text:
What provides the energy allowing membrane proteins to facilitate diffusion across the membrane?
We know that facilitated diffusion requires no energy.  But the transport proteins, when bind to a molecule from ECF (extra-cellular face) , the proteins rotate and deliver the molecule into ICF (intra-cellular face). Now my question is that--- Which force provides with the torque to rotate and how it works with out energy?

[...]how it works with out energy!

The transport proteins can change conformation in a stochastic manner. Even if the conformation opening to the extra-cellular side is a bit different in energy than the one opening to the intra-cellular side, there is no net change in energy/entropy/Gibbs energy in a complete cycle. The only thing that changes is that the cargo molecules are on the other side, a process for which the change in Gibbs energy is negative.
This argument is the same for any catalyst. If the catalyst is recovered in the same state as it was at the beginning of the reaction (and it has to be, otherwise it would not be a catalyst), the catalyst does not change the thermodynamic quantities of reactants or products. As a consequence, it speeds up the reaction, but does not influence the equilibrium constant.
If the catalyst has a high activation energy to do what it needs to do, then it is not an efficient catalyst. After all, the role of the catalyst is to lower the activation energy. So the transport proteins must be able to change from one to the other conformation both with cargo and without cargo in a facile manner. Otherwise, they could not go through the cycle of 1) picking up cargo 2) switching conformation 3) releasing cargo 4) switching conformation again.

Which force provides with the torque to rotate [...]

Random forces associated with thermal energy are responsible for the conformational change. The binding constant for cargo has to be moderate so the the cargo dissociates easily. One the other hand, the binding constant for non-cargo has to be so low that nothing other than cargo binds.

As another answer correctly points out, the redistribution of energy is enough to keep a protein in constant random motion, including motions that result in transport of a solute from the outside to the inside of the cell by attaching temporarily to the protein. A passive diffusion process does not require bond breaking/formation as is usual in a directed process.
Passive diffusion depends on a concentration gradient. So unless the concentration (or, more accurately, activity) of the diffusing solute is lower inside the cell, there will not be net transport into the cell. The point is not that facilitated diffusion does not require energy — it does — just not directed motion (work) based on active chemical modification of the protein or its environment. Random thermal motion (energy) will do.
This is really a rewording of what another answer explains, namely that for net passive diffusion into the cell to occur, the chemical potential (or free energy) of the solute inside the cell must be lower than that outside. The chemical potential gradient results from some other process that does not involve the protein (it was created perhaps by removing or exhausting the solute inside of the cell, or by increasing the concentration of solute outside the cell). The origin of the potential energy gradient is not a result of chemical modifications (work) at or around the protein itself. Instead it is only a result of the solute concentration difference, and it is therefore we refer to the process as "passive" diffusion.


text:
Is it possible to electroplate from a gallium anode?
I have been looking at some papers on electro-deposition of iron gallium alloys and these all use solutions made up of Fe and Ga sulphates.
I am just curious as to why perhaps an anode made of one of the materials might not have been used as part of the process? I mention gallium specifically because the cost of the Ga sulphates is quite high, while the Ga metal is relatively cheap.
My impression is that it is typically difficult to electro-deposit gallium, but I think this was in specific reference to plating from solutions of the Ga salts, I have not yet come across plating from a Ga anode, and I'm curious about this.

Electroplating is a still an art rather than science. Too many magical additives are required to get the desired results as you can see here in the Table for electrodeposition og Ga-Fe alloys https://www.sciencedirect.com/science/article/pii/S0924424714005214. Ga is perhaps too soft and two low of a melting point (30 C). Like aluminum it is prone to form insoluble hydroxides during anodic dissolution, if the pH is just not right.

@M.Farooq thank you for your comment, I got that impression for electroplating more art than science. The paper you reference is based of a couple earlier works where they got the recipe for the electrolyte. I guess it's more often based on experience what they choose you use as additives.

Yes, it is possible to electrodeposit gallium. The difficulties are: efficiency due to the need for low electrodeposition potentials, formation of extremely fine hydrogen bubbles which spoils the surface finish, the requirement to use extremely low temperatures and the melting/alloying of gallium with the metal it's plated on, and poor repeatability of the process. There doesn't seem to be a patent describing using a pure gallium anode.
The difficulties are described in these patents:

US20110226630A1 - "Gallium electroplating methods and electrolytes employing mixed solvents":

... Gallium electrodeposition electrolytes and electrodeposition methods for solar cell manufacturing processes have many more stringent and special requirements than the electrodeposition methods and solutions employed for many other commonly plated metals such as Cu, Ni, Co, Pb, Sn, Ag, Au, Pt, and their alloys, etc. This stems from the facts that;
i) Ga is one of the lowest melting point metals in existence, with a melting point of about 30° C.,
ii) Ga has a high negative electrodeposition potential and thus Ga electrodeposition efficiency is naturally low since high electrodeposition potentials cause hydrogen generation, in addition to Ga deposition, at the cathode surface in aqueous electrolytes,
iii) hydrogen bubbles generated on a cathode surface form defects such as un-deposited regions unless such bubbles could immediately be removed from the surface,
iv) Ga has a tendency to form low temperature melting alloys with many alloy-partner materials such as In, Cu, Ag, Pb, Sn, etc. Furthermore, such alloys may form during electrodeposition of Ga onto surfaces comprising any of such alloy-partner materials.
Electrodeposition solutions employing glycerol are very viscous and difficult to handle. The viscosity of glycerol at room temperature is 1500 centipoise (cP) compared to the viscosity of water, which is 1 cP. Gas bubbles such as hydrogen bubbles formed on the electroplated (cathode) surface during Ga plating in viscous electrolytes cannot be easily removed from that surface and therefore cause voids and other defects in the electrodeposited films. Such defects may be acceptable for some applications of thick electrodeposited Ga globules. However, they cannot be tolerated in electronic device applications such as solar cell absorber formation applications where they cause compositional non-uniformities, morphological non-uniformities, and pinholes etc., all of which negatively impact the device performance.
Glycerol based plating solutions become more viscous as their temperature is lowered and therefore the problems cited above may get worse at lower temperatures. One other important point about the electrodeposition process for Ga is its sensitivity to the nature of the substrate surface on which the electrodeposition is performed. For example, to form a Cu/In/Ga precursor stack, the Ga film needs to be electrodeposited on an In surface. To form a Cu/Ga/In precursor stack, on the other hand, Ga plating needs to be performed on a Cu surface. One Ga electrodeposition solution that performs well for plating Ga on a Cu surface may not perform well for electrodepositing Ga on an In surface because the electrodeposition efficiency of Ga on one surface may be very different from its electrodeposition efficiency on another surface.
As mentioned above, gallium is a low melting point material with a melting temperature of around 30° C. As a result, when electrodeposited out of aqueous electrodeposition solutions kept at about room temperature (20-25° C.), it often forms rough films comprising molten surface features, especially at high electrodeposition current densities such as current densities greater than about 5 mA/cm$^2$. This is because even though the electrodeposition solution may be at a temperature lower than the melting point of Ga, the local temperature on the cathode surface may actually exceed this melting point due to the heat generated by the electrodeposition current. As further mentioned above, when Ga is electrodeposited on surfaces of materials that easily form alloys with Ga, molten droplets of Ga alloys with low melting temperatures may be formed on such surfaces. If the Ga film is electrodeposited over In and/or Cu, the local heating and Ga melting may actually promote alloying between the plated Ga film and the underlying In and/or Cu because there are low melting alloy phases between Ga and these materials such as In—Ga alloy phases and CuGa$_2$ alloy phase. As a result, the surface roughness of the deposit may further be increased due to the above mentioned reaction and the formation of molten alloy phases. For example, Mehlin et al. (Z. Naturforsch, vol. 49b, p.1597 (1994)) attributed the rough morphology of their electroplated Ga layers to the alloying of the electrodeposited Ga with the underlying Cu surface of the cathode and the formation of a molten CuGa$_2$ alloy.
Gallium may be electrodeposited from the electrodeposition solution at temperatures below −10° C., preferably below −20° C., most preferably below −30° C., so that local melting of the deposited Ga and its possible reaction with the materials on the cathode surface are avoided. Furthermore, at these low temperatures, the electrodeposition current densities may be increased to levels above 5 mA/cm$^2$, preferably above 10 mA/cm$^2$ and even above 20 mA/cm$^2$ without causing melting and/or alloying on the cathode surface. As a result, the electrodeposition rate and therefore the process throughput may be increased while, at the same time, the deposited film roughness is reduced. All of these benefits are important for the successful use of electrodeposited Ga layers in thin film solar cell manufacturing. For example, the melting point of methanol is −97° C. and the freezing point of a methanol/water mixture is a function of the ratio of methanol to water in the electrodeposition solution. A mixture of 75% methanol and 25% water, for instance, has a freezing point of −82° C. (−115° F.). This means that a Ga plating electrodeposition solution comprising 75% methanol and 25% water may be operated at a temperature as low as about −70-80° C., thus avoiding the melting, reaction and surface roughness problems described above.
...

US7507321B2 - "Efficient gallium thin film electroplating methods and chemistries":

... Gallium is a difficult metal to deposit without excessive hydrogen generation on the cathode because Ga plating potential is high. Hydrogen generation on the cathode causes the deposition efficiency to be less than 100% because some of the deposition current gets used on forming the hydrogen gas, rather than the Ga film on the substrate or cathode. Hydrogen generation and evolution also causes poor morphology and micro defects on the depositing films due to the tiny hydrogen bubbles sticking to the surface of the depositing film, masking the micro-area under them, and therefore impeding deposit on that micro-area. This causes micro-regions with less than optimum amount of Ga in the film stack.
Poor plating efficiencies inherently reduce the repeatability of an electrodeposition process because hydrogen generation phenomenon itself is a strong function of many factors including impurities in the electrolyte, deposition current densities, small changes on the morphology or chemistry of the substrate surface, temperature, mass transfer etc. As at least one of these factors may change from run to run, hydrogen generation rate may also change, changing the deposition efficiency.
Electrodeposition of Ga out of low pH aqueous electrolytes or solutions may suffer from low cathodic efficiencies arising from the presence of a large concentration of H+ species in such electrolytes. Therefore, hydrogen gas generation may be expected to lessen at higher pH values. However, as the pH is increased in the solution, Ga forms oxides and hydroxides which may precipitate as reported in the literature. Only at extremely alkaline pH values these oxides/hydroxides dissolve as soluble Ga species. Therefore, it becomes possible to electrodeposit Ga in a bath of pH>14 containing Ga salts as was done in prior-art techniques using high concentrations of KOH and NaOH in the bath formulation.
High concentrations of alkaline species, however, cause corrosion problems for the equipment as well as the cathode material itself. There is also a limit of the Ga amount that can be dissolved in the form of acidic Ga salts (GaCl3, Ga(NO3)3 etc) in such solutions before Ga starts to precipitate. Therefore, the pH needs to be adjusted again by further addition of alkaline species such as NaOH and KOH.
As pointed out above, solutions comprising a large molar amount of caustics are difficult to handle and they also have high viscosity. High viscosity makes the hydrogen bubbles formed on the cathode stick more to the cathode making it very difficult to remove them by stirring or other means of mass transfer. As explained above, such gas bubbles on the cathode surface increase defectivity of the deposited Ga layer.
...

