text:
Which rate, the forward or reverse rate of acid dissociation, is more strongly affected when diluting acetic acid in aqueous solution?
Acetic acid is a weak acid. It is in equilibrium with acetate and hydronium ions in aqueous solution:
$$\ce{CH3COOH(aq) + H2O(l) <=> CH3COO-(aq) + H3O+(aq)}$$
Ostwald's law states that the degree of ionisation of a weak acid increases upon dilution (in quantitative terms: as the square root of the volume).
Consider two scenarios:

We start out with 1.001 moles of water and 1.000 moles of acetic acid (so water is the solvent and acetic acid is the solute), and dilute it with water to twice the volume.
We start out with 100 mM acetic acid, and dilute it with water to
twice the volume.

How do the rates of the forward and reverse reaction change in these two scenarios, and how do these changes help to explain Ostwald's dilution law? (I know that once equilibrium has been reached after dilution, the forward and reverse rates will be equal again, so I am asking about the rates after dilution, but before changes to concentrations by the reaction going towards equilibrium. If mixing is slower than the acid-base reactions and you can't do the actual experiment, treat it as a thought experiment.)
This question was inspired by Why degree of dissociation/ionisation affected by dilution? . I am specifically interested in addressing the following ideas from the OP of that question:

as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

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since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

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since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

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What motivates or pushes [the acid] to dissociate more?

If it helps, it would be fine to discuss what happens with a single acetic acid molecule and a single acetate anion upon dilution.

How about Le Chatelier's principle? To dilute, you need to add water to LHS of the equilibrium equation.

Yes, but this is not the textbook Le Châtelier situation where you change just one concentration. By adding water, all of the concentrations change. In fact, in scenario 2 the concentration of water is the one that changes the least.

Let's consider case 2. To make the math easier, I'm going to assume that acetic acid and any mixture of acetic acid/water has the same density as water. Thus, the molarity of pure acetic acid is approximated as 16.7 mol/L. That isn't exactly right, but conceptually that doesn't change anything. Also, since the mole fraction of water is a bit less than 1, let's use its actual molarity of approx 55.6 M.
So, to start, we have 0.1 M of acetic acid in water, which means only ~6 mL/L  water, and we can assume the water is still 55.6 M. And assume a total volume of 1 L. At equilibrium, we know that the forward and reverse rates of ionization are the same, so
$k_f[\ce{H2O}][\ce{AcOH}]=k_r[\ce{AcO-}][\ce{H3O+}]$
Based on the published $K_a$, we can calculate that [AcOH]~0.1 M and [AcO-]=[H3O+]=0.0013 M.
If we instantly add an additional 1 L water to double the volume, the AcOH, AcO- and H3O+ concentrations all drop by half, but the water concentration stays at about 55.6 M. Intuitively, it should be clear that the result is a much greater drop in the rate of reverse reaction than the rate of the forward reaction. More rigorously, we relative rate of the forward reaction is (55.6x0.05)/(55.6x0.1)=0.5, while that of the reverse reaction is (0.00065)^2/(0.0013)^2=0.25. Since the rates were equal before dilution, the forward rate is now approx twice that of the reverse, and in order to return to equilibrium, the AcOH concentration must get even lower and the ion concentrations must come back up.
Conceptually, this comes about because the chance of an acetate ion and a hydronium ion encountering each other diminishes as the solution gets more dilute. The dissociation of the acid in bulk water, on the other hand, is just a function of how many undissociated molecules are present.
With regard to the quotes you included, some brief responses are

as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

The concentration of water is not the same as H3O+, which is decreased by dliution. Water self-ionization will partially compensate, but not very much.

since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

The dielectric constant of water does have an influence on the $K_a$, but it doesn't change when you add more water.

since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

If we start with dilute acetic acid in water, we assume that the acetate ions are far enough apart from each other that they do not interfere with each other's solvation, so additional dilution doesn't factor in.

What motivates or pushes [the acid] to dissociate more?

I think it's more appropriate to say that it reassociates less rather than dissociates more, but that's mostly semantics.

I think you are considering case 2  (you said 1). I don't understand why you say the concentration of water goes from 50 M to 53 M when the (total) acetic acid concentration merely decreases from 100 mM to 50 mM. Otherwise, excellent points.

Thanks for pointing that out. I was being dumb there and counting 100 mL of acetic acid instead of 100 mM. The 100 mM acetic acid is only ~ 6 mL, so the water is essentially 55 M in both. I'll fix that and the case 1 typo

Let's consider case 1. First, we will calculate the concentrations before and after dilution. Then, we will estimate the activities of the solutions. Finally, we will discuss what will happen to the rates of dissociation and association.

For simplicity, let's say we are starting with an equimolar mixture of water and acetic acid. This corresponds to about 77% of acetic acid by mass, a solution with a density of about 1.07 g/mL 1. The concentration of acetic acid is about 13.7 M, and the concentration of water is the same (equimolar). If we dilute with water to twice the volume, the concentration of acetic acid will be about half, i.e. 6.8 M (if volumes are roughly additive). The concentration of water will increase significantly to about 34 M. So the dilution cuts the acetic acid concentration in half while it almost triples the water concentration
According to Hansen et al (1955), the activity coefficients of acetic acid and water are both about 1.3 for an equimolar mixture (with the pure liquid as the reference state where the activity coefficient is one). For the diluted solution (molar ratio of acetic acid to water of about 1:5), the activity coeffient for acetic acid is about 1.6 and the one for water about 1.05. Because the other two relevant species, hydronium and acetate, are present in low concentrations and there are no other ionic species at high concentration, their activity coefficients probably are close to what they are in a dilute aqueous solution, but I have not found experimental data supporting this assumption.
Different from case 2, where the forward rate was predicted to
decrease by a factor 2, in case 1 we can predict that the forward
rate will increase. The concentration of acetic acid drops by a
factor two, but the concentration of water increases by more than a
factor two. Similar to case 2, the reverse rate is predicted to
decrease by a factor of about four.

What motivates or pushes [the acid] to dissociate more?

In case 1, it is correct to say acetic acid dissociates faster because the concentration of water is significantly higher after dilution. This contributes to the higher degree of dissociation upon dilution. At the same time, the rate of association drops by a factor of about four, a much larger change.

According to an NMR study, the pH of water acetic acid mixtures is minimal at about equimolar ratio.


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How are octet-rule models supported by thermodynamics and quantum theory?
I understand that the four quantum numbers dictate how many electrons can "fit in a shell," but I don't understand how gaining a full octet is particularly energetically favorable.
The example of the ionization of sodium seems to make sense: the atom loses a single 3s electron, and the energy of the system is lowered significantly down to the 2p energy level.
However, the ionization of chlorine isn't as obvious to me: it is indeed granted its "full octet," but filling the last available space in the 3p suborbital would not lower the energy of the system. Rather, it would raise it slightly, as the newly added electron would experience some electron/electron repulsion.
Is the main drive for this ionization "reaction" the drop in energy in the formation of the cation? Do I have to consider molecular orbital theory to answer this question for covalent interactions?
I'm mainly wondering why there's a drive to form a full octet, particularly for the halogens, chalcogens, and pnictogens.

You have it backwards. No element gives up an electron willingly. Those which achieve a full octet by doing so (like Na) are just the least unwilling. On the contrary, Cl and many others would take an extra electron quite eagerly.

Could you elaborate? This doesn’t really explain the “why,” even if I have it backwards.

Well, not much can be said besides the obvious. One atom wants an electron really badly, and another is like "alright, dude, I'd give you mine, but I sorta need it myself". Then comes a huge energy of crystal lattice formation, and that's the main drive.

This question is off topic or not explained fully if you are asking 'why does it quantum theory or thermodyanamics affect stability of electronic configuration' then you must know a lot about quantum mechanics or thermodynamics or studying it but your explanation does not reflect that. You are comparing your opinions based on 'old theories' with quantum mechanics. You have to consider quantum mechanics if you want answer based on quantum mechanics same for thermodyanamics. Also no explanation is given in MOT or any 'old theories' it is just an observation.

@SauravSingh The question is certainly on topic on [Chemistry.se], even if there are misconceptions within it.

Related: Hypervalency and the octet rule

Possible duplicate of Why are atoms with eight electrons in the outer shell extremely stable?

This is imho a great question (even if probably a duplicate) and Andrew does a good job of answering. The intuition seems right, you have to consider what the electron was doing before it entered the Cl atoms shell. The chlorine $is$ excited when it accepts the electron, relative to the configuration with the electron at infinite distance, but that is countered by the electron being lower in energy residing closer to the nucleus thanks to the electrostatic interaction.

I suggest reading https://chemistry.stackexchange.com/questions/26833/whats-the-relation-between-the-octet-rule-and-quantum-exclusion-and-how-is-it-r and the answer by Wildcat

Reading carefully thus rather than too broad should have been closed as a duplicate.

I was very aware of similar problems being asked, but they all focused on how quantum numbers allow 8 electrons in a shell. I was instead curious about why rudimentary electron orbital diagrams and basic approaches to ionization didn't seem consistent with literature ionization energies and electron affinities. This is why I didn't think my question was a duplicate and why I appreciate Andrew's answer—the electrostatic potential was something I overlooked and clears up the energetics I was confused about. Wildcat's answer on a related post was also helpful in outlining binding energy effects.

I also would like to say I tagged quantum mechanics just because I wanted to apply what I've been learning in QM to a rule of thumb that freshmen are taught to blindly accept. I was curious about how quantum mechanics would go about explaining the drive for ionization in order to satisfy the octet rule. Maybe I'll post this another time.

