text:
Why does the pH of very strong NaOH solutions not make sense?
I'm wondering why the pH of extreme alkaline solutions, such as the case with $\pu{10 M}$ $\ce{NaOH}$ does not make practical sense and I'm wondering why this is the case? Refer to my calculations below:
$$\ce{NaOH ->[H2O] Na+(aq) + OH-(aq)}$$
$$\mathrm{pH} + \mathrm{pOH} = 14$$
$$\mathrm{pOH} = -\log{\ce{[OH-]}} = -\log(\pu{10M}) = -1$$
Therefore,
$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - (-1) = 15(?)$$
This doesn’t make sense as I thought the pH scale maxes out at 14?
This is even stranger when considering that most commercial NaOH solutions are sold as bulk $\pu{1000 L}$ aqueous $\pu{50 w/w\%}$ NaOH solutions corresponding to approximately $\pu{19 M}$... much higher than $\pu{10 M}$... does the industry not use pH either when it comes to extreme solutions? What am I missing here?
PS Refer below to my calcs which got the $\pu{19 M}$ number in case any of you wanted to check my proof (all symbols used are standard symbols i.e. masses (m), molar masses (M), density ($\rho$), standard gravity (SG) etc.):
Assuming $\pu{50 wt\%}$ caustic (i.e. alkalinity), in 100 g of Solution; $\pu{50 g}$ $\ce{NaOH}$, $\pu{100 g}$ solution.
$\mathrm{SG}_\mathrm{solution} = \pu{1.525 g/mL}$ (from spec at $\pu{20 ^\circ C}$, $M_\ce{NaOH} = \pu{39.997 g/mol}$
\begin{align}
\mathrm{M}_\ce{NaOH} &= \frac{n_\mathrm{solute}}{V_\mathrm{solution}} \\
n_\mathrm{solute} &=\frac{m_\mathrm{solute}}{M_\mathrm{solute}} \\
V_\mathrm{solution} &= \frac{m_\mathrm{soln}}{\rho_\mathrm{soln}}\\
\rho_\mathrm{soln} &= \frac{\mathrm{SG_\mathrm{soln}}}{\rho_\ce{H2O}}\\
\end{align}
Solving with said values we get,
$$\mathrm{M}_\ce{NaOH} = \pu{19.12 mol/L} \approx \pu{19 M} $$
For context, I am trying to better understand this behaviour of extreme alkaline systems because I want to keep track of the pH of other solutions I'm making by mixing highly alkaline caustic solutions with sodium silicate systems. Sometimes these systems form precipitates when they've mixed, other times they don't. I thought understanding the pH of everything is a good place to start, hence why I'm here.

It does make sense, but you will probably not find sensors to measure it, unless incorrectly, with strong trend and short life. Activity coefficients of 10M solutions is a "rocket science".

I thought the pH scale maxed out at 14 and that anything higher doesn't make sense? Can you elaborate on how the activity coefficients play a role in this example being very complex?

pH scale is not limited to 0-14 range. But value out of this range are not that useful and there is experimental trouble in measuring them. Typically, most of glass oH electrodes do not like much pH > 12, which affects their responses and progressively damages them. Reasonable chemists do not put their pH electrodes to 10 M NaOH.

https://en.wikipedia.org/wiki/Activity_coefficient // https://en.wikipedia.org/wiki/PH

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@Poutnik, I see, I will look more into activities which seems to be what will make this all practical. Hopefully there'll be more detailed answers posted cheers!. Also, I have used LaTeX here for my equations, aside from the picture i posted because I had done it on word equation a while ago and thought it easier/quicker with the formatting. Regardless, the picture is still clear. What's the issue with formatting which you're pointing out?

Element symbols are typographically never in italic. CH SE uses MathJax mhchem extension by default, involving \ce{} and \pu{} Mathjax syntax. More in links. // Use $\ce{NaOH(aq) <=> Na+(aq) + OH-(aq)}$
to get $\ce{NaOH(aq) <=> Na+(aq) + OH-(aq)}$ // But there is no NaOH(aq), all is dissociated.

pH and pOH are not in italic either, use \mathrm{} to get them upright. Generally, simplified, only variables are in italic.

What's a good way to measure pH in these situations then (i.e. highly alkaline caustic conditions)? Either some sort of complex model with activity coefficients or something along the lines of p(NaOH)?

There is no good way and it is not usually used. IF it is used, there is determined pOH, as pH would be highly  temperature dependent. For high concentration of strong acids or bases, there is usually used molar or mass concentration, or density.


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What is the significance of charge balancing when analysing system speciation (carbonate system given as an example)?
Following on from my previous question and accepted answer, How do I quantify the carbonate system and its pH speciation?, I also calculated a charge balance ($CB$) on the system where $z$ is the species' ionic charge:
$CB = \sum{z\times[\textrm{positive ions}]}=\sum{z\times [\textrm{negative ions}]}$
The CB is thus;
$\ce{ [H+] = [OH-] + [HCO3-(aq)] + 2[CO_{3}^{2-}(aq)]}$
Usually, I'd assume this number to equal zero for complete balance which makes sense. The system indeed shows this when looking at this balance (which I've done by getting the difference in negative and positive ionic charge) I see that the charge balance becomes unbalanced (i.e. >0) at around pH = 12.5 signifying a net negative charge as seen in the graph below:

Here are my main questions concerning this:

(For a better personal understanding) What is the practical significance of including these charge balance equations?

As the charge balance is not in integer values (I.e. it can’t be [z] or the quantised [e-]), what is the units of the charge balance?

What are the practical differences in doing the same calculations with activities instead of concentrations (I.e. not assuming infinite dilution)? I have read in other questions that the standard pH model is not the best to use at extreme pH values and activity is the better model.

What does the charge balance impact on for the chemical species? Is it to infer species' reactivity? Does this infer that around pH= 12.5 the system becomes unstable, highly reactive and thus more likely to form an unwanted precipitate?

Can someone please help me understand this?
(For context, I am trying to better understand this behaviour of extreme alkaline systems because I want to keep track of the pH of other solutions I'm making by mixing highly alkaline solutions with sodium silicate systems. Sometimes these systems form precipitates when they've mixed, other times they don't. I thought understanding the pH and charge balancing of everything is a good place to start, hence why I'm here.)

My comments have been deleted, relevant have been moved to my answer.

(For a better personal understanding) What is the practical significance of including these charge balance equations?

There are variables for concentration of $\ce{H+}$, $\ce{Na+}$, $\ce{CO2(CO2+H2CO3)}$, $\ce{HCO3-}$, $\ce{CO3^2-}$, $\ce{OH-}$, i.e. 6 variables.
There are 3 equations for equilibrium (2 for CO2, 1 for H2O).
There is 1 equation for charge balance. $[\ce{H+(aq)}] + [\ce{Na+(aq)}] = [\ce{OH-(aq)}] + [\ce{HCO3-(aq)}] + 2[\ce{CO3^2-(aq)}]$
There are 2 equations for the molar amount inventory ( 1 for Na+, 1 for total CO2 )
So 6 equations for 6 variables, therefore solvable. Without CB equation, there would be infinite number of solutions, having still 1 degree of freedom. You could deliberately ( within a range) choose a value of 1 of 6 variables and still being able to solve the equations set for values of remaining 5 variables.
You have omitted in your analysis the essential presence of other cations, like of alkali metals. You cannot have $\mathrm{pH}>7$ (at $\pu{25^{\circ}C}$) and to have $\ce{H+(aq)}$ as the only cation, as then $[\ce{H+(aq)}] \lt [\ce{OH-(aq)}]$. If the system consists just of $\ce{H2O}$ and $\ce{CO2}$ in any of its dissolved forms, without addition of a base like e.g. $\ce{NaOH}$, $\ce{Na2CO3}$ or $\ce{NaHCO3}$, so having truly just $\ce{H+(aq)}$ cations, then your chart should end at $\mathrm{pH}=7$, as you would not be able to reach alkalic $\mathrm{pH}$.

As the charge balance is not in integer values (I.e. it can’t be [z] or the quantised [e-]), what is the units of the charge balance?

The unit of the charge balance is the molar density of elementary charges - $\pu{1 mol/L}$, equivalent $\approx \pu{96490 C/L}$

What are the practical differences in doing the same calculations with activities instead of concentrations (I.e. not assuming infinite dilution)? I have read in other questions that the standard pH model is not the best to use at extreme pH values and activity is the better model.

Involving activities instead of concentrations is kind of opening of a can of worms or even the Pandora's box. It has several aspects:

Increasing ionic strength $I = \sum_i{c_iz_i^2}$ leads to decreasing of activity coefficients and decreasing activities, what leads to higher equilibrium concentration of respective ions.
Addressing solutions with significant ionic solutions. For Ionic Strength $I \le \pu{0.1 M}$ is applicable Debye–Hückel equation, for more concentrated solutions Pitzer equations
Addressing solutions with extreme pH values and related extreme ionic concentrations. For these ighly concentrated solutions, you can safely throw all pH calculations out of the window.

What does the charge balance impact on for the chemical species? Is it to infer species' reactivity? Does this infer that around pH= 12.5 the system becomes unstable, highly reactive and thus more likely to form an unwanted precipitate?

Having 1 mol of extra ions out of charge balance would blow solution from the vessel, yourself from the room, many things from the house and possibly making house to spread out like a big firework. As nearly 100000 Coulombs is like 1000 times bigger charge than the longest 10 km long CG+ lightnings with potential difference up to a billion Volts.

Many questions are already answered in resources-for-learning-chemistry.
The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It helps preventing  others to tell you what you already know or what you could easily find yourself.
Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.
Check also site:chem.libretexts.org acid base equilibrium or any other suitable keywords of interest.

I see, thanks for the great reply! To follow up then, the non-zero charge balance occurring when $pH \geq 13$ is explainable in this scenario (as you suggest) due to the inapplicability of my conventional pH calculations from this point onwards as it becomes a highly concentrated solution?

I have moved comments to the answer.

Great answer! The feedstock I usually use is sodium silicate mixed with sodium hydroxide. I was thinking that only the speciation of the silicates was important but you correctly point out that I need to also focus on the Na cation. I have never looked into sodium speciation but it seems I will have to in order to get to the bottom of this. Thanks for the insight! :)

[OP] I am trying to better understand this behavior of extreme alkaline
systems because I want to keep track of the pH of other solutions I'm
making by mixing highly alkaline solutions with sodium silicate
systems.

You are missing sodium ions in your model. To get alkaline at all, you need some carbonate or bicarbonate salt. To get past ph 13 or so, you need to add hydroxide directly, e.g. in the form of NaOH. So there will be substantial concentration of sodium, fixing your charge balance.

[OP] What is the practical significance of including these charge balance equations?

