text:
Gibbs free energy of formation vs. Gibbs free energy of reaction?
Consider the following reaction:
$$\ce{A(s) + B(s) <=> AB(s)}.$$
I know there might be kinetic limitations but I am only interested in thermodynamic relations.
In general, the Gibbs free energy of reaction reads:
$$\Delta G = \Delta G^⦵ + RT\ln{\prod a_i^{\nu_i}}$$
and for the given reaction:
$\Delta G = \Delta G^⦵ + RT\ln{\frac{a_\ce{AB}}{a_\ce{A}\cdot a_\ce{B}}} \label{eqn:1}\tag{1}$
Since all components are pure solid substances, all activities equal 1 and therefore, $\Delta G = \Delta G^⦵$.
If $A$ and $B$ are elements, I found the following formula:
$$\Delta G_\mathrm{f}^⦵ = N\cdot RT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)} \label{eqn:2}\tag{2}$$
where $N$ is the total number of atoms and $x_\ce{A}$ and $x_\ce{B}$ are the mole fractions. To me, this latter formula implies, that the activities of $\ce{A}$ and $\ce{B}$ inside the product $\ce{AB}$ are of interest.
I have two questions:

What is the origin of the second formula $\eqref{eqn:2}$? In almost every source, the first type of Gibbs energy is used and I was not able to find any derivation. Also I think in principle, equation $\eqref{eqn:1}$ and $\eqref{eqn:2}$ should produce the same, but it looks very different!?
How does it come, that in one case the activity of the whole product $\ce{AB}$ is important and in another the single activities of the components of the product?

--------------------Update--------------------
Based on comments I'll try to reformulate my problem to make it more clear.
As I have learned, there should be no difference between the reaction product AB and the mixture of A and B with the corresponding stoichiometry (1:1). Is this valid in general or just for reactions like this (cf. my water-example in the comments)?
The transformation of equation (2) (in NightWriter's answer) shows, that it is an identity of the Gibbs free energy of mixing but in terms of moles of compound AB rather than in terms of moles of mixture.
If equation (1) describes the reaction (i.e. formation) of AB and equation (2) the mixing of A and B, and there is no difference between a product of reaction and the corresponding mixture of the two components to form that product, there must be some mathematical relation between equation (1) and equation (2). Furthermore, I am not certain, whether the standard-state symbol in equation (2) is correct or not.
After I read the answer and the comments again, I concluded the following: equation (1) describes the formation of AB and equation (2) the mixture of A and B at arbitrary compositions. So, if I set $x_A = x_B =0.5$, it should be possible to replace $\Delta G^⦵$ in equation (1) by $\Delta G_f^⦵ = N\cdot RT\ln{a_A^{0.5}a_B^{0.5}}$. $N$ must be chosen properly, since the two expressions are related to different things (mixture and compound). Is this correct?

I don't think your reaction will reach equilibrium - it will go to completion (all products if the Gibbs free energy of reaction is negative, all reactants if it is positive). Only if the Gibbs free energy happens to be exactly zero, both reactants and products will be present, and the extent of reaction will be undefined.

What is the origin of the second formula (2)?

Equation (2) evidently refers to the free energy of formation (at standard pressure) of a solid solution of A and B, in which case neither element is in its standard state as a product (so that $a_i\neq 1$). You can rewrite expression (2) as follows:
$$\Delta G_\mathrm{f}^\circ = nRT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)} = RT\ln{\left(a_\ce{A}^{x_\ce{A}}a_\ce{B}^{x_\ce{B}}\right)^n}= RT\ln{\left(a_\ce{A}^{nx_\ce{A}}a_\ce{B}^{nx_\ce{B}}\right)}= RT\ln{\left(a_\ce{A}^{n_\ce{A}}a_\ce{B}^{n_\ce{B}}\right)}$$
$$ = n_\ce{A}RT\ln{a_\ce{A}+n_\ce{B}RT\ln{a_\ce{B}}}$$
where $n$ is the total number of moles ($n=n_A+n_B$).
Since the activity of the elements in pure solutions is 1 we can finally write
$$\Delta G_\mathrm{f}^\circ = n_\ce{A}\Delta G_{mA}+n_\ce{B}\Delta G_{mB}$$
where $\Delta G_{mi} = G_{mi}-G_{mi}^\circ= n_\ce{i}RT\ln{\left(\frac{a_\ce{i}}{a_\ce{i}^\circ}\right)}$.
However the most useful way to express this is as follows:
$$\Delta G_\mathrm{f}^\circ = (n_\ce{A}G_{mA}+n_\ce{A}G_{mB})-(n_\ce{B}G_{mA}^\circ+n_\ce{B}G_{mB}^\circ)$$
In words, the free energy change corresponds to mixing of $n_\ce{A}$ moles of pure A and $n_\ce{B}$ moles of pure B (the pure components having activity $a_i^\circ=1$) to form a mixture where the components have activities $a_i$.

How does it come, that in one case the activity of the whole product AB is important and in another the single activities of the components of the product?

Equation (1) refers to a molar free energy of formation of $\ce{AB}$ from reagents $A$ and $B$, all under the same constant (P,T, composition) conditions, whereas (2) refers to the free energy of formation of a solid solution of $n_A$ moles of A and $n_B$ moles of B from pure components. Equation (1) refers to combination of A and B at a 1:1 mole ratio, or equivalently reaction to form 1 mole of $\ce{AB}$ from $n_A=n_B=\pu{1 mol}$. Reaction (2) refers to mixture of A and B at any arbitrary ratio or total number of moles. Therefore equation (2) is in a way more general. Also, equation (1) refers to a differential process (transformation to form 1 mole of product under contant conditions) whereas (2) refers to an integral (mixing) process.

For the given reaction:
  $$\Delta G = \Delta G^⦵ + RT\ln{\frac{a_\ce{AB}}{a_\ce{A}\cdot a_\ce{B}}} $$
  Since all components are pure solid substances, all activities equal 1 and therefore, $\Delta G = \Delta G^⦵$.

A comment on this: the fallacy here is to assume that $a_i=1$ at equilibrium. Only for the pure reagents and products in their standard states it is strictly required by definition that $a_i=1$.
Note also that since $\Delta G^\circ = G_{mAB}^\circ-G_{mA}^\circ-G_{mB}^\circ$ and  $G_{mi} = G_{mi}^\circ+RT\ln{\left(\frac{a_\ce{i}}{a_\ce{i}^\circ}\right)}$ we can write
$$\Delta G = G_{mAB}-G_{mA}-G_{mB}$$
In words (and repeating myself), $\Delta G$ in this case is for the process of converting 1 total mole of $A$ and 1 mole of $B$ into 1 mole of $AB$, all at constant T, P and composition (or, equivalently, in a sufficiently large mixture, such that a 1 mole change in the amount of the substances does not change the properties).
Schematically, this is how I interpret the two processes, with equation 2 for mixing in the top, and the reaction of A and B to form AB in the bottom:

So the second formula does not describe the reaction at all but a different process, that of mixing two miscible solids.

@KarstenTheis I think what you call a solid mixture versus a solid reaction product (with activity=1) is subject to some potential ambiguity unless you are careful with your definitions.

The reaction given by the OP shows a 1:1 stoichiometry. Formula two works for any mixing ratio. That is one of the differences between a compound and a mixture of elements.

@KarstenTheis Yes, that's true. That's pretty much included in my answer.

@NightWriter Is your first calculation correct? Shouldn't it read $\Delta G_f^\circ=N_A RT\ln{a_A}+N_B RT\ln{a_B}$, since $N_i=N\cdot x_i$ and $n_i=\frac{N_i}{N_{Av}}$? So, the first case is a reaction and the second case a mixture? How do I decide which case is the correct one for a given "reaction". For example two metals forming a intermetallic compound. In wide ranges of composition there is a mixture of A and B and at a specific stoichiometry there is a reaction?

@user76122 I used $n_i$ to signify number of moles (your $N_i$), so to your first question I would answer yes.

Isn't there a difference, in particular since the questions seems to revolve around the difference between molar and particle-based definition? The original formula, which I wrote in the OP specifically states N=number of atoms. Is this not an issue?

You might want to ask yourself, what is the difference between the product of a reaction, and a simple mixture? Take specifically the case where you have an equal number of moles (or molecules) of A and B.

Regarding n versus N, I changed my answer to be more clear and consistent with standard usage of these labels. But note the main point: the important point is that one of the equations refers to free energy per amount of substance, while the other refers to free energy change for a given total amount.

Edit comment: To elaborate a little more it would help if you could provide some links or sources of your equations, or more specific examples. I provided a rather generic explanation. Thermodynamics does not require you to provide an atomic level model, only the total amounts of substances (in moles, molecules, whatever consistent quantity unit you choose). I'll try to work this into my answer if I find a good way to do so, but probably it'll add unnecessary confusion.

Ok, thanks! I will improve the original question and try to make clear what my problem is. Regarding your comment about the difference between product of reaction and mixture: I asked myself this question several times and I think there is none. What would be the correct answer to this question?

I have another thought regarding the difference between reaction product and mixture: There is a difference between a gas-mixture of hydrogen and oxygen (in the correct composition) and gaseous Water, isn't it? It would be wrong to say $\Delta G_f^0 (H_2O(g)) = n_H RT\ln{a_H} + n_O RT\ln{a_O} = n_H RT\ln{\frac{p_H}{p^0}} + n_O RT\ln{\frac{p_O}{p^0}}$, wouldn't it?

@NightWriter I saw your updated answer. Basically I think, that I've understood the difference if a solid solution of solid or liquid mixtures is considered. There must be a difference between several variations of what is called "chemical reaction". Please see my (perhaps dull) water example. Maybe my problem arises also from a language barrier... I would really appreciate a discussion in chat :)

Let us continue this discussion in chat.


text:
What is the difference between a configuration state function and a Slater determinant?
I do not understand the distinction between configuration state functions and Slater determinants. Is not every Slater determinant a CSF ?
Please explain the difference and the use/need of configuration state functions.