Maybe you should rather read https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding

You can't really compare the energy of $\ce{Cl-}$ to $\ce{Cl.}$, since the two states have different composition. Instead, you're interested in the difference in energy between $\ce{Cl-}$ and $\ce{Cl. + e-}$, where the electron is completely free and resting in a vacuum with no interactions with anything else. We give that electron a reference potential energy of 0. Then the question becomes, how much potential energy is lost by interacting with various atoms, and the thermodynamically favorable state is the one with the most negative potential energy.
If we consider nuclei to be point positive charges and electrons to be negative charge distributed in a much wider space around the nucleus, we find that as that free electron gets closer to the nucleus, it experiences an electrostatic interaction with the positively charged nucleus even if the atom overall is neutral. We can simplistically estimate the magnitude of the potential energy loss based on basic principles of electrostatics, where potential energy is a function of $\frac{q^+q^-}{r}$.
Where molecular orbital theory comes in is when we calculate the charge and the distance r. Simplistically, orbitals that are higher in energy on the same atom are higher in energy because they have a longer average distance from the nucleus (denominator in PE term is bigger) and because there is more electron density from other electrons between them and the nucleus, so the "effective" positive charge of the nucleus is reduced (numerator in PE is smaller). This is called "shielding".
Slater's rules are simple heuristics that provide a fairly good approximation of the effective nuclear charge experienced by an electron based on the atomic number and orbital.
Conceptually, then, the answer to your question is that adding an electron to a neutral atom like $\ce{Cl.}$ is most favorable when the electron experiences the best combination of effective nuclear charge and short average distance to the nucleus. This happens when adding the last electron to an octet because that is the element of that period with the greatest nuclear charge. Adding an electron to the next neutral element Argon is much less favorable because it has to go in the next shell (n=4) meaning that it has a larger r (larger denominator) and has much more electron density between it and the nucleus, so the effective nuclear charge is less (smaller numerator) even though the atomic number of Ar is one more than Cl.
For $\ce{Na+}$, it isn't correct to say that the system has lower energy after an electron is removed, but rather that it takes a lot less energy to remove that electron than an electron from neutral neon, which hopefully makes sense based on the discussion above.

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Dissociation behaviour of polyelectrolytes
Assuming a solution of 100 polycationic molecules, each of them carrying 100 amino groups and an apparent pKa of 6.5. Would it mean that, at a pH of 6.5,

50 molecules will be fully protonated
50 amino groups of each molecule will be protonated  or
50% of the total amount of amino groups present in solution will be
protonated?

The third option would mean that each molecule can have any charge number, from 0 to 100, with 50% of the 10.000 amino groups being protonated.
In other words, is it safe to state that, at a pH below 6.5, all molecules will be mostly protonated?
Edit due to ambiguity:
In other words, is it safe to state that, at a pH below 6.5, all molecules will have more protonated amino groups than unprotonated?
Many thanks in advance!
Ben

Do a thought experiment: Into your polycationic molecule, insert very large spacers in between the amino groups so that they are far away from each other in terms of covalent connection and in terms of spatial distance. Now, break a bond in the middle of each spacer. Do you think this will influence the protonation state of the amino groups?
If the $\mathrm{pH} = \mathrm{p}K_\mathrm{A}$ for your $100^2$ independent amino groups, half of them will be protonated and half deprotonated. If you randomly pick 100 amino groups and turn them into a molecule by repairing the spacers in our thought experiment, you get a distribution of charges (a Gaussian bell curve centered on 50).
If you lower your pH by one unit, about 91% of amino groups will be protonated and 9% deprotonated, and again there will be a distribution of charges for the polycationic molecule/ion.
If, instead of taking a snapshot at a given time, you ask about average protonation state over a couple of milliseconds, you will have a molecular (or ionic, I guess) charge of 50+ for pH 6.5, and about 91+ for pH 5.5.
If the amino groups are not separated by spacers, they will "talk to each other", like the carboxylate groups in citric acid, for example. For citric acid, three pKa values are given, 3.13, 4.76 and 6.40, even though each group has similar chemistry and two are identical. So in that case, things are more complicated. It also gets more complicated if you bundle species with identical charge when figuring out pKa values theoretically (for citrate, there is one neutral species and one 3- species, but three 1- and three 2- species).

This is exactly what I was thinking. Given the random distribution of charges and your example in the last paragraph which basically states that each amino group will have its own pKa (that's why I was speaking of "apparent" (=average) pKa), it is therefore impossible to state that all molecules will be mostly protonated below pH 6.5 since there is the possibility of fully deprotonated molecules, isn't it?

And with that said, actually one could only say THAT (some) amino groups (which are not protonated yet) within one particular molecule become (also) protonated at pH 6.5, right?

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text:
Why is phosphoric acid preferentially used for dehydrating alcohols during synthesis of alkenes?
I have also seen concentrated sulphuric acid being used, but what makes phosphoric acid preferable to sulfuric acid?

Sulphuric acid can also act as an oxidant, phosphoric cannot.

Ohh, thanks! Also, does the concentration of acid matter?

Phosphoric acid is most often used as polyphosphoric acid which is a thick syrup. This is essentially neat phosphoric acid, which can be made stronger by addition of P2O5.


text:
Boltzmann factor vs. graph of Maxwell–Boltzmann distribution
In discussions of the Arrhenius equation describing how rates change with temperature, there is often a reference to collision theory.
In the Arrhenius equation,
$$ k = A \exp\left(-\frac{E_\mathrm A}{RT}\right)$$
there is a Boltzmann factor interpreted as the fraction of collisions of sufficient energy to overcome the activation barrier. If you plot the Boltzmann factor $\mathrm e^{-E_\mathrm{A}/RT}$ against temperature, it looks like a simple exponential decay (all quantities in the exponent are positive). However, most textbooks use a graph of the Maxwell–Boltzmann distribution (i.e. distribution of speeds of gas molecules), which is not a simple exponential decay but has a maximum near the average speed.
So how is a graph of the Maxwell–Boltzmann distribution relevant to the Arrhenius equation?

When the particles are about to react, in fact you are only interested in the distribution of one component of the velocity vector (say, $v_x$). This distribution is a simple exponential decay. The same is true for $v_y$ and $v_z$. Now if you try to convert that into the distribution of $|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}$, you'll get the Maxwell–Boltzmann thing.

Draw a vertical line on the the Maxwell-Boltzmann distribution. To the right of this  is the fraction of molecules with  sufficient energy to equal or exceed the reaction barrier and hence may react.

@porphyrin I question whether all of the molecules to the right of your line would have collisions of sufficient energy to exceed the activation energy. As Ivan Neretin pointed out, only the component perpendicular to the collision plane is relevant. If two fast particles don't have a head-on collision but something analogous to a side-by-side car collision, there will be no reaction. Also, I don't think the fraction of molecules corresponds to the Boltzmann factor.

The Arrhenius eqn is only a v crude model, sure molecules have to collide in the correct way, even if this is inline the wrong part may still collide depending on the molecule's size. Also vibrational energy can aide or impede collisions even if every thing else is ok. Have a look at RRKM, Tolman's stat mech model or better still Kramer's model of reactions.

Does the second half of this, or the appendix of porphyrin's answer there, address your question? The TL;DR is that porphyrin's first comment is pretty much correct, the area under the graph does generate an $\exp(-E/kT)$ form, although the maths is slightly more nuanced than simply doing the integral from $E_\mathrm A$ to $+\infty$. It also takes collision geometry into account, which seems to be one of your major points of contention. There are more references in my answer if that helps.

@orthocresol  Those two answers are great, and I was not aware that there are so many possible pre-exponential factors depending on what distribution (speed, particle energy, momentum, collision energy) one considers. What would be really helpful would be a set of graphs with those distributions with accurate axis labels, and a sense how different the tail of the distribution (for example at an energy corresponding to an activation energy of 80 kJ/mol) would be at two or three different temperatures attainable in a student lab.

I’m looking for an explanation and a graph that is better than what is currently in the first year college textbooks .

Rationale for posting an "answer" to my own question

[Comment from OP] I’m looking for an explanation and a graph that is better than what is currently in the first year college textbooks .

The comments to the question contain some excellent insight and references but there is no comprehensive answer to far, so I am collecting what was posted here. Maybe the depth of the comments and the lack of a highly rated answer reflects that it is impossible (at the first year undergraduate level at least) to make a meaningful quantitative connection between Maxwell-Boltzmann distribution of kinetic energy in a ideal gas on the one hand and the Arrhenius equation on the other hand.
Textbook treatment of the question
In first-year college textbooks, the Arrhenius equation is often rationalized for gas phase reactions via the collision theory. The quotes here are from OpenStax Chemistry as hosted by Libretext.org, but other textbooks have similar material:

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, $e^{−E_a/RT}$, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

This is great because it references collisions providing sufficient energy.
Then they go on to explain:

The Arrhenius equation (Equation 12.5.6) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of $E_a$ results in a smaller value for $e^{−E_a/RT}$, reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller $E_a$ has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of $e^{−E_a/RT}$, a larger rate constant, and a faster rate for the reaction.

This is pretty much correct if you don't over-interpret "fraction of molecules with enough energy to react". What is relevant is the energy made available by collisions, not the energy of the molecules, but if the molecules have higher kinetic energy, the collisions will be of higher energy. (There might be some energy associated with rotation and vibration, but since collision theory starts with a mono-atomic ideal gas and we are shooting for the simplest explanation, those will be ignored).
Then they show a figure (and this figure is fairly representative of the corresponding figures in other textbooks and those found online from various sources):

This figure suggests that the fraction of molecules with kinetic energy higher than the activation energy will react. It also suggests that if you integrate over the shaded area, you get the Arrhenius equation.
If you take a close look at the curve, you will see that the slope at the origin is too small. It looks like the x-axis is the velocity rather than the energy, and the graph is accordingly distorted.
As an aside, there are two aspects of this figure that are not representative of what first-year students will experience and observe in the lab.

The activation energy shown is on the order of a couple RT. This would be a reaction too fast to measure in an undergraduate lab, and it gives the impression that maybe 1 out of 10 collisions leads to a chemical reaction.
The two temperatures differ by almost a factor two. If $T_1$ is room temperature, the second experiment would have to be done at 586 K, i.e. in an oven where most solvents would evaporate.

Relevant distribution for collision energy
The comments were very insightful. Ivan Neretin addressed the relevant degrees of freedom:

When the particles are about to react, in fact you are only interested in the distribution of one component of the velocity vector (say, vx). This distribution is a simple exponential decay.

I made a diagram to illustrate what happens in an inelastic collision:

So you can't simply take the kinetic energy of one molecule (or add both kinetic energies together) but instead have to consider one degree of freedom each. The maximum energy provided by a collision is not the sum of the kinetic energies of the particle because the center of mass of the two particles combined continues on its path, so part of the kinetic energy is not available.
The wikipedia article on the Maxwell–Boltzmann distribution states:

Since the energy is proportional to the sum of the squares of the three normally distributed momentum components, this distribution is a gamma distribution; in particular, it is a chi-squared distribution with three degrees of freedom.
By the equipartition theorem, this energy is evenly distributed among all three degrees of freedom, so that the energy per degree of freedom is distributed as a chi-squared distribution with one degree of freedom.