Typically, your system will be underdetermined if you leave it out. In your case, you probably set the pH to a certain value, but in reality, a mixture has a pH and there is no dial (other than adding stuff) to change it.

Great answer thank you! I was thinking that only the speciation of the silicates was important but you correctly point out that I need to also focus on the Na cation. I have never looked into sodium speciation but it seems I will have to in order to get to the bottom of this. Thanks for the insight and please let me know if you have any info on this, cheers! :)

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text:
Formation of ester during preparation of Fluorescein
I was going through PYQ of JEE where, I encountered the following question:
Q) Phthalic acid reacts with resorcinol in the presence of concentrated $H_2SO_4 $ to give :
In the solution provided I realised that conc. $H_2SO_4$ was acting as a dehydrating agent and leads to the formation of Fluorescein.

I have learnt that in presence of an acid catalyst esterification reaction is feasible, then what are the factors that prevent it from happening here ?
Thanks.

Phenolls are not alcohols.

@user55119 Thanks for the comment , I get it now that it is not possible for phenols to attack the carboxylic acids like alcohols as their nucleophilicity is lesse.

Certainly the para-acylation is faster.

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text:
Why are there FOUR $sp^3$ hybridized orbitals in methane?
(This may be a stupid question, but I've only learnt about hybridization yesterday.)
I will take the example of methane.
The valence shell electron configuration of ground state of carbon is $2s^2 2p^2$. There are 2 unpaired electrons, both in the $2p$ orbitals. But we need 4 unpaired electrons to form 4 bonds. So one of the electrons from the $1s$ orbital is excited to the 3rd $2p$ orbital. So, there is an $s$ orbital with an unpaired electron, and 3 $p$ orbitals with unpaired electrons, which can all participate in bonding, so now we can form 4 bonds. In the end, there will be 4 $sp^3$  hybridized orbitals.
My question is, why are there going to be 4 hybridized orbitals? I imagine the hybridization as "there is an $s$ orbital and 3 $p$ orbitals, hence $sp^3$." I thought a single $sp^3$ hybridized orbital is the combination of a single $s$ orbital and 3 $p$ orbitals. So how can there be 4 $sp^3$ orbitals in the end if the valence shell of carbon only has a single $s$ orbital and three $p$ orbitals? Intuitively it seems to me that 4 $sp^3$ orbitals would require 4 $s$ orbitals and 12 $p$ orbitals.

The sum of the exponents in $\mathrm{sp^3}$ is $1 + 3 = 4$.
The first $\mathrm{sp^3}$ is made with $(1/2)^2 = 1/4$ of the $\mathrm{2s}$ orbital, one quarter of the $\mathrm{2p_x}$ orbital, one quarter of the $\mathrm{2p_y}$ orbital and one quarter of the $\mathrm{2p_z}$ orbital. Total $4$ quarters = $1$ hybrid orbital. This process is repeated four times, taking care to change twice the sign + to minus according to the character table. The definition of the four hybrid orbitals are :
$(1/2)[\mathrm{2s + 2p_x + 2p_y + 2p_z}$]
$(1/2)[\mathrm{2s - 2p_x - 2p_y + 2p_z}$]
$(1/2)[\mathrm{2s - 2p_x + 2p_y - 2p_z}$]
$(1/2)[\mathrm{2s + 2p_x - 2p_y - 2p_z}$]
This choice of plus or minus signs allows the hybrid orbitals to have a maximum density in the four directions of the tetrahedron.

Or 4 of the possible 8 corners of a cube (with the center at the origin of the 3D cartesiona coordinate system), making sure you don't pick adjacent corners. This explains the pattern ++++, +--+, +-+- and ++-- in the four definitions. It also explains how you can build a methane molecule using a cube, see https://chemistry.stackexchange.com/a/129129/72973

Hybrid orbitals are linear combinations of unhybridised orbitals:
$$\mathrm{sp^3} = a \mathrm{s} + b\mathrm{p}_x + c\mathrm{p}_y + d\mathrm{p}_z$$
Now, you seem to have picked up the notion that $a = b = c = d = 1$. This is partly true in that for a perfectly tetrahedral molecule like methane you have that $a = b = c = d$; but recall that each orbital only has space for 2 electrons,* one spin-up and one spin-down. When you add up four orbitals like that, you'll find that you have space for way  more electrons, which is not quite right. An orbital should be normalised, in that it only has space for 2 electrons, one spin-up and one spin-down. Or, mathematically, we require that
$$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1,$$
(the modulus is required because in general these numbers may be complex, and it's $1$ not $2$ because we don't count the electrons of opposite spin). Now, if you set $a = b = c = d = 1$ you see that you've completely overshot the target of $1$. Instead of doing that, you need to scale down of the coefficients so in fact we have that $a = b = c = d = 1/2$.
Now, that looks like it should only be enough to make two hybrid orbitals, but it's a curious fact of quantum mechanics that only $|a|^2$ of the s-orbital (i.e., $(1/2)^2 = 1/4$) goes into each of the hybrid orbitals. So you do have enough for four hybrid orbitals.
The other $\mathrm{sp^3}$ orbitals also have that $|a| = |b| = |c| = |d| = 1/2$, but they are combined with different signs. See e.g. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_polyatomic_molecules/5.2%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2D%3A_sp3_Hybridization.

* OK, if you're really clever at chemistry you know this should really be $1$, and that what I'm talking about here is really a spatial orbital, and that's why the equation below says $1$ and not $2$.


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Why and how do aqueous silica systems precipitate and can its occurrence be mitigated by quantifying its speciation correctly?
Background/Context:
For context, I have been doing a lot of work on experimental systems in geopolymer chemistry lately with a focus on "activator" solutions which are essentially mixtures of sodium hydroxide, sodium silicate and water used to enhance the reactivity of aluminosilicate materials upon mixing all the components together (simplified explanation). When making these "activator" solutions I usually change the $\frac{Si}{Na}$ ratios around in the hope of optimising the reaction with the aluminosilicate material which I've been measuring through other means experimentally.
The Ingredients I use are the following:

Commercial Sodium Hydroxide (Aqueous):

49.9 wt% $NaOH$;

50.1 wt% $H_2O$;

Commercial Sodium Silicate (Aqueous):

30.5 wt% $SiO_2$;

16.7 wt% $NaOH$;

52.8 wt% $H_2O$;

"Extra" (Distilled) Water (Liquid):

100% wt% $H_2O$;

Most of the time, the activator solution I make is clear and good to go. Due to the exothermicity of $NaOH$, I usually leave the solution to cool down for easier handling and so that I can be sure everything has mixed well and reacted together to a steady state. Here is a picture of what the solution looks like in this case with an electric overhead stirrer constant at 200rpm. Note also the beaker in the background which is another activator solution which has stirred a bit more and become clear and "good to go" made with the following feedstocks:

97.94 g Commercial Sodium Hydroxide
171.63 g Commercial Sodium Silicate
68.3 g Extra Water

This corresponds to the following mols:

0.87 mols $Si$
1.94 mols $Na$
11.55 mols $H_2O$

However, I've noticed that sometimes when I make it at specified $\frac{Si}{Na}$ ratios I seem to get precipitation occurring after I've made the solution and allowed it to cool down as seen in the picture below. This is NOT good as it becomes hard to physically handle (like a viscous gel) and limits reaction with the aluminosilicate source; it essentially becomes useless to me and my experiments. This precipitated mixture is made with the following feedstocks:

97.29 g Commercial Sodium Hydroxide
171.67 g Commercial Sodium Silicate
59.47 g Extra Water

This corresponds to the following mols:

0.87 mols $Si$
1.93 mols $Na$
11.04 mols $H_2O$

I've done some reading and consulted some colleagues and I still do not have a clear answer of what's going on and why it's precipitating. I've had a few ideas as to why this may be the case including the following which I explore in the rest of the question:

Atmospheric $CO_2$ Interaction.
Soluble Silica Speciation.

Atmospheric $CO_2$ interaction?:
I first wondered if it was $CO_2$ getting in and causing precipitation, so I tried two identical solutions (which I knew had previously precipitated) and exposed only one to atmosphere and they both precipitated at (almost) exactly the same time (the capped one took a bit longer to cool down). So I figured that $CO_2$ ingress is not a severely detrimental factor.
Speciation Curves:
Moving on and thinking it may be related to pH, I've been focusing on chemical speciation in the hope to work this problem out. To start, I reviewed some literature first here and there to see if there were existing silica speciation curves. I then made these speciation curves based on the literature as seen below:

I figured that the less stable silica species, like $Si(OH)\circ O_{3(aq)}^{3-}$ are more reactive due to their unstable charge and thus likely to precipitate and used a pH meter to find that the solution was extremely alkaline (i.e. $pH \geq 14$). Note that the above speciation curves were made with the following equilibria calculations seen below in Equations 1-5 which I back calculated from the mentioned literature assuming the basis of aqueous dissolution of amorphous silica
(i.e. $SiO_{2(s)} + 2H_2O_{(l)} \rightleftharpoons Si(OH)_{4(aq)}$):

Where, in general, the transitioning of silica species can be defined by Equations 6 below:

Each of the reactions above had both their product (with subscripts $y_1$ and $y_2$) and reactant (with subscripts $x_1$ and $x_2$ for silica and a for protic species quantified by linking with their equilibrium values as per the Equation 7 below:

I then worked on calculating the idealised carbonate system speciation to make certain I understood the maths and chemistry fundamentals. When applying the same logic to the silica system with the above equations and equilibria constants I got these curves:

These curves are entirely different to what I've seen in literature. I'm not sure what I'm doing wrong here or what other things I could try to quantify the system better so that I can mitigate precipitation occurring with the silica system. I thought understanding the pH of everything would be a good place to start, hence why I'm here.
---UPDATE 24/11/22---
I've been trying to understand this better from the POV of the following silicate solubility ternary plot. I've noticed that some of my solutions precipitate outside of precipitation zones (i.e. in the middle of zone 3 and not zone 2), I believe this is because of the different ways of manufacturing sodium silicate in industry. Unfortunately I have not found a link to the plot without a paywall :/ .

Please clarify: 1) The picture(s) show an electric stirrer which is not on a hotplate. Is this the correct item? 2) What sodium silicate (liquid? solid?) are you using, and what are the amounts of water, NaOH and the sodium silicate in your pictured mixture? 3) Is there any aluminum in the mixtures shown?

Apologies. I have edited the question with more information. (1), it is an electric stirrer no hotplate. (2) Liquid, refer to edited parts. (3) No aluminium at all :)

The diagrams do nor consider silicate condensation, especially at acidic pH and or higher concentration. I would expect rather prediminance diagrams with regions of dominant silicate form.