Slater determinants are not eigenfunctions of the $\hat S^2$ operator, but CSFs are.
The Hamiltonian commutes with the operators for total and projected spin
\begin{align}
  [\hat H, \hat S^2] &= 0 \\
  [\hat H, \hat S_z] &= 0
\end{align}
Therefore, a set of common eigenfunctions to all three operators exists. As Slater determinants are eigenfunctions of $\hat S_z$, but not $\hat S^2$, using them as a basis for the electronic wave function (=eigenstate of $\hat H$), will not guarantee the (approximate) solution to be an eigenfunction of $\hat S^2$. The spin multiplicity of the found solution may not be a pure Singlet (or Doublet, Triplet, etc.).
As an example, consider a 2 electron system, with both electrons occupying different spatial orbitals. The situation with same spin for both electrons can be represented as the Slater determinant $|\alpha\alpha\rangle=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\alpha(2)\beta(1)]$. The spatial part is omitted here, to keep the notation short. The corresponding eigenvalue equations are then
\begin{align}
    \hat S_z |\alpha\alpha\rangle &= 1 |\alpha\alpha\rangle \\
    \hat S^2 |\alpha\alpha\rangle &= 2 |\alpha\alpha\rangle
\end{align}
Thus we have $S=1$, from $S(S+1)=2$, and the spin multiplicity is $2S+1=3$, i.e. a Triplet. Here the Slater determinant directly corresponds to a CSF.
For the case of opposite spin, we have
\begin{align}
    \hat S_z |\alpha\beta\rangle &= 0 |\alpha\beta\rangle \\
    \hat S^2 |\alpha\beta\rangle &= |\alpha\beta\rangle + |\beta\alpha\rangle
\end{align}
and therefore $|\alpha\beta\rangle$ is not an eigenfunction of $\hat S^2$. The expectation value for total spin would yield $\langle\hat S^2\rangle=1$, which is neither a Singlet nor a Triplet (not even a Doublet where $S(S+1)=0.75$).
This can be fixed by taking suitable linear combinations of Slater determinants. In this example we have 2 options:
\begin{align}
    |^1\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right) \\
    |^3\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)
\end{align}
This yields the eigenvalue equations
\begin{align}
    \hat S^2 |^1\Psi\rangle &= 0 |^1\Psi\rangle \\
    \hat S^2 |^3\Psi\rangle &= 2 |^3\Psi\rangle
\end{align}
which correspond to a Singlet and a Triplet state respectively.
Overall, this yields 1 Singlet component $|^1\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right)$, and the 3 Triplet components $|^3\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)$, $|\alpha\alpha\rangle$ and $|\beta\beta\rangle$.
For the arithmetic on how to apply the $\hat S_z$ and $\hat S^2$ operators to multi-electron systems, see Chapter 2.5 in Modern Quantum Chemistry by A. Szabo and N. Ostlund.

The kets are shorthand for determinants i assume ? So |ab> = |a(1)b(2)> - |a(2)b(1)>. And are alpha and beta spinorbitals or do you assume some factorization of spin and spatial part ? Sorry for the poor format of my comment but how can do line breaks in comments and can i do mathjax in the comments ?

@HansWurst Mathjax can be used in comments, e.g $|ab\rangle$ yields $|ab\rangle$. Linebreaks are unavailable, however you can do display math, e.g. $$|ab\rangle$$ yields $$|ab\rangle$$

@HansWurst yes, the ket is a Slater determinant here. $\alpha$ and $\beta$ refers to the spin of one orbital each, so spin orbitals. The wave function can be factorized into spatial and spin part, however I omitted the spatial part.

Do the spatial orbitals alone also form a determinant or do they factorize as a simple product ? Like $$ |^1 \Psi \rangle = \phi_1(r1) \phi_1(r2) \cdot (|\alpha \beta \rangle - |\beta \alpha \rangle) $$ or does it take some other form ?

@HansWurst its always a Slater determinant, otherwise you get problem with the Pauli principle.

source: Stack Exchange

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

sort: votes


text:
How to give different Basis set to different atoms in molecules using GAMESS-US
I am new to GAMESS-US, I am trying to do UHF for DyC10H10 while giving different basis set to Dy, C, H to reduce the computational cast.
Until I tried with following input using manual.

OUTPUT
 *** ERROR, ILLEGAL BASIS FUNCTION TYPE=SPKRDZC  NGAUSS=   0
 ERROR OCCURED READING ATOM NO.   1, INPUT ATOM NO.    1, NAME=DYSPROSIUM
 CHARGE= 66.0 X,Y,Z=     0.66154900    -0.08688200    -0.43858100

Please include the input file as text and not as image, so that the calculation could be reproduced. It would also be nice, if you added the version you are using.

I'm voting to close this question as off-topic because it's about how to use software and not chemistry.

It appears you can't access the SPK basis sets in this manner. To quote from the GAMESS Input Documentation:

Because they are stored in an external file supplied with GAMESS, these (the SPK basis sets) can only be accessed via GBASIS in this group, not by using them in-line in $DATA.

Yeah, you right. I was including basis set in a wrong way. Now I copied basis from EML library and past here and it worked. BTW, Thank you.

@Aman not a problem. Consider upvoting as well if the answer was useful to you.

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


text:
Pseudoelementary reactions in the Briggs-Rauscher oscillatory reaction mechanism
In this paper on the Briggs-Rauscher mechanism (JACS, 1982, paywalled), Noyes and Furrow propose an explanation of the oscillatory behaviour of the reaction based off of about eleven reactions from a selection of processes they consider "might contribute". The authors, however, call these processes "pseudoelementary reactions" and some of them, such as
\begin{equation*}
\ce{IO3- + HIO2 + 2H+ <=> 2IO2 + H2O}
\end{equation*}
seem to involve the collision of four molecules at the same time, which I was always taught is not very plausible. Many of these reactions are trimolecular, like
\begin{equation*}
\ce{HIO2 + I- + H+ -> 2HIO}
\end{equation*}
which I also find strange (these two reactions that I picked out are specifically reactions they say postulate that are important for the oscillatory behaviour).
I had never come across such a thing and I am nor really sure what to make of this. Are these reactions not actually elementary? What do they mean when they say the reactions are pseudoelementary? Are reactions involving the collision of four or three molecules more common than I thought? When they write down the equations for these reactions they write them as if the stoichiometric coefficients were the order with respect to each reagent, which seems to imply that the reactions are somehow elementary, which further confuses me. For instance, for the first reaction (which they call I3 in the paper) they explicitly write
\begin{equation*}
\frac{\mathrm{d}[\ce{HIO2}]}{\mathrm{d}t} = k_{I3} [\ce{H+}]^2 [\ce{IO3-}][\ce{I-}]
\end{equation*}

Of course they aren't... Also postulated mech. for oscillatory reactions tend to be made up - having little with reality, but adjusted to some experiments, so that can work as models.

What do they mean when they say the reactions are pseudoelementary?

Here is a portion of the first page:

So for the reaction (I added an H+ to the product to balance it)
$$\ce{IO3- + HIO2 + 2H+ <=> 2IO2 + H2O + H+},$$
they assume that the two protons attach in a fast reversible step (not sure to which of the reactants), and then there is a bimolecular reaction between the two iodine-containing intermediates X and Y:
$$\ce{IO3- + HIO2 + 2H+ <=>[fast] X + Y -> 2IO2 + H2O + H+},$$
The rate law for this set of reactions taken together is the one given in the question.

Are reactions involving the collision of four or three molecules more common than I thought?

No, they are rare (and slow). They could have written out the separate elementary reactions, but they probably also did not know where the protons attach and in what order, and they are not claiming to get the mechanism right. They just want to show a plausible set of reactions and rate constants that leads to the observed oscillations.

When they write down the equations for these reactions they write them as if the stoichiometric coefficients were the order with respect to each reagent, which seems to imply that the reactions are somehow elementary, which further confuses me.

It does not imply that. From a rate law alone you can't distinguish between mechanisms. Take these two mechanisms for example:
$$\ce{ A + H+ <=>[fast] AH+}$$
$$\ce{ AH+ ->[slow] B + H+}$$
on the one hand and
$$\ce{ A + H+ -> B}$$
on the other hand. Both will be first order in $A$ and in $\ce{H+}$.


text:
How to confirm a powder is Sodium Hydroxide?
Long-story-short, my mother found a bag of white powder with some soap making supplies.  We're about 90% sure it's powdered lye, but I thought it would be fun to confirm this somehow.
Is there a way to do so using easily-sourced supplies?  A simple pH test would at least confirm it's caustic, and I've read burning the powder might indicate the sodium content, but I'm not sure what else might help.  Thanks!

If it is a bag that has been sitting on a shelf for a long while, then its contents is most likely not sodium hydroxide anymore, even if it was initially.

@IvanNeretin Why, in a tightly closed plastic bag NaOH is going to be just fine.

If it had been kept sealed for all these years and then cut open just now, then sure, it may still be fine.

Ift it was perfectly sealed it might be NaOH. If it has alcaline reaction,  it yellows even a stove flame, and it shows deliquescency then you are almost sure it is.

It could be also anhydrous sodium carbonate, even if the decahydrate is more common as the washing soda. It is also possible it got dehydrated in dry air. I would try an acid test, .e.g. with vinegar, how much CO2 is released.

Tests for sodium hydroxide:

Do a pH test. Sodium Hydroxide is basic.
Do the flame test. A bright yellow flame shows that sodium ions are present.
A few drops of dilute sodium hydroxide solution react to form a white precipitate with aluminium ions, calcium ions and magnesium ions. However, in excess sodium hydroxide solution

The aluminium hydroxide precipitate dissolves to form a colourless solution
The calcium hydroxide precipitate is unchanged
The magnesium hydroxide solution is unchanged

Adapted from [1] and [2].

Quora is as good as a source just as a sieve is good as a container for water. Judging from how broad the answer is, I'm pretty confident you could cite any introductory analytical chemistry textbook instead.

@andselisk I've added info and a more reputable source.

I never heard of BBC Bitesize, but BBC as news network has horrible reputation, especially after NPC journalism episodes. Maybe I'm too old, but I don't see how either Quora or bbc.co.uk can be a reputable source.

@andselisk I'm in the UK where the BBC (British Broadcasting Corporation) is based, and BBC bitesize is a well-known and well-recommended resource to help students revise for exams.

Not all food that is easy to digest and is widely recommended is good for you. Same goes with online studying platforms. Maybe the material is good, maybe not. No way to tell for sure in the absence of transparent referencing, peer reviews and editorial board consisting of the specialists in the field.


text:
pH of CH3CH2NH2 buffer solution

For $\pu{550.0 mL}$ of a buffer solution that is $\pu{0.170 M}$ in $\ce{CH3CH2NH2}$ and $\pu{0.150 M}$  in $\ce{CH3CH2NH3Cl}$, calculate the initial pH and the final pH after adding $\pu{0.020 mol}$ of $\ce{HCl}$.

The following is my attempt on solving this problem:
Since Kb for ethylalamine = $5.6 \times 10^{-4}$.
pKb = $-\log( 5.6 \times 10^{-4}) = 3.25$
pOH = $3.25 + \log (0.150/0.170)$
pOH = $3.196$
pH = $14 - 3.196 = 10.80$ for Initial
Final pH:
$\pu{0.170 M} * \pu{0.55 L}= 0.0935 mol$ $\ce{CH3CH2NH2}$
$\pu{0.150 M} * \pu{0.55 L} = \pu{0.0825 mol} \ce{CH3CH2NH3Cl}$,
$0.0825 - 0.02 = \pu{0.0625 mol} \ce{CH3CH2NH3Cl}$,
$0.0935 + 0.02 = \pu{0.1135 mol} \ce{CH3CH2NH2}$
pOH = $3.25 + \log(0.0625/0.1135)$
pOH = $2.991$
pH = $14 - 2.991 = 11.01$ for final
However, these values turn out to be wrong. Please help, thank you!