So maybe the distribution of energies in a collision is a problem with two degrees of freedom ($v_x$ of either particle). The chi-square distribution with two degrees of freedom happens to be a pure exponential lacking a T-dependent pre-exponential factor, matching the Arrhenius equation.
Orthocresol gave a reference to the rigorous derivation of the Arrhenius equation from the Maxwell-Boltzmann distribution in one of his comments:

The TL;DR is that porphyrin's first comment is pretty much correct, the area under the graph does generate an exp(−E/kT) form, although the maths is slightly more nuanced than simply doing the integral from EA to +∞.

In this answer, orthocresol sketches the derivation, with two references that show the details.
Relevance to teaching at the first-year college level
Most reactions students will see and do will be in solution, where things are different than in the gas phase, but the Arrhenius equation still is a good description of rate vs. temperature. On the other hand, as porphyrin state in one of his comments, things are usually more complicated than a derivation of rates vs. temperature via the collision model suggests:

The Arrhenius eqn is only a v crude model, sure molecules have to collide in the correct way, even if this is inline the wrong part may still collide depending on the molecule's size. Also vibrational energy can aide or impede collisions even if every thing else is ok. Have a look at RRKM, Tolman's stat mech model or better still Kramer's model of reactions.

So maybe it is best to treat the Arrhenius equation as a definition of the activation energy in an empirical sense.

A graph to illustrate collision energy distribution at multiple temperatures might be helpful.

Following the logic of deriving the Arrhenius equation from the collision model, this would simply be a graph of $e^{−E_a/RT}$ at different temperatures. For values of temperature relevant to first year experiments (eg. room temperature vs water bath), the two traces would be very close to each other, and almost zero at energies equal to a typical activation energy obtained in a first year lab (e.g. 80 kJ/mol for iodination of acetone). To see anything, you have to add an inset zooming in near the activation energy of interest.

This graph is very different from the textbook graph. It shows the distribution of collision energies rather than the kinetic energy of individual particles, and the two temperatures chosen are explicitly given and more relevant to an undergraduate lab experience. It shows that the energy distribution is not so different between room temperature and body temperature (the average energy of a molecule is proportional to T, after all, and this is just a 6% difference in temperature). However, at typical activation energies (insets at 40 kJ/mol and 80 kJ/mol), this smaller difference results in a large difference in rate constants. It also shows that of the collisions of sufficient energy, most have an energy quite close to the activation energy (fast drop-off of probability with increasing energy).

I haven't read your answer thoroughly but just wondered if you've looked at introductory pchem books, eg Atkins does a pretty good job presenting this. I suspect though that your approach may be more useful if you are aiming for intuition.

@Night Writer My question is what is appropriate in General Chemistry. Once you reach PChem, there are more tools and skills you can rely on.

Ok, but as soon as you move away from a reasonably mathematical explanation it suffices to wave your hands and say the tail of the Boltzmann distribution increases with T in such a way as to result in an exponential T dependence of collision activation rates. Anyway, I still need to look carefully at your answer!

From a quick look I'd say that the figure 12.5.4a is simply wrong. When the activation barrier is lower more of the molecules in the tail (larger area of the distribution) have sufficient energy.

I think bringing in the reaction coordinate model with vectors etc is overly complicated because then you need to invoke an actual reaction mechanism. A model based on a Boltzmann distribution of the translational energy does not require any such mechanism, even though it requires some approximations (for instance, the assumption of thermal equilibrium). Also, the simple Arrhenius model distinguishes between factors such as the orientation of reagents (if relevant) and the translational energy. What is difficult is describing the first part, but the Bolzmann distribution remains useful.

@Night_Writer - Yes, you are right, the labels in the figure I posted should be switched.

@Night_Writer - The newest edition has this figure correct. I will edit my answer to include that. The libretext.org version I used was not updated yet.

libretext has some great stuff but as with most things that are "free" in life, you have to be careful (it's bad enough with some textbooks).

Well , the relation between the Boltzmann factor and the Maxwell distribution curves is not that intuitive.
By looking at the Arrhenius equation, $$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}} $$
where the term  $\operatorname{e}^{\frac{-E_\text{a}}{RT}}$ represents the fraction of collisions that are effective (fraction of molecules that have sufficient energy).

If we increase the temperature, the fraction of collisions that are effective ( fraction of molecules having sufficient energy) increases exponentially.

Now let's look at the Maxwell Boltzmann curve.

Here the fraction of molecules is plotted against the kinetic energy at a fixed temperature $T$. Let's plot the graph at another temperature $T+ 10$ :

As we can clearly see , the fraction of molecules having kinetic energy greater than a particular value increases with the increase of temperature that is the fraction of collisions that are effective increases which is what the Boltzmann factor in the Arrhenius equation tells us.

Figure are from http://ncert.nic.in/ncerts/l/lech104.pdf

IMHO, one should not consider  speed nor energy distribution in 3D context, but from 1D context of energy and velocity component a single degree of freedom. The former has the Maxwell-Boltzmann distribution with maximum at the most probable speed, the latter has the Boltzmann distribution with the maximum at zero energy of velocity component.

@Karsten Theis  Does the answer address your question or you feel it  is incomplete ?

@Starboy I was looking for a quantitative treatment (I now realize that for a first-year college course, a complete rigorous treatment is over-ambitious, but at least the quantitative parts shown should be correct). In your answer, I think that equating collisions of sufficient energy with molecules of sufficient energy is not correct. Also, I don't think the textbook images you are showing are actual graphs of the Maxwell Boltzmann distribution but rather artists renditions. The shape is odd, and the areas under the two curves seem different (should be one).


text:
What would be different about a heavy-water ocean world?
The water on Earth is primarily made of $\ce{H2O}$. I'm wondering about the conditions in a hypothetical $\ce{D2O}$ ocean world.
Would it be stable over geologic timescales (~$\pu{0.1–2 Ga}$)? What would be expected to be different, if anything? Would it affect the chemistry of its atmosphere, or appreciably change the incident radiation from its host star?

I wouldn't expect much of a difference.

As long as somehow all the hydrogen on the world is deuterium, no problem. Deuterium is stable. But if there is protium in the rocks, weathering will mix the isotopes over time.

The Oklo phenomenon would have looked different.

@Loong - true, and a number of biological reaction pathways would be a bit different to account for the isotope effects...

A deuterium-rich world would make some important gasses substantially heavier (e.g. $\ce{D2}$, $\ce{D2O}$, $\ce{CD4}$) which would affect their concentration profile with altitude in the atmosphere ("scale heights") and could have some significant effect over geological timescales, such as significantly less $\ce{D2}$ escaping Earth compared to $\ce{H2}$ and making the atmosphere more reducing.

The entire chemistry where hydrogen is involved could be different. Deuterium bond energy and length are different than those in hydrogen-1. It could affect the entire biochemistry. Consider also physical properties like viscosity and... ice made from heavy water SINKS in normal water.

$\ce{D2O}$ has a higher melting point (3.8 C) than the of $\ce{H2O}$ of 0 C.  On chemical side, expect reactions with hydrogen to proceed slowlier (heavy isotope effect) than normal.

@JakubMuda There’d be little reason for sea ice and sea water to have different kinds of water though (right?). I’d expect sea ice to come from the ocean, so they’d have similar compositions.

It would be easier to answer general question like how deuterium would change chemistry if all hydrogen-1 would "magically" transform into deuterium :)

While for example https://chemistry.stackexchange.com/questions/71894/is-it-true-that-heavy-water-is-not-blue looks OK for the site, I think this question is far beyond the scope of the site.

I think the OP's multiple questions are not actual questions, but examples to help a community member answer the question.

Life would be much more difficult because biochemical synthesis of almost anything involves many sequential steps many of which would be just a little slower due to isotope effects. Although each step would be slower buy only a small amount when multiplied over a sequence of steps the effect will be  be significant.

I doubt that - most biochemical reactions are tuned to specific rates and subject to various regulatory mechanism beyond the catalytic mechanism, therefore while things might slow down due to the higher mass of deuterium, it is likely organisms could find ways to compensate as necessary.

I don't agree, things would be more difficult just because each reaction is that bit slower, natural selection is very nearly optimal, on average, and with D instead of H, synthesis would be slower, not impossible. Side reactions would play a greater part lowering yields. Why would organisms find a better way when they  could do so at present?

@porphyrin nature is what it is because that's what it had to fool around with, not because it had a choice.... if it had to work with deuterium, it might find different solutions to the same problems, you would not regard them as better or worse. You might as well ask what would happen if the mass of the proton were lower, what would organisms look like? (such a world might be impossible due to physics, but not due to chemistry or biology)

yes it 'might find different solutions' but it might not, that is my point, you cannot simply assume that it will work because it is natural selection; this cannot be divorced from chemistry.

For one thing, it would be hugely more unlikely to exist.  We are more likely to have worlds where there is a little more deuterium in the hydrogen, not 100%.
Most of the hydrogen in our Universe is protium (hydrogen-1) because the vast majority of unfused nuclear particles in stars are protons.  Neutrons must either fuse with something or they decay to make protons with a half-life of ten minutes.  Deuterium and tritium are formed initially when the protons are fused, but they are just intermediate towards the formation of more stable helium-4 nuclei (alpha particles) and so do not accumulate to large concentrations.
If there were a heavy water world, life would have to adapt to it.  Prolonged consumption of pure heavy water is deadly to life on Earth.

As a planetary scientist, I know well that a D2O water world would be extremely unlikely for the reason you give. What I'm looking for is less how likely it would be and more what would be different.