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See e.g. researchgate.net: The Stumm and Morgan diagram where silica species present in equilibrium with amorphous silica depending on pH and silica concentrations

See also Aqueous Silica and Silica Polymerisation

A lot of data is available online from commercial suppliers. PQ offers a handbook (free: Ref 1) which may explain some of the effects that you are observing.
Since you know the composition of the solution you are starting with, you can locate your mixture on Figure 3:

The ratio numbers are SiO2/Na2O by weight, and are for commercial liquids.
Another figure (#5, below) shows a very interesting viscosity minimum for certain concentrations and SiO2/Na2O ratios. While the data brochure does not address temperature as a variable, or precipitation, because that is undesired in a commercial product, you could very well assume that a sharp upward trend of the viscosity suggests a boundary line between two different behaviors, perhaps between clear and fluid versus much less fluid (gel-like) and perhaps also not so clear. Throw in the temperature variable, and you may be able to imagine what is happening to your mixture as you cool it.

The variability of your results indicates that you may be very close to a boundary line, or a metastable boundary region, so stirring or scratching or cooling might "precipitate" a phase change.
Ref 1: https://www.pqcorp.com/docs/default-source/recommended-literature/pq/sodium-silicate-solids/sodiumsilicates.pdf?sfvrsn=394ebc05_3

Thanks for the great reply! My solutions both are around the SiO2/Na2O [wt%/wt%] value of 0.9. My supplier does not have original viscosity measurements. Regardless, I'll have to go about doing these viscosity measurements myself on the solutions I've created to check whether this is the answer which explains the precipitation. It'll be hard to see the ball drop amongst the white cloudiness so I may have to do the height difference of the whole beaker by listening to the ball drop to the bottom. Any other suggestions on this?

Viscosity suggestion: start with a tared glass rod; immerse it a certain distance into the mixture; pull it out and let it drain for 15 sec +/-; weigh it. The weight increase is a measure of viscosity; this can be done to compare viscosities. A similar method uses a cup/beaker: fill it, empty it, drain it, weigh it. The increase (stuck to the inside of the container) is a measure of viscosity. You need to make sure there are no special effects (like, don't use a non-stick teflon beaker).

That wouldn't have the same units as the viscosity in the graph though so how could I compare it to the graph without making a height or time measurement and only a mass?

The viscosity has value only because it can be compared to some other viscosity. So, if the falling ball method will not work to give a time in seconds because you cannot see the ball, take a (clear) mixture you can analyze that way, get a time in seconds, then test it by a weighing test on a rod or a cup. If you make another mixture with double the time to fall, the weight test will also increase (maybe not exactly double, but you can establish a correlation). The viscosity of these mixtures is complex. You should be able to get +/- 10 - 20% estimate of real viscosity. But not on gels.

Doing some more research this seems to be one part of quite a big picture. One main thing I've been working on to get to the bottom of this is manually plotting solutions on the ternary diagram (which seems outdated as not all the precipitation points align) that I've put in my latest edit to the question.

I wonder if equilibration is slow at room temperature so that metastability is common. This would suggest that the method of preparation (e.g., time at temperature, rather than temperature alone, or even the rate of cooling) could determine the final state (clear or turbid). And mixing. There are so many ways to do the experiment that you will have to visualize what's happening from a molecule's eye view!

Interesting that you mention that! I did an experiment with identical solutions on the ternary diagram but on the "knifes edge" of precipitation. I added the sodium silicate in exp 1 at ambient temperature when it cooled down and at 80% of $T_{Max}$ in exp 2. Exp 1 precipitated within 12-24 hrs, Exp 2 did not. This is driving me crazy trying to understand this system! More info on the temperature aspect of the system is here https://chemistry.stackexchange.com/questions/169462/how-can-the-heat-equation-be-used-to-model-temperature-vs-time-in-a-solubility-e

I have since read that equilibrium is in fact slow at room temperature for silica systems such as this. For QAQC, I might have to just have fixed time periods where I stop mixing the solution. Although the system will not be at equilibrium it should, hopefully, be a consistent solution (within reason) to the point of repeatability; despite not being at the final equilibrium if you know what I mean. What are your thoughts?

A fixed (relatively short) time period could give repeatable results, but follow at least one experiment out till no further change occurs, to insure that you are selecting 1) a useful time (i.e., remains clear) and 2) not very sensitive to variations (i.e., if you are at 80-90% of cool equilibrium, any further variations should be small). This allows you to do many experiments, but if they must remain clear for-ever, you must run the chosen recipe out for a long time (maybe not for-ever), for proof of stability. Learn how to identify the knife-edge!

Sounds great! I can plot solutions for a good visualisation of repeatability. Is there anything else to consider aside from choosing specific times like (1) $t_i$ at $T_i$ (ambient temperature) where NaOH is introduced into water and the exothermic reaction begins, (2) the addition of the sodium silicate ($t_{SS}$) to the aqueous NaOH (I'm thinking when it gets to 80% of $T_{Max}$ and (3) I stop mixing the solution ($t_f$) and use it in my following experiments? Any other considerations you'd recommend? Is the rate of sodium silicate addition an important consideration because it's polymeric?

Aaaaghhh! Don't ask me how many variables you have! But, after you have identified all the variables and established a perfectly reproducible recipe, do something different which does not violate the recipe, just to see how robust your recipe really is. There may be one more (or more!) variables that are hidden. E.g., mixing. Incorporation of air. Light (photons). Gravity (nah? Do it in orbit?) Impurities that act as catalysts for rearrangement? (Maybe low mw silicates, borates, phosphates, alcohols phenols, carbonate - CO2 from air?) Exclude some, if not all...


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What is the best ratio of salt to water when using a salt solution in the electrolysis of water?
I am using the electrolysis of water to provide middle school students with a phenomena to explore chemical and physical change. The set up is producing much more hydrogen gas than oxygen gas and I am wondering if this is caused by the salt water solution I am using? I was looking to get a 2;1 ratio of hydrogen to oxygen.

Dump the salt and use sodium sulfate or magnesium sulfate. You do not want chlorine being produced from the electrolysis of saltwater.

$\ce{NaCl}$ is not a good choice for showing the electrolysis process to high school students. The electrolysis of $\ce{NaCl}$ starts by producing $\ce{Cl2}$ (and a little $\ce{O2}$) at the anode, plus $\ce{NaOH + H2}$ at the cathode. But $\ce{Cl2}$ is a bit soluble in water, so its gaseous volume is not consistent with the stoichiometric proportions. Furthermore, after some time the dissolved $\ce{Cl2}$ reacts with the dissolved $\ce{NaOH}$ according to :  $$\ce{Cl2 + 2 NaOH -> NaClO + NaCl + H2O}$$ As a consequence the solution contains now a new substance, sodium hypochlorite $\ce{NaClO}$ or its ions $\ce{Na^+ + ClO^-}$. The gaseous volume observed at the anode is much less than $1/2$ times the volume of $\ce{H2}$ produced at the cathode.
So if you want to show that the gaseous volumes produced at the cathode is twice what is produced at the anode, it is better to use a solution of sodium sulfate $\ce{Na2SO4}$, which produces exactly the gaseous volumes in a proportion $2 : 1$. The secondary reactions observed with $\ce{NaCl}$ do not occur with sodium sulfate.

As @Maurice pointed, $\ce{Na2SO4}$ is the best option. However, your results might be also affected by the electrodes material.  This is discussed in a recent article from JCE : Kihyang Kim et al., Water Electrolysis Accompanied by Side Reactions, Journal of Chemical Education 2021 98 (7), 2381-2386.


text:
Common-ion effect vs complexation
Quoting from Wikipedia,

The common-ion effect refers to the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. This behaviour is a consequence of Le Chatelier's principle for the equilibrium reaction of the ionic association/dissociation.

Let us consider an example of $\ce{AgCl}$ in aqueous solution of $\ce{NaCl}$. The following equilibrium exsist in the solution,
$$ \ce{ AgCl(s) <<=> Ag+(aq) + Cl-(aq)} $$
According to the common-ion effect, as we increase the concentration of $\ce{NaCl}$ in the solution the equilibrium should shifts towards left side i.e., solubility of $\ce{AgCl}$ decreases with increase in the concentration of $\ce{Cl-}$ ions. But what I think here is that, can't the excess $\ce{Cl-}$ ions forms a soluble complex with the $\ce{AgCl}$ precipitate? That reaction is as follows,
$$ \ce{ AgCl + Cl- <=> [AgCl2]-} $$
If that's the case then, how to judge the solubility with these two competing factors?

If by 'judge' you mean 'calculate', then the answer is: make a system with all the equilibrium expressions involved, meaning the solubility product, equilibrium constant for the complex, and whatever else is needed, and solve. This is actually a rather frequent situation in inorganic analytical chemistry. You can stop certain rather insoluble salts from precipitating, by forming complexes. The effect of pH on precipitation is a related example. Just take care of handling the case where the precipitate disappears, because then you no longer have the solubility product.

BTW, just remembered that this happens with silver and cyanide: https://en.wikipedia.org/wiki/Silver_cyanide#Reactions . As you add more cyanide, the initially formed precipitate redissolves.

If there is some serious complexation (high K) than common ion is irrelevant.

OTOH, there can be precipitation of Ag+ and AgX2^-, or Hg^2+ and [HgX4]^2-, if the excess of the complexant was just temporary. E.g. during progressive addition of metal salt solution to the solution of the complexant.

If that's the case then, how to judge the solubility with these two competing factors?

Solubility is how much of a single substance you can dissolve in pure solvent. For the common ion effect, you add a second solute, so at that point it does not make sense to talk about solubility anymore. Thus, the Wikipedia definition cited is a little bit awkward and could be improved. Here is another definition that avoids the problem:

The common-ion effect is used to describe the effect on an existing equilibrium by the addition of a second substance that contains an ion common to the equilibrium. If several salts are present in a system, they all ionize in the solution.

Source: https://ecampusontario.pressbooks.pub/genchemforgeegees/chapter/6-1-common-ion-effect/
Instead, you can discuss how much solid precipitated, as an easy quantitative description of what is going on.
If you want to demonstrate the common ion effect for the reaction $$ \ce{ AgCl(s) <<=> Ag+(aq) + Cl-(aq)} $$
you can add some silver nitrate. This will increase the mass of precipitated silver chloride as predicted from the common ion effect.
Apart from complexation, precipitation of other species also lead to results you might not expect at first. If you add sodium bromide (no common ion), it will precipitate some silver bromide, and some silver chloride will dissolve.

@Poutnik I think a simple and limited answer is a first step. A second answer delving into the complexity of complexation would be great, and I invite everyone (except me) to write it.

