]           # VBmax in mL
set yrange [0:14]                   # pH 0 to 14
set xtics 10                        # tics at each 10 mL
set ytics 1                         # tics at each 1 pH unit

set output filename . ".pdf"        # Create the pdf file

plot $Titration_Data using 2:1 with lines linestyle 1 linewidth 2

# ---------- Concentrations curves ------------------------------------------------
filename = "Concentrations_Curve" . sufix

set table $ConcB_Data
    plot [pH0:pHmax] CB(pH)
unset table

set table $ConcA_Data
    plot [pH0:pHmax] CA(pH)
unset table

set ylabel "Concentration / mol L^{–1}"
set yrange [0:CA0]
set ytics .001
set format y "%.3f" # number with 3 decimal places

# Labels
set label 1 "{/:Bold [HB^{+}]}" at 20.0, 0.003 tc ls 1 font ',20' offset 0,0.75
set label 2 "{/:Bold [A^{–}]}"  at 05.0, 0.003 tc ls 2 font ',20' offset 0,0.75

set output filename . ".pdf"

plot \
    $ConcB_Data u (1000*VB($1)):2 w l ls 1 lw 2 ,\
    $ConcA_Data u (1000*VB($1)):2 w l ls 2 lw 2

unset for [i=1:2] label i   # Clean the labels
unset format                # Clean the format

# ---------- Individual conductivity curves --------------------------------------
filename = "Individual_Conductivity_Curve" . sufix

set table $ConducA_Data
    plot [pH0:pHmax] kappaA(pH)
unset table

set table $ConducB_Data
    plot [pH0:pHmax] kappaB(pH)
unset table

set table $ConducH_Data
    plot [pH0:pHmax] kappaH(pH)
unset table

set table $ConducOH_Data
    plot [pH0:pHmax] kappaOH(pH)
unset table

set ylabel "{/:Italic κ} / S m^{–1}"
set yrange [0:2]
set ytics .2
set format y "%.2f"

set label 1 "{/=24 {/:Italic κ}_{/=12 A^{–}}}"      at 95, 0.70 center tc ls 1
set label 2 "{/=24 {/:Italic κ}_{/=12 HB^{+}}}"     at 95, 1.50 center tc ls 2
set label 3 "{/=24 {/:Italic κ}_{/=12 H_{3}O^{+}}}" at 07, 0.90 center tc ls 3
set label 4 "{/=24 {/:Italic κ}_{/=12 OH^{–}}}"     at 95, 0.15 center tc ls 4

set output filename . ".pdf"

plot \
    $ConducA_Data  u (1000*VB($1)):2 w l ls 1 lw 2 ,\
    $ConducB_Data  u (1000*VB($1)):2 w l ls 2 lw 2 ,\
    $ConducH_Data  u (1000*VB($1)):2 w l ls 3 lw 2 ,\
    $ConducOH_Data u (1000*VB($1)):2 w l ls 4 lw 2

unset for [i=1:4] label i
unset format

# ---------- Total conductivity curves -------------------------------------------
filename = "Total_Conductivity_Curve" . sufix

set table $ConducSolution_Data
    plot [pH0:pHmax] kappaSol(pH)
unset table

set yrange [0.75:3]
set ytics .25
set format y "%.2f"

set output filename . ".pdf"

plot $ConducSolution_Data u (1000*VB($1)):2 w l ls 1 lw 2
unset format

# ---------- Qsp curves ----------------------------------------------------------
filename = "Qsp_Curve" . sufix

set table $Qsp_Data
    plot [pH0:pHmax] Qsp(pH)
unset table

set yrange [0:0.000100]
set ytics 0.00001
set ylabel "{/:Italic Q}_{sp}"
set format y "%h" # scientific notation format

set label 1 "{/:Bold {/:Italic Q}_{sp}}"\
    at 32.0, 3.0E-5 tc ls 1 font ",20" offset 0,0.75
set label 2 "{/:Italic Q}_{sp} = {/:Italic K}_{sp}" \
    at 70.0, Ksp    tc ls 1 font ",20" offset 0,0.75

set output filename . ".pdf"

plot $Qsp_Data u (1000*VB($1)):2 w l ls 1 lw 2

What software did you use to plot the graphs? It looks really clean.

@Blaise I used gnuplot.

@GRSousaJr when $Q_{sp}>K_{sp}$ precipitation is favoured and so the conductivity should decrease to some extent. So when adding more base to the solution, although it may form more $\ce{HB+}$ ions, the extent to which its concentration increases would be limited and so shouldn't it be slightly concave down after equivalence point in the conductivity curve as opposed to concave up in your diagram? Is there a possible explanation for this or can someone point out my misunderstanding?

Would the conductivity curve differ for hypothesis 1?

@sabhoque When $Q_\ce{sp}  > K_\ce{sp}$ the chemical intuition say that there will be precipitation, and the amount of precipitate will be calculated by (15) equation.
I'm still working on understanding of this equation and its consequeces on titration and conductivity curves.
How soon as possible I'll complete the answer with some conclusions.

Strangely enough you have accounted for the loss of conductivity due to the formation of precipitate when you were calculating $\ce{[A-]}$ and $\ce{[HB+]}$. When you say: "In this case, the plot of the concentrations of [HB+] and [A−] through titration is" do you mean for hypothesis 1, 2 or 3?

With regard to hypothesis 1, I believe you are correct in saying the pH value can be calculated from (10) equation and the titration curve don't be affected because the precipitation reaction does not change the concentration of $\ce{H+}$ (as I am aware of). It is mainly the conductivity curve I am struggling to get a grasp of. I will test out your derivations and everything after my upcoming exams

@sabhoque I'm refer to hypothesis #3.  I made some edits on answer to clarify. Additionally, I included some graphs on hypothesis #1.

@GRSousaJr if you don't mind could you please send the GNUplot code that you used to render the titration curves? I am unsure of how to graph implicit functions in gnuplot

I will if you'll give me a time to clean up the code.
It's really a mess.
Meanwhile you can search by gnuplot on SO.

@sabhoque I included the gnuplot code to create almost them graphs. I hope it be useful.


text:
Structure and oxidation state of bridging complexes
I am confused about the structure of bridging complexes with the bridging ligands as $\ce{OH-}$ or $\ce{NH2-}.$ Aren't they both monodentate ligands?
And if they form a simple coordinate bond, then the negative charge on them must be lost, but then the oxidation states of the metal atoms doesn't match. For example:

In this compound, please explain what bonds $\ce{OH}$ and $\ce{NH2}$ are actually forming, and the charges on them. Also, how do we find the oxidation state of platinum?