Some studies (e.g. https://www.ncbi.nlm.nih.gov/pubmed/10535697) have shown that some life on Earth can adapt to D2O consumption and be fine! It is toxic to other forms of life in varying degrees, however. I've also heard that if humans were to very slowly replace their H2O intake with D2O, the mild toxicity effect would be reduced or perhaps eliminated — but I haven't seen a study that supports this.


text:
How did Noe et al. derive $k_{AB}$ in their PNAS 2009 paper?
This question is regarding Constructing the equilibrium ensemble of folding pathways from short off-equilibrium simulations which uses Markov State Model to find the equilibrium protein folding pathways from off-equilibrium simulations.
I find the notation a bit hard to follow. How does Frank Noe and others derived equation 1?

In the supplementary information they don't really explain and aren’t careful with the math notations:

I do see how a rate equals $k_{AB} = P(A^C|A)/P(B^C \cap A^C)$, but I don’t see how $\sum_{i=1}^m \pi_i q_i^{-}$ gives $P(B^C \cap A^C)$ which is the probability to be outside of $A$ and not in $B$.

Let us start by defining all the terms in Eq. 1 carefully, as the OP does not do so. I have always found it easier to interpret such models in the context of particles flowing from state to state, and that is the picture that I will try to present here.

$T_{ij}$ measures the probability of going from state $i$ to state $j$ within a time $\tau$.
$\pi_i$ is the equilibrium probability of being in state $i$.
$q_i^+$ is the probability that, starting at state $i$, a particle will reach the set of states $B$ before the set of states $A$.
$q_i^-$ is the probability that, ending at state $i$, a particle has more recently encountered $A$ than $B$. (At equilibrium, all trajectories are reversible, so we can alternately think of $q_i^- = 1 - q_i^+$ as the probability that, starting at state $i$, a particle will reach $A$ before $B$, but that is irrelevant for this problem.)
$k_{AB}$ is the normalized rate of particle flow from $A$ to $B$. In other words, given particles that start out in $A$, $k_{AB}$ is the fraction of particles per unit time that end up in $B$.

Now we simply need to interpret the numerator and denominator of Eq. 1. To think about these quantities in the particle picture, let us multiply top and bottom by the number of particles $N$ evolving in the system, defining $N_i = N\pi_i$, so that $$k_{AB} = \frac{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+}{\tau\sum_{i\in\text{all states}}N_iq_i^-}.$$ In the numerator $$\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+,$$ the quantity $N_iT_{ij}$ is the number of particles that leave $A$ (going from $i$ in $A$ to $j$ not in $A$) in a time $\tau$. A fraction $q_j^+$ of these particles will end up in $B$. Now, in steady state, this must also be the number of particles that are arriving at $B$ from $A$, or otherwise we would have an accumulation of particles in the system.
In the denominator $$\tau\sum_{i\in\text{all states}}N_iq_i^-,$$ the quantity $\sum_{i\in\text{all states}}N_iq_i^-$ is the number of particles in the system that started out in $A$, which serves to normalize the number of particles in the numerator, and $\tau$ is the time interval in which this all occurs.
Putting all this together, then, $$k_{AB} = \underbrace{\frac{1}{\sum_{i\in\text{all states}}N_iq_i^-}}_\text{normalized}\underbrace{\frac{1}{\tau}}_\text{rate of}\underbrace{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}}_\text{particle flow from A}\underbrace{q_j^+}_\text{to B.},$$ as desired.

Can you explain this phrase:
"$q_i^-$ is the probability that, ending at state $i$, a particle has more recently encountered $A$ than $B$."

@0x90: In other words, $q_i^-$ is the probability that a trajectory currently at state $i$ started out in a state associated with $A$.

So here is where this formalism isn't sound  $q^{−i}=1−q{+i}$ defined where the particle will end up and $q^{+i}$ where it will start. These events are not complementary.

@0x90: I have addressed this point in my answer. In equilibrium, where the trajectories are reversible, the events are indeed complementary.


text:
Why does the proton NMR of ferrocenecarboxaldehyde appear to be four singlets?

One singlet stems from the aldehyde H, and another from the Cp ring with 5 hydrogens attached.
However, the ring with the aldehyde group attached has hydrogens in two chemical environments which are just three bonds apart from each other. Why do they not couple? (400 MHz, CDCl3)

I'm thinking: A is the aldehyde H, D is the 5 H's on the Cp ring without the aldehyde, B the 2 H's closer to the aldehyde on the ring and C the remaining two, right?

Related: 1H NMR of Acetylferrocene I generally agree with your assignments. As long writes in the linked question: due to magnetic inequivalence, this is an AA'XX' system where the spectrum will exhibit second order effects. Why this fine structure isn't manifested, though, I don't know.

Judging from your residual chloroform peak, your sample is highly concentrate. Also, for good mesure, it is at least $\pu{0.1 ppm}$ broad ($\pu{40 Hz}$ since you used $\pu{400 MHz}$ machine). That means your peaks looks broad singlets, instead of having splitted.

@MathewMahindaratne that explanation is very reasonable, but if you look at a c13 spectrum (eg Aldrich) you only see one peak for the substituted ring. What's that all about?

@Try Hard: I posted some date to answer your question a few days ago. I apologize for the delay.

I like to address some interesting NMR features of ferrocene derivatives. Try Hard's question about $\mathrm{^{13}C-NMR}$ of ferrocenecarboxaldehyde showing one signal for the substituted cyclopentyl ring prompt me to find some solid evidence. However, it seems like most of ferrocene derivatives give similar results regardless of how strong magnetic field you used. Therefore, Aldrich NMR spectra may not be the best spectra with high definition (e.g., they use $\lt \pu{300 MHz}$ machine for $\mathrm{^{1}H-NMR}$).
Following $\mathrm{^{13}C-NMR}$ values from very recent paper, which published NMR data of few ferrocene derivatives using a $\pu{400 MHz}$ (but $\pu{100 MHz}$ for $\mathrm{^{13}C}$) machine (Ref.1):

The values in left hand side is for $\mathrm{C_p}$ ring (substituted) while single value at the bottom is for all 5 $\ce{C}$s of unsubstituted  $\mathrm{C_{p'}}$ ring. If you look closely, you'd see monosubstituted rings have so close values that they may overlap in spectra taken in low resolution machines.
Reference:

N. D'Antona, R. Morrone, G. Gambera, S. Pedotti, “Enantiorecognition of planar “metallocenic” chirality by a nitrile hydratase/amidase bienzymatic system,” Org. Biomol. Chem. 2016, 14, 4393-4399 (DOI: 10.1039/C6OB00689B).


text:
How is traditional thermite powder stored such that it doesn’t corrode?
I watched some guy talk about redox reactions. Then he was going to do the good old thermite demonstration. He just had a bottle of thermite mixture. I’m wondering, why didn’t it have a chance to corrode before the demonstration? I’ve always thought that aluminum metal gets a thin layer of corrosion in air.

Thermite powder is a heterogeneous mix and consists of fuel and oxidant, in you case aluminium $\ce{Al}$ and rust $\ce{Fe3O4}$. Separately, those can be corroded over time all right: aluminium can be oxidized and form aluminium(III) oxide, hydroxides and even some salts depending on what atmosphere thermite has been kept in. As you probably know, aluminium, being a reactive metal, doesn't corrode under ambient conditions as it's surface is protected by the thin layer of $\ce{Al2O3}$. Iron(II) is also capable of further oxidation, although rust is already the most stable form under ambient conditions.
However, it seems like you are using the term "to corrode" quite liberally to denote the act of burning of the thermite, e.g. the primary exothermic reaction
$$\ce{8 Al + 3 Fe3O4 → 4 Al2O3 + 9 Fe}$$
which only occurs when enough external energy (say, in a form of heat) is supplied. The thing is, not every reaction that is permitted by thermodynamics is also permitted by kinetics. Burning of a thermite is a heterogeneous process requiring to overcome activation barrier. Simply put, there are numerous solid particles or two phases being barely in contact, plus there is a protective layer on one phase ($\ce{Al2O3}$). Here, you need energy at least to increase thermal motion (promoting mass transfer between the phases) and compact the mix (ideally, melt to increase contact area).
That's why thermite won't easily go off, and it's a good thing making its storage safe and the working process predictable. There are only a few documented accidents [1, p. 62] resulting in ignition of thermite-alike mixtures, but none of them were spontaneous; instead, they were caused by the supply of substantial amount of external energy (heat, electricity, sparks, kinetic energy etc.), or third-party chemicals acting like catalysts:

Aluminium-painted rusty surfaces are only hazardous if the coating has
  been heated to 150 °C: on cooling, the hazard remains (Eckart et al., 1971).
  Paints with cellulose nitrate as an additive are always hazardous. [...]
Hawkins Colliery, 1950. The impact of a hand-held drilling machine cased
  in Elektron C alloy (94% Mg) upon rusty steel scaffolding ignited
  firedamp, burning 10 men, one fatally. This incident led to the initiation
  of a major research programme into the dangers of light alloys by the
  Safety in Mines Research Establishment. [...]
SS Esso Durham, 1961. An explosion in one of the ship's tanks on 29th January was attributed to the impact of a brass object, of weight 361 b, onto a magnesium anode on the wall of the tank after a fall of 16 ft. From simulations of possible alternatives, it was considered likely that a grazing impact on the rough, dirty magnesium surface would have smeared the brass, which would then have reimpacted the anode a fraction of a second later, causing a thermite reaction and an explosion. The oxide necessary was probably provided by dirt in the area (ignitions were not obtained when impacting a clean magnesium surface). The possibility of ignition from impact onto hardened steel after a 40 ft fall was eliminated. [...]
An aluminium anodizing works, Bletchley, Buckinghamshire, 1971. A ferrous metal object caused a spark on an aluminium grinding and polishing machine; the abrasive belts were lubricated by mineral oil, which caught light. The fire spread to the oil tank and extract ducting, causing £65000 worth of damage. [...]
Horden Colliery, 1955. A steel weight was dropped on a friction prop with
  aluminium frictional surfaces, leading to ignition of firedamp.
Glyncorrwg Colliery, 1955. A rusty steel wedge was struck from an aluminium roof bar, leading to ignition.
United Kingdom, 1971. Propane was vented into a workshop during the filling of 140 lb propane cylinders when one was accidentally overfilled. The propane was ignited, and two people were injured in the subsequent explosion. The ignition source was believed to have been a spark from rusty steel cylinders being loaded outside, which, on examination, were
  shown to have aluminium smears on them. An aluminium roller conveyor was being used in the loading operation. A very light impact on such a smear is capable of generating a spark through a thermite reaction. [...]
Ruhr Refinery, 1976. Two explosions within two hours of each other occurred on the 2nd December whilst switch-loading diesel into two tankers of carbon steel fabrication using an aluminium loading arm. The explosions were attributed either to a static electricity discharge or to a thermite reaction. [...]
SS British Industry, 1963. The tanker had cathodic protection by incorporating a series of magnesium anodes in seven tanks. The cause of an explosion on the tanker was attributed to the impact of loose anode materials onto rusty deposits in the tank in which the explosion occurred. On examination after the explosion, many of the anodes were seen to be almost detached from their angle-iron bases, and bits had broken away. It
  was deemed that detachment could have been effected by a washing machine whipping violently in the tank, and that there would have been flammable concentrations of crude oil vapour in the tank.