Sure. But we need not extremes.  :-)


text:
Why are cyanamides more electrophilic than nitriles?
According to https://www.researchgate.net/publication/6642597_A_generally_applicable_method_for_assessing_the_electrophilicity_and_reactivity_of_diverse_nitrile-containing_compounds, cyanamides (compounds with a N-C≡N group) are more electrophilic than aliphatic nitriles (C-C≡N).
For reference, see the electrophilicity scale on Figure 3 on the link provided. In this scale, you can see a cyanamide marked with an 8, and an aliphatic nitrile as the first structure to the left.
Now, shouldn't the presence of an electron-donating N atom make the nitrile less electrophilic? Compare this to the amide x ketone example usually taught in organic chemistry. Also, notice on Figure 3 that the aromatic rings get progressively more electron-deficient as we move from left to right.
I have thought of two possible explanations:

The presence of the N leads to delocalization, which in turn would make the cyanamide a "softer" electrophile (important: this scale on Figure 3 refers to a reaction with a thiol).
A CN group has two pi bonds, so the nitrogen's +R resonance effect happens on one pi bond, while the nucleophilic attack happens on the other, effectively making the N a -I group. In other words, while a nitrile would have two degenerated LUMOs, the cyanamide has one with less energy (due to the +R effect from N) and one with more energy (due to the -I effect).


text:
Adiabatic Venting of Propane Tank
A few weeks ago, a question was submitted regarding adiabatic venting of a chlorine cylinder:
Issues with Chlorine Liquefaction
The member submitting the question was interested in why the stream exiting a chlorine tank became cold, particularly why the exit valve became cold.  @poutnik presented and excellent answer to these questions, and also referred to the case of propane tanks which display similar behavior.  I presented a similar discussion for the case of an ideal gas.
Subsequently, I became interested in how to quantitatively predict this kind of behavior, say for the typical case of venting a propane tank. Does anyone have any ideas on how to thermodynamically model such a situation?  In such a model, the liquid propane and propane vapor would be gravitationally segregated, and only propane vapor would be able to exit the tank through a valve situated at the top.  I was thinking of applying the open system (control volume) version of the 1st law of thermodynamics to solve for the temperature as a function of the mass of propane remaining in the insulated tank. I envision the model to include relative changes in the amounts of liquid and vapor in the tank as a result of liquid vaporization and vapor loss through the vent.  I would also envision he model to treat the vapor as non-ideal because of the relatively high pressures initially (say, > 10 bars). So, in my judgment, use of thermodynamic tables on propane would be preferable over analytic representations of thermodynamic properties.
Thoughts?

Since I have not yet received any answers to my question, I will post the answer I arrived at using the methodology I alluded to in the question.
The analysis makes use of the open system (control volume) version of the 1st law of thermodynamics, which, for the present transient adiabatic venting situation, reduces to $$d(mu)=h_Vdm=(u_V+Pv_V)dm\tag{1}$$where m is the mass of propane within the tank at any time, u is the average internal energy per unit mass (averaged over vapor and liquid), P is the equilibrium vapor pressure of the propane in the tank, and $v_V$ is the specific volume of the vapor.  Eqn.1 tells us that the differential changes in internal energy of the liquid/vapor combined mass in the tank are equal to the differential amount of vapor enthalpy exiting the tank through the valve.
In addition to Eqn. 1, we have $$u=u_Lx+u_V(1-x)\tag{2}$$and $$v_Lx+v_V(1-x)=\frac{V}{m}\tag{3}$$where x is the mass fraction of liquid in the tank, the u's and v's are the specific internal energies and specific volumes, respectively, and V is the volume of the tank.
I combined the above 3 equations into a single differential equation (by eliminating the mass fraction x), and, after considerable mathematical manipulation, arrived at the following differential equation for the changes in propane mass in the tank along the equilibrium venting trajectory:$$d\ln{m}=\frac{dh_V}{\lambda v_V}+\frac{V}{mv_
V}d\ln{(\lambda/T)}-d\ln{(\lambda v_V)}\tag{4}$$where $$\lambda=\frac{h_V-h_L}{v_V-v_L}$$and where the h's are enthalpies.
I integrated Eqn. 4 numerically for the test case of a 33.4 liter insulated propane tank initially at 300 K and 10.0 bars and initially containing 8.4 kg propane, using the thermodynamic data presented in the following table (Thermopedia):

The results of the calculations are shown in the figure below.

The beginning of venting is at 300 K and 8.4 kg propane, the upper right point of the line.  The line shows the trajectory of temperature vs propane amount remaining in the adiabatic tank at any time. These results seem consistent with the description provided by @poutnik in the original thread.


text:
Why is aluminum a less active metal than magnesium despite having a lower first ionization energy?
I was thinking about the general trend in metal activity (reactivity increases down a group and decreases across a period) when I started thinking about aluminum and how it has a lower first ionization energy than magnesium because its highest-energy electron is in a high-energy 3p orbital. Despite this, magnesium is the more active metal. Is this because activity measures the potential of the oxidation of Al t0 Al3+, not Al to Al+? Although it may not take much energy to remove the first electron from Al, oxidizing Al+ to Al3+ requires removing electrons from the lower energy 3s orbital, making the overall oxidation less favorable than the oxidation of Mg to Mg2+. Is this thinking correct?

Because neither aluminum nor magnesium forms many compounds by contributing one electron to a bond, whether ionic or covalent. So first ionization energy really does not have much to tell.
Neither do second or third ionization energies. The reactivities of aluminum and magnesium are hard to compare on the basis of any ionization energies because the elements have different common valences and aluminum does not form very many predominantly ionic compounds. No matter what you try to do with ionization energies, you're unavoidably comparing apples and olives.

Another reason is the fact that metallic aluminum is always covered by a thin film of alumina (aluminum oxide $\ce{Al2O3}$). This film is transparent, continuous and waterproof. It is formed spontaneously in the contact of air, and prevents the contact with reagents. If this film is removed by scratching, it will be immediately redone by atmospheric air. This is why, apparently the metallic aluminum objects do not react with many usual ionic solutions. For example, the reaction with dilute $\ce{HCl}$ solutions is relatively slow to start, because alumina film is rather resistant to acidic corrosion.
The production of this film can be prevented by mercury chloride $\ce{HgCl2}$. If a piece of aluminum is dipped into a solution of $\ce{HgCl2}$, some mercury chloride is able to cross this film and react with aluminum to produce mercury $\ce{Hg}$ (which soon makes an aluminum amalgam $\ce{Al(Hg)}$) according to the reaction : $$\ce{2 Al + 3 HgCl2 -> 2 AlCl3 + 3 Hg}$$
Of course, the aluminum amalgam $\ce{Al(Hg)}$ is soon oxidized in air, but the alumina film does not adhere on the outer surface of the amalgamated aluminum. As a consequence, the oxidation proceeds in a couple of minutes and produces long white filaments of alumina which are "growing" on the surface of the aluminum sample. This phenomena lasts as long as there is still some mercury on the outer surface of the aluminum sample.

Does this mean that the reactivity series is determined empirically, since we can say that aluminum is less reactive due to the observed slower reactions (caused by the oxide layer)?

I guess the reactivity series has been determined by comparing redox potentials.


text:
How does xenon react with O₂F₂?
This reaction was asked in my college entrance exam. I searched through the entire Internet, but I couldn't find any papers/sources.
I suspect since O₂F₂ is a strong fluorinating agent, one of the xenon fluorides (XeF₂, XeF₄, or XeF6) will form, but I am not sure. Reaction conditions were not given in the question.

This was asked in JEE advance this year (right ?). I gave it too btw.

$$\ce{XeF4 + O2F2 ->[143K] XeF6 + O2}$$ source:- https://ncert.nic.in/ncerts/l/lech107.pdf page 40 in the pdf

From my research, it really depends on the experimental conditions and procedure. One paper in 1965 found that primarily $\ce{XeF2}$ was formed when xenon reacted with $\ce{O2F2}$ at low temperatures$^1$. In contrast, another paper found a mixture of $\ce{XeF2}$ and $\ce{XeF4}$ was formed, and prolonged exposure to $\ce{O2F2}$ generated mostly $\ce{XeF4}$ but no $\ce{XeF6}$ was observed$^2$.
Sources:

https://pubs.acs.org/doi/10.1021/ic50027a038
https://pubs.acs.org/doi/10.1021/ic00334a037

How does a Xenon fluorine bond work exactly? I thought Xenon is a noble gas so it wouldn't react

@SidharthGhoshal, Xenon is the least-noble of the non-radioactive noble gasses, and fluorine is extremely reactive.  This lets the fluorine bond to the outer-shell electrons of xenon.

@SidharthGhoshal This should answer your question: https://chemistry.stackexchange.com/questions/31845/why-are-there-more-fluoride-compounds-formed-with-xenon


text:
How do I get a pH of 2 with 37% HCl?
I am following a guide on seed germination of a rare wild species. Others who have worked with the seeds suggest soaking at pH 2 (adjusting with HCl). I have a stock solution of 37% HCl and I want to make 250 ml of solution total. How much distilled water and how much HCl should I add to get a pH of 2?

pH is equal to $2$ if the concentration is equal to $0.01$ mol/L (= $0.365$ g HCl/L)

Thank you for your help @Maurice. I found online that 37% HCl = 12 mol/L.  So I could essentially pipette 0.12 mL HCl to every 1 mL of water? So if I wanted 250 mL total solution I would multiply 0.12 * 250 = 30 mL HCl?

No, it means you need to dilute the acid 1200 times.

No, that's not how all of it works. You grab 1 L of your 37% acid; that's 12 mol. You divide that by 12; that's 1 mol. You divide it further by 100, that would be 0.01 mol, and that's how much (or how little, actually) of it you need per 1 L of water.

Thank you for your patience with me, LOL! So if I read this right, Mithoron and Ivan Neretin are saying the same thing. 1 part HCl  to 1200 parts water. It's crazy that such a small amount of acid still has such a low pH!!!

Just a side note regarding this sentence "suggest soaking at pH 2 (adjusting with HCl)". Usually, when people say 'adjusting with', they mean that you measure the pH and adjust it to the desired value by adding something, in this case HCl. But I don't know the details or constraints of your experiment. In any case, be warned that given the relatively low (0.01 M) concentration of HCl in your solution, if you then add to it a significant amount of something that contains bases, the pH can vary, it's not buffered. E.g. even a very small amount of ammonia would dramatically alter the pH.

Putting together all the helpful comments into a single answer! Thanks to @Maurice we know that 2 pH = 0.01 mol/L). We can use that in @Ivan Neretin comment who says to divide 1 L of HCl by 12 (the molarity of 37% HCl), then divide that again by 100. This concurs with the comment by @Mithoron who suggests diluting the HCl 1200 times.
So, to get 250 mL of solution with a pH of 2 using 37% HCl, I would divide 250 by 1200 = 0.21 mL HCl.