US8545689B2 - "Gallium electrodeposition processes and chemistries" and
WO2012028416A2 - "Gallium and indium electrodeposition processes":

"... Current processes for electroplating metal containing thin films such as gallium and gallium alloys present numerous problems. These problems include, among others, low cathodic deposition efficiency due to excessive hydrogen generation, poor repeatability of the process, partly due to the poor cathodic efficiency, and the poor quality of the deposited films such as their high surface roughness and poor morphology.
Gallium is generally considered a difficult metal to deposit without excessive hydrogen generation on the cathode because gallium plating potential is relatively high. Hydrogen generation on the cathode causes the deposition efficiency to be less than 100% because some of the deposition current gets used on forming the hydrogen gas rather than for forming the gallium film on the substrate or cathode. Moreover, hydrogen generation and evolution is a causal factor for introducing porosity into the deposited film, thereby contributing to increased surface roughness and microdefects.
The plating efficiencies inherently reduce the repeatability of an electrodeposition process because hydrogen generation itself is a strong function of many factors including impurities in the electrolyte, deposition current density, small changes on the morphology or chemistry of the substrate surface, temperature, mass transfer, and the like.
...
Summary:
The present invention is generally directed to solutions and methods for electrodeposition of a substrate. In one aspect, the solution for electrodeposition of the metal containing film comprises a metal salt, wherein the metal is selected from the group consisting of gallium, indium, a combination thereof, and a combination of any of the preceding with copper; an acid selected from the group consisting of an alkane acid, an alkene acid, an aryl acid, a heterocyclic acid, an aromatic sulfonic acid, an aromatic sulfuric acid, hydrochloric acid, perchloric acid, and nitric acid; optionally a metalloid compound wherein the metalloid compound is selected from the group consisting of arsenic, antimony, bismuth, a combination thereof; and a solvent to dissolve said metal salt, wherein the pH of the solution is in a range selected from the group consisting of from about zero to about 2.6 and greater than about 12.6 to about 14, and wherein the solution is free of a complexing agent.
The method for electrodepositing a substrate comprises contacting
(i) a solution comprising a metal salt, wherein the metal is selected from the group consisting of gallium, indium, a combination thereof, and a combination of any of the preceding with copper; optionally an inorganic metalloid compound; further optionally an organic additive having at least one sulfur atom; and a solvent to dissolve said metal salt, wherein the solution is free of a complexing agent; and
(ii) a substrate; adjusting pH of the solution to a range selected from the group consisting of from about zero to about 2.6 and greater than about 12.6 to about 14; and applying a current to electroplate the substrate with a metal containing film.
...

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text:
Reaction of Br2/KOH with succinic anhydride

What products will be formed when $\ce{Br2/KOH}$ reacts with the following compound in water at room temperature?

Here is what I think is possible:

$\ce{OH-}$ will take up the acidic hydrogen.
$\ce{OH-}$ can attack the carbonyl group.
$\ce{OBr-}$ can attack the carbonyl group.
$\ce{Br+}$ can attack the lone pair of the doubly bonded oxygen and break the carbonyl bond.

Your first event ($\alpha$-deprotonation) is highly possible.

Succinic anhydride hydrolyzes readily to give succinic acid: https://en.wikipedia.org/wiki/Succinic_anhydride

so you are saying that $Br_2$ will not participate in the reaction.

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text:
Sound or light reversible super glue possible?
Sorry if this is a stupid question and help me to improve it.
Super glue is meant to be quick and mostly universal but we can stick to ceramics and/ or wood.
Imagine you put parts with the glue together.
Then later on a specific light or sound frequency would weaken the glue bonds.
Total nonsense, reality or research topic?

Could be done with light - a lot of UV could degrade the cyanoacrylate, but would be impractical (takes too long, releases toxic fumes, needs access to the glued layer so at least one object must be UV-transparent).

USC has a temperature reversible super glue for treating battlefield eye injuries.


text:
Planarity of given molecules in all conformations

Which of these have all its atoms in one plane in all the possible conformations (if any):
A) buta-1,3-diene
B) but-1-en-3-yne
C) $\ce{CH2=C=O}$
D) $\ce{CH2=C=CH2}$

While I was attempting this question, I got stuck in option B!
I was sure about option A that since it has conformational change possible between C2-C3 bond, atoms will be present in more than one single plane.
Next, I am also convinced that option D can't be the answer as allenes are non planar and they have H-C-H planes perpendicular to one another.
Then, for option C also, I know that since all the atoms are $sp^2$ hybridised, it has to be a planar straight molecule.
But, I am having a doubt in option B; how should we determine if the molecule is planar or not and whether all of its atoms are present in a single plane?

Do you have a modeling kit to build it in 3D?

No, I don't have.

All the atoms includes hydrogens, consider the angles that the H atoms have wrt the plane of the c atoms around double  and  triple bonds.

Perhaps I am still sleeping but if a molecules hss all its atoms there is no reason to say in all possible conformations. Question should be which molecules has just one confirmation, or even better which one doesn't have conformers. Not pedantic, it is really conceptual. ...

Yes, it's obvious that a molecule has all its atoms...but the question is not just about whether it has a conformer or not; rather, it also questions if the atoms lie in a single plane provided, that molecule has conformers.

The question is ratger nonsensical. It  should be which molecule is planar. It is at least a very convoluted way to check if someone is aware of Csp3 geometry.

It's just the other way of writing the same question, and there shouldn't be any convolution...bcz they mean the same (especially if one is clear about the definition of planarity of molecules).

To answer this you need to consider the hybridization and geometry around each atom but also think about all of the single bonds present. Single bonds will (generally) allow rotation along their axes as this can happen to a sigma bond without breaking the orbital overlap. If rotation about a single bond moves other atoms in space then you must consider conformational changes that may take atoms out of the plane.
You are correct with A and D. Option C is a ketene and the central C is sp hybridized, it is very similar to the allene in D. However, the terminal oxgyen has no other atoms bonded to it so it doesn't suffer from atoms out of the plane like D. So, C is completely planar (so far as atoms are concerned!), but not for the reason you stated.
For B, which you have shown, it is planar as drawn. Now consider what would happen if you rotated any of the single bonds. Would you break planarity? If not, it is also completely planar. Interestingly, if you rotate the C2-C3 bond, you rotate the alkyne group compared to the alkene but, as the alkyne group has infinite rotational symmetry along its axis, you can't tell the difference, all atoms remain on the plane!

@Karsten Theis. Nice JSmol!


text:
Why does nose grease knock down the head on beer and soda?
Freshman year, my first semester, I was taught many things.  One important trick I learned was when I was passed a beer in a 12 oz. plastic cup, I could eliminate the fresh foamy head by wiping my nose and touching it to the top of the foam.
I thought to do this recently with soda (as it seems soda head is more persistent).  The trick works there too.
What is the chemistry behind this?  Is it the salinity?  Could I use something less revolting than nose grease to do the same trick?

Reduced surface tension. Many craft beer drinkers want exactly the opposite effect: they try to keep their classes so-called "beer clean" in order to retain the head. Soap and greasy residue are two big foam-killers.

The proper beer glass should be wet before filling. As indirect consequence, it helps keeping the foam of the Czech beer..

@Poutnik True, filling a dry glass foams like hell and thus takes forever. By the time it lands on your table, it's stale.

Inspired by your family name: in Austria serious barmen might take up to five minutes or so for crafting an impressively thick and persistent head (called Krone, crown). Prost!

Any oil should work.
Naturally occurring surfactants, such as lecithin, make bubbles or foam because one end of the detergent molecule is hydrophilic while the opposing end is hydrophobic (not rabietic, though). Without oil, the hydrophobic end prefers air to water, encapsulating the air. With oil added, though, the water-repellent end is attracted to the oil, forming micelles, instead.
You could put vegetable oil or mineral oil on your finger, but then the server might wonder why you stick your finger in the beer.
You might also try making a beer-powered boat! Tell your server, and ask for a free beer to prove that this works.

And pop goes your foam (Schaumkrone (Crown) or Blume (flower) in German. God are we a poetic people.). I can see the freshman humour in trying this on someone else's beer, but your own? :-|

BTW, my uncle, when he was 8 years old or so, would get beer in a stein for his father at the pub downstairs (in ~1915, nobody cared). He'd talk to the bartender for a while, the bloom (well, head) would be off the beer, and he would say "My father will not be happy with me bringing him less than a full stein." The bartender wasn't fooled, but he always refilled to the top for the kid.


text:
Is there a mathematical basis behind the HSAB theory?
I learnt about the HSAB concept about a year ago, and found that it helped rationalise many of the solubilities/reactivities that “pure” Lewis acid/base theory could not.
Is there a mathematical basis behind Pearson’s theory, or was it just an empirical observation?

Be warned, HSAB theory can be deeply flawed: https://onlinelibrary.wiley.com/doi/full/10.1002/anie.201007100

Why do you have to start every question with 'Title.'? I have already asked you not to do that.


text:
What is double titration?
I an currently studying acid-base titrations. There is a concept of double titration. Can someone provide me an idea on what actually it is? Also what resources can I refer to?

Double titration also often refers to Gilson's titration method (also known as "Gilman color tests" or "Gilman double titration") for determining the concentration of organometallic compounds in hydrocarbon solvents.
At the first stage the total concentration of the base in the solution is determined.
At the second stage the residual amount of alkali after the organometallic compound reacts with the halide is established, allowing to calculate the concentration of metalloorganic compound. [1]

For the double titrations, a 3 ml aliquot of the solution to be analyzed was
withdrawn by pipette using a suction bulb and added to 10 ml of ether.
The solutionwas hydrolyzed with 10 ml of distilled water and titrated with standard acid, using phenolphthalein as indicator, to give the total present.
A second 3 ml aliquot was withdrawn and added 10 of ether, containing 1 ml of the halide, under an atmosphere of nitrogen.
The solution was swirled gently and allowed to stand for 2 min, after which it was hydrolyzed with distilled water and titrated immediately with standard acid, again using phenolphthalein as indicator.
In titrating the latter solution the end point is easily overstepped, since the
aqueous layer decolorizes before the ether layer. Vigorous shaking near the end point
is recommended.
There is the possibility of interference from hydrolysis of the halide, so that the titration should be carried out within a reasonable length of time.

An updated cumulative summary for the method can be found in Wakefield's Organolithium Methods [2, pp. 17–18]:

For detecting the presence of an organolithium compound (for example for checking whether any remains at the finish of a reaction) Gilman's Colour Test 1 is convenient:
$$
\begin{align}
\ce{(4-Me2NC6H4)2C=O + RLi &-> (4-Me2NC6H4)2C(R)OLi}\\
\ce{ &->[(i) H2O][(ii) I2, AcOH] [(4-Me2NC6H4)2CR]+ I-}
\end{align}
$$
A sample (0.5—1 ml) of the test solution is added to an equal volume of a I solution of Michler's ketone, 4.4'-bis(dimethylamino)benzophenone, in dry benzene.
A few drops of water are added, and the mixture is shaken.*
A few drops of an 0.2% solution of iodine in glacial acetic acid are added,
resulting in the formation of a blue or green colour.
Gaidis proposed two modifications, to distinguish between alkyl- and
aryllithium compounds (15).
(a) Following hydrolysis, the solution is buffered to pH 9 by the addition of 20% aqueous catechol (1—2 ml).
A few drops Of iodine in benzene are added and the solution is shaken.
A green to blue colour in the organic layer indicates the presence of an aryllithium compound.
The mixture is then acidified with acetic acid to reveal the presence Of an alkyllithium compound.
(b) Following hydrolysis the solution is acidified with acetic acid (5—15
drops).
20% aqueous sodium bisulphite is added; aryl- but not alkyllithium
compounds give a green colour.
Gilman's Colour Test 2 (16) which also distinguishes alkyl- from aryllithium compounds, depends on the rapid metal—halogen exchange of the former with 4-bromo-N,N-dimethylaniline:
$$
\begin{align}
\ce{4-BrC6H4NMe2 + RLi &<=>> 4-Li2C6H4NMe2 + RBr}\\
\ce{4-Li2C6H4NMe2 &->[(i) Ph2CO][(ii) H3O+] [Ph2CC6H4NMe2]+}
\end{align}
$$
The solution to be tested (0.5-1 ml) is added to an equal volume of a solution of 4-bromo-N,N-dimethylaniline in dry benzene.
A 15% solution of benzophenone in dry benzene (1 ml) is added.
After a few seconds, water is added and then the mixture is acidified with concentrated hydrochloric acid.
A positive result is indicated by a red colour in the aqueous layer.