References

Bond, J. Sources of Ignition: Flammability Characteristics of Chemicals and Products; Butterworth-Heinemann: Oxford; Boston, 1991. ISBN 978-0-7506-1180-0.


text:
Why doesn't chemical equilibrium lead to uncertainties?
Let's consider a reaction
$$\ce{a A + b B <=> c C + d D}$$
We will also assume that the concentrations are low and the activity coefficient is one. Now let's say initially one has $p_0$, $q_0$, $r_0$, $s_0$ as the concentrations of $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, respectively. Let the equilibrium constant be $K$. Let's investigate about the equilibrium composition.
If $x$ equivalents of reactants reacted, then the equation for $x$ would be:
$$K = \frac{(r_0 + cx)(s_0 + dx)}{(p_0 - ax)(q_0 - bx)}.$$
Now, my question is how do we know all the time such equations have proper roots for $x$? I mean all the conditions $x$ needs to satisfy, like $ax < p_0$ and first of all, how do we know the equation has roots in first place? It's a quadratic and it need not have real roots necessarily. This is just a simple case of 2 reactants and 2 products.
I would like to see some rigorous proof that such equations always contain proper (roots which satisfy) all the conditions or a counter example of such a chemical equation. If such counter examples exist, how does the chemical reaction proceed?

As long as K is between rs/pq and infinity, x is 0 or greater.

@JonCuster Okay, now I see something. Thanks !

I think you are seeing this from the wrong end: we know that chemicals behave this way. There is always one equilibrium position, and there is a thermodynamic reason for that. And yes, formally we can most cases describe this equilibrium with simple equations like you just presented, but this equation is derived from several approximations, so if you find cases when it doesn't work, then it doesn't work. It is like proving that e.g. r0 is always positive: No, we know that r0 is positive, we don't prove that mathematically, and we care about only the examples which do not contradict this.

Well, assuming an elementary reaction you have the wrong equation. The equilibrium is given by $$\ce{K = } \dfrac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$

@MaxW oops my bad.

The reaction quotient Q approaches zero when at least one of the product concentration approaches zero (as the reaction goes to completion in the reverse direction and the extent of reaction x is at the lowest possible value), and it approaches infinity when at least one of the reactant concentrations approaches zero (as the reaction goes to completion in the forward direction and the extent of reaction x is at the highest possible value). Also, the reaction quotient increases monotonically with x.
No matter what K is, there is always a Q that matches K because Q assumes all values between zero and infinity.
The only real case where this argument does not work (real as in concentrations have to be non-negative real numbers) is when at least one reactant and one product concentration is zero. In that case, there will be no reaction at all, and Q is not defined.

@Esha_Manideep Congrats regarding the Maths olympiad!

Sometimes, if you allow initial product concentrations to be non-zero, the reaction has to go backwards to reach equilibrium, and x will be negative. (Non of the concentrations will become negative though because the reaction has reached its limit when one species is depleted.) With negative values of x, the fraction assumes values in the range of zero to r s / p q. Bottom line: for some values of r, s, p, q, and K, a negative value of x is a physically (or perhaps chemically) meaningful result.


text:
What's this glassware?
Can anybody tell me what this glassware with a stem, bulb and long tube is?

The glassware shown is a drying tube. The bulb holds cotton or glass/quartz wool while the tube part holds a drying media such as anhydrous calcium sulfate to keep moisture out of the vessel during moisture sensitve reactions.


text:
Why does NH3 hybridize at all?
So in Chemistry class I've been taught that hybridization is a way we can explain things such as how $\ce{CH4}$, for example, forms four, equally strong bonds.
However at the same time I'm told that the Nitrogen in  $\ce{NH3}$ hybridizes and forms sp3 orbitals. When I draw the orbital diagram for $\ce{NH3}$ this is what I get:

and so looking at this, I see that $\ce{NH3}$ should be able to form 3 equal bonds in the 2p subshell which matches our observations.
So my question is— why do we then explain the bonding orbitals of nitrogen in $\ce{NH3}$ using hybridization? It seems that all of our observations of $\ce{NH3}$ can be explained without the idea of hybridization?

Re: "It seems that all of our observations of NH3 can be explained without the idea of hybridization?" // So how do you explain the geometry?!?

Let's consider the alternatives to the pseudo-pyramidal structure we observe in reality (not sure how, probably neutron diffraction or rotational spectroscopy). First alternative: a $\text{sp}^2$-hybridization with a lone pair in a p-orbital. This already assumes hybridization, but is not good energetically because a lone pair in a p-orbital is higher in energy than one in an orbital with partial or full s-character.
Second alternative: a $\ce{PH3}$-like with H-N-H angles close to 90°, putting the lone pair in the s-orbital. This does not occur because the s and p-orbitals are close enough in energy that they will participate in the valence bonds, that is, they will hybridize. The net repulsion between the N-H bonds would outweigh the energy gained by putting the lone pair in the s-orbital.
Of course, these are post-fact explanations and they are not backed by numbers here. However, using electronic structure theory methods (computational/quantum chemistry) we can calculate the relative energies of the structures described and even carry out a search for (local) minima of energy with respect to molecular structure. They would show the same thing, albeit without explanation.

Note that the lone pair in PH3 is not "in the s-orbital", nor is it correct that the s and p orbitals of NH3 are closer together in energy than the s and p orbitals of PH3. What is correct is that the bonding MOs of NH3 have more s character than in PH3 because there is less s/p mixing because s and p are farther apart in NH3.  NH3 is thus closer to sp2 in its bonding MOs than PH3 is, and its structure is correspondingly closer to a flat plane than PH3's structure (though still quite far from flat).

@Andrew I guess you have a typo there, but put that aside, you say it is not correct that s and p orbitals are closer together in nitrogen than in phosphorus; then you say there is less sp mixing in NH3 (where I assume you mean PH3) because s and p are further apart. From how I understood it, that is a contradictory statement. Could you please elaborate. Also from electronic structure theory, the lone pair orbital in PH3 has about 90% s character; I'd say you could consider this as an s orbital.

@Martin - there's no typo - less mixing in NH3 means less s character in the lone pair, since the starting state of the lonepair orbital is pure p. It's well described here: https://chemistry.stackexchange.com/questions/38599/why-is-the-inversion-barrier-larger-in-ph3-than-it-is-in-nh3. I don't know where you're getting your 90% number for PH3 lone pair s character. Jan says 50% in his answer, and I have an (admittedly quite old) review that has it at 14% (Dixon, Adv Inorg Chem Radiochem vol 29 p.41). 50% would be the maximum with an essentially sp type orbital.

@Andrew I apologise, I had a wrong number in my mind. Coulson's law would predict about sp4 orbitals for the PH bond, ergo 40% s for the lone pair. G09/NBO6 DF-BP86/def2-svp results in 57% s / 43% p character of the lone pair (or sp0.75 if you prefer); and 14% s / 85% p (or sp6) for P in the PH bonds. For comparison: NH3: LP 30% s / 70%p (sp2.4); BD 23% s / 76% p (sp3.3). Now you might call that less mixing, I wouldn't. But then again... that's all not real anyway.


text:
What is the benefit of using a pH indicator reagent over pH strips or a meter?
I've recently started getting into chemistry a bit as a hobby. Nothing serious, just amateur level stuff mostly, primarily just to educate myself.
Whenever I need to determine the pH level of a solution, I typically use either a pH meter I have, or sometimes the pH test strips. And while I do have a "universal indicator" solution, I've never used it before. Mainly because it seems to me that adding an additional chemical to the solution you're working with just to determine the pH seems a bit counter intuitive to me.
A couple of reasons I can think of off the top of my head:

Even if the indicator seems somewhat benign, you've now added some extra volume that isn't essential to the reaction or solution.
Some pH indicators use chemicals that are somewhat hazardous or carcinogenic.

Any time I've seen someone use a pH indicator reagent, it's always been in some type of demonstration of a reaction, where the yield or result of the reaction is just discarded in the end. This gives me the impression that pH indicator reagents are useful for neat chemistry tricks (like here, where the chemist uses bromophenol blue to demo dichromatism), but not so much for instances where one plans on utilizing the resulting solution for any reason.
Thanks in advance for any input!

If you are doing titrations you need a quick response to sometimes drop amounts that could be too much. That means you have to keep your sample under constant stirring so it mixes very well. If you had to stop after every drop and use a strip it would take for ages. And I don't know how fast pH-meters actually react to changes. They are also expensive and indicators are much older and well known. There are also indicators or titrations that do not base on pH but on redox-reactions for instance.

In a general chemistry class giving every student a pH meter would be very expensive. Plus the pH electrode is expensive too and delicate. A pH electrode is easily smashed if banged against the side of a flask or with a stirring bar. Lastly a pH meter setup won't give any better results than the proper indicator.

And you don't have to smash a pH electrode to break it. Too alkalic is bad, drying is bad, basically just using it isn't too good. And it doesn't turn green or purple if it wants to be left in a beaker with buffer solution for a few days to recuperate.

I should also add that while swirling a solution with an indicator you can get a good idea of how close you are to the end point due to the inhomogeneity of the solution.