The typical way of dealing with these is to treat it as an X ligand for one metal and an L ligand for the other. You should alternate so that each metal see one L and one X ligand in the bis-bridged case.


text:
Is scopolamine same as scopolamine hydrobromide?
Is scopolamine extracted with methanol or acetone from Datura or Brugmansia scopolamine free base or scopolamine hydrobromide?
If you take a look at the Wikipedia page for hyoscine, the Synonyms part lists scopolamine hydrobromide as a synonym for hyoscine even though neither molecular structure nor chemical formula contains bromine.

Is HBr involved anywhere in the process?

Scopolamine, not scolopamine. Please note it is a terrible idea to use this drug recreationally. Do yourself a favor and satisfy your curiosity in other ways.

As indicated by you, there is a form called "fee base", and an other "hydrobromide".  In the later case, one basic functional group (e.g., amine) was protonated (e.g., $\ce{|NR2 -> N+HR2 + Br-}$.  This may alter where, how, and how quickly the active ingredient enters the body (anabolism) and may influence the kinetics how it leaves the organism again (catabolism).  Hence, both chemically, as well as from medicinal perspective, treat them as "different".  If not prescribed by a physician, though, do not use such drugs for recreation.

I dont have the drug and I have no intent to use it. Asking just out of curiosity. The drug is called The devils breath and rumor is that it can zombify you by disabling free will or something. It is extracted from datura or Brugmansia as scopolamine. But some websites say the devils breath is Hyoscine ( scopolamine) hydrobromide. Do you have such information?

If the extraction procedure does not involve HBr, then I would doubt it is the hydrobromide because:

hydrobromide salts don't dissolve well in acetone;
hydrobromide salts are rare in nature.

According to the work by Bracci et al. [1], the major component of MeOH extraction of Brugmansia is atropine.
Śramska et al. [2] used ethyl acetate extraction and isolated scopolamine and atropine.
No mention of hydrobromide salts in either.
References

Bracci, A.; Daza-Losada, M.; Aguilar, M.; De Feo, V.; Miñarro, J.; Rodríguez-Arias, M. A Methanol Extract of Brugmansia Arborea Affects the Reinforcing and Motor Effects of Morphine and Cocaine in Mice. Evidence-Based Complementary and Alternative Medicine 2013, 2013, 1–7. https://doi.org/10/f96zpc.
Śramska, P.; Maciejka, A.; Topolewska, A.; Stepnowski, P.; Haliński, Ł. P. Isolation of Atropine and Scopolamine from Plant Material Using Liquid-Liquid Extraction and EXtrelut® Columns. J. Chromatogr. B Analyt. Technol. Biomed. Life Sci. 2017, 1043, 202–208. https://doi.org/10/f9wvv8.

Any information on whether extracted scolopamine is the devils breath? By the what I think extraction is acetone. PubChem says hyoscine is soluable in acetone and doesnt mention acetone in atropine or hyoscyamine

This article says it is https://www.newsweek.com/does-devils-breath-drug-really-exist-332465 but until I see the LCMS and NMR of the extracted fractions I don't regard it as definitive. Atropine will be soluble in acetone

I am confused. If you take a look at the Wikipedia page for hyoscine. In synonyms part it says scopolamine hydrobromide is a synonym for hyoscine even though neither moleculer structure nor Chemical Formula contains bromine?

Hyoscine is the commercial name for the medical product containing scopolamine that is used to treat motion sickness etc. It is sometimes formulated as the hydrobromide to make it more water soluble and easier to take.


text:
Why anhydrous copper sulphate is acidic in nature but pentahydrate copper sulphate is amphoteric?
$\ce{CuSO4}$ is acidic salt because it is made up of $\ce{Cu(OH)2}$ (weak base) and $\ce{H2SO4}$ (strong acid). But why $\ce{[CuSO4.4H2O]H2O}$ is amphoteric?

How is the pentahydrate amphoteric?

Before you ask "why", ask "if".

Hint: you may substantiate your question by addition of experimental evidence.  Did you dissolve the same quantity (referring to moles) of either $\ce{CuSO4}$ or its pentahydrate in the same quantity of water and determine the pH value of these solution with pH electrodes?  Maybe you have better equipment at disposition, but let us assume the typical error of such a determination is less or equal to 0.05 pH units, how many pH units the two solutions differ from each other?

The formula weight of $\ce{CuSO4}$ is 159.609 g/mol, that of $\ce{CuSO4 . 5 H2O}$ is 249.685 g/mol and the molecular weight of $\ce{H2O}$ is 18.015 g/mol. So suppose you have two 500 mL beakers and you put 200.0 mL of water in each one. To the first beaker, add 15.9609 g of $\ce{CuSO4}$ and 9.008 g of water. Stir until all the copper sulfate has dissolved. The resulting blue solution has some pH, whatever it is. Now to the second beaker, add 24.9685 g of $\ce{CuSO4 . 5 H2O}$. Stir until all the copper sulfate pentahydrate is dissolved. The two solutions have the same intensity of blue color, because their $\ce{Cu^2+}$ concentrations are the same, and their pH values are the same. So, like @Oscar Lanzi says, how is the pentahydrate amphoteric?

Exactly. The original poster seems to have a major misconception regarding the concept of pH in water.


text:
Lewis structure of CN radical
I need to draw the Lewis formula for $\ce{CN}$. However, I am not sure how to draw the single free electron.
My idea would be as follows:
$$\Large\ce{\llap{·}C#N\rlap{:}}$$
Does that make sense?

This is an unusual question. Generally, we refer to cyanide as the $\ce{CN-}$ anion.

That's what I thought too. I wasn't able to find anything related online.

@JohnyDow Well yes, there is indeed the cyanido radical, e.g. https://en.wikipedia.org/wiki/Cyano_radical

It has been made in the lab using $\ce{CH3CN + N2 -> CN + ...}$: http://articles.adsabs.harvard.edu//full/1995A%26A...304L...5K/L000006.000.html

Yes, this makes sense.  For radicals, the lewis structure usually gives the single electron to the element with the least electronegativity, while trying to make a structure that is as close as possible to maintaining the octet rule.  This can involve resonance, as seen in nitrogen dioxide.  Finally, the total valence electrons in this structure add up to the total number of valence electrons that should exist in the cyanido radical (4 from carbon and 5 from nitrogen makes 9).