Hi! This site has a policy that homework-like questions have to show some effort to solve them. Hence, could you please edit your post and add, how you would solve it and what exactly is your difficulty here? Thank you!

Hi! Thank you for that info. I edited my question and added my attempt, but the values that I calculated turns out to be wrong.

"$0.0825 - 0.02 = \pu{0.0625 mol} \ce{CH3CH2NH3Cl}$",$$0.0825 + 0.02 = \pu{0.1025 mol} \ce{CH3CH2NH3Cl}$$

"$0.0935 + 0.02 = \pu{0.1135 mol} \ce{CH3CH2NH2}$" $$0.0935 - 0.02 = \pu{0.0735 mol} \ce{CH3CH2NH2}$$

Part A
-   Calculating  the initial $\pu{pH}$(before adding $\pu{0.02 mol}$ of $\ce{HCl})$
The equilibrium here is
$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+  + OH−}$$
For  $\ce{ CH3CH2NH2} , \mathrm{K_b} = 4.6\times{10^{−4} }$
$$\mathrm{K_b} = 4.6\times{10^{−4} }=\frac{[\ce{CH3CH2NH3+}][\ce{OH−}]}{ [\ce{CH3CH2NH2}]}$$
The nominal concentrations of the base and conjugate acid  are respectively $$\mathrm{C_b} =\pu{0.170 M}~\text{ and } \mathrm{ C_a } =\pu{0.150 M}$$.
The mass balance expressions are:
$$ [\ce{CH3CH2NH3+} ]+ [\ce{CH3CH2NH2}] = \mathrm{C_a} + \mathrm{C_b} = \pu{0.32M}~\text{ and}~ [\ce{Cl−}] = \mathrm{C_a}=\pu{0.150M}$$
and the charge balance is given by:
$$[\ce{Cl−}] + [\ce{OH−}] = [\ce{H+}] + [\ce{CH3CH2NH3+}] \approx{[\ce{ CH3CH2NH3+}]}$$
in which the approximation shown above is justified  that the solution will be weakly alkaline.
The equilibrium concentrations of the conjugate species are then:
$$[\ce{CH3CH2NH3+}] = 0.150 + [\ce{OH−}]$$
\begin{array}{  }
[\ce{CH3CH2NH2}] &= &(\mathrm{C_a} + \mathrm{C_b}) &-&([\ce{ CH3CH2NH3+} ])\\ &= &(0.32) &–&([\ce{ CH3CH2NH3+} ])\\ &= &0.320 &- &(0.150 + [\ce{OH−}])\\&=&0.170 &- &[\ce{OH−}]\end{array}
Because $\mathrm{C_a}$ and $\mathrm{C_b}$ are large compared to $[\ce{OH−}]$ (the solution is not expected to be strongly alkaline),
the $[\ce{OH−}]$ terms in the above expressions can be dropped and the equilibrium expression becomes
$$\mathrm{K_b}=5.6\times{10^{-4}} =\frac{ (0.150)[\ce{OH−}]}{(0.170)}$$
from which we find $[\ce{OH−}] = \pu{6.347\times{10^{-4}} M}$ , $\pu{pOH} = 3.2$ and $\pu{pH} = 14- 3.2 = 10.8$
Part B : Calculating  the final $ \pu{pH}$(after adding $\pu{0.020 mol}$ of $\ce{ HCl}$)

Calculate the initial amount of $\ce{ CH3CH2NH3+}$ and the initial amount of $\ce{CH3CH2NH2}$:
$$\begin{align}n_\ce{ CH3CH2NH3+} &=n_\ce{ CH3CH2NH3Cl}\\ &=V_\text{(solution)}\times{[\ce{ CH3CH2NH3Cl}]_\text{I}}\\ &=
\frac{\pu{550.0 ml}}{\pu{1000 ml/L}}\times{\pu{0.150 mol/L}}\\
&=\pu{0.083 mol}
 \end{align}$$
$$\begin{align}n_\ce{ CH3CH2NH2} &=V_\text{(solution)}\times{[\ce{ CH3CH2NH2}]_\mathrm{I}}\\ &=
\frac{\pu{550 ml}}{ \pu{1000 ml/L}}\times{\pu{0.170 mol/L}}\\
&=\pu{0.094 mol}
 \end{align}$$
Addition of the strong acid $\pu{ 0.020 mol}$ of $\ce{ HCl}$ , consumes this same amount of $\ce{ CH3CH2NH2}$ and produces $\pu{ 0.020 mol}$ of $\ce{ CH3CH2NH3+}$ ion as explained in the following table:

\begin{array}{c | c c c c c c c}
\text{rxn}&\ce{ CH3CH2NH2} &+ &\ce{HCl} &-> &\ce{CH3CH2NH3+} &+ &\ce{Cl-}\\\hline
 \text{I}&(0.094)&  &(0.020)&  &(0.083)& &(0.083)\\
\text{C} &(0.094-0.020)& &(0.020-0.020)&   &(0.083+ 0.020)&
   &(0.083+ 0.020)  \\
\text{F} &(\pu{0.074 mol})&  &(\pu{0.000 mol})&  &(\pu{0.103 mol})&  &(\pu{0.103 mol})
 \end{array}

We can continue using the approximations for the concentrations of the conjugate species
that we developed in the preceding part A.
The values of C$_\mathrm{b }$ and C$_\mathrm{a }$ are now

$$\mathrm{C}_\mathrm{b }= [\ce{CH3CH2NH2}]_\mathrm{I} =\frac{n_\ce{CH3CH2NH2}}{V}=\frac{\pu{0.074mol}}{\pu{0.550 L}}=\pu{0.133 M}\approx{[\ce{CH3CH2NH2}]_\mathrm{e}}$$
and
$$\mathrm{C}_\mathrm{a }= [\ce{CH3CH2NH3+}]_\text{I} =\frac{n_\ce{CH3CH2NH3+}}{V}=\frac{\pu{0.103 mol}}{\pu{0.55 L}}=\pu{0.187 M}\approx{[\ce{CH3CH2NH3+}]_\mathrm{e}}$$

Substituting these into the equilibrium constant expression yields
$$\begin{align}[\ce{OH−}] &=(\mathrm{K_b})\frac{[\ce{CH3CH2NH2}]} {[\ce{CH3CH2NH3+} ]}\\
&=(5.6\times{10^{-4}})\times{\frac{\pu{0.133 M}} {\pu{0.187M}}}\\&=\pu{3.983\times{10^{-4}} M }
\end{align}$$
the new $\pu{pOH}= 3.40$ ,and the new $\pu{pH}=14- 3.40=10.60$, so addition of the acid has changed the $\pu{pH}$, a decrease of only $0.20~~\pu{pH}$ unit.


text:
Kinetic and thermodynamic control of sulphonation of toluene
Sulfonation of toluene with conc. $\ce{H2SO4}$ at room temperature gives ortho- and para-toluenesulphonic acid (with $95 \%$ yield), whereas sulfonation of toluene with conc. $\ce{H2SO4}$ at $\pu{150-200 °C}$ gives meta-toluenesulphonic acid (with $95 \%$ yield). Why is this so?
What I thought is that at room temperature, the reaction is kinetically controlled, and the ortho-/para- directive effects of methyl group operate leading to ortho- and para-toluenesulphonic acid.
At high temperature, the reaction would be thermodynamically controlled. So the more stable product would be formed, which happens to be meta toluene sulfonic acid. Why is this more stable?
Does it have anything to do with the equilibrium of the sulfonation reaction?
Related : Thermodynamic vs Kinetic Sulphonation of Naphthalene


text:
Why is manganate ion paramagnetic
When I tried to figure out why manganate ion is paramagnetic, given that its oxidation number is +6, I am not able to understand why there are unpaired electrons. Is it the case that six of the seven valence electrons of Mn are taken by oxygen by ionic means, and a unpaired electron remains?
Secondly, for manganese to make 7 bonds, I believe an electron should be promoted from the 4s orbital. Where does this electron move to, 5s or 4p?

Could you please edit your question to make it more intelligible, using punctuation where appropriate?

@NightWriter Hi, I just am asking about its electron sharing while formation of manganate ion. I find difficulty to imagine how manganese has managed to make 6 bonds! Where and to which orbital does the electron jump to. Kindly answer it by providing diagrammatic explaination of orbitals

You might want to consider using ligand Field Theory.

We've already had an answer from the point of view of crystal field theory. This is actually more or less equivalent to your solution that six electrons are taken by ionic means (which is an oversimplification of the meaning of oxidation number +6, but not a bad place to start thinking about this) leaving one unpaired electron to explain the paramagnetism.
At the other extreme of oversimplification, you could model this using a simple valence-bond model based on the Lewis structure (see Wikipedia), which I think is what you may have in mind in your second question. In this model, since there are formally two single and two double bonds, you would say that there are four sigma bonds formed by Mn $4s3d^3$ hybrid orbitals, and two pi bonds formed by the remaining two Mn $3d$ orbitals. These six orbitals together contain seven electrons from the Mn and four from the four O atoms, for a total of eleven electrons, again leaving one unpaired.
In practice, for what it's worth most research chemists would use neither of these pictures, but instead a molecular orbital model.

I’ll try and rationalise this using only crystal field theory, for the sake of simplicity.
In the presence of a tetrahedral ligand field, the metal d ions split into a E set and a T2 set. (The E set involves the x^2 - y^2 and z^2 orbitals, and the T2 set involves the other three).
The T2 set is raised in energy, and the E set is lowered in energy. The splitting is particularly low (O2- is a pi donor ligand, and it’s a tetrahedral complex) so this would be high-spin.
In other words, there would be one electron populating the E set, so it’s paramagnetic.
Hope this helped.

Even if you are restricting yourself to crystal field theory, I am not sure talking about high and low spin helps for a nominally $d^1$ complex.

Fair enough. I Guess the HS/LS just came out naturally since I was too used to invoking it.


text:
Why is maximum buffer capacity for some diprotic buffers not when pH = pKa?
According to Rajkovic et al. [1]:

$$β = \frac{[\ce{H+}]}{K_\mathrm{w}} + 2.303\cdot\frac{[\ce{H+}]\cdot K_\mathrm{a}\cdot c}{[\ce{H+}] + K_\mathrm{a}}$$
where $K_\mathrm{a}$ is the dissociation constant of the acid; $c$ is the concentration of the acid in all forms; and $K_\mathrm{w}$ is the ionization constant of water $(1·10^{–14})$. It can be seen that the buffer capacity of the acid is greatest when $[\ce{H+}]$ equals $K_\mathrm{a}$ (or when the $\mathrm{pH}$ is equal to the $\mathrm{p}K_\mathrm{a}$ of the acid). This relationship can be used for mixtures of monoprotic acids and many diprotic acids (by considering them to be made up of two monoprotic acids). Unfortunately, this is not true for most of the diprotic acids found in wines because the second dissociation is not completely independent of the first.