And each indicator has a particular looking at about equivalence point which tells a lot to an experienced operator. To get the same feeling one should plot real time the ph meter data. Now feasible of course but there is quite history here. You can image dedicated glasswares easy to swirl and incorporating micro electrodes etc but all will be expensive. Etc.


text:
Polymer Chemistry NMR question
I am extremely new to both polymer chemistry and NMR.
I am struggling with the following question:
A proton NMR is used to attempt to quantify the molecular weight of a poly(ethylene oxide) molecule with methyoxy end groups at each terminus. If the integration of the methyl protons relative to the methylene protons gave a ratio of 1:20, what can  you say about the molecular weight?
Thank you. All help is appreciate including resources for me to be able to read up on my lacking knowledge.

Note that "all" NMR does in this case is provide you with the relative amounts of methyl and methylene group hydrogen atoms, so while NMR is particularly useful for such a determination, it's really a question about understanding molecular structure and some math.

The longer the polymers, the more challenging they can be in NMR, because of their relaxation and diffusion times. You absolutely need to talk to your NMR specialist, expose the molecules you are studying and ask him/her: "Can I use the standard proton settings? If not, how can I make sure the spectra lines are actually quantitative?"

@SteffX It's not so much about the size, but the solubility. Diffusivity is largely irrelevant, and relaxation times don't change once you are in the range of actual macromolecules. In all cases, I would however run a short saturation recovery $T_1$ experiment. Takes little extra time, and then you know where you are with respect to quantitativeness.

I hope this is not a homework question. Suppose your polymer has $n$ reapeating units and capped with methyl groups at the end as you described. Thus, it should looks like following figure:

Thus it has $n$ $\ce{(-CH2CH2-)}$ units and $2$ $\ce{(-CH3)}$ units. Thus your $\ce{(-O-CH2-)}$ signal is accounted for $4n$ protons while $\ce{(-O-CH3)}$ signal accounted for only $6$ protons. Thus,
$$\frac{4n}{6} = \frac{20}{1}$$
You may find the value of $n$ from this equation and you can calculate the molecular weight of the polymer in hand.

Importantly, almost all polymer samples are a mixture of molecules with different chain lengths, and quantification by NMR provides an average value for the whole sample, more specifically the number-average chain length/molecular weight.

Also important, it might not be the case tough, but polymers tends to have different dynamics and respond differently to nmr, to the point that quantitative evaluation of this kind might be fallacious. Not sure in this case. Sure with polymers having pendants attached. The latter can be seen while the backbone no, for instance.

@Alchimista Very unlikely in solution. If you use standard recycle delay settings for 1H NMR (20 s or so), the end groups will definitively give a quantitative response, and the less mobile middle part of the polymer chain has an even shorter $T_1$. What do you mean by "pendants"?

Pendants as functional groups consisting of mobile alchyl chains terminated with big groups. My comment might over emphasises the situation in solution but it is what I have observed at least with conjugated polymers. Might also be related to the fact that true solutions are hard to attain with many polymers in general. But I am just suggesting a possibility and I do not think is strictly related to the Q. @Karl

@Alchimista I can imagine conjugated backbones to be a problem. Might be so stiff even in solution that their nuclear spins don't find the way back to equillibrium. I'd love to see T1-correlated spectra on such stuff.

@Karl not a specialist at all on nmr. But I see colleagues apply special routines in non conjugated polym. as well.  I think viscosity is one of the major problem. But again all this exceed the scope of the Q. Perhaps I shouldn't have mention that.

@Nicolau Saker Neto: I completely agree with you. However, I was rush to put the answer on Friday night so it contained some clerical errors, which I corrected today. But do not want to touch the rest because it was on hold now.


text:
Why 1-ethenyl-3-ethyl-5-ethynylbenzene has that name?
I would priorized these funtional groups based on two options (following the ciclohexane rules):

Molecular weight
Representing each pi bond as a alkane branch. e.g. Ethenil as a isopropil backbone.

But, none of these option works. Why this molecule has that name?

Ordering substituents by complexity has been abandoned long ago (before 1979) in favor of the much simpler alphabetic order. Here, ethenyl < ethyl < ethynyl.
Note, however, that simple substituents are ordered alphabetically before adding a multiplicative prefix, which means that "dimethyl" is ordered as "methyl".
Complex substituents are ordered by their first letter, e.g. isopropyl (I), dimethylphenyl (D)

almost.... but why ethyl is prioritized over ethynyl or why ethenyl is not prioritized? With prioritization I refer to assign numbers to the substituent groups on the ring.

@Another.Chemist Here whatever way you go (starting from any substituents) you will always get no. $1,3$ and $5$ as the locants number. So, alphabetical priority is the only determining factor for nomenclature.


text:
Aromaticity fails?
Why is cyclopentadiene ($\mathrm{p}K_\mathrm{a} = 16$) less acidic than
A) 2,4-pentanedione $(\mathrm{p}K_\mathrm{a} = 9);$
B) phenol $(\mathrm{p}K_\mathrm{a} = 10);$
C) water $(\mathrm{p}K_\mathrm{a} = 15.7)?$
Cyclopentadiene forms an aromatic compound as conjugate base and is quasiaromatic, while phenol has just a negative charge near an aromatic benzene ring, nothing special.
Pentanedione has got active methylene but still how can resonance (even equivalent resonance) can dominate over the aromatic effect?
Finally, why is it even less acidic than $\ce{H2O}?$
$\ce{H2O}$ gives $\ce{OH-}$ which has a localised electron oxygen atom; how is it better than aromatic delocalisation?
It has defied most of my concepts of organic chemistry.

Two hints: a) cyclopentadiene is still just a hydrocarbon, b) hydrogen bonds.

Oxygen beats aromaticity.

Because aromaticity isn't the only factor that determines how stable something is, maybe?

For phenol, think about which bond is easier to break in the solution phase(that is, is more polar),O-H or C-H?

@YUSUFHASAN H-F is more polar than H-I but stronger

@orthocresol Yeah,but I think that happens because we are moving down a group, whereas C and O are of the same period. So, I think my statement was valid as we move across a period

Aromaticity does not rule over everything else.  It merely impacts a species relative to others that have formally similar bonding.
To take an example involving a simpler aromatic species:  3-chlorocyclopropene transfers a chloride ion to a Lewis acid, such as $\ce{SbCl5}$, more readily than 3-chloropropene or chlorocyclopropane because the 3-chlorocyclopropene forms an aromatic cation.  But sodium chloride, in which the bonding to chlorine is radically different from chloroalkenes or chloroalkanes, still reacts even more readily and with more Lewis acids than 3-chlorocyclopropene.
Likewise, when cyclopentadiene is compared with other candidates for acids, it will be stronger than most hydrocarbons due to the aromatic conjugated base.  But it does not achieve equality with bonding the hydrogen to, say, oxygen or chlorine.


text:
I am not able to get the exact definition of a solution
A substance which is in larger proportion by mass is called solvent and which is in lesser proportion is called solute.
   What if the volume of the substance with lesser mass is more. Will it still remain same?

At room temperature you can dissolve about 1000 grams of cesium acetate ( asolid) into 100 grams of water (a liquid). I'd still consider water to be the solvent and the cesium acetate to be the solute.

The definition given by the IUPAC Gold Book

A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the  ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

So this means that the substance which has more mole fraction is considered solvent irrespective of the volume occupied right? (eng is not my first language)

No, the mole fraction is only mentioned with respect to dilute solutions. The answer to your question is covered by the "for convenience" part, which essentially means: nobody cares, not even IUPAC.

@Pruthvi — Yes usually, though it is less clear in extreme cases.

More appropriately, a solvent is the one which has the same physical state as the solution. For example when you dissolve a gas in liquid then the solution of gas in liquid resembles the same physical state as the solvent i.e. liquid.
If both solvent and solute(s) are liquids then generally proportions are taken into account i.e. one with higher proportion is solvent.
But rarely some other criterion is used you will understand when you see it.


text:
Why does MgCl2 become Mg^2+ and Cl^-?
I'm taking a biology class, and it's one-chapter intro to chemistry chapter had been driving my study group nuts. On a recent quiz, it asked the question:

In water, $\ce{MgCl2}$ dissociates into $\ce{Mg^{2+}}$ and $\ce{Cl-}$. Based on this information what type of bond is involved in the formation of $\ce{MgCl2}$?

I guessed (and got correct) that the bond was ionic, since the only thing that changed in their symbols was the ionization ($\ce{Mg}$ has +2 electrons, $\ce{Cl}$ is -1 electron). What it doesn't explain (it literally doesn't even attempt to explain), is how do these break apart to form two ions? At first I thought this had to do with the stability of common $\ce{Mg}$ and $\ce{Cl}$, but since they both are missing electrons in their outermost shell, that didn't seem to be the way to go. So at this point, we're all lost, and the textbook/professor's explanation of another question which made it sound like the periodic table measures atomic mass in grams instead of Da makes us have no confidence in what it's explaining.

Hello and welcome to Chemistry Stack Exchange! You may go through the tour page (https://chemistry.stackexchange.com/tour) and the help center (https://chemistry.stackexchange.com/help). While your question is not entirely correct, I believe that there may be a typo mistake in your title ($\ce{Mg}$ instead of $\ce{Na}$). Also, please use mhchem (LaTeX commands for chemistry formatting) to make chemistry terms stand out in your text clearly.

The Compound MgCl2 consists of magnesium 2+ ions and Cl- ions already. No electrons are transferred in the process of getting those ions into solution.

From reading your question, maybe it's best to begin with an introductory chemistry course for even younger students (not to demean you or disrespect you in any way) so that you may learn better. I'd suggest Khan Academy - Atoms, compounds, and ions. This should show you the way through whatever chemistry you need to study biology.
Regarding your question specifically, don't worry too much.

In water, MgCl2 dissociates into $\ce{Mg^{2+}}$ and $\ce{Cl-}$. Based on this information, what type of bond in involved in the formation of $\ce{MgCl2}$?

Think of what the question suggests. It tells us clearly that the substance $\ce{MgCl2}$ has dissociated into two ions $\ce{MgCl2->Mg^{2+} + Cl-}$. This could happen only if ions were present in the substance to begin with. Hence, the substance must have had ions, which would be held together by ionic bonds.
(Regarding the atomic mass being measured in grams and not Daltons, don't worry. That is irrelevant to this topic, and probably came up due to some other discussion.)
I further repeat my advice, please go through an introductory level chemistry course which will make you comfortable with basic chemistry and the principles which you will need to apply for biology. I speak from experience, having faced a similar condition a couple of years ago...