* Aerial oxidation sometimes leads to formation of the blue-green at this stage.

References

Gilman, H.; Cartledge, F. K.; Sim, S.-Y. The Analysis of Organic-Substituted Group IVB Lithium Compounds. Journal of Organometallic Chemistry 1963, 1 (1), 8–14. DOI: 10.1016/S0022-328X(00)80039-X.
Wakefield, B. J. Organolithium Methods; Best synthetic methods; Academic Press: London; San Diego, 1988. ISBN 978-0-12-730940-8.

Andselisk, You have given a modern version of the meaning of double titration. In Indian textbooks/test preparation books, a century old meaning is still used. The answer by Wasnik reflects that meaning. That is also historically correct.

Double titration is a method of determining the amount of substance present in the form of a solution along with another solution. It can be used in the case of a mixture of bases involving $\ce{NaOH}$, $\ce{Na2CO3}$ or $\ce{NaHCO3}$, $\ce{Na2CO3}$. These bases are titrated against strong acids with proper indicators. Generally a single indicator is not sufficient so two indicators are used.
Procedure:
Firstly Indicator 1 (generally phenolphthalein) is added to the solution and the solution is titrated against a strong acid of known normality which on complete neutralization gives endpoint 1. Then Indicator 2 (generally Methyl Orange) is added to the remaining solution and is titrated against a strong acid which gives endpoint 2.
For example: In case of titration of $\ce{NaOH}$, $\ce{Na2CO3}$ at the first endpoint only half of $\ce{Na2CO3}$ will be converted to $\ce{NaHCO3}$
At the second endpoint remaining half gram equivalent of $\ce{Na2CO3}$ and $\ce{NaHCO3}$ is neutralized. Thus by equating the gram equivalents acid 1 with half equivalent of $\ce{Na2CO3}$ and half equivalent of $\ce{Na2CO3}$ plus equivalent of $\ce{NaHCO3}$ to equivalent of acid 2 we can get the amount of $\ce{NaHCO3}$ and $\ce{Na2CO3}$ in the solution.

Another option is titration with the pH meter.

I once faced the following problem in my textbook: "When 10 g mixture of Na2CO3 and NaOH was titrated against HCI by using phenolphthalein, followed by adding methyl orange after first equivalence point, 7.8 g of HCI was consumed in total. Find the mass of Na2CO3 in the mixture." Its solution involved the use of equivalence that rather confused me. Actually it turns out the problem simply asks to find the mass of Na2CO3 in a 10g NaOH-Na2CO3 mixture that requires 7.8 g HCl to get neutralized. And this can solved using elementary mole concept.

This is a problem with two unknowns. Needs two equations. 1) The initial mixture contains $a$ mol $\ce{NaOH}$.  and $b$ mol $\ce{Na2CO3}$. The mass of this mixture is $40 a + 106 b = 10$. This is first equation. 2) This mixture needs $7.8$ g $\ce{HCl}$ to be titrated, or $7.8/36.5 = 0.214$ mol $\ce{HCl}$ The second equation is : $a + 2b = 0.214$. So now, go on !


text:
DFT Exchange-Correlation Two Electron Integrals
Are there any electronic structure programs that print the exchange-correlation (XC) integrals for a given DFT functional? These integrals are of the form $$\langle pq|w|rs\rangle$$
where $w$ is an XC kernel specific to a given functional and $pqrs$ label arbitrary MOs.
I'm trying to make a Python code that can do TDHF/KS, with the prior SCF done using an existing electronic structure package. I have it working fine for TDHF, but TDKS requires these XC integrals and I can't find any program that prints them (or the atomic orbital basis equivalent). Ideally, if this option exists in Gaussian or Psi4, that would be most helpful to me, but I would be interested to see if any program will print these.

Just a guess, but psi or pyscf should at least be modifiable to print it, if it is not possible yet.

Also just an idea, the output keyword in Gaussian provides some options to interface to other programs. Not sure how helpful that can be (and there is probably an iop for printing options, too). For these technical things you also might have some luck on the CCL. Usually I wouldn't commend this, but there you would be a lot closer to the source...

@Martin-マーチン I've been using Gaussian output files to get other integrals. I've been looking around the "Interfacing to Gaussian" page on their website, but have yet to come across what I'm looking for. I will have to check out the CCL.

It seems that the only integrals Gaussian can print is this one:  IOp(3/33)
Integral package printing.
0
No integrals are printed.
1
Print one-electron integrals.
3
Print two-electron integrals in standard format.
4
Print two-electron integrals in debug format.
5
Combination of 1 and 3.
6
Combination of 1 and 4.

@Martin-マーチン Why do you discourage people from using CCL? I've seen lots of useful answers on there.

@OhLook That's true, and I don't discourage anyone to use it (I use it). I don't usually recommend it here; when the question is very much on topic, it only doubles the effort (avoid cross posting) for everyone. On a more general note, I find the CCL very hard to search through and sometimes you only get a very quick reply (as it happened in this case, iirc someone said to contact Gaussian support).

Not so much an answer, but an update. To my knowledge, no electronic structure program explicitly forms these XC integrals. While some have the tools in place in their code to do the numerical integration, this isn't how they evaluate linear response equations for DFT.


text:
Binding energy question
The electrons in the 1s orbital of chlorine have a binding energy of 273 MJ/mol, but the 1s electrons in sulfur have a binding energy of 239 MJ/mol. Why is this?

HINT - Consider $\ce{{16}S}$ and $\ce{{17}Cl}$

Chlorine has 17 protons and Sulfur has 16. In the 1s orbital there is no ineer electron to provide shielding effect of repulsion. So greater the positive charge, greater is the attraction and better is the binding.


text:
Equivalents of Oxalic acid

In this question the answer key given is A and C. But i think that D should also be wrong because equivalents = molecular mass / n factor = $\frac{90}{2} = 45 \neq 40 $
I believe I am correct but I may be wrong so can someone please help me clarify my question.

The reaction (ii) is incorrectly enumerated in the first place. The wording of (D) does not make sense and the whole question has inferior quality.

+1 for the OP bringing a problematic question here and asking for guidance. I can't understand the down vote.

I agree with Poutnik's point that this is a poor question.
Oxalic acid has two acidic protons so its equivalent weight is $\frac{1}{2}$ its molecular weight. Thus:

Answer A is false.
Answer B is true.
Answer C is false since there are 2*100*0.2 = 40 milliequivalents of acid, but only 100*0.2 = 20 milliequivalents of base.

Answer D requires some extra work. First it is not clear if Answer D applies to both reaction (i) and reaction (ii)

For reaction (i), in 100 ml of 0.2 M oxalic acid there are 40 milliequivalents of acid
For reaction (ii), the reaction is a redox reaction, so the number of electrons exchanged by oxalic acid needs to be determined.
Here is another sloppy point about the question. Reaction (ii) is not balanced.
However to figure out the number of electrons exchanged by oxalic acid, we only need the half cell reaction which is:

$\quad\quad\ce{2CO2(g) + 2 H+ + 2e− <=> HOOCCOOH(aq)}$
$\quad\quad$ Since oxalic acid exchanges two electrons in the redox reaction,
its "redox" equivalent$\quad\quad$ weight is $\frac{1}{2}$ its molecular weight.Thus there are also there are 40 milliequivalents of$\quad\quad$ oxalic acid redox wise too.

Answer D is false, because there are 40 milliequivalents of oxalic acid, not 40 equivalents

Answer D is true only if it states that "milliequivalents of oxalic acid in..."

@MathewMahindaratne - Thank you for the correction. I fixed the answer.

Okay but isn't oxalic acid's meq = 45 because 90/2 @MaxW.

@user198885 - No, 100 ml of 0.2 molar solution is 20 millimoles of oxalic acid. In a redox reaction because there are 2 electrons per molecule of oxalic acid that is 40 milliequivalents of oxalic acid. Answer D refers to 40 equivalents not 40 milliequivalents.


text:
What are molecular species?
I am in high school learning about adsorption for the first time. My school textbook   says:

"The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed adsoption" (Definition 1).

And I started thinking what does molecular species mean. So I checked out Wikipedia:

Adsorption is the adhesion of atoms, ions or molecules from a gas, liquid or dissolved solid to a surface (Defiinition 2).

So does the word molecular species in definition 1 refer to atoms,ions and molecules.(Definition 2)?
Also definition 1 mentions that this phenomenon takes place on the surface of a solid or liquid; why does it not mention gases?

Can you imagine molecules of gas get adsorbed on the surface of other gas ?

A species (or chemical species) is an atom, ion, or molecule, basically anything that can appear in a chemical equation as reactant or product. It seems weird to include atoms in molecular species, but from the context of the definition, I would think they wanted to include atoms in their definition.

@Poutnik no i can't imagine that but adsorption can happen at the interface of a gas and solid i guess so can i say that the solid is getting adsorbed.

@KarstenTheis so molecular species would usually mean molecules??

No, the solid does not get adsorbed, it stays where it is.

I don't know what molecular species would mean, probably molecules and molecular ions.

So does the word molecular species in definition 1 refer to atoms,ions and molecules.(Definition 2)?

It is most probably referring to molecules.

Also definition 1 mentions that this phenomenon takes place on the surface of a solid or liquid why does it not mention gases?

That's because gases as such, do not have what you would call a "well-defined surface" unlike solids and liquids.

I start with an example of the use of the term 'molecular species' in practice, to quote from 'Periodic Trends among Interstellar Molecular Species: The Case of Oxygen- and Sulfur-Containing Species':

In the present work, the effect interstellar hydrogen bonding on the variation of the S/O abundance ratio with respect to the cosmic S/O ratio is examined using high-level quantum chemical simulations. The binding energy between water molecule on the surface of the dust grains and the O- or S-containing molecule gives inside about the level to which the interstellar abundance of such molecule is affected. There are 59 O-containing and 19 S-containing interstellar species; for 16 that are S and O analogues, there is no order regarding their astronomical observations, i.e., in some cases, the O-containing species was observed before the S-containing and vice versa. Thus, the observation of one always gives information about the presence and the possible detectability of the order. In the light of this, the known molecules from this S/O group whose corresponding analogues are not yet observed are examined for their possible detectability. These species are subjected to the effect of interstellar hydrogen bonding.

The above context appears to be consistence with this Wikipedia description for chemical species:

Generally, a chemical species can be defined as an ensemble of chemically identical molecular entities that can explore the same set of molecular energy levels on a defined time scale.

So, it appears that terms 'molecular species' and 'chemical species' are pretty much synonymous.


text:
Fluorine isotopes and isomers
I was looking up fluorine isotopes out of curiosity. The Wikipedia entry gives info on them but adds there are also two isomers which are $\mathrm{^{18m}F}$ and $\mathrm{^{26m}F}$. What is an isomer of an atomic nucleus please someone?

Basically an isomer is a nucleus that is in an excited state and upon relaxation emits a gamma ray. Similar to (in principle) to how electrons get excited in phosphorescence and released hours later. A cool example is Tantalum 180 is radioactive with a half-life of 8 hours. Tantalum 180m is stable.