You drew the radical, which is in normal cases very unstable.
The cyanide ion has an ion pair instead of the single electron at the carbon atom.


text:
Why the term 'unified' in "unified mass unit"?
Why different unit symbols - amu, u and dalton are used for representing atomic mass? And what does the term 'unified' mean in "unified mass unit"?
Thank you.

This site typically gives answers that can be given in a paragraph or two.  the history of units standardization is long and involves arbitrary changes. read the Wikipedia article https://en.wikipedia.org/wiki/Atomic_mass_unit.

Related: Units of mass on the atomic scale

Your question is historically important. Nobody has the energy to write an answer which would require several pages of an article. In short, it is a can of worms. For almost 5-6 decades chemists and physicists used a different scale for the atomic masses, so the atomic masses in the publications/books used by physicists did not exactly match what the chemists used for their own publications. Note that it is just not a matter of academic egos, the entire chemical industry is dependent on atomic weights (for calculations and trade). Money matters. This is why both the governments and the scientific community wanted to sort this out.
Both parties were not willing to budge because that would have affected their millions of pages of research. However a compromise was reached to use a common scale by two three sensible men i.e., they proposed to choose C-12 in the 1960s instead of oxygen. This process sort of "unified" the community of physicists and chemists and it did not affect the values of either party by much.
More history is given here, The AME2003 atomic mass evaluation (I). Evaluation of input data, adjustment procedures, Nuclear Physics A 729 (2003) 129–336
There are scores of other articles from the 1960-70s on the choice of C-12. Use Google Scholar. The original work is in French I guess, I did not check
T.P. Kohman, J.H.E. Mattauch and A.H. Wapstra, J. de Chimie Physique 55 (1958)
393.

Actually C-12 isn't the gold standard anymore. In 2018 the kilogram and Avogadro's number were redefined as fundamental constants. So the number of moles of C-12 in 12 grams of C-12 is now an experimental value. It will be very close to 1, but not 1 exactly.

@MaxW Actually, the SI revision was proposed in 2018 but didn’t take effect until 20 May 2019.


text:
Mechanism of 5‐(iodomethyl)oxolan‐2‐one synthesis from 4‐pentenoic acid and iodine in acetonitrile
There is a question in my organic chemistry textbook with an answer. What it doesn't explain (and neither does my theory textbook, I checked) is the mechanism involved in the reaction. The reaction is as follows:

Can someone help me?

If the reaction was simply iodine adding to a double bond do you know how it works?

Well, yes. A Iodonium ion is formed, then another iodine attacks the ion from the other side, is that not right?

Well, as product suggests, something different attacked...

@ClaudioLancia Well I- is not a great nucleophile, and internal nucleophilic attack goes better than external nucleophilic attack. Carboxylate may not be a great nucleophile but it is internal, and the formation of 5-membered rings is favoured. See where this is going?

Ok, so does the carboxylate form the ring before the Iodine attack?

https://en.wikipedia.org/wiki/Iodolactonization

Oh, I get it now! Thanks a lot for the help. I didn't know the name of the reaction (especially since the exercise was in the alkene section, and my book didn't mention anything about Halolactonization), so this is a great help!

@ClaudioLancia Now that you understand, could you write up your own answer to summarize what you learned?

https://en.wikipedia.org/wiki/Iodolactonization
I+ reacts with the CC-π electron donor. A cyclic iodonium three-membered ring intermediate is formed. The carboxylate group opens the electrophilic intermediate to form the product lactone. 5-membered rings (gamma-lactones) https://en.wikipedia.org/wiki/Lactone are prefered due to optimum entropic and enthalpic parameters of activation.

Uh ... Why the -1?

@OscarLanzi Well, it's not my downvote, but there are many bad details - links to improve, weird wording/terminology, e.g. it's simply cyclic iodonium not "epi-iodium" etc.

In this case, the regioselectivity (to form the 5-membered instead of the 6-membered ring) also has a stereoelectronic factor: https://chemistry.stackexchange.com/q/106846/16683


text:
Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?
While writing this answer to the question Transmitter receiver coil separation for Electromagnetic Terrain Conductivity Measurement I ran across this large PDF file of a book Soil and Environmental Analysis; Physical Methods Soil and Environmental Analysis; Physical Methods, 2nd Ed. K. A. Smith and C. E. Mulllins, Eds. 2000 Dekker, New York.
Figure 5. on page 29 is shown below. It shows a plot of the real and imaginary part of the permittivity of water in solid (ice) and liquid forms.
The shapes are the roughly the same; the real part is around 80 to 90 at low frequency and drops to perhaps 2 to 1 at high frequency, with the imaginary part reaching a peak as the slope of the real part is maximum as one would expect.
The difference however is that the peak in the absorption is around 3 kHz for ice and 20 GHz for liquid!

Is the physics behind this strong absorption peak the same in both cases?
If so, what is it about the transition from solid to liquid that can move it seven orders of magnitude?

Wow, great question! Maybe this paper will shed some light: J.H. Jiang, D.L. Wu, Ice and water permittivities for millmeter and sub-millimeter remote sensing applications, Atmos. Sci. Let., 5 (2004) 146-151. The authors are/were at JPL at CalTech. The paper is not paywalled. I hope to read (and understand) a good answer here!

@EdV Wow that looks quite helpful for the GHz line at least (radar doesn't usually go down to kHz); I'll give it a read, thanks! It's open access here and also posted at JPL here

@uhoh From https://physics.nist.gov/cgi-bin/MolSpec/mole.pl?prefix=tri&molecule=H2O one may find examples about water and absorptions, typically GHz though (and not kHz)  -- only accessing their references will tell if it was liquid, or solid water probed.

Please note: This is a relaxation spectrum. The transitions do not come from a resonance (absorption, like in NMR, IR, etc.), but an applied electric field has a lossy (at low frequencies) or elastic (at higher f.) effect on some mode of molecular orientation of electrical dipoles in your sample.

https://en.wikipedia.org/wiki/Dielectric_spectroscopy

@Karl reading now...

@Karl still stuck, so I've asked separately; Difference between a relaxation spectrum and an absorption line spectrum?

I am not an expert in this, but IMHO, it is quite obvious the imaginary parts have their local maxima near the middle of the transition of the real parts. It is the transition region where dipole angular motion is still partially able to follow the frequency, but with difficulties and with a phase shift between external intensity and overall dipole polarization. So the question rather is, why is such big frequency difference for the permittivity drop for water and ice, with the probable answer involving motion restictions in solids.