Water molecules have what is called a dipole. Basically oxygen atoms are more capable of attracting electrons to them than hydrogen atoms(oxygen is more electronegative than hydrogen). This results in the oxygen end of the water molecule having a partial negative charge and the hydrogen end having a partial positive charge. So what happens when you add magnesium chloride to water is that the hydrogen end of water is attracted to the negative chloride ions in the lattice so water molecules surround the chloride ions using their hydrogen end. The opposite occurs for the magnesium ions where now the oxygen end of thewater molecules is attracted to the magnesium ions in the lattice so water molecules surround magnesium ions using their oxygen end. The magnesium ions and chloride ions are now separate and can move freely in water.This is known as hydration.

This pic is of sodium chloride but it's the same as magnesium chloride.
I didn't really get your question but I hope this answers it!


text:
Why does Coca-Cola Zero Sugar Contain Sodium Despite no Sodium Ingredients?
According to the Coca-Cola website and every label I read, Coca-Cola Zero Sugar is a mixture of organic compounds:

CARBONATED WATER (OC(OH)₂), CARAMEL COLOR (Various sugars), PHOSPHORIC ACID (H₃PO₄) , ASPARTAME (C₁₄H₁₈N₂O₅), POTASSIUM BENZOATE (C₇H₅KO₂), NATURAL FLAVORS, POTASSIUM CITRATE (C₆H₅K₃O₇), ACESULFAME POTASSIUM (C₄H₄KNO₄S), CAFFEINE (C₈H₁₀N₄O₂).

The label also says 40 mg sodium. There's a perceptible salty taste in the beverage, which suggests to me that something activates the taste receptors for saltiness, but a potassium cation can do that just fine.
There is no sodium in any of those molecules. That means "Natural Flavors" must contain the sodium... but in what form? The essential oils in things like OpenCola seem to all be hydrocarbons without sodium. And if it's plain old NaCl table salt, why wouldn't they just list it as "salt"?
Where's the sodium in Coke coming from?
EDIT: The discussion has centered around the sodium content of "still" drinking water; what about the process of carbonating the water? Some processes add a sodium carbonate to the solution; could that be the source, contained in the ingredients list as "carbonated water"?

You forgot water, it's not like they're using twice distilled.

They probably are; it would be the only way to control flavor. Dasani gets filtration and purification and is then re-flavored.

The claimed sodium content is far too uniform to explain this by saying it's in the water. Dasani is bottled by Coca Cola bottlers.

I keep forgetting to mention that Dasani's sodium content is reported as "0", meaning in the U.S., between 0 and 4 mg. No; not twice distilled, but yes; reverse osmosis and other filtration are almost certainly in use.

And, I'm made thoughtful by the idea that I considered ingredients but not process while thinking about it. See my edit in the question: Could the sodium come from the carbonization process?

Basically, the Sodium could come from any or all of the ingredients used to manufacture Coca Cola. The ingredients list is based on the raw-materials used in the product, but these ingredients are not necessarily pure, even if they are Food Grade. As an example: The Phosphoric acid used is probably a 75% H3PO4, a common commodity in the industry, but the remaining 25% is not pure water. It may contain a number of impurities and by-products, such as Sodium, but as long as they are considered harmless they will not be listed anywhere.
The Sodium-content in the Cola, is on the other hand based on an actual analysis of the final product (at some point during product development).

So if I'm understanding it correctly I'm considering the fact of food grade ingredients containing other ingredients that might be sodium, implying that the FDA permits impurities that add up to real nutrient content, measured separately from a different analytical method. That's decent but the sodium content is itself too uniform on that label for every Coca-cola produces carbonated beverage, worldwide. That implies some kind of regulated process inserts or balances it.

I have no inside information about the production of Coca Cola, but I imagine that the company has very strict specifications for the ingredients they use, above and beyond what FDA requires. Also, at least in my region of the world, Cola is made from concentrate that is diluted at a local brewery. All in all, this results in a very uniform product. But furthermore, Sodium is not analyzed and specified (on the label) for each batch. It is stated as a “typical value” and it is only required that it is monitored as part of the company’s own quality control.

Both answers are correct; it's "water and other stuff", but this one said so specifically. I think what I ran into is a difference in analytical methods (in the Coke, not the answers). A lesson not to expect purity from real life. :-) Thank you all for contributing.

Two sources: water and natural flavors
According to the Wisconsin department of healt natural sodium levels in water is $\pu{40-60 mg/L}$ meaning a $\pu{20 fl oz (\! 591 mL)}$ Coke has about $\pu{25 mg}$ of sodium from water. Thus the other $\pu{50 mg}$ of sodium must come from natural flavors. This makes sense since Coca-Cola uses plant products like coca leaves which would naturally add sodium to the carbonated sugar water.

coca leaves, are you sure, generally to use these  these is illegal.

@porphyrin Yes they do use coca leaves still, though by the time Coca Cola receives them the cocaine is not present. https://en.wikipedia.org/wiki/Coca#Legal_status

They probably are using purified water. It would be the only way to keep flavors uniform between regional bottlers. Dasani has no sodium. My area reports about 0.01 mg/g, or around 6 mg per 20 oz., around a tenth of WI figures. But the bottles are labeled consistently everywhere I've been.

@RobPerkins Taste generally isn't uniform and water has like smallest effect on it. They even use different receipts in different countries.

Uhm, you probably meant "the other 20 mg of sodium". I didn't realize coke still contained extract from coke leaves, but what guarantee that this is the source of the sodium? Fascinating, but seems like guesswork.

The most important fact here is that potable domestic water supplies can contains 40mg/L of sodium. This puts the presence in Coke in perspective and suggests it isn't really a lot of sodium.

Sure. But the nutrient content "sodium" is still uniform across all their products. It's a specific ppm in the solution. As I pointed out in my question edit, I'm leaning toward "it's in the water" but there's this strong indication that they put it there during bottling, not that they received it from a stock ingredient.

Both answers are correct; it's "water and other stuff", but the other one said so specifically. I think what I ran into is a difference in analytical methods. (in the Coke, not the answers). A lesson not to expect purity from real life. :-) Thank you all for contributing.

Note that this doesn't mean Coca-Cola typically contains 40 mg sodium per serving.  Instead, that's likely an upper limit designed so that, with normal variation, the product will still be in compliance.
FDA regulations specify that that, for "Third Group" nutrients like sodium, the actual amount must not exceed the label amount by more than 20%.  Thus companies are required to take enough samples of their product to determine the variation, and use a label amount high enough to ensure that, even with the variation, the actual amount doesn't exceed the label amount by 20%.
Here's a screenshot from FDA's website (https://www.fda.gov/food/guidanceregulation/guidancedocumentsregulatoryinformation/labelingnutrition/ucm063113.htm) showing an example of how this might be done for sodium.  Note that the mean, found from an analysis of 96 samples, is 95.81 mg per serving.  However, based on the stdev, a one-sided 95% prediction interval estimates that (within this interval) the maximum amount of sodium would be 120 mg/serving, so that's what is listed on the label.  Given this, the product would be considered in compliance if the actual amount of sodium is <= 144 mg/serving.


text:
How does hyperconjugation lead to stability?
I know that hyperconjugation leads to delocalization of electrons between filled and unfilled orbitals and thus greater stability (that's what most textbooks said).
However, I think that it reduces the bond strength, because by saying that the electrons delocalized would mean that the electron density between two atoms decreases and hence the 'glue' (bond) that hold the two atoms together becomes weaker.
Any misconception here? Because my statement seems to contradict with the fact that hyperconjugation increasing stability. Why reduction in bond strength increase stability? (or does it actually decrease the bond strength?)

The little ionic character does not weak any bond..... You don't have to see the sacrifice of a covalent bond as leading to unbound atoms!

Think of salts for instance!

What you are ignoring is the magnitude of the effect.
While hyperconjugation does decrease the electron density of a localized bond and slightly increases the energy, the overlap with a high-energy, empty orbital is at least very poor. Think that the orbitals are very far from each other and have different symmetries. The total electron density of this bond would remain largely unaffected.
On the other hand, the difference between zero electrons and partial electron density for an empty orbital of high energy is significant, and it increases (to a moderate extent) the stability of, say, a carbocation. Besides, a single orbital often receives electron density by hypercongujation with more than one bond.

Hyperconjugation effect also improves stability by distributing the positive charge or the radical electron on the involved atoms . Also for alkenes after hyperconjugation effect the pie bond shifts between other atoms thus decreasing the bond lengths of Sigma bonds( involved in the effect) and thereby strengthening their bond.  Thus overall it  stabilizes the compound.

This requires molecular orbital theory, which (it seems) is something organic chemistry classes absolutely refuse to teach.
Basically, it goes along these lines:
In the case of CH3+, you only have the 3 sigma bonds (A1’ and E’), meaning you have 6 electrons delocalised along the entire molecule and holding it together.
But in the case of CH2CH3+, the added methyl group can form a pi bond with The Central carbon; this forms a pi bond that stabilises the ion.
I recommend either An Introduction to Molecular Orbitals or Orbital Interactions in Chemistry (which I just bought recently-good book.)


text:
Dominant contribution of v = 0 → v = 1 transition in vibrational spectroscopy
Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.
After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($\Delta v = \pm 1$), my professor's notes state that:

Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $v = 0 \to v = 1$ transition and is called fundamental transition.

I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $v = 0 \to v = 1$ will take place?

A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $\nu$ (nu). The Greek letter $\nu$ is used to refer to a frequency; so we can have a transition for which $\Delta E = h\nu$, for example.