@Mithoron I agree that some editing is required. I returned the post to its original state.


text:
How would I go about determining the concentration of ammonia in Windex Original Glass Cleaner, post titration?
so as seen in the table below, we did our 4 titres plus 1 sighter with Hydrochloric Acid (HCl) (.1M) successfully as the Windex solution with 3 drops of methyl-orange already in, (our indicator) turned to the colour we wanted and were all within .1.
The whole point of this investigation is to measure the concentration of Ammonia in Windex Original Glass Cleaner as a % for the 4 Titres and then of course get an average percentage out of the 4.
However as "Windex" does not have a chemical equation, its not as simple as measuring acetic acid in vinegar as a %, so I dont how to go about my molar/percentage steps.
Did I have all the required information? Or do i need more?
So I am unsure of how to go about finding this? Any info would be great :)

It IS as simple as determining acetic acid in vinegar which does not have "chemical equation" either. You mean chemical formulas, CH3COOH for acetic acid, NH3 for ammonia. For both, you need to involve sample volume.

Please give us the volume of the solution to be titrated by HCl 0.1 M. Without this data, nothing can be done. And what is the meaning of "Sighter" ?

You do need to assume that Windex can be approximated as a solution of ammonia in water with nothing else that interferes with the titration. That's probably reasonable.

@Maurice I guess the sighter is rough quick shot to have idea about the expected volume. The next fine titrations can go fast until they are approaching the expected volume.

@Maurice, yes sorry, the volume of the windex solution is 20mL or .02L to be titrated by HCl. Sighter is just a rough estimate about the expected volume

Methyl Orange changes color from yellow to red as pH drops from 4.4 to 3.1.
Unless you are titrating a solution with other ingredients that buffer at a higher pH, or react with HCl, your results will be accurate for amonnia.
You can get an MSDS sheet or ingredient list for Windex to make sure.
If any other ingredients interfere with the analysis, purification steps can be done.
With Windex, a chemist might separate the ingredients on a Gas Chromatography column, and test vs a standard solution (of known NH3 concentration) using a detector that works for ammonia.
Another way would be to titrate with a pH meter, rather than an indicator.
You will see different pH "plateaus" for different functional groups, such as carboxyl, or phosphate, if they are present.

Just a few information to help you solve this problem. $\ce{NH3}$  reacts with $\ce{HCl}$ according to :$\ce{NH3 + HCl -> NH4^+ + Cl^-}$. So, your tested solution ($20$ mL) must contain the same amount $\ce{NH3}$ (in moles) as $\ce{HCl}$ from the buret ($\ce{13.5 mL HCl 0.1 M}$). So do calculate this amount of $\ce{HCl}$ (in moles) and divide by $0.02$ L.


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Which gases(or gas) are evolved in the leaching of Galena(PbS) with conc. Nitric acid?
I want to know the gases produced(primarily nitrogen oxides) in leaching of Galena in nitric acid and the effect of concentration of nitric acid on the gases evolved.
I found papers about the efficiency of leaching in different concentrations of nitric acid but none mentioned which gases are produced.

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It helps preventing  others to tell you what you already know or what you could easily find yourself.

Some information about a similar dissolution : pyrite $\ce{FeS2}$ in acidic media. Strangely enough, the reaction pyrite + acids is not yet well known. Whatever the proposed equation, the temperature, the concentration the nature of the acid ($\ce{HCl, HNO3, HClO4}$), and the dimension of the grains, the ratio Sulfur/Iron is NEVER equal to 2 at the end of the reaction. It changes from report to report, but it is usually about 1.5. Some sulfur is always lost somewhere. Of course, Sulfur may be produced in elementary form $\ce{S8}$, in a gaseous form $\ce{SO2}$ or in various polyoxygenated ions.

$62$  publications have been written about the problem of pyrite dissolution in acidic media. and analyzed, then compared by M. Descostes et al., in a $10$ pages report : Pyrite Dissolution in Acidic Media, Geochimica Cosmochimnica Acta, Vol. 68. No. 22, p. 4559 - 4569, 2004. Nineteen equations have been compared. None is satisfactory. Descostes has also analyzed the system by numerous methods (spectroscopy, X-ray, solution chemistry, ion chromatography, etc.) In vain. Sulfur is lacking in all explanations.

Nitric oxide: https://chemistry.stackexchange.com/questions/51543/finding-the-amount-of-nitrous-oxide-produced-in-the-reaction-between-lead-sulfid

The answer is hydrogen sulfide with dilute nitric acid, otherwise little to none using the concentrated acid.
Here is a source: "Leaching of Lead Sulfide with Nitric Acid", quoting:

Galena, lead sulfide, is leached with nitric acid in order to prepare higher value-added products such as lead nitrate, lead carbonate, lead basic carbonate, etc. Leaching behavior has varied with the concentration of nitric acid. At the lower concentration of nitric acid, galena is dissolved into lead nitrate and hydrogen sulfide. However, at the higher concentration of nitric acid, galena is not dissolved but it is changed into lead sulfate as precipitate. The leaching efficiency is strongly dependent upon temperature and nitric acid concentration. The optimal conditions are found to be 40°C and 4–8 N nitric acid where about 95% of lead can be dissolved from lead sulfide.

I hope this helps.
For those students interested in the advanced underlying chemistry surrounding the action of hydrogen sulfide on nitric acid and associated nitric oxides, see this source: "Potential Biological Chemistry of Hydrogen Sulfide (H2S) with the Nitrogen Oxides".

Adding to the previous answer (with a fume hood recommendation) is that $\ce{NO2}$ and $\ce{SO2}$ can also evolve using the concentrated acid at higher temperatures.
It is really interesting that $\ce{H2S}$ (gas) can be produced with dilute $\ce{HNO3}$ conditions because hydrogen carries a very slightly stronger positive charge than $\ce{Pb^2+}$, $\ce{NO3-}$ serve as counter ions, and the oxidation rate of sulfur is very low.
With more concentrated nitric acid at higher temperatures, the oxidation of sulfur rate becomes higher than the evolution of $\ce{H2S}$ gas as:
$\ce{HNO3 -> NO2 (g) + H2O}$
$\ce{ [S] atoms -> $\ce{SO2}$ (g) -> SO4^2-}$
resulting in $\ce{PbSO4}$ salt.  The loss of $\ce{H2S}$ and $\ce{SO2}$ during the reaction explains the less than 2:1 ratio of $\ce{S}$ to $\ce{Pb}$ after the galena leaching reaction is complete.

Robert, I would amend your reply as your 'evolution' link leads to a white paper noting sulfur and nitric acid reaction. However, you answer largely remains correct as per my advanced source reference, to quote: "Electrochemical evidence indicates −SH acts as a weaker two electron reducing agent (forming So)", where here the  −SH ion is created from H2S/H2O. However, the direct action of HNO3 with −SH ion appears to be NO which can even further react with H2S,

@AJKOER noted.  The "tipping" point where higher conc. HNO3/heat increases rate of oxidation of H2S is of interest to me, as there are similarities to silver recovery reactions I have done.


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Definition of activation energy
There seems to be contradictions in the definitions of activation energy ($E_a$) stated by different sources. I found two such popular definitions. Before that, I state my understanding of how an elementary reaction generally occurs, as it will be used to ask the main question.
The reactants have a certain potential energy $U_i$ per molecule (which is constant for each molecule unless the molecules come too close). The reactants must pass through the transition state in order to form products, and the configuration in the transition state has a potential energy $U$ (per molecule) associated with it. Thus the goal is to increase the potential energy of the reactants from $U_i$ to $U$. In order to do this, the configuration of the molecules of the reactants must change, which can either be achieved by physical distortion of the molecule, or changing their location in space (because electrostatic interactions are present). The former can be achieved by collisions, and the latter by the molecules just approaching each other and coming too close. In any case, the increasing potential energy must be compensated by a decrease in the molecules' kinetic energies. If the molecules have kinetic energies $K_1$ and $K_2$, then those two molecules would be able to convert into products if $K_1 + K_2 \geq 2(U - U_i)$. If I assume $K_1 = K_2 = K$ in order to simplify things for the following discussion, then the condition to form products would become $K \geq U - U_i$.

Definition 1:

Activation energy is the minimum amount of energy that must be
provided to the reactants for them to form products. (source:
Wikipedia) (not stated exactly)

This implies that $E_a$ is the additional energy that needs to be supplied to the reactants in order to form products. According to this definition, $E_a = U - U_i$. I can rephrase that into this (using the relation $K \geq U - U_i$): $E_a$ is the minimum kinetic energy a molecule should possess in order to form products.
Definition 2:

The activation energy is the minimum energy reactants must have in
order to form products. (source: Physical Chemistry by Peter Atkins)

This definition suggests that $E_a = U$, which is not in agreement to the first one. Someone could argue that energy in this definition means kinetic energy, and then the definition is equivalent to the first one. But in the first definition, energy was interpreted as potential energy, and that should be the case with this definition too. If I instead interpret it as kinetic energy, then again these two definitions contradict each other. Also, in the text preceding this definition, kinetic energy was not mentioned anywhere, so it is natural to think they are talking about potential energy.

Why does it matter so much?
Activation energy appears in the Arrhenius equation.
$\begin{equation}
k = Ae^{-E_a/RT}
\end{equation}$
The two definitions would give different values of $k$, so obviously one of the definitions is wrong.
(I am currently studying chemical kinetics)

So what should I answer?
I would like you to clarify all this and state a final definition along with interpretation (and verifying my understanding of a reaction will be much appreciated). Also, I have come across the term threshold energy while going through a few sources, and they define it to be $U$ (i.e., potential energy of the activated complex). This leads me into thinking that $E_a = U - U_i$, and so the second definition might be wrong.

Distortion may happen and may be necessary for some reactions. But it is not necessary generally and may not be even possible.

It may depend on what you mean by distortion. Consider 2 O -> O2. PE increase may be based on initial electrostatic repulsion, without distortion in sense of change of atomic nuclei relative placement.

Chemistry is the part of physics about electrostatic interactions of atoms. The Coulomb law is the basis of quantum chemistry.

Is the reason for all this you didn't got that if you "provide" energy to reactants then they "have" energy? Because it sounds like that...

@Mithoron I do not quite understand what you mean

Then read your definitions again.

It seems as if it is a matter of where the zero of energy is taken to be. Definition 1 assumes that the energy is initially zero, and definition 2 assumes that it is the current thermal energy. Both are correct although neither states things clearly.

@Mithoron Obviously if you provide energy then the reactants have energy. But there is a difference between "amount of energy provided" and "amount of energy they have".