Related: How do I correctly typeset metastable radionuclide symbols?

Link to that wiki https://en.wikipedia.org/wiki/Isotopes_of_fluorine#List_of_isotopes and no, they don't have any chemical relevance.

Disagree with closure but have no votes - if gamma radiation is considered relevant to chemistry so should this

@Karl https://en.wikipedia.org/wiki/Fluorine-18

Again disagree with this being put on hold - see my example of Mossbauer spectoscopy which is a direct chemical use of isomers

Just like electrons nuclei can exist in a number of quantised states. We see evidence of this when, during radioactive decay, a gamma ray is produced. This is a nucleus in an excited state decaying to the ground state, and the excess energy being radiated away as a gamma ray photon. This is precisely analogous to an electron in an atom falling from a high energy state to a lower energy one, and thus producing radiation which we can see in an atomic spectrum. So an isomer of a nucleus is simply that nucleus in a state which is not the ground (lowest energy) state.
Most isomers are very short lived, and thus the gamma radiation appears to be more or less coincident with any other radiation that is produced during a radioactive decay. For instance the $\ce{^{18m}F}$ isomer you mention has a lifetime of 162ns according to https://en.wikipedia.org/wiki/Isotopes_of_fluorine . However the odd one is much longer lived, and for a remarkable case of one that is observationally stable take a look at $\ce{^{180m}Ta}$. 0.012% of natural Tantalum consists of this isomer, and it has never been observed to decay to the ground state - as is usual in such cases symmetry forbids the transition. And what I find even more remarkable is that the ground state itself is unstable with respect to both electron capture and beta emission, decaying to isotopes of Hafnium and Tungsten with a half life of just over 8 hours! See https://en.wikipedia.org/wiki/Isotopes_of_tantalum for details.
Sorry - slightly belated edit, but it occurred to me that isomers do have one very direct use in Chemistry, namely Mossabuer spectroscopy, see https://en.wikipedia.org/wiki/M%C3%B6ssbauer_spectroscopy. In essence very small changes in the energy of the gamma ray that is emitted when the isomer decays to the ground state can be detected by this technique, and can be related to changes in the chemical environment within which the atom finds itself.

The nucleus of an atom contains a mass of quarks which combine to form a larger particle which is the nucleus. It is possible to assemble these quarks in different ways, for example sometimes the different excited states of a nucleus have different nuclear spins. This is due to the quarks being arranged differently.
Chemistry can be defined as the study of the interaction of atoms with each other and the substances which can formed by doing things with the atoms. We can regard the whole of chemistry as a special area of electrostatics and the study the things which can ´be formed in this way. The whole of chemical bonding can be understood by considering the electrostatic effects between nuclei and electrons. While there is a set of rules which are well understood in chemistry (MO theory) for how the bonding in a molecule works. I am sure that in atomic nuclei that there are rules for the order of the filling of some shells which are similar in concept to the atomic / molecular orbitals.
Now after learning about atomic orbitals and MO theory it is easy to understand the idea of an excited electronic states. For example it is possible to move an electron from the antibonding HOMO of di beryllium into a higher energy orbital which is bonding. Thus allowing us to have diberylium only in an excited state.
Equally if we excite an atomic nucleus we could move quarks from their positions in the ground state of the nucleus into higher energy shells. When this happens the energy of the system is adsorbed by moving the quark into a higher energy shell. The energies of excited states of nuclei clearly are only at well defined energy levels. Consider for a moment the energy levels of a nickel-60, there are three commonly used levels which defines the fact that the excited nickel-60 nuclei from the beta decay of cobalt-60 emit gamma photons at 1.1 and 1.3 MeV. Very occasionally a nucleus jumps from the + 2.4 MeV level to the ground level emitting an extra high energy gamma photon.
One of the biggest bits of evidence for the nuclear shell model is the fact that there are nuclei with magic numbers which have special stability. For example carbon-12 is very stable (but not magic). There is a nuclear reaction which is similar to photoionization of a molecules / atom. This is the photonuclear reaction where the adsorption of a gamma ray causes an atom to eject a neutron. For C-12 to eject a neutron a staggering 18.722 MeV photon would be required. For C-13 to eject a neutron only a 4.946 MeV photon will be needed.
If we take the oxygen isotopes (O-16 is magic), then it is clear that when we are dealing with a magic +1 neutron then the effect is much stronger.
To eject a proton from oxygen nuclei
O-15 requires a 13.22 MeV photon.
O-16 requires a 15.66 MeV photon.
O-17 requires a 4.14 MeV photon.
O-18 requires a 8.04 MeV photon.
This shows that like an alkali metal a nucleus with an extra neutron above one which has both the magic number of protons and neutrons is special. In an alkali metal it is very easy to release that single electron while the energy required to release the extra neutron is lower.
While the question might be formally nuclear physics I do not see a clear line which separates chemistry from physics. Many radioactive decay processes and nuclear reactions do look like something out of chemistry.


text:
Cis trans isomer and cyclic compounds
8 Calcitriol is a steroid hormone found in human blood.
Calcitriol shows geometrical isomerism.
Give the number of geometrical isomers of calcitriol, including calcitriol.
I’m confused I think it doesn’t have a single geometric isomer because rotation is restricted in cyclic compounds! What is the concept behind it!

The question is not asking about conformers. It's about geometrical isomers.

for calcitriol:

Assuming only Z and E isomer 64 isomers are possible .For rest of isomers (3 ,ZZ,EZ,EE)
each will have 64 isomers.
For Z and E isomer 64 isomers are :

enter image description here

Quoting
MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE SIXTH EDITION
Michael B. Smith ,Jerry March

Out of curiosity, how did you produce the structures for all 64 isomers?

@andselisk: If you consider all stereo centers, there should be $2^6 \times 4= 256$ different structures. Therefore, I gave only 4 ($EE, EZ, ZZ, ZE$) possible $E/Z$-isomers. I thought giving all 256 isomers would be too much for OP. However, Chakravarthy Kalyan did a great job arranging at least 64 of them. :-)

@MathewMahindaratne I see. Still 64 is an impressive number and I don't think it's been done manually in ChemDraw or something. There must be some way to do this programmatically. As the author remains silent, the tension raises:)

@andselisk i have used good ol paper an pencil with chemdraw to convert into reality.I got all 256 isomers but time constraints did not allow me.I will update.Thankyou.

@ChakravarthyKalyan Oh. So you've drawn 64 isomer structures in ChemDraw manually, and now you are about to add 192 structures on top of that. You left me speechless standing here applauding.

The following diagram depicts the four possible geometrical isomers of calcitriol:

This is more complex than simple cis/trans-isomerism. This is E/Z-isomerism. If you look at all four structures, no structural changes have been made. All about geometry of trisubstituted double bonds. You may read: https://www.chemguide.co.uk/basicorg/isomerism/ez.html.


text:
Do electrons go to even lower ground states after exothermic reactions?
I am a bit confused about what happens to the energy levels during an exothermic reaction.
I understand that during an exothermic reaction, the energy of the electrons decreases.
Is the potential energy of electrons related to the energy level diagram, shown below?

So if the potential energy decreases, does that mean the ground state for the new molecules formed will be different from the old molecules?

Related: https://chemistry.stackexchange.com/questions/114347/what-is-meant-by-electrons-moving-to-lower-potential-energy/114350?noredirect=1#comment214923_114350

Yes, electron energy levels are like that, especially for single atoms.
Yes, forming molecular bonds creates different electron energy levels, with exothermic reactions leading to chemical bonds with lower ground levels then the reagents had.
Note that by an electron ground energy level is meant the minimal energy of  energy levels unocuppied by other electrons.
As an electron is forbidden by quantum rules to be in the same quantum state as any other electron within an atom or molecule.
Thermal, photon or collision excitations may keep part of electrons in excited state.
Much more info you can get reading about

molecular orbitals
molecular orbitals diagrams
molecular orbital theory

on Wikipedia.

So when the exothermic reaction finishes, will the electrons end up at the ground energy level?

Usually yes, even if sometimes they can even temporarily stay excited.

AFAIK,  reaction of NO with O3 produces excited NO2, that later emits a photon.


text:
Can sodium carbonate-sodium bicarbonate buffer be used for EDTA titration of calcium?
I am planning to measure general/total hardness (calcium and magnesium) of aquarium water using EDTA titration with Eriochrome Black T indicator. From what I understand, the unknown solution must be kept basic, at pH 10, for the EDTA to work in chelating the calcium ions off of the dye indicator complex. The procedures I can find online all use ammonia-ammonium chloride buffer. I would prefer using another buffer if possible as I find ammoniacal solutions difficult to handle and relatively hard to source.
Can I use a sodium carbonate-sodium bicarbonate buffer? Will it interfere with the titrant: will the EDTA form chelates with sodium ions?
From a similar titration procedure for exclusively determining calcium ion concentration using EDTA but with a Patton-Reeder indicator (calconcarboxylic acid), pH was raised to 12.5 using sodium hydroxide to precipitate magnesium off the solution. I suppose sodium will not interfere then?
Thank you!

Did you check the $K_{sp}$ of $\ce{CaCO3}$?

That made me thinking. I think that's already the answer... sodium carbonate will precipitate the calcium being measured via the common-ion effect? @MaxW care to post that as an answer?

I'm not sure what you mean by common ion effect, but the precipitation of calcium carbonate and magnesium carbonate would interfere with the EDTA titration.

@MaxW I remembered something about sodium soaps not working in saltwater because the equilibrium is shifted to prevent the dissociation of sodium from the soap in the presence of excess sodium chloride. In contrast potassium soap will work. From that I drifted into Wikipedia and under the "common-ion effect", they have listed an example of precipitating calcium carbonate off solution by adding sodium carbonate.

@shimofuri Rather, soap does not work in saltwater(=seawater) due content of calcium and magnesium, forming insoluble salts of fatty acids, like in very hard fresh water.

Sodium hydroxide $(\ce{KOH})$ works as $\mathrm{pH}$ modifier (measure the $\mathrm{pH}$ by adding drop by drop of hydroxide until $\mathrm{pH}$ is approx. $10$ to $12).$
Carbonate ion will be combined with calcium(2+) cation to form calcium carbonate. You can measure the calcium content by precipitating the calcium using soda ash light solution. Good luck.


text:
Isobaric-isothermal ensemble with degeneracy
I've been studying statistical mechanics for a while, the topic is very far from what I've learned till now so my questions may be superficial. When trying to see what my entropy is in an isobaric-isothermal ensemble, which turns out to be applicable in reactive systems, if I write my partition function $\Delta$ without the degeneracy factor $\Omega$, the answer I get from
$$S = \sum p_i\ln p_i$$
is
$$G = -T\ln\Delta$$
(with $T$ in energy units, $k_\mathrm{B} = 1$). However, if I take my systems to be in favor of a state or degenerate, I get
$$G + T\ln\Omega = -T\ln\Delta$$
Anywhere I read about this ensemble, the relationship for $G$ is in the first form and not the second, can someone give hints to some reading for what I'm missing here?


text:
How to calculate pH of basic salt solution mixed with strong base?
Calculate $\mathrm{ pH }$ of $\pu{ 0.05 mol }$ of $\ce{Zn(CH3COO)2}$ and $\pu{ 0.025  mol }$ of $\ce{ NaOH }$ dissolved in $\pu{ 1 L }$ of distilled water?
I tried to use the systematic method by forming the following equations to extract cubic equation which can  be solved by computer :
\begin{array}{   }
[\ce{ CH3COO- }]_I &= 2\times {0.05} = \pu{ 0.1 M }\\
[\ce{Zn^{2+}}] &=\pu{ 0.05 M }\\
[\ce{Na+}] &=\pu{ 0.025 M }\\
0.1 = [\ce{CH3COO-}]_I &= [\ce{CH3COO-}]_e + [\ce{CH3COOH}]_e\\
2\ce{[Zn^{2+}] + [Na+] +[ H+]_e &= [CH3COO-]_e +[ OH-]_e} \\
K_\mathrm{a} &=\frac{ [\ce{ H+}]_e\cdot{ [\ce{CH3COO-}]_e}}{[\ce{CH3COOH}]_e}\\
[\ce{OH-}]_e &=\frac{ K_\mathrm{w}}{[\ce{H+}]_e}
\end{array}
First question: Are the equations correct?
Second question: Is there a simple approximate method to solve the above problem? How?
I appreciate any help?

$\large \text{ --------- Maxw Notes ---------}$
Editing the answers with all the markup is painfully slow, so I added different analysis in a number of different answers to this problem. No doubt that will be confusing to someone looking at this problem. So let me summarize the answers.

Note: I am going to carry "extra" significant figures in the  calculations to gauge the mathematical exactness, but this is of course is ridiculous for the chemistry since the equilibrium constants are not known to that precision. If the true solution values are within 5% of the mathematically calculated values that's really good.

Exact Solution
Setup an ICE table and solved the nasty linear equations with Wolfram Alpha to get an "exact solution."
Iterative Solution
Setup an ICE table and then simplified the problem to create a "Simplified ICE Table". Went through some head pounding to create an iterative solution. Realized the original ICE table could be used for iterated solution so calculated that solution too.
Reducing the Problem to a Single Variable
I had tried to solve the problem first with what I am calling the Iterative Solution. I could tell all the species other than $\ce{[Na+]}$ were independently pH controlled, but I couldn't figure out how to iterate the solution. Using the charge balance finally occurred to me so I tabulated the data to show that the approach would work.
Solution with Cubic Equation
Used charge balance of solution, and one assumption to solve the simplified problem with a cubic equation with $\ce{[OH-]}$ as the variable. Solved the cubic equation with Wolfram Alpha.
Assumption:

$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$

Solution with Quadratic Equation
Used charge balance of solution, and two assumptions to solve the simplified problem with a quadratic equation with $\ce{[OH-]}$ as the variable. Solved the quadratic equation with Wolfram Alpha, but is reasonable to assume that solution could have been done by hand.
Assumptions:

$\ce{[CH3COO-]_{equil} \gg [OH-]_{equil}}$
$\ce{[CH3COO-]_{equil} \gg K_b }$
That the normalization tweak works. (I know that it did from the exact solution, but the calculations showed that it worked too.)

You algebraic equations look fine. My main concern is that whoever wrote this question for you does not know basic chemistry. Zinc acetate will react with NaOH, it will first form a hydroxide, that hydroxide will dissolve in base forming zincate ions. I don't know what will be the final pH. These equilibrium calculations assume no side reaction.

Yes if you know chemistry the problem is pretty simple. Zinc hydroxide has a $K_{sp}$ of $3.0\cdot10^{−17}$.

There isn't enough base to form an appreciable amount of the zincate anion ($\ce{Zn(OH)4^{2-}}$. It is reasonable to assume that $\ce{Zn(OH)2}$ forms quantitatively. // An acetate solution would have a pH somewhere near 9 with no $\ce{CO2}$, somewhere between 7 and 8 with $\ce{CO2}$. // So how much chemistry are you supposed to know?

@MaxW thanks. you neglect the $\ce{Zn(OH)2}$ contribution,and consider $ [\ce{CH3COO-}]=\pu{ 0.05 M}$ in calculating $\mathrm{pH}=9$

@MaxW Correct my solution.thanks in advance.

Exact Solution

I've fouled up this problem badly. Here is a more carefully thought out solution.

So let's start at the beginning with the ICE table. Making the normal assumptions:

Concentrations will be used instead of activities.
The $\ce{Zn(CH3COO)_2}$ dissociates completely in aqueous solution, thus the $\ce{Zn^{2+}}$ and $\ce{CH3COO-}$ ions can be considered separately.
The $\ce{NaOH}$ dissociates completely in aqueous solution, thus the $\ce{Na+}$ and $\ce{OH-}$ ions can be considered separately.
The $\ce{Na+}$ is a spectator ion and is left out of the table.

The problem only calls for the pH of the final solution, but a solution to the whole table will be calculated as a check.
Initial ICE Table
Let's let:

$x = $ the moles of $\ce{Zn(OH)2}$ formed. (The numerical value is the same as the change in molarity of $\ce{Zn^{2+}}$ since there is 1.0 liter of solution.)

$y = $ the molarity of the $\ce{CH3COOH}$.

\begin{array}{c | c c c c c}
               &\ce{[Zn^{2+}]}     & \ce{[CH3COO-]}        & \ce{[CH3COOH]}   & \ce{[OH-]}                         & \ce{Zn(OH)2(s)}  \\\hline
\text{I}  &\pu{0.050000 M}     & \pu{0.100000 M}           & \pu{0.000000 M}               & \pu{  0.025000 M}               & \pu{0 moles}  \\
\text{C} &  -x                           &  -y                               &   y                              & \pu{- 2x + y M}                  & \pu{x moles}  \\
\text{E} &\pu{0.050000 - x M} & \pu{0.100000 - y M}     &   y                              & \pu{0.025000 - 2x + y M}   & \pu{x moles}   \\
\end{array}
Applicable equilibriums
Wikipedia gives the $K_{sp}$ of zinc hydroxide is $K_{sp}=3.0\cdot10^{−17}$. So:
$K_{sp}=3.000\cdot10^{−17} = \ce{[Zn^{2+}][OH-]^2}$
Wikipedia gives the $\pu{pK_b}$ for sodium acetate as 9.25, which yields $K_b = 5.623\cdot10^{−10}$.  So:
$K_b = 5.623\cdot10^{−10} = \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
Exact Solution
Starting with the $K_{sp}$ equation for zinc hydroxide and making the appropriate substitutions for the final concentrations:
$K_{sp} = \ce{[Zn^{2+}][OH-]^2}$
$3.000\cdot10^{−17} = (0.050000 - x)\cdot(0.025000 - 2x + y)^2$
Starting with the $K_b$ equation for the acetate anion:
$K_b =  \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
$5.623\cdot10^{−10} = \dfrac{(y)\cdot(0.025000 - 2x + y)}{0.100000 - y} = \dfrac{0.025000y - 2xy + y^2}{0.100000 - y}$
Wolfram Mathworld's solution is:

x = 0.0134623
y = 0.00192466

$\ce{[Zn^{+2}]_{equil}} = 0.0500000 - 0.0134623 = 0.0365377$
$\ce{[OH-]_{equil}} = \sqrt{\dfrac{ K_{sp}}{\ce{[Zn^{2+}]_{equil}}}} = \sqrt{\dfrac{3.000\cdot10^{−17}}{0.0365377}} = 2.86543\cdot10^{-8}$
$\pu{pOH} = 7.542810$
$\pu{pH} = 14.000000 - 7.542810 =  6.457190 \ce{->[Round to 2 SF]} 6.46$
$\ce{[CH3COO-]_{equil}} = 0.100000 - 0.00192466 = 0.09807534$
There are two equations and two unknowns. Solving the equations yielded an exact solution. This is no problem with a computer but a mess even for a good calculator.

Finally !!!

I've never known a text book problem to require more than a quadratic equation. I finally figured out how to do that.

Solution with Quadratic Equation
Let's start with the simplified problem.
So the notion for the second ICE table is that to 1L decarbonated distilled water there is added 0.037500 moles $\ce{Zn(CH3COO)2}$ and 0.025000 moles $\ce{Na(CH3COO)}$.
Simplified ICE Table
Let's let $y = \ce{[CH3COOH]_{equil}}$
Assume:

$\ce{[CH3COOH]_{equil} \gg 1\cdot10^{-7}}$
Thus the autoionization of water will contribute a negligible quantity of $\ce{OH-}$ which is precipitated by the zinc.

Obviously each molecule of $\ce{CH3COOH}$ must create an anion of $\ce{OH−}$ in the solution. The precipitation $\ce{Zn{OH}_2}$ removes two $\ce{OH−}$ anions from the solution.
Thus $\Delta\ce{[Zn^{2+}]} = \frac{1}{2}\cdot y$

\begin{array}{c | c c c c c}
     &\ce{[Zn^2+]} & \ce{[CH3COO-]} & \ce{[CH3COOH]} & \ce{[OH-]'} & \ce{Zn(OH)'_2(s)}  \\\hline
\text{I} &\pu{0.037500 M} & \pu{0.100000 M} &  \pu{0.000000 M} &  &\pu{0 moles} \\
\text{C} & \pu{-y/2} & \pu{-y} & \pu{y} &  & y/2 \ \pu{moles}\\
\text{E} & \pu{0.037500 - y/2 M} & \pu{0.100000 - y M} & \pu{y} &  & y/2 \ \pu{moles} \\
\end{array}
The fraction of the acetate species which is acetic acid is
$\large f( \normalsize \ce{CH3COOH}\large ) \normalsize \  = \ \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}}$
and the equilibrium concentration of $\ce{CH3COOH}$ is going to be:
\begin{align}
\ce{[CH3COOH]_{equil} &= [CH3COO-]_{init}\cdot} \large f( \normalsize \ce{CH3COOH}\large ) \normalsize \\
 &= 0.1000000\cdot \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}} \\
\end{align}
We know that $\ce{[Zn^{2+}]_{init} = 0.0375}$ and that $\ce{[Zn^{2+}]_{equil}} = \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}}$ thus:
\begin{align}
\Delta \ce{[Zn^{2+}] &= [Zn^{2+}]_{init} - [Zn^{2+}]_{equil}} \\
 &= 0.0375 - \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}}\\
\end{align}
Starting with $\Delta\ce{[Zn^{2+}]} = \frac{1}{2}\cdot y$, then substituting and factoring:
$ 0.0375 - \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}} = \dfrac{1}{2}\cdot 0.100000\cdot \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}}$
$ 0.0375 - \dfrac{\ce{K_{sp}}}{\ce{[OH]^2_{equil}}} = 0.050000\cdot \dfrac{\ce{K_b}}{\ce{K_b + [OH-]_{equil}}}$
multiplying by $\ce{[OH-]^2_{equil}}$ and rearranging we get:
$ 0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.05\cdot \dfrac{\ce{K_b}} {\ce{K_b + [OH-]_{equil}}}\cdot \ce{[OH]^2_{equil}} - \ce{K_{sp}} = 0 $
Now if we multiply the equation by $\ce{K_b + [OH-]_{equil}}$, then we'll end up with a cubic equation. But let's be a bit creative and factor the equation thus:
$ 0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.05\cdot \dfrac{\ce{[OH]_{equil}}} {\ce{K_b + [OH-]_{equil}}}\cdot \ce{K_b} \cdot\ce{[OH]_{equil}} - \ce{K_{sp}} = 0 $
We know the pH is going to be near neutral so $\ce{[OH-] \approx 1\cdot10^{-7}}$ and we know that $\ce{K_b} = 5.623\cdot10^{-10}$ so let's assume that $\ce{[OH-]_{equil} \gg K_b}$ which means that:
$\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}} \approx 1 $
so the equation reduces to a simple quadratic:
$ 0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.050000\cdot \ce{K_b} \cdot \ce{[OH]_{equil}} - \ce{K_{sp}} = 0 $
Solving with Wolfram gives $\ce{[OH]_{equil}} =\pu{2.86616e-8}$.
The known value of $\ce{[OH]_{equil}} = \pu{2.86543e-8}$
Although the assumption was "good enough" to get nearly four figure agreement with our known value, one can't really just look at the quadratic and instinctively understand the implications. Now we can see that the assumption was not quite true. So we'll calculate a normalization factor:
$\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}} \approx  \dfrac{\pu{2.86616e-8}}{\pu{0.05623e-8 + 2.86616e-8 }} = \dfrac{\pu{2.86616e-8}}{\pu{2.92239e-8}} = 0.980759$
Since that is about a 2% error in the factor we'll calculate a  new quadratic:
$ 0.0375 \cdot \ce{[OH-]^2_{equil}} - 0.049038\cdot \ce{K_b} \cdot \ce{[OH]_{equil}} - \ce{K_{sp}} = 0 $
Solving with Wolfram gives $\ce{[OH]_{equil}} =\pu{2.86543e-8}$.
Since the value $\pu{2.87e-8}$ is in agreement for both calculations it gives some confidence that the value is legitimate.

Taking into my consideration to the above comment, I solved the problem.
I used the following table to determine the amount of each species at the end of the reaction:
\begin{array}{c | c c c c c c}
       \text{RXN}  &\ce{Zn(CH3COOH)2}_\text{(aq)}       & + &\ce{2NaOH}_\text{(aq)}      &\ce{->}& \ce{2CH3COONa}_\text{(aq)} &+& \ce{Zn(OH)2}_\mathrm{(s)}  \\\hline
\text{I} &(\pu{0.05 mol})  &   &(\pu{0.025 mol})  &        &(\pu{0 mol}) &&(\pu{0 mol}) \\
\text{C} &(\pu{\frac{-0.025}{ 2}  mol }) &   &(\pu{-0.025 mol}) &        &(\pu{0.025 mol}) &&(\pu{0.0125 mol})\\
\text{F} &(0.05-0.0125)=\pu{0.0375 mol}  &   &(\pu{0 mol})  &        &(\pu{0.025 mol})&&(\pu{0.0125 mol})
\end{array}
Ignoring the contrbition of the insoluble $\ce{Zn(OH)2}$ ,and taking  into account the  ionization of  both $\ce{Zn(CH3COO)2}$ and $\ce{CH3COONa}$ ,so: $$[\ce{CH3COO-}]_\mathrm{I}= 2\times{0.0375} + 0.025 = \pu{0.1 M }$$.
The acetate ion reacts with water as the following :
$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
so, the equilibrium expression as the following :
$$K_\mathrm{b} =\frac{K_\mathrm{w}}{K_\mathrm{a}}=\frac{[\ce{CH3COOH}]\cdot{[\ce{OH-}]}}{[\ce{CH3COO-}]}$$
Ignoring the autoionization of water,so :
$$[\ce{OH-}] = [\ce{CH3COOH}]$$
As $K_\mathrm{b}$ of acetate ion equal $5.55\cdot{10^{-10}}$ it is aweak base, so ,we assume :
$$[\ce{ CH3COO-}]_\mathrm{e} =[\ce{ CH3COO- }]_\mathrm{I}=\pu{ 0.1 M }$$
Substitute in the above equilibrium expression to find :$$[\ce{ OH- }]=\pu{ 7.45\cdot{ 10^{ -6 } } M}\quad\to\quad \mathrm{pOH}= 5.13 \quad\to\quad \mathrm{pH}\approx{8.9}$$

@MaxWF: final(at the end of the reaction). The amount of reacted" zinc acetate" $=\pu{0.0125 mol}$, and $\pu{ 0.0375 mol }$ ionized in solution. May I have a misconception that $\pu{0.0125 mol}$ of zinc was quantitatively ppt' ed as $\ce{Zn(OH)2}$.

@MaxW really, thanks to your interest and for your fruitful knowledge.The above problem for the group of a chemistry teacher.

Reducing the Problem to a Single Variable
$\ce{Na+}$ is a spectator ion and a constant, but $\ce{[OH-]}$, $\ce{[CH3COO-]}$, and $\ce{[Zn^{2+}]}$ are all pH dependent. We can define the charge balance, $\pu{C.B.}$ as:
\begin{align}
\pu{C.B.} &= \pu{cations - anions} \\
  &= \ce{([Na+] + 2*[Zn^{2+}]) - ([OH-] + [CH3COO-])} \\
\end{align}
If there are too much cationic charge then the C.B value is positive, and if there are too much anionic charge the C.B value is negative. This will allow a one-dimensional search of pH vs C.B looking for pH value where C.B. = 0.
We can simplify the CB equation by noting that $\ce{[CH3COO-] \gg [OH-]}$ so:
$ \pu{C.B.} \approx \ce{([Na+] + 2*[Zn^{2+}]) - ([CH3COO-])} $
Just to show that this works the values were calculated in a spreadsheet. Showing the results to 6 decimal places gives:
\begin{array}{c | c | c | c | c| c |}
\pu{pH} &\ce{[OH-]} & \ce{[CH3COO-]} & \ce{[Na+]} & \ce{[Zn^{2+}]} & C.B \\\hline
5.0000 & 0.000000 & 0.064008 & 0.025000 & 30.000000 & 59.960992 \\\hline
5.2000 & 0.000000 & 0.073812 & 0.025000 & 11.943215 & 23.837618 \\\hline
5.4000 & 0.000000 & 0.081709 & 0.025000 &  4.754680 & 9.452650 \\\hline
5.6000 & 0.000000 & 0.087624 & 0.025000 &  1.892872 & 3.723120 \\\hline
5.8000 & 0.000000 & 0.091817 & 0.025000 &  0.753566 & 1.440314 \\\hline
6.0000 & 0.000000 & 0.094676 & 0.025000 & 0.300000 & 0.530324 \\\hline
6.2000 & 0.000000 & 0.096574 & 0.025000 & 0.119432 & 0.167291 \\\hline
6.4000 & 0.000000 & 0.097810 & 0.025000 & 0.047547 & 0.022283 \\\hline
6.6000 & 0.000000 & 0.098607 & 0.025000 & 0.018929 & -0.035750 \\\hline
6.8000 & 0.000000 & 0.099117 & 0.025000 & 0.007536 & -0.059045 \\\hline
7.0000 & 0.000000 & 0.099441 & 0.025000 & 0.003000 & -0.068441 \\\hline
7.2000 & 0.000000 & 0.099646 & 0.025000 & 0.001194 & -0.072258 \\\hline
7.4000 & 0.000000 & 0.099777 & 0.025000 & 0.000475 & -0.073826  \\\hline
7.6000 & 0.000000 & 0.099859 & 0.025000 & 0.000189 & -0.074481 \\\hline
7.8000 & 0.000001 & 0.099911 & 0.025000 & 0.000075 & -0.074761 \\\hline
8.0000 & 0.000001 & 0.099944 & 0.025000 & 0.000030 & -0.074885 \\\hline
8.2000 & 0.000002 & 0.099965 & 0.025000 & 0.000012 & -0.074942 \\\hline
8.4000 & 0.000003 & 0.099978 & 0.025000 & 0.000005 & -0.074971 \\\hline
8.6000 & 0.000004 & 0.099986 & 0.025000 & 0.000002 & -0.074986 \\\hline
8.8000 & 0.000006 & 0.099991 & 0.025000 & 0.000001 & -0.074996 \\\hline
9.0000 & 0.000010 & 0.099994 & 0.025000 & 0.000000 & -0.075004 \\\hline
\end{array}
Using very conservative starting guesses of $\ce{[OH-]}$ $1\cdot10^{-9}$ and $1\cdot10^{-5}$  (pH 5 and 9), for the search of the sign change of C.B., the final solution could be obtained (somewhat painfully...). Rather than a binary search, it would be worth the extra effort to interpolate the subsequent guesses.  So using a calculator or four place log tables you could get an overly precise value for the final pH.

Thanks for your powerful effort and the tangible knowledge

Solution with Cubic Equation
As noted in a previous answer, $\ce{[OH-]}$, $\ce{[CH3COO-]}$, and $\ce{[Zn^{+2}]}$ are all pH dependent. So we can define the charge balance, $\pu{C.B.}$ as:
$\pu{C.B.} = \pu{cations - anions} = \ce{([Na+] + 2*[Zn^{2+}]) - ([OH-] + [CH3COO-])}$
Obviously in the equilibrium solution the C.B. = 0. Thus:
$0 = \ce{([Na+] + 2*[Zn^{2+}]) - ([OH-] + [CH3COO-])}$
or
$\ce{[Na+] + 2*[Zn^{2+}] = [OH-] + [CH3COO-]}$
But the equilibrium is going to be "near" neutral so $\ce{[CH3COO-] \gg [OH-]}$ and thus the $\ce{[OH-]}$ term can be neglected.
$\ce{[Na+] + 2*[Zn^{2+}] \approx [CH3COO-]}$
rearranging
$\ce{2*[Zn^{2+}] \approx [CH3COO-] - [Na+]}$
But:
$\ce{[Zn]_{equil} = \dfrac{K_{sp}}{\ce{[OH-]^2_{equil}}}}$
$\pu{Fraction(\ce{CH3COO-})} = \dfrac{\ce{[OH-]}}{K_b + \ce{[OH-]}}$
$\ce{[CH3COO-]_{equil}} = \ce{[CH3COO-]_{init}}\cdot\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}}$
$\ce{[Na+] = 0.025}$
making the substitutions:
$2\cdot\dfrac{K_{sp}}{\ce{[OH-]^2_{equil}}} = 0.1\cdot\dfrac{\ce{[OH-]_{equil}}}{\ce{K_b + [OH-]_{equil}}} - 0.025$
working through the factoring yields
$\ce{0 = 0.0375[OH-]^3 - 0.0125\cdot K_b\cdot[OH-]^2 - K_{sp}\cdot[OH-] - K_b\cdot K_{sp}}$
Solving with Wolffram Mathword's again: $\ce{[OH-] = 2.86543\cdot10^{-8}}$
Sometimes when solving cubic equations in chemistry you know about what the unknown is, and you can simplify by eliminating the cubic term. But in this case that is clearly not possible since the cubic term is the only positive one, and $\ce{OH-}$ must be real and positive. Given the Wolffram answer, the terms evaluate to  $\pu{8.823e-25, - 0.058e-25, -8.596e-25,}$ and $\pu{-0.169e-25}$.

Note that the cubic equation above only assumes that $\ce{[CH3COO-] \gg [OH-]}$.

Iterative Solution
In the Exact Solution There are two equations and two unknowns. Solving the equations yielded an exact solution. This is no problem with a computer but a mess even for a good calculator. So let's try to simplify the problem.
Initial ICE Table
Let's let:

$x = $ the moles of $\ce{Zn(OH)2}$ formed. (The numerical value is the same as the change in molarity of $\ce{Zn^{2+}}$ since there is 1.0 liter of solution.)

$y = $ the molarity of the $\ce{CH3COOH}$.

\begin{array}{c | c c c c c}
               &\ce{[Zn^{2+}]}     & \ce{[CH3COO-]}        & \ce{[CH3COOH]}   & \ce{[OH-]}                         & \ce{Zn(OH)2(s)}  \\\hline
\text{I}  &\pu{0.050000 M}     & \pu{0.100000 M}           & \pu{0.000000 M}               & \pu{  0.025000 M}               & \pu{0 moles}  \\
\text{C} &  -x                           &  -y                               &   y                              & \pu{- 2x + y M}                  & \pu{x moles}  \\
\text{E} &\pu{0.050000 - x M} & \pu{0.100000 - y M}     &   y                              & \pu{0.025000 - 2x + y M}   & \pu{x moles}   \\
\end{array}
Considerations for Simplification
Comparing $\ce{[Zn^{2+}]_{init}}$ and $\ce{[NaOH]_{init}}$, $\ce{Zn^{2+}}$ is clearly in excess. Let's assume that the $\ce{[NaOH]_{init}}$ is quantitatively precipitated as  $\ce{Zn(OH)2}$.
So the notion for the simplified ICE table is that to 1L decarbonated distilled water there is added 0.037500 moles $\ce{Zn(CH3COO)2}$ and 0.025000 moles $\ce{Na(CH3COO)}$.

This ignores the fact that the solution would also have a precipitate of $\ce{Zn(OH)2}$.

Pronation of acetate anion and the dissolution of $\ce{Zn(OH)2}$ both make the solution more basic. Thus it is assumed that the precipitate of $\ce{Zn(OH)2}$ would contribute a negligible quantity of $\ce{Zn^{2+}}$ or $\ce{OH-}$. (This could be solved exactly using a quadratic equation if it was really needed.)

Use a "?" for the initial concentration of $\ce{OH-}$ but it will be near $\pu{1e-7}$.

The precipitation of $\ce{Zn(OH)2}$ is trying to make the solution acidic. Pronation of the acetate anion is trying to make the solution basic. Considering that $\ce{Zn(OH)2}$ is very insoluble, and that the acetate anion is a weak base, at equilibrium the solution should be slightly acidic.
Simplified ICE Table
\begin{array}{c | c c c c c}
     &\ce{[Zn^2+]'} & \ce{[CH3COO-]'} & \ce{[CH3COOH]'} & \ce{[OH-]'} & \ce{Zn(OH)'_2(s)}  \\\hline
\text{I} &\pu{0.037500 M} & \pu{0.100000 M} &  \pu{0.000000 M} &\pu{?} \approx \pu{ 1.000e-7} &\pu{0 moles} \\
\text{C} & \pu{-x'} & \pu{-y'} & \pu{y'} & \pu{ y' -2x'} & x' \pu{moles}\\
\text{E} &\pu{0.037500 - x' M} & \pu{0.100000 - y'M} & \pu{y'} &\pu{? + y' -2x'} & x'\ \pu{moles} \\
\end{array}
Ignoring the problem with the initial concentration of $\ce{OH-}$ for the moment, at equilibrium there are four species , $\ce{[Zn^{2+}], [CH3COOH] , [CH3COO-]}$ and  $\ce{[OH-]}$ to consider. A direct solution would require a quartic equation.
Let's simplify this further by assuming that each acetate anion which is converted to acetic acid creates a hydroxide anion which is precipitated by zinc. Thus  $x' = \frac{1}{2}y'$. This reduces the problem to a cubic equation which is still painful to solve by hand. So let's use an iterative solution.
Now we assume that $\frac{y'}{2}$ is small and we can assume 0.0375 for $\ce{[Zn]_{equil}}$ in the $\ce{K_{sp}}$ formula to calculate the $\ce{[OH-]}$.
$\ce{[OH-]^2_{equil}}  = \dfrac{K_{sp}}{\ce{[Zn]_{equil}}} = \dfrac{3.00000\cdot10^{-17}}{0.0375}= 2.828427\cdot10^{-08}$
$\pu{Fraction(\ce{CH3COOH})} = \dfrac{\ce{K_b}}{K_b + \ce{[OH-]}}$
$\ce{[CH3COOH]} = 0.1\times \dfrac{\ce{K_b}}{K_b + \ce{[OH-]}} = 0.001949278$
$\Delta\ce{Zn^{2+}} = \dfrac{\Delta\ce{[CH3COOH]}}{2} = 0.000974639$
But now
$\ce{[Zn^{2+}]_{init}} = \ce{[Zn^{2+}]_{equil}} + \dfrac{y'}{2} = 0.037500 + 0.000974639 = 0.038474639$.
So let's use a correction factor CF to normalize $\ce{[Zn]_{init}} + \dfrac{y'}{2}$ to 0.037500.
$\ce{CF} = \dfrac{0.03750000}{\ce{[Zn]_{init}} + \frac{y'}{2}}$
The table below shows the results of the iterative solution. The first value for $\ce{Zn^{2+}}$ of 0.0365377 is the exact solution from Wolfram, which wouldn't be known of course. The %Err column is calculated based on the true value from Wolfram.

The second group of calculations with $\ce{Zn^{2+}}$ from 0.037500 to 0.0365377 is for the simplified ICE Table.

The third group of calculations with $\ce{Zn^{2+}}$ from 0.0500000 to 0.0365377 is for the original ICE Table.

\begin{array}{c | c c c c c}
\ce{[Zn^{+2}]} guess & \ce{[OH]} & \ce{[CH3COOH]} & \ce{CF} & \%Err \\ \hline
0.0365377 & 2.865431E-08 & 0.001924590 &  & \\
\\
0.0375000 & 2.828427E-08 & 0.001949278 & 0.974668 & 0.026337 \\
0.0365501 & 2.864947E-08 & 0.001924909 & 0.999667 & 0.000338 \\
0.0365379 & 2.865425E-08 & 0.001924594 & 0.999996 & 0.000005 \\
0.0365377 & 2.865431E-08 & 0.001924590 & 1.000000 & 0.000000 \\
\\
0.0500000 & 2.449490E-08 & 0.002244066 & 0.733539 & 0.368450 \\
0.0366769 & 2.859987E-08 & 0.001928183 & 0.996253 & 0.003811 \\
0.0365395 & 2.865360E-08 & 0.001924637 & 0.999951 & 0.000050 \\
0.0365377 & 2.865430E-08 & 0.001924591 & 0.999999 & 0.000001 \\
\end{array}
Answer
$\ce{[OH-]_{equil}} = \sqrt{\dfrac{3.0000\cdot10^{-17}}{0.0365377}}2.865\cdot10^{-8}$
$\pu{pOH} = 7.5428$
$\pu{pH} = 14.0000 - \pu{pOH} = 14.0000 - 7.5428 = 6.4572 \ce{->[Round to 2 SF]} 6.46$
Comparison Wolfram Mathword's solution
For the original problem Wolffram Mathword's solution is:

$ x = 0.0134623$
$ y = 0.00192466$

In comparison to the starting problem,$x' = x - 0.0125$ and $y' = y$ so the simplified problem gets:

$ x = x' + 0.0125 = 0.000962 + 0.0125 = 0.01346229$
$ y = y' = 0.001924590$


text:
Solubility of Iodine in Some Organic Solvents
The solubilities of iodine in $\ce{CCl4,n-Hex,Et2O,PhCH3}$ were given respectively as 19 g/kg, 13 g/kg, 337 g/kg, and 182 g/kg. These results puzzle me for the following reasons:
1) Iodine, being a non-polar substance, should dissolve the most in nonpolar solvents. Yet diethyl ether the best solvent for iodine.
2) Toluene and n-Hexane, both being hydrocarbons, should have similar solubilities of iodine. Yet the solubility of Iodine in Toluene almost 20 times that of n-Hexane.
How do we qualitatively predict these results?
Note: This question's source is from HBCSE INCHO 2012, a chemistry test for high school students, without fancy tools such as the Abrahams' equation.

Note that the values order or differences will look different once the density of the solvent or iodine/solvent is accounted. Your data are expressed in mass to mass.  At the denominator I am not sure what to be intended, mass of the solution or mass of the solvent. Probably the last so one can proceed with knowing individual densities.  I think this might impact your question though there is not an order of ten in density values difference.

Iodine forms charge transfer complexes, most notably with starch.
In these complexes, a fraction of the electronic charge is transferred to the other molecule.
The formation of these complexes increase the solubility of iodine.
This has also been called solvation, see for instance this article.
One effect of the formation of these complexes is the change in colour.
When there is no interaction with the solvent, the colour of the solutions is violet.
This colour changes to red-violet and brown when there is increasing charge transfer.
The order of solubility in the different solvents can be rationalized to some extent looking at the relative permittivity (also known as dielectric constant):
compound               solubility  permittivity   colour
------------------------------------------------------------
carbon tetrachloride       19          2.24       violet
hexane                     16          0.44       violet
diethyl ether             337          9.83       brown
toluene                   182          2.4        red-violet

Of course, carbon tetrachloride seems an outlier.
What happens here is that the dielectric constant is high due to the high $\delta+$ on the carbon.
But, since the $\delta-$ is shared by four chlorine atoms and the molecule is symmetrical, there is no effect on the iodine.

Iodine as a halogen has the strong electron affinity, what is the best met by the free electron pair of diethylether, then by $\pi$ electron density of toluene.
Note that iodine is also well soluble in ethanol, commercially used in medical treatment of skin injuries.


text:
pH of a Calcium Hydroxide vs Magnesium Hydroxide solution

In an AQA 2017 Chemistry paper, a $0.0131 \pu{mol}\cdot\pu{dm^-3}$ saturated solution of calcium hydroxide is prepared and then the question asks to predict the pH of a similar solution of magnesium hydroxide which is stirred until no more solid dissolved.

The answer (stating that  $\ce{Mg(OH)2}$ will be less alkaline) only references the comparative solubilities of the two but says to ignore any reference to concentration.
My initial thought was that 'concentration' of the hydroxide ions is important in getting to this result as $\ce{Mg(OH)2}$ is less soluble and therefore since less of the solid will dissolve than in the case of $\ce{Ca(OH)2}$, the actual concentration of the hydroxide ions could be say 0.029 (for example and assuming complete dissociation in both solutions) and therefore the solution is more acidic. Is this logic wrong even though I would get to the same result as the mark scheme?

see also solubility data page

This is a simplistic question which doesn't require a great deal of analysis. When ionic salts dissolve the typical assumption is that they completely ionize. So:
$$\ce{Ca(OH)2 -> Ca^2+ + 2OH-}$$
$$\ce{Mg(OH)2 -> Mg^2+ + 2OH-}$$
The assumption is that the calcium hydroxide will dissolve completely. However the magnesium hydroxide is less soluble and will dissolve incompletely. Hence the calcium solution will be more basic, and have a higher pH.
Now if the solutions are both 0.0131 molar, then the maximum hydroxide concentration can only be twice that or 0.0262 molar, not 0.029 molar.
I'm not sure what sort other data, or data lookup you could do with this question. It seems you just supposed to know that magnesium hydroxide is less soluble, and that a 0.0131 molar solution would exceed the solubility of magnesium hydroxide.

Hi MaxW, thanks for your answer, essentially I'm asking is will the concentration of the hydroxide ions that are actually in solution for the Mg case be less than that of Calcium's (given they both completely ionise)?

According to Wikipedia at 25 °C the solubility of magnesium hydroxide is 0.0064 g/L or 0.00011 molar. So if the concentration was less than that, then yes both the magnesium hydroxide and the calcium hydroxide solution would have the same pH. // Just for chuckles, Wikipedia states that at 20 °C the solubility of calcium hydroxide is 1.73 g/L or 0.023 molar.

@user77287 - I changed the wording of my answer a bit to note that complete ionization happens when the salt dissolves. So if it dissolves incompletely then the molarity of the solution is obviously less than the moles per liter added.

Thanks very much for the clarification, I was only doubting the molarity side as the mark scheme said to ignore any reference to concentration/molarity but I guess it's an obvious follow-on

@MaxW"However the magnesium hydroxide is less soluble and will dissolve incompletely"
Which equation represents the above statement?
$$1)~\ce{Mg(OH)2_\mathrm{(s)} -> Mg^2+ + 2OH-}$$
$$2)~\ce{Mg(OH)2_\mathrm{(s)} <=> Mg^2+ + 2OH-}$$
$$3)~\ce{Mg(OH)2_\mathrm{(s)} -> Mg(OH)+ + OH-}$$

@AdnanAL-Amleh - If the $\ce{Mg(OH)2}$  doesn't dissolve, it stays solid $\ce{Mg(OH)2}$. // Yes I suppose some Mg would form a complex like $\ce{Mg(OH)^+}$ in high concentrations of $\ce{OH^-}$, but it would have a very small concentration relative to $\ce{Mg^{2+}}$ when $\ce{Mg(OH)2}$  dissolves.


text:
Derive expression for heat flow from oil bath to metal sphere

A spherical ball of iron is put into a hot bath of oil. The iron ball is initially at a low temperature and heat transfer from the oil to the iron will increase the temperature of the iron ball.
Use the concept of heat transfer coefficients to express the heat flow (i.e. the quantity with units J/s) from the oil to the ball. Denote your variables according to the notation in Table 1. Notice that not all variables in Tab. 1 might show up in your expression. Variables not listed in Table 1 must not be used in your final expression.
Table 1: Notations
$$
\begin{array}{ll}
\hline
c_{p, f} & \text{ heat capacity oil} \\
c_{p, s} & \text{heat capacity iron} \\
D & \text{diameter iron ball} \\
h & \text{heat transfer coefﬁcient} \\
k_f & \text{thermal conductivity oil} \\
k_s & \text{thermal conductivity iron} \\
Q & \text{heat ﬂow oil to iron (J/s)} \\
R & \text{radius of the iron ball} \\
T_s & \text{temperature iron ball} \\
  & \text{(can be assumed homogeneous)} \\
T_0 & \text{bulk temperature oil} \\
ρ_s & \text{density iron} \\
ρ_f & \text{density oil} \\
\hline
\end{array}
$$

I'm really unsure of how to solve this problem. I know that the heat flow for a sphere with an inner and outer radius is
$$Q = \frac{4πkR_o}{1 - R_o/R_a}(T_0 - T_a),$$
but here I only have the radius of the whole sphere. So, how do I write the expression for the heat flow then?

If you let the outer radius approach infinity, you will be able to estimate the heat flow.

Newton's law of cooling is
$$q = A h (T_0 - T_S )$$
With the area of the sphere being
$$A = 4 \pi R^2$$


text:
Why is casein hydrophobic?
According to this source hydrophobic behaviour occurs when a molecule does not have a charge or is neutral and therefore cannot be attracted to the negative nor positive parts of water. Casein has an isoelectric point of 4.6 and milk has a pH of about 6.7. As a result, the casein proteins would have a negative charge, hence wouldn't the casein be attracted to the positive part of water?

In milk, casein exists as a colloidal aggregate (casein micelle) with the properties you state, but purified casein is different. It is rich in proline (shown below), has no disulfide bridges, and hence little tertiary structure to hide the hydrophobic residues in the core of the protein with the hydrophilic residues on the surface to bind to water. In addition to having a hydrophobic side chain, proline stiffens the aminoacid backbone, hindering tertiary structure. The net effect is a molecule with too much exposed hydrophobicity to be water soluble. Casein micelles are complex structures and quite different from lipid micelles.
PROLINE

Why does a lack of tertiary structure result in hydrophobic behaviour?

Folding can allow a protein to segregate hydrophobic residues in the core and hydrophilic resides on the surface.


text:
Friedel Crafts Acylation: Reaction of acyl halide with catalyst
To form an electrophile for this reaction to occur, the acyl halide interacts with the catalyst. However, on reviewing the conditions for the reaction, I have encountered that the 2 equivalence of the acyl halide is required.
In the case of the acyl chloride, there are 2 ways the acyl chloride can react with the catalyst.

(Sorry for the sloppy diagram)
In the case of the first reaction, Cl is good leaving group, but what is the reasoning of the second reaction to occur?

Strength of the O-Al bond is a significant factor

@Waylander, could it be that Al has vacant p-orbital where the 1 lone pair from oxygen atom is donated, to create a bond forming the adduct (similar as in BF3, since both are in the same group)?

That is exactly what occurs


text:
The equivalent weight of HNO3 in the following reaction
The equivalent weight of HNO3 in the following reaction is ‘x’, then the value of x/21 is
3Cu+8HNO3⟶3Cu(NO3)2+2NO+4H2O
in this I atempted by using redox formulas but my answer is comming 1 but correct answer is 4

The term equivalent weight is used in conjuction with normality (which is quite obsolete, except in the industry). Anyway, whenever you determine equivalent weight, you have to know beforehand whether you are considering a given "molecule" as an acid or as an reducing/oxidizing agent. This is where the confusion sets in.
Nitric acid as an acid is a monoprotic acid: It's equivalent weight is the same as its formula weight
Calculating the equivalent weight of nitric acid as an oxidizing agent is another story, it depends on how the redox equation is written.
From your equation nitric acid is acting as an oxidizing agent because there is no nitrate ion on the right hand side and the oxidation state of N is different on both sides:
$\ce{HNO3}$ is converting to NO. In the answer above, you already have the equation
$\ce{ NO3−}$ +3e => NO
In this case, the equivalent weight of nitric acid is 1/3 (formula weight).
As you can see equivalent weight is very much dependent on the way you write the equation. This will come with practice.

I'd agree that this is a redox reaction, but the OP gives the answer as equivalent weight factor as 4, so the person who made the answer key was considering the half cell reaction as an acid-base reaction.

The student's answer of 1 is correct. 63/3 = 21, x/21 gives us 21/21 =1; whoever wrote this question made a typo in the answer.

Ah, yes. When I read the question I thought the student misstated the problem so I jumped to 4H+ vs 3e- confusion. The question itself was weird. I realize now that you are right. The answer has to be an integer, so 1 is the only possibility.

"Equivalent weight" has been depreciated by IUPAC because it is ambiguous in some cases. To me this is one of them. The relevant half cell reaction is:
$$\ce{NO3−(aq) + 4H+ + 3e−  <=> NO(g) + 2H2O(l)}\quad\quad E = +0.958$$
Thus there are $\ce{4H+}$ but only $\ce{3e-}$. So is the equivalent weight 1/4 or 1/3 as much as the molarity?

can you explain it more clearly

I don't see how. What is confusing about my answer to you?

but the answer is different

You mean 4 vs 3? Did you follow the link and read the definition of equivalent weight?


text:
Can the van der Waals coefficients be negative in the van der Waals equation for real gases?

A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?

My approach
$$P(\text{ideal}) = P(\text{real}) + P(\text{changes due to intermolecular forces})$$
So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.
Questions

The answer to the question is given as True. So I want to understand what is wrong about my approach.
Since the intermolecular forces depend on $a$, for attractive nature $a$ is positive. Is $a$ negative for repulsive nature of forces?

Upon further thought, I have an explanation of why what I am thinking is wrong. The constant 'a' only describes attractive forces while the constant 'b' describes repulsive forces. So by definition 'a' cannot be negative. Can anyone tell me if this is correct? I'm still unsure about the 1st doubt.

Rather, b does not address repulsive forces, but the volume of molecules that is neglected in the ideal gas equation. As consequence, the gas has higher pressure, as collisions with volume walls are more frequent.

@Poutnik Isn't repulsive forces and volume neglected the same thing? I think that volume occupied is due to repulsive forces between atoms. I may be wrong, can you clarify?

No, it is not due repulsive forces. Neither for ideal gas, neither for real gas. Bigger own volume = lower fre volume = more frequent wall collisions.

Related https://chemistry.stackexchange.com/questions/60432/when-z1-are-repulsive-forces-dominant-or-are-attractive-forces-dominant

This question requires a simplistic notion of real gas behavior.
The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is
$$(P + a/V_\mathrm{m}^2)(V_\mathrm{m} -b) = RT$$
If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.
The rest of the story...
The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)

Is it correct to say that if we ignore 'a' then we are ignoring all the attraction forces and if we ignore 'b' then we are ignoring all the repulsive forces?

If "a" is ignored then the attraction forces within the gas are ignored. If "b" is ignored then the actual volume of the gas molecules is ignored.

Do we account for repulsive forces in the van der Walls equation then?

There aren't any "repulsive" forces in the van der Walls model. The "a" term would have to be negative and I do not know of any gas that has a negative "a" term.

I think I understand what I am doing wrong. I am confusing other models with this one. We have also been taught the graph we make for compressibility factor vs. Pressure. In that case, we take repulsive forces for cases in which Z>1.

You have to understand that the ideal gas model and the van der Walls model are supposed to work from a temperature of zero to infinity and pressure from zero to infinity. That is nuts of course. If you fit data from just a small pressure and temperature range to get the best fit "a" and "b" constants for the van der Walls equation then you could easily get a negative "a" term.

The repulsive forces might be simply reflect in a further V reduction. If the molecules repel each others, there is apparently less Volume for them to span, ie collisions happen at a longer distance. They will have the apparent effect of increasing the proper volume of the molecules.

The constant $a$ only describes attractive forces while the constant $b$ describes repulsive forces. So by definition $a$ cannot be negative. Can anyone tell me if this is correct?

It is correct.
Neither parameter $a$ or $b$ is regarded as negative, because a negative parameter would be opposite to its original meaning within the van der Waals theory. In agreement with this meaning, as pointed out by MaxW, when the theory is compared to data for real gases, the parameters are consistently positive.
As you correctly point out, the $b$ parameter encodes the repulsive interaction when gas particles come into close contact. Excluded volume refers to the same thing but can be encoded in different mathematical language by different models. The parameters in the van der Waals equation can be related to the second virial coefficient $B_2$ as follows:
$$ B_2 = b - \frac{a}{R T} \tag{1}$$
 as can the parameters describing a square-well, Lennard-Jones, or other similar potential. All of these models can generate equivalent virial equations of state to first order, provided appropriate values are chosen for the coefficients (in some cases both attractive and repulsive terms are specified in the potential function). Note that equation (1) includes a second term, $-a/RT$ , that reflects the attractive nature of the interaction implied by the parameter $a$. When $B_2>0$ repulsions dominate. When $B_2<0$ attractions dominate.

So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.

Absolutely. The vdW equation can be written as
$$P = \frac{RT}{(V_\mathrm{m} -b)}- \frac{a}{V_\mathrm{m}^2}$$
Therefore for constant temperature and volume, the repulsive parameter $b$ results (intuitively) in a larger pressure, whereas the attractive parameter $a$ results in a reduced pressure. Equation (2) can be expanded as follows:
$$\begin{align} P &= \frac{RT/V_\mathrm{m}}{(1 -b/V_\mathrm{m})}- \frac{a}{V_\mathrm{m}^2}\ \\ &\approx \frac{RT}{V_\mathrm{m}}\left(1+\frac{b}{V_\mathrm{m}}\right)- \frac{a}{V_\mathrm{m}^2} \\ &= \frac{RT}{V_\mathrm{m}} + \frac{RTb-a}{V_\mathrm{m}^2} \end{align} $$
which shows that (to first order) when $RTb=a$ the gas behaves as if ideal. Consistently, in that case you also have that the second virial coefficient $B_2=0$. Depending on the relative magnitudes of $a$ and $b$ either attractive or repulsive interactions will predominate.

The answer to the question is given as True. So I want to understand what is wrong about my approach.

Nothing is wrong with your approach except. It is more usual that the repulsive term associated with $b$ is dominant, so that the pressure in a real gas is typically greater than in an ideal gas at the same $T$ and $V_m$. It is not impossible for the opposite to occur, though (see this problem, for instance, but beware the different naming convention of the parameters used there).