@Poutnik yes, the question has two parts; 1) "Does water really have..." and 2) "Why the huge shift?" The first part requires some authoritative confirmation that the illustration I've shown (it can't be actual data, at best it's some simple parameterization) and the second part is your "the question rather is". I'm going to implement some question re-bountying because... enquiring minds want to know! :-)

Intuitively I know the answer, bit it is not authoritative.


text:
Methocel Hydration
Need help understanding impact of temperature and content of water/glycol on strength  (crush, attrition) of methocel (methyl cellulose) extrudates.
In a given process, methocel is mixed at 150oF at 2wt% with a sand-water mixture (50:50) and 2% glycol. The mixture is then dried to extrudable moisture (20-30%) and extruded into spaghetti-like pills.
Recently the pill strength has been poor, with high level of dusting.
Is the temperature of dispersion too low (150oF)?
Does methocel need to ‘hydrate’ to provide strength?
Does the mixture need residence time and/or low temperature (<100oF) to hydrate? What is the ideal temperature?
Does the methocel loose strength if it dehydrates? Is the process reversible?
At what temperature does methocel dehydration take place?
Does high level of glycol prevent methocel hydration?
Can methocel be extruded with glycol only and still provide strength?
Does presence of certain contaminants (electrolytes, organic acids) prevent hydration?
What is the ideal temperature of extrusion?
Thank you!

For the customs of ChemSE, the question is rather broad; perhaps even too broad to provide you specific compositions, times and temperatures to improve your process of producing extrudates.
On the other hand, you already collected a number of plausible parameters influencing the target property, which you named as «pill strength».  As general picture, you are heading to a systematic analysis -- preferably by design of experiments.

Aim to reformulate «pill strength» by a quality you may quantify easily like a scale, e.g. temperature, mass.  This could be the Young modulus, which may be defined for either compression, or traction.  For one batch of your extrudates, determine this «monitor property», to determine how well this property may be determined (arithmetical mean value $\mu$, standard deviation $\sigma$).  Determine in advance where you like to see your process -- sometimes, it is not a single property describing your optimum.
Assuming the Young modulus were the sole criterion to describe the optimum, you may determine in advance how many experiments $N$ per factor level are to be performed to recognize a relevant effect $\Delta\mu$ with a known standard deviation $\sigma$ by
$$ N \approx 60 \left(\frac{\sigma}{\Delta\mu}\right)^2$$
and with a screening with $m$ combinations in a screening, each of them would be $n$ times realized
$$ n = \frac{N}{m} $$
Because of the number of parameters already compiled, it is useful to screen their influence on «pill strength» in a two-level factor experiment.  Think of them like a light switch, either «low level / off», or «high level / on».  This applies to (i) temperature of dispersion, (ii) residence time and temperature in the extruder, (iii) high / low level of glycol perhaps preventing the methocel hydration.  Run batches of extrusion with these parameter levels, and determine «pill strength» accordingly.
This allows you to determine which of the parameters influence most the target property, and subsequently focus the optimization on a lesser number of them.

There are many publications about this large subject, both general textbooks (e.g., Box, Hunter, Hunter), in chemistry (e.g., ref, ref, ref, or ref), or entries on SECrossValidated (e.g., here).


text:
Synthetic approach for precursor 1,1'-(bromomethylene)bis(1H-pyrazole) in ligand synthesis
Relevance
During my Bachelor's project I focused on the synthesis of iron(II) complexes comprising a tetrapodal pentadentate ligand from the ligand family $\text{PY5R}_2\ (\text{R = H, Me, OH, OMe})$. [1-4] This ligand consists of one axial pyridine unit and four equatorial pyridine units. I investigated the activity of these complexes in the oxidation of alkenes by hydrogen peroxide.
While doing literature research I discovered several means of tuning the properties and chemical reactivity of these complexes.
For example, the cone angle for steric accessibility is larger for ligands with five-membered pyrazolyl rings compared to those with similarly substituted six-membered pyridyl rings. See the ligand $\text{pz}_{4}\text{lut}$. [5]
Furthermore, one could also influence the donor-strength of the ligand by replacing the axial pyridine unit by a $N$-heterocyclic carbene (NHC) unit. See the ligand $\text{PY4Im}$. [6]
So I thought, why not combine both features? This leads to a tetrapodal pentadentate ligand with an axial NHC unit and four equatorial pyrazole units, abbreviated as $\text{pz}_{4}\text{Im}$.
Figure 1. Tetrapodal pentadentate ligands.

By modifying the procedure employed in the synthesis of $\text{PY4Im}$ [6], the synthesis of the iron complex comprising $\text{pz}_{4}\text{Im}$ would look like this
Figure 2. Synthetic approach towards $[(
\text{pz}_{4}\text{Im})\text{Fe(MeCN)}](\text{PF}_{6})_{2}$.

Problem
My actual problem is related to the starting material.
In the synthesis of $\text{PY4Im}$ [6], 2,2'-(bromomethylene)bis(pyridine) is used as starting material. It is known how to synthesize this compound.
In this synthesis, that of $\text{pz}_{4}\text{Im}$, 1,1'-(bromomethylene)bis(1$H$-pyrazole) I would choose as starting material. However, I cannot find any literature on the synthesis of this particular compound.
Figure 3. 1,1'-(bromomethylene)bis(1$H$-pyrazole) (left) and 2,2'-(bromomethylene)bis(pyridine) (right) - starting materials.

Question
So my question is: Does anyone know a way to synthesize the starting material 1,1'-(bromomethylene)bis(1$H$-pyrazole) for the synthesis of the iron(II) complex comprising the pentadentate NHC-pyrazole ligand $\text{pz}_{4}\text{Im}$?
References

Eur. J. Org. Chem. 2004, 21, 4319-4322.
Tet. Lett. 2012, 53, 1, 54-55.
ChemSusChem. 2016, 9, 10, 1178-1186.
Inorg. Chem. 2006, 45, 15, 6048-6055.
Inorg. Chem. 2008, 47, 17,7468-7470.
Inorg. Chem. 2010, 49, 23, 11223-11230.

I am sorry if my english is not perfect but your question does not make any sense to me!

@SteffX How to synthesize the precursor 1,1'-(bromomethylene)bis(1H-pyrazole) (figure 3, left).