A few basic ideas: A molecule, compound, atom, etc. is stable [possibly only fleetingly so] because it is at a relative energy minimum. For a moiety to react energy must be available to break a bond or attraction. This energy can come from the molecule itself either from the initial KE of the molecules that collide and form a temporary attraction or from being in such a high rotational or vibrational state that the molecular bond is broken [possibly from photoinitiation]. At any temperature above zero Kelvin there is a distribution of energies; the higher the temperature the greater the distribution and the more molecules that either have enough individual energy to become unstable or in collisions with enough total kinetic energy to either break bonds or to cause orbitals to readjust and form orbitals of lower energy [not necessarily of lower energy than the original bonds, transition states can lead to reactive intermediates]. This latter also requires a method to release the excess energy.  The activation energy is not added energy it is inherent in the distribution of molecular energies that is a function of the temperature. The molecules with the higher KEs might have it.
The idea is that there is a required energy for a reaction to happen, the activation energy. The activation energy comes from the KINETIC energy in the heat content of the molecules. The reaction energy comes from the different POTENTAL energies from the different bond strengths between the reactants and products in other words how close the electrons can get to the various nuclei. How precise this value is I have no idea, but precisions in kinetics and calculations are usually in the low parts per thousand. The energy or enthalpy change is measured or calculated, and the activation energy is calculated from the kinetics at several temperatures.
The various definitions seem, to me at least, to be fumbled attempts to explain something relatively simple in a more complicated manner.


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How does carbon dioxide in the atmosphere keep heat trapped?
In the classic description of how greenhouses gases are increasing the temperature of Earth's surface, shorter wavelengths of light like visible and ultraviolet (UV) penetrate the atmosphere, hit the ground, and come back as infrared light that cannot pass through the carbon dioxide in the atmosphere. According to my research, $\ce{CO2}$ absorbs wavelengths around $15 \,\pu{\mu m}$ or infrared best. When molecules absorb energy, it does so by bumping electrons to higher energy levels.
I have the following questions now:

Why can $\ce{CO2}$ not absorb arbitrarily high amounts of energy?

Why is the wavelength of light when it hits earth different than what is spit out when an electron goes back to ground?

Your following sentence is wrong : "When a molecule absorbs energy it does so by bumping electrons to higher energy levels." This is only true for photons in the visible and UV range. In your case (infrared), molecules absorb energy by increasing their vibration energy, without changing electronic energy.

See: https://chemistry.stackexchange.com/questions/81242/what-properties-of-carbon-dioxide-make-it-a-greenhouse-gas

$\ce{CO2}$, like all other matter, can only absorb photons of energies corresponding to some kind of quantum transition with a high transition probability, whether that is a rotational, vibrational, electronic, or nuclear transition. These possible transitions are not continuously distributed over all energies. Within the IR region of light, only a few vibrational transitions of $\ce{CO2}$ can absorb light with high probability.
The light released from the earth's surface, just like the sun's surface, is blackbody radiation. The blackbody radiation spectrum is temperature dependent. Higher temperatures allow for release of more photons of higher energy. The sun is much hotter than earth so it releases more visible blackbody radiation while the low-temperatures of Earth's surface gives a peak in the blackbody radiation spectrum squarely in the IR.

In our daily experience, we are used to different colors of solid objects. The absorption spectra of solids often have broad peaks, whereas the absorption spectra of gases typically have narrow peaks. This is because gases lack the intermolecular interactions that broaden the peaks in solids.

Your first question. A CO2 molecule can absorb many quanta one after the other provided the wavelength is correct and there are sufficient number of photons /second. This can happen if you excite with a v powerful ir. laser. Normally, in the atmosphere, the number of photons is so small that the CO2 has radiated the extra energy away or lost it in a collision.
Second question. If the wavelength striking the earth is in the visible this causes an electronic transition in any molecule absorbing it, say chlorophyll in a plant. This energy is used to make biomass but the process is not 100% efficient so some energy is lost and this becomes heat in the surroundings. In rocks and soil, electronic transitions also occur but now the excited state decays by transferring energy to the surroundings in non-radiative (i.e. 'dark') transitions to the ground state and again heating nearby molecules.
Third question. The CO2 does not of itself keep the heat trapped in the atmosphere but clearly is involved in doing so. The CO2 absorbs ir radiation emitted from warm molecules on the earth's surface, warmed by the processes described above. After absorption a warmed CO2 can radiate into space or back the the ground again and this process can be repeated many times so most photons would eventually get radiated away if nothing else happens. However, most of the atmosphere is O2 and N2 (CO2 is at $\approx 0.04$% so tiny) and as a result any CO2 suffers numerous collisions/second which can remove any excess vibrational/rotational energy present as a result of absorbing ir radiation. The O2/N2 are now hotter than the were but cannot radiate away this energy (as CO2 can) because they have no dipole, and so by colliding with other similar molecules spread the energy and the atmosphere warms.


text:
What kind of solution is formed when alkali metals are dissolved at high concentration in liquid ammonia?

When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution. $$\ce{M + (x + y) NH3 -> M+(NH3)_x + e^-(NH3)_y}$$
The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is blue in colour.
At a higher concentration (3 M), clusters of metal ions are formed. This causes the solution to attain a copper–bronze colour and a characteristic metallic lustre.
Source: byjus

So when the concentration of a solution of liquid ammonia is above $\pu{3 M}$, the alkali metal ions form clusters which give the solution a copper-bronze color.
Can I call such a solution an associated sol (solid in liquid colloid) similar to micelle formation in water?
Textbook I am using "Sudarshan Guha's J. D. Lee Concise Inorganic Chemistry"

Comments are not for extended discussion; this conversation has been moved to chat.

Search for articles or work where Pavel Jungwirth  contributed. He worked a lot on various aspects of alkali metals + liquid ammonia or water chemistry. Few of many the Nature article about Coulombic explosion or also about metallic like solutions. See. e.g. metallic water

https://chemistry.stackexchange.com/questions/82874/solutions-of-group-1-and-group-2-metals-in-ammonia

Does this answer your question? How to think of solvated electrons?

Are you speaking of neutral metallic sodium $\ce{Na}$ (as in the title), or of ions $\ce{Na+}$  as in the first sentence ? Because ions are not soluble in liquid ammonia.

metallic sodium, but doesn't the metallic sodium leave ions and electrons when dissolved in liquid ammonia ? @Maurice

There are solvated metallic ions and electrons, with the latter causing dark blue or bronze colour, slowly reducing ammonia to hydrogen and amide.

Please give also the name of the book's editor, its year of publication, and maybe the ISBN number.

Before asking what the nature of the solution is, let's fact-check the unsupported statements made:

[OP] when the concentration of a solution of liquid ammonia is above 3 M, the alkali metal ions form clusters which give the solution a copper-bronze color.

The phrase "solution of liquid ammonia is above 3 M" is unclear. From the context, it should probably read "solution of alkali metals in liquid ammonia with a concentration above 3 mol/L". The statement "the alkali metal ions form clusters" is misleading; according to Zahn (DOI: 10.1039/C7RA11462A), single metal ions are surrounded by ammonia and amide ($\ce{NH2-}$), forming clusters. This is different from metal clusters, which have metal-metal interactions.
Finally, if there are no metal-metal interactions, it is unlikely that the color is related to them. A source referencing the different colors in this Libretext page, which mentions a threshold of 3 mol/L for a color and conductivity change. The conductivity is due to the solvated electrons, so maybe the colors are also related to them rather than the metal ions.
This paper concludes about the changes at high concentrations:

The present study shows that the electrolyte-to-metal transition in increasingly concentrated alkali metal–liquid ammonia solutions is a gradual process rather than an abrupt first-order transition, which is in line with previous suggestions (1). From the molecular point of view, this transition may be understood in a simplified way as gradual coalescence of individual solvated electrons and dielectrons upon increasing alkali metal doping, with the metallic behavior appearing around the percolation threshold.

For readers outside of the specialty (like me), the paper is difficult to read without further study of the topic and the cited literature, but the quoted conclusion shows that it is complicated, and seems to imply that it is the combination of electrons and metal cations that results in the metallic behavior.

[OP] Can I call such a solution an associated sol (solid in liquid colloid) similar to micelle formation in water?

No, the metal ions are more like hydrated ions, with solvent molecules forming a first and maybe second shell of solvation. Micelles have multiple solutes "clumped" together, so that is not the best analogy. The metal ions are not solids, and the particles (clusters?) are not large enough to be called colloids. Liquids containing solids are not translucent (like milk or blood), with particles whose size is larger than the wavelength of light.

From content by Rachel Bain, at the University of Wisconsin, to quote:

When sodium metal is added to ammonia, some of the sodium dissolves. Each of the dissolving sodium atoms loses an electron and becomes a cation. Both the cation and the free electron are solvated by ammonia molecules.

$\ce{Na(s) + (x + y) NH3(l) → Na(NH3)^x+(aq) + e^-(NH3)y(aq) (blue)}$

The blue color is characteristic of a solution containing solvated electrons; as the solution becomes more concentrated, it takes on a bronze color. The bubbles appearing as the reaction proceeds are hydrogen gas that is formed in a second reaction between the sodium and ammonia:

$\ce{2 Na(s) + 2 NH3(l) → 2 NaNH2(aq) + H2(g)}$

And, further in a narration section:

A small piece of sodium is cut to expose a fresh surface. The sodium is dropped into liquid ammonia at a temperature of approximately -33 degrees Celsius. Some of the sodium dissolves, forming sodium cations surrounded by ammonia molecules and electrons surrounded by ammonia molecules. The solvated electrons give the blue color to the solution. Because of the mobility of the electrons, the solution is a good electrical conductor. Bubbles of hydrogen gas are formed by a second reaction that also produces sodium amide. More concentrated solutions appear bronze-colored and have a conductivity similar to metals.

which is all in accord with provided comments above.

Please don't just reference comments, rather include relevant information from the comments into your post. Comments may be deleted.


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How does heating alkali metals with graphite change its conductivity and density?
My textbook says when heating potassium, rubidium, cesium with graphite, the metals react by invading the space between the layers. They form highly colored interstitial carbides that are non stoichiometric in nature (due to variable composition).
Do the intercalated metal atoms increase the conductivity ( maybe as alkali metal ions give away an extra valence electron) and density (because of interstitial defects) of graphite?

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As you've already noted, heating the alkali metals on graphite causes the alkali metals to enter the interstitial spaces within the graphite layers.
An example of this would be $\ce{(CaC6)_n}$, whose structure is given below. The grey balls signify carbon and purple ones signify calcium

Do the intercalated metal atoms increase the conductivity ( maybe as alkali metal ions give away an extra valence electron) and density (because of interstitial defects) of graphite?