Is it true that for some diprotic acids maximum buffer capacity is not reached when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$? If so, why is it true? Also, then when is buffer capacity maximum? Can it only be determined experimentally?
Or did I just misinterpret the article? Is it saying that this equation of buffer capacity is invalid for some diprotic acids instead of the fact that buffer capacity is max when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$?
References

Rajkovic, M.; Novakovic, I.; Petrovic, A. Determination of Titratable Acidity in White Wine. Journal of Agricultural Sciences, Belgrade 2007, 52 (2), 169–184. https://doi.org/10.2298/JAS0702169R.

Well, the last sentence more or less sums it up: "Unfortunately, this is not true for most of the diprotic acids found in wines because the second dissociation is not completely independent of the first."

@NightWriter How is the second dissociation not being completely independent of the first a reason why maximum buffer capacity for some diprotic buffers is not when pH=pKa? I feel like I'm missing some intermediate steps of understanding this concept.

You can think of it qualitatively as a tug of war between the two groups. Each has a different associated buffering region and pulls the pH into that region. If they each pull the pH into separate regions, nobody wins. Admittedly not a quantitative explanation. The best is probably to work out an example.

A illustration of @Poutnik's explanation can be found here and here.

Any aqueous solution increases its buffer capacity toward $\mathrm{pH} = 0$ or $\mathrm{pH} = 14$, as more and more acid or basis is needed to change $\mathrm{pH}$ by a given interval. It happens even without presence of any buffering substance, aside of strong acids or bases.
If an acid, monoprotic or diprotic, is added, a peak(s) of maximum buffer capacity is added. But if $\mathrm pK_\mathrm a$ is small, the peak of maximum buffer capacity does not lay on a steady baseline of the capacity function. It lays on a sloped baseline. If the maximum absolute peak slope is smaller than the absolute baseline slope, the peak does not manifests itself  as a peak(=maximum capacity), but rather as just as a shoulder on the slope, without a maximum.
This phenomena is not specific for chemistry, but rather for mathematics and mathematical analysis. If you have a sinusoid function on a large enough slope, its peaks would disappear at some slope value, becoming just a shoulders.
The similar phenomena occurs in case of multiple $\mathrm{p}K_\mathrm{a}$ in the solution, of the same or of different compounds. On the background of non steady capacity function, a particular peak may not have a shape of a peak.
Additionally, the presence of close peak neighbors affects position of the peak maximum.
If 2 peaks of the similar height are close enough each other, their maxima shifts towards each other, and with enough closeness they merge into a single peak.


text:
Why is dichloromethane immiscible in water?
We know that like dissolves like. And dichloromethane is a polar solvent and water is also a polar solvent. Also there ought to be a strong hydrogen bonding between the chlorine and hydrogen atoms. So why is it immiscible?

The quick way to answer is that DCM isn't polar enough, as well as the strength of the H-bond it might form in water. A detailed answer involves the enthalpy end entropy terms analysis of the dissolution but I am afraid it will remain rather qualitative. And see also https://chemistry.stackexchange.com/questions/38137/dichloromethane-solubility-in-water where interestingly an azeotrop is mentioned. It is quite possible that ideally mixing results in a ordered phase, so that the heat of solubilisation doesn't compensate for the reduced entropy.

@Alchimista What do you mean by ordered phase?

@Alchimista. Dipole moment of DCM is 1.47 and that of water is 1.85.

It can be that entropy is lower as for the water might have to arrange more strictly to accommodate the dcm molecules. This would give a positive delta G as for the intermolecular forces are weaker for sure.

And if you insist on polarity, dcm is heavier than water. Still if you let a drop on tge bench,it simply disappears in seconds. Thus, H-bonding is much weaker as compared to water.

Forget. I wanted to say that it has a low boiling point and evaporates very fast. It was a drop on the bench, not a layer in a funnel.  I wild try to better explain later, but if you didn't get a grasp of the situation yet, probably my answer won't help much. Wait for another user.

Do you know about frontier MO theory? if yes, check out this to understand why DCM resists solvation by nucleophilic solvents like water

For an immiscible compound, DCM is quite soluble in water (~20g will dissolve in a litre of water). As will chloroform (~10g/L) which is enough to make a recreational drink (or was in Victorian times before we worked out how toxic it was).

@matt_black it depends on temperature right? 25.6g/L at room temperature..

@Sreetamaghoshhazra yes temperature matters. I was approximating.

@YUSUFHASAN DCM dissolves in benzene right?

@matt_black data shows with temperature solubility increases.. Is it only since the entropy is also increasing.. And hence the solvation enthalphy...

I add this comment for further clarification. Somehoe cunterintuitive, is the occurrence of hydrogen bonding (whatever extent it happens) that, together with the endothermic character of the process, unfavours mixing. This h-bond occurrence reduce the positive entropy term associated with mixing. A hypothetical compound with comparable size and dipole as DCM but not suitable for H-bond would be much more soluble.

@Alchimista hypothesis will be true if aprotic solvent is used. H bonding favors solubility here. isn't?

Also. But most of the h-bonding  DCM to solvent involves its hydrogens rather than chlorine, I guess.  But don't try to make a predictive rule out of my example. It served to let you consider deltaG in finer details.

In the following figure the free energy computed from solubilities available in the wikipedia are plotted against T. The positive values of $\Delta G$ imply that formation of a 1 molal solution from the pure components is an endergonic process (requires energy input), consistent with the limited solubility of DCM in water.

From the slope and intercept we see that the enthalpy of solubilization of DCM in water is negative (exothermic) while the entropy change is also negative. The negative entropy change, as explained in the comments, opposes mixing. The exothermic character means that increasing T actually discourages solubilization (the negative entropic effect is emphasized at higher T).
This does not address the mechanism of solubilization, of course. However, it supports the argument in the comments that formation of water cages or conformationally restrictive hydrogen bonding to the solute imposes an entropic penalty upon solubilization.

Nice formulation for what I tried to convey. I am just surprised by the fact that the mixing has an exothermic character, as for I would have expected water to water interaction to be stronger as well, so that the entropy term could impose at the reltively small concentrations also mentioned in comments.

Thanks @Alchimista This is admittedly just an extension of the argument you raised. I did not come across a more systematic description of the solvation process in this particular system, the details are missing.

Dichloromethane is non-polar.
The physical causes of solubility of substances in each other can semiquantitatively be expressed by the Hansen solubility parameters.
$\begin{array}{}
\text{Water}:&&\delta_d=15.6&\delta_p=16.0&\delta_h=42.3&R_0=47.8\\
\text{Dichloromethane}:&&\delta_d=18.2&\delta_p=6.3 & \delta_h=6.1&R_0=20.2
\end{array}$
$$R_0=20.2,\ \ \ R_a=37.8$$
$$RED=\frac{R_a}{R_0}=1.87$$
RED < 1 would show solubility of both substances in each other. But RED > 1, so both substances are little soluble in each other.
The solubility parameters above show, the energies from polar and from hydrogen bridging bond intermolecular attractions are higher in water.
Generally, chlorinated hydrocarbons have lower energies from polar and hydrogen bridging bond intermolecular attractions.
The strongest hydrogen bridging bonds are formed between the atoms of the strongly electronegative elements $\ce{F}$, $\ce{O}$ and $\ce{N}$. But the atoms of other electronegative elements, e.g. $\ce{Cl}$-, $\ce{S}$- and $\ce{C}$-atoms, are capable of weak hydrogen bridging bonds at suitable molecular structure.
The atomic radii of the third period elements ($\ce{P}$, $\ce{S}$, $\ce{Cl}$) are much larger than those of the second period elements ($\ce{N}$, $\ce{O}$, $\ce{F}$). $\ce{Cl}$-atoms have lower charge density therefore. They are weaker proton acceptors therefore.
Hydrogen bridging bonds can be proved e.g. by molecular spectroscopy.
Li Bian: Proton Donor Is More Important Than Proton Acceptor in Hydrogen Bond Formation: A Universal Equation for Calculation of Hydrogen Bond Strength. J. Phys. Chem. A 107 (2003) (51) 11517–11524
Weinhold, F.; Klein, R. A.: What is a hydrogen bond? Mutually consistent theoretical and experimental criteria for characterizing H-bonding interactions. Mol. Phys. 110 (2012) (9-10) 565-579
Gilli, G.; Gilli, P.: The Nature of the Hydrogen Bond: Outline of a Comprehensive Hydrogen Bond Theory. Oxford University Press, Oxford, 2013


text:
Eutectic Phase Diagram of Al-Si Analysis

Hello,
How can I know from this phase diagram:
1. The composition of the alloy Al-Si that allows precipitation to occure?
2. The range of temperatures where the Al-Si is soluble?
3. The range of temperatures in which precipitation can take place?
Regarding the 1st and 2nd question- I assume that 1.6%wt Si at 557 degrees Celsius is the max solubility limit of this alloy, and that you can't have precipitation if there are higher percentages of Si in the aluminium.
But what about precipitation temperatures? Which conditions allow having precipitation?

Please clarify a bit about the precipitation.

precipitation=age hardening


text:
No sign flipping while figuring out the emf of voltaic cell?
I learnt that for a voltaic cell, the value for the $E_\text{cell}^\circ$ when the reaction is spontaneous is given by
$$E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ, \label{eqn:1}\tag{1}$$
so that the difference in the right gives us a positive value for $E_\text{cell}^\circ$.
But suppose we are given two half-reactions:
$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$
When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation.

But when determining the $E_\text{cell}^\circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $\eqref{eqn:1}$ to figure out the $E_\text{cell}^\circ$? Shouldn't we do that because we reversed the second equation?
My book tells me to keep the $E_\text{half-cells}^\circ$ as they are written in the tables and simply put them in $\eqref{eqn:1}$. But why?

Take a look at the two half reactions:
$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.

No sign flipping while figuring out the emf of voltaic cell?

Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.
$$E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ$$
If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)

My book tells me to keep the E∘half-cells as they are written in the
  tables and simply put them in

Your book is then one of the few books which teaches electrochemistry properly. The sign of the electrode reduction potential is invariant. If reflects the sign of the electrostatic charge of the electrode with respect to hydrogen electrode. I don;t know if you like history or not, long time ago in the 1950s-60s showing the electrostatic sign of a cell by means of a specially designed electroscope was a standard experiment in physics.
Imagine if I say H2O (l) --> H2O (g) at 100 oC
Does this mean reversing the reaction
H2O(g) --> H2O (l)   will be at -100 oC?
You can see logical fallacy in reversing the sign of electrode potentials.
Yes, there was a lot of confusion in electrode signs for more than 100 years but now it has been sorted out.

The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.


text:
What is the formula for theoretical buffer capacity for a diprotic buffer system?
According to Chembuddy, the formula for theoretical buffer capacity for a monoprotic buffer system is as follows:

$$β = 2.303\left(\frac{K_\mathrm{w}}{[\ce{H+}]} + [\ce{H+}] + \frac{C_\mathrm{buf}K_\mathrm{a}[\ce{H+}]}{(K_\mathrm{a} + [\ce{H+}])^2}\right)$$

where $C_\mathrm{buf}$ is the total concentration of buffer and $K_\mathrm{w}$ is the water ionization constant.
However, I have conducted a lab in which I mix $\ce{C4H6O4}$ (tartaric acid; diprotic) with its double salt $\ce{NaKC4H4O4}$ to create a 1.0 M potassium sodium tartrate buffer solution where there are two simultaneous buffer systems. Also, $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ of tartaric acid are quite similar (around 1.6 difference), so I believe that this alternative equation that provides the theoretical buffer capacity for a system with multiple buffers:

$$β = 2.303\left(\frac{K_\mathrm{w}}{[\ce{H+}]} + [\ce{H+}] + \sum\frac{C_\mathrm{buf}K_\mathrm{a}[\ce{H+}]}{(K_\mathrm{a} + [\ce{H+}])^2}\right)$$

will not be correct in my case.
Is this right? If so, is there a proper way to theoretically calculate the buffer capacity of the composite system?

The logic is the same used for a monoprotic weak acid, considering the two ionization steps, as indicated by @MaxW.
For a diprotic weak acid we have:
The first ionization
\begin{align*}
\ce{H2A + H2O &<=> H3O+ + HA-} &K_{\ce{a}1}=\frac{\ce{[H3O+][HA-]}}{\ce{[H2A]}}
\end{align*}
and the second ionization
\begin{align*}
\ce{HA- + H2O &<=> H3O+ + A^{2-}} &K_{\ce{a}2}=\frac{\ce{[H3O+][A^{2-}]}}{\ce{[HA-]}}
\end{align*}
To make further calculations easier let's assume that the strong base added is monoprotic:
\begin{align*}
\ce{BOH &-> B+ + HO-}
\end{align*}
Charge balance of the solution is given by the equation
$$\ce{[H3O+]} + \ce{[B+]} = \ce{[OH-]} + \ce{[HA-]} + \ce{2[A^{2-}]}$$
When combined give us formula for the amount of the strong base (for short $\ce{[H3O+]}$ was written as $\ce{[H+]}$)
$$C_{\ce{B}} =
\frac{K_{\ce{w}}}{\ce{[H+]}}-
\ce{[H+]}+
\frac{C_{\ce{H2A}}K_{\ce{a}1}\left(\ce{[H+]}+2K_{\ce{a}2}\right)}
{\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}}$$
when $K_\ce{w}$ is the water ionization equilibrium constant, $C_{\ce{H2A}}$ is the concentration of diprotic weak acid, $K_{\ce{a}1}$ and $K_{\ce{a}2}$ are the acid dissociation constants, and $C_{\ce{B}}$ is the concentration of strong base added.
The buffer capacity of a diprotic weak acid-conjugate bases buffer is defined as the
maximum amount strong base that can be added before a significant change in the pH will occur.
\begin{align*}
\beta &= \frac{dC_{\ce{B}}}{d\ce{pH}} = \frac{dC_{\ce{B}}}{d\ce{[H3O+]}}\frac{d\ce{[H3O+]}}{d\ce{pH}}\\
\end{align*}
Differentiating the equation we should then find that
\begin{align*}
\beta &= \left[-\frac{K_{\ce{w}}}{\ce{[H+]}^2}-1 -
\frac{C_{\ce{H2A}}K_{\ce{a}1} \left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)}
{\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right]
\left(-\ce{[H+]}\ln{10}\right)
\end{align*}
So finally buffer capacity is given by
\begin{align*}
\beta &= 2.303\left[\frac{K_{\ce{w}}}{\ce{[H+]}}+\ce{[H+]}+
\frac{C_{\ce{H2A}}K_{\ce{a}1}\ce{[H+]}\left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)}
{\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right]
\end{align*}
you can then plot $\beta$ vs $\ce{pH}$ and from this you should be able to find what you want.
Look a example based on this question. I hope it be useful.


text:
Where does Free Energy go in a Redox Reaction?
This doubt came to my mind while answering the following question:

Why does a voltaic cell not operate unless the two separate compartments are connected by an external circuit?

The answer seems to be that only on connecting the two halves of a Redox reaction via an external circuit, does one allow the electrons to be transferred from one species to another. And while the electrons move towards their goal (to a higher potential), they do work in the way (light a bulb or a motor or something).
Another way to view this process is that the free energy of the Redox reaction is used to do useful electrical work.
But then my brain started to wonder and there came up a question - what happens to all this energy when none of it is actually used to do electrical work, i.e, when the Redox reaction is let to occur spontaneously, where does all the free energy go? Does it get converted to heat, thus warming the solution? Or does it simply stay stored in the solution in some form?

Free energy is a state function. No matter how you run the reaction, if you start with a certain state (set of concentration) and end with a certain state (set of concentrations at equilibrium), the change will be the same.
The free energy of reaction is the maximal amount of non-pV work (electrical work in this case) the reaction can do. If it does not do any work, that energy will be released as heat instead. For example, if you short the battery, the wire and the electrochemical cell will heat up while the reaction is going on.
If you want an argument from first principles, you should look at the first law of thermodynamics. The change in energy of a system is equal to the sum of the heat and the work exchanged with the surrounding. When making optimal use of the free energy, you maximize the work (even sometimes transferring heat from the surrounding to do even more work).

All depends on the cell chemistry and geometry. It is usually a combination of keeping the energy and conversion of the free energy to thermal energy, when reagents diffuse and happen to meet each other. Or, some side reaction with solvent or auxiliary components may occur, like for $\ce{Li-ion}$ cells.
Some primary lithium cells last many years, while some early generation $\ce{NiMH}$ cells selfdischarge in few months.


text:
How Can I Test Something for the Presence of Iodine?
I want to test a supplement for the presence of iodine.
Iodine should turn starch blue-black. So, can I powderize the supplement tablet, add water to it, drop it on a piece of bread, and if it turns blue-black, that would be an accurate way to test?

I'm not too sure that'd work very well. Is the tablet colourised? Does it have an ingredients label? Assuming (giant IF here) that it contains iodine, and the amount is great enough that adding it to water results in an aqueous solution of iodine, AND that there's nothing else in the tablet that may react with either reactant, then yes, you may observe it.

Just keep in mind that only elemental iodine will cause a blue-black coloration. Nobody add elemental iodine to a supplement just like nobody can eat chlorine out of table salt (NaCl). The typical compound they add is KIO3, which will not color starch right away, until and unless you reduce it to KI and then oxidize KI to free I2. In any case, search natural foods which contain iodine.


text:
Why are N-H stretching vibrations often sharp and not broad?
It is known that hydrogen bonding results in the broadening of peaks in the infrared spectrum of a molecule. Hydrogen bonding results in a spectrum of different bond lengths with different stretching frequencies. This results in a the functional group giving peaks of a whole continuous range of vibrational frequencies and hence, we see a broad peak. Broad peaks are most often observed from $\ce {2500}$ to $\ce {3000 cm^{-1}}$ for the stretching of carboxylic acid functionalities. Similarly, hydrogen bonding also results in alcohol $\ce {O-H}$ stretching giving broad peaks in the range of $\ce {3200}$ to $\ce {3600 cm^{-1}}$, albeit the degree of broadening is much less for alcohols. Amines are also able to form rather strong hydrogen bonds. However, amine $\ce {N-H}$ stretches often are not seen as broad peaks and are in fact, often sharp. Why is this so?

The sharpness or broadness of a stretch in IR spectra depends on extent of Hydrogen bonding present in the molecule. Basically, if it undergoes immense intermolecular hydrogen bonding, the peaks tend to be broader and the lesser the hydrogen bonding becomes, the sharper the peaks get in the spectra.
Now, if you recall the criteria needed for stronger hydrogen bonding, the more electronegative the atom attached to hydrogen is the better is the extent of hydrogen bonding as the more electronegative atom will be able to create more $\delta +$ve cahrge on hydrogen which is better for stronger hydrogen bonding. $\ce{O}$ is far more electronegative than $\ce{N}$. $(\chi_o = 3.44, \chi_N = 3.04)$. So, the extent of hydrogen bonding is much much lesser with $\ce{N}$ being the attached atom than $\ce{O}$. (If you want to get an idea how strong the hydrogen bonding is in case of $\ce{O}$ than $\ce{N}$, simply consider $\ce{H2O}$ and $\ce{NH3}$, due to hydrogen bonding the former has its boiling point at $100^o C$, while that for the latter is $-33.4^oC$.) But anyways Hydrogen bonding is there in amines, only the extent is lesser.
However if you can somehow increase the $\delta+$ charge on hydrogen, by attaching some electron withdrawing group to the nitrogen, like in the case of amides, there you will be again able to see broad $\ce{N-H}$ stretches as now it can undergo stronger hydrogen bond formation, as $\ce{N}$ pulls the electron density more towards itself due to resonance., and $\ce{H}$ becomes electron deficient, and peak becomes broad. Here is the IR spectra of $\ce{N-methylacetamide}$ from the book Introduction to Spectroscopy which shows broad $\ce{N-H}$ stretches due to the reason mentioned above.

In this spectra you can also see the $\ce{C=O}$ stretch is also relatively broader than observed generally. The reason is again extent of intermolecular hydrogen bonding, which is greater due to higher electronic density on the $\ce{O}$. But apart from that, the general reason for broadening the spectra is the extent of intermolecular hydrogen bonding, which is somewhat lesser for normal amines.

I think with regards to amides, the reason for the broadening may be the "tautomerism" rather than hydrogen bonding. Due to strong delocalisation of the N lone pair, there may be some non-negligible tautomerism going on.


text:
Why is it not possible for seven close copper atoms to come together to gain a noble-gas configuration of valence electrons?
My textbook says the following about metallic bonding:

The metal copper has one valence electron. The valence electrons of nearby copper atoms are shared in a similar way to covalently bonded solids, but it is not possible for seven close copper atoms to come together to gain a noble-gas configuration of valence electrons as occurs in a covalent bond. Instead, the electron sharing that occurs in metals is not as restricted as in the covalent solid.

The problem is that this section doesn't adequately (in sufficient detail) explain why it is not possible for seven close copper atoms to come together to gain a noble-gas configuration of valence electrons as occurs in a covalent bond.
I would greatly appreciate it if people could please take the time to clarify this in further detail.

It is not possible for any element. Why should be copper the exception ? Or, why do you think it should be possible ?

@Poutnik Well the textbook, as I wrote, says that such a thing occurs in covalent bonds, so either the textbook is incorrect or such a thing is possible.

Sure, but too many is too many. More than 6 bonds is very rare, and it never happens for elements. It requires a big central cation and small coordinating anions. It is very challenging for difficulties in geometry, orbital overlapping and saturation of bonding abilities.

@Poutnik But notice that it says "as occurs in a covalent bond", so the author implies that it happens in certain cases. Or is the author incorrect?

It is rather misinterpretation. There is not said an atom cannot reach for total count of 8 shared electrons. E.g. fluorine 7+1, oxygen 6 + 21, nitrogen 5+31. But no element bonds with itself by 1+7*1 pattern.

@Poutnik Hmm, are you saying that an atom can or can't share 8 electrons? For fluorine, which has 7 valence electrons, it would share 1 electron with hydrogen, by covalent bond, to form a 7 + 1 configuration, right? But it is only sharing 1 electron, not 8. I'm confused as to what the author is trying to say here.

Take it the other atoms do not fit to the close enough position to make a bound, plus the central atom is not willing to create 7 covalent bonds. Note also 7 bonds are not known for covalent bonds, but for coordinate bonds, what is little different case.

Let us continue this discussion in chat.

Thanks to @Poutnik, I think I understand what it is saying now: "Yes, it is about the electron configuration. It does not say all 8 electrons are shared, but that 8 electron count can be reached by sharing."


text:
What decides the nature of emulsifying agents?
I read

The emulsifying
  agent forms an interfacial film between suspended particles and the
  medium. The principal emulsifying agents for O/W emulsions are
  proteins, gums, natural and synthetic soaps, etc., and for W/O, heavy
  metal salts of fatty acids, long chain alcohols, lampblack, etc.

What specific property of each type of emulsifying agent makes it suitable for the particular application for O/W or W/O emulsion?

An emulsifier is a surfactant. A surfactant is characterized by the fact that its molecule has one or more hydrophilic parts and one or more hydrophobic parts. The ratio between the molar mass of the hydrophilic parts and the molar mass of the hydrophobic parts in the molecule is expressed by its HLB value (Hydrophilic-lipophilic balance).
Higher HLB means higher water solubility, and lower HLB means lower water solubility.
The emulsifier should have a certain solubility in the outer phase of the emulsion to be distributed in the emulsion.
An emulsifier for forming o/w-emulsions should have higher water solubility, that means higher HLB, and an emulsifier for forming w/o-emulsions should have lower water solubility (= higher oil solubility), that means lower HLB. The corresponding range of HLB is given in the Wikipedia artivle above.


text:
Spin state change during crystallisation for [NiCl2(PPh3)2]
$\ce{NiCl2(PPh3)2}$ is a complex that is borderline between tetrahedral(high spin) and square planar(no unpaired electron, high splitting parameter) because $\ce{Cl-}$ is a weak field ligand and $\ce{PPh3}$ is a strong field ligand. Apparently, the tetrahedral and square planar states of the compound are interconvertible. The tetrahedral high spin state is blue, and produced directly by reacting hydrated nickel chloride and triphenylphosphine in alcohol. But when the complex is crystallised out from a cholrinated solvent like dicholoromethane, it converts to the red square planar complex.
Is there any theoretical explanation as to
1) Why the square planar form is favoured in the crystalline state?
2) What role does the chlorinated solvent play in isomerising the complex to its sq. planar form?
And are there any other examples like this?

I made both isomers as a PhD student many years ago. The red diamagnetic form crystallises out when a concentrated CH2Cl2 solution is cooled in a dry ice-acetone bath, and you can quickly filter off the red product. But on leaving the solid at room temperature, it reverts in a couple of days at room temperature to the blue, paramagnetic form. So it's a thermal equilibrium.


text:
Deducing the shell when transition occurs such that two photons are discharged

Find $n$ (quantum number) corresponding to the excited state of $\ce{He+}$ ion if on transition to ground state the ion emits two photons in succession with wavelengths $\pu{108.5 nm}$ and $\pu{30.4 nm}$.

Solution:
Using Rydberg's equation where $n_1$ is the shell from which the electron transitions to shell $n = 1$ (ground state). $(1)$. $n_2$ is the excited state $n$ from which the electron transitions to $n_1$. $(2)$
When finding $n_1$ the wavelength $\pu{30.4 nm}$ is used, while the wavelength $\pu{108.5 nm}$ is used for the other transition.
My Question: How are the wavelengths coupled with those transitions specifically?
My Guess: Since the absolute value of energy in a transition is inversely proportional to the wavelength of the photon and since the energy to remove electrons from outer shells is lower, the smaller wavelength corresponds to the transition closer to the nucleus (i.e : to the ground state). Is this the reason for this correlation?

You've got yourself all twisted around with 1's and 2's.
The problem states that "the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm." So the first photon emitted has a wavelength of 108.5 nm and the second photon has a wavelength of 30.4 nm.
Furthermore the problem stated that "on transition to ground state." So after the second photon was emitted the helium ion is in its ground state which means that the single electron is in the 1s subshell.
The gist here is that you have to work backwards. Since you know for the 30.4 nm photon was from a transition that ends up in the 1s subshell, then did the electron start in the 2s subshell, or the 3s subshell and so on. So once you figure out the two subshells for the 30.4 nm photon, then you repeat the procedure for the 108.5 nm photon.
The relative wavelengths of the two photons isn't the important clue here. In other words you can't always say that the shortest wavelength will always be to the ground state for some electronic transition.


text:
Intensive and Extensive Properties--Resistance
Is resistance an intensive quantity or an extensive one? And why?
Let me present 2 scenarios.

Resistance of a wire:
The resistance of a wire depends on area and length. This implies it depends on the "quantity" of the wire taken. Does this mean it is extensive?
Resistance of a cell:
As we know,
resistance of a cell is $R = \dot G/\sigma$, where $\dot G$ is the cell constant and the property of the cell, and $\sigma$ is the conductivity. Now, this won't depend on the amount of electrolyte taken in the cell. Does this imply resistance is an intensive quantity?

So, my final question is, whether the resistance is an intensive or an extensive quantity and why?

Welcome to Chem.SE -- thanks for posting your question here! Can you link to the reference where that cell constant $\dot G$ is defined? It would help to understand your argument in point (2). Thanks!

Ah, nevermind, found a reference.

There's already a good answer, so supplementing it. Consider resistance and resistivity. Resistance is extensive while resistivity is intensive. The resistance of a cell or a wire is a function of its geometry and its resistivity.

To the extent that it's appropriate to treat it as a thermodynamic quantity, given that it's generally a description of a specific, macroscopic system (and thus violates the statistical-mechanical large-numbers assumption of thermo), resistance is extensive, always.

Resistance of a wire: The resistance of a wire depends on area and length. This implies it depends on the "quantity" of the wire taken. Does this mean it is extensive?

Yep! You're spot on on this.

Resistance of a cell: As we know, Resistance of a cell is $R=\dot G/\sigma$ where $\dot G$ is the cell constant and the property of the cell, and $\sigma$ is the conductivity. Now, this won't depend on the amount of electrolyte taken in the cell. Does this imply resistance is an intensive quantity?

Actually, this resistance measure will depend on the amount of electrolyte in the cell. Per here, $\dot G$ is a function of the dimensions of the active portion of the cell, in particular the sizes and shapes of the electrodes, and how far they are from one another. These geometrical characteristics then affect how much electrolyte must be used in order to fill the cell. Thus, the resistance $R=\dot G/\sigma$ is still extensive.

But one thing, suppose now i have a cell which is completely immersed in the electrolyte. If i add more electrolyte, of the same concentration, this won't change the resistance of the cell, right? Because the resistance still remains Ġ/σ and the cell constant won't change since the area and the separation between the plates doesn't change.

@user226375 See my edit, first.  Then: , this a situation where the assumption that resistance can be treated as a thermodynamic quantity can start to break down. However, I would argue that that extra electrolyte you're adding in this situation isn't actually part of the electrochemical cell--it's "spectating", in a certain sense, because little to no current will be passing through it, since it's being added in a region of the apparatus where it's not a functional part of the system.


text:
Acetone Enolate-HaloBenzene Reaction
A reaction I am trying to perform says to use a strong base like Sodium Amide with Acetone and Iodo-Benzene as reactants with DMSO as the solvent to form Phenylacetone. The main idea is to use the Acetone Enolate to form Phenylacetone when the Halogen is removed from the Halobenzene.
Can I just put NaOH in Acetone with a Halobenezene?

you are far more likely to form mesityl oxide with those conditions

I could be wrong but I think this reaction works by forming benzyne, which then undergoes enolate attack:

I don't think hydroxide is basic enough to do this, so no I don't believe your proposed reaction conditions will work. Is there any reason you can't just use NaNH2, or BuLi, LDA etc.?

BuLi will probably do lithium-halogen exchange, but I agree hydroxide doesn't work

Thank you for the suggestion. I had NaOH readily available, and wondering if that would work just as well. NM2 will take longer to get a hold of.

I'm almost certain that this reaction will not proceed for both reasons outlined above. NaOH is also not soluble in acetone. Additionally NaOH is not basic enough to deprotonate acetone to form the lithium enolate so even if you could do SnAr, you would just end up catalysing the formation of mesityl oxide. The enolate must be formed all at once to stop this happening. So in short, no you cant just do as you suggest.
I've never seen this done via benzyne (although I haven't looked) but there are rather a lot of palladium catalysed reactions of this type in the lit. Also some cool procedures from acetylacetone using Cu(I) (C-C acitvation!).
Hope this helps x


text:
Equilibrium constant for esterification process
Teacher told us that the concentration term for pure liquids and solids is always $1$. But online I found this reaction:
$$\ce{CH3COOH(l) + CH3CH2OH(l) <=> CH3COOCH2CH3(l) + H2O(l)}$$
According to this page on LibreTexts (search for "esterification" to quickly find it),
$$K_c = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$
Shouldn't $K_c = 1$ as all reactants and products are liquids? The above expression even contains $[\ce{H2O}]$.

The key point is pure liquids. In context of the esterification equilibrium, it considers the reactant concentration in the common solvent - which may or may not be  one of the reactants.
Note that if the solvent is at the same time a reactant or a product, which concentration (activity) can be considered constant, then there can be defined the equilibrium constant for this solvent, with the solvent excluded from the equilibrium.
$$K_{aqua} = \frac{[\ce{CH3COOCH2CH3}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$
$$K_{ethanol} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}]}$$
$$K_{ethylacetate} = \frac{[\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$
$$K_{acetic acid} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3CH2OH}]}$$

Doesn't (l) signify pure liquid as opposed to (aq) which is used for solutions?

generally yes, but not in the reaction equilibrium contexts, as all compounds are miscible, with exclusion of ethylacetate, in any ratio. By other words, if you take these 4 compounds in "(l)" pure state and shake them, they are not pure any more, but you get 1 or 2 liquid phases. Then the equilibrium gets established in each of the phases.

I have encountered several examples where a reactant or product is excluded from the equilibrium constant expression because it was a solid or liquid (denoted by s and l respectively). Is there a rule to decide when to include liquids and when to not?

The rule is, that the compound exists in separated phase in pure form, so its state is constant,not depending of ongoing reactions.

It is typical for the solubility constants.

To be more precise, those distinct phases which remain pure during the reaction (e.g if a solid participates in the reaction, the actual composition of the solid does not change, while that of the surrounding fluid might. Note that it is composition, not quantity.
Purer a phase, closer is its activity (kinda the degree to which it deviates from "normal") is 1, which is what we are using as the "1" in the equation.

Your teacher is correct. But sometimes it is difficult to figure out when the "standard state" condition applies.
For the reaction
$$\ce{CH3COOH(l) + CH3CH2OH(l) <=> CH3COOCH2CH3(l) + H2O(l)}$$
You have missed the context. The starting reactants are acetic acid and ethanol. Some water and some ethyl acetate are produced. The assumption here is that the solution stays as one phase. But there is so little water that the solution is not an "aqueous solution."
If the reaction was happening in an aqueous solution then the mole fraction of water should predominate. The reaction could be denoted as such:
$$\ce{CH3COOH + CH3CH2OH <=>[H2O] CH3COOCH2CH3 + H2O}$$
or
$$\ce{CH3COOH(aq) + CH3CH2OH(aq) <=> CH3COOCH2CH3(aq) + H2O(aq)}$$
With water predominating, the equilibrium would be forced far towards the left.


text:
Why does buffer capacity of tartrate buffer generally decrease as the buffer pH increases?

From what I see, the buffer capacity of tartrate buffer is generally decreasing as buffer pH increases. Why is this? Also, does this mean that tartrate buffer doesn't have a maximum buffer capacity at pH=pKa?
Or is there a more reliable source where I can get the buffer capacity vs. buffer pH graph for buffers of tartaric acid? This image is from a blog, and I can't find the real source of this image, even with Google image search.

See: https://www.researchgate.net/publication/274826340_Determination_of_titratable_acidity_in_white_wine

@MaxW I have seen that source, but for some reason the equation they give for their monoprotic buffer capacity is very different from what is given here (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffer-capacity) and the equation for their diprotic buffer capacity gives extremely high values as [H+]/Kw, which is a part of the equation, gives, for example, 10^10, when pH = 4.0. I'm not sure why there's such a big difference from other sources, seeing that the unit are both mol/L/pH. Or am I mistaken about the units being the same?

Generally, the maximum buffer capacity is at $\ce{pK_a}$ . The tartaric acid is somewhat special for 2 reasons:

It is a diprotic acid with both $\ce{pK_a}$ very close, with the $\ce{pK_{a1}}$ rather low, being affected by the reason 2. :
$$\ce{pK_{a1}}=2.89,\ce{pK_{a2}}= 4.40 (L+)$$
The solution buffer capacity (not limited to presence of specific buffer substances) generally increases toward $\ce{pH}=0$. It means, the buffer capacity is not given by concentration of  of conjugated acid and base, but also by concentration of $\ce{H+}$ itself.
That in large extent masks the $\ce{pK_a}$ maximum. It means, at a slope, a local peak must be big enough to be a peak.

So can the maximum buffer capacity of tartaric acid only be determined experimentally? I've seen a lot of sources that state that tartaric acid is special because of the first reason you've mentioned, but no source actually states what the buffer capacity would look like graphed against buffer pH. Also, the second reason is slightly confusing. Why does buffer capacity generally increase toward pH=0? Could you expand a bit?

It can be computed. But you may realize there is no maximum.

Sorry, I'm not very good with buffers -- could you explain why there's a maximum "given by the solubility of the tartaric acid?"

How would you compute it?

I deleted the note, as I missed the graph legend. Imagine a slope. If the slope is tilted enough, and if there is a peak small enough, there is no maximum ( or the maximum is not significant) .

It might be me, but I'm having a hard time imagining that in my head. I do wish there was a way to upload images in the comments. Is the slope you've mentioned a slope of that buffer capacity vs buffer pH graph or a tartaric acid titration graph? I'm imagining a infinite linear slope, so I'm not sure where the "peak" is supposed to be either.

Notice also the citric acid, it is the same case. It does not have maximum for $pK_{a1}$. It is on the slope toward low $pH$, that is common for all solutions. And yes, I mean the slope on the left on the buffer capacity versus pH graph.

Ah, thank you. I think I'm getting it. I'm wondering, however, if there's a way to generate a better graph through computation or if there's a more reliable source that has the graph for buffer capacity vs. buffer pH.

It can be created by computation. See also updated answer.

Could I do the computation myself? I need to be able to generate a more reliable graph. Could you point me to a source? Thank you so much for your patience, by the way.

pH dependency of buffer capacity - Let's make it simple. Buffer capacity is defined as the moles necessary to change pH +/- 1 pH unit. So if we have a two solutions of HCl in water, one with a pH of 2 and another with a pH of 5. It is going to take a lot more acid to raise the pH 2 solution to 1, than it will take to raise the pH 5 solution to 4.

For a diprotic acid see http://www.aqion.de/site/184 You'll need to plug $a_1$ and $a_2$ from formulas (2.3b) and (2.3c) into formula (5.1a)


text:
Equilibrium Constant and Reaction Mechanism
How is it possible for the equilibrium constant to not depend on a reaction's mechanism?
For the elementary reaction
$$aA + bB \rightleftharpoons dD + eE$$
I understand that the rate of the forward reaction is $k_f [A]^a[B]^b$ and that the rate of the reverse reaction is $k_r[D]^d[E]^e$, and from that it follows that the equilibrium constant is $$K_c = \frac{k_f}{k_r} = \frac{[D]^d[E]^e}{[A]^a[B]^b}$$
What bothers me, however, is that my chemistry textbook says that for ANY reaction, elementary or not, of the form
$$aA + bB \rightleftharpoons dD + eE$$
the equilibrium constant will be
$$K_c = \frac{[D]^d[E]^e}{[A]^a[B]^b}$$
just as in the elementary reaction. Wouldn't this imply that the reaction rate of this forward reaction is $k_f [A]^a[B]^b$ (just as in the elementary reaction), and that the reaction rate of the reverse reaction is $k_r[D]^d[E]^e$ (also the same as the elementary reaction), which might not necessarily be true depending on the reaction mechanism? Or is the math behind calculating the equilibrium constant for a "complex"/multistage reaction different than the math behind a regular equation?

@KarstenTheis I saw that question; however, the answer didn't help me very much because to me that explanation seems more like "lucky" algebraic manipulation, instead of explaining how to reconcile the equation for the rate constant with the idea that in chemical equilibrium the rates of the forward and reverse reactions are equal. If the reaction is multistage, then how do you reconcile the equation for the equilibrium constant with the idea of equal forward/reverse reaction rates?

@KarstenTheis I guess another way of asking my question is this: why when you multiply out the rate constants for each step, as Philipp did, does this always equal (D^dE^e)/(A^aB^b)? At my current level of understanding, Philipp's math seems just "lucky" in the specific explanation he provided, and I fail to see how to generalizes to any equation. Sorry if I am missing something obvious here

You are already convinced that the equilibrium constant expression can be derived from the rate laws for elementary reactions. A multistage reaction is a combination of elementary reactions where all the intermediates cancel out of the net equation. Its equilibrium constant is the product of all the elementary steps, and the intermediates will cancel out of that product as well, leaving you with the equilibrium expression as given in textbooks. The cancelling is not luck, it always happens that way.

Also, you can derive the equilibrium constant expression without invoking any kinetics by asking at which concentrations the Gibbs free energy of reaction is zero. Again, you get the stoichiometric coefficients as the exponents in the equilibrium constant expression. So lucky, elegant, or wonderful, the kinetic derivation and the thermodynamic derivation give the same result.

I wrote up the general argument as answer for the previous, similar question.

@KarstenTheis This is exactly what I needed. Thanks so much!

@KarstenTheis - Thanks for the correction. I deleted my answer which was wrong.


text:
Reactivity of Ni Vs. Pd - available oxidation states
I'm hoping someone can help me tighten up my understanding of the origins of the differences between Ni and Pd catalysis.
I understand that Ni has higher charge density and this is the reason for it's larger capacity for oxidative addition for say an aryl halide. It also has access to the oxidtaion states (0), (I), (II), (III) and sometimes (IV). Palladium conversely only shuttles between (0) and (II) in a standard cross-coupling reaction. I believe the lower first ionisation energy can account for the increased tendency of Ni to do SET but I don't know how best to explain why all the additional oxidation states are available to Ni and not for Pd.
thanks in advance, Cat


text:
Please explain the mystery of the equilibrium constant expression for chemical reactions
Given a reversible chemical reaction, e.g.
$$\ce{CuCl4^2-(aq) <=> Cu^2+(aq) + 4Cl-(aq) + \text{heat}}$$
occurring in a medium, the equilibrium constant is defined as
$$K = \frac{[\ce{Cu^2+}][\ce{Cl-}][\ce{Cl-}][\ce{Cl-}][\ce{Cl-}]}{[\ce{CuCl4^2-}]}$$
where [X] is the molarity or molar concentration of X = amount$_X$ / volume.
Then, as you can read anywhere (e.g. google 'equilibrium constant equals 1' to see (1st hit) - 'The Equilibrium Constant, K - Chemistry LibreTexts' -

If K is about equal to 1, the reaction will reach equilibrium as an
  intermediate mixture, meaning the amounts of products and reactants
  will be about the same.

Yet, if the 'amounts of products and reactants' are the same, that means to me the sum of the molarities of the products would equal the sum of the molarities of the reactants, but K = 1 means that the products of molarities are the same.
I can't attach any intuitive meaning to the product of the molarities.
Note:  the first answer is exactly what I needed.  This question was a pedagogical question - as noted above the first google hit gives a textbook with incorrect info on the equilibrium constant.  That had me confused.  Also, I note that my kid's 10 grade honors chemistry text defines the equilibrium constant as the ratio the product of the molarities of the products divided by the product of molarities of the reactants when the system is in equilibrium, and not as in answer 1 below as the ratio of forward and reverse reaction coefficients, and the latter is a much better definition.

Equilibrium is not when the concentrations of products and reactants are about the same. It is when they don't change anymore even though none of them are zero (i.e. reaction does not go to completion). K can be a million or a millionth, and the reaction will still be able to reach equilibrium.

Yes, so the Chemistry LibreTexts quoted in the question is ... wrong.

There is some wiggle room. K about 1 might mean between a millionth and a million. Amounts of products and reactants will be about the same might mean their concentrations are within the same couple of orders of magnitude.

Here are two correct statements. (1) If we start with all concentrations at standard state (1 M for solutes), the reaction will go forward if K > 1 and reverse if K < 1. (2) Everything else being the same, the higher the equilibrium concentration the more product at equilibrium. Both statements are based on Q vs K. For some more information why the textbook statement is a bit misleading, have a look at my answer to a previous question.

I think the most intuitive way to think about this is to view it through a kinetic lens. That is, recall that at equilibrium, the rate of the forward reaction and the rate of the reverse reaction must be equal. When we write expressions for the rates of reactions, we multiply the reactants because the probability of collision is a function of the product of the two concentrations, not the sum. This only really makes sense for an elementary step, but the result applies to the whole reaction.
Let's consider the simpler reaction $\ce{A + B <=> C}$.
If there's is only one step in the reaction, the rate of the forward reaction will be given by $k_\mathrm{fwd}[A][B]$ and the rate of the reverse reaction by $k_\mathrm{rev}[C]$.
At equilibrium, these are equal, so
$$k_\mathrm{fwd}[\ce{A}][\ce{B}] = k_\mathrm{rev}[\ce{C}].$$
Rearranging,
$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]},$$
and we see the product of the concentrations in the equilibrium constant.
You can also go the route of deriving the expression from the free energies and Boltzmann distribution, but I don't think that's nearly as intuitive as thinking in kinetic terms.

Thanks.  Is there anything that can be concluded if the equilibrium constant equals 1, other than that the products(*) of the concentrations of the products and reactants are equal?

Yes, K=1 means that the Gibbs free energy of the product state is exactly equal to that of the reactant state when all concentrations are 1 M. (And if there are gases, they are at 1 bar. Or more technically, the activities of all species are 1.)

K = 1 is nothing fundamentally special because it depends how we choose the standard state. Alien chemists not aware of IUPAC conventions might be using c = 3.1415 mol / L as their standard state for solutes (because that is equal to 1 in the units they happen to be using), but after converting from one to the other standard state, the measurements on earth would be entirely consistent with the alien measurements.

I can't attach any intuitive meaning to the product of the molarities.

The answer by Andrew gives a good rationale for the product of concentrations in the equilibrium constant expression. In my answer, I will try to give an intuitive argument why the sum of concentrations would not work out.
When a reaction is at equilibrium, forward and reverse rates are non-zero, i.e. the reaction is still going on at the particular level. That means all reactants and products have to be present when the reaction reaches equilibrium. This is in contrast to a reaction that goes to completion, where the limiting reactants are depleted, and the reaction stops.
So let's say you measure the reactant and product concentrations of the reaction $$\ce{A + B <=> C}$$ at equilibrium. Then, using your "sum equilibrium expression", you take the sum of the product concentrations and divide those by the sum of the reactant concentrations to arrive at the "equilibrium sum-constant".
I could now go ahead and make a mixture containing A and C only, with the concentration of C equal to that you measured at equilibrium, and the concentration of A equal to the sum of concentrations of reactants you measured at equilibrium. This reaction would have the same ratio of concentration sums, so it should be at equilibrium. Yet, the concentration of B is zero, so there can't be any forward reaction, and it can't be at equilibrium, leading to a contradiction.
Another contradiction is that we expect if we start at equilibrium and dilute the reaction mix, C should dissociate a bit (Ostwald's dilution law, see this question). With your "sum equilibrium expression", dilution would affect the numerator and denominator equally, so the equilibrium would not be disturbed.


text:
Does propylene glycol dissolve plastic?
According to Médecins Sans Frontières propylene glycol can dissolve plastic:

As propylene glycol can dissolve plastic... (Source).

Is this claim true? This seems rather spurious to me. If it matters, for a specific "plastic" I am particularly interested in evaluating the claim in regards to polypropylene and polyethylene as both are materials common for making syringes.

The kind of plastic almost always matters, I doubt this source meant PE/PP - they aren't particularly soluble in solvents.

I understand that maybe based on the medical source of the question this might be downvoted, but otherwise I don't really understand. If a presumably reputable source makes a farfetched statement and someone posts it here I think that is a good question. Perhaps meta addresses why this is not a worthy question?

Thank you, Mithoron and @NightWriter for the comment. I am asking this question because of a question on Medical Sciences Beta. If you read more, you will realize that I actually doubt the claim and am asking this question as the claim comes from a medically reputable source.

@Mithoron From what I could find, plastic syringes are made from PE/PP. Of course, please correct me if I am wrong.

As propylene glycol can dissolve plastic...

This claim is absurd. Propylene glycol cannot dissolve poly(olefin) polymers especially polypropylene or polyethylene. It is possible for glycols to cause paneling of containers made to such, but not dissolve; and certainly wouldn't happen in the amount of time it takes to perform an injection.

Out of curiosity, would you say that poly(olefin) polymers could leach out from the syringe into the propylene glycol?

@BarryHarrison No, though there could be unreacted monomer, but far below any concentration of concern. Think about how unclean it is to even fill a car with petrol.

Got it. Do you mind clarifying your analogy?

@BarryHarrison Sorry, no analogy I was just trying to juxtapose the minuscule amount of monomer potentially present in a plastic syringe versus the concoction of vapors present at an activity most people perform regularly.

Could you comment on propylene glycol leaching/dissolving latex and/or natural rubber? Thanks!

@BarryHarrison To keep this G-rated I will be obscure, but my research shows that it would be about the same situation for latex as for HDPE and PP as propylene glycol is used with a certain latex product for men. Also my google adds are going to be a bit wonky the next few days.

I am sorry to create you any inconveniences! Thank you so much for looking into this.

Plastic syringes are made of polypropylene and the document linked by @A.K. states that polypropylene is not compatible with antifreeze.
When referring to the plastic containers antifreezes are sold in the most common material used I have seen is HDPE and there is no data available on compatibility according to the Geotech data sheet linked.
The thing is that elsewhere in the data sheet it states that polypropylene is compatible with propylene glycol when it can be used as antifreeze as an alternative to ethylene glycol.


text:
Would a salt bridge work through which the ions could not diffuse into the half cells?
Instead of covering the ends of the salt bridge in an electrochemical cell with a membrane, one could also plug them on both sides with a conductive non-porous material, for example gold, which doesn't go into solution. Would the accumulation of the ions at the plugs and their electrostatic force field be enough to keep the cell running?

In doing so, you will effectively split your electrochemical cell in two  independent cells, each having gold as one of their electrodes. Then you may apply the common rules to see if these cells will work.

I think the gold plugs would not act as electrodes - gold ions would not go into solution, and there is no wire between the gold and the other electrode. The plugs do not necessarily have to be a metal - I want them to be non porous and able to let the electric field through that the accumulating charges have. I think a closed glass tube would also work.

It is a very good thought experiment as to why we don't cap the salt bridge ends with gold plugs. By capping, you are essentially providing an interface which can source and sink for electrons in the cells, thus each end will act as an electrode in the cell. One end of the salt bridge will act as a cathode and one end will act as an anode (strange as it may sound!)
As to your comment above "I want them to be non porous and able to let the electric field through that the accumulating charges have. I think a closed glass tube would also work". No a close tube will not work. You need the ions or electrons to move in order to observe current flow. A closed glass tube will have an infinite resistance and hence no current can flow!

The ions inside of the closed glass salt bridge could move - and would polarize the solution inside the closed glass salt bridge by opposite charged ions accumulating in opposite ends.The opposite charges in the solution on the other side of the glass would mirror the charges and push up against the salt bridge. The resulting electrostatic field would be zero, I suppose. And that's what a salt bridge should accomplish. I am not sure I am right, so that is why I ask the question.

I upvoted your answer sometime ago. Of course, you are correct for the glass rod. I just did a very rough calculation assuming a "closed glass salt bridge" with electrolyte filling, etc., resulting in about 0.25 pF equivalent capacitance. Even if charged to 1 V, this is only about 1.5 million electrons worth of charge. The charging would take place very quickly and it would be a fairly finicky to measure. So, case closed as far as I can see!

Could you add your rough calculations as a separate answer? I think it be very useful. Thanks.


text:
Simple tests for presence of CO₂
Am aware that one method for testing for the presence of $\ce{CO2}$ is passing the gas through limewater, with it going cloudy if positive.
Are there any other convenient methods?

https://www.youtube.com/watch?v=SldWCgRasNM

There is not many simple tests. One of possible tests would be passing gas through neutral solution of suitable acidobasic colour indicator, e.g bromthymol blue, turning green to yellow while getting slightly acidic.
But it would not be specific, any acidic gas would do that.
You have not mentioned context, why do you need to know ? Or is it just homework-like question ?

I wanted to test a bottle of argon gas. Some unscrupulous sellers mix CO2 in them. I just wanted to compare methods

Another, not as simple method, would be IR spectroscopy.


text:
Solubility of metal halides
Are LiBr and LiCl ionic or covalent? Are they soluble in water? Some texts say they are partially soluble in water and in organic solvents also. Or can it be believed that they have more of a covalent character than ionic character ?

You are asking too many questions. I'm gonna answer about the covalent character in $\ce{LiCl}$ & $\ce{LiBr}$

$\ce{LiCl}$ and $\ce{LiBr}$ are ionic compounds with a significant covalent character. This is due to the strong polarising power of $\ce{Li^{+}}$ cation. This polarising power is due to its large charge to mass ratio.
$\ce{Li^{+}}$ pulls the electron cloud of the halogen, hence resulting in a covalent character. The covalent character is more in $\ce{LiBr}$ than $\ce{LiCl}$ as $\ce{Br^-}$ is larger in size, hence valence shell is held comparatively loosely.

This is what essentially happens.

$\ce{LiCl}$ is hygroscopic. If stored in not well tight container, it gradually absorbs water from air up to the point it creates a solution. I suppose $\ce{LiBr}$ can be similar.
The ion $\ce{Li+}$ is very small with strong electrostatic field that attracts electron pairs of water to form the coordinate bonds.


text:
Hydrolysis and Hydration
Does Ammonium Acetate, a salt of a weak base and weak acid, undergo both hydration and hydrolysis in aqeous solution? What is meant by the statement that " Some hydration reactions are hydrolysis"

Hydrolysis and hydration are quite totally unrelated.

:-) Quite or totally ? :-) Both involve water.

Hydrolysis is reaction with water, forming usually 2 different reaction products. It can be also named decomposition by water.
Hydration is (usually) reversible chemical bonding of water. It can be also named water addition.
One action of above can preceed or follow the other. If the hydration step preceeds hydrolysis in the reaction kinetics, the distinction may look vaque from some points of view.
Ammonium acetate undergo hydrolysis.

$$
\begin{align}
\ce{CH3COONH4 &-> CH3COO- + NH4+} \\
\ce{CH3COO- + H3O+ &<=> CH3COOH + H2O } \\
\ce{CH3COO- + H2O &<=> CH3COOH + OH- } \\
\ce{NH4+ + H2O &<=> NH3 + H3O+} \\
\ce{NH4+ + OH- &<=> NH3 + H2O} \\
\ce{2 H2O &<=> H3O+ + OH-}
\end{align}
$$

Please note that $\ce{<->}$ arrow denotes resonance. For equilibrium or reversible reactions in general $\ce{<=>}$ is more suited (also it won't hurt to align a set of related reactions about the arrow marks:) ).

Yes, I know, I have forgotten to use =.
Also, it does not hurt if it is not aligned on the phone. :-) The markdown script is not available.

Well, you just confirmed the fact that phones are poorly suited for content creation:)

I have never needed that confirmation :-) But I am not going to publish my answer either. Additionally, the web editor alone is poorly suited either. :-)

I have never remembered until now, what is right, \align{begin}, or \begin{align}