These are my thoughts on this as I don't have database access to check the reactions. 1-Chloromethylpyrazole is commercially available so I would start with this. Reaction with pyrazole anion will give the 1,1'-methylene bispyrazole. I would try a radical bromination of this to access the desired material

@Waylander I've been thinking about that too, but would this also work?
https://imgur.com/9jIUDjC

looks an entirely reasonable route to try


text:
Nucleophilicity of amines

I think the bridge head nitrogens won't react.The answer is B C A. Can someone help me, please

What type of reaction is it? What are the products? How will the two reactants approach to make the new bond? Please edit to share how far you got in making and argument. Why do you think the bridge nitrogens won't react.

Am not sure. I thought it was a nucleophillic reaction.  Can you please help me with what type of reaction it is and why bridge head amine is more reactive.

@Karsten Theis , this problem is from MS Chowhan , in chapter amines and their reactivity. This book is generally used by students preparing for IIT entrance exam in India.

I'm voting to close this question as off-topic because it lacks context. Could you edit the question to show what the nucleophile and what the electrophile is, what mechanism you expect and what products you anticipate?

Is this the reaction: https://en.wikipedia.org/wiki/Menshutkin_reaction

https://www.masterorganicchemistry.com/2018/05/07/nucleophilicity-of-amines/

@ Nivedita , In the current form ,your question may be closed.SE expects you to include your thoughts regarding your question in your post.

@Nivedita  the bridgehead amines do react. Consider which amines have most steric hindrance

A chemical reaction (here an SN2 reaction) is just zillions of molecules hitting each others in a certain direction and with a certain kinetic energy. If the angle is right and the energy is right, then the reaction can occur. With A, there is a large chance that the ethyl groups (which are free to move) just get in the way, but not for B and C (imagine them like A has its hands cuffed in its back). However, I have always been told that C is more nucleophilic than B.


text:
How does sodium carbonate raise pH?
Sodium carbonate is used in pools to raise pH. From my understanding, it would dissociate in water like so:
$$\ce{Na2CO3 <=> 2 Na+ + CO3^2-}$$
To raise pH, doesn't it have to absorb hydrogen ion/s? If so then wouldn't it just form carbonic acid that would then just dissociate again?

It's all about equlibrium. Try to write the corresponding equation(s) for hydrolysis and see what dissociates better, sodium hydroxide or hydrocarbonate. Also, the term "to absorb" looks weird in this context.

This whole thing is silly. An aqueous solution of sodium carbonate has basic pH because carbonate ions hydrolize to yield hydroxide ions plus bicarbonate ions. Even just starting with sodium bicarbonate (baking soda), the pH is basic: around 8.3 for reasonable concentrations. So why is this  bountied question at all?

@EdV In similar tune to @Maurice, I equally wonder about the bounty set.  The answer by @LilBluey basically adds $K_b$'s, «comment like».  I set a flag for the moderators, maybe then @Haha Hahaha adds a comment why the bounty was set.

@Buttonwood I think you can use the notification @hahaha to 'contact' the user setting the bounty. (It will not autofill though.) I believe this is possible because s..he is an editor of the post, see help center, but please don't put a gun to my head if that information is not accurate. You can only notify one (additional) person per comment though.

@Martin-マーチン Well, meanwhile ... I identified comment (below https://chemistry.stackexchange.com/questions/127027/why-doesnt-nitrogen-react-with-oxygen-at-room-temperature/127036#comment244243_127036) of @HahaHahah which offered an autofill where I asked him / her about the motivation for this bounty. We will see «what happenz».  Thank you.

Sodium carbonate indeed dissociates in water : $$\ce{Na2CO3 <=> 2 Na+ + CO3^2-}$$ But that is just beginning, as in solutions with carbonates, bicarbonates or dissolved carbon dioxide happen multiple chemical equilibriums:
( See a lot of info at Carbonic acid on Wikipedia):
Carbonate anion partially hydrolyzes in water: $$\ce{CO3^2- + H+ <=> HCO3-}, \mathrm{p}K_\mathrm{a2}=10.33$$
or
$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$
Consumption of $\ce{H+}$ and production of $\ce{OH-}$ is complemented by water autodissociation $\ce{H2O <=> H+ + OH-}$ with $\mathrm{p}K_\mathrm{w}=14$
As a result, $\mathrm{pH}$ gets more alkalic.
Another hydrolytic step, that occurs significantly only in acidic and about neutral solutions is:
$$\ce{HCO3- + H+ <=> H2CO3}$$
respectively
$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$
with $\mathrm{p}K^{*}_\mathrm{a1}=6.35$, $\mathrm{p}K_\mathrm{a1}=3.6$.
$\mathrm{p}K^{*}_\mathrm{a1}$ pretends all carbon dioxide converts to carbonic acid.
$\mathrm{p}K_\mathrm{a1}$ considers the equilibrium of formation of carbonic acid.
And finally:
$$\ce{H2CO3 <=> CO2(aq) + H2O}$$
$$\ce{CO3(aq) <=> CO2(g)}$$

You have the equilibrium in water: $$\ce{H2O <=> H+ + OH-}$$
The protons are removed from the equilibrium: $$\ce{H+ + CO3^2- <=> HCO3-}$$
Now you have more $\ce{OH-}$ in the solution which increases the pH.

How can we do more than repeat what Deuti said in July '19 ? Could Haha Haha explain us what he wants and what he does not understand ? Where is the problem ?

The reaction system can be described as:
$\ce{H2O <=> H+ + OH-}$
$\ce{H+ + CO3^{2-} <=> HCO3-}, \mathrm{p}K_\mathrm{a2}=10.33$
Note, the net of the first two reactions imply a rise in pH in the presence of carbonate. Further, with time and carbon dioxide exposure:
$\ce{H2O + CO2 <=> HCO3- + H+}$
$\ce{H2CO3 <=> H2O + CO2}$
And, with the introduction of bicarbonate, the $\ce{Na2CO3}{/}$NaHCO3 equilibrium apparently progresses to a pKa of 10.33, which makes it a poor pH 9 buffer.
The reaction system, upon reaching equilibrium, has been succintly presented as:
$\ce{H2CO3 <=> HCO3- + H+ <=> H2O + CO2(g)}$
which could move to the right with warming.
[EDIT] A more quantitative (and useful) illustrative tool of the pH effect is given, in logarithmic form, as the Henderson-Hasselbalch Equation, see (4) in this source, presented below:
$\ce{pH = pK  + log([CO_2]/[HCO3^-])}$
However, the above assumes an equilibrium situation, and in the case of sodium carbonate added to a swimming pool with continuing exposure to air containing carbon dioxide, in time perhaps, such an equilibrium could be established measuring at the same temperature.


text:
Redox reactions redefined
I saw in my textbook (NCERT class 11 part2 chemistry) that oxidation is defined as

gain of oxygen or more electronegative element by a substance.