Let's talk about the more intuitive one first, density. The best way to understand this is by thinking about what exactly happens.
Graphite is a carbon allotrope that is made up of layers of sheets of carbon (graphene) where each layer has carbon atoms arranged in a trigonal planar manner. There are two types of graphite depending on how the graphene is stacked - α$-$graphene where the layers are placed in a hexagonal nature and β$-$graphene where the layers are placed in a rhombohedral nature but I digress.
Now, when an interstitial carbide is formed, the alkali metals are seen to occupy the spaces between the carbon atoms. Now, what happens if an atom replaces empty space? The mass increases volume remains the same and therefore density increases.
Now, time to talk about the conductivity. Here I refer to Carbon 1966, 4 (1), 125–127, which talks about the electrical conductivity of different types of potassium carbides.

TABLE $1$. RELATIVE CHANGES IN RESISTANCE WITH COMPOUND FORMATION
\begin{array}{lc}
\text{Compound} & R/R_o \\ \hline
\text{Initial Graphite} & 1.0\\
\ce{KC60} & 0.156\\
\ce{KC36} & 0.093\\
\ce{KC24} & 0.08\\
\ce{KC8} & 0.037\\ \hline
\end{array}

Here $R$ refers to the resistance of the compound and $R_o$ refers to the resistance of the interstitial carbide. Therefore, as you can see the conductivity increases with the amount of alkali metal being added to the spaces between the layers.

Now, what happens if an atom replaces empty space? Hmm, that assumes the layer distance does not change....


text:
Why does lead gains electrons in the cathode?
In lead(II) bromide electrolysis, lead is attracted to the cathode and gains electrons. Why does it gain electrons when losing them would be easier since its charge is +2 meaning it would need to lose electrons to obtain an entire outer shell. This would also apply to bromine, as its charge is -1

Less favoured operations often happen if they are forced to happen.

At equilibrium of the open circuit state, there is ongoing the dynamic chemical equilibrium:
$$\ce{Pb(s) <=> Pb(aq)^2+ + 2 e-}$$
When the circuit is closed and the electrode potential is externally forced to change from the equilibrium value, then:

If the potential is lowered, e.g. becoming the cathode during electrolysis, then the backward electrode reaction gets faster rate and the lead 8s being deposited.
If the potential is raised, e.g. becoming the anode in the galvanic mode, or if electrolysis is reversed, then the forward electrode reaction gets faster rate and the lead is being dissolved.


text:
Total entropy change for reversible reactions and its relationship to equilibrium constant
I'm in year 12 of school doing A2 chemistry learning about entropy and its relationship to reversible equilibria. I learned about the following three ideas from my course book (verbatim), for a reaction at equilibrium:

$\Delta S_{\text{total}} = 0$
$\Delta S_{\text{total}}$[forward reaction] = $\Delta S_{\text{total}}$[backward reaction].
$\Delta S_{\text{total}} = R\ln K$, where $R$ is the gas constant and $K$ is either $K_c$ or $K_p$.

I'm confused about points 1 and 3 and would like to know if my understanding of 2 is correct.
For point 2, does it imply that at equilibrium the total entropy change from both the forward and backward reactions are zero, but if the forward reaction is say $\Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 22 + (-22)$, then the backward reaction has to be $\Delta S_{\text{sys}} + \Delta S_{\text{surr}} = (-22) + 22$ (as opposed to something like  $(-60) + 60$ which is also zero but the individual magnitudes are different)?
For points 1 and 3, if they're both agreed to be true then $\Delta S_{\text{total}} = 0=R\ln K \implies K = 1$, no matter the reaction. This result is just absurd because it's implying all reversible reactions have a position of equilibrium close to the middle. What exactly is going wrong? Later on in the book, there is another contradiction where it provides the following reaction at equilibrium:
$$\ce{SO2(g) + 1/2O2(g) <=> SO3(g)}\quad\Delta S^\circ_{\text{total}}=238.3 JK^{-1}mol^{-1}$$
How come the total entropy change is nonzero during equilibrium? My guess is that it has something to with the fact that this entropy change is standard whereas the one in point 1 is not. Similarly, I think the entropy change in point 3 should also be standard. If that's the case, what exactly is the difference between $\Delta S_{\text{total}}$ and $\Delta S^\circ_{\text{total}}$?
Note, I don't know what Gibbs free energy is but from the expression I've seen online, it seems to be equal to $-T\Delta S_{\text{total}}$ so I'd appreciate if the explanations of my questions are in terms of entropy changes.
Edit: the book is Edexcel IAL Chemistry Student Book 2 (ISBN 13: 9781292244723), page 72 as requested in the comments.

What is the book? Could you edit the question to include the title and author, or the ISBN?

@Karsten I've added it at the bottom.

When you say that $Delta S=0$ or $Delta S=R\lnK$, please precisely state the initial and final states for these changes, in terms of the temperatures and pressures of the reactants and products (and if the reactants and products are pure, or not).

For points 1 and 3, if they're both agreed to be true then $\Delta S_{\text{total}} = 0=R\ln K \implies K = 1$, no matter the reaction.

The entropy change of a reaction is concentration-dependent:
$$\Delta S = \Delta S^\circ - R \ln Q\tag{1}$$
For the standard state (indicated by the $^\circ$), the reaction quotient $Q$ is one. At equilibrium, $Q = K$. So the entropy change is different at standard state and a equilibrium. Point 3 should indicate standard state to make sense, like this:
$$\Delta S^\circ_\mathrm{total} = R \ln K\tag 2$$In fact, you can derive (2) by setting $$\Delta S = 0 $$ and $$Q = K$$ (true for equilibrium) and substitute into (1).

Later on in the book, there is another contradiction where it provides the following reaction at equilibrium:
$$\ce{SO2(g) + 1/2O2(g) <=> SO3(g)}\quad\Delta S^\circ_{\text{total}}=\pu{238.3 J K^-1 mol^-1}$$

The double-harpoon in the chemical equation does not indicate that we are at equilibrium, it just says this reaction shows some forward and reverse reaction at the same time. The entropy change at standard state is non-zero, meaning there will be a net change in the forward direction until we reach equilibrium. At this point, the entropy change of a net reaction will be virtually zero, so there will be no further net reaction (i.e. forward and reverse rate will be equal). So this statement is fine once you figure out the meaning of the chemical equation.

How come the total entropy change is nonzero during equilibrium? My guess is that it has something to with the fact that this entropy change is standard whereas the one in point 1 is not. Similarly, I think the entropy change in point 3 should also be standard. If that's the case, what exactly is the difference between $\Delta S_{\text{total}}$ and $\Delta S^\circ_{\text{total}}$?

Without the $^\circ$, concentrations could be anything. With the $^\circ$, concentrations are all at standard state, and $Q = 1$.

$\Delta S_{\text{total}}$[forward reaction] = $\Delta S_{\text{total}}$[backward reaction]

Point 2 is written in a way that confuses me. "Total" implies it covers everything that goes on, so separating it into forward and backward reaction is surprising. I asked the OP for the source of this statement in the comments. Maybe the context in the book would shed some light on the intention of the textbook author.

I'm starting to understand but I just need some clarification on somethings if you don't mind. My understanding of standard state is that temperature and pressure get specified and are kept constant (298K and 1 atm). What does "concentrations are all at standard state" mean? Why does the reaction quotient = 1?

"At this point, the entropy change of a net reaction will be virtually zero". So if the standard change of entropy keeps decreasing as the reaction reaches equilibrium, then don't we again arrive the problem of K equaling 1?

The standard entropy change is constant, it refers to a specific set of concentrations. The actual entropy change becomes smaller and smaller, approaching zero as equilibrium is approached. For the definition of standard state (rather than STP), see https://en.wikipedia.org/wiki/Standard_state#Conventional_standard_states

Sorry if I sound like a broken record but my confusions just aren't clearing up... If at standard state, the standard entropy change is constant, doesn't that contradict "The entropy change at standard state is non-zero, meaning there will be a net change in the forward direction"? As in, if it is always constant, even during equilibrium, it should be a constant so it doesn't actually indicate whether the forward reaction currently has a greater rate? It appears to me that standard entropy change is actually just a number attributed to a reaction derived from K, is that correct?

Sorry, comments might not be the right space for this. Anyway, the standard state values only apply to the standard state. Even if you start at the standard state, concentrations will change, so most of the reaction will have a state different from the standard state, and thermodynamic quantities different from the standard state. You are welcome to ask a question directly about this.

(1) The entropy reaches a maximum at equilibrium, thus any change will depart from equilibrium, but this cannot happen if we are at equilibrium unless some other parameter changes such a temperature or pressure thus any change in entropy is zero.
If the reaction is $A+B\rightleftharpoons C$ if we start with one mole of A and one of B the formation of C can occur only for as long as the entropy can continue to increase. If we start with just C then A+B will be produced until equilibrium is again formed, i.e. some composition of A,B and C that maximises the entropy.
(2) I'm not clear what is meant here, possibly entropy of mixing?
(3) The expression for the Gibbs free energy is $\Delta G^0 = -RT\ln(K_P)$ where $\Delta G^0$ is the free energy at standard conditions ( 1 mol of vapour at 1 atm. pressure for a gas and 1 mol for a liquid independent of pressure)  and $K_p$ is the equilibrium constant.  The free energy is defined as $\Delta G=\Delta H-T\Delta S$ so (3) assumes that any change in $H$ is zero which is not generally true, i.e only accidentally so. The free energy $G$ is a minimum at equilibrium.
(That $\Delta G^0$ is not zero at equilibrium is because it is the slope of free energy $G$ (not standard /reference value $G^0$) wrt. extent of reaction $\xi$ that is zero, i.e $dG/d\xi=0$ at constant $T,P$.)

Thanks for the response. I understand your (1) but (3) is completely flying over my head unfortunately. I mentioned in the OP I have yet to study free energy G and I don't know what $\xi$ is. It is not part of my current syllabus of A levels. If it's not too much to ask, could you address my questions directly: why the result $K=1$ is wrong and why $\Delta S^\circ_{\text{total}} > 0$ at equilibrium? If it must require the knowledge of free energy then maybe can you give a conceptual explanation like (1) (to get me through until I study the required topics)?