The reference should really be to the Boltzmann distribution where the  population is proportional to $\exp(-E_n/(k_\mathrm BT))$  where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_\mathrm BT$ at room temperature $\approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $n\gt 0$.


text:
Solubility of Potassium Dichromate in Water
According to Wikipedia, the solubility of Potassium Dichromate increases very significantly with the temperature.
At 20°C, 13g/100ml, and at 100°C, 104g/100ml.
Therefore, it should be possible to dissolve about 520g of potassium dichromate in 500 ml of boiling water, but only about 65g in the same water, but at room temperature.
Today, I acquired about one kg of technical potassium dichromate and tried recrystallizing it, by weighing 507g of raw dichromate and dissolving it in about 500ml of water, room temperature. I expected it to no dissolve fully until the water was heated to boiling.
However, as soon as the water got slightly warm (45°C-50°C) everything dissolved. I added more raw material, and it kept dissolving until in the end everything (~970g) was added and dissolved before reaching boiling temperature.
Now, it is very slowly cooling down, and at about 35°C I seeded some small crystals and they're growing.
What went wrong in my calculations of the solubility? Anything I should've considered and missed?
Thanks.
PS.: Things I've already considered and ruled out:
Crystallization water: As far as I reseached, potassium dichromate doesn't have crystallization water, unlike, for example, copper (II) sulphate.
High humidity: My raw material was a free flowing powder, < 1mm particle size, uniformly deep orange colored. Had I high humidity, it'd clumping and not free flowing.
Contamination with potassium chromate, sodium chromate: Any of those wouldn't be soluble enough to explain what I observed.

This may be due to equilibrium with chromate, which is better soluble in low temp. If that's the case, you should acidify water.

There are lots of images online of the color difference between dichromate and chromate solutions. Maybe you can tell from the color if that equilibrium was a big factor?

Nice idea. I will try acidifying the water, but I believe there was little to no chromate, since the color is a deep orange, not the strong yellow of chromates. Granted, the yellow may be masked by the orange. Collected crystals formed overnight (~25°C) and they weight at around 260g, still humid, guessing 230g dry.

What happened was very simple.
I've been duped.
The seller sold me (intentionally or not) sodium dichromate and not potassium dichromate, or at least very contaminated with sodium dichromate.
According to online source (https://en.wikipedia.org/wiki/Solubility_table#S), the solubility of sodium dichromate in water is about 198g/100ml of water at 30°C, and a very high 415g/100ml at 100°C.
That explains clearly why everything dissolved promptly while the water was only slightly warm (40°C? Didn't come to mind put a thermometer in since I was expecting it to dissolve only near 100°C), since 500ml of solution would be able to dissolve 970g (my raw material) at only 30°C.
Also, when I crystallized some of it overnight on a cold (~15°C?) water bath, the solubility would drop to about 175g/100ml, causing the precipitation of about 100g of salt.
Finally, to settle matters about the presence of sodium dichromate, I dropped some saturated potassium chloride on a small sample of the solution in a test tube, and observed the instant formation of an orange precipitate.
The same occurred when I picked some of the crystals I separated before and made a small amount of saturated solution - therefore the crystals were made of sodium chromate and not dichromate - since they have much higher solubility they shouldn't've precipitated at all before potassium dichromate.
Also, I picked some potassium dichromate of known procedence and made some saturated solution and added a few drops of a saturated solution of potassium chloride, and nothing happened.
In the end, everything said, for what I intend to make, sodium dichromate or potassium dichromate are interchangeable after adjusting quantities, but I'll have to execute extra steps to purify it from this solution.
As an piece of advice (for me and everyone else), I'd test any new potassium/sodium dichromate from unknown sources before doing anything with it, simply by making some saturated solution and adding saturated potassium chloride.
I'll have to devise a procedure for separating one from another given unknown proportions of either.
Thanks for everyone time.

RE: "... I'll have to devise a procedure for separating one from another given unknown proportions of either. ..." -- The best way would be with an ion exchange column. I'll add though that I did a lot of x-ray fluorescence. You never get just a K or Na salt alone. There is always a detectable amount of the other cation.

Sure, truly pure Na or K salts are hard to obtain, if their original production source involved both. For me, if it's > 99% of one or another, it is fine. Careful multiple recrystallizations are able to achieve that. Ion exchange columns are probably just outside of the reach of an amateur lab.


text:
Why might the yield of my synthesis of triphenylmethanol be so poor?
I performed the classic Grignard reagent formation leading into the preparation and purification of triphenylmethanol using phenylmagnesium bromide as my Grignard, and benzophenone as the reagent for the subsequent reaction.
In the formation of the Grignard reagent, I used Mg turnings to react with bromobenzene to produce phenylmagnesium bromide with excess bromobenzene as Mg was my limiting reagent.
In the second reaction, I reacted the Grignard reagent with benzophenone, my limiting reagent, to form an alkoxide intermediate, to which I added excess 6M HCl to protonate the intermediate to form crude triphenylmethanol.
I then performed a recrystallization of the crude triphenylmethanol (with biphenyl as a by-product) with isopropyl alcohol.
Aside from poor technique with the laboratory procedures, what can contribute to a poor yield of purified, recrystallized triphenylmethanol?
I did some searching online, and found that certain individuals mentioned a low yield being attributed to "competing side reactions", but with no further details. Is this referring to the biphenyl side-reaction between the phenylmagnesium bromide and bromobenzene, or another external factor?

Too little Grignard reagent. Evaporate the mother liquors of from the crystallization for more trityl alcohol and run a tlc. Formation of trityl chloride in your workup.

Use a milder workup such as aq NH4Cl. 6N HCl pretty well guarantees formation of the trityl cation from trityl alcohol.


text:
Acidity of carbonate ions creates paradox?
Whilst I understand why salts can be acidic or basic, the carbonate reaction confuses me (carbonate being a basic ion).
$$\mathrm{{CO_3}^{2-} + H_2O \ce{<=>} {HCO_3}^- + OH^-} \\
\mathrm{{HCO_3}^{-} + H_2O \ce{<=>} {H_2CO_3} + OH^-} $$
The second reaction produces an acid plus a base, so why don't they neutralise immediately? Also, I can see why it is a basic salt by releasing $\ce{OH-}$ but doesn't it also release carbonic acid which would disassociate to give off $\ce{H+}$ ions which would just react with the OH and give no net change in pH then? So basically, I'm confused as to why the acid and base don't neutralize immediately and the carbonic acid just "stores" protons.

The first reaction also produces an acid plus a base. Then again, have you heard about the autoionization of water? $$\ce{H2O<=>H+ +OH-}$$

The concept of equilibrium between reagents and products is important.

It is more complicated than you indicated. Read about carbonic acid at Wikipedia. So it is really the various equilibria between dissolved $\ce{CO2, H2CO3, HCO3^-}$ and  $\ce{CO3^{2-}}$

I think it might be more clear if you wrote the second reaction as $\ce{HCO3- + H3O+ <=> H2CO3 + H2O}$. Does that help you make sense of it? Basically, as you've written it, the equilibrium lies far to the left. The right side products are only favored if there is a high concentration of protons and low concentration of hydroxide ions, ie at low pH.

There are two concepts you need to know.
What is equilibrium?
Not all reactions occur until one of the reactants is consumed completely. Some of them occur until a certain point, where the system doesn't continue to evolve; rather, the concentrations of all the species remain constant in what is called a chemical equilibrium. This was first observed experimentally and then some theories started to appear in order to explain this.
The most important variable is the equilibrium constant, a quotient of the equilibrium concentration. If you think of an equilibrium as two opposite reactions, then this constant measures which one is prevalent.
If we have:
$$
\ce{aA + bB <=> cC + dD}
$$
Then the equilibrium constant K is:
$$
K = \frac{
[C]^c \times [D]^d
}{
[A]^a \times [B]^b
}
$$
This means that if you mix A and B, or C and D, or those two with more reactants added, no matter what the initial result is, the final concentration of everyone has to be such that at the end, K is the same.
So, if K is big, it means at any given time there is more of C and D than A and B. The direct reaction (to the right) is prevalent. If K is small, then the opposite reaction (to the left) is prevalent.
So yeah, it is possible to have at the same time the reactants and the products of a reaction simultaneously in the same place. How? That is concept number two.
Equilibrium is dynamic
I told you to picture an equilibrium as two opposite reactions. That is because in reality, those two reactions are happening continuously! Indeed, in your example:
$$
\ce{CO3^2- + H2O <=> HCO3^- + OH-}
$$
A certain amount of carbonate molecules are reacting with water to give $\ce{HCO3^-}$ and $\ce{OH-}$, while the exact same amount of other, different $\ce{HCO3^-}$ molecules are reacting with $\ce{OH-}$ to give water and carbonate.
Now, equilibrium isn't instantaneous. What happens is that when you throw reactants together, the more favorable reaction happens at light speed while the opposite almost doesn't happen. As products start to accumulate, the speeds start to become more moderate until they are exactly the same and the overall reaction stops. The concentrations stop changing even though there are still molecules reacting all the time.
Conclusion
To simplify, you can think of the process you described as a two-step process. First, $\ce{CO3^2-}$ gives $\ce{HCO3^-}$ and $\ce{OH-}$ and achieves equilibrium (equation 1 that you wrote). Then, the $\ce{HCO3^-}$ reacts and reaches equilibrium, and as initial conditions we take the products of the first step.

Carbonate in water
$$
 \ce{CO3^2- + H2O <=> HCO3^- + OH-}\qquad K_1 = 2.13 \times 10^{-4}
 $$
Okay, so here we see that most $\ce{CO3^2-}$ remains as-is, producing only a small amount of $\ce{HCO3^-}$ and $\ce{OH-}$.
Everybody reacts
$\ce{HCO3^-}$ has many options. It can undergo the same reaction as before, reversed:
$$
 \ce{HCO3^- + OH- <=> CO3^2- + H2O}\qquad K_2 = 1/K_1 = 4700^\mathbf{*}
 $$
It can also further react with water:
$$
 \ce{HCO3^- + H2O <=> H2CO3 + OH-}\qquad K_3 = 2.22 \times 10^{-8}\\
 \ce{HCO3^- + H2O <=> CO3^{2-} + H+}\qquad K_4 = 4.7 \times 10^{-11}
 $$
Here we see that the only significant reaction would be to reverse the first one (although the others also happen, but to a minimal extent). But we said before that an equilibrium consists of two simultaneous opposite reactions. Thus, the concentration remains largely the same, unaffected by the others which have very small K.

The amount of $\ce{OH-}$ produced is bigger than what was originally on water,
$$\ce{H2O <=> H+ + OH-} \quad K_w = 10^{-14}$$
due to the value of the equilibrium constants ($K_w = 10^{-14} \ll K_1 = 2.13 \times 10^{-4}$).
So $\ce{CO3^2-}$ is indeed a basic salt.

$\mathbf{^*}$ Easy to understand if you look at the generic expression for K.