After it was redefined and said as

oxidation is loss of electron(s) by any species.

Again it is being redefined and stated as

oxidation process is visualised as a decrease in electron density.

I can't understand why they are changing the definitions again and again. I mean what things they are taking into consideration, each time they define the term

Well, oxidation and reduction don't necessarily take place in defined substances, but it can simply involve ions in a solution, for instance (I am thinking about a simple Daniell cell here). Here the definition involving loss or gain of elements would fail, but that of electrons would be correct.

IUPAC definition
The IUPAC definition is the following:

The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').
An increase in the oxidation number of any atom within any substrate.
Gain of oxygen and/or loss of hydrogen of an organic substrate.

All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation
  of an organic substrate that can be rationally dissected into steps or primitive changes
  . The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons
  or hydroxide ions, or by nucleophilic
  substitution by water or its reverse and/or by an intramolecular molecular
  rearrangement
  . This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation
  transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1.

So that is the official definition, and it also gives multiple rules (and admits that the third rule does not always work). In order to use the second definition, you also need to look up their definition of oxidation state.
Definitions in your textbook

I can't understand why they are changing the definitions again and again.

They are changing the definition to apply the concept to more and more reactions.
In the explanation of the IUPAC definition I posted above, they mention "the original idea of oxidation (combination with oxygen)". That's where your first textbook definition comes from. So the following reaction is an oxidation because there is a reaction with dioxygen:
$$\ce{4Fe^2+(aq) + O2(g) + 4H+(aq) -> 4Fe^3+(aq) + 2 H2O}$$
If I want to call the similar reaction with fluorine an oxidation as well, I have to generalize the definition to loss of electrons (your second definition). So this would be an oxidation of iron(II) as well:
$$\ce{2Fe^2+(aq) + F2(g)  -> 2Fe^3+(aq) + 2 F-}$$
Finally, if I want to generalize further to reactions where I want to discuss loss of electrons in atoms that are part of a covalent molecule, I need some accounting scheme to assign valence electrons to individual atoms (this is called oxidation number or oxidation state). Once that is in place, I can discuss reactions such as the following in terms of oxidation (and reduction):
$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$
Here, your first definition does not apply (no oxygen) and your second definition is difficult to apply (how do we know which species lost electrons - they are all neutral, and they all have different composition?). So that is where you need some rule similar to your third rule ("decrease in electron density"). Usually, though, this is stated using oxidation states so that you are back to gaining or losing an integral number of electrons (i.e. oxidation state changed by +1 rather than "the electron density decreased slightly).
At this point, it becomes a bit fuzzy and there will be situations where the oxidation state is ambiguous. You gain clarity in the context of electrochemistry when you separate reactions into half reactions in electrochemical cells, and can actually measure the transfer of electrons that pass through the wire connecting the two cells.

Why they are changing the classical definition to make it hold in some more reactions. I think instead of redefining the older ones, new terms should have been introduced, which can be applied to the reactions such as reaction of iron with fluorine, etc.

@user36956 Maybe you are right, but science is not top down, with one entity making decisions. Instead, new words start being used, or old words are used in new ways. I wish Lewis acid were called something else without the term acid in it, but it is too late to change that.

Hmmmm.... It makes sense!

So, if i consider any of the first two IUPAC definitions, am i right? The only difference is that i am allowed to use any of the two with which i feel easier with. Hmm?

For $\ce{Fe^2+ vs Fe^3+}$, you can use either. For methane vs carbondioxide, for example, you need to use the second definition. @user36956

So, at last i am right if i consider that the definitions of redox reactions were broadened to cover more type of reactions. I could say that any reaction involving addition of oxygen is oxidation. If it follows this, it surely follows the other two definitions. But if any reaction is judged by change in oxidation numbers, it is no sure that it follow the other two. So, at last it should be considered that the 'oxidation number' involving definition is the updated one and must be  followed instead of the other two.

These statements rather clumsily explain oxidation, but none of them are really "wrong".
If you want a single solid definition of oxidation, you can consider it as an increase in the oxidation number. To calculate the oxidation number for a given atom, you calculate the number of more electronegative atoms bounded to it minus the number of less electronegative atoms bounded to it.
Oxidation is typically (in nature, at least) done by forming a new bond to oxygen, since it is abundant in the air and rather reactive. Because oxygen is normally (again, in nature) more electronegative than whatever it is bounded to, it drains electronic density, which is equivalent to a "loss" of electrons. However, in chemistry, the situation is a bit more complex.
Different functional groups will influence the electronic density. For example, a carbon O-bonded to an ester will likely be more positively charged than if it were O-bonded to an alcohol. The reason is that the oxygen in the alcohol is relatively electron rich, which the ester oxygen delocalizes part of its electronic density in the carbonyl, making it poorer.
In other words, the oxidation number is a simple and robust way of defining oxidation. But to fully understand the reactivity of molecules, you need to "see" the electronic density.

This is a bit of a circular argument. The oxidation number is an accounting scheme to keep track of oxidation and reduction. So of course oxidation numbers go up when a species undergoes oxidation.

As per IUPAC gold book , reduction and oxidation are defined as follows$^1$ $^2$.

A reduced species can be formed also through the gain of electrons (either at the negative electrode in a cell or through transfer from another atom, ion or group of atoms in a chemical reaction).
Oxidation consist in removal of one or several electrons from the substrate.

The above equation can be divided as oxidation half reaction and reduction half reaction.

In the reduction half , Silver in $+1$ oxidation state gains $2$ electrons to form $\ce{Ag}$. This is reduction.

$$\ce{2Ag^+(aq)  -> 2 Ag(S}$$

In the oxidation half , Copper in $0$ oxidation state looses $2$ electrons to form $\ce{Cu^2+}$.

$$\ce{Cu(S) -> Cu^2+(aq)}$$
In , organic chemistry , Ethyne conversion to Ethene is reduction. Further Ethene conversion to Ethane is also reduction.

Further , the following series are also taken to be reduction and oxidation (in organic chemistry).

References
$^1$ IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
$^2$ IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


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Calculating bond dissociation enthalpy of F-F bond

$$\ce{N2 + 3 F2 -> 2 NF3} \quad ΔH = \pu{-263 kJ mol-1}$$
The bond dissociation enthalpies of the nitrogen triple bond and nitrogen–flourine bond are $\pu{946 kJ mol-1}$ and $\pu{272 kJ mol-1}.$
  Find the bond dissociation enthalpy of the $\ce{F-F}$ bond.
A. $\pu{-423 kJ mol-1}$
B. $\pu{-393 kJ mol-1}$
C. $\pu{-141 kJ mol-1}$
D. $\pu{+141 kJ mol-1}$
E. $\pu{+423 kJ mol-1}$

I know that since bond dissociation enthalpy is endothermic it has to be one of the last two.
I tried substracting bond dissociation enthalpies of products from reactants and equating that to the enthalpy change but my answer is nowhere near the given ones.
Is my method wrong?

If you can't wrap your head around getting the answer directly, just calculate the enthalpy of reaction for the two bond dissociation energies you are considering.
Assume bond dissociation energy is 141 kJ/mol
$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{141 kJ/mol} - 6 \cdot \pu{272 kJ/mol} =  \text{negative value}$$
Assume bond dissociation energy is 423 kJ/mol
$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{423 kJ/mol} - 6 \cdot \pu{272 kJ/mol} = \text{positive value}$$
So it can't be 423 kJ/mol because the reaction is exothermic. If you complete the math for a bond energy of 141 kJ/mol, the numerical answer is -263 kJ/mol for the reaction enthalpy.
What if you actually wanted to estimate a bond energy?
In this case, you could call the bond energy $x$, and solve for it:
$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot x - 6 \cdot \pu{272 kJ/mol} = \pu{-263 kJ/mol}$$
Solving for $x$ would give you the bond energy.


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How can I know the composition of a gas pocket entrapped in a plastic sheet?
I have a 1 mm thick plastic sheet, the plastic is not opaque.
Some gas pockets, I would say some bubbles, are inside the sheet; the size of the bubbles is about 1 mm.
What methods are available to know the composition of the gas that is inside the pockets?
Methods can destroy the sample.

Gas chromatography, possibly even GC/MS. Sample a bubble using a small syringe and put the sample through a GC/MS set up to identify common gases.


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Is the rate of electrolysis the same throughout as long as voltage is same
In an electrolytic cell, where the electrolysis of dilute sulfuric acid is taking place, would the rate of electrolysis be constant always, even at the beginning?
By rate of electrolysis, I mean the rate of gas production or the rate of charge delivery. And would this rate be established as soon as I turn on the voltage, or does it slowly increase to a constant level?

No, no and no. // The rate depends on area of electrodes. So for a given voltage, bigger electrodes have a higher reaction rate. // The instantaneous rate is high because there is no double barrier. // Immediately after the cell starts, a double layer forms and the reaction rate lowers and becomes controlled by diffusion to the electrodes from the bulk solution.

Furthermore, if the compounds to be electrolyzed is a salt like CuSO4, copper metal will slowly gets deposited on the cathode. So the concentration of copper will slowly decrease. It means that the rate of copper deposition decreases with the time. It will even tend to zero when there is no copper any more in solution.

The rate of electrolysis (decomposition of water) is defined by the current that passes. If you apply a constant voltage, factors arise as suggested in the comment by MaxW.
At constant current, the rate of hydrogen (and oxygen) evolution would be somewhat different initially because on a microscopic level, hydrogen atoms have to cover the surface to some extent to form H2; H atoms may reduce some oxides on the electrode surface, and H2 has to agglomerate to form bubbles (and dissolve in water to saturate it with H2). And then the bubbles have to rise to the surface.


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What is non-hückel double bond of Aromaticity?
Chemistry of Aromaticity. What's non hückel double bond ?
What are the difference between hückel and non-hückel double bond ?
[ Here, in figure the green one is non hückel double bond ]

There is no intrinsic difference. Hückel's rule counts the number of pi electrons in a closed ring. It says nothing about other double bonds.

In the given molecule Pyrene , the π  bond shown in green is in cross conjugation (shown in green in figure 1). This bond( figure 2) is excluded from interaction with rest of conjugated system. Hence , the central $\ce{C=C}$ isn't hosted by the largest p orbital loop, so it's not counted towards Hückel's rule for aromaticity.
The outer periphery (shown in blue in figure 3), is a loop of $\ce{14 π}$ electrons , therefore aromatic. This may be possible hückel double bond you are referring to.
The π bond in cross-conjugation (shown in green) maybe non-hückel double bond.

Beware: The central pi bond does not count towards the Hückel rule, however that does NOT mean that bond is not part of the aromatic system!


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Why do we use Ḡ ("G bar") as a notation for partial molar free energy?
For partial molar entropy there is a bar that is positioned above the $G$ with a subscript I that is sometimes added to the $G~(\bar{G}).$ The bar I read sometimes denote the "partial nature of the $G$".
However, is it not rhetorical. I am looking for the explanation as to what is so partial about this quantity with regard to it's occurrence in the equation.

The bar denotes we took the derivative with respect to composition. Molar free energy would just be the free energy divided by "n" the composition. For a pure component these are the same, but not for a mixture.
Not everyone uses a bar also...Like all notation, you have to pay attention with each new article you read, what they mean.

Thermodynamic properties are functions of composition, which can be represented as $(\chi _1,...,\chi _N)$ where $\chi _i$ is the mole fraction of component $i$ present in the system. Now imagine a change in the composition such that the amount of component $i$ is increased by a very small quantity $dn _i$ (alternately, imagine the system to be infinitely large), and a thermodynamic property $X$ is changed by $dX$. Then (using the notation you describe) we define the partial molar property as $$\bar{X}=\left(\frac{\partial X}{\partial n_i}\right)_{p,T,n_j}$$ This notation means that we make a limiting, infinitely small change in $n_i$, such that the change in $X$ is for all practical purposes nil, ie $\Delta X=0$. The partial molar quantity describes the slope of the property $X$ (which can be represented as a surface in a multidimensional manifold) along the composition coordinate $n_i$, at a specific composition $(\chi _1,...,\chi _N)$.


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Number of optically active compounds among the products of ozonolysis

The number of optically active compound(s) obtained upon complete ozonolysis of
  the following optically active compound is

In the question, I broke all the double bonds and added a $\ce{-CHO}$ group according to the question (ozonolysis).
Checking for the optically active compounds I found two of them:

But the solution has given only one compound! Can you please help me with this?

Ozonolysis of the given compound gives $\ce{I}$ , $\ce{II}$ , $\ce{III}$ and $\ce{IV}$.

$\ce{I}$ and $\ce{II}$ are Homomers and have same configuration "S" and therefore identical (as shown in the figure).
$\ce{III}$ and $\ce{IV}$ are achiral . Therefore , the total number of optically
active compounds formed after ozonolysis is $\ce{= 1}$.

The given answer is correct. The product scheme you drew is correct as well. However, as I marked in your scheme (see below), products A and B are essentially the same enanthiomer (both have (2S)-configuration). The compounds you have drown in right-hand side are also an identical compound, which is not optically active. Therefore, ozonolysis has given only one optically active compound.


text:
What would be the concentration of products in this calcium producing system?
I wanted to figure out a way to make calcium metal with potassium oxide. $\ce{K2O}$ undergoes disproportionation at $\pu{350-430 °C}:$
$$\ce{2 K2O -> K2O2 + 2 K}$$
Now, if I mixed in $\ce{CaO}$ (1:2 molar ratio with $\ce{K2O})$ it would react with the potassium formed in the first reaction, forming calcium metal and $\ce{K2O}$ that would feed back into the first reaction. Eventually I should be left with mostly $\ce{K2O2}$ and calcium metal, but since these two react together, this can't be as simple.
Reaction of these two would lead to the same mixture as I started with. Is there a way to figure out what the concentration of Ca will be when the reaction reaches equilibrium?

If i could reduce the pressure enough to lower the melting point of K2O2 below 430 degrees, the formation of calcium metal would be favoured since it would float to the top, correct?

T l, Dr:  despite what you see from the aqueous electromotive series, potassium does not reduce calcium oxide in this setting.
You may be assuming that potassium, which lies above calcium in the electromotive series for aqueous solutions, therefore lies above calcium in all other settings regardless of chemical environment or temperature.  It doesn't.  The electromotive series we see can change in different environments and at different temperatures.
For instance, in this Ellingham-Richardson diagram you see calcium oxidation at a very low oxygen potential, corresponding to a great tendency towards oxidation and thus a very high position for calcium in the "electromotive series" for anhydrous oxide formation.  Where is potassium?  Much higher up in free energy (starting around -650 on the free energy scale), corresponding to a less stable oxide and thus a lower tendency for potassium oxidation.  In the absence of water solvent and at elevated temperatures (or even without elevated temperature but still in the absence of water), potassium does not displace calcium, or even manganese, from oxide formation.  The order is very different from the aqueous one.
Apart from lithium, all of the alkali metals are notoriously poor at oxide formation, compared with what we expect from aqueous solutions.  Look again at the diagram; sodium isn't so hot compared with calcium, magnesium or aluminum either.

Calcium is made industrially by either electrolysis or reduction of calcium oxide with aluminum.  In the latter reaction aluminum still forms a less stable oxide than calcium, but unlike potassium it can form ternary calcium-aluminum oxides that are still more stable than just calcium oxide.  Thus part of the calcium from the lime is converted to these ternary compounds while the rest, not fitting in the ternary compounds, comes off as vapor.

Terrific answer (+1)! And a higher resolution version (easier on my eyes!) of the Ellingham-Richardson figure is available at http://web.mit.edu/2.813/www/readings/Ellingham_diagrams.pdf.

Could potassium reduce a non-oxide, like CaCO3 or Ca(OH)2, despite sitting lower in the ellingham diagram?

@Francis changing the anion could change the series again, but hydroxides and carbonates probably get you nowhere since these compounds with calcium decompose to oxides upon heating.  HSAB suggests that chlorides or heavier halides could work, but these are more amenable to an electrolytic route.


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Boiling chips for concentrating sulfuric acid
I am concentrating sulphuric acid. What should I use as boiling chips? I have tried broken mug pieces, the solution became cloudy, and I have tried broken glass (retrieved from a drinking class, not crystal) and the solution became yellow.

Clean, non-dyed glass won't react

The_Vinz If that is true then what could have caused the solution to turn yellow?

I have no idea, probably residual dirt or dye? You could try with another piece of glass. Also, since I suppose that you did it in a glass container, maybe the container itself was dirty?

Anyway, unless you are doing this in a proper lab setting, note that this is a nasty procedure: the acid has to be heated a lot, and it can make nasty fumes. Some time ago I had to vacuum distill sulfuric acid in the lab, and I can't avoid suggesting great care

The_Vinz Yes I have been treating the acid with the utmost respect. The beaker was clean and so were the glass pieces. Keep in mind when I boiled the acid using porcelain chips and the same beaker, the solution did not turn yellow but instead cloudy. I guess I should try again with another source of glass? Or is the glass most definitely not the problem here?

A better question is why? It's probably safer, more accurate, and more cost effective to buy concentrated acid and dilute it.

@Zhe "Science isn't about why, it's about why not." :)

You could break the beaker and use the shards as boiling chips - if you have another beaker.

@Karsten Theis Ehy, Pasteur pipettes exist for one reason! :P

Clean porcelain or glass shouldn't react with concentrated sulfuric acid.
For boiling chips you want to use homogeneous material with a substantial/developed surface:

Porous unglazed ceramic.
Raschig rings, which are also widely used industrially for the similar distillation purposes, are probably the ideal choice.
An alternative for a consumer market would be porous ceramic rings for aquarium filters, which is basically the same thing (stick with the white ones).

The majority of glass types (borosilicate glass, quartz, …).
Broken pieces of chemical glassware or light bulbs would do.
For higher efficiency, use glass capillaries or sintered glass pieces.

Potential cause of cloudy solution when you used mug pieces might be the glaze on top of porcelain or the porcelain itself if it hasn't been properly thermally treated (remaining binder), and, of course, porcelain dust particles leftovers may float around forming a suspension.
Yellow color of sulfuric acid when you used glass pieces is likely an indicator that acid came in contact with organics (e.g. oil, grease).
Pure colorless glass won't cause this coloration.
On a side note, there is no need to use grease at all as sulfuric acid itself is quite viscous and can serve as a sealant for the joints (if you are using any).
Notes, advises

If the liquid has been cooled down, make sure to add at least one new chip before heating again to the boiling point (prevents superheating).
Avoid using boiling chips for vacuum distillation.
Instead, use mechanical stirring or bubble inert gas (argon, nitrogen) through a submerged glass capillary.
Boiling chips don't work under vacuum.
Never add boiling chips to a hot solution as it will result in rapid vaporization and the liquid is going to boil over or bump.
Keeping a substantial amount of sodium carbonate nearby might come in handy in neutralizing spilled acid, if there is any.

Why should one avoid using boiling chips for vacuum distillation?

@Hans Because boiling chips are pretty much useless under vacuum.