At equilibrium there is no change in entropy then $\Delta S=0$, if this is true then $K=1$ but this cannot be general. $\xi$ is simple, it is the extent of reaction and varies from 0 (no product) to 1 (all product) . So the  energy is a minimum (at equilibrium) when plotted against $xi$, i.e. at some composition of product and reactant. continued.

continued. You cannot avoid free energy,  it is simple anyway. In a reaction $\Delta G=\Delta G^0+RT\ln(P_a/P_b)$ where $P$ is pressure of species a and b, this ratio is also  called the quotient $Q$ and then $K_p$ at equilibrium. Divide by $T$ then $\Delta S=\Delta S^0+R\ln(K) $ assuming $\Delta S=\Delta G/T$ and at equilib $\Delta S^0=-R\ln(K)$ where the superscript indicates standard state.


text:
What will be the products of the reaction between a metal (i.e. zinc) and a solution containing two dilute acids i.e. HNO3 and H2SO4?
The metal is dipped into a beaker that has the two acids, sulphuric (VI) acid and nitric (V) acid in equal concentrations.
In my preliminary research, I found that there might be several possibilities. I disregarded the case of the nitrating mixture since in this case the acids are dilute. There might be formation of a double salt, zinc sulphate nitrate, or formation of the individual salts, zinc sulphate and zinc nitrate with preference to zinc sulphate, the stronger acid. I would like to ask which of these, if any, describe the fate of the reaction.

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Zinc will dissolve and stay in the solution. That's all that can be said. Which salts are there in the solution is a question without meaning. There are no salts in the solution, only ions. Some of the ions are $\ce{Zn^2+}$.

Furthermore, metallic zinc $\ce{Zn}$ reacts slowly with nitrate ions $\ce{NO3^-}$, producing nitrite ions $\ce{NO2^}$

The reaction of zinc with dilute sulphuric acid
Adding zinc plate to dilute sulphuric acid results in metal dissolution with the formation of colourless solution and the release of gas bubbles according to the following reaction: $$\ce{Zn(s) + H2SO4(aq) -> ZnSO4(aq) + H2(g)}$$
The reaction of zinc with dilute nitric acid
Products of the reaction between zinc and nitric acid directly depends on the concentration of the latter. Being very dilute nitric acid dissolves zinc without releasing gas: $$\ce{4Zn(s) + 10HNO3(aq) -> 4Zn(NO3)2(aq) + NH4NO3(aq) + 3H2O(l)}$$
A slight increase in the concentration of nitric acid may lead to the liberation of a gas: $$\ce{4Zn(s) + 10HNO3(aq) -> 4Zn(NO3)2(aq) + N2O(g) + 5H2O(l)}$$
References

Ronald L. Rich. Inorganic Reaction in Water (1st ed.): "Oxidized nitrogen", p. 290.
C.E. Housecroft, A.G. Sharpe. Inorganic Chemistry: "21.13 Group 12: zinc", p. 771.
Demonstration #1 of the reaction between zinc and dilute sulphuric acid.
Demonstration #2 of the reaction between zinc and dilute sulphuric acid.
Demonstration #3 of the reaction between zinc and dilute nitric acid.

I find this question merits discussing as there is some difference in the usual simple academic provided answer and actually what occurs in reality.
This likely high school test level question is based on the usual ascribed simplistic renditions of the underlying chemistry. Namely, these dilute acids do not interact with each other, the action of sulphuric with zinc metal results in sulfate and just molecular hydrogen (not exactly accurate) and the nitric acid renders nitrate and somehow (not generally discussed) creates also Laughing Gas. However, this likely suffice as an acceptable answer at this level.
The more accurate advanced answer involves surface chemistry (obviously not likely appropriate at this academic level) where the hydrogen is chemisorbed onto the zinc metal surface as an active hydrogen atom or so-called the $\ce{.H}$ radical, where radical-based chemistry is also an advanced topic. The latter $\ce{.H}$ continues via surface interaction beginning with the aqueous nitrate ion leading to nitric oxides ($\ce{.NO2}$ and $\ce{.NO}$) formation in dilute solutions, and finally the $\ce{.HNO}$ radical. The latter can self react creating hyponitrous acid ($\ce{H2N2O2}$), which being unstable, further breaks down liberating the generally only final cited gaseous product (for simplicity) namely, just Laughing Gas ($\ce{N2O}$).
Interestingly, the interaction of hydrogen and zinc metal is also of concern in corrosion studies, to quote: "Corrosion of the zinc generated the hydrogen causing the embrittlement". Also,  $\ce{.H}$  radical functionally behaves, per its seemingly reversible formation reaction: $\ce{e- + H+ = .H}$ as apparently a $\ce{(e-,H+)}$ pair acting on ions. For example, per 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', from Page 818, to quote:

$\ce{ PbS + 2 •H = Pb + H2S (5) }$

as a proposed commercial reductive leaching reaction and, in the case of zinc with acid acting on aqueous nitrate, one of the reactions:

$\ce{ NO3- + •H = OH- + •NO2 }$

in accord with Maurice's comment.


text:
What conclusions can I draw from this graph?
I've been doing an experiment where we would have copper chloride in an aqueous solution, and then mix aluminium foil in it which produces copper. We then modified that experiment to do the same thing, but investigate the yield of copper with the solution at 10°C, 24°C, 30°C and 50°C.
I expected it to be a linear trend, with copper increasing as you increase in temperature, but with our results, I got this trend:

My research question is this: 'How does changing the temperature of an aqueous solution of copper chloride (at 50°C, 30°C, 24°C and 10°C) in a reaction with aluminium affect the reaction rate and hence result in a greater yield of copper in grams?'
In my conclusion, I need to provide an answer to my research question and link scientific theory to my results. I'm not quite sure how I'm supposed to do that with the results I've got.
Thanks in advance for your help.

I would guess that the yield is constant and about equal to 0.8 g at all temperatures, if the validity domains are taken into account.

Let's start by confronting the obvious: there's no trend to be found in your data. Your error bars are so large that it's impossible to draw any real conclusion from it. (Science is like that sometimes, perhaps even most of the time; and the lack of reproducibility in your results may well not be your fault. But it would be your fault if you tried to discover some trend where there wasn't any.)

Repeat experiment but use CuSO4! The chloride ion is very particular with respect to the protective coating of Al2O3 on the Al. Some of what you may be observing relates to this surface chloride chemistry which may not be precisely even your topic of interest (or make it a topic of interest to explain the deviation). Relatedly, add some HCl to the CuCl2/Al experiment.

What are the sources of error (how did you dry the copper, and how did you separate it from other solids)? What was the color of the solution, and how fast did it change? The yield of product does not sound like a good way to measure a rate. Do you have any time-dependent data? Could you edit the question to give some more context about your data, and how your were hoping to answer the research question based on the outcome of the experiment?

What is the theoretical yield of copper? What is the limiting reactant, the copper salt or the aluminum foil? Could you edit your question to add that information?

I agree with @orthocresol. Fitting a quadratic (3 parameters) seems too much, fitting a line (2 parameters) would probably give a horizontal line within margins of error. You could plot all the yields (not the averages), and use the old-school method of enclosing them in a rectangle where the diagonals are the lowest and highest reasonable slope. Taking the average (1 parameter) seems the best option, quoting the standard deviation for the entire data set.

As mentioned by Karsten, it would be nice to see more experimental details (from Material and Methods - this is why this is included in reports, so people can gauge whether things were done properly, not just to repeat the experiment). How did you compute uncertainties? Did you stir the solution during the experiments?

First, the liberation of copper metal with a presence of Al enables a secondary electrochemical galvanic corrosive of the Al (see https://www.researchgate.net/figure/Comparison-of-the-Galvano-Fenton-and-the-Galvano-Fenton-with-pre-immersion-in-terms-of_fig4_353697572). At low temperature possible more dissolved O2 with solvated electrons forming superoxide and very acidic .HO2 at pH <5.  But warmer solutions, less O2 and also  less .HO2 but further raising the temperature offsets with an increased Cu formation, see reference. This explains the curve!


text:
Is Silica Gel fit for Ultra-High Dessication?
I had the (perhaps wrong) notion that silica gel wasn't that strong of a dessicant, especially compared to CaO, MgO, CaSO4, H2SO4, KOH, Mg(ClO4)2), but I've been exposed to information* that might give me wrong.
Can silica gel beat most of those dessicants in terms of residual water in a dried solid? Can silica gel really be used to dry salts down to only twice the residual water level that can be achieved with Mg(ClO4)2 and half the residual water level than can be achieved with KOH as dessicant? Or does that only apply to air/gases?
Will anhydrous silica gel allow to dessicate compounds that have high affinity to water down to very low residual water levels? (when used at say 5-20C at atmospheric pressure on salts pre-dried with MgSO4)
What are the conditions that give silica gel its greatest capacity to dessicate salts to ultra low residual water levels?

*Merck Memento, p14, says:
Residual humidity in mg H2O /L of air after dehydration at 25°C:

White CuSO4 1,4
Molten ZnCl2 0.8
CaCl2 0.14-0.25
CaO 0.2
Molten NaOH 0.16
MgO 0.008
Anhydrous CaSO4 (plaster of Paris) 0.005
Concentrated H2SO4 (95-100%) 0.003-0.3
Dry Al2O3 0.003
Molten KOH 0.002
Silica gel (SiO2) 0.001
Anhydrous Mg(ClO4)2 0.0005
P2O5 below 0.000025

I've always thought that silica gel was one of the best dessicants in terms of its capacity, but not necessarily it's affinity, for water absorption.  In other words, I thought it was preferred in many applications because it can suck up a lot of water from fairly wet materials before becoming saturated, not because the residual water it leaves behind under optimal conditions was so low.  But maybe I've always been wrong!

@CurtF. I was under a similar impression (i.e. I thought it absorbed more water per weight than the likes of KOH but with much higher residual water levels) but these numbers seem to contradict that. I don't know if the numbers are correct and if they are, if this order holds true for residual water in dehydrating solids / salts as well.

Without you telling what you want to dry, there is no good answer.

@Karl: https://chemistry.stackexchange.com/questions/167576/can-choline-chloride-be-dehydrated-with-calcium-sulfate

Silica gel is really one of the best desiccants known. But it has to be thoroughly dehydrated at high temperature. Usually silica gel contains a small amount of cobalt chloride $\ce{CoCl2}$ : blue when anhydrous, and red when hydrated. Silica gel should not be used if reddish. It has to be blue to act as desiccant.

Thank you for your answer Maurice! Will thoroughly dehydrated silica gel dessicate salts to half the residual water level as Mg(ClO)4 would?

The residual water amount in air from my tables (from Germany) are not the same as yours. The maximum amount of water in air is $0.006$   mg/L for silica gel, and    $0.0005$ mg/L for magnesium perchlorate. So Magnesium perchlorate is more efficient, in my tables.

Magnesium perchlorate is not $\ce{Mg(ClO)4}$, but it is $\ce{Mg(ClO4)2}$

Thank you! Sorry for the mistake (I can't correct the comment anymore though). Your numbers are more in line with what I would expect. I wonder how Merck came up with a 6x (!) smaller figure. It can hardly be a mistake because it's not just the figure that's very different, but with it the ranking of silica gel in that list which becomes completely different from what seems to be common expectations amongst